E ectively Closed Sets - People · 2013-09-13 · E ectively Closed Sets 0 1 Classes PARTIAL DRAFT JANUARY 2012 Douglas Cenzer Department of Mathematics, University of Florida Gainesville,

E�ectively Closed Sets

�01 Classes

PARTIAL DRAFT

JANUARY 2012

Douglas Cenzer

Department of Mathematics, University of Florida

Gainesville, FL 32605-8105, USA

email: cenzer u .edu

Je�rey B. Remmel

Department of Mathematics

University of California, San Diego

February 20, 2012

ii

Contents

A Computability Theory and �0

1Classes 1

I Background 3

I.1 Trees . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

I.2 Topology and Measure . . . . . . . . . . . . . . . . . . . . . . . . 5

I.3 Structures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

I.4 Orderings and Ordinals . . . . . . . . . . . . . . . . . . . . . . . 7

II Computability Theory 9

II.1 Formal de�nitions of the computable functions . . . . . . . . . . 9

II.1.1 Turing machines . . . . . . . . . . . . . . . . . . . . . . . 12

II.2 Basic results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

II.3 Computably enumerable sets . . . . . . . . . . . . . . . . . . . . 18

II.4 Computability of real numbers . . . . . . . . . . . . . . . . . . . 20

II.5 Turing, many-one, and truth-table reducibility . . . . . . . . . . 22

II.6 The jump and the arithmetical hierarchy . . . . . . . . . . . . . . 26

II.7 The lattice of c. e. sets . . . . . . . . . . . . . . . . . . . . . . . . 29

II.8 Computable ordinals and the analytical hierarchy . . . . . . . . . 35

II.9 Inductive De�nability . . . . . . . . . . . . . . . . . . . . . . . . 41

II.10 The hyperarithmetical hierarchy . . . . . . . . . . . . . . . . . . 46

IIIFundamentals of �01 Classes 57

III.1 Computable trees and notions of boundedness . . . . . . . . . . . 57

III.2 De�nition and basic properties of �01 Classes . . . . . . . . . . . 60

III.3 E�ectively Closed Sets in the Arithmetic Hierarchy . . . . . . . . 68

III.4 Graphs of Computable Functions . . . . . . . . . . . . . . . . . . 71

III.5 Computably enumerable sets and �01 Classes . . . . . . . . . . . 74

III.5.1 Separating classes . . . . . . . . . . . . . . . . . . . . . . 75

III.5.2 Subsimilar classes . . . . . . . . . . . . . . . . . . . . . . 77

III.6 Retraceability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79

III.7 Reducibility . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83

III.8 Thin and minimal classes . . . . . . . . . . . . . . . . . . . . . . 86

III.9 Mathematical Logic . . . . . . . . . . . . . . . . . . . . . . . . . 90

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iv CONTENTS

IVMembers of �01 Classes 95

IV.1 Basis theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95

IV.2 Special �01 classes . . . . . . . . . . . . . . . . . . . . . . . . . . . 99

IV.3 Measure, Category and Randomness . . . . . . . . . . . . . . . . 106

IV.4 Mathematical Logic: Peano Arithmetic . . . . . . . . . . . . . . . 110

IV.4.1 Peano Arithmetic . . . . . . . . . . . . . . . . . . . . . . . 111

V The Cantor-Bendixson Derivative 115

V.1 Cantor-Bendixson derivative and rank . . . . . . . . . . . . . . . 115

V.2 Basis results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121

V.3 Ranked Points and Rank-Faithful Classes . . . . . . . . . . . . . 122

V.4 Rank and Complexity . . . . . . . . . . . . . . . . . . . . . . . . 123

V.5 Computable Trees with One or No In�nite Branches . . . . . . . 133

V.6 Logical Theories revisited . . . . . . . . . . . . . . . . . . . . . . 136

VI Index Sets 139

VI.1 Index sets for �01 classes . . . . . . . . . . . . . . . . . . . . . . . 141

VI.2 Cardinality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 146

VI.3 Computable Cardinality . . . . . . . . . . . . . . . . . . . . . . . 156

VI.4 Index Sets and Lattice Properties . . . . . . . . . . . . . . . . . . 161

VI.5 Separating Classes . . . . . . . . . . . . . . . . . . . . . . . . . . 165

VI.6 Measure and Category . . . . . . . . . . . . . . . . . . . . . . . . 169

VI.7 Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 174

VI.8 Index Sets for Logical Theories . . . . . . . . . . . . . . . . . . . 177

VIIReverse Mathematics 181

VII.1Subsystems of Second Order Arithmetic . . . . . . . . . . . . . . 182

VII.1.1Recursive Comprehension . . . . . . . . . . . . . . . . . . 183

VII.1.2Weak K�onig's Lemma . . . . . . . . . . . . . . . . . . . . 185

VII.1.3Arithmetic Comprehension . . . . . . . . . . . . . . . . . 187

VII.2Mathematical Logic . . . . . . . . . . . . . . . . . . . . . . . . . 188

VIIIComplexity Theory 191

VIII.1Complexity of Trees . . . . . . . . . . . . . . . . . . . . . . . . . 194

VIII.2Complexity of Structures . . . . . . . . . . . . . . . . . . . . . . 203

VIII.3Propositional Logic . . . . . . . . . . . . . . . . . . . . . . . . . . 207

B Applications of �0

1Classes 213

IXAlgebra 217

IX.1 Boolean algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . 219

X Computer Science 227

X.1 Non-monotonic Logic . . . . . . . . . . . . . . . . . . . . . . . . . 227

CONTENTS v

XIGraphs 229XI.1 Matching problems . . . . . . . . . . . . . . . . . . . . . . . . . . 230XI.2 Graph-coloring problems . . . . . . . . . . . . . . . . . . . . . . . 233XI.3 The Hamiltonian circuit problem . . . . . . . . . . . . . . . . . . 236

XIIOrderings 241XII.1Partial orderings . . . . . . . . . . . . . . . . . . . . . . . . . . . 241XII.2Linear orderings . . . . . . . . . . . . . . . . . . . . . . . . . . . 249XII.3Ordered algebraic structures . . . . . . . . . . . . . . . . . . . . . 251

XIIIIn�nite Games 255

XIVThe Rado Selection Principle 261

XVAnalysis 263XV.0.1Computable continuous functions . . . . . . . . . . . . . . 269

XV.1Symbolic Dynamics . . . . . . . . . . . . . . . . . . . . . . . . . . 277XV.1.1Undecidable subshifts . . . . . . . . . . . . . . . . . . . . 278XV.1.2Symbolic Dynamics of Computable Functions . . . . . . . 280

XVIFeasible versions of combinatorial problems 283

C Advanced Topics and Current Research Areas 295

XVIIThe Lattice of �01 classes 297

XVII.1Countable thin classes . . . . . . . . . . . . . . . . . . . . . . . . 298XVII.2Initial Segments of the Lattice . . . . . . . . . . . . . . . . . . . 301XVII.3Global Properties of the Lattice . . . . . . . . . . . . . . . . . . . 301XVII.4Almost complemented classes . . . . . . . . . . . . . . . . . . . . 301XVII.5Perfect thin classes . . . . . . . . . . . . . . . . . . . . . . . . . . 301

XVIIIDegrees of Di�culty 303XVIII.1Reducibility . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 303XVIII.2Completeness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 306XVIII.3Separating Classes . . . . . . . . . . . . . . . . . . . . . . . . . . 308XVIII.4Measure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 313XVIII.5Randomness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 314XVIII.6Thin Classes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 315

XIXRandom Closed Sets 317XIX.1Martin-L�of Randomness of Closed Sets . . . . . . . . . . . . . . . 319XIX.2Members of Random Closed Sets . . . . . . . . . . . . . . . . . . 324

vi CONTENTS

Preface

E�ectively closed sets have been a central theme in computability theory, al-gorithmic randomness and applications to computability and e�ectiveness inmathematics. This book is intended to be a self-contained introduction to thetheory and applications of e�ectively closed sets, or �0

1 classes. It may be usedfor a graduate-level course and also as reference for researchers in computabilitytheory and related areas.

Part A begins with some basic facts from computability theory which willbe needed. The members of a �0

1 class are real numbers, often represented byin�nite strings of natural numbers, or by sets of natural numbers. Backgroundis taken from the classic book of Soare [198] on computably enumerable (c.e.)sets and degrees. The fundamental problem, going back to work of Kleene [108]in the period 1940-1960, is to determine the complexity of the members of a �0

1

class, as measured by the Turing degree, or by the de�nition in the hyperarith-metic hierarchy, or by the amount of resources in time and space required. TheKleene basis theorem showed that every �0

1 class contains a member which isrecursive in some �11 set and the Kreisel-Shoen�eld basis theorem [188], whichshowed that every c. b. �0

1 class contains a member of degree < 00. Two fun-damental papers in this area are [99, 98] by Jockusch and Soare. They show,among other things, that there is a �0

1 class with no recursive members andsuch that any two members have mutually incomparable Turing degree.

The Cantor-Bendixson derivative which reduces a closed set to its perfectkernel, plays an important role here going back to the 1959 paper of Kreisel[116], who �rst noticed that the degree of a member x of a �0

1 class is relatedto the Cantor-Bendixson rank of x in P and that any countable class has acomputable member. Countable �0

1 classes were closely examined by Soare andothers [23, 46] in the 1980's. �0

1 classes are given an enumeration as P0; P1; : : :and index sets for families of �0

1 classes are then studied in the manner thatindex sets for c.e. sets are studied in [198]. These can measure the complexity ofcertain properties of �0

1 classes, related in particular to cardinality and measure.�01 classes may be de�ned as sets of in�nite paths through computable trees.Part B presents some applications of �0

1 classes in logic, mathematics andtheoretical computer science. The solution sets of many mathematical problemsmay be represented by �0

1 classes and the complexity of the problem can thenbe determined. The more di�cult representation problem is to show that every�01 class (or every bounded or c. b. �0

1 class) can represent the solution set of a

vii

viii CONTENTS

certain problem. For example, in 1960, Shoen�eld [189] showed that the familyof complete consistent extensions of an axiomatizable theory is a c. b. �0

1 classand Ehrenfeucht [69] showed that any c. b. �0

1 class can represent such a family.The family of complete consistent extensions of an axiomatizable theory is

of course closely related to the Lindenbaum algebra of the theory and Booleanalgebras are an important topic for �0

1 classes. A number of articles in thearea use the notion of a computably enumerable ideal of the computable denseBoolean algebra as an equivalent notion to that of a �0

1 class. This concept willbe discussed in detail in the section on Boolean algebras.

Non-monotonic logic [135] is a general form of reasoning where certain \de-fault" assumptions are made and may later be rescinded. The set of stablemodels of a logic program is a non-monotonic generalization of the (unique)closure under consequence of a set of axioms and rules. Di�erent versions ofa logic program may be used to represent c. b., bounded and unbounded �0

1

classes. Another area of theoretical computer science where �01 classes have

application is the study of !-languages. This refers to a sets of in�nite wordswhich is accepted, in some fashion, by a program.

The surjective matching problem of Philip and Marshall Hall [82] was ana-lyzed by Manaster and Rosenstein, who showed that the set of bijective match-ings in a symmetrically highly recursive society is always a c. b. �0

1 class, andcan represent an arbitrary c. b. �0

1 class. Bean [12] showed in 1976 that thefamily of k-colorings of a highly computable graph is a c. b. �0

1 class and Rem-mel [174] showed that any c. b. �0

1 class can represent, up to a permutation ofthe colors, such a family.

The reason that �01 classes arise so naturally in the study of recursive combi-

natorics is that many combinatorial theorems about �nite graphs and partiallyordered sets (posets) can be extended to countably in�nite graphs and posetsby applying K�onig's Lemma, which states that every in�nite �nitely branchingtree T has an in�nite path through it. Now K�onig' Lemma, and also the so-called Weak K�onig's Lemma play an important role in the Reverse Mathematicsprogram of Friedman and Simpson [192]. Thus the study of �0

1 classes can berelated to the study of K�onig's Lemma. For example, Simpson [192] showedthat Lindenbaum's lemma (that every countable consistent set of sentences hasa complete consistent extension) and G�odel's completeness theorem are bothequivalent to Weak K�onig's Lemma over a certain subsystem (RCA0) of secondorder arithmetic. For another example, Hirst [88] showed that a version of Hall'ssymmetric matching theorem is equivalent to K�onig's Lemma over RCA0.

The role of �01 classes in computable algebra and computable analysis is also

presented.Part C examines recent results on the family of �0

1 classes. One very im-portant topic is the connection between e�ectively closed sets and algorithmicrandomness, as developed by many researchers from Kucera [117, ?, 119] toLewis [?, 11, 10] and surveyed in the books of Downey-Hirschfeldt [63] and Nies[162]. The lattice E� of �0

1 classes under inclusion is compared and contrastedwith the lattice E of c.e. sets under inclusion. This includes results of Downeyand others [25, 51, 50] on thin classes and automorphisms and work of Cenzer

CONTENTS ix

and Nies [32, 33] on intervals and on de�nability in E�. The degree of di�cultyof a class was de�ned by Medvedev [149] and refers to the di�culty of �nding amember of the class. The Medvedev lattice of degrees of di�culty was studiedlater by Sorbi [200] and then the study of the Medvedev and also the relatedMuchnik degrees of �0

1 classes was developed further by Simpson [193] and oth-ers. Here we also examine �0

1 classes which arise from trees with a speci�edcomplexity, such as polynomial time computable.

Acknowledgements

The authors owe a great debt to their advisors, Peter Hinman and Anil Nerode,to the founders of the subject of computability theory, in particular StephenKleene, and also to Carl Jockusch and Robert Soare, whose seminal papersestablished the subject of �0

1 classes back in the 1970's. We would also liketo thank some others for enlightening discussions on the subject over manyyears, including Peter Cholak, Peter Clote, Rod Downey, Andre Nies, RichardShore, Stan Wainer, Rebecca Weber. The �rst author would like to thankseveral people for their careful reading of drafts. This includes Chris Alfeld,Paul Brodhead, Ali Dashti, Sergey Melikhov, Diego Rojas, Rebecca Smith andFerit Toska.

x CONTENTS

Part A

Computability Theory and

�01Classes

1

Chapter I

Background

This chapter contains some of the de�nitions and notations needed for the studyof e�ectively closed sets. We begin with objects under study: numbers, func-tions, sequences (or strings) and trees.

The set f0; 1; 2; :::g of natural numbers is denoted by N and also by ! whenwe view N as an ordered set. Here n = f0; 1; : : : ; n � 1g is identi�ed with theset of smaller natural numbers. Lower-case Latin letters a; b; c; d; e; i; j; k; l;m; ndenote integers; p; q; r; s; t denote rational numbers; u; v; w; x; y; z denote realnumbers. The letters f; g; h (and occasionally other lower-case Latin letters)denote total functions from Nk to N for k � 1; the Greek letters �; ; � (andoccasionally other lower-case Greek letters) denote (possibly) partial functionson Nk (functions whose domain is a subset of Nk for some k). Lower case Greekletters �; �; �; � denote �nite sequences of natural numbers; �; �; �; denote or-dinals. Upper-case Latin letters A;B;C;D;E; I; J;K;L;M denote subsets of N;S; T denote trees; P;Q;U; V;W;X; Y; Z denote sets of real numbers. Upper-caseLatin letters F;G;H denote total functions of real variables (with domain andrange included in Nm�<n); Upper-case Greek letters �;;� (and occasionallyothers) denote (possibly) partial functions of real variables. In our usage, a setusually refers to a set of natural numbers.

The composition of two functions f and g is denoted by f �g; fn denotes thefunction f composed with itself n times. For a partial function �, �(x) # denotesthat �(x) is de�ned and �(x) " denotes that �(x) is not de�ned. dom(�) = fx :�(x) #g and ran(�) = f�(x) : x 2 dom(�)g denote the domain and range of�, respectively. If F : X ! Y , then F [U ] denotes fF (x) : x 2 Ug for U � Xand F�1[V ] denotes fx : F (x) 2 V g for V � Y . �A denotes the characteristicfunction of A, which is often identi�ed with A and written simply as A(x). �dmdenotes the restriction of A to x.

For two sets X and Y , X � Y denotes the direct product of X and Y , thatis, the set of ordered pairs (x; y) with x 2 X and y 2 Y . The direct productX1 �X2 � : : :�Xk of a sequence X1; : : : Xk of sets is similarly de�ned. Xk isthe product of k copies of X.

The power XY of two sets denotes the set of (total) functions with domain

3

4 CHAPTER I. BACKGROUND

Y and range a subset of X. In particular, f0; 1gN is the usual Cantor spaceand may be identi�ed with the family of subsets of N. NN is the Baire space.< denotes the space of real numbers. The Cantor space may be identi�ed witha (compact) subset of < and the Baire space may be identi�ed with the set ofirrational numbers. For us a class refers to a subset of < (or of the Cantor spaceor Baire space). A class in the Cantor space may be called a \class of sets"since its elements are the characteristic functions of sets of natural numbers.

I.1 Trees

Let � be a set of symbols (an alphabet), usually an initial segment of N. Then fora natural number n, �n denotes the set of strings � = (�(0); �(1); : : : ; �(n� 1))of n letters from �; the length n of � is denoted by j�j. The empty string haslength 0 and will be denoted by ;. �� (or sometimes �<!) denotes the set[n2!�n and �! denotes the set of in�nite sequences. Strings may be codedby natural numbers in the usual fashion. First let [x; y] denote the standardpairing function 1

2 (x2 + 2xy + y2 + 3x + y) and in general [x0; x1; : : : ; xn] =

[[x0; : : : ; xn�1]; xn]. Then we can code strings of arbitrary length n > 0 byh�i = [n; [�(0); �(1); : : : ; �(n� 1)]] and also h;i = 1. A string may be identi�edwith its code, so that functions on N� are represented by functions on N. Aconstant string � of length n will be denoted kn. For m < j�j, � � m is thestring (�(0); : : : ; �(m � 1)); � is an initial segment of � (written � � �) if� = � jm for some m. Initial segments are also referred to as pre�xes. Similarly� is said to be a su�x of � if j� j � j�j and, for all i < j� j, �(j�j� j� j+ i) = �(i).The concatenation �_� (or sometimes � � � or just ��) is de�ned by �_� =(�(0); �(1); : : : ; �(m � 1); �(0); �(1); : : : ; �(n � 1)), where j�j = m and j� j = n;in particular we write �_a for �_(a) and a_� for (a)_�. Thus we may alsosay that � is a pre�x of � if and only if � = �_� for some � and that � is asu�x of � if and only if � = �_� for some �.

For any x 2 �� and any �nite n, the initial segment xdn of x is (x(0); : : : ; x(n�1)). We write � � x if � = xdn for some n. For any � 2 �n and any x 2 ��,we have �_x = (�(0); : : : ; �(n� 1); x(0); x(1); : : : ).

For a sequence a0 < a1 < � � � < an, we denote by ba0; : : : ; anc the string� 2 f0; 1gan such that �(k) = 1 if and only if k = ai for some i < n. Thusba0; a1; : : : ; anc = 0a010a1�a0�11 � � � 0an�1�an�2�110an�an�1�1:

For any x; y 2 NN, the join x� y = z, where z(2n) = x(n) and z(2n+ 1) =y(n). For two classes P and Q, the product P Q = fx � y : x 2 P & y 2Qg. An in�nite sequence x0; x1; : : : may be coded as hx0; x1; : : :i = y, wherey(hm;ni) = xm(n). For an in�nite family fPi : i 2 !g of sets, the product maythen be de�ned as fhx0; x1; : : :i : (8i)xi 2 Pig. We can also de�ne the disjointunion P �Q = f0_x : x 2 Pg [ f1_y : y 2 Qg.

A tree T over � is a set of �nite strings from �� which is closed under initialsegments. The set � is sometimes called an alphabet. We say that � 2 T is animmediate successor of a string � 2 T if � = �_a for some a 2 �. Since ouralphabet will always be countable and e�ective, we may assume that T � N�.

I.2. TOPOLOGY AND MEASURE 5

For any tree T and any �, T (�) = f� : � � � or � � �g.A tree T is said to be a shift if it is also closed under su�xes.

Example I.1.1. De�ne T � f0; 1g� so that � 2 T if and only if � does not have3 consecutive 0's, that is, if � has no consecutive substring of the form (000).Clearly if � does not have 3 consecutive 0's then no initial segment of � canhave 3 consecutive 0's either. Furthermore, if � has no consecutive substring(000), then no su�x of � can have a consecutive substring (000). Thus T is ashift.

We say that a tree T is �nite-branching if for every � 2 T , there are only�nitely many immediate successors of � in T . Certainly any tree T over a �nitealphabet is �nite-branching.

Example I.1.2. De�ne the tree T � N� so that for strings � of length n,� 2 T () �(n � 1) � 1 + �(0) + �(1) + : : : �(n � 2). Then for any � 2 T ,�(0) � 1, �(1) � 2, and by induction �(n) � 2n; it follows that � can have atmost 2n immediate successors.

We will see later that a tree T is �nite-branching if and only if there is afunction f such that for all strings � 2 T of length n, � has at most f(n)immediate successors. The problem of computing the function f will be a veryimportant one. More generally, we will look at the problems of computing listof these successors, or an upper bound on the size of the successors, or an upperbound on the number of successors.

I.2 Topology and Measure

The topology of the real line has a basis of open intervals (x; y) = fu : x < u <yg where x = �1 and y = 1 are allowed; [x; y] denotes the closed intervalfu : x � u � yg; [x; y) and (x; y] are similarly de�ned. The topology on thespaces �N, where � is either a �nite alphabet or equals N, is determined by abasis of intervals I(�) = fx : � � xg and has a sub-basis of sets of the formfx : x(m) = ng for �xed m;n. Notice that each interval is also a closed set andis therefore said to be clopen and that the clopen subsets of the Cantor spacef0; 1gN are just the �nite unions of intervals.

For a tree T � ��, we de�ne the set [T ] of in�nite paths through T by letting

x 2 [T ] () (8n)x � n 2 T:

A subset P of NN is closed if and only if P = [T ] for some tree T . Thisjusti�es the description of a �0

1 class as an e�ectively closed subset of NN. Afunction F : X ! Y is continuous if F�1[V ] is open for every open set V � Y .Then a function F : NN ! NN is continuous if, for all m;n, fx : F (x)(m) = ngis open.

Let X be either <, NN or f0; 1gN. A subset Y of X is dense in an interval Iif it meets every subinterval of I; Y is nowhere dense if it is dense in no interval.


Y is meager (�rst category) if it is a countable union of nowhere dense sets; Yis non-meager (second category) if it is not meager. Y is comeager (residual) ifY is meager.

An element x 2 Y is isolated in Y if there exists an open set U such thatY \U = fxg. A closed, non-empty set Y is perfect if it has no isolated elements.Each of the spaces <, NN and f0; 1gN are perfect.

De�nition I.2.1. The Cantor-Bendixson derivative D(P ) of a compact set Pis the set of nonisolated points in P .

Note that D(P ) is empty if and only if P is �nite.

The iterated Cantor-Bendixson derivative D�(P ) of a closed set P is de�nedfor all ordinals � by the following trans�nite induction.

D0(P ) = P ; D�+1(P ) = D(D�(P )) for any �; D�(P ) =T�<�D

�(P ) forany limit ordinal �.

The Cantor-Bendixson (C.B.) rank of a closed set P is the least ordinal �such that D�+1(P ) = D�(P ). If � is the C-B rank of P , then D�(P ) is theperfect kernel of P and is a perfect closed set. For an element x 2 P which isnot in the perfect kernel, the Cantor-Bendixson (C.B.) rank of x in P is theleast ordinal � such that x =2 D�+1(P ).

The standard Lebesgue measure � on f0; 1g! is determined by letting �(I(�)) =2�j�j. A product measure on NN may be de�ned (with �(NN) = 1) by set-ting the measure of fx : x(m) = ng to be 2�n�1, so that I(�) has measure2�(m0+m1+��+mk�1+k).

I.3 Structures

We shall use the logical symbols &, _, :,! and () to denote as usual \and",\or", \not", \implies" and \if and only if". The symbols 9 and 8 denote thequanti�ers \there exists" and \for all". In addition, (9m < p) and (8m < p)denote bounded quanti�ers where the range of the quanti�er is restricted tonumbers less than p, and (91x) denotes \there exist in�nitely many x suchthat".

As usual, a �rst-order language L is given by a set fRigi2S of relation sym-bols, a set ffjgj2T of function symbols, and a set fcigi2U of constant symbols,together with functions m(i) and n(i) such that Ri is an m(i)-ary relation sym-bol and fi is an n(i)-ary function symbol. We assume here that S, T and Uare subsets of !. The language also includes variables and both existential anduniversal quanti�ers using these variables. The set of terms of L and the setSent(L) of sentences of L are de�ned as usual by induction. A propositionallanguage is given by a set of 0-ary relation symbols, or propositional variables.The reader is referred to Shoen�eld [190] for details.

We shall consider structures over an e�ective �rst-order language

L = hfRm(i)i gi2S ; ffn(i)i gi2T ; fcigi2U i;

I.4. ORDERINGS AND ORDINALS 7

where S, T and U are initial segments of !, for all i 2 U , ci is a constant symboland there are partial recursive functions s and t such that, for all i 2 S, Ri isan s(i)-ary relation symbol and, for all i 2 T , fi is a t(i)-ary function symbol.

Let � be some complexity class of sets (and functions), such as partial re-cursive, primitive recursive, exponential time, polynomial time (or p-time). Wesay that a set or function is �-computable if it is in �.

Amodel or structure, A = (A; fRAi gi2S ; ffAi gi2T ; fcAi gi2U ), for the languageL is given by a set A together with interpretations of the relation, function andconstant symbols.

De�nition I.3.1. (a) A structure (where the universe A of A is a subset of��) is a �-structure if

(i) A is a �-computable subset of ��

(ii) for each i 2 S, RAi is a �-computable relation on Am(i).

(iii) for each j 2 T , fAj is a �-computable function from An(j) into A.

(iv) If S = !, then there is a �-computable relation R such that, for all i 2 Sand all (x0; : : : ; xm(i)),

RAi (x0; : : : ; xm(i)) () R(i; hx0; : : : ; xm(i)i):

(v) If T = !, then there is a �-computable function f such that, for all j 2 Tand all (x0; : : : ; xn(j)),

fAi (x0; : : : ; xn(j)) = f(i; hx0; : : : ; xn(j)i):

For any complexity class �, we say that two structures A and B are �-isomorphic if there is an isomorphism f from A onto B and �-computable func-tions F and G such that f = F dA (the restriction of F to A) and f�1 = GdB.

I.4 Orderings and Ordinals

The results of this book are all theorems of Zermelo-Fraenkel Set Theory withthe Axiom of Choice. The (Generalized) Continuum is not assumed.

Our set-theoretic conventions are standard and we refer the reader to (for ex-ample) Jech [90] for further background. The inclusion relation X � Y denotes(8x)(x 2 X ! x 2 Y ) and X � Y denotes X � Y and X 6= Y . The symbols[, \ and n denote the binary operations of union, intersection and di�erence;A denotes the complement of A.

A set X is transitive if (8y)(y 2 X ! y � X) and X is an ordinal (number)if X and all of its elements are transitive. For ordinals � and �, � < � if andonly if � 2 �. For any ordinal �, � + 1 = � [ f�g is the successor ordinalof �. � is a limit ordinal if it is neither 0 nor a successor, which implies that(8� < �)(�+1 < �). For any set X of ordinals, infX denotes the least elementof X and supX denotes the least ordinal greater than or equal to every elementof X.


An ordinal � is said to be a recursive ordinal if there is a recursive well-ordering of ! of order type �. The least non-recursive ordinal is denoted by!C�K1 , and was introduced by Church and Kleene [52].

The natural, or Hessenberg sum, � � �, of two ordinals � and �, may bede�ned as follows. Let � = ! 1a1+!

2a2+ � � �+! kak and � = ! 1b1+! 2b2+

� � � + ! kbk be the Cantor normal forms of � and �, where we have insertedai = 0 and bj = 0 to obtain expressions with the same powers of !. Then

�� = ! 1(a1 + b1) + ! 2(a2 + b2) + � � �+ ! k(ak + bk):Thus we treat ordinals as polynomials over ! with natural number coef-

�cients. This natural addition is commutative. For any ordinals � and �,�+ � � �� . See [121] (p. 253) for details.

An ordinal � is a cardinal number if there is no one-to-one correspondencebetween � and any � < �. It follows from the Axiom of Choice that for everyset X, there is a unique cardinal � and a one-to-one correspondence between Xand �; � is the cardinality (Card(X))of X. The natural numbers are exactlythe �nite cardinals and ! is the least in�nite cardinal. A set X is countable ifCard(X) � ! and countably in�nite if Card(X) = !. The in�nite cardinal ! isalso denoted by @0 and the least uncountable cardinal by @1.

For any set X, P(X) denotes the power set of X, the set of all subsets of Xand 2� denotes Card(P(�). Since there is a one-to-one correspondence betweenP(N) and the continuum <, Card(<) = 2@0 .

A relation R on a set X is a subset of X �X; the domain of R is dom(R) =fx : (9y)(x; y) 2 Rg and the range is ran(R) = fy : (9x)(x; y) 2 Rg. R(x; y)and also xRy are sometimes used in place of (x; y) 2 R. R is re exive if R(x; x)for all x and is irre exive if :R(x; x) for all x. R is symmetric if R(x; y) impliesR(y; x) for all x; y and is antisymmetric if R(x; y) & R(y; x) implies y = x forall x; y. R is transitive if R(x; y) & R(y; z)) implies R(x; z) for all x; y; z. R istotal or connected if R(x; y) _ R(y; x) for all x; y. R is an equivalence relationif it is symmetric, re exive and transitive.

R is a pre-partial-ordering if it is re exive and transitive. A pre-partial-ordering R is a pre-linear-ordering if it is total. A pre-partial-(linear-)orderingis a partial (linear) ordering if it is antisymmetric.

R is is well-founded if every subset A of X has a minimal element, thatis, some m such that for all x, R(x;m) ! R(m;x). Assuming the Axiom ofDependent Choice (DC), this is equivalent to the following

(8f 2 NX)[(8m)(R(f(m+ 1); f(m))! (9m)R(f(m); f(m+ 1)):

A (pre-)linear ordering is a (pre-)well-ordering if it is well-founded.

Chapter II

Computability Theory

In this chapter, we present some basic de�nitions and results from classicalcomputability theory which are needed for the study of �0

1 classes. The keynotion here is that of a computable functional, or function with domain a subsetof NN.

We begin with a brief review of computable functions and computably enu-merable (c. e.) sets. Formal de�nitions of the set of computable functions havebeen given in many di�erent ways. The computable functions are the functionsmapping natural numbers (or more generally �nite strings of symbols taken froma �nite alphabet) which are computable by a Turing machine, register machine,or other idealized computer. These are the functions which can be computedby a program in Maple, or Matlab, or some other �xed programming language.The set of computable functions is the smallest which includes certain basicfunctions and is closed under primitive recursion, composition, and unboundedsearch.

All of these approaches are known to lead to the same family of functions,and Church's Thesis proclaims that any other attempt to formalize the notionof a computable function will lead to the same family of functions.

We refer the reader to Soare [198] and to Odifreddi [163] for full details onthe basic de�nitions and results of computability theory.

II.1 Formal de�nitions of the computable func-

tions

Since index sets will be a central topic in our work, we will give a de�nitionin the spirit of Kleene [110] and Hinman [87] based on the index or code fora computable function. We will give the general de�nition for a computablefunction or functional with both natural number inputs and real number inputs(that is, functions from NN). It is crucial that our functions may be partial, thatis, de�ned on a proper subset of Nk � (NN)l. The second crucial observation

9

10 CHAPTER II. COMPUTABILITY THEORY

is that the (partial) computable functions may be enumerated as �0;�1; : : : sothat the universal function U(e;�!m;�!x ) = �e(

�!m;�!x ) is itself partial computable.An index e = hi; k; `; : : :i for a computable function is the code for a function

�e of k natural numbers and ` real numbers. �e is a function on natural numbersif ` = 0 and will then be denoted also by �e. Here �!m = (m0; : : : ;mk�1) and�!x = (x0; : : : ; x`�1).

The basic indices and functions are the following:

(0) Constant Functions: �e(�!m;�!x ) = n when e = h0; k; `; ni.

(1) Projection Functions: �e(�!m;�!x ) = mi when e = h1; k; `; ii and i < k.

(2) Successor Functions: �e(�!m;�!x ) = mi + 1 when e = h2; k; `; ii and i < k.

(3) Application Functions: �e(�!m;�!x ) = xj(mi) when e = h3; k; `; i; ji, i < k

and j < `.

The primitive recursive functions are obtained from the basic functions byclosure under composition and primitive recursion, which are de�ned as follows.

(4) Composition: �e(�!m;�!x ) = �a(�b1(

�!m;�!x ); : : : ;�br (�!m;�!x )) when e = h4; k; `; a; b1; : : : ; briwhen (a)1 = r, (a)2 = 0 and, for each t, (bt)1 = k and (bt)2 = `.

(5) Primitive Recursion: �e(0;�!m;�!x ) = �a(

�!m;�!x ) and, for each n, �e(n +1;�!m;�!x ) = �b(�e(n;

�!m;�!x ); n;�!m;�!x )) when e = h5; k+1; `; a; bi, (a)1 = k,(a)2 = `, (b)1 = k + 2 and (b)2 = `.

A set A � Nk is primitive recursive if the characteristic function is primi-tive recursive. It is worth noting that the set of indices for primitive recursivefunctions is itself a primitive recursive set. Thus we may de�ne an enumeration�e of the primitive recursive functions by letting �e(

�!m;�!x ) = �e(�!m;�!x ) if e is

a primitive recursive index and otherwise �e(�!m;�!x ) = 0.

Lemma II.1.1. There is a partial recursive function � such that for each e,�e = ��(e).

Details are left to the exercises.The computable functions are obtained from the basic functions by closure

under composition, primitive recursion and search, which is de�ned as follows.Here we let \(least p)R(p)" denote the least p such that R(p).

(6) Search: �e(�!m;�!x ) = (least p)�a(p;

�!m;�!x ) = 0 when e = h6; k; `; ai, wherethis means as usual that �e(

�!m;�!x ) = q if �a(q;�!m;�!x ) = 0 and for all

p < q, �a(p;�!m;�!x ) is de�ned and not equal to zero.

If �e(�!m;�!x ) is de�ned by the above, we say that �e(

�!m;�!x ) converges andwrite �e(

�!m;�!x ) #. If �e(�!m;�!x ) is not determined by this de�nition, then

�e(�!m;�!x ) is unde�ned. We say that �e(

�!m;�!x ) diverges and write �e(�!m;�!x ) ".

II.1. FORMAL DEFINITIONS OF THE COMPUTABLE FUNCTIONS 11

If e is not an index of a computable function, then of course �e(�!m;�!x ) " for all�!m;�!x , so that �e is the empty function.

If we replace the real variables xj with �nite sequences �j , then the de�nitionof �e(

�!x ;�!� ) is obtained as above when we begin with �e(�!m;�!� ) = �j(mi)

provided that mi < j�j j.Then the computation of �e(

�!m;�!x ) = q is coded by c = he;�!m;�!� ; qi, where�j is the shortest initial segment of xj needed.

We will next de�ne the notions of a computation tree and a derivation for acomputation.

For the constant, projection and successor functions, the computation tree

of �e(�!m;�!x ) = n has a single node he;�!m;�!; ; ni and this is also the derivation.

For the application function, the computation tree for �e(�!m;�!x ) = xi(mj) =

n also has a single node he;�!m;�!; ; (xi(0); : : : ; xi(mj));�!; ; ni and this is the

derivation.

The other cases are more complicated.

(4) Composition:

The computation tree for �e(�!m;�!x ) = �a(�b0(

�!m;�!x ); : : : ;�br�1(�!m;�!x )) =

q has a top node c = he;�!m;�!� ; qi and has immediate predecessors c0; : : : ; cr�1; c0,where ct is the top node of the computation tree for �bt(

�!m;�!x ) for t < r, and c0

is the top node of the computation tree for �a(�b0(�!m;�!x ); : : : ;�br�1

(�!m;�!x )).For each j, �j is the union of the initial segments of xj used in ct. The derivationis hd1; : : : ; dr�1; d; ci where dt is the derivation of �bt(

�!m;�!x ) for t < r and d isthe derivation of �a(�b0(

�!m;�!x ); : : : ;�br�1(�!m;�!x )).

(5) Primitive Recursion:

The computation tree for �e(0;�!m;�!x ) = �a(

�!m;�!x ) = q0 has top node d0 =he; 0;�!m;�!� ; q0i with a single immediate predecessor c0 = ha;�!m;�!� ; q0i. Thederivation is hc0; d0i.

The computation tree for �e(n + 1;�!m;�!x ) = �b(�e(n;�!m;�!x ); n;�!m;�!x )) =

qn+1 has top node dn+1 = he; n+ 1;�!m;�!� ; qn+1i with two immediate predeces-sors, the top node dn of the computation tree for �e(n;

�!m;�!x ) and the top nodecn of the computation tree for �b(qn; n;

�!m;�!x )). For each j, �j is the union ofthe initial segments of xj used in cn and in dn. The derivation is hdn; cn; dn+1i.

(6) Search:

The computation tree for q = �e(�!m;�!x ) = (least p)�a(p;

�!m;�!x ) = 0 has topnode d = he;�!m;�!� ; qi with immediate predecessors c0; : : : ; cp where ct is the topnode of the computation tree for �a(t;

�!m;�!x ) for t � p. For each j, �j is theunion of the initial segments of xj used in ct for some t � p. The derivation ishc0; : : : ; cp; di.

We will often write �ye(�!m;�!x ) for �e(�!m;�!x ; y) and refer to the function �ye

as being computable from the oracle y.

Lemma II.1.2. The set of derivations is primitive recursive and, furthermore,the relation T (e; h�!m;�!� i; d) which indicates that d is the derivation of �e(

�!m;�!� )is also primitive recursive.


Sketch. The set of derivations may be de�ned by course-of-values recursion usingcoding and decoding of �nite sequences, all of which is primitive recursive. Thenthe values of e, �!m, �!� and �e(

�!m;�!� ) can be obtained from the last entry of the�nite sequence coded by the derivation d. See Chapter II of Hinman [87] fordetails.

II.1.1 Turing machines

The classic Turing machine, de�ned by Alan Turing, provides a very useful ap-proach to computable functions. It has a simple elegant format but neverthelesshas a strength equal to any other model of computing.

Our model of the Turing machine will be as follows. Let �0 be a �nitealphabet, let B denote the blank symbol (not included in �0), and let � =�0 [ fBg. A Turing machine tape consists of a potentially in�nite sequence ofsquares, on which symbols from the alphabet � may be stored, and possiblyerased or written over during a computation.

Each tape comes equipped with a pointer or reading head, which will bepointing at one of the entries during any step of a Turing machine computation.The entries on a tape are ordered as a0; a1; : : : beginning with a leftmost square.Initially each reading head points at the leftmost square of its tape. Turingmachine computations are based on two fundamental operations, the following.Say that the pointer on a tape is located over ai. The Turing machine can replacethe symbol ai with any other symbol. Then it can move from the current squareto ai+1 or to ai�1 (if i > 0) or remain at the current square.

A Turing machine M which de�nes a function 'M : �k0 ! � for some �nitek will have k input tapes, an output tape, and a �xed �nite number m of workor scratch tapes. The inputs �0; : : : ; �k�1 are written on the input tapes atthe start of the computation and the other tapes are initially empty. We willassume that the input tapes are read-only, that is, M does not ever write overany symbol on the input tapes and does not write any new symbols onto theempty squares of an input tape. The output tape is assumed to be write-only,that is, once a symbol is written onto the output tape, it cannot be changed.

The instructions for a Turing machine M to compute the function 'M aregiven by a �nite set Q of states, including some initial state s and a haltingstate h, together with a transition function

�M : Q� �k+m+1 ! Q� �m+2 � f ;!;`gk+m+1:

The state of the machine together with the symbols on the scanned squares,are used via the transition function to determine the operation of the machineas follows. Let the tapes be numbered so that tapes 0 through k � 1 are theinput tapes, tapes k through k +m � 1 are the scratch tapes, and tape k +mis the output tape. Suppose that M is in state q and that, for each i < k +m, pointer on tape i is scanning the symbol ai. Let �M (s; a0; : : : ; ak+m) =(q0; b0; : : : ; bk+m; X0; : : : ; Xk+m), where each Xi 2 f ;!;`g. Here we assumethat, for i < k, bi = ai and that, if ak+m+1 6= B, then bk+m+1 = ak+m+1. We

II.1. FORMAL DEFINITIONS OF THE COMPUTABLE FUNCTIONS 13

also assume that if bk+m 6= B, then Xk+m =! and otherwise Xk+m =`. Thenthe symbol ai is replaced on tape i by the symbol bi. The pointer on tape imoves right if X =!, moves left if X = and it is not the leftmost squarewhich is being scanned, and otherwise remains pointing at the same square.Finally, the machine transitions into state q0. If q0 = h, then the computationis �nished and the output 'M (�0; : : : ; �k�1) is the sequence of entries on theoutput tape. The length of the computation is the number of steps until thehalting state is reached, if any, and also represents the amount of time used inthe computation for the purpose of complexity theory. The amount of spaceused is the total number of squares on the work tapes which were ever writtenon during the computation.

Example II.1.3. Natural numbers are usually represented in reverse binaryform, so that 6 is represented as 011. (This is due to having a leftmost squareon each tape.) The function '(x) = x + 1 may be computed by the followingTuring machine M . M has three states, s, q and h and just two tapes, the inputtape and the output tape. The transition function has the following values.

�(s; 0; B) = (q; 0; 1;!)

�(s; 1; B) = (s; 1; 0;!)

�(s;B;B) = (h;B;B;`)�(r; 0; B) = (r; 0; B;!)

�(r; 1; B) = (r; 1; B;!)

�(r;B;B) = (h;B;B;`)

Here we omit any transition where the output tape is not scanning a blanksquare, since that situation cannot occur.

The computation '(101) = 011 (that is, 5+1 = 6) takes three steps, remain-ing in state s after the �rst step, moving to state r after the second step and�nishing in the halting state h after scanning the blank at the third step.

Frequently, we use computations to test whether a given input � meetscertain criteria, that is, belongs to some set A. Then our Turing machine Mmight output Yes or No if the input does or does not meet the criteria, or Mmight halt if � meets the criteria and not halt otherwise. In the �rst case,M demonstrates that the set A is computable, and in the second case, Mdemonstrates that A is computably enumerable.

Example II.1.4. Let A = f� 2 f0; 1g� : (9n)�(n) = 0 = �(n + 1)g. Wecan show that A is computably enumerable with the following simple Turing


machine. Here we do not need any work tapes or even an output tape.

�(s; 0) = (q;!)

�(s; 1) = (s;!)

�(s;B) = (s;!)

�(q; 0) = (h;`)�(q; 1) = (s;!)

�(q;B) = (s;!)

If the input string � is in A and n is the least such that �(n) = �(n+1) = 0,then the Turing machine takes n+1 steps to read through the �rst n+1 entriesof � and then halts. If � is not in A, then the machine take j�j + 1 steps toread through � (without �nding 00 and �nd the blank at the end of �. Then itsimply continues to read blanks and thus never halts.

Example II.1.5. Let A = f0n1n : n 2 Ng. We will give an informal descriptionof a Turing machineM which outputs Y if � 2 A and otherwise outputs N . Themachine M has one work tape where it copies the 0s from the input tape untileither a 1 or a B is read. The reading head on the work tape will be pointing tothe �nal 0. When a 1 is read on the input tape, M transitions to a new stateand begins erasing the 0s from the work tape. When a B is now read in theinput tape, M checks to see whether there is a B or a 0 on the work tape. If itis a B, then � is accepted by writing Y on the output tape. If it is a 0, then �is rejected by writing N on the output tape (in this case there are not enough 1sto match the initial sequence of 0s). If M �nds a 0 after some sequence of 1s,then again � is rejected. For the remaining case, if B is read on the input tapeafter a sequence of 0s but before any 1s are read, then � is also rejected.

Exercises

II.1.1. Show that the set of primitive recursive indices is itself a primitive re-cursive set. (You may assume here that the coding functions mapping(a0; a1; : : : ; an) to a = ha0; a1; : : : ; ani and the decoding functions (a)i = aiare primitive recursive.)

II.1.2. Prove Lemma II.1.1.

II.1.3. Show that the universal sequence f�ege2! is not uniformly primitive re-cursive, that is, the function f de�ned by f(e;m) = �e(m), is not itselfprimitive recursive.

II.2 Basic results

In this section, we state a number of results about computable functions whichwill be needed later. Most proofs are omitted; the reader is referred to Hinman

II.2. BASIC RESULTS 15

[87] and Soare [198]. For simplicity of expression, we will generally write �e(m)for a function of k variables rather than �e(m1; : : : ;mk) or �e(

�!m). Thus theresults given here apply to functions taking any number of variables.

Lemma II.2.1 (Padding Lemma). Each partial computable function �e has anin�nite set of indices, and furthermore, there is a primitive recursive, one-to-onefunction f such that, for all e and n, f(e; n) is an index for �e.

Sketch. Let f(e; n) be an index for the function which �rst computes �e(m),then adds n to the output, and �nally subtracts n from the output.

Theorem II.2.2 (Normal Form Theorem). (Kleene) There is a primitive re-cursive predicate T1(e;

�!m;�!� ; q) and a primitive recursive function U such that

�e(�!m;�!x ) = U(least qT1(e;

�!m;�!x dq; q))Sketch. Let the T predicate be given by Lemma II.1.2 and de�ne the predicateT1 so that, for any e;�!m;�!� ; q, T1(e;�!m;�!� ; q) if and only if there exists initialsegments �j of each �j such that T (e; h�!m;�!� i; q) and U outputs �e(

�!m;�!� ) fromthe derivation q.

Theorem II.2.3 (Enumeration Theorem). For any k; ` < !, there is a partialcomputable function � such that, for all e, �!m and �!x , �(e;�!m;�!x ) = �e(

�!m;�!x ).Proof. Just let �(e;�!m;�!x ) = U(least qT1(e;

�!m;�!x dq; q)), where T1 and U aregiven by Theorem II.2.2.

The �nite approximation �e;s at stage s of a partial computable function�e is de�ned as follows.

De�nition II.2.4. (i) �e;s(�!m;�!x ) = p if and only if

(9q < s)[T1(e;�!m;�!x dq; q) & U(q)) = p]:

(ii) �e;s(�!m;�!x ) converges (written �e;s(

�!m;�!x ) #) if �e;s(�!m;�!x ) = p for somep and otherwise �e;s(

�!m;�!x ) diverges ( �e;s(�!m;�!x ) "). Similar de�nitionsapply for �e;s(

�!m;�!� ).(iii) �e(

�!m;�!� ) = �e;s(�!m;�!� ), where s = j�!� j.

The following results are immediate from the de�nitions and the NormalForm Theorem above. For simplicity of expression, the results are written onlyfor a function of one real variable but applies to functions of several variablesas well.

Theorem II.2.5 (Master Enumeration Theorem). fhe;�!m;�; si : �e;s(�!m;�) #gand fhe;�!m;�; p; si : �e;s(�!m;�) = pg are both primitive recursive sets.

Theorem II.2.6. (Use Principle)

(a) �e(�!m;x) = n =) (9s)(9� � x)�e;s(�!m;�) = n.


(b) �e;s(�!m;�) = n =) (8t � s)(8� � �)�e;t(�!m; �) = n.

(c) �e;s(�!m;�) = n! (8x � �)�e(�!m;x) = n.

Theorem II.2.7 (s-m-n Theorem). For every m;n � 1, there exists a one-to-one primitive recursive function Smn such that, for all e; i1; : : : ; im; j1; : : : ; jn,

�Smn (e;i1;:::;im)(j1; : : : ; jn;�!x ) = �e(i1; : : : ; im; j1; : : : ; jn;

�!x )Proof. For m = 1, we want S11(e; i) to be the index for the function � suchthat �(j1; : : : ; jn;

�!x ) = �e(i; j1; : : : ; jn;�!x ). Let u be given by the Enumeration

Theorem so that �u(e; i; j1; : : : ; jn;�!x ) = �e(i; j1; : : : ; jn;

�!x ). Let Ck denotethe constant function Ck(

�!m;�!x ) = k and let Pi denote the projection functionPi(�!m;�!x ) = mi, both with n number and ` real variables. Then

�(j1; : : : ; jn;�!x ) = �u(e; i; j1; : : : ; jn;

�!x )= �u(Ce(

�!j ;�!x ); Ci(�!j ;�!x ); P0(�!j ;�!x ); : : : ; Pn�1(�!j ;�!x ));

so that

S11(e; i) = h4; n+ 1; `; u; h0; n; `; ei; h0; n; `; ii; h1; n; `; 0i; : : : ; h1; n; `; n� 1ii:Then Sm+1n may be de�ned recursively by

Sm+1n (e; i0; : : : ; im) = Smn (S1m+n(e; i0); i1; : : : ; im):

This result is very useful. Here is an example.

Proposition II.2.8. There is a primitive recursive function g such that, forall a and b, Wg(a;b) =Wa [Wb.

Proof. Let �(a; b;m) = U((least q)[T1(a;m; q)_T1(b;m; q)]) and let � have indexe. Then let g(a; b) = S21(e; a; b).

More importantly, we will need the following.

Theorem II.2.9 (Substitution Theorem). There is a primitive recursive func-tion f such that, for all e;m;A such that �Ab is total, �e(m;�

Ab ) = �f(b;e)(m;A):

Proof. Let R(e; b;m; �) if �e(m;�) #; R is primitive recursive by the MasterEnumeration Theorem. Now let g(e; b;m;A) = (least s)R(e; b;m;Ads) and

�c(e; b;m;A) = Adg(e; b;m;A);where we identify a �nite sequence with its code. Then

�e(m;�Ab ) = �e(m;�c(e; b;m;A)) = �d(e; b;m;A);

whered = h4; 3; 1; h1; 3; 1; 2i; ci:

Now apply the s-m-n Theorem to get f(b; e) = S21(d; e; b):

II.2. BASIC RESULTS 17

Theorem II.2.10 (Recursion Theorem). For any partial computable function�, there exists an index e such that, for all �!m, �e(

�!m;�!x ) = �(e;�!m;�!x ). Fur-thermore, there is a primitive recursive function g such that if � = �i, thene = g(i).

Proof. Given �, let �b(a;�!m;�!x ) = �(Sk+11 (a; a);�!m;�!x ) and let e = Sk+11 (b; b).

Then

�e(�!m;�!x ) = �b(b;

�!m;�!x ) = �(Sk+1(b; b);�!m;�!x ) = �(e;�!m;�!x ):

This leads to the following.

Theorem II.2.11 (Fixed Point Theorem). For any computable function f ,there exists an index e such that �e = �f(e). Furthermore, there is a primitiverecursive function h such that if f = �i, then e = h(i).

Proof. Let �(a;�!m;�!x ) = �f(a)(�!m;�!x ) and let e be given by the Recursion

Theorem such that �e(�!m;�!x ) = �(e;�!m;�!x ).

Corollary II.2.12. For any computable function f , there exists an index e suchthat We = Wf(e). Furthermore, there is a primitive recursive function h suchthat if f = �i, then e = h(i).

De�nition II.2.13. A function F : (NN)` ! NN is (partial) computable (orcomputably continuous) if there is a (partial) computable functional � suchthat, for all �!x and n, �(n;�!x ) = F (�!x )(n).Theorem II.2.14. Let F : (NN)` ! NN be total. Then F is continuous if andonly if F is computable in some oracle A � N.Proof. ( �). We give the proof for ` = 1. Let A be the given oracle andsuppose that F (x)(m) = �(m;x;A). It su�ces to show that, for any m andn, fx 2 NN : F (x)(m) = ng is an open set. Suppose that F (x)(m) = n. Bythe Use Principle (Theorem II.2.6) there is some s and some �nite � � x and�nite � � �A such that �e;s(m;�; �) = n. It follows that for all x 2 I(�),�e;s(m;x; �) = n and hence I(�) � F�1(fy : y(m) = ng).

(�!): Let F : NN ! NN be continuous. Then for each m and n, Um;n =fx : F (x)(m) = ng is open and thus for each x 2 Um;n, there exists a �nite� � x such that I(�) � Um;n. Let

A = fhm;n; �i : I(�) � Um;ng:

To compute F (x)(m) from x simply �x m and search for the least hm;n; �i 2 Asuch that � � x; then F (x)(m) = n.

Exercises


II.2.1. Use the s-m-n Theorem to prove Lemma II.1.1.

II.2.2. Use induction to prove Theorem II.2.6 (a).

II.2.3. Show that there is a primitive recursive function g such that, for all a andm, m 2Wg(a) () (9p)hm; pi 2Wa.

II.2.4. Use the Recursion Theorem to show that the Fibonacci sequence 1; 1; 2; 3; 5; 8; : : :is computable.

II.3 Computably enumerable sets

De�nition II.3.1. (i) A subset A of N is computably enumerable ( c. e.) ifA is the domain of some partial computable function.

(ii) The c. e. sets can be enumerated in the form

We = fm : �e(m) #g = fm : (9q)T (e;m; q)g:

(iii) We;s = fm : �e;s(m) #g:The following lemma is immediate from Theorem II.2.5.

Lemma II.3.2. fhe;m; si : m 2We;sg is primitive recursive.

There are several equivalent de�nitions.

De�nition II.3.3. A set A � Nk is �01 (resp. �B1 ) if there is a computable rela-tion R (resp. computable in B) such that, for all �!m, �!m 2 A () (9p)R(p;�!m).

Theorem II.3.4 (Normal Form Theorem for c. e. sets). A set A is c. e. if andonly if it is �01.

The proof is left as an exercise. Observe that any computable set is trivially�01 and hence also is computably enumerable.

Theorem II.3.5 (Quanti�er Contraction Theorem). If W is a c. e. set, thenfm : (9p)hp;mi 2Wg is a c. e. set.

Proof. Let V = fm : (9p)hm; pi 2 Wg. Then m 2 V () (9q)hm; (q)0i 2We;(q)1 . Thus V is c. e. by the Normal Form Theorem for c. e. sets.

The intended meaning of the term \computably enumerable set" is thatthere is an e�ective listing a0; a1; : : : of the set.

Theorem II.3.6 (Listing Theorem). A set A is c. e. if and only if either A = ;or A is the range of a total computable function.

II.3. COMPUTABLY ENUMERABLE SETS 19

Proof. ((=): If A = ;, then A is c. e. If A = f�e(m) : m 2 Ng where � is atotal computable function, then

n 2 A () (9n)(9s)�e;s(p) = n:

Thus A is c. e. by Theorem II.2.5 and the Quanti�er Contraction Theorem.(=)): Let A = We 6= ; and choose a 2 A. Then A is the range of the

following computable function.

f(hm; si) =(m; if m 2We;s+1 nWe;s;

a; otherwise:

Theorem II.3.7 (Complementation Theorem). A set A is computable if andonly if both A and N nA are c. e.

Proof. ((=): If A is computable, then NnA is also computable and hence bothsets are c. e..

(=)): Suppose that A =Wa and N nA =Wb and let �(m) = (least s)[m 2Wa;s _ m 2 Wb;s]. Then � is a total computable function and m 2 A ()m 2Wa;�(m), so that A is computable.

There are natural noncomputable c. e. sets.

De�nition II.3.8. (a) K = fe : e 2Weg;(b) K0 = fhm; ei : m 2Weg.

Proposition II.3.9. K and K0 are noncomputable c. e. sets.

Proof. It follows from Lemma II.3.2 that K0 and K are c. e. sets. Supposenow that K were computable, so that N nK is c. e., by the ComplementationTheorem, and choose a such that N nK =Wa. Then, for any m,

m 2Wm () m 2 K () m =2Wa

and when m = a we obtain the contradiction

a 2Wa () a 2 K () a =2Wa:

Now a 2 K () ha; ai 2 K0, so that K would be computable if K0 werecomputable.

Exercises

II.3.1. Prove lemma II.3.2 and the Normal Form Theorem for c.e. sets. Hint:Use the corresponding results for partial computable functions.


II.3.2. Show that a partial function is partial computable function if and only ifthe graph is �01.

II.3.3. Use the s-m-n Theorem to obtain a primitive recursive function such thatfor any e, fm : (9p)hm; pi 2Weg =Wf(e).

II.3.4. Show that if A is a �01 relation and B = fhm; pi : (8n < p)hm;ni 2 Ag,then B is also �01.

II.4 Computability of real numbers

Any set A of natural numbers represents a real number rA 2 [0; 1] where rA =Pn2A 2

�n�1. For every real r in [0; 1], there exists A � N such that r = rAand r has a unique representation except for dyadic rationals r, which haveexactly two such representations. The real rA is said to be computable if A is acomputable set. For an arbitrary real x, we have x = i+ r, where i is an integerand r 2 [0; 1], so we will say that x is computable if and only if r is computable.

The unit interval [0; 1] � R has a natural linear ordering and this correspondsto the lexicographic ordering on f0; 1gN.De�nition II.4.1. For x; y 2 NN, x <lex y if x(n) < y(n) where n is the leastsuch that x(n) 6= y(n).

It is easy to see that <lex is a linear ordering on NN. We sometimes saythat \x is left of y" if x <lex y, since this �ts the picture of the tree N�. Forx; y 2 f0; 1gN, if rx 6= ry, then rx < ry () x <lex y. If r is a dyadic rational,then there are two representations x 6= y such that rx = ry = r and these aresuccessors under <lex.

Another useful way of determining the complexity of a real number is bymeans of Dedekind cuts of rationals. Rational numbers may be represented asquotients of integers and thereby as �nite sequences of natural numbers. Thuswe may view the set Q of rational numbers as a computable structure equippedwith a computable ordering and computable operations of addition, subtraction,multiplication and division. The Dedekind cut L(r) of a real number is de�nedby

L(r) = fq 2 Q : q � rg:It turns out that the complexity of the Dedekind cut is quite useful in com-putable analysis.

Proposition II.4.2. For any real r, r is computable if and only if L(r) iscomputable.

Proof. It su�ces to consider r 2 [0; 1], so let r = rx for some x 2 f0; 1gN. Ifr is rational, then both r and L(r) are computable. So let r be irrational andsuppose �rst that r is computable. Then, for any rational q,

q < r () (9n)q <nXi=0

x(i)2�i�1

II.4. COMPUTABILITY OF REAL NUMBERS 21

and

q > r () (9n)q > 2�n�1 + r () (9n)q > 2�n�1 +

n�1Xi=0

x(i)2�i�1:

Next suppose that L(r) is computable. Then we can recursively de�ne x so thatr = rx as follows. Let x(0) = 0, if r < 1

2 and x(0) = 1 otherwise. Given x(n),let

x(n+ 1) =

(0; if r < 2�n�2 +

Pi=0n x(i)2

�i�1;

1; otherwise.

There is a nice characterization for �01 and �01 Dedekind cuts.

Proposition II.4.3. (a) L(r) is �01 if and only if r = limn qn, where fqngn2!is a computable, increasing sequence of rationals.

(b) L(r) is �01 if and only if r = limn qn, where fqngn2! is a computable,

decreasing sequence of rationals.

Proof. (a) Suppose �rst that r = limnqn where fqngn2! is a computable, in-creasing sequence of rationals. Then, for any rational q, q < r () (9n)q < qn.

Suppose now that L(r) is �01 and let L(r) have a computable enumerationas p0; p1; : : : . Then we can de�ne a computable nondecreasing sequence qn ofrationals with limit r by

qn = maxfp0; p1; : : : ; png:

It is routine to convert this into an increasing sequence.(b) If L(r) is �0

1, then L(1� r) is �01, since q < 1� r () r < 1� q. Thus1�r = limnpn where fpngn2! is a computable, increasing sequence of rationals.It follows that r = limn(1�pn) is the limit of a computable, decreasing sequence.Conversely, if r = limnqn where fqngn2! is a computable, increasing sequenceof rationals, then 1� r is the limit of a decreasing sequence so that L(1� r) is�01 and hence L(r) is �0

1.

The following notions are important in computable analysis.

De�nition II.4.4. Let r be a real number. Then

(a) r is lower semicomputable if it is the limit of an increasing computablesequence of rationals;

(b) r is upper semicomputable if it is the limit of a decreasing computablesequence of rationals;

(c) r is weakly computable if it is either lower semicomputable or upper semi-computable.


It would be natural to say that rx is �01 if the set with characteristic function

x is �01, but there is no corresponding equivalence as in Proposition II.4.2.One direction only holds. In example II.5.3 below we will construct a lowersemicomputable real r which is not the characteristic function of a c. e. set.

Proposition II.4.5. (a) If A is �01, then L(rA) is �01;

(b) If A is �01, then L(rA) is �

01.

Proof. (a) If A is �nite, then of course rA is rational and therefore L(rA) is com-putable. Suppose therefore that A is �01 and in�nite and let A have computableenumeration a0; a1; : : : without repetition. Then for any rational q,

q < rA () (9n)q <nXi=0

2�ai�1:

(b) If A is �01, then N nA is �01 and rNnA = 1� rA, so that L(1� rA) is �01 and

therefore L(rA) is �01.

Exercises

II.4.1. Show that a real number r is computable if and only if there is a com-putable sequence qn of rationals such that jqn � rj < 2�n for all n.

II.5 Turing, many-one, and truth-table reducibil-

ity

De�nition II.5.1. (i) A is many-one reducible (m-reducible) to B (A �mB if there is a computable function f such that a 2 A () f(a) 2 B

(ii) A is one-one reducible to B (A �1 B if there is a one-to-one computablefunction f such that a 2 A () f(a) 2 B

(iii) C is m-complete (or �01 complete) if A �m C for all c. e. sets A.

For example, any c. e. set is m-reducible to K0, since m 2We () hm; ei 2K0; here the function f is given by f(m) = hm; ei. Thus K0 is m-complete.The following useful lemma is left as an exercise.

Lemma II.5.2. If every �01 (respectively �01) set is m-reducible to A, then A

is not �01 (resp. �01).

Example II.5.3. Let K be a noncomputable c. e. set and let A = f2n : n 2Kg[f2n+1 : n =2 Kg. Then A is a di�erence of c. e. sets and is m-complete forboth �0

1 and �01 sets. It follows from Lemma II.5.2 that A is not �01, but rA is the

limit of the nondecreasing computable sequence fqsgs2! de�ned as follows. Let

II.5. TURING, MANY-ONE, AND TRUTH-TABLE REDUCIBILITY 23

K = [sKs where Ks is a uniformly computable �nite subset of f0; 1; : : : ; s� 1gand let

qs =Xf2�2n�1 : n 2 Ksg+

Xf2�2n�2 : n < s & n =2 Ksg:

Observe that if n 2 Ks nKs�1, then 2�2n�1 is added to the �rst part of qs and2�2n�2 is subtracted from the second part, so that qs�1 < qs.

De�nition II.5.4. (i) A �m B if A �m B and B �m A.

(ii) A �1 B if A �1 B and B �1 A.Proposition II.5.5. Suppose that A �m B. If B is c. e., then A is c. e. andif B is computable, then A is computable.

The proof is left as an exercise.

De�nition II.5.6. A is computably isomorphic to B (written A � B) if thereis a computable permutation � of N such that �[A] = B.

The following is an e�ective version of the classic Cantor-Schr�oder-BernsteinTheorem.

Theorem II.5.7 (Cantor-Schr�oder-Bernstein Theorem). Let A and B be setsand let f and g be injections, f : A ! B and g : A ! B; then there exists anisomorphism h : A! B.

Banach's version of the Cantor-Schr�oder-Bernstein Theorem adds the re-quirement that, for all a 2 A, either h(a) = f(a) or h(a) = g�1(a).

Theorem II.5.8 (Myhill Isomorphism Theorem). A � B () A �1 B.Proof. The direction (=)) is trivial. Suppose therefore that A �1 B via f andB �1 A via g. We will de�ne � in stages �s = fhm0; n0i; : : : ; hm2s�1; n2s�1igso that for all m < s, m 2 Dom(�s) and m 2 Ran(�s) and such that

mi 2 A () ni 2 B:

We begin with �0 = ;.Stage s + 1: Let �s be given as above and let m = m2s be the least

m =2 Dom(�s). �(m) is computed as follows. First compute b0 = f(m) andcheck if b0 2 Ran(�s). If not, then b0 = �s+1(m). If so, then compute b1 =f(��1s (b0)) and again check whether b1 2 Ran(�s) and let b1 = �s+1(m) ifnot and b2 = f(��1s (b0)) if so. Observe that after we reach b2s�1, Ran(�s) =fb0; b1; : : : ; b2s�1g is exhausted, so that b2s = �s+1(m).

Next let n = n2s+1 be the least not in Ran(�s) [ f�s+1(m)g and similarlyde�ne a0 = g(n), a1 = g(�s(a0)), and so on to obtain ��1s+1(n).

We now show that, in the setting of m-reduciblity, Banach's version of theCantor-Schr�oder-Bernstein Theorem is not e�ective,


Theorem II.5.9. There exist computable injections f : N! N and g : N! Nsuch that, for any computable permutation h of N, there is some i such thath(i) 6= f(i) and h(i) 6= g�1(i).

Proof. Let K be a noncomputable c. e. set and let be a one-one computablefunction with range K. We de�ne injections f and g as follows.

(1) f(2m(2n+ 1)) =

8><>:2n+ 1; if (m� 2) = n;

2m(2n+ 1); if (9j < m� 2) (j) = n;

2m+1(2n+ 1); otherwise:

(2) g(2m(2n+ 1) =

(2m+1(2n+ 1); if (9j � m� 2) (j) = n;

2m(2n+ 1) otherwise:

Now by way of contradiction, let h be a computable permutation such that,for all i, either h(i) = f(i) or h(i) = g�1(i). We claim that

n =2 K () h(2n+ 1) = 2n+ 1:

This would contradict the assumption that K is not computable. It remains toverify the claim. Suppose �rst that n =2 K. Then g(2n+1) = 2n+1 and 2n+1is not in the range of f , so that h(2n+1) = g�1(2n+1) = 2n+1. Next supposethat n 2 K and let n = (m). Then 2m+2(2n + 1) is not in the range of g, soh(2m+2(2n+ 1)) = f(2m+2(2n+ 1) = 2n+ 1, so that h(2n+ 1) 6= 2n+ 1.

Let Sent be the set of propositional sentences on variables a0; a1; : : : . Thereis a an e�ective enumeration 0; 1; : : : of these sentences, so that we mayidentify the sentence n with n in context. For any set B � N and any sentence 2 Sent, we say that B j= if is true under the truth assignment whichmakes ai true if and only if i 2 B. Let Btt denote the set of 2 Sent such thatB j= . Then we say that A is truth-table reducible to B (A �tt B) if A �m Btt.Equivalently, A �tt B if and only if there exists a computable relation R and acomputable function f such that for any n, n 2 A () R(hB � f(n)i). It isimmediate that many-one reducibility implies truth-table reducibility.

Theorem II.5.10. (Trakhtenbrot-Nerode [211, 157]) A �tt B if and only ifthere is a total, computable function � : f0; 1gN ! f0; 1gN such that �(B) = A.

Proof. Suppose �rst that there is a total computable function � such that�(B) = A. We will de�ne a function f : N! Sent such that a 2 A () B j=f(a). Given a, use the Master Enumeration Theorem II.2.5 to compute theleast n such that �(a; �) # for all � 2 f0; 1gn. For each � such that �(a; �) = 1,

let '� be the conjunctionVn�1i=0 bi, where bi = ai if �(i) = 1 and bi = :ai if

�(i) = 0. Thus B j= '� if and only if, for all i < n, �(i) = 1 () i 2 B.Finally, let ' be the disjunction of f'� : �(a; �) = 1g. Then

a 2 A () �(a;B) = 1 () B j= ':

II.5. TURING, MANY-ONE, AND TRUTH-TABLE REDUCIBILITY 25

Next suppose that A �tt B and let f be given so that a 2 A () B j= f(a).Then in general, de�ne � so that

�(c;X) = 1 () X j= f(c)

Here we can compute for each propositional variable ai occuring in f(c), whetherX j= ai immediately from X and then use truth tables to check whether X j=f(c).

This leads naturally to Turing reducibility, where we allow partial com-putable functions.

De�nition II.5.11. (i) A is Turing reducible to B (written A �T B) ifthere is a functional � such that, for all m, A(m) = �(m;B).

(ii) A is Turing equivalent to B if both A �T B and B �T A.(iii) The Turing degree a of A is the equivalence class of A under Turing equiv-

alence.

Informally, this means that A �T B if A can be computed using B asan oracle. It follows from Theorem II.5.10 that truth-table reducibity impliesTuring reducibility. It is easy to see that �T is an equivalence relation. TheTuring degrees are partially ordered by �T with least element 0, which is theTuring degree of a recursive set.

It is possible that A �T B but A is not truth-table reducible to B. (Seethe exercises below.) However, there is a family of sets B for which the tworeducibilities are equivalent.

De�nition II.5.12. A function g 2 NN is almost computable (or hyperimmune-free) if for all f �T g, there exists a computable function h such that f(n) � h(n)for all n.

Theorem II.5.13. Suppose g is almost computable. Then for all f , if f �T g,then f �tt g.Proof. Suppose that y is almost computable and that f(n) = �(n; g) where �is computable. De�ne u(n) to be the least k such that �(n; gdk) #. Then u iscomputable from g and hence there is a computable function h such that g(n) �h(n) for all n. It follows that f(n) = �(n; gdh(n)) for all n, so that � can beextended to a total function by letting (n; x) = 0 whenever �(n; xdh(n)) ".Hence f is truth-table reducible to g, as desired.

Exercises

II.5.1. Prove Lemma II.5.2 and show that A is �01 complete if and only ifK �m A.

II.5.2. Prove Proposition II.5.5.

II.5.3. Show that �m and �T is transitive.


II.5.4. Show that the two de�nitions of truth-table reducibility are equivalent.

II.5.5. Give an example to show that A �T B but not A �tt B. (Hint: let A bethe set of e such that �e de�nes a total functional Fe, where y = Fe(x)means that y(m) = �e(m;x). Then let e 2 B () (e 2 A & �e(e;A) =0).

II.6 The jump and the arithmetical hierarchy

De�nition II.6.1. 1. WAe = fm : �e(m;A) #g.

2. B is c. e. in A if B =WAe for some e.

3. The jump of A is KA0 = fhe;mi : m 2WA

e g and is denoted by A0.

4. A(n) is the nth jump of A, that is, A(0) = A and A(n+1) = (A(n))0.

The following two theorems generalize from results of Section II.3 and II.5.

Theorem II.6.2. The following are equivalent:

(a) B is c. e. in A;

(b) B = ; or B = Ran(�Ae ) for some e;

(c) B is �A1 .

Theorem II.6.3. B �T A if and only if B and N nB are both c. e. in A.

Theorem II.6.4 (Jump Theorem). (a) A0 �T A.

(b) B is c. e. in A if and only if B �1 A0.(c) B �T A if and only if B0 �1 A0.

Proof. Parts (a) and (b) relativize from results of sections II.3 and II.5. Forpart (c), suppose �rst that B �T A and let B(n) = �b(n;A). Then for anye, �e(m;B) = �e(m;�

Ab ) and by Theorem II.2.9, there is a primitive recursive

function such that �e(m;B) = �f(e)(m;A). Thus he;mi 2 B0 () hf(e);mi 2A0. For the other direction, suppose that B0 �1 A0. Then B and NnB are bothc. e. in A and therefore B �T A by Theorem II.6.3.

In particular, there is an in�nite hierarchy of degrees 0(n) = deg(;(n)).The arithmetical hierarchy of sets of natural numbers may be de�ned as

follows.

De�nition II.6.5. Let R � Nk � (NN)` and let n > 0 be a natural number.

1. R is �00 if it is computable.

2. R is �0n if Nk � (NN)` nR is �0n.

II.6. THE JUMP AND THE ARITHMETICAL HIERARCHY 27

3. R is �0n+1 if it is the projection of a �0n set, that is, if there exists a �0

n

relation B such that, for all �!m and �!x :R(�!m;�!x ) () (9j)B(j;�!m;�!x ):

4. R is �0n if it is both �0n and �0

n.

Note of course that the �01 sets are just the computably enumerable sets.These de�nitions can be relativized to any oracle C to de�ne the �0n[C], �

0n[C]

and �0n[C] sets and relations.

Here are some basic facts about the arithmetical hierarchy. Part (e) refersto bounded quanti�cation. See [198] for proofs.

Theorem II.6.6. (a) A 2 �0n [�0n and m > n implies A 2 �0

m;

(b) A;B 2 �0n(�0n) =) A \B;A [B 2 �0n(�0

n);

(c) If R 2 �0n for n > 0 and A = fm : (9p)R(m; p)g, then R is �0n;

(d) If B �m A and A 2 �0n, then B 2 �0n;(e) If R 2 �0n and A = fhm; pi : (8i < p)R(i;m; p)g, then A 2 �0n.An important result is the following.

Theorem II.6.7 (Post's Theorem). For any subset A of N:

(a) A is �0n+1 if and only if it is c. e. in ;(n).(b) A is �0

n+1 () A �T ;(n).Proof. The proofs are by induction on n. For n = 0, both parts are immediate.Now suppose by induction that (a) and (b) are true for n and for all subsets ofN.

First we show that ;(n+1) is �0n+1. That is, by induction assume that ;(n)is �0n. Then

he;mi 2 ;(n+1) () (9s)(9�)[� � ;(n) & �e(m;�) #]:But in general, � � C if and only if (8i < j�j)[�(i) = 1 () i 2 C], so that forC = ;(n), this condition is a disjunction of �0n and �0

n clauses. It follows thatthe quanti�ed condition is �0

n+1 and hence ;(n+1) is �0n+1 by Theorem II.6.6.

Now suppose that A is c. e. in ;(n). Then A �1 ;(n+1) and therefore A is�0n+1.

Conversely, suppose that A is �0n+1 and let R be �0n such that m 2 A ()

(9p)R(m; p). Then NnR is �0n and hence c. e. in ;(n�1) by induction. It followsfrom the Jump Theorem that N n R �1 ;(n) and therefore R �T ;(n). Since Ais c. e. in R, it follows that A is also c. e. in ;(n).

For part (b), A is �0n+1 if and only if both A and N nA are �0n+1, which is

if and only if A and N n A are c. e. in ;(n) (by (a)). But this is if and only ifA �T ;(n) by the Jump Theorem.


De�nition II.6.8. A is said to be �0n (�0n) complete if A is �0n (�0

n) and every�0n (�0

n) set is m-reducible to A.

It follows from Post's Theorem that ;(n) is �0n complete for all n > 0.�02 sets and functions will be of particular interest.

De�nition II.6.9. Let ffsgs2! be a sequence of total functions from NN.

(i) lims fs = f means that, for all m, there exists s such that f(m) = ft(m)for all t � s;

(ii) h is a modulus of convergence for ffsgs2!, if, for all m and all s � h(m),fs(m) = f(m).

Given a uniformly computable sequence ffsgs2! of functions with limit fand modulus of convergence h, f is always computable in h.

Lemma II.6.10 (Modulus Lemma). If A is c. e. and f �T A, then there is auniformly computable sequence ffsgs2! such that lims fs = f and a modulus ofconvergence h �T A.Proof. Let A =Wi be c. e., let �s =Wi;sds and let f = �Ae . Now let

fs(m) =

(�e;s(m;�s); if convergent;

0; otherwise.

andh(m) = (least s)(9z � s)[�e;s(m;�sdz) # & �sdz = Adz]:

Then ffsgs2! is a computable sequence with limit f and h is a modulus ofconvergence which is computable from A.

In particular, this implies that any �02 function is the limit of a computable

sequence.

Lemma II.6.11 (Limit Lemma). f �T A0 if and only if there exists an A-computable sequence ffsgs2! such that f = lims fs.

Proof. (=)): This follows from the Modulus Lemma relativized to A, sincef �T A0 if and only if f is c. e. in A.

((=): Let f = lims fs and let h(m) = (least s)[(8t � s)ft(m) = fs(m)].Since ffsgs2! is computable in A, it follows that h �T A0, so that f �T A0 aswell.

For real numbers, we say that rx is computably approximable if x is �02, so

that r is computably approximable if and only if it is the limit of a computablesequence of rationals.

For any �02 set A, it follows from the Jump Theorem that ;0 �T A0 �T ;00.

De�nition II.6.12. Let A �T ;0; A is low if A0 = ;0; A is high if A0 = ;00.

II.7. THE LATTICE OF C. E. SETS 29

Clearly any computable set is low, whereas ;0 is high. A c.e., noncomputablelow set is constructed in Soare [198] (p. 111).

Exercises

II.6.1. The di�erence B n C of two c. e. sets is said to be a d. r. e. set. Moregenerally, a set C is n-r.e. if there is a computable sequence fAsgs2Nsuch that A = limsAs and such that (i) A0 = ; and (ii) for each m,card(fs : As+1(m) 6= As(m)g) � n. Show that for each n, there is an(n+ 1)-r.e. set which is not n-r.e.

II.7 The lattice of c. e. sets

The lattice E of c. e. sets is ordered by inclusion and has the natural operationsof union and intersection. The lattice E� is the quotient of E under equalitymodulo �nite di�erence.

In this section, we consider properties of c. e. sets related to the lattice.

De�nition II.7.1. (i) A set is immune if it is in�nite but contains no in�-nite c. e. set;

(ii) A c. e. set A is simple if N nA is immune.

Simple sets were �rst constructed by Post [168] as a partial solution to Post'sProblem, which was to �nd natural intermediate c. e. degrees. It is easy to seethat simple sets are neither computable nor m-complete.

De�nition II.7.2. Let R be a property of c. e. sets, that is R � E.(i) R is lattice-theoretic or invariant in E (E�) if it is invariant under all

automorphisms of E (E�).(ii) R is elementary lattice-theoretic or de�nable in E (respectively, E�) if

there is a �rst-order formula ' with one free variable in the languagef�;_;^; 0; 1g of lattice theory such that R(A) if and only if E j= '(A)(resp. E� j= '(A)).

Clearly any de�nable property is also invariant.

Lemma II.7.3. The properties of computability and of �niteness are both de-�nable in E.Proof. A is computable if and only if

E j= (9y)[A _ y = 1 & A ^ y = 0]:

A is �nite if and only if

E j= (8y)[y � A �! A is computable]:


In the lattice E�, A is simple if, for all B 6= 0, A\B 6= 0. Thus, the propertyof being simple is elementary lattice-theoretic in E�. The following lemma willimply that simplicity is also de�nable in E .Lemma II.7.4. If a property R is preserved under �nite di�erences, then R isde�nable in E if and only if R is de�nable in E�.Proof. Let R be preserved under �nite di�erences. If R is de�nable in E by aformula ', then the same formula works in E�. Suppose next that R is de�nablein E�. Since �niteness is de�nable in E , the relation =� of equality modulo �nitedi�erence is also de�nable in E . Thus the de�nition from E� may be rewrittenin E� by replacing all occurrences of = with =�.

De�nition II.7.5. (i) If e =Pki=0 ei2

i, then De = fi � k : ei = 1g. (ThusD0; D1; : : : e�ectively enumerates the �nite sets of natural numbers.)

(ii) A sequence fFngn2N of �nite sets is a strong (weak) array if there is acomputable function f such that Fn = Df(n) (Fn =Wf(n)).

(iii) An in�nite set B is hyperimmune (h-immune) (respectively, hyperhyper-immune (hh-immune)) if for any pairwise disjoint strong (respectively,weak) array, Fn \B = ; for some n.

(iv) A c. e. set A is hypersimple (h-simple) (respectively, hyperhypersimple(hh-simple)) if N nA is h-immune (respectively, hh-immune).

It is easy to see that hh-simple implies h-simple and that h-simple impliessimple.

De�nition II.7.6. (i) A function f majorizes a function g if f(n) � g(n)for all n and f dominates g if f(n) � g(n) for all but �nitely many n.

(ii) The principal function pA of an in�nite set A is de�ned by pA(n) = an,where a0 < a1 < � � � enumerates A is increasing order.

The following result is easy to prove.

Theorem II.7.7. An in�nite set A is hyperimmune if and only if no computablefunction majorizes pA.

Dekker used this characterization to show that every nonzero c. e. degreecontains a h-simple (hence simple) set.

A degree a is said to be hyperimmune-free if it does not contain any hyper-immune sets. A set A is sometimes said to be almost recursive if its degree ishyperimmune-free, that is, any function computable in A is dominated by somerecursive function.

We will need the following result from Soare [198] (p. 85) in our study of�01 classes.

Theorem II.7.8. For any noncomputable c. e. set B, there is a simple, nonhy-persimple c. e. set A �T B. Furthermore, for any c. e. set C and any in�niteset D �T C, if A \D = ;, then B �T C.


Proof. (based on [198], p. 85). The requirements are fourfold:

(i) A is simple;

(ii) A �T B;

(iii) A is not hypersimple;

(iv) B �T A.

Let f be a one-to-one computable function with rangeB and letBs = ff(0); : : : ; f(s)g.A is enumerated in stages As, beginning with A0 = ;. Let N nAs = fas0 < as1 <: : : g. There are two actions which may be taken at stage s+ 1.

Step 1. Here we take action to satisfy requirements (i) and (ii). For all e � s,attention is required if We;s \As = ; and

(9n)[n > 3e & n 2We;s & f(s+ 1) < n]:

In this case, put into As+1 the least such n corresponding to e. If no such eexists, do nothing.

Step 2 Here we take action to satisfy requirements (iii) and (iv) by puttinga3f(s+1)+1 into A.

We show that the requirements are satis�ed.

(i) Suppose by way of contradiction thatWe is in�nite andWe\A = ;. Thenhere is an algorithm for testing m 2 B, that is, �nd s and n > maxfm; 3egsuch that n 2 We;s; then by Step 1, f(t) � n > m for all t > s, so thatm 2 B () m 2 Bs.

(ii) We claim that Bsdn = Bdn implies Asdn = Asdn. To see this, let m < nand m 2 At+1 n At. Then in Step 1, we have m > f(t + 1), and in Step 2,we have m = as3f(t+1)+1 > f(t + 1), so that in either case f(t + 1) < n and

f(t+ 1) 2 Bt+1 nBt. Thus to test m 2 A, just compute from B a stage s suchthat Bsdm+ 1 = Bdm+ 1 and then m 2 A () m 2 As.

(iii) Note that jA \ [0; 3e]j � 2e, since at most e elements � 3e are put intoA under Step 1 (one from each Wi, i < e) and at most e elements under Step 2(one for each i 2 B, i < e, since a3f(s+1)+1 = a3i+1 > 3i for i = f(s+ 1) 2 B).Thus N nA is majorized by the function 3x and A is not h-simple.

(iv) To test m 2 B, use A to compute s such that as3m+1 = limt at3m+1; then

m 2 B () m 2 Bs. Thus B �T A.Now let C be a c. e. set and let D be any in�nite set such that D �T C and

A \D = ;. Let D = fd0 < d1 < : : : g. By the Modulus Lemma, there exists auniformly computable double sequence fdsigi;s2N such that lims d

si = di for all i

with a modulus of convergence computable in C.

Let s(e) = f�1(e) and use the Recursion Theorem to de�ne a computablefunction h(e) such that Wh(e) = ; if e =2 B and otherwise

Wh(e) = fds(e)0 ; ds(e)1 ; : : : ; d

s(e)3(h(e))g:


We may assume without loss of generality that h(e) > e for all e. Use C to

compute a function r(e) such that dr(e)i = di for all i � g(h(e)). Let

B = fe 2 B : e =2 Br(e)g:

There are two cases.Case 1. B is �nite. Then clearly B �T C.Case 2. B is in�nite. Here is the procedure to test whether b 2 B using the

function r. First �nd e 2 B such that 3(h(e)) > b. Since e 2 B, r(e) < s(e)

and therefore ds(e)i = di for all i � g(h(e)), so that Wh(e) � D � N n A. It

follows that Wh(e) contains an element u > 3h(e); let sb be a stage such thatu 2Wh(e);sk . It follows from Step 1 that f(s) � u > b for all s � sk, so that

b 2 B () b 2 Bsb :

Since sb can be computed from C uniformly in b, it follows that B �T C.We will obtain a lattice-theoretic characterization due to Lachlan [123] of

hhsimple using the following two lemmas. For any c. e. set C, let L(C) denotethe lattice (under inclusion) of c. e. supersets of C.

Lemma II.7.9. (Lachlan) For any c. e. set C, if L(C) is a Boolean algebra,then C is hh-simple.

Proof. Suppose that C is not hhsimple as witnessed by the disjoint weak arrayfWf(n)gn2N and let A = C [Sn(Wn\Wf(n)). Suppose by way of contradictionthat Wn is the complement of A in L(C) and choose m 2 Wf(n) n C. Sincem =2 C, it follows that m 2 A () m 2Wn, a contradiction.

Theorem II.7.10 (Owings Splitting Theorem). Let C � B be c. e. sets suchthat B nC is not co-c. e. Then there exists disjoint c. e. sets A0 and A1 (whoseindices may be obtained uniformly from those of B and C) such that

(i) B = A0 [A1;

(ii) Ai n C is not co-c.e. for i = 0; 1;

(iii) For any c. e. set W , if C [ (W nB) is not c. e., then C [ (W nAi) is notc. e. for i = 0; 1.

Proof. Let f be a 1:1 computable function with range B and let fCsgs2N beany computable enumeration of C. We try to meet the requirements

Ps : f(s) 2 A0;s [A1;s; s 2 N

andRhe;ii : Ai n C 6= N nWe; for i = 0; 1:

Requirement Rhe;ii requires attention at stage s + 1 if f(s + 1) 2 We;s andf(s+ 1) � g(e; i; s).


Stage s = 0: Put f(0) 2 A0 and set g(e; i; 0 = 0 for all e; i.

Stage s+ 1:

Step 1: If there exists x � g(e; i; s) such that x 2 We;s \ (Ai;s n Cs), setg(e; i; s+ 1) = g(e; i; s). Otherwise g(e; i; s+ 1) = s+ 1.

Step 2: Let y = f(s+1) and choose the least he; ii such that Rhe;ii requiresattention at stage s + 1. Then put y 2 Ai;s+1. If no such e; i exist, then puty 2 A0;s+1.

Let Ai = [sAi;s. Clearly B = A0 [A1.

To prove (ii), assume that Ai n C = N nWe. We must show that B n C isco-c. e. to obtain a contradiction. For he0; i0i < he; ii, there are two possibilities.Either lims g(e

0; i0; s) = ze0;i0 <1 so that after some stage s, we never put anyy 2 Ai for the sake of Rha;ii, or lims g(e; i; s) =1. Let z be the maximum of theze0;i0 and choose s0 large enough so that for all he; ii of the �rst type, g(e0; i0; s)has already converged to ze0;i0 and such that f(s) > z for all s � s0. De�ne thec. e. set

Ve = fm : (9s � s0)[m 2We;s nBs & m � g(e; i; s)g:Now Ve nB =We nB since lims g(e; i; s) =1, so that in fact N nB � Ve, thatis, if m =2 B, then also m =2 Ai, so that by our assumption, m 2 We and thusm 2 Ve nB.

Also Ve \ (B n C) = ; (and hence Ve � C [ (N n B)) by the following. Letm 2 Ve \ B and take s � s0 such that m � g(e; i; s) and m 2 We;s n Bs. Thenm 2 B n Bs so that m = f(t) for some t > s0. Now at stage t+ 1, there existshe0; i0i � he; ii such that limr g(e

0; i0; r) =1 and m is put into Ai0 at stage t+1.But m cannot be a permanent witness for he0; i0i, and therefore m 2 C.

It follows that

N n (B n C) = C [ (N nB) = C [ Ve:

Theorem II.7.11. (Lachlan) For any c. e. set C, C is hhsimple if and only ifL(C) is a Boolean algebra.

Proof. The direction ( �) follows from Lemma II.7.9. For the other direction,suppose that B is not complemented in L(C), that is B nC is not co-c. e. ApplyTheorem II.7.10 to obtain A0 and A1. Let Wg(0) = A0 and apply TheoremII.7.10 to A1 and C \A1 to obtain A

10 and A

11. Set Wg(1) = A1

0 and continue inthis fashion to obtain a pairwise disjoint sequence of c. e. sets Wg(n) such thatWg(n) n C is not co-c. e. and hence is non-empty for all n. Finally, it is easyto uniformly de�ne �nite subsets Wf(n) � Wg(n) such that Wf(n) n C 6= ; foreach n. That is, given a c. e. set W = Wg(n) such that W n C 6= ;, let Wf(n);s

contain i if i 2 Ws and (8j < i)[j 2 Ws ! j 2 Cs]. (Thus if j 2 Ws n C, thenno i > j can enter Wf(n) after stage s, so that Wf(n) is �nite, as desired. If j isthe least element of W nC, then j will be put into Wf(n) as soon as all elementsi < j of W have come into C.)


De�nition II.7.12. An in�nite set C is cohesive if there is no c. e. set Wsuch that W \ C and C nW are both in�nite. A c. e. set A is maximal if forany c. e. set B � A, either B is co�nite or B nA is �nite.

Thus A is maximal if and only if its complement is cohesive.From the lattice viewpoint, A is maximal if it is as large as possible in E�

without being trivial. Friedberg �rst constructed a maximal c. e. set in [70].The proof is based on the following notion.

De�nition II.7.13. The e-state of a number m is fi � e : m 2 Wig and thee-state at stage s is fi � e : m 2Wi;sg.

The e-states are ordered lexicographically so that m has a higher e-statethan n if there is some j < e such that m 2 Wj but n =2 Wj and for all i < j,m 2Wi () n 2Wi.

Theorem II.7.14 (Friedberg). There exists a maximal c. e. set A.

Proof. Let �(e;m; s) denote the e-stage of m at stage s. We de�ne the maximalset A in stages As so that

N nAs = fas0 < as1 < � � � <g:Then ai = lims a

si and A = N n fai : i 2 Ng.

Initially a0i = i for all i, since A0 = ;. The construction proceeds in stageswith the goal of making �(e; ai) � �(e; aj) for all e �(e; ase; s+1) For this e, choose the least such i and let a

s+1e = asi .

For j < e, as+1j = asj and for all j, as+1e+j = asi+j . If no such e exists, then

as+1i = asi for all i.Claim 1: For every e, lims a

se = ae exists.

Proof of Claim 1: The proof is by induction on e, so we may suppose thatit holds for all i < e and take s0 so that asi = ai for all i < e and all s � s0.Then for any s > s0, if a

s+1e > ase, it follows that �(e; a

s+1e ; s+ 1) > �(e; ase; s).

But there are only 2e+1 di�erent e-states, so this can happen at most 2e+1 � 1times after stage s0.

It follows from Claim 1 that A is coin�nite. It is clear from the constructionthat a 2 A () (9a)a 2 As, so that A is a c. e. set.

Claim 2: For all e < i, �(e; e) � �(ae; ai).Proof of Claim 2: Assume by induction that the Claim holds for all d < e.

Fix e s such that �(e; e; t) =�(e; e) and �(e; i; t) = �(e; i). Then at stage t + 1, the construction will forceat+1e 6= ate, a contradiction.

Claim 3: For all e such that A �We, either We nA is co�nite or We nA is�nite.

Proof of Claim 3: Fix e such that A �We and suppose thatWenA is in�nite.Now given any i > e, we may choose j > e such that aj 2 We and it followsfrom Claim 2 that �(i; i) � �(i; j), so that ai 2We. Hence We is co�nite.

II.8. COMPUTABLE ORDINALS AND THE ANALYTICAL HIERARCHY35

Martin proved the following result connecting high degrees and maximal setsin [143].

Theorem II.7.15. (Martin) A degree d is high if and only if there exists amaximal c. e. set A of degree d.

Exercises

II.7.1. Prove Theorem II.7.7.

II.7.2. Show that any hypersimple set is simple.

II.7.3. Prove that any maximal set must have high degree. Hint: Show that theprincipal function pa is dominant, that is, for any computable function f ,f(m) � pa(m) for almost all m.

II.8 Computable ordinals and the analytical hi-

erarchy

De�nition II.8.1. (i) A relation P is said to be �10 (and also �10) if P is

arithmetical.

(ii) A relation P is said to be �1n+1 if there is a �1n relation R such that

P (�!m;�!x ) () (9y)R(�!m;�!x ; y):

(iii) A relation S is �1n if the complement if �1n.

(iv) S is �1n if it is both �1n and �1

n.

The relativized notions of �1n[z] and �1n[z] are similarly de�ned. A relation

is said to be analytic (resp. coanalytic, Borel) if it is �11[z] (resp. �11[z], �

11[z])

for some z.There are two classic examples here.

Example II.8.2. A set A � N may code a partial ordering �A, where m �An () hm;ni 2 A. Also, we write m <A n () m �A n & :n �A m.

WO = fA :�A is a well-orderingg:Let LO be the set of linear orderings. It is easy to see that LO is �0

1. Note forexample that �A is transitive if and only if

(8i)(8j)(8k)[(i �A j & j �A k)! i �A k)];which is a �0

1 condition. Now a linear ordering is a well-ordering if and only ifit is well-founded, that is, has no in�nite descending chain. Thus WO is a �1

1

class, since �A is well-founded if and only if

(8x)[(8m)(x(m+ 1) �A x(m))! (9m)(x(m) �A x(m+ 1))]:


We similarly de�ne the �01 class

PWO = fA :�A is a pre-well-orderingg:We may also de�ne the following �11 relation A - (�)B to mean that �B isa (pre)-linear ordering and �A is isomorphic to a (proper) subordering of �B.Here the subordering property can be expressed as

(9x)(8p)(8q)[p �A q () x(p) �B x(q)]:

For the proper subordering add the following clause

(9r)(8p)(x(p) <B r):

Observe that if �A and �B are linear orderings and B 2WO and A - B, thenA 2WO. A similar result holds for pre-orderings.

The order type kRk of a well-ordering in WO is the unique ordinal � suchthat (Fld(R);�R) is isomorphic to the standard ordering (�;2), where Fld(R) =dom(R) [ ran(R). For a pre-well-ordering, the norm kRk of R is the uniqueordinal � such that there is an order-preserving map from Fld(R) onto �.

In a certain sense, the ordering relation - on PWO is �11. Let kAk =

@1 if A =2 WO. The proof of the following result, sometimes known as thePrewellordering Theorem, is left as an exercise.

Theorem II.8.3. For all pre-linear orderings A and B, if either A or B is inWO, then

(i) A - B _B =2W () [A 2W & kAk � kBk] () :B - A;

(ii) A � B _B =2W () [A 2W & kAk < kBk] () :B � A.A related example is the following.

Example II.8.4. T � N is said to be a tree if f� 2 N� : h�i 2 Tg is closedunder initial segments.

Then the set WF of well-founded trees is �11 since

T 2WF () (8x)(9n)xdn =2 T:An ordinal may be associated with a well-founded tree by means of the Brouwer-Kleene linear ordering �KB on N�, where

� �KB � () (� � �) _ (9j)[�(j) < �(j) & (8i < j)�(i) = �(i)]:

Now given a well-founded tree T , let

F (T ) = fhh�i; h�ii : � 2 T & � 2 T & � �KB �g:Lemma II.8.5. T is a well-founded tree if and only if F (T ) is a well-ordering.


Proof. It is easy to see that �KB is a linear ordering on N� (see the exercises).If T is not well-founded, then there is an in�nite path y through T and fydn :n 2 Ng provides an in�nite descending �KB chain in F (T ). On the other hand,suppose that F (T ) is not well-founded and let f�igi2! be a descending �KBchain in F (T ). Then we can de�ne by recursion an in�nite path through T .For all i > 0, j�j > 0 and �i+1(0) � �i(0) since �i+1 �KB �i. Thus y(0) =limi �i(0) exists. Now if �j(0) = y(0), then we have �j+1(1) � �j(1), so thaty(1) = limi �i(1) also exists. Proceeding by recursion we can de�ne a sequencey(n) = limi �i(n). Now for each n, there is some j such that y(i) = �j(i) for alli < n, so that ydn � �j and therefore ydn 2 T for all n.

For a computable well-founded tree, F (T ) is of course a computable ordinal,since �KB is a computable relation.

The well-ordering F (T ) of a well-founded tree T is closely related to theheight ht(T ), de�ned inductively as follows.

De�nition II.8.6. For any non-empty well-founded tree T � N� and any � 2T , the height htT (�) of � 2 T is given by

htT (�) = supfhtT (�_i) + 1 : �_i 2 Tg:Then the height ht(T ) = htT (;). If T is not well-founded, then ht(T ) =1.

Lemma II.8.7. For any non-empty well-founded tree T , ht(T ) � kF (T )k �!ht(T ) + 1.

Proof. These inequalities are proved by induction on ht(T ). For the base case,ht(T ) = 0 if and only if T = f;g, which is if and only if kF (T )k = 1. Now forh(T ) > 0, the ordering F (T ) consists of ! blocs T ((0)); T ((1)); : : : followed bythe largest element ;. By induction ht(T ((i))) � kF (T ((i))k � !ht(T ((i))) + 1for each i. The �rst inequality is immediate. For the second, we have

kF (T )k � (!ht(T ((0))) + 1 + !ht(T ((1))) + 1 + : : : ) + 1:

There are two cases.(Case 1): There is a �xed m such that ht((T ((i))) � ht(T ((m))) for each i,

so that ht(T ) = ht(T ((m))) + 1 = � + 1. Then kF (t((m)))k � !� + 1 for eachm and hence

kF (T )k � ! � !� + 1 � !�+1 + 1:

(Case 2): There is no maximum ht(T ((m))). Let ht(T ) = � and, for each m,let ht(T ((m))) = �m, so that � = supf�n : n 2 Ng. For each m, there existsn > m such that �i < �n for all i < n and thus

kF (T ((0)))k+kF (T ((1)))k+� � �+kF (T ((n)))k � !�0+1+� � �+!�n = !�n < !�;

so thatkF (T )k � !� + 1 = !ht(T ) + 1:


The next result follows from the Enumeration Theorem II.2.5.

Theorem II.8.8. For each n, there is universal �1n+1 and a universal �1n+1

set of numbers.

The sets WO and WF are both many-one complete for �11 sets. To see this,

we �rst need a normal form for �11 and �11 relations.

Lemma II.8.9. For any �11 relation S, there is a computable relation R such

that

S(�!m;�!x ) () (9y)(8n)R(ydn;�!m;�!x ):

Proof. It is clear that the family of relations expressible in this form includes thecomputable relations and it will su�ce to show that this family is closed undernumber quanti�cation and under existential function quanti�cation. Given S inthis form,

(9i)S(i;�!m;�!x ) () (9z)(8n)R(z(0); hz(1); : : : ; z(n)i;�!m;�!x ):

Also

(8i)S(i;�!m;�!x ) () (9z)(8n)R(i; hz(2i); : : : ; z(2i(2n+ 1))i;�!m;�!x ):

Finally,

(9u)S(�!m;�!x ; u) () (9z)(8n)R(hz(1); : : : ; z(2n� 1)i;�!m;�!x ; (z(0); z(2); : : : )):

De�nition II.8.10. An ordinal � is computable if there is a computable well-ordering A such that kAk = �. W = fe : �e = �A for some A 2 WOg andPW = fe : �e = �A for some A 2 PWOg. The least noncomputable ordinalis denoted by !CK1 or just !1.

Later on, we will need the concept of a system of notations for a computableordinal.

De�nition II.8.11. A system of notations for a computable ordinal � is a mapo from ! n f0g to �+ 1 such that each of the following relations is recursive:

(i) o(a) is a limit ordinal;

(ii) o(b) = o(a) + 1;

(iii) o(a) < o(b).

Lemma II.8.12. Any computable ordinal � possesses a system of notations.


Proof. Note that a computable well-ordering induces a mapping from ! ! �with property (iii) but not with the other two properties. First observe that thecomputable ordinals form an initial segment of the ordinals (See the exercises.)

Now let � be an in�nite countable ordinal and let � be the largest limitordinal < �. Let � = ! � and � = �+ n for some ordinal and some �nite n.Let �G denote a computable well-ordering of type with domain G � ! andlet a 2 G such that jajG = . Let

C = fhg; ii : g 2 G & i 2 N & (a = � ! i � n)g:C is an in�nite computable set, so computably isomorphic to !. Thus it su�cesto de�ne the desired map o from C to � by

o(hg; ii) = ! � jgjG + i:

To verify that this de�nes a system of notations, observe that for a = hg; ii andb = hh; ji in C,(i) o(a) is a limit ordinal if and only if i = 0.

o(b) = o(a) + 1 if and only if g = h and j = i+ 1.

o(a) < o(b) if and only if either g = h and i < j or if g <G h.

Theorem II.8.13. For all relations P , P is �11 if and only if P �m WO, which

is if and only if P �m WF . Furthermore, if P � N, then P is �11 if and only

if P �m W .

Proof. It is clear that the �11 relations are closed under �m, which gives one

direction. Now let R be �11 and let P be a computable relation so that

P (�!m;�!x ) () (8y)(9n)R(ydn;�!m;�!x ):We may assume without loss of generality that

R(�;�!m;�!x ) & � � � ! R(�;�!m;�!x ):by using bounded quanti�cation if necessary. Then we may de�ne a computablefunctional F such that

F (�!m;�!x ) = f� : :R(�;�!m;�!x )g;and we claim that

P (�!m;�!x ) () F (�!m;�!x ) 2WF:

It is clear that a witness y to the fact that :P (�!m;�!x ) will also be a witness tothe fact that F (�!m;�!x ) is not well-founded, and vice versa. Thus P �m WF .

It remains to check that WF �m WO. It follows from Lemma II.8.5 thatT is well-founded if and only if F (T ) is a well-ordering. Now for P � N,let F be a computable function such that m 2 P () Rm 2 WO, whereRm(p) () F (p;m) = 1 and otherwise F (p;m) = 0. Let f be a computablefunction such that �f(m)(p) = F (p;m). Then m 2 P () f(m) 2W .


The proof of this theorem also shows that PWO is also m-complete for �11

sets.

Corollary II.8.14. None of the sets W , PW , WO, PWO and WF are �11.

Proof. Since these sets are m-complete, if they were �11, then all �11 sets would

be �11. (See exercise 4 below.

Theorem II.8.15 (Selection Theorem). Any �11 relation R has a partial selec-

tion function SelR with a �11 graph such that

(9n)R(n;�!m;�!x ) () R(SelR(�!m;�!x );�!m;�!x ) () SelR(

�!m;�!x ) # :Proof. Let R be reducible to W by the function F and let

SelR(�!m;�!x ) = b () R(b;�!m;�!x ) & (8a)[F (a;�!m;�!x ) � F (b;�!m;�!x )! b � a]:

Theorem II.8.16 (Boundedness Theorem). (Spector, [201])

(i) If S is a �11 subset of PWO, then supfkAk : A 2 Sg < !1;

(ii) If S is a �11 subset of PW , then supfk�ek : e 2 Sg < !1.

Proof. Suppose that S were a counterexample to (i). Then

e 2W () (9c)[c 2 S & �e � �c];contradicting Corollary II.8.14.

If S were a counterexample to (ii), then f�c : c 2 Sg would be a counterex-ample to (i).

Let WO� = fR 2 WO : kRk < �g and W� = fc : �c 2 WO�g. Similarly,PWO� = fR 2 PWO : kRk < �g and PW� = fc : �c 2 PWO�gTheorem II.8.17. For all R � Nk � NNl,(i) R is �1

1 if and only if R �m WO� for some � < !1.

(ii) R is �11 if and only if R �m W� for some � < !1.

Proof. Let B 2WO such that kBk = �. Then

A 2WO� () A - B () (A 2WO & :(B - A):

Thus WO� is �11.

Now let R � N be �11, let F be a computable functional such that R(x) ()

F (x) 2WO and de�ne the �11 set Q by

Q = fF (x) : R(x)g:Then Q �WO, so by the Boundedness Theorem, there exists � < !1 such thatQ �WO�.

The proof of (ii) is similar.

II.9. INDUCTIVE DEFINABILITY 41

Here is a surprising corollary to the Boundedness Principle.

Theorem II.8.18. (i) For any �11 pre-well-ordering relation R, kRk < !1;

(ii) There is a �11 well-ordering of order type !1.

Proof. (i) Let kRk = �. Then A 2 W�+1 () A - R and it is easy to seethat this is a �11 de�nition of W�+1, implying that � < !1 by the BoundednessPrinciple.

(ii) De�ne the �11 setW

� to contain a unique index c with k�cg = � for each� < !1. That is,

c 2W � () c 2W & (8d < c)[:�c - �d _ :�d - �c]:

Then letR(c; d) () c 2W � & d 2W � &:�d - �c:

Exercises

II.8.1. Show that the computable ordinals form an initial segment of the count-able ordinals.

II.8.2. Show that the property of coding a linear ordering is in fact �01.

II.8.3. Show that the Brouwer-Kleene ordering is a linear ordering.

II.8.4. Prove the Enumeration Theorem II.8.8 for �11 sets and show that the

universal �11 set cannot be �

11.

II.8.5. Prove Theorem II.8.3.

II.8.6. The de�nition of a computable ordinal may be relativized to computabilityfrom a �xed oracle A. Give appropriate de�nitions for WA and !A1 andprove relativized versions of Theorems II.8.16, II.8.17 and II.8.18.

II.9 Inductive De�nability

Inductive de�nitions play a fundamental role in many areas of mathematics.We have already seen that the set of computable functions is de�ned induc-tively and of course the set of terms and formulas of a given language are alsode�ned inductively. The formal notion of inductive de�nability was �rst givenby Spector [203] and is fully developed by Moschovakis in his book [154].

An operator � over a set X is a a function from P(X) to P(X). � is said tobe inclusive if Y � �(Y ) for all Y � X. � is said to bemonotone if �(Y ) � �(Z)for all Y � Z � X. � is said to be inductive if it is either inclusive or monotone.The operator � inductively de�nes a subset Cl(�) as follows. A sequence �� ofsubsets of X is de�ned recursively by �0 = ;, ��+1 = �(��) and �� =

S�<� �

�

for limit ordinals �. The closure of � is Cl(�) =S� �

�. A set Y is said to be a�xed point of � if �(Y ) = Y .


Lemma II.9.1. For any inductive operator �,

(i) For any ordinal �, if ��+1 = ��, then �� = �� = Cl(�) for all � � �.(ii) There exists � such that Card(�) � Card(X) such that ��+1 = ��.

Proof. (i) is easily proved by induction on �.For (ii), let � be a cardinal and suppose that ��+1 n �� is non-empty for all

� < �. Then clearly Card(�) � Card(��) � Card(X).

Now we can de�ne the closure ordinal j�j of � to be the least ordinal � suchthat ��+1 = ��. The following is immediate.

Corollary II.9.2. For any inductive operator � over X, Card(j�j) � Card(X)and Cl(�) = �j�j.

In particular, for X = N, j�j < @1.Example II.9.3. The �1

1 set W can be given by a �01 monotone inductive

de�nition. First de�ne the computable function � so that ��(c;n) de�nes a re-striction of �c to elements below n in the following sense. Recall that i �c jmeans that �c(i; j) = 1 and i <c j means that i �c j but not j �c i. Let

��(c;n)(i; j) =

(�c(i; j); if j <c n

0; otherwise.

Now letc 2 �(A) () Wc = ; _ (8n)�(c; n) 2 A

It is clear that c 2 �1 if and only if Wc = ;, which is if and only if c 2 Wand kck = 0. It follows by induction that c 2 �m+1 () kck � m and hencec 2 �! () kck < ! and in general, c 2 �� () (c 2W & kck < �).

Thus we see that j�j = !1 and Cl(�) =W .

Note that Cl(�) is a �xed point of �.

Theorem II.9.4. For any monotone operator � over a set X, Cl(�) is theleast �xed point of �, that is,

x 2 Cl(�) () (8Z)[�(Z) = Z ! x 2 Z]:Proof. Let U = fx : (8Z)[�(Z) = Z ! x 2 Z]g. Then x 2 U implies x 2 Cl(�),since Z = � satis�es �(Z) = Z. For the other direction, let Z be any set suchthat �(Z) = Z. It can be seen by induction that �� Z for all �. Thatis, certainly �0 = ; � Z. Then supposing �� Z, we have ��+1 = �(�� (Z) = Z. Finally, if �� Z for all � < �, then �� Z. It follows that�� U for all � and hence Cl(�) � U .

The monotone operator � is said to be �nitary if for all x and Y , if x 2 �(Y ),then there is a �nite Z � Y such that x 2 �(Z). For example, the operatorwhich de�nes the set of formulas of propositional logic is �nitary, since each newformula is generated by either one or two previously generated formulas.


Lemma II.9.5. If � is a �nitary monotone operator on X, then j�j � !.Proof. Suppose x 2 �(�!) and let Z � �! be a �nite set such that x 2 �(Z).Then there exists n < ! such that Z � �n and hence x 2 �n+1. Thus �!+1 =�!.

The complexity of an operator � on N is given by the complexity of therelation fhm;Ai : m 2 �(A)g. We will also consider operators on N with realparameters. That is, for example, a family f�x : x 2 NNg of operators over Nwill be computable if fhm;x;Ag : m 2 �x(A)g is a computable relation.Lemma II.9.6. Any �01 operator is �nitary.

Proof. Let � be a �01 operator and let R be a computable relation so that

m 2 �(A) () (9k)R(m;Adk):Suppose m 2 �(A) and let k be given as above so that R(m;Adk). Now letZ = fi : i < k & i 2 Ag. Then R(m;Zdk) so that m 2 �(Z).Theorem II.9.7. If � is a �01 monotone operator over N, then j�j � ! andCl(�) is a c. e. set.

Proof. Let � be a �01 operator and let R be a computable relation so that

m 2 �(A) () (9k)R(m;Adk):It follows from lemmas II.9.5 and II.9.6 that j�j � !.

For the closure, we have the followingCLAIM: m 2 �(We) () (9k; s)R(m;We;sdk):Proof of Claim: If m 2 �(We), then (9k)R(m;Wedk) and thus R(m;We;sdk)

where s is large enough so that We;sdk = Wedk. If R(m;We;sdk), then m 2�(We;s) and since � is monotone, m 2 �(We).

It follows that there is a computable function � such that �(We) = W�(e).Now we can recursively de�ne a function such that �n = W (n) by letting (0) = 0 and then (n + 1) = �( (n)). Finally, we have m 2 Cl(�) ()(9n)m 2W (n):

Theorem II.9.8. For any n > 0 and any �1n monotone operator � over N,

Cl(�) is �1n.

Proof. By the improved version of Theorem II.9.4 (see Exercise 1 below), wehave

m 2 Cl(�) () (8Z)[�(Z) � Z ! m 2 Z]() (8Z)[(8m)(m 2 �(Z)! m 2 Z)! m 2 Z];

which is a �11 de�nition.

In particular, the closure of any �01 monotone operator is �

11. This can be

reversed up to many-one reduction.


Theorem II.9.9. For any �11 P � N�NN, there is a uniformly �0

1 monotoneoperator �x and a computable function f such that

P (m;x) () f(m) 2 Cl(�x):Proof. Let R be a computable relation such that

P (m;x) () (8y)(9k)R(m; ydk; x):Let

hm;�i 2 �x(A) () R(m;�; x) _ (8n)hm;�dni 2 A:Then it is easy to check that P (m;x) () hm; ;i 2 Cl(�x):Theorem II.9.10. For any �0

1 inductive operator �, j�j � !1 and Cl(�) is �11.

Proof. Let R be a computable relation so that, for all m and A,

m 2 �(A) () (8k)R(m;Adk):We want to uniformly de�ne the levels �� of the inductive de�nition � as

follows. Let i <c j denote �c(i; j) = 1 & �c(j; i) = 0, so that if c 2W and �c isthe characteristic function of the set C, then i <c j means that i <C j; we alsolet kikc = kikC . Now de�ne

S(c; P ) () (8m; i)(P (m; i) () (9j)[j <c i & m 2 �(fn : P (n; j)g)]):It follows by induction that if c 2W and S(c; P ), then, for all i 2 Fld(R),

P (m; i) () m 2 �kikc :Now we have

m 2 �!1 () (9i; c)[c 2W & (8P )(S(c; P ) ! P (m; i))]:

Thus �!1 is a �11 set, which we denote as Q. For each c 2 W , let Qc denote

�kck, so that as above, we have a �11 de�nition

m 2 Qc () (9i)(8P )[S(c; P ) ! P (m; i)];

and we also have a �11 de�nition

m 2 Qc () (9i; P )[S(c; P ) & P (m; i)]:

We note that this part of the argument applies to any �11 operator.

It remains to show that �(Q) � Q. Suppose therefore that m 2 �(Q), sothat

(8k)R(m;Qdk):and de�ne the �11 set V by

a 2 V () (9k)(8c)[(c 2W & R(m;Qc))! �a - �c]:

It follows that V � W and hence supfk�ak : a 2 V g = � < !1. Then(8k)R(m;��dk), so that m 2 ��+1 and thus m 2 Q.


A natural problem in connection with Theorem II.9.4 is the nature of thefamily of �xed points of an inductive operator. Of course we have �(N) = N bythe assumption that X � �(X), so that we will not have a unique �xed point.However, we can re�ne the previous result to show that

The Boundedness Principle for inductive de�nability is needed for the analy-sis of the Cantor-Bendixson derivative in Chapter V. The following prewellorder-ing theorem for inductively de�nable sets is due to Kunen; see [154], p. 27.

The prewellordering associated with any inductive operator � is induced bythe following norm:

jxj� =((least �)x 2 ��+1; if x 2 Cl(�);1; otherwise.

The following Stage Comparison Theorem is due independtly to P. Aczeland K. Kunen; see [154] for a more general result.

Theorem II.9.11. Let � be a �11 monotone inductive operator. Then the

following relations are both �11.

R(m;n) () jmj� � jnj� & m 2 Cl(�);S(m;n) () jmj� < jnj� & m 2 Cl(�):

Proof. De�ne a �11 monotone operator � by

(0;m; n) 2 �(A) () m 2 �(fi : 1; i; n) 2 Ag);(1;m; n) 2 �(A) () n =2 �(fj : (0;m; j) =2 Ag):

It can be checked that R(m;n) () (0;m; n) 2 Cl(�) and S(m;n) ()(1;m; n) 2 Cl(�).Theorem II.9.12. For any �1

1 monotone inductive operator � and any �11 setA � Cl(�), there is a computable ordinal � such that A � ��.

Proof. Let A be a �11 subset of Cl(�) as described. Then by Theorem II.9.11,the prewellordering RA de�ned by

RA(m;n) () jmj� � jnj� & n 2 Ais a �11 prewellordering. It now follows from Theorem II.8.18(a) that jRAj <!CK1 .

Corollary II.9.13. For any �11 monotone inductive operator �, j�j � !C�K1 .

Exercises

II.9.1. Improve Theorem II.9.4 by showing that Cl(�) =TfZ � X : �(Z) � Zg.

II.9.2. Show that a non-monotone �01 operator need not have a �01 closure andin particular can have a �0

n complete closure for any n.


II.9.3. Show that a non-monotone �02 operator may have a closure which is not

�11.

II.9.4. Show that there is a �01 inductive operator � such that the set of truesentences of arithmetic is many-one reducible to Cl(�).

II.9.5. Show that if � is a �11 inductive operator and A � �!1 is �11, then for

some � < !1, A � ��. Then show that if � is �01, then Cl(�) is �

11 if and

only if j�j < !1.

II.10 The hyperarithmetical hierarchy

In our study of the Cantor-Bendixson derivative D(P ) of a �01 class P , we will

see that the iterated derivative D�(P ) for an in�nite, computable ordinal �is a hyperarithmetical, or e�ectively Borel set. Since index sets for �0

1 classesand for hyperarithmetical sets will be important in this work, we will presenta de�nition of the hyperarithmetical sets based on indices. This approach isbased on that given by Hinman ([87], p. 163).

Informally, a set is �0! if it is the union of an e�ective sequence An ofarithmetical sets and more generally a set is �0� for a computable ordinal � ifit is the union of an e�ective sequence of sets, each of which is �0� for some� < �. As for the arithmetical hierarchy, a set is �0

� if its complement is �0�and is �0�+1 if it is the union of an e�ective sequence of �0

� sets.The following is an inductive de�nition of the hyperarithmetic sets He, taken

essentially from Hinman ([87], p. 163).First we de�ne a set of ordinal notations.

De�nition II.10.1. H is the smallest subset of N such that for all a,

(i) h7; ai 2 H;

(ii) if �a(n) 2 H for all n, then a 2 H.

We observe that H is the closure of a �01 monotone inductive operator �H

and thus each a 2 H is assigned an ordinal � = kakH so that a 2 H�+1 �H�.It follows from Theorem II.9.8 that H is �1

1 and that each ordinal kakH iscomputable.

Then each a 2 H is assigned a hyperarithmetic set by the following. (Fora =2 H, we let Ha = ;.)De�nition II.10.2. Let a 2 H. Then

(i) If a = h7; bi, then Ha =Wb;

(ii) if �a is total, then Ha = [nN nH�a(n).

For example, let B be a �02 set and let R be a computable relation such thati 2 B () (9n)(8m)R(i;m; n). Let An = f(i;m) : :(8m)R(i;m; n) so thatB = [nN n An. Let (i;m) = (least n):R(i;m; n) and de�ne � by the s-m-n

II.10. THE HYPERARITHMETICAL HIERARCHY 47

Theorem so that ��(m)(i) = (i;m). Then An = W�(m) for each n, so that if�a(m) = h7; �(m)i, then Ha = B.

A subset of N is said to be hyperarithmetical if it equals Ha for some indexa 2 H. The hyperarithmetical hierarchy is de�ned as follows.

De�nition II.10.3. For all � and all A � N,(i) A is �0� if A = Ha for some a 2 H�;

(ii) A is �0� if N nA is �0�;

(iii) �0� = �0� \�0

�.

It follows that for limit ordinals �, A is �0� if and only if A is �0� for some� < �. It is easy to see that this de�nition agrees with the arithmetic hierarchyfor � < !. Note that for in�nite ordinals �, some authors (e.g. Hinman [87])denote this hierarchy as the �0(�) sets and let the �0� sets be e�ective unions of

�0(�) sets. The de�nition we use here is for uniformity of results concerning �01

classes.There is a natural set PH = fha;mi : a 2 H & m 2 Hag which is com-

plete for the family of hyperarithmetical sets. However, this set is �11 and not

hyperarithmetical.

Lemma II.10.4. Every hyperarithmetical set is �11.

Proof. We give a monotone �02 inductive de�nition of the following set:

V = fh0; a;mi : a 2 H & m 2 Hag [ fh1; a;mi : a 2 H & m =2 Hag:

That is, let

(0) h0; a;mi 2 �(X) () [(a = h7; bi ^ m 2 Wb) _ (9n)(�a(n) #^ h1; �a(n);mi 2 X)].

(1) h1; a;mi 2 �(X) () [(a = h7; bi ^ m =2 Wb) _ (8n)(�a(n) #^ h0; �a(n);mi 2 X)].

We then show by induction on � that �� = V �. For � = 0, both sets areempty and for � = 1,

�1 = V 1 = fhh7; bi; 0;mi : m 2Wbg [ fhh7; bi; 1;mi : m =2Wbg:

Now suppose that �� = V � for all � < �. If � is a limit ordinal, then clearly

�� = [�<�� = [�<�V � = V � :

Next suppose that � = �+ 1 for some �.If ha; i;mi 2 ��+1, then, for all p, (9j � 1)h�a(p); j; i 2 ��, so that by

induction �a(p) 2 H� for all p and hence a 2 H�+1. For i = 0, there existsp such that h�a(p); 1;mi 2 �� and hence by induction m =2 H�a(p). It follows


that m 2 Ha and therefore ha; 0;mi 2 V �+1. For i = 1, it follows similarly thatm =2 Ha.

Now suppose that ha; i;mi 2 V �+1. Then a 2 H�+1, so that for all p,�a(p) 2 H�. For i = 0, m 2 Ha and therefore there exists p such that m =2H�a(p) and hence by induction, h�a(p); 1;mi 2 ��. It follows that ha; 0;mi 2��+1. The argument for i = 1 is similar.

Since � is an arithmetical monotone inductive operator, it follows from The-orem II.9.8 that V is a �1

1 set and hence the hyperarithmetical set Ha is �11 for

each a.

It follows from the proof of Lemma II.10.4 that the set PH is �11. We can

now prove the Spector-Gandy Theorem, due independently to Spector [202] and

Gandy [76]. We say that a relation R � Nk �NN` is �11HY P if and only if thereis an arithmetical relation P such that

R(�!m;�!x ) () (9y 2 �11[�!x ])P (�!m;�!x ; y):

Theorem II.10.5 (Spector-Gandy Theorem). A relation R � Nk � NN` is

�11HY P

if and only if it is �11.

Proof. We give the proof without the real parameters �!x and with just onenumber variable m. We may assume that the 9y quanti�er ranges over f0; 1gN.It follows from the proof of Lemma II.10.4 that we can de�ne a �11 relation Ssuch that for any a 2 H, S(a; y) () y = Ha. That is,

y = Ha () (8n)(8i < 2)[(a; i; n) 2 V ! y(n) = i]:

Now we have

(9y 2 �11)(P (m; y)) () (9a)[a 2 H & (8y)(y = Ha ! P (m; y))]:

This demonstrates that any �11HY P

relation R is in fact �11.

For the reverse direction, it clearly su�ces to show that the �11 complete

relation W is �11HY P

.It follows immediately from the �rst part of our proof that the set f0; 1gN\�1

1

is itself �11. Then by Theorem II.8.13, there is a computable function F such

that, for all z,z 2 �1

1 () F (z) 2W:Now �1

1\f0; 1gN cannot be a �11 set, by the following argument. Choose a0 2W

and let

Q(a; y) () a 2W & [(y 2 �11 & F (y) = a] _ (y =2 �1

1 & a = a0)]:

Then Q is �11 and therefore has a selector SelQ with �1

1 graph by TheoremII.8.15. Since SelQ is total, the graph is actually �1

1. Now if �11 were itself

�11, then the image of �1

1 would be a �11 subset of W and hence bounded byTheorem II.8.16.


It follows from Theorem II.8.17 that the range of F is unbounded in W andtherefore

a 2W () (9z 2 �11)�a � F (z):

Before establishing the reverse implication of lemma II.10.4, we will needseveral technical lemmas from [87].

Lemma II.10.6. For all �, �0� and �0� are e�ectively closed under many-one

reduction, that is, there is a primitive recursive function g such that for all m; e:

m 2 Hg(a;e) () �e(m) 2 Ha:

Proof. Let h be a primitive recursive function such that

g(a; e) = h7; bi; where �b(m) = �(a)1(�e(m)); if a = h7; bi;

and otherwise de�ne g by the Recursion Theorem so that

�g(a;e)(p) = g(�a(p); e):

Lemma II.10.7. For all � > 0, the family of �0� relations is e�ectively closedunder computably enumerable union and �nite intersection; that is, there existsprimitive recursive functions f and g such that

(i) If �a(p) 2 H� for all p, then f(a) 2 H� and

m 2 Hf(a) () (9p)m 2 H�a(p):

(ii) if a; b 2 H�, then g(a; b) 2 H� and

Hg(a;b) = Ha \Hb:

Proof. (i) We will de�ne �f(a)(r) in two cases and then use the s-m-n theoremto de�ne f(a). In either case,

m 2 Hf(a) () (9r)m =2 Hf(a)(r):

Let (r)0 = p and (r)1 = q.Case I: If �a(p) =2 H1, then

�f(a)(r) = ��a(p)(q):

If there exists such an r with m =2 Hf(a)(r), then

(9p)[(9q)m =2 ��a(p)(q)];


so that as desired(9p)m 2 H�a(p):

This argument is clearly reversible for �a(p) =2 H1.Case II: If �a(p) = h7; bi, then de�ne h(a; r) so that

Wh(a;r) = N nWb;q

and let �f(a)(r) = h7; h(a; r)i.If m =2 H�f(a)(r), then m =2 Wh(a;r), so that m 2 Wb = H�a(p) and hence

(9p)m 2 H�a(p) as desired. Again the reverse direction is clear.

(ii) There are four cases.(1) If a = h7; di and b = h7; ei, then g(a; b) = h7; ci, where

�c(n) = �d(n) + �e(n):

(ii) If a = h7; di and b =2 H1, then g(a; b) = c may be de�ned by part (i) sothat

H�c(p) = H�a(p) [ (N nWd):

(iii) The case when a =2 H1 and b = h7; ei is similar to (ii).(iv) If a; b =2 H1, then g(a; b) is de�ned by (i) so that

H�g(a;b)(p) = H�a(p) [H�b(p):

Lemma II.10.8. For all � > 0, the family of �0� relations is e�ectively closedunder existential number quanti�cation (and thus the family of �0

� relationsis e�ectively closed under universal number quanti�cation); that is, there isa primitive recursive function h such that for all a 2 H�, h(a) 2 H� andm 2 Hh(a) () (9p)hp;mi 2 Ha:

Proof. Let h be a computable function such that �h(p)(m) = hp;mi. Then(9p)hp;mi 2 Ha () (9p)�h(p)(m) 2 Ha:

Then taking g from Lemma II.10.6, we have

(9p)hp;mi 2 Ha () (9p)m 2 Hg(a;h(p)):

Now letting ��(a)(p) = g(a; h(p)) and taking f from Lemma II.10.7, we have

(9p)hp;mi 2 Ha () m 2 Hf(�(a));

so we let h(a) = f(�(a)).

Theorem II.10.9. Let � be a �0k (resp. �

0k) inductive operator, let � be a limit

ordinal and let n < !. Then �n is �0kn (resp. �0kn), �

� is �0�+1 and ��+n is�0�+kn+1 (resp. �0�+kn+1).


Proof. We give the proof for a �0n operator and leave the other case to the

reader. First we show that there is a primitive recursive function g such thatN n �(He) = N nHg(e) and furthermore if e 2 H�, then g(e) 2 H�+n. We givethe proof for k = 1 and leave the general result as an exercise. Let R be acomputable relation such that

i 2 �(A) () (8j)R(i; Adj):

Then

i =2 �(He) () (9j):R(i; (N nHe)dj):Now

R(i; (NnHe)dj) () (9� 2 f0; 1gj)[R(i; �) & (8t < j)(�(t) = 0 () t 2 He)]:

Since the last clause makes both positive and negative reference to He, it followsthat we can de�ne primitive recursive functions fp and fn such that

R(i; (N nHe)dj) () (i 2 Hfp(e;j) ^ i =2 Hfn(e;j)):

Then

i =2 �(He) () (9j)(i =2 H�fp(e)(j)) _ (9j)(i 2 H�fn(e)(j)

):

It follows that

N n �(He) = Hfp(e) [Hf(fn(e));

where f is the function from Lemma II.10.7. Now the set Hg(e) = Hfp(e) [Hf(fn(e)) is �

0�+1 by Lemma II.10.7 and thus �(He) is �

0�+1 as desired.

Now �x c 2 W with kck = � and use the Recursion Theorem to de�ne aprimitive recursive function h such that

�kikc = N nHh(i):

(1) If kikc = 0, then Hh(i) = N;

(2) If kjkc = kikc + 1, then Hh(j) = Hg(h(i);

(3) If kjkc is a limit, then Hh(j) =TfHh(i) : i <c jg.

It follows by induction that if n is �nite and � is a limit, then �n is �0n, that

�� is �0�+1 and that ��+n is �0�+n+1.

The next result gives a uniform inductive de�nition of the hypararithmeticsets and can be used to de�ne the trans�nite jumps 0�.

Theorem II.10.10. There is a �01 inductive de�nition � such that, for all a

and m, m =2 Ha () h4; a;mi 2 Cl(�) and furthermore, if a 2 H�, thenm =2 Ha () h4; a;mi 2 ��.


Proof. There are several clauses in the de�nition. We assume that �0 is theempty function and omit the inclusive part of each clause (that i must be in�(A) if it is in A.)

(1) h1; a;mi 2 �(A) () m =2Wa.

(2) h2; ai 2 �(A) () h1; 0; 0i 2 A & (8m)h1; a;mi =2 A.The result of these two clauses is that �a is total if and only if h2; ai 2 Cl(�),

which is if and only if h2; ai 2 �2.

(3a) h3; h7; aii 2 �(A)(3b) h3; b 2 �(A) () h2; bi 2 �(A) & (8n)h3; �a(n)i 2 A.These two clauses ensure that, for all a and for all ordinals �,

h3; ai 2 �� () a 2 H�.

(4a) h4; h7; ai;mi 2 �(A) () m =2Wa.

(4b) h4; b;mi 2 �(A) () h3; bi 2 �(A) & (8n)h4; �b(n);mi =2 A.

These �nal two clauses complete the de�nition. The theorem follows byinduction on � as follows.

For � = 1, if b = h7; ai 2 H1, then h3; bi 2 �1 by clause (3a) and, by clause(4a):

m =2 Hb () m =2Wa () h4; b;mi 2 �1:For � � 1 and b 2 H�+1 �H�, �b must be total, so we have h2; bi 2 ��+1 andthen

hm =2 Hb () (8n)m 2 H�b(n) () (8n)h4; �b(n);mi 2 �� () h4; b;mi 2 ��+1:

This theorem has two important corollaries. Note that we have alreadyde�ned 0(n) for �nite n.

De�nition II.10.11. For any computable ordinal � � !, let

0(�) = fha;mi : a 2 H� & m 2 Hag:

Theorem II.10.12. For each computable ordinal � � !,1. 0(�+1) is �0�+1 complete, and

2. any set A is �0�+1 if and onlyl if it is �01 in 0(�).

Proof. (1) It follows from Theorems II.10.9 and II.10.10 that 0(�+1) is �0�+1.The completeness is immediate from Theorem II.10.10.

(2) is left as an exercise.


The �0� and �0� sets may be characterized in terms of inductive de�nability.

The next result follows directly from Theorems II.10.9 and II.10.10.

Theorem II.10.13. For any A � N and any computable ordinal �, A is �0� if

and only if A is m-reducible to �� for some �01 inductive de�nition �.

Finally, we can characterize the �0�+1 sets as relative c. e. over the jumps.

Theorem II.10.14. For any computable ordinal � and any A � N, A is �0�+1if and only if A is c. e. in 0(�) and A is �0

�+1 if and only if A is computable in

0(�).

Proof. Let B = 0(�) = Hb for some b 2 H�. Suppose �rst that A is �0�+1.Then A = Ha for some a 2 H�+1. Thus we have

m 2 A () (9n)m 2 H�a(n) () (9n)h�a(n);mi 2 B:For the other direction, suppose that A is c. e. in B. Then for some e, we have

m 2 A () �e(m;B) # () (9t)�e(m;Bdt) # () (9�)[� � B & �e(m;�) #]:By Lemma II.10.7, it su�ce to show that � � B is �0�+1. But we have

� � B () (8i < j�j)(�(i) = 0! i 2 B) & (8i < j�j)(�(i) = 1! i =2 B):Now \i 2 B" is �0� and therefore �0�+1 and i =2 B is clearly �0�+1, so the resultfollows, again by Lemma II.10.7.

Monotone inductive de�nitions are frequently used and there is a �ner resultfor the complexity of the levels.

Theorem II.10.15. For any ordinal �, any n 2 N and any monotone �01

inductive operator �, �!�� is �02� and �!��+n+1 is �02�+1.

Proof. The proof is similar to that of Theorem II.10.9, with the additional ideathat if A is �0

� , then �(A) is also �0� , since by monotonicity,

i 2 �(A) () (8j)R(i; Adj) () (8j)(8C � n)[A � C ! R(i; C)]:

It follows that �n is �01 for all n and that if �� is �0� , then ��+n is �0

�+1.Making use of Lemmas II.10.6, II.10.7, II.10.8 and the Recursion Theorem, theproof follows as above. Details are left to the reader.

Theorem II.10.16. (Souslin-Kleene) A subset of Nk� (NN)l is �11 if and only

if it is hyperarithmetical.

Proof. The direction ( �) is a routine generalization of Lemma II.10.4. For theother direction, it su�ces by Lemma II.10.6 and Theorem II.8.17 to show thatW� is hyperarithmetical for any computable ordinal �. Recall the �0

1 monotoneinductive de�nition � of W given in Example II.9.3 such that such that W� =��, It now follows from Theorem II.10.9 each W� is hyperarithmetic.


The next application of inductive de�nability is part of a theorem of Chen[48].

Theorem II.10.17. Let � > 1 be a computable ordinal and let n � 1 be anatural number. Then PW!�� is �02� complete and PW!��+n is �0

2�+1 complete.

Proof. We will just demonstrate the upper bound on the complexity. A partialcomputable function �c represents the pre-ordering Rc if it is the characteristicfunction, that is

Rc(i; j) () �c(hi; ji) = 1

and

:Rc(i; j) () �c(hi; ji) = 0:

The restriction of Rc to elements below n may be given by

�h(c;n)(i; j) = �c(i; j) � (1� �c(n; j)):

Note that if �c is total, then �h(c;n) is total for all n.

There is a natural �01 monotone inductive de�nition of PW given by

c 2 �(A) () (8n)h(c; n) 2 A:

Then it is easy to see that for a total function �c which represents a pre-linear-ordering,

c 2 PW� () c 2 Tot & c 2 �� :The condition that �c is total is �

02 and the condition that �c is the characteristic

function of a pre-linear-ordering is �01. Thus the upper bound on the complexity

follows from Theorem II.10.15.

The proof of the other direction is omitted.

Theorem II.10.18. Let T be a computable tree and de�ne a prewellorderingR on T so that k�kR = htT (�) (where ht(�) = 1 if � =2 Ext(T ).) Then forany ordinal �, and any n 2 N, f� 2 T : htT (�) < ! � �g is �02� and �!��+n+1

is �02�+1.

Proof. The proof is left as an exercise.

Computable trees with a unique in�nite branch were studied by Clote [53].It is well-known that for every hyperarithmetic set A, there exists a computabletree with a unique in�nite branch x such that A is Turing reducible to x. Wewill prove this below in Chapter V.

Theorem II.10.19. ([53]) Let T be a computable tree T with a unique in�nitebranch x. Then for any � 2 T �Ext(T ), htT (�) < !1. If htT (�) < ! �� for all� =2 Ext(T ), then x is Turing reducible to a �02� set.


Proof. The �rst part follows from Lemma II.8.7. For the second part, note thatthe set A = f� 2 T : htT (�) < ! � �g is �02� by Lemma II.10.18. Then x maybe computed recursively from A by

x(n+ 1) = (least i)[(x(0); : : : ; x(n); i) =2 A]:

Exercises

II.10.1. Show that for �nite n, the de�nition of the hyperarithmetical sets agreeswith the previous de�nition of the arithmetical sets.

II.10.2. Give an alternate proof of Lemma II.10.4 using the Prewellordering The-orem II.9.11.

II.10.3. Give the following improvement of Theorem II.10.9. Suppose that � isa �0

k (respectively, �0k) inductive operator and that �1 is �0k for some

m < k. Show that for each n, �n+1 is �0kn+m (resp. �0kn+m).

II.10.4. For computable ordinals � < �, �0� [�0� � �0

� .

II.10.5. Show that a set A is �0�+1 if and onlyl if it is �01 in 0(�).

II.10.6. Give the details in the proof of Theorem II.10.15.

II.10.7. Let T be a computable tree and de�ne a prewellordering R on T so thatk�kR = htT (�) (where ht(�) = 1 if � =2 Ext(T ).) Show that for anyordinal �, and any n 2 N, f� 2 T : htT (�) < ! � �g is �02� and �!��+n+1

is �02�+1.

II.10.8. Show that PH is �11.

II.10.9. Show that there can be no universal hyperarithmetical set ( so in partic-ular PH is not hyperarithmetical.)

II.10.10. For each computable ordinal �, both the �0� and the �0� relations are

closed under �m reducibility.

II.10.11. The de�nition of the hyperarithmetical hierarchy is easily extended tosubsets of NN and in general to relations R � Nk � (NN)`. Give thedetails.


Chapter III

Fundamentals of �01Classes

This chapter contains the formal de�nition of a �01 class as well as the set of

in�nite paths through a computable tree, well as some equivalent formulations.In particular, the notation \�0

1" indicates that a �01 class may be represented

in arithmetic by a formula having one universal quanti�er, ranging over naturalnumbers. We will explain the connection between these notions. Some notionsof boundedness for trees are examined, including highly computable and �nite-branching trees. This leads to computably bounded and bounded �0

1 classes.Decidable �0

1 classes are de�ned, corresponding to computable trees with nodead ends. Products and disjoint unions of trees and classes are studied. Thenotion of compactness is examined together with K�onig's Lemma. The family of�01 classes is shown to have the dual reduction and separation properties. Strong

�0n classes are de�ned. Computable and continuous functions on N

N and f0; 1gNare studied in connection with �0

1 classes. Classes of separating sets for a pair ofdisjoint c.e. sets are studied and in particular diagonally non-computable sets.The connection between retraceable and hyperimmune sets and the �0

1 class ofinitial subsets of a co-c.e. set is given. Several notions of reducibility betweenvarious classes are examined. For example, every c. b. class is computablyhomeomorphic to a subclass of f0; 1gN, every bounded �0

1 class is computablyhomeomorphic to a strong �0

2 class of sets and every �02 class can be put in

one-to-one degree-preserving correspondence with a �01 class.

We also introduce the applications of �01 classes by considering the repre-

sentation of logical theories.

III.1 Computable trees and notions of bound-

edness

Recall that a tree T is a subset of N� which is closed under initial segments.Such a tree is said to be !-branching, since each node has potentially a countablyin�nite number of immediate successors. We identify an element � of N� with

57

58 CHAPTER III. FUNDAMENTALS OF �01 CLASSES

its code h�i 2 N and say that T is computable if the set of codes h�i such that� 2 T is a computable set.

De�nition III.1.1. (i) For a given function g : N� ! N, a tree T � N� issaid to be g-bounded if for every � 2 T and every i 2 !, if �_i 2 T , theni < g(�).

(ii) T is �nite branching if each node � of a tree T has �nitely many immediatesuccessors �_i.

There are other equivalent formulations.

Lemma III.1.2. For any tree T � N�, the following are equivalent:

1. T is �nite branching;

2. T is g-bounded for some g.

3. There is a function f : N� ! N� such that, for each � 2 T , f(�) =(i1; : : : ; ik), where i1 < � � � < ik enumerates fi : �_i 2 Tg;

4. There is a function f 0 : N� ! N such that, for each � 2 T , � has at mostf 0(�) immediate successors;

5. There is a function h such that �(i) < h(i) for all � 2 T and all i <j�j.

The proof of this lemma is left as an exercise.

De�nition III.1.3. (i) A tree T is computably bounded (c. b.) if it is g-bounded for some computable function g.

(ii) A computable tree T is said to be highly computable if it is also computablybounded.

(iii) T is highly computable in z if it is computable in z and also g-boundedby some function g computable in z.

Lemma III.1.4. For any computable tree T � N�, the following are equivalent:(a) T is highly computable;

(b) There is a computable function h such that �(i) < h(i) for all � 2 T andall i < j�j.

(c) There is a computable function f : N� ! N� such that, for each � 2 T ,f(�) = (i1; : : : ; ik), where i1 < � � � < ik enumerates fi : �_i 2 Tg.

Proof. (a)! (b): Given the function g, let h(0) = g(;) and recursively computeh by h(n+ 1) = maxfg(�) : � 2 f0; 1; : : : ; h(n)gn \ Tg.

(b)! (c): Given the function h, the sequence i1 < � � � < ik can be computedfrom � by testing in turn whether �_i 2 T for i < h(j�j+ 1).

(c) ! (a): For a given �, use the function f to compute i1 < � � � < ik asindicated and then g(�) = ik +1 will be an upper bound for fi : �_i 2 Tg.

III.1. COMPUTABLE TREES AND NOTIONS OF BOUNDEDNESS 59

Note that the argument that (c) implies (a) does not use the computabilityof T but the other two arguments do. We leave it as an exercise to de�ne(noncomputable) trees which possess computable bounding functions of type(a) but do not possess bounding functions of type (b) and similarly for (b) and(c).

Observe that we have omitted here the condition from Lemma III.1.2 thatthere is a computable upper bound on the number of successors of �. It is anexercise to show that any highly computable tree possesses such a function, butthere exists a computable tree which is not highly computable but possessessuch a bounding function.

Example III.1.5. Here is an example of a computable tree which is boundedbut not computably bounded. Put 0e_s + 1 2 T if and only if �e(e) convergesat stage s. Then each node � 2 T has at most two extensions, so that T is�nite branching. Now suppose by way of contradiction that T were computablybounded by the function g. Then �e(e) # () �e;g(0e)(e) #, which would makefe : �e(e) #g a computable set.

There is a further notion of boundedness.

De�nition III.1.6. 1. T is almost bounded by g if there is some k 2 Nsuch that for all � with j�j > k and for all i, if �_i 2 T , then i < g(�).

2. T is almost bounded (a. b.) if it is almost bounded by some g.

3. T is almost computably bounded (a. c. b.) if it is almost bounded by somecomputable function g.

Note that these notions are not equivalent to the existence of a (computable)function h and a �nite k such that �(i) < h(i) for all � 2 T with j�j > k andall i < j�j.Example III.1.7. Let T = fnk : n; k 2 !g. Then T is a. c. b. but has, foreach k and n, strings � = nk+1 2 T with �(k) = n. It is clear that T is not�nite branching.

Two trees may be joined together in various ways.

De�nition III.1.8. For two trees S; T � N�

1. S � T = f0_� : � 2 Sg [ f1_� : � 2 Tg.2. For strings � and � with j�j = j� j = n, � � � = (�(0); �(0); : : : ; �(n �

1); �(n�1)); if j�j = n+1 and j� j = n, then �� = (�(0); �(0); : : : ; �(n�1); �(n� 1); �(n)).

3. S T = f� � � : � 2 S & � 2 Tg.Clearly S � T is bounded if and only if both S and T are bounded and

similarly for the other notions of boundedness.

Exercises


III.1.1. Prove Lemma III.1.2.

III.1.2. Find a (non-computable) tree T with a computable function h such that�(i) < h(i) for all � 2 T and all i < j�j, such that there is no computablefunction f which enumerates the immediate successors of � 2 T .

III.1.3. Given a highly computable tree T , �nd a computable function f such that,for each � 2 T , � has � f(�) immediate successors in T .

III.1.4. Show that S � T and S T are trees.

III.2 De�nition and basic properties of �01 Classes

In this section, we examine �01 classes with various boundedness conditions. We

begin with some general facts.

Lemma III.2.1. A subset K of NN is closed if and only if K = [T ] for sometree T .

Proof. Suppose �rst that K is closed and let T = f� 2 N� : K \ I[�] 6= ;g. Wewill verify that K = [T ]. If x 2 K, then for any n, x 2 K \ I[x � n], so thatx � n 2 T . It follows that x 2 [T ]. Conversely, suppose that x =2 K. Since Kis closed, there must be some basic interval I[x � n] such that K \ I[x � n] = ;.But then x � n =2 T and hence x =2 [T ].

De�nition III.2.2. 1. For any closed set P � NN, let TP denote the treef� 2 N� : P \ I[�] 6= ;g.

2. For any tree T , an in�nite path through T is a sequence (x(0); x(1); : : : )such that x � n 2 T for all n. Let [T ] be the set of in�nite paths throughT .

3. A subset P of NN is a �01 class (P is e�ectively closed if P = [T ] for some

computable tree T .

4. A subset P of NN is a decidable �01 class if TP is a computable set.

It is important to note that for a �01 class P , the set TP need not be com-

putable.

Example III.2.3. Let A be an arbitrary c.e. set and let P = f0n1! : n =2 Ag.Then P is a �0

1 class but TP is not computable. (Details left as an exercise.)

The notions of boundedness for trees from the previous section carry overto notions of boundedness for closed sets.

De�nition III.2.4. A subset K of NN is (topologically) bounded if there is afunction h such that, for all x 2 K and all n, x(n) � h(n).

III.2. DEFINITION AND BASIC PROPERTIES OF �01 CLASSES 61

In the study of �01 classes, the term \bounded" has the e�ective version

given below, so that we use the modi�er \topologically" to distinguish the twonotions.

De�nition III.2.5. 1. A �01 class P is bounded if there is a computable

tree T such that P = [T ] and a function h (not necessarily computable)such that �(n) � h(n) for all � 2 T .

2. A �01 class P is computably bounded (c. b.) if P = [T ] for some highly

computable tree T .

3. A �01 class P is almost bounded if P = [T ] for some a.b. tree T .

4. A �01 class P is almost computably bounded (a. c. b.) if P = [T ] for

some a. c. b. tree T .

In the next chapter, we will show that for any hyperarithmetical real x 2 NN,there exists y �T x such that fyg is a �0

1 class. Here we give the special casewhen x is a �02 set.

Example III.2.6. Here is an example of a �01 class P which is topologically

bounded, but not bounded. That is, P = fgg for a particular function g andhence P is bounded by g itself. At the same time, P = [T ] for some computabletree T . However, the tree T is not itself bounded by g, or even �nite branching.

Let A be a �02 set which is not �02. Let R be a computable relation so that

e 2 A () (9m)(8n)R(e;m; n):

De�ne the (non-computable) function f in two cases as follows.

(Case I): If e 2 A, then f(e; 0) = 1, f(e; 1) is the least m such that(8n)R(e;m; n) and f(e; 2) = hn0; : : : ; nf(e;1)�1i where for each i, ni is the leastn such that :R(e; i; n).

(Case II): If e =2 A, then f(e; 0) = 0 and f(e;m+ 1) is the least n such that:R(e;m; n).

Then de�ne g by g(2e(2m + 1)) = f(e;m). Observe that for each e, thevalues of g(2e(2m+ 1) tell us whether e 2 A and also verify the answer.

We claim that fgg is a �01 class. That is, fgg = [T ] for the computable tree

T de�ned as follows.

� 2 T if and only if, for all e with 2e < j�j, one of the following.

(0) �(2e) = 0 and, for all i with 2e(2i + 1) < j�j, :R(e; i; �(2e(2i + 1)), andfor all j < �(2e(2i+ 1)), R(e; i; j).

(1) �(2e) = 1 and, for all n < j�j, R(e; �(3 � 2e); n). For all i > 2, �(2e(2i+1)) = 0. Finally, �(5 � 2e = hn0; n1; : : : ; n�(3�2e)�1i where for all i <�(3 � 2e), ni is the least n such that :R(e; i; n).


Clearly g 2 [T ]. Now suppose there were some other function h 2 [T ]. Fix eand consider two cases as above.

(Case I): h(2e) = 0. Then for each i, :R(e; i; h(2e(2i+1)) and hence e =2 A.It follows that, for each m, h(2e(2m + 1)) is the least n such that :R(e;m; n)and hence h(2e(2m+ 1)) = f(2e(2m+ 1) for all m.

(Case II): h(2e) = 1. Then for all n, R(e; h(3 � 2e); n) and hence e 2 A.Furthermore, for each i < h(3 � 2e), h(5 � 2e) witnesses that :(8n)R(e; i; n), sothat h(3 �2e) is the least m such that (8n)R(e;m; n); that is, h(3 �2e) = f(3 �2e).It follows that h(2e(2i + 1) = h(3 � 2e) = f(3 � 2e) = f(2e(2i + 1) for all i > 2.Finally, h(5�23) must code the sequence of least witnesses n such that :R(e; i; n)for each i < h(3 � 2e) and hence h(5 � 2e) = f(5 � 2e) as well.

It follows that h = f and hence [T ] = ffg as desired.Now we claim also that this class does not have a �0

2 bounding function h.Suppose by way of contradiction that there were such a function h. Then

e =2 A () (8m < h(2e3))(9n):R(e;m; n):This would give a �0

2 de�nition of A. To see this, let

�(e) = (least n)(8m < h(2e3))(9n0 < n):R(e;m; n0):Then � is computable in 00 and N nA = Dom(�), so that A is a �0

2 set.Now the �0

1 class fgg is certainly topologically bounded, but it follows fromLemma III.2.7 that it is not bounded.

Lemma III.2.7. (a) A closed subset K of NN is (topologically) bounded ifand only if there is a �nite-branching tree T such that K = [T ];

(b) A �01 class P is bounded if and only if there is a �nite-branching com-

putable tree T such that P = [T ]. Furthermore, the bounding functionmay always be taken to be computable in 00.

(c) A decidable �01 class P is e�ectively bounded if and only if it is bounded.

Proof. We leave the proof of part (a) to the reader. Let T be a computable treesuch that K = [T ]. Suppose �rst that K is bounded and let h be a functionsuch that �(i) � h(i) for all � 2 T and all i < j�j. It is immediate that T is�nite-branching, since for any � 2 T , if �_j 2 T , then j � h(j�j). On the otherhand, if T is �nite-branching, then it is easy to see by induction that T \Nk is�nite for all k and hence we may de�ne the bounding function h by

h(k) = maxfi : (9� 2 T \ Nk)�_i 2 Tg:It is clear that h is computable in 00.

(c) Let P = [T ] where T is computable and has no dead ends. If T is �nite-branching, then P is bounded by (a). Now suppose that P is bounded by afunction h. Since T has no dead ends, it follows that T is also bounded by h.Thus T must be �nite-branching.


Proposition III.2.8. A �01 class P is computably bounded if and only if there

is a computable function h such that x(n) � h(n) for all x 2 P .Proof. Let P = [T ] for the computable tree T . If P is computably bounded,then there is a computable function h such that �(n) � h(n) for all � 2 T andall n < j�j and it follows that x(n) � h(n) for all x 2 P . For the other direction,let h be given as described. Then we can de�ne a tree S � T by having

� 2 S () � 2 T & (8n < j�j)�(n) � h(n):It is clear that [S] = [T ] = P so that P is computably bounded.

De�nition III.2.9. (i) � is an extendible node of T if I(�) \ [T ] 6= ;, thatis, if � has an in�nite extension which belongs to [T ]; Ext(T ) is the setof extendible nodes of T .

(ii) A node � such that � =2 T , but all � � � are in T , is a dead end of T .

Observe that TP has no dead ends and is the unique tree without dead endssuch that P = [T ]. The following lemma gives an alternate de�nition of thenotion of a decidable �0

1 class. The proof is left as an exercise.

Lemma III.2.10. For any �01 class P , the following are equivalent:

(a) P is decidable;

(b) There is a computable tree T with P = [T ] such that Ext(T ) is computable.

(c) There is a computable tree T with no dead ends such that P = [T ].

It is a fundamental property of the real line that a subset is compact if andonly if it is closed and bounded.

Theorem III.2.11. (a) A subset K of NN is compact if and only if it is closedand (topologically) bounded.

(b) A �01 class P is bounded if and only if there exists a computable tree T

with P = [T ] and a function h such that �(n) � h(n) for all � 2 T andall n < j�j. Furthermore, h may be taken to be computable in 00.

Proof. (a) Suppose �rst that K is compact. Then K is certainly closed. Foreach n, K � Sifx : x(n) = ig and it follows from compactness that there existssome in such that K � fx : x(n) � ing. Then the function h(n) = in satis�esthe condition above. Suppose next that K is closed and bounded and let h bea bounding function for K. Then K � Qn2!f0; 1; : : : ; h(n)g and is thereforecompact since it is a closed subset of a compact space.

(b) If P = ;, then this is obvious. Thus we let P be a nonempty �01 class

and let T be a computable tree such that P = [T ]. Suppose �rst that T is�nitely branching. Then we may de�ne the bounding function h by letting h(n)be the maximum of f�(m) : � 2 T; mk�j � ng. It is clear that h is computablein 00. Conversely, suppose that P = [T ] and that h is any bounding functionsuch that �(n) � h(n) for all � 2 T . It is immediate that T must be �nitelybranching.


A crucial result for bounded classes is K�onig's Lemma, which follows fromthe compactness of bounded classes.

Lemma III.2.12 (K�onig's Lemma). Any in�nite, �nite-branching tree has anin�nite path.

Proof. Let T be an in�nite, �nite-branching tree and let P = [T ]. An in�nitepath x through T is de�ned as follows. Let x(0) be the least i such that T con-tains in�nitely many extensions of (i) and for each n, similarly let x(n+1) be theleast i such that T contains in�nitely many extensions of (x(0); x(1); : : : ; x(n); i).Since T is �nite-branching, it follows by induction that such an i always ex-ists.

Notice that the path de�ned in the proof of K�onig's Lemma is not necessarilycomputable, despite the \recursive" de�nition. The complexity of this path willbe considered further below in Chapter IV.

We can use K�onig's Lemma to determine the complexity of Ext(T ) for acomputable tree T .

Theorem III.2.13. Let T be a computable tree in N�.

(a) T , Ext(T ) is �11.

(b) For a �nite-branching tree T , Ext(T ) is �02.

(c) For a highly computable tree T , Ext(T ) is �01.

Proof. In general, we have

� 2 Ext(T ) () (9x)[� � x & (8n)x � n 2 T ]:For a �nite branching tree, K�onig's Lemma implies that

� 2 Ext(T ) () (8n)(9� 2 N�)[j� j = n & �_� 2 T ]:Part (c) is left as an exercise.

We now present the fundamental basis result is due to Kleene.

Theorem III.2.14 (Kleene). For any tree T such that the �01 class P = [T ] is

nonempty, P contains a member which is computable in Ext(T ).

Proof. The in�nite path x through T can be de�ned recursively by letting x(0)be the least n such that (n) 2 Ext(T ) and, for each k, letting x(k + 1) be theleast n such that (x(0); : : : ; x(k); n) 2 Ext(T ).

Combining this with Theorem III.2.13, we get the following:

Theorem III.2.15. For any nonempty �01 class P � NN:

(a) P has a member computable in some �11 set;


(b) if P is bounded, then P has a member of �02 degree (hence computable in000);

(c) if P is c. b., then P has a member of c.e. degree (hence computable in00);

(d) if P is decidable, then P has a computable member.

Proof. Part (a) and the parenthetical consequences in parts (b) and (c) areimmediate from Theorems III.2.13 and V.2.3. The existence of a member of c.e.�02 degree in (b) and of a member of c.e. degree in (c) are left as an exercise.

The following corollary is very useful.

Corollary III.2.16. Any isolated element of a computably bounded �01 class is

computable.

Proof. Without loss of generality, Let x 2 NN and let P = fxg be a c.b. �01

class. Let P = [T ] where T is computable and computably bounded and let fbe given so that �(m) < f(n) for all � 2 Nn and all m < n. Then Ext(T ) is�01 by Theorem III.2.13. But for any � 2 Ext(T ), � = xdj�j and is the unique

member of Ext(T ) with length j�j. Thus we also have

� 2 Ext(T ) () (8� 2 f0; 1; : : : ; f(n)gj�j)[� 2 Ext(T ) =) � = �]=

This shows that Ext(T ) is also c. e. and is therefore computable, so that x iscomputable by Theorem III.2.15.

Part (a) is the Kleene basis theorem [108] and part (c) is the Kreisel basistheorem [115].

For two �01 classes P = [S] and Q = [T ], de�ne the amalgamation of P

and Q, P Q, by P Q = fx � y : x 2 P & y 2 Qg. Then it is clear thatP Q = [S T ]. More generally, de�ne the in�nite amalgamation iSi to bethose strings � such that for each i, (�([i; 0]); �([i; 1]); : : : ; �([i; j])) 2 Si, wherej is the maximum such that [i; j] < j�j. Then [iSi] is isomorphic to the directproduct �i[Si].

We also wish to consider the disjoint union. For two �01 classes P = [S] and

Q = [T ], P � Q = f0_x : x 2 Pg [ f1_y : y 2 Qg. It is easy to see thatP �Q = [S � T ].

Another consequence of compactness is the dual reduction property of �01

classes in f0; 1gN. Some de�nitions are needed.De�nition III.2.17. A family � of subsets of some set X has the ReductionProperty if, for any A and B in �, there exist A1 and B1 in � such that

(i) A1 � A and B1 � B;(ii) A1 [B1 = A [B;(iii) A1 \B1 = ;.


A standard result from classical computability theory is the following.

Proposition III.2.18. The family of c.e. subsets of N satisfy the reductionproperty.

Proof. Given c.e. sets A and B, let f and g be computable functions whichenumerate A and B, respectively. Then de�ne

A1 = ff(n) : (8m < n)f(n) 6= g(m)gand

B1 = fg(m) : (8n � m)g(m) 6= f(n)g:That is, i = f(n) is enumerated into A1 as long as it has not already appearedin B and similarly j = g(m) is enumerated into B1 as long as it does not comeinto A by stage m.

We can modify this to show that the �01 classes in NN also have the reduction

property.

Proposition III.2.19. The family of �01 classes in NN has the reduction prop-

erty.

Proof. Since A and B are the complements of �01 classes, it follows from The-

orem III.3.2 that there exist computable functions f and g such that A =Sn I(�f(n)) and B =

Sm I(�g(m)) where �f(n1) and �f(n2) are incompatible for

n1 6= n2 and similarly for g. Now de�ne computable sequences Un and Vn ofclopen sets by

Un = I(�f(n)) n[m<n

I(�g(m))

andVm = I(�g(m)) n

[n�m

I(�f(n)):

Then let A1 =Sn Un and B1 = [mVm.

De�nition III.2.20. A family � of subsets of some set X has the Dual Re-duction Property if, for any P and Q in �, there exist P1 and Q1 in � suchthat

(i) P � P1 and Q � Q1;

(ii) P1 \Q1 = P \Q;(iii) P1 [Q1 = X.

This is the dual of the usual reduction property and was �rst studied byHerrmann. It will be important in the study of the lattice E� in Chapter XVII.It is easy to see that a family � will satisfy the reduction property if and only ifthe family of complements of � satis�es the dual reduction property. (See theexercises.)


Corollary III.2.21. The family of �01 classes in f0; 1gN has the dual reduction

property.

De�nition III.2.22. A class � of subsets of some set X has the SeparationProperty if for any P and Q in �, if P \Q = ;, then there exists R such thatboth R and X nR are in � and P � R � X nQ. R is said to separate P and Q.

If � is the family of �01 classes in f0; 1gN (or in general, the family of closed

subsets of f0; 1gN), then R and f0; 1gN n R are both closed if and only if R isclopen.

The following result is left as an exercise.

Proposition III.2.23. If � has the dual reduction property, then it also hasthe separation property.

Corollary III.2.24. [Separation Property] The family of �01 classes has the

separation property.

It is well-known that the c.e. sets do not satisfy the separation property.(See section III.5 below.) This also carries over to the �01 classes.

Proposition III.2.25. The family of �01 classes does not have the separationproperty.

Proof. Let A and B be disjoint, computably inseparable c.e. sets and let U =[n2AI(0n1) and V = [n2BI(0n1). Suppose by way of contradiction that G isa clopen set such that U � G and V \G = ;. Since G is closed, 0! 2 G. SinceG is open, there must be some �nite m such that I(0m) � G. But then theremust be some n 2 B with n > m, so that 0n1! 2 V \G.

Exercises

III.2.1. Prove Lemma III.2.7(a).

III.2.2. Show that TP is the least tree T such that P = [T ], that is, the intersectionof all such trees.

III.2.3. Explain why the computable tree T in Example III.2.6 cannot be �nite-branching. Where does the in�nite branching occur?

III.2.4. Show that P as de�ned in Example III.2.3 is a �01 class and that TP is

not computable.

III.2.5. Let P = f0; 1gN. Show that there are in�nitely many computable trees inf0; 1; 2gN such that P = [T ] and continuum many non-computable treesT in f0; 1; 2; 3gN with P = [T ].


III.2.7. Show that Ext(T ) is a �01 set for a highly computable tree T .


III.2.8. Complete the proof of Theorem III.2.15. Hint: If T is a computablybounded tree, then the leftmost in�nite path of T is a c.e. real, as de�nedin Section II.4.

III.2.9. Verify that [S � T ] = [S]� [T ] and that [S T ] = [S] [T ].

III.2.10. Show that � has the reduction property if and only if ~� = fX nS : S 2 �ghas the dual reduction property.

III.2.11. Prove Proposition III.2.23

III.2.12. Show that the family of all open sets in NN has the reduction propertyand hence the class of closed sets has both the dual reduction propertyand the separation property.

III.2.13. Improve Proposition III.2.25 by showing that the �01 classes de�ned in theproof cannot be separated by any decidable �0

1 class.

III.3 E�ectively Closed Sets in the Arithmetic

Hierarchy

The following lemma makes the connection between trees and quanti�ed rela-tions precise.

Proposition III.3.1. For any class P � NN, the following are equivalent:

(a) P = [T ] for some computable tree T � N�;(b) P = [T ] for some primitive recursive tree T ;

(c) P = fx : (8n)R(n; x)g, for some computable relation R;

(d) P = [T ] for some �01 tree T � !<!;

Proof. : [(a) ! (b)]: Suppose that P = [T ], where T is a computable treeand let �e be a total f0; 1g-valued computable function such that � 2 T ifand only if �e(�) = 1. De�ne the primitive recursive tree S by � 2 S ()(8n < j� j):�e;j� j(�dn) = 0. Clearly T � S, so that [T ] � [S]. Suppose nowthat x =2 [T ]. Then for some n, xdn =2 T . Thus we have some m such that�e;m(xdn) = 0. Then for any k > maxfm;ng, we clearly have xdk =2 S. Itfollows that x =2 [S].

[(b) ! (c)]: Suppose that P = [T ], where T is a primitive recursive tree.De�ne the relation R by R(n; x) () xdn 2 T . then we have x 2 [T ] ()(8n)xdn 2 T () (8n)R(n; x):

[(c)! (d)]: Suppose that x 2 P () (8n)R(n; x) where R is a computablerelation, that is, there is a computable functional � = �e such that R(n; x) ()�(n; x) = 1 and :R(n; x) () �(n; x) = 0. By the Master EnumerationTheorem II.II.2.5, we have �(n; x) = i if and only if �(n; xdm) = i for some m.De�ne the tree T by

III.3. EFFECTIVELY CLOSED SETS IN THE ARITHMETIC HIERARCHY69

� 2 T () (8n < j�j)�(n; �) #! �(n; �) = 1:

It is clear that P = [T ].[(d) ! (a)]: Suppose that the tree T is a �0

1 subset of !<!, so that thereis a computable relation R such that � 2 T () (8n)R(n; �). De�ne thecomputable tree S � T by � 2 S () (8m;n � j�j)R(m;�dn). It is easilyveri�ed that [T ] = [S].

Standard topology tells us that a closed set may be de�ned as the comple-ment of an open set, and that an open set in NN is a countable union of intervalsI(�). Since NN is completely disconnected, this countable union can be madedisjoint. The e�ective version of this fact is the following. Let �0; �1; : : : bean e�ective enumeration of N�. (This can be done in order �rst by the sum�(0) + �(1) + � � �+ �(j�j) and then lexicographically.)

Theorem III.3.2. (a) If P � NN is a �01 class, then there is a computable

set W such that NN n P =Sn2W I(�n). Furthermore, the set W may be

chosen so that I(�m) and I(�n) are disjoint for m 6= n.

(b) For any c.e. set W � N�, NN nSn2W I(�) is a �01 class.

Proof. (a): Let P be a �01 class and let T be a computable tree such that

P = [T ]. Let W = fn : �n is a dead end of Tg. Then in fact W is a computableset and clearly NN n P =

Sn2W I(�n). That is, if x 2 I(�n) for some n 2 W

and k = j�nj, then x � k = �n =2 T , so that x =2 P . On the other hand, if x =2 P ,then there is a least k such that x � k =2 T . Take n so that x � k = �n. Thenn 2 W and x 2 I(�n). To check the disjointness condition, suppose that �mand �n are both dead ends of T . If they were comparable, then without loss ofgenerality, �m � �n, which contradicts the assumption that both are dead ends.

(b) Given the c.e. set W � N�, let P = NN n Sn2W I(�) = [T ] and de�nethe �0

1 tree T by� 2 T () (8� � �)� =2W:

Then NN nSn2W I(�) = [T ]. It follows from Proposition III.3.1 that P is a �01

class.

Part (d) of Proposition III.3.1 may be used to de�ne an enumeration of the�01 classes. That is, let we could let Pe = NN n

Sn2We

I(�n), where �0; �1; : : : isthe standard enumeration of f0; 1g� in length-lexicographic order. By applyingProposition III.3.1 uniformly, we obtain the following.

Theorem III.3.3. There is a uniformly primitive recursive sequence hTeie2!of trees such that h[Te]ie2! enumerates the �0

1 classes.

Proof. Simply let

� 2 Te () (8n < j�j)[�n � � ! n =2We;j�j:


The o�cial enumeration for the �01 classes will be de�ned below in Chapter

VI and is based directly on primitive recursive trees.There is another notion equivalent to being a computably bounded �0

1 class.

Proposition III.3.4. Let P � NN and h a computable function such thatx(n) < h(n) for all x 2 P . Then P is a �0

1 class if and only if TP is a �01 set.

Proof. If TP is a �01 set, then P is a �0

1 class by Proposition III.3.1 sinceP = [TP ]. Now suppose that h is a computable function and x(n) < h(n)for all x 2 P and all n. Then by K�onig's Lemma,

� 2 TP () (8k > j�j)(9� 2 h(0)� h(1)� : : : h(k))[� 2 T & � � � ]:

We will also consider in general the families of �0n classes and �0n classes.

Analogous to the de�nition of �0n sets, we have the following.

De�nition III.3.5. Let R be a relation on Nk �NN and let n > 0 be a naturalnumber.

1. R is �00 if it is computable.

2. R is �0n+1 if there is a �0n relation Q � Nk+1 � N such that

R(a1; : : : ; ak; x) () (9i)Q(i; a1; : : : ; ak; x):

3. R is �0n+1 if Nk � NN nR is �0n+1.

4. R is �0n+1 if R is both �0n+1 and �0

n+1.

Note that a computable class in NN is both open and closed. These de�-nitions can also be relativized to a set oracle C, but the results are not quiteanalogous to those for sets.

De�nition III.3.6. (i) A subset of NN is a strong �0n+1 class if P = [T ]

for some tree T computable in ;(n).(ii) A strong �0

n+1 class P is highly bounded if P = [T ] for some tree T

computable in ;(n) and a bounding function f also computable in ;(n)such that �(n) � f(n) for all � 2 T .

Proposition III.3.7. For any class P � NN, the following are equivalent:

(a) P = [T ] for some tree T computable in 0(n).

(b) P = [T ] for some �0n+1 tree T .

(c) P = [T ] for some �0n tree T .

Furthermore, if P � f0; 1gN, then T � f0; 1g�.

III.4. GRAPHS OF COMPUTABLE FUNCTIONS 71

Proof. The equivalence of (a) and (b) follows from a relativization of PropositionIII.3.1. Clearly (c) implies (b). It remains to be shown that (b) implies (c). LetT be a �0

n+1 tree such that P = [T ]. Then there is a �0n relation R such that

� 2 T () (8i)R(i; �);so that

x 2 P () (8m)(8i)R(i; x � m):

Now de�ne the �0n tree S by

� 2 S () (8m � j�j)(8i � j�j)R(i; � � m):

It is easy to check that S is a tree and that P = [S].

Exercises

III.3.1. Complete the proof of Proposition III.3.7 by showing that S is a tree andthat P = [S].

III.4 Graphs of Computable Functions

In classical computability theory, computable functions and computably enu-merable sets are the two primary objects of study. In this context, the twoare naturally related. For any (partial) computable function, the domain, therange and the graph are all c.e. sets. Furthermore, every c.e. set is the domainof a computable function and every nonempty c.e. set is the range of a totalcomputable function. In addition, any partial function with a c.e. graph is nec-essarily computable. For our purposes, computably continuous functions and�01 classes are the primary objects of study. In this section, we explore possible

analogues of the classical results.Recall from the Master Enumeration Theorem II.II.2.5 that a (partial) com-

putable function � : NN ! NN may be approximated by maps on sequences. Let�(�;m) = n if � computes output n on input m using only oracle informationfrom � and in j�j or fewer steps; we may assume that n < j�j. Let �(�) = �denote the partial function on strings as before.

Before giving a characterization of a computably continuous function, we�rst consider arbitrary continuous functions.

Lemma III.4.1. A function F : NN ! NN (respectively, F : f0; 1gN ! f0; 1gN)is continuous if and only if there is a function f : N� ! N� (resp. f : f0; 1g� !f0; 1g�) such that

1. for all � � � , f(�) � f(�);2. for all x 2 NN (f0; 1gN), limn!1 jf(x � n)j =1;

3. for all x 2 NN (f0; 1gN), [nf(x � n) = F (x).


Proof. Since F is continuous, it follows that F�1(I(�)) is open for each � 2 N�.De�ne f(�) to be the unique longest � such that I(�) � F�1(I(�)) [equivalently,F (I(�)) � I(�)].

Then f is certainly monotonic.Fix x 2 NN and let Y = F (X). For each n, X belongs to the open set

F�1(I(Y � n)) and hence there is some basic open set I(�) such that X 2I(�) � F�1(I(Y � n)). It follows that Y � n � f(�) and of course � = x � mfor some m. Then for all t > m, jf(x � t)j � n. Thus limt!1jf(x � t)j =1, asdesired.

It now follows that [nf(x � n) = F (x).

The fundamental notion of computable analysis is the computable versionof Lemma III.4.1.

Lemma III.4.2. A function F : NN ! NN (respectively, F : f0; 1gN ! f0; 1gN)is computably continuous if and only if there is a computable function f : N� !N� (resp. f : f0; 1g� ! f0; 1g�) such that

1. for all � � � , f(�) � f(�);2. for all x 2 NN (f0; 1gN), limn!1 jf(x � n)j =1;

3. for all x 2 NN (f0; 1gN), [nf(x � n) = F (x).

Proof. Given such a representation f for F , compute y(n) for y = F (x) from xby computing f(x � k) for su�ciently large k.

Given a computable function F , de�ne the representation f as follows. Oninput � of length n, compute the values of � = f(�) for each i < n by applyingthe algorithm for F for n steps, using oracle �. The length of � will be the leastk < n such that �(k) does not converge in n steps.

Remark : The modulus of convergence function � of F : NN ! NN as de-�ned from the approximation map f : N� ! N� may be de�ned by �(x; n) =(least s)jf(x � n)j > s. For a total computable function F , this modulus func-tion is also computable. The following lemma will be useful. The proof is leftas an exercise.

Lemma III.4.3. For any computable function F : f0; 1gN ! f0; 1gN, there is acomputable uniform modulus function � : N ! N such that for all x 2 f0; 1gN,jf(xd�(n)j > n.

Theorem III.4.4. Let � : NN ! NN be a (partial) computable function. Thenthe graph of � is a �0

2 class. Furthermore, if � is total, then the graph is adecidable �0

1 class.

Proof. Let � be a representing function for �. In general, we have,

�(x) = y () (8m)(9k)[�(x � k)(m) = y(m)]:

III.4. GRAPHS OF COMPUTABLE FUNCTIONS 73

For a total function, de�ne the computable tree T with graph(�) = [T ] byputting � � � 2 T if and only if � is consistent with �(�), that is, for any m, if�(�;m) = n, then �(m) = n. Then Ext(T ) is �01 and hence computable, since

� � � 2 Ext(T ) () (9�)[� � � & � � �(�)]:

To see this, note that if � � �(�) and � � �, then for any x 2 I(�), � �F (x).

Example III.4.5. The graph of a partial computable function need not beclosed. De�ne the partial function � : NN ! NN by �(x)(m) = 0�[(least n)x(n) =1]. For each n, let xn = 0n_1_01, so that limnxn = 01. Then, for each n,F (xn) = 01, whereas F (01) is unde�ned.

For total functions on f0; 1gN, there is a converse.Theorem III.4.6. A function F : f0; 1gN ! f0; 1gN is computably continuousif and only if the graph is a �0

1 class.

Proof. One direction follows from Theorem III.4.4. Next suppose that F :f0; 1gN ! f0; 1gN and let T be a computable tree such that graph(F ) = [T ].De�ne a computable function f on strings by letting f(�) be the common partof f� : � � � 2 Tg.Theorem III.4.7. A subset D of NN is a �0

2 class if and only if D is thedomain of some partial computable function � : NN ! NN.

Proof. Suppose �rst that D is the domain of �. Then

x 2 Dom(�) () (8n)(9k)[j�(x � k)j > k]:

Next suppose that D is a �02 class and let R be a computable relation such that

x 2 D () (8n)(9k)R(n; xdk):

Then D is the domain of the partial computable function � de�ned by

�(x)(n) = (leastk)R(n; xdk):

We next examine the complexity of the image of a �01 class under a com-

putably continuous function. The classical results is that the image of anycompact set under a continous function is compact and that the image of aclosed set is an analytic set.

Theorem III.4.8. Let F be a computably continuous function on a �01 subclass

P of NN and let F [P ] = fF (x) : x 2 Pg. Then1. F [P ] is a �11 class;


2. if P is bounded, then F [P ] is a strong �02 class;

3. if P is computably bounded, then F [P ] is a computably bounded �01 class

and, furthermore, if P is decidable, then F [P ] is decidable.

Proof. (1) We have y 2 F [P ] () (9x)(x 2 P & x� y 2 graph(F )).(2) Suppose that T is a �nite branching, computable tree and let S be

a computable tree such that graph(F ) = [S]. Then it follows from K�onig'sLemma that F [P ] = [R], for the �nite branching �01 tree R de�ned by

� 2 R () (9�)[� 2 T and � � � 2 S]:(3) Now suppose that T is computably bounded and let F be represented

by the computable function f : N� ! N�. Then the de�nition of R above in (2)becomes computable since the (9�) quanti�er becomes bounded.

To �nd a bound for the possible value of �(n) for � 2 R, compute the leastmsuch that jf(�)j > n for all � 2 T of length m. Then we compute the maximumvalue h(r) of f(�(n)) for all � 2 T of length n. Thus R is seen to be highlycomputable.

This result has a converse for f0; 1gN.Theorem III.4.9. A �0

1 subclass P of f0; 1gN is the computably continuousimage of f0; 1gN if and only if it is decidable.

Proof. Let P = T , where T is a computable tree with no dead ends. Wewill de�ne the computable map f : f0; 1g� ! f0; 1g� which represents a mapF : f0; 1gN ! f0; 1gN such that y = F (x) is some element of P which is nearestto x. Let f(�) = � if � 2 T and, if � =2 T , let � be the longest initial segmentof � which is in T and let f(�) be the lexicograpical least extension of � whichis in T and has length j�j.

Exercises


III.4.2. A mapping f from N� (or f0; 1g�) into N� is a tree homomorphism if � � �implies f(�) � f(�) for all � and � . Show that for any tree homomorphismf , T = f� : (9�) : f(�) � �g is a tree and that if f is one-to-one andcomputable, then T is a computable tree and [T ] is a perfect �0

1 class.

III.5 Computably enumerable sets and �01 Classes

There are numerous connections between computable functions, �01 and c.e.

subsets of N, and �01 classes. We will consider in particular the class S(A;B) of

separating sets for a pair of c.e. sets A;B, and the class I(C) of initial subsets ofa �0

1 set C. Then we will take a �rst look at the notion of forbidden words andsets and their connection with �0

1 classes and with subshifts. In this section,the notion of retraceability

III.5. COMPUTABLY ENUMERABLE SETS AND �01 CLASSES 75

III.5.1 Separating classes

The most basic example here is that, for any �01 set C, the power set P(C) is a

�01 class. That is, we have

x � C () (8n)[x(n) = 1! n 2 C]:More generally, consider the notion of separating sets.

De�nition III.5.1. Let A and B be in�nite disjoint c.e. sets and let C � N.1. C is a separating set for A and B if A � C and B \ C = ;.2. A and B are said to be computably (or recursively) inseparable if there is

no computable separating set C for A and B.

3. The class of separating sets for A and B is denoted by S(A;B).

4. P is a separating class if P = S(A;B) for some c.e. sets A and B.

Of course, C 2 S(A;B) if and only if C 2 P(N n B) and N n C 2 P(N n A).The notion of computably inseparable sets was introduced by Kleene in [109].Shoen�eld showed in [189] that every non-computable c. e. degree containsa pair of computably inseparable sets. Shoen�eld observed in [189] that theclass S(A;B) of separating sets for A and B is a c. b. �0

1 class. We note thatS(A;B) is �nite if and only if A[B is co�nite, in which case A and B are bothcomputable and every separating set is also computable. Otherwise, S(A;B) isa perfect set and thus has the cardinality of the continuum. In either case, boththe c. e. set A and the co-c. e. set N nB are of course separating sets for A andB.

The classic example of a separating class comes from the notion of diagonallynon-computable sets. Here a function f 2 f0; 1gN is diagonally non-computableif f(e) 6= �e(e) whenever �e(e) converges. Let Ki = fe : �e(e) = ig. Then inparticular, K0 and K1 are c.e. non-computable sets and any separating set forK0 and K1 has a diagonally non-computable characteristic function. (See theexercises.) Separating classes are important for the study of reverse mathematicsand so will be examined further in Chapter VII. The complexity of the membersof separating classes will be studied in Chapter IV.

For �; � 2 f0; 1g�, let � � � mean that, for all i � minfj�j; j� jg, �(i) = 1implies �(i) = 1. De�ne � [ � to be the sequence � of length maxfj�j; j� jg suchthat �(i) = maxf�(i); �(i)g and similarly de�ne � \ � to be the sequence � oflength maxfj�j; j� jg such that �(i) = minf�(i); �(i)g

Recall that for any closed subset P of NN, TP = f� 2 f0; 1g� : I(�)\P 6= ;g.Lemma III.5.2. Suppose that P is a closed subset of f0; 1gN.

1. P is closed under subsets if and only if for every � � � , if � 2 TP , then� 2 TP .

2. P is closed under supersets if and only if for every � � � , if � 2 TP , then� 2 TP .


3. P is closed under union if and only if, for every � and � in TP , �[� 2 TP .4. P is closed under intersection if and only if, for every � and � in TP ,

� \ � 2 TP .Proof. We prove (1) and (3) and leave (2) and (4) to the reader.

(1) Suppose P is closed under subsets. Let � � � and � 2 TP . Then thereexists B 2 P such that � � B. Let C = fi 2 B : �(i) = 1g. Then C � B soC 2 P by assumption and clearly � � C. On the other hand, suppose that TPis closed under �. Let B 2 P and C � B. Then for any n, Cdn � Bdn andBdn 2 TP , so that C 2 P .

(3) Suppose P is closed under union and let � and � be in TP . Then thereexist A and B in P such that � � A and � � B. It follows that � [ � � A [Band A [ B 2 P by assumption, so that � [ � 2 TP . Suppose next that TP isclosed under union and let A;B 2 P . Fix n and let � = A � n and � = B � n.Then (A [B) � n = � [ � 2 TP since both �; � 2 TP . Hence A [B 2 P .Lemma III.5.3. Let P be a closed subset of f0; 1gN. Then P = P(A) for someA if and only if P is closed under subsets and P is closed under union.

Proof. If P = P(A), then P is certainly closed under subsets and union. Sup-pose that P is closed under subsets and union and let A = fi : (9� 2 TP )�(i) =1g. Suppose �rst that B 2 P . If i 2 B, then �(i) = 1 for � = Bd(i + 1) 2 TP ,so that i 2 A. Hence B � A. Next suppose that B � A. Then for each i 2 B,there exists �i 2 TP such that �i(i) = 1. Fix n and de�ne � 2 f0; 1gn by� = [f�i : i < n & i 2 Bg. Then � 2 TP by Lemma III.5.2 since P is closedunder union. Also, Bdn � �, so that Bdn 2 TP again by Lemma III.5.2, sinceP is closed under subsets. It follows that B 2 P .

Observe that if P is actually a �01 class, then TP is a �0

1 set and the setA de�ned in the proof of Lemma III.5.3 is in fact a �0

1 set. Thus we have thefollowing.

Proposition III.5.4. For any nonempty �01 class P of sets, the following are

equivalent:

1. P is the class of subsets of a �01 set A;

2. P is the class of subsets of some set A;

3. P is closed under subsets and under union.

There is a similar result for supersets, which is left to the exercises.Let us say that a class P of sets is closed under between-ness if, for any

sets X;Y; Z, if X � Y � Z and X;Z 2 P , then Y 2 P . It is clear that anyseparating class is closed under between-ness.

Proposition III.5.5. For any �01 class P , the following are equivalent.

1. P is the class of separating sets of some pair A;B of r. e. sets.

III.5. COMPUTABLY ENUMERABLE SETS AND �01 CLASSES 77

2. P is the class of separating sets of some pair A;B

3. P is closed under union, intersection and between-ness.

Proof. It is immediate that (1) implies (2) and (2) implies (3). Suppose thereforethat P is closed under union, intersection and between-ness. De�ne the �0

1 classQ to be the family of subsets of sets in P . That is, for � 2 f0; 1gn,

� 2 TQ () (9� 2 f0; 1gn)� � � & � 2 TP :It is clear that Q is closed under subsets and under union, so it follows fromProposition III.5.4 that Q = P(C) for some �0

1 set C. Let B = N n C.Similarly de�ne the �0

1 class R to be the family of supersets of sets in P .That is, for � 2 f0; 1gn,

� 2 TR () (9� 2 f0; 1gn)� � � & � 2 TP :It follows that R is the class of supersets of some c.e. set A.

We claim that P = Q \ R = S[A;B]. Suppose �rst that X 2 P . Thencertainly X 2 Q \ R and therefore A � X and X \ B = ;. Next suppose thatX 2 S[A;B]. Then A � X, so that X 2 R and therefore Y � X for someY 2 P . Also, X \B = ;, so that X � C and X 2 Q, which means that X � Zfor some Z 2 P . It now follows by the between-ness property that X 2 P .

The proof of the following corollary is left as an exercise 6.

Corollary III.5.6. For any subset A of N, if fAg is a �01 class, then A is a

computable set.

III.5.2 Subsimilar classes

The notions of a subsimilar set (or subshift) and of forbidden words providesanother link between c. e. sets and �0

1 classes and is also closely related tosymbolic dynamics.

De�nition III.5.7. 1. A �nite string � 2 N� is a factor of another string� if � = �0

_�_�1 for some strings �0; �1.

2. Similarly � 2 N� is a factor of x 2 NN if x = �_�_Y for some strings �and some Y 2 NN.

3. x 2 NN avoids � if � is not a factor of x and x avoids a set S of stringsif x avoids each � 2 S.

4. AV (S) � NN is the set of reals which avoid S.

Similar de�nitions apply for strings in f0; 1g� and in�nite words in f0; 1gNand also for �nite and in�nite sequences any other alphabet �. In symbolicdynamics, strings are referred to as words and in�nite sequences as in�nitewords. The set S may be thought of a set of forbidden words for AV (S).

The shift function Shift is de�ned as follows.


De�nition III.5.8. 1. For a �nite string � , Shift(�) = (�(1); �(2); : : : ; �(j� j�1).

2. For an in�nite sequence x, Shift(x) = (x(1); x(2); : : : ).

3. A set Q � NN is a subshift (or subsimilar) if Shift(x) 2 Q for all x 2 Q;that is, Q is closed under the subshift function.

Informally, the shift function simply removes the �rst symbol of a �nite orin�nite sequence.

Lemma III.5.9. For any set S, AV (S) is closed; if S is c. e. , then AV (S) isa �0

1 class.


Theorem III.5.10 (Dashti [24]). 1. A closed set Q is a subshift if and onlyif Q = AV (S) for some set S.

2. A �01 class Q is a subshift if and only if Q = AV (S) for some set c. e. set

S.

Proof. For the �rst part, it is clear thatAV (S) is a subshift, so we will just sketchthe reverse implication. Suppose that P � f0; 1gN is subsimilar and closed, andlet S = f0; 1g� � TP . If x =2 P , then for some n, x � n 2 S, so that x =2 AV (S).On the other hand, suppose that x =2 AV (S) and let x = (x � n)_�_y for somen < ! and some � 2 S. Then � =2 TP and thus �_y =2 P . Since P is subsimilar,it follows that x =2 P . A similar argument works for any alphabet �.

The proof of the e�ective version of this proposition is left as an exercise.

Exercises

III.5.1. Show that the diagonally non-computable functions form a �01 class in

f0; 1gN.III.5.2. Show that S(K0;K1) is a �0

1 class of sets with no computable members,that is, K0 and K1 are computably inseparable.

III.5.3. Suppose that P = [T ] where T is a tree with no dead ends. Show thefollowing.

(a) P is closed under supersets if and only if, for every � 2 T and every� such that � � � , � 2 T .

(b) P is closed under intersection if and only if, for every � and � in T ,� \ � 2 T .

III.5.4. Let P be a closed subset of f0; 1gN. Then P is the class of supersets ofsome set A if and only if P is closed under supersets and P is closed underintersection.

III.6. RETRACEABILITY 79

III.5.5. Show that, for any nonempty �01 class P of sets, the following are equiv-

alent:

(i) P is the class of supersets of a �01 set A;

(ii) P is the class of supersets of some set A;

(iii) P is closed under supersets and under intersection.

III.5.6. Use Proposition III.5.5 to show that for any subset A of N, if fAg is a �01

class, then A is a computable set.

III.5.7. Show that a �01 class Q is subsimilar if and only if Q = AV (S) for some

c. e. set S.

III.6 Retraceability

For any in�nite set A, recall that the principal function pA enumerates theelements a0 < a1 < � � � in increasing order and that A is hyperimmune if, forany computable function f , there is an n such that an > f(n). A c. e. setis hypersimple if its complement is hyperimmune. A is said to be retraceableif there is a partial computable function � such that �(an+1) = an for all n.Retraceable sets were introduced by Dekker and Myhill [56], who proved thatany retraceable noncomputable �0

1 set A is hyperimmune. For �01 sets A, a

stronger characterization can be given.

Theorem III.6.1. The following are equivalent for any in�nite �01 set A:

(a) A is retraceable

(b) There is a total computable function � such that, for all n, �(an+1) = anand, for all y, fx : �(x) = yg is �nite.

(c) There is a total computable function such that, for all n, (an) = nand fx : (x) = ng is �nite

Proof. Let A = fa0 < a1 < � � � g be an in�nite �01 set and let A be the decreasing

intersection of uniformly computable sets As.(a)! (b): Let � be a partial computable retracing function for A. Assume,

without loss of generality, that �(a0) = a0. Then for any x, we de�ne �(x) asfollows. Look for the least s such that either x =2 As, or such that �s(x) = y � xconverges and, for all z with y < z < x, z =2 As. In the former case, we let�(x) = x and in the latter case, we let �(x) = �(x). Note that if x =2 A, thenthe former case will obtain and if x 2 A, then the latter case will obtain, so that� is total and is a retracing function for A. It follows from the de�nition thatfor every x, �(x) � x and there are no elements of A between �(x) and x. Nowfor any y, let a be the least such that a > y and a 2 A. Then �(x) = y impliesthat x � a, so that fx : �(x) = yg is �nite, as desired.

(b) ! (c): Let � be given as described. Then we de�ne (x) to be lengthn of the chain x > �(x) > �(�(x)) > � � � > �n(x) = a0, if there is such an


n-chain, and (x) = x if �i+1(x) = �i(x) for some i. Thus for an 2 A, weobtain (an) = n. To complete the proof, we show by induction that, for eachn, there are only �nitely many n-chains x > �(x) > � � � > �n(x) = a0 of lengthn. For n = 1, this follows from the assumption that �(x) = a0 for only �nitelymany x. Suppose now that there are only �nitely many such n-chains of lengthn. Then any n + 1-chain must extend one of these and, by our assumption,there are only �nitely many ways to extend each chain. Thus there can be only�nitely many n+ 1-chains.

(c)! (a): Let be given as described. Then for a = an+1 2 A, (a) = n+1and the retracing function �(an+1) = an may now be computed by searchingfor the least s such that exactly n+1 elements of As are less than a and takingan to be the largest of those.

We say that an in�nite set A = fa0 < a1 < � � � g is second-retraceable if thereis a (total) computable function � such that, for any m < n, �(am; an) = m. Ingeneral, A is k-retraceable if there is a computable � such that �(am1

; am2; : : : ; amk) =

m1 for any m1 < m2 < � � � < mk. Of course, any k + 1-retraceable set is alsok-retraceable.

A subset F of the set fa0 < a1 < � � � g is said to be an initial subset of A ifan+1 2 F implies an 2 F for all n. Thus the initial subsets of A are A togetherwith the �nite sets fa0; : : : ; an�1g for each n. Let I1(A) denote the class ofinitial subsets of A. In general, the k-initial subsets Ik(A) are the subsets F ofA such that for any elements a < b1 < b2 < � � � bk of A, if b1; : : : ; bk 2 F , thena 2 F .Theorem III.6.2. For each �nite k, the set A = fa0 < a1 < : : : g is �0

1 andk-retraceable if and only if the class Ik(A) of k-initial subsets of A is a �0

1 class.

Proof. Suppose �rst that A is k-retraceable via the function � and that A is a�01 set. Let As denote the computable approximation to the set A at stage s,

so that A = \sAs. Now de�ne the computable tree T as follows.bb0; b1; : : : ; bn; sc 2 T ()1. (8i � n)(bi 2 As) and2. if n � k, then �(bn�k+1; bn�k+2; : : : ; bn�1; bn) = n� k + 1.

It is easy to check that [T ] = Ik(A), so that Ik(A) is a �01 class.

Now suppose that Ik(A) is a �01 class and let T be a computable tree so

that Ik(A) = [T ]. We will explain how to compute a k-retracing function�. Given b1 = am1

< b2 = am2< : : : < bk = amk , observe that there is

only one possible string � = ba0; a1; : : : ; am1�1; b1; b2; : : : ; bkc_1 of the formbc0; c1; : : : ; cr; b1; b2; : : : ; bkc_1 which has an extension in T ; �(b1; : : : ; bk) = m1

is then easily computed from �. To �nd �, we just search through all strings oflength m > bk until we �nd m large enough so that all strings � in T of lengthm and with �dbk + 1 of the desired form, start with the same initial segment(�) of length bk + 1.

To see that A is a �01 set, recall that Ext(T ) is �

01 and observe that

a 2 A () (9�)[j�j = a+ 1&� 2 Ext(T )&�(a) = 1]:

III.6. RETRACEABILITY 81

We can now give a quick proof that any retraceable non-computable �01 set

is hyperimmune.

Theorem III.6.3. [Dekker-Myhill] If A = fa0 < a1 < � � � g is a retraceablenon-computable �0

1 set, then A is hyperimmune.

Proof. By Theorem III.6.2, P (A) is a �01 class. Now suppose by way of con-

tradiction that f were a computable function which dominated pA, that is,f(n) > pA(n) for all n. Then the set fAg would be the intersection of I(A) withthe following �0

1 class:fB : (8n)(card(B \ f0; 1; : : : ; f(n)g) � ng.Thus fAg would be a �0

1 class, so that A would be computable (this is seenbelow in Exercise V.1.11. This contradiction now demonstrates the result.

For any k-retraceable �01 set A, the �

01 class Ik(A) is provides an example

of a class with C-B rank k.

Theorem III.6.4. For any set A, Dk(Ik(A)) = fAg.Proof. It is easy to see that D(I(A)) = fAg and that, for each k, D(Ik+1(A)) =Ik(A).

It follows that if A is k-retraceable, then A has rank k in Ik(P ) and thus hasrank � k.

We next give a result which shows how to de�ne a retraceable �01 set by

�01-recursion.

Theorem III.6.5. Suppose that the set A = fa0 < a1 < : : : g is de�ned recur-sively by a �0

1 relation Q(x; y) such that, for all n and x, x = an () Q(x;<a0; : : : ; an�1 >). Then A is a �0

1 set and is retraceable.

Proof. De�ne the �01 relation R(n; x) by

R(n; x) () (9x0 < � � � < xn�1 < xn = x)(8i < n)Q(xi; < x0; : : : ; xi�1 >).Then the set A is �0

1 sincea 2 A () (9n � a)R(n; a).De�ne the uniformly computable relation Rs(n; x) as in the de�nition of R

above with Qs in place of Q.The counting function such that (an) = n may of course be de�ned by

the fact that n is the unique y such that R(y; an). Since, given a 2 A, there isjust one n � a such that a = an, (a) = n may be computed by searching foran s large enough so that Rs(n; a) for only one number n � a.

This result can now be applied to give a quick proof of the following theoremof Dekker and Myhill [56] (Theorem T3).

Theorem III.6.6. [Dekker-Myhill] Every r.e. set B is Turing equivalent to aretraceable �0

1 set A.


Proof. Let the c.e. set B be the union of uniformly computable sets Bs andde�ne the set A by �0

1 recursion as follows. There are two cases in the de�nitionof an. If n =2 B, then an = an�1 +1 and if n 2 B, then an is the least s > an�1such that n 2 Bs. It is clear that this is a �0

1-recursion, so that A is a �01

retraceable set. The de�nition also shows that A is computable in B. On theother hand, for any n, we have n 2 B () n 2 Ban , so that B is computablein A.

De�nition III.6.7. The Cantor-Bendixson (C-B) rank of a set A is the leastordinal � such that A has rank � in some �0

1 class P � f0; 1gN.It follows from Theorems III.6.4 and III.6.6 that every non-zero r.e. degree

contains a set A of C-B rank one. A slightly better result was obtained in [25]by Cenzer, Downey, Jockusch and Shore, hereafter abbreviated as C-D-J-S.

Theorem III.6.8. [C-D-J-S] Every c.e. non-computable set B is Turing equiv-alent to a hypersimple c.e. set E of rank one; furthermore there is a computabletree U with no dead ends such that D([U ]) = fEg.Proof. Let A = a0 < a1 < � � � be the �0

1 retraceable set de�ned in TheoremIII.6.6 and let A be the intersection of the uniformly computable, decreasingsequence As. De�ne the computable tree S to be a slight extension of the treeT de�ned in Theorem III.6.2. That is, we let � be the retracing function givenby Theorem III.6.1 for A so that �(an+1) = an and so that fx : �(x) = yg is�nite for each y, and de�nebc0; : : : ; cn; cn+1c 2 S () c0 = a0 & (8i � n)[ci 2 Acn & (i > 0 ! �(ci) =ci�1)].

S has no dead ends because for any string � 2 S, it is clear that �_0 2 S.We leave it to the reader to check that D(P ) = fAg.To obtain the c.e. set E, we note that the complement function F (C) = NnC

is a computable homeomorphism of f0; 1g! to itself, so that the c.e. set E = NnAhas rank one in the �0

1 class of complements fF (C) : C 2 Pg. Finally, anyretraceable noncomputable �0

1 set is hyperimmune by Theorem III.6.3, so thatE is hypersimple.

Exercises

III.6.1. Verify that D(I(A)) = fAg and that, for each k, D(Ik+1(A)) = Ik(A).

III.6.2. Show that D(P ) = fAg, in the proof of Theorem III.6.8.

III.6.3. For any computable function f , the set K(f) = fx : (8n)x(n) � f(n)g isa c. b. �0

1 class. Show that K(pA) is also a �01 class, where A is a in�nite

�01 set and pA is the principal function de�ned above.

III.6.4. Show that for the total retracing function � of Theorem III.6.1, if g(x) =card(fy : �(y) = xg) is computable, then A is a computable set. However,show that for the retraceing function from Theorem III.6.6, there is alwaysa computable upper bound for the cardinality.

III.7. REDUCIBILITY 83

III.7 Reducibility

In this section, we consider some connections between subclasses of f0; 1gN,computably bounded �0

1 classes, bounded classes and �0n classes. Our goal is

to reduce every class to a class of sets.Computably bounded �0

1 classes play a fundamental role and occur fre-quently in the applications. A very useful result which simpli�es the theoryof c. b. �0

1 classes is that every such class is computably homeomorphic to asubclass of f0; 1gN.De�nition III.7.1. Classes P and Q are computably homeomorphic if there isa (total) computably continuous functional F : NN ! NN such that F maps Pone-to-one and onto Q.

Notice that we do not require that F be one-to-one onto NN or that it mapNN onto NN.

Lemma III.7.2. If there is a partial computable function � mapping a subsetof NN to a subset of NN such that � is total on P and maps P one-to-one andonto Q, then P and Q are computably homeomorphic.

Proof. Let � have a representation � : N� ! N�; that is, let �(�) be thelongest sequence of the form (�(0; �);�(1; �); : : : ;�(n�1; �)). Let P = [T ] andQ = [S] for computable trees tree S and T , De�ne a new mapping f : N� ! N�

by letting f(�) = �(�) for � 2 T and letting f(�_n) = f(�)_n if �_n =2 T .Then f represents a computably continuous function F such that F (x) = �(x)for x 2 P and F (x) = �(x � k)_(x(k); x(k + 1); : : : ) if x =2 P and k is the leastsuch that x � k + 1 =2 T .Theorem III.7.3. Any c. b. �0

1 class P is computably homeomorphic to a �01

class Q of sets.

Proof. Let T be a highly computable tree such that P = [T ] and let h be acomputable function such that �(n) < h(n) for all � 2 T and all n < j�j.

The homeomorphism � will be de�ned by

�(x) = 0x(0)10x(1) : : :

and the class Q � f0; 1gN is simply f�(x) : x 2 Pg.The functional � is clearly one-to-one and maps P onto Q. � is com-

putably continuous since it is represented by the computable map taking � =(�(0); : : : ; �(n � 1)) to 0�(0)1 : : : 0�(n�1). Q = �[P ] is a �0

1 class by TheoremIII.4.8.

More speci�cally, Q = [S], where

0x(0)10x(1) : : : 0x(k�1)10i 2 S () (x(0); : : : ; x(k � 1)) 2 T & i < h(k)g:


This result can be relativized to an oracle. Also, we can apply the sameargument to �0

1 classes which are not computably bounded.

Theorem III.7.4. (a) Any strong �02 class P which is highly computable in

00 is computably homeomorphic to a strong �02 class Q of sets.

(b) For any �01 class P , there exists a �0

1 class Q � f0; 1gN and a one-to-onedegree-preserving correspondence between the non-computable members ofP and the non-computable members of Q.

Proof. (a) This is just the relativization of Theorem III.7.3 to the oracle 00.(b) Consider the representation of the mapping from Theorem III.7.3. We

can use the same mapping as above, but in this case Theorem III.4.8 only tellsus that the image is a strong �11 class. If we look at the de�nition of the tree Ssuch that Q = [S], we have to remove the condition i < h(k). This potentiallyintroduces computable elements 0x(0)10x(1) : : : 0x(k�1)0! into Q. However, anynon-computable element has in�nitely many 1's and hence will be the image ofan element of P .

If the same technique is applied to a bounded �01 class, we get one half of

the following result from [93].

Theorem III.7.5. [Jockusch-Lewis-Remmel]

(a) Any bounded �01 class P is computably homeomorphic to a strong �0

2 classQ of sets.

(b) For any strong �02 class P which is highly computable in 00, there is a

bounded �01 class Q and an e�ective one-to-one degree-preserving corre-

spondence between P and Q.

Proof. (a) This is left as an exercise.(b) Let P = [T ], where S is highly computable in 00. By Theorem III.7.4,

we may assume that T is a binary tree. It now follows from Proposition III.3.7that T may be assumed to be a �01 tree. Thus there is a computable relationR � N� f0; 1g� such that

x 2 P () (8m)(9n)R(n; x � m):

Now we may de�ne Q by

z = x� y 2 Q () (8m)[R(y(m); x � m) & 8i < y(m):R(i; x � m)]:

Then for each x � y 2 Q, we have x 2 P and for each x 2 P , there is aunique y such that x� y 2 Q and that y is de�ned so that y(m) is the least nsuch that R(n; x � m). Thus y is computable in x and therefore x � y has thesame degree as x.

The proof of part (b) can be modi�ed for an arbitrary �02 class to give a

theorem from [94]. The proof is left as an exercise.

III.7. REDUCIBILITY 85

Theorem III.7.6 (Jockusch-McLaughlin). For any �02 class P , there is a �0

1

class Q and an e�ective one-to-one degree-preserving correspondence between Pand Q.

Jockusch and Soare showed in Theorem 1 of [98] that an arbitrary �01 class

P with no computable members can be represented by a c. b. �01 class Q in the

sense that the degrees of members of P are a subset of the degrees of membersof Q. We give this result together with a relativized version. Let D(P ) denotethe set of degrees of members of the class P .

Theorem III.7.7. (a) For any �01 class P � NN, there is a �0

1 class R of setssuch that (1) D(P ) � D(R) and (2) there is a one-to-one correspondencebetween the computable members of P and the computable members of R.(So that R has no computable members if P has no computable members.)Furthermore, there is a primitive recursive function k such that for P =Pe, we have R = Pk(e).

(b) For any �01 class P � NN with no members computable in 00, there is

a strong �02 class R of sets with no members computable in 00 such that

D(P ) � D(R). Furthermore, there is a primitive recursive function h suchthat for P = Pe, we have R = P 2

h(e).

Proof. (a) Let Q be a �01 class of sets with no computable member. Let P = [S],

let Q = [T ] and assume without loss of generality that S � (N n f0; 1g)�.It su�ces, by Theorem III.7.3 to obtain a class R which is computably

bounded and otherwise meets the requirements of the conclusion. De�ne thetree U to be the set of strings

mu = �1 � �1 � �2 � �2 � � � � � �n � �nsuch that �i 6= ; for i > 1 and �j 6= ; for j < n, such that s(�) = �1�� n 2 Sand �j 2 T for all j, and such that �(k) � k + 1 for all k.

We claim that R = [U ] satis�es the requirements of the theorem. Note �rstthat the construction is uniformly computable in the tree T , so that there is aprimitive recursive function k such that the for T = Te, the tree U = Tk(e).

The tree U is �nite-branching by the restriction that �(k) � k + 1. ThusR is a computably bounded �0

1 class. Now for any x 2 P we can de�ne z 2 Rwith the same degree as x as follows. First of all, de�ne a computable sequence; = t0; t1; : : : such that tj is the lexicographically least string in T of length j.Now given x, let

zx = ti0 � (x(0)) � ti1 � x(1)) � � � ;where for each n, in is the least such that

x(n) � jti0 � (x(0)) � � � � � tin j+ 1:

Then zx is computable in x by the de�nition, and x is computable in zx, sinceit is the subsequence of zx consisting of the entries zx(n) > 1.


Now let z be any element of R which is not of the form zx for any x. Thereare two cases.

(Case 1): Suppose that z(i) > 1 for in�nitely many i and let i0; i1; : : :enumerate fi : z(i) > 1g. De�ne x by x(n) = z(in). Then x is computable in zand that x 2 P , so that x is not computable and therefore z is not computable.

(Case 2): Suppose that z(i) > 1 for only �nitely many i and let m bethe largest such that z(m) > 1. De�ne y by y(n) = z(m + n). Then y iscomputable in z and that y 2 Q, so that y is not computable and therefore z isnot computable.

(b) The proof is just a modi�cation of the proof of (a). Let Q = [T ] inthis case be a strong �0

2 class of sets with no member computable in 00. Thenwe de�ne a tree U computable in 00 with �(k) � k + 1 as above so that R =[U ] is a c. b. strong �0

2 class with the desired properties and apply TheoremIII.7.5(a).

We will consider members of �01 classes in detail in Chapter IV.

Exercises

III.7.1. Show that there exist �01 classes P and Q which are computably home-

omorphic, but for which there can be no homeomorphism F of NN ontoitself which maps P one-to-one and onto Q.

III.7.2. Show that if P = [T ] where T is a �nite-branching, �01 tree, then P ishighly computable in 00.

III.7.3. Show that there is a computably continuous map on NN such that theimage is not even a closed set.

III.8 Thin and minimal classes

A �01 class P is said to be thin if, for every �0

1 subclass Q of P , there is a clopenset U such that Q = U \ P . An in�nite �0

1 class C is said to be minimal ifevery �0

1 subclass Q of C is either �nite or co�nite in C. Thus the notion of aminimal �0

1 class is the analog of the notion of a co-maximal �01 subset of !.

In particular, if C is a co-maximal set, then the class of subsets of C containingeither one or no elements is an example of a minimal �0

1 class which is not thin.(See the exercises.)

We have seen that any isolated element of a computably bounded �01 class

must be computable. For a thin �01 class, the converse also holds. (Exercise 2).

It follows that a perfect thin class has no computable members.

The �rst construction of a thin �01 class is due to Martin and Pour-El [144].

Theorem III.8.1. (Martin{Pour-El) There exists a perfect thin �01 class with

no computable member.

III.8. THIN AND MINIMAL CLASSES 87

Proof. Let Pe = [Te] be the e'th �01 class as in Theorem III.3.3 and let �e be the

e'th partial recursive function from ! into f0; 1g. We will construct a recursivetree S with corresponding �0

1 class P = [S] and a homeomorphism F fromf0; 1g! onto P . F will be constructed by means of a map f : f0; 1g<! ! S suchthat � � � () f(�) � f(�); then for x 2 f0; 1g!, F (x) = [nf(xdn).

To ensure that P is thin, we construct f to satisfy the following requirementfor each e.

Re: For each � 2 f0; 1ge+1, if f(�) 2 Te, then (8�)(� � � ! f(�) 2 Te).To see that this makes P thin, let U = [fI(f(�) : j�j = e+ 1 & f(�) 2 Teg

and observe that if Pe � P , then Pe = P \ U .The map f is de�ned in uniformly computable stages fs, beginning with f0

as the identity function.(Stage s + 1): Look for e < s and � 2 f0; 1ge+1 and � � � with j� � s + 1

such that fs(�) 2 Te, but fs(�) =2 Te. If such e, � and � exist, then we takethe least such e and the lexicographically least � and � for that e. Then we letfs+1(�) = fs(�) and in general, for any � we let

fs+1(�_�) = fs(�

_�) andfs+1(�) = fs(�) for � incomparable with �.If no such e, � and � exist, then we just let fs+1 = fs.It is easy to see by induction on j�j that for each �, fs(�) converges to a limit

f(�). Then we see by induction on e that the requirements Re are satis�ed.

Countable thin classes were studied in [25].The connection between thin and minimal classes is given by the following.

Theorem III.8.2. (C-D-J-S) The following are equivalent for any �01 class P .

(a) P is thin and D(P ) is a singleton.

(b) P is minimal and has a non-computable member.

Proof. (a) ! (b): Suppose that P is thin and that D(P ) = fAg. Then A isnon-computable by Exercise 2. Let Q be a �0

1 class such that Q � P . ThenQ = U \ P for some clopen U , so that P n Q is also a �0

1 class. If both Qand P nQ were in�nite, then both would contain limit points, contradicting theassumption that P has only one limit point.

(b) ! (a): Suppose that P is minimal and has a nonrecursive member A.Then A 2 D(P ) by Corollary III.2.16. For any B 6= A in P , let U be an intervalsuch that A 2 U and B =2 U . Then U \ P is in�nite, and therefore P n U mustbe �nite, which implies that B =2 D(P ). Therefore D(P ) = fAg. Now let Q beany �0

1 subclass of P . For any B 6= A in P , let U(B) be an interval such thatU(B) \ P = fBg. Since P is minimal, there are two cases.

Case 1: Q is �nite. Then Q = P \ [B2QU(B).Case 2: P nQ is �nite. Then Q = P \ [2! nSB2PnQ U(B)].We next construct a minimal, thin class.

Theorem III.8.3 ([25]). There exists a minimal, thin �01 class P ; furthermore,

P is decidable.


Proof. Let Pe = [Te] be the e'th �01 class as above. We will construct a set A,

a sequence �0 � �1 � : : : of strings with A = [i�i and a �01 class P such that

(1) D(P ) = fAg.(2) For any e and any extension B 2 P of �e, if A 2 [Te], then B 2 [Te].Properties (1) and (2) imply that P is minimal, by the following argument.Note �rst that, for all B 2 P , if B 6= A, then the set B is isolated in P by

property (1), so that there exists a clopen set U(B) such that P \U(B) = fBg.Suppose now that [Te] is a subset of P . Then there are two cases.

(Case 1) If A =2 Pe, then, since A is the only limit point of P and everyin�nite class has a limit point, it follows that Pe is �nite.

(Case 2) If A 2 Pe, then it follows from property (2) that every extensionof �e is also in Te. Now the set P n I(�e) of paths through T which are notextensions of �e is a closed set and has no limit point (since A is the only limitpoint of P ). Thus P n I(�e) is �nite and, since P n Pe � P n I(�e), P n [Te] isalso �nite.

It also follows from properties (1) and (2) that A is not computable. Tosee this, suppose by way of contradiction that A were computable. Then fAgwould be a �0

1 class, so that fAg = Pe for some e. Now by property (2), wehave P \ I(�e) � Pe, which makes A isolated in P , contradicting property (1).It now follows from Theorem III.8.2 that P is thin.

It remains to construct the set P . The construction will proceed in stages.At stage s we will have, for e � s, �nite sequences �se such that, for all e < s,�s_e 1 � �se+1. The construction will ensure the existence of the limits �e =lims �

se for each e. The point A will the union of f�e : e 2 !g. At the same

time we will be de�ning a sequence k(0) < k(1) < � � � so that s � k(s) andconstructing a computable tree T in stages T s. At stage s, we will have decidedwhether each �nite sequence of length k(s) is in T . This will ensure that T iscomputable. We will always put �_0 into T whenever � is in T . This will implythat xe = �_e 0! 2 P for all e; since A is non-computable and therefore in�nite,there are in�nitely many distinct xe, so that A 2 D(P ). This also implies thatP is decidable, that is, T has no dead ends. To obtain D(P ) = fAg, we dothe construction so that, whenever �s+1e = �se , then there are no new branchesadded below �se . Thus once we have reached a stage s such that �se = �e andcounted the number n of distinct branches of T s not passing through �se , thenwe know that all but n points of P will pass through �e. Now suppose thatsome path B is in D(P ) but is di�erent from A. Just let k be the least numbersuch that A(k � 1) 6= B(k � 1) and let e be least such that Adk � �e. Then noextension of Bdk passes through �e. It follows that the set of extensions of Bdkin P is �nite, so that B is isolated in P . This will take care of property (1).

In order to satisfy property (2), we want the construction to ensure thefollowing requirements for each e.

(Re): If �e 2 Te, then every extension of �e which is in T is also in Te.We begin the construction by setting k(0) = 1, putting (0) and (1) in T 0

and setting �00 = ;.Now suppose we have completed the construction as far as stage s. At stage

s + 1, we look for the least number e � s such that �se 2 T s \ Te but �se has

III.8. THIN AND MINIMAL CLASSES 89

some extension � 2 T s which is not in Te. If such an e exists, then we acton requirement Re at stage s + 1, as follows. Let � be the lexicographicallyleast extension of �se of length k(s) which is in T s n Te. Then let �s+1e = � .For i < e, let �s+1i = �si . For i � s � e + 1, let �s+1e+i = �_1i. Now letk(s+1) = k(s)+ s� e+1 and de�ne T s+1 to be the union of T s with the set ofthe following strings. First, for any � 2 T s of length k(s) and any i � s� e+1,the extension �_0i. Next, for any i � s� e+ 1, and any j � s� e+ 1� i, theextension �_(1i)_(0j).

If there is no such e, just let �s+1i = �si for all i � s and let �s+1s+1 = �ss_1.

Let k(s+1) = k(s)+1 and let T s+1 be the union of T s with the set of all strings�_0 where � 2 T s and the string �s+1s+1 .

Observe that in either case, we have extended all nodes in T s by at least onenode in T s+1, so that T will have no dead ends.

Claim 1: For every e, the sequence �se converges to some limit �e.Proof of Claim 1: This is by induction on e. Suppose therefore that Claim 1

is proved for all i < e and that we have reached a stage s such that �si = �i forall i < e. There are two cases. If � re = �se for all r > s, then the limit �e = �seand we are done. Otherwise, let r > s be least such that � re 6= �se . It followsfrom the construction that we must have � re =2 Te. After stage r, there is no waythat � te can be di�erent from � re . Thus the limit �e = � re .

Since �se � �se+1 for all s and e, it follows that �e � �e+1 for all e. Thus wecan de�ne the set A to have characteristic function [e�e.

Claim 2: For any e and any s, if �s+1e = �se , then there are no new branchesin T s+1 n T s which do not pass through �se .

Claim 2 is immediate from the construction.

It now follows that, for any e, all but �nitely many points of P pass through�e. It follows from the discussion preceding the construction that D(P ) = fAg.

Claim 3 If �e 2 Te, then every extension of �e which is in T is also in Te.

Proof of Claim 3: Suppose by way of contradiction that �e 2 Te but that �ehas some extension � 2 T such that � =2 Te. Consider a stage s > lh(�) suchthat �si = �i for all i � e and � 2 T s. Then at stage s + 1, we have � 2 T s sothe construction dictates that we act on requirement Re and make �s+1e = � ,contradicting the assumption that �se = �e.

This establishes property (1) and (2) above and thus completes the proof

We close the section with two important properties of thin classes.

Lemma III.8.4. Let Q � f0; 1gN be a thin �01 class. Then

1. Every �01 P � Q is thin;

2. For any computable � : f0; 1gN ! f0; 1gN, �[Q] is thin.Proof. (1) is trivial. For (2), suppose that R � �[Q]. Then ��1[R] is a �0

1

subclass of Q and hence ��1[R] = U \Q for some clopen U . Thus R = �[U ]\Qand �[U ] is clopen since f0; 1gN is compact.


Exercises

III.8.1. Show that for any maximal c.e. set A, the class containing the empty settogether with all singletons fmg where m =2 A, is an example of a minimal�01 class which is not thin.

III.8.2. Show that any computable element of a thin �01 class must be isolated.

III.9 Mathematical Logic

In this section, we set up the framework for the representation and applicationof �0

1 classes, using the area of logical theories. There is a very close connectionbetween �0

1 classes and logical theories and we will return to this topic in latersections as we develop the theory of �0

1 classes.Recall the arbitrary �rst-order e�ective language L described in Chapter I.

Let Sent(L) be the set of sentences of L. For any subset � of Sent(L), the setCon(�) of consequences of � is the closure of � under logical deduction andthe set Ref(�) of refutations of � is the set of negations of the consequencesof �. A subset � of Sent(L) is a �rst order logical theory if � is closed underlogical deduction. � is said to be a set of axioms for � if � = Con(�) and �is (computably) axiomatizable if � has a computable set of axioms; the mod-i�er (computably) will normally be omitted. It is not hard to see that � isaxiomatizable if and only if � is computably enumerable. A theory is said to bedecidable if it is computable. It follows from Post's Theorem that a completeaxiomatizable theory is decidable.

The usual idea for the application of �01 classes is that the set of solutions to

some computable problem should correspond to a �01 class. The problem here is

to �nd a complete consistent extension of a given computable or axiomatizabletheory. The classical result here is the Completeness Theorem of G�odel that anyconsistent theory has an extension to a complete consistent theory and followsas usual from Zorn's Lemma. The other fundamental result is the CompactnessTheorem, which states that if all �nite subsets of a theory � have a model, then� has a model; this follows from K�onig's Lemma.

Shoen�eld observed in [189] that in general, the family of complete, consis-tent extensions of an axiomatizable �rst order theory can be represented by a�01 class. Now the undecidability of arithmetic was discovered by Turing and

Church (independently) in 1936, following soon after G�odel's incompletenesstheorem. This result stated that there is no decidable complete consistent ex-tension of Peano Arithmetic and also showed, in our terminology, that there isa nonempty c. b. �0

1 class with no computable member. This led to the de�ni-tion of an essentially undecidable theory as a theory with no decidable completeconsistent extension. Now if � is any consistent complete extension of a theory�, then � separates the set T of consequences of � from the set R of refuta-tions of �. A theory is said to be separable if the consequences and refutationscan be separated by a computable set and is otherwise said to be inseparable.Rosser[180] observed also in 1936 that Peano arithmetic is an inseparable theory

III.9. MATHEMATICAL LOGIC 91

and that any inseparable theory is essentially undecidable. This also providedthe �rst example of computably inseparable c. e. sets. Ehrenfeucht showed in1961 [69] that there are separable theories which are essential undecidable. Hisconstruction, using theories of propositional calculus, also shows that every �0

1

class may be represented as the set of complete consistent extensions of a theory.A complete, consistent extension of Peano Arithmetic is of course just the thetheory of some (possibly non-standard) model of Peano arithmetic. The theoryof Peano arithmetic is of great interest in mathematical logic, due in part to theconnection with G�odel's Incompleteness Theorem, and has been developed inthe papers of Jockusch and Soare [99, 98], Knight [111], Marker [142] and manyothers.

Theorem III.9.1. (Shoen�eld [189] For any c. e. theory � of an e�ectivelanguage L, both the class of consistent extensions of � and the class of completeconsistent extensions of � can be represented as �0

1 classes. Furthermore, if �is a decidable theory, then these classes can be represented by computable treeswith no dead ends.

Proof. Let L be an e�ective �rst-order language and let S = Sent(L) have ane�ective enumeration as 0; 1; : : :. Then the sentence i may be identi�ed withthe number i, so that a theory � is represented by the set fi : i 2 �g, and a classof theories is represented by a class in f0; 1g!. Let � `s i be the computablerelation of which means that there is a proof of i from � of length s. Thenthe class P (�) of complete consistent extensions of � may be represented bythe set of in�nite paths through the computable tree T de�ned so that for any� = (�(0); : : : ; �(n� 1)), � is in T if and only if the following conditions hold.

(1) For any i < n, if � `n i, then �(i) = 1.

(2) For any i; j < n, if � `n i ! j and �(i) = 1, then �(j) = 1.

(3) For any i; j; k < n, if k = ( i & j), �(i) = 1 and �(j) = 1, then �(k) = 1.

(4) For any i; j < n, if �(i) = 1 and j = : i, then �(j) = 0.

(5) For any i; j < n, if j = : i, then either �(i) = 1 or �(j) = 1.

Let x be an in�nite path through T and let � = f i : x(i) = 1g. Condition(1) ensures that � � �, while conditions (1), (2), and (3) ensure that � is atheory. Condition (4) ensures that � is consistent and condition (5) ensuresthat � is complete. To represent the class of consistent extensions of �, simplyomit the �nal clause (5).

If � is decidable, then in each case we can modify the clauses given above asfollows to get a tree S with no dead ends which has the same class of in�nitepaths. First, combine the �rst three clauses into the statement:

(10) : For any k < n, if � ` ^f i : i < n & �(i) = 1g ! k, then �(k) = 1.

Next, replace clause (4) with


(40) It is not the case that � ` [^f i : i < n & �(i) = 1g ! ( 0&: 0)].

It follows that for any � 2 S, � [ f i : i < j�j & �(i) = 1g [ f: i :i < n & �(i) = 0g is consistent and therefore has an extension to a completeconsistent theory �(�) which will be represented by an extension of �. Thus Shas no dead ends.

We can now apply Theorem III.2.15 to logical theories.

Theorem III.9.2. For any consistent, axiomatizable �rst-order theory �:

(i) � has a complete consistent extension which is computable in 00.

(ii) If � is decidable, then � has a complete, consistent, decidable extension.

Next we turn to the other direction of our correspondence, that is, represent-ing an arbitrary �0

1 class by the set of complete consistent extensions of someaxiomatizable theory.

Theorem III.9.3. Any c. b. �01 class P may be represented by the set of

complete, consistent extensions of an axiomatizable theory � in propositionallogic. Furthermore, if P is a decidable �0

1 class, then � may be taken to be adecidable theory.

Proof. We give the proof due to Ehrenfeucht [69]. Let the language L consist ofa countable sequence A0; A1; : : : of propositional variables. For any x 2 f0; 1gN,we can de�ne a complete consistent theory �(x) for L to be Con(fCi : i 2 !g),where Ci = Ai if x(i) = 1 and Ci = :Ai if x(i) = 0. It is clear that everycomplete consistent theory of L is one of these. Thus for any �0

1 class P �f0; 1gN, we want a theory � such that �(P ) = f�(x) : x 2 Pg is the set ofcomplete, consistent extensions of �.

For each �nite sequence � = (�(0); : : : ; �(n�1)), let P� = C0^C1^� � �^Cn�1,where Ci = Ai if �(i) = 1 and Ci = :Ai if �(i) = 0. Let the binary tree T begiven such that P = [T ] and de�ne the theory �(T ) to consist of all P� ! Ansuch that � 2 T and �_0 =2 T and all P� ! :An such that � 2 T and�_1 =2 T , where j�j = n. We claim that �(P ) is in fact equal to the setof complete consistent extensions of �(T ). Suppose �rst that x 2 P and letCon(fCi : i 2 !g) = �(x). Now any 2 �(T ) is of the form P� ! �Aifor some � 2 T ; say that j�j = n. There are several cases. If � 6= xdn, then�(x) ` :P�, so that we always have �(x) ` P� ! �An. Thus we may supposethat � = xdn. If �_0 =2 T , then of course x(n) = 1, so that Cn = An 2 �(x)and therefore �(x) ` P� ! An. Similarly, if �

_1 =2 T , then �(x) ` P� ! :An.Thus �(x) is a complete consistent extension of �(T ). On the other hand, let� be a complete consistent extension of �(T ). Then, for each i, we have either� ` Ai or � ` :Ai; let Ci = Ai if Ai 2 � and Ci = :Ai otherwise. De�nex 2 f0; 1g! so that x(i) = 1 if and only if � ` Ai. Then clearly � = �(x). Itremains to be shown that x 2 P . Now if x =2 P , then there is some n such that� = xdn+1 =2 T and xdn 2 T . Then P� = C0 ^ � � � ^Cn�1, so that � ` P�, and

III.9. MATHEMATICAL LOGIC 93

P� ! :Ci 2 �(T ), so that � is not consistent with �(T ). This contradictionproves that � = �(x).

Now suppose that P is a decidable class, so that the tree T has no deadends. Let a sentence = (A0; : : : ; An�1) of the language L be given. We claimthat �(T ) ` if and only if

VfP� ` : � 2 T & j�j = ng; that is, if and onlyif P� ` for all � 2 T with j�j = n. This claim clearly implies that �(T ) isdecidable.

We argue by the contrapositive. Suppose �rst that �(T ) 6` . Then thereis some x 2 [T ] such that �(x) ` : . Since only depends on A0; : : : ; An�1,it follows that P� ` : , where � = xdn 2 T . Thus P� ` is clearly false,making it also false that

VfP� ` : � 2 T & j�j = ng. Suppose next thatVfP� ` : � 2 T & j�j = ng is false. Then P� ` is false for some �xed � 2 T ,which means that P� ` : (since depends only on A0; : : : ; An�1). Since Thas no dead ends, there is some x 2 P such that � � x and therefore �(x) ` : and therefore �(T ) 6` .

This proof can be adapted to �rst order logic; see Exercise 1 below.This representation theorem has the following corollary.

Theorem III.9.4. There is a consistent axiomatizable �rst-order theory Gammawhich has no computable consistent complete extension.

The perfect thin class constructed by Martin and Pour-El (see TheoremIII.8.1) was designed to produce a certain type of axiomatizable theory. Atheory � is said to be Martin{Pour-El if every axiomatizable extension of � isgenerated by a single proposition. It follows from the proof of the next theoremthat an axiomatizable theory � is a Martin{Pour-El theory if and only if theclass of complete consistent extensions of � is a thin �0

1 class.

Theorem III.9.5. There exists an axiomatizable, essentially undecidable theoryT such that each axiomatizable extension of T is a �nite extension of T .

Proof. Let P be a perfect thin class with no computable members and let theaxiomatizable theory T be given by Theorem III.9.3 such that P representsthe family of complete consistent extensions of T . Now suppose that � is anaxiomatizable extension of T . Then the family of complete consistent extensionsof � is represented by a �0

1 subclass Q of P . Since P is thin, there is a clopenset U such that Q = U \P . Let U = I(�1)[I(�2)[� � �[I(�n) for some distinct�nite sequences �i all having the same length k and let �i = P�i as in the proofof Theorem III.9.3. Then the complete consistent extensions of � are exactlythose complete consistent extensions of � which satisfy P1 _ � � � _ Pn.

Exercises

III.9.1. Show that any c. b. �01 class may be represented as the set of complete

consistent extensions of a �rst order logical theory in the language of onebinary relation R Jockusch and Soare in ([99], p. 54). Hint: the underlying


axioms assert that R is an equivalence relation and that, for any n, thereare either one or two equivalence classes consisting of exactly n members.The propositional statement An in the proof above is replaced by thestatement that there is exactly one equivalence class with n elements.

III.9.2. Let the propositional language L have variables A0; A1; : : : . Variables andtheir negations :Ai are said to be literals. Show that a consistent theory� for L is c. e. if and only if the set C(�) of conjunctions of literals,consistent with �, is co-c. e. and that � is decidable if and only if C(�)is computable. Show that this is not true for the set of literals consistentwith �.

Chapter IV

Members of �01Classes

In this chapter, we study the complexity of members of �01 classes. We present

some \basis theorems" and \anti-basis theorems". The class � � NN is said tobe a basis for a family � of subclasses of NN if every nonempty class from � hasa member from �. For example, the class �0

0 of computable reals is a basis forthe family of open subclasses of NN. This is an example of a \basis theorem".We have already given the simple positive result V.2.3 that the class P = [T ] ofin�nite paths through the tree T contains a member computable from Ext(T ),the set of nodes of T which have an in�nite extension in P . Recall in particular,that if T is computably bounded, then P has a member computable in 00 andif P is decidable, then P has a computable element. We will give several morebasis results, including the Low Basis Theorem of Jockusch and Soare [99] isgiven.

On the other hand, the class of computable reals is not a basis for the familyof closed subclasses of NN since every singleton is a closed class. This is anexample of an \anti-basis theorem". One result given is that the set of Booleancombinations of c. e. sets is not a basis for the c. b. �0

1 classes. Any c. b.�01 class with no computable members is perfect and has a set of continuum

many mutually Turing incomparable elements [99]. There is a c. b. �01 class of

positive measure which has no computable element.

IV.1 Basis theorems

One of the most cited results in the theory of �01 classes is the Low Basis

Theorem of Jockusch and Soare [99]. We will introduce the method of forcingwith �0

1 classes in connection with this theorem. First we consider the notiona generic real.

De�nition IV.1.1. An element x 2 NN is said to be 1-generic if, for every �01

class P , there exists n such that either I(x � n) � P or I(x � n) \ P = ;.The existence of a generic real is obtained by forcing. Let P = [T ]. The idea

95

96 CHAPTER IV. MEMBERS OF �01 CLASSES

here is that the �nite sequence � = xdn \forces" x =2 P if � =2 Ext(T ), that is,� x =2 P if no extension of � is in P , and similarly � x 2 P if I(�) � P ,that is, if every extension of � is in P . Then we write x 2 P ( x =2 P ) ifthere is some � such that � x 2 P (� x =2 P ). With this notation, x is1-generic if, for every �0

1 class P , either x 2 P or x =2 P , or equivalentlyx 2 P implies x 2 P . (See Section III.6 of Hinman [87] for a presentation ofarithmetical forcing.)

A subset D of f0; 1g� is said to be dense if, for any �, there exists a � � �such that � 2 D. Now let D = fDi : i 2 Ig be a family of dense sets. Theelement x of f0; 1gN is said to be D-generic if, for each i, x � n 2 Di for some n.

The standard forcing theorem shows that any countable family of dense setspossesses a generic set. We observe that this is an e�ective version of the BaireCategory Theorem.

Lemma IV.1.2. If D = fDi : i < !g is a sequence of subsets of N� uniformlycomputable in 00, there exists a D-generic x �T 00.Proof. Let m be the least such that �m 2 D0 and let �0 = �m. Then for eachn, �nd the least m such that �m is a proper extension of �n and let �n+1 = �m.Then x = [n�n will be the desired generic real. This construction is computableusing an oracle for the sequence D.

Recall that by Theorem III.3.3, there is a uniformly primitive recursive enu-meration of trees Te such that Pe = [Te] is the eth �0

1 class. Then the standardfamily of dense sets is now Di = f� : � =2 Ti _ (8� � �)� 2 Tig. Observeeach Di is dense and that the sequence of sets is uniformly �0

1. The elementx 2 f0; 1gN is 1-generic if it is generic for this sequence of dense sets. Then theremarks above imply the existence of a 1-generic real. The crucial property ofa 1-generic real is given by the following well-known fact.

Theorem IV.1.3. For any 1-generic x 2 f0; 1gN, if x �T 00, then x0 �T x�00.Proof. Let x be 1-generic. For each e, let the �0

1 class Pe = fy : �ye(e) "g, sothat Pe = [Ue], where � 2 Ue () ��e (e) ". Thus e 2 x0 () x =2 Pe. Ife 2 x0, then of course there is some n such that xdn =2 Ue. Since x is 1-generic,if e =2 x0, then there is some n such that (8� � xdn)� 2 Ue. Let f(e) be theleast n such that either xdn =2 Ue or (8� � xdn)� 2 Ue. Then f is computablein x� 00. But then we have e 2 x0 if and only if xdf(n) =2 Ue, so that x0 is alsocomputable in x� 00.

It follows that if a 1-generic real x is computable in 00, then x0 = 00, thatis, x is low. For the low basis theorem of Jockusch and Soare, a modi�cation ofthis argument is used.

Theorem IV.1.4 (Low Basis Theorem). (a) Every nonempty c. b. �01 class

P contains a member of low degree.

(b) There is a low degree a such that every nonempty r. b. �01 class contains

a member of degree � a.

IV.1. BASIS THEOREMS 97

Proof. (a) We may assume that P � f0; 1g!. Let P = [T ] and, as above,let � 2 Ue () ��e (e) ". We will de�ne, computably in 00, a sequenceT = S0 � S1 � S2 � � � � of in�nite subtrees of T and show that any member xof \e[Se] has low degree. There are two cases in the de�nition of Se+1.

1. If Se \ Ue is �nite, then Se+1 = Se.

2. If Se \ Ue is in�nite, then Se+1 = Se \ Ue.

Observe that this construction is computable in 00, in that there is a functionf �T 00 such that Se = Uf(e) for each e. This is because the determinationof whether Se \ Ue is �nite can be made using a 00 oracle. Since each Se isin�nite by the construction, it follows that each [Se] is nonempty, so that \e[Se]is nonempty. Now suppose that x 2 \e[Se]. Then, for any e, we have e 2 x0 ifand only if x =2 [Ue], and it follows from the construction that x =2 [Ue] if andonly if Se \ Ue is �nite. It follows by the observation above that x0 �T 00.

(b) It su�ces to prove the result for classes in f0; 1gN. Let Pe = [Te] be ane�ective enumeration of the �0

1 in f0; 1gN and let pe be the e'th prime number.Let the tree T be the amalgamation of the nonempty �0

1 classes, in the followingsense. Let � 2 T if, for each e such that Pe is nonempty and each k such thatpke < j�j, (�(pe); �(p2e); : : : ; �(pke)) 2 Te. Since we can test computably in 00

whether Pe is nonempty, the tree T is computable in 00. Thus the constructionabove can be carried out to produce a member x of P = [T ] of low degree. Thenany nonempty class Pe has a member (x(pe); x(p

2e); : : : ) computable in x.

The same technique can be used to prove other basis results. For example, ageneralization shows that if the c. b. class P contains no computable member,then for any degree b, P has a member A of degree a such that a� 00 = a0 = b0.It follows (for b = 00) that any �0

1 class P has a member A of degree a suchthat a� 00 = a0 = 000. Now a degree a is said to be high if a � 00 and a0 = 000.It will be shown later that not every r. b. �0

1 class contains a member of highdegree.

The following result is from Jockusch and Soare [98].

Theorem IV.1.5 (Jockusch and Soare). Every nonempty c. b. �01 class P

contains a member of hyperimmune-free degree, that is, contains an almost com-putable member.

Proof. We sketch the proof indicated in Soare [198] (p. 109). Let P = [T ], whereT is an in�nite computable binary tree. Recall that A is hyperimmune-free ifevery function f computable in A is majorized by some computable function.Thus we want to �nd A 2 P such that �Ae whenever total, is majorized bya computable function. We de�ne the decreasing sequence Se of computablesubtrees of T beginning with S0 = T . Then for each e and i, let U ie = f� 2 Se :��e (i) "g. There are again two cases in the de�nition of Se+1.

1. If U ie is �nite for every i, let Se+1 = Se.


2. If U ie is in�nite for some i, choose such an i and let Se+1 = Se \ U ie.

Suppose now that A 2 \eSe and that f = �Ae is total. If the second caseapplied in the de�nition of Se+1, then �

Ae (i) ", so that �Ae is not total. If the

�rst case applied, then �Ae is de�ned for all A 2 [Se+1]. Since [Se] � P , [Se]is computably bounded and it follows from Theorem III.4.8 that �e[Se] is alsobounded by some computable function h, which thus majorizes �Ae .

We will show in the next section that a nonempty �01 class P � f0; 1gN need

not contain any sets which are c. e. or co-c. e.. However, it always containselements which corresponds to c. e. and co-c. e. Dedekind cuts.

Theorem IV.1.6. Any nonempty �01 class P � f0; 1gN contains elements x

and y such that the Dedekind cut L(rx) is c. e. and the Dedekind cut L(ry) isco-c. e..

Proof. Let P = [T ] where T is a computable tree. Let x be the \leftmost"element of P under the lexicographic order. For each n, let �n be the leftmostnode in T \ f0; 1gn and let qn = r� =

Pni=0 �(i)2

�i�1. It is clear that fqngn<!is an increasing sequence and hence has limit r such that L(r) is c. e. set byProposition 2.II.4.3. We claim that x = limn�n converges to a path in P andthat r = rx. For each n, �n(0) � �n+1(0), since otherwise �n+1dn <lex �n.Thus the sequence �n(0) converges to some x(0). Once �n(0); : : : ; �n(k � 1)have all converged, then �n(k) becomes increasing and thus also converges tosome limit which we call x(k). It remains to show that x 2 P . Fix n and chooses > n such that �m(i) = x(i) for all i < n and all m � s. Then xdn � �s andhence xdn 2 T .

A similar argument shows that the \rightmost" element Y of P correspondsto a real with a �0

1 Dedekind cut.

Some further basis results are given below in Chapter V.

Exercises

IV.1.1. The Baire Category Theorem for NN states that the countable intersectionof a sequence of dense open sets is nonempty. Use forcing to prove thistheorem.

IV.1.2. Show that x is 1-generic if and only if it belongs to every �01 co-meagerset (equivalently, every non-meager �01 set).

IV.1.3. Show that if the c. b. class P contains no computable member, then forany degree b, P has a member A of degree a such that a� 00 = a0 = b0.Then show that any �0

1 class P has a member of degree a such thata� 00 = a0 = 000.

IV.1.4. Show that if x is an isolated member of P , then x is computable. (De�nea decidable subclass of P and use Theorem III.2.15.)

IV.2. SPECIAL �01 CLASSES 99

IV.1.5. Show that any nonempty �01 class which is almost recursively bounded

must contain an element computable in 00.

IV.1.6. Show that for any x and y with x <lex y such that L(rX) is �01 and L(ry)

is �01, there is a �0

1 class with leftmost element x and rightmost elementy.

IV.2 Special �01 classes

A c. b. �01 class is said to be special if it has no computable members.

We have seen in Section 3.III.5(Exercise (3)) that the diagonally non-computablesets form a �0

1 class with no computable element. We give an improvement ofthis result due to Jockusch [91]. Recall that a set A is immune if it has no in�-nite recursive subset; A is said to be bi-immune if both A and NnA are immune.It is elementary that any in�nite c. e. set has an in�nite recursive subset, andit follows that the di�erence of two c. e. sets cannot be bi-immune and then byinduction that no Boolean combination of c. e. sets can be bi-immune.

Theorem IV.2.1 (Jockusch). There is a nonempty �01 class of sets containing

only bi-immune sets.

Proof. Let We be the e'th c.e. set and let Dn be the n'th �nite set. Let be apartial recursive function such that, whenever jWej � e+3, then (e) is de�nedand D (e) � We and jD (e)j = e + 3. De�ne the �0

1 class P = \ePe, whereA 2 Pe if and only if, if (e) is de�ned, thenA\D (e) 6= ; and (NnA)\D (e) 6= ;.Any element A of P is clearly bi-immune. To see that P is nonempty, note thatfor each e, f0; 1gNnPe has measure� 2�e�2. (ForA =2 Pe, either all e+3 elementsof D (e) are in A or all e+ 3 elements are not in A, which allows only 2 of the

2e+3 possibilities.) It follows that f0; 1gN n P has measure �Pe 2�e�2 = 1

2 , sothat P 6= 0.

This immediately implies the following.

Theorem IV.2.2. (Jockusch) There is a nonempty c. b. �01 class with no

member a Boolean combination of r.e. sets.

A set A is said to be e�ectively immune if there is a computable function gsuch that for any e, ifWe � A, then jWej � g(e). For the �0

1 class P constructedin the proof of Theorem IV.2.1, it is clear that any set A 2 P is e�ectively bi-immune via the function g(e) = e+3. This yields the following corollary, whichwe note is essentially exercise 4.2 on page 87 of [198].

Theorem IV.2.3. There is a nonempty c. b. �01 class such that if a is the

degree of a member of P and b is a c.e. degree and a � b then b = 00.

Proof. Let P be the �01 class de�ned in the proof of Theorem IV.2.1. Then

every member of P is e�ectively immune. Now suppose that P had a memberC of c. e. degree. We claim that C must have degree 00. By the Modulus


Lemma (Soare [198], C = lims Cs with a modulus function m(i) computablein C such that s � m(i) implies that i 2 C () i 2 Cs. We may assumewithout loss of generality that each Cs is in�nite and let c0;s; c1;s; : : : enumeratein increasing order the elements of Cs. Let C = fc0; c1; : : : g enumerate C inincreasing order, so that for each n, cn = lims cn;s. Now let the complete c. e.set K have enumeration Ks and let the partial recursive function � be de�nedso that �(i) = s if and only if s is the least such that x 2 Ks.

By the recursion theorem, de�ne the computable function h so that Wh(i) =;, if i =2 K and otherwise Wh(i) = fc0;�(i); c1;�(i); : : : ; cg(h(i));�(i)g.Let r(i) be the least s such that, for all j � g(h(i)) and all t � s, cj;t = cj .Then the function r can be computed from C using the modulus function m. Ifr(i) � �(i), then Wh(i) � C, so that C has g(h(i)) + 1 elements, contradictingthe hypothesis on g. It follows that �(i) < r(i) for all i, so that K is computablefrom C, as desired.

It follows that the only possible c. e. degree of a member of P is 00. Thereis a more general version of this result, Theorem 5 of [98].

Theorem IV.2.4. For any c. e. degree c, there is a nonempty c. b. �01 class P

such that the c. e. degrees of member of P are precisely the c. e. degrees abovec.

Proof. For c = 0, this is trivial. For c 6= 0, let A be the simple but not hyper-simple c. e. set of degree c from Theorem II.7.8. Let Df(n) be a disjoint strongarray such that Df(n) n A 6= ; for all n and assume without loss of generalitythat Card(Df(n) nA) � 2. Now de�ne the �0

1 class P by

D 2 P () D \A = ; & (8n)(D \Df(n) 6= ;):

For any D 2 P of c. e. degree, it follows from Theorem II.7.8 that A �T D.Now let E be a set of any degree d � c and �nd D 2 P such that D �T E

andCard(Df(n) nD) = 1 () n 2 E:

Clearly E �T D. Since A �T D, we can compute from D a sequence of pairshan; bni with an; bn both in Df(n) n A. Let E = fe0 < e1 < : : : . Then we cancode D into E by letting Df(ei) n D = faeig if i 2 D and Df(ei) n D = fbeigotherwise. Thus D will be a member of P with degree d.

We next show that every special c. b. �01 class satis�es a weakened form of

Theorem IV.2.3.

Theorem IV.2.5. (Jockusch-Soare [98]) For any special �01 class, there exists

a non-zero c. e. degree a such that P has no members of degree � a.

Proof. We give the proof from [98]. Let P = [T ] where T is a computable tree.We will construct a simple (and hence noncomputable) c. e. set A in stage As,such that �Ae =2 P for any e.

Two binary computable functions will be used in the construction.


p(e; s) = maxfn � s : �e(Asds)dn) 2 Tg;q(e; s) = (least j) : j�e(Asdj)j � p(e; s)g.Initially, A0 = ;. At stage s + 1, let es+1 be the least e � s such that

We;s \ As = ; and We;s contains u � maxfq(e0; s) : e0 � eg [ f2eg. LetAs+1 = As [ fug, where u is the least such number for e = es+1. (If no such eexists, let As+1 = As.)

Claim 1 No element of P is computable in A.

Proof of Claim 1: Suppose by way of contradiction that y = �Ae 2 P (and isthus a total function). It follows that limsp(e; s) = 1. For the contradiction,we will show that y is computable. Let t be a stage such that for all e0 < e,if We \ A 6= ;, then We;t \ At 6= ;. Then for all s � t, either es is unde�nedor es � e. To compute y(n), �nd sn � t such that n < p(e; sn), so that�e(Adsn; n) is de�ned and in T . We claim that y(n) = �e(Adsn; n). To showthis, it su�ces to prove that no number u < q(e; sn) enters A after stage sn.But if u 2 As+1 �As and s+ 1 > sn, then es+1 � e and so u � q(e; sn) by theconstruction.

Claim 2 For �xed e, q(e; s) is bounded over all s.

Proof of Claim 2:Let n be the least such that �Ae dn =2 T and let t be as in the proof of Claim

1. It follows that p(e; s) � n for all s � t. Choose j such that �Ae dn � �e(Adj).It follows that q(e; s) � j for all su�ciently large s.

Claim 3 A is simple.

Proof of Claim 3: For each c. e. set We, A contains at most one memberu � 2e fromWe, so that A contains at most e elements � 2e and the complementof A is in�nite. Fix e and let j = maxfq(e0; s) : e0 � e; s 2 Ng. Let t be givenfrom Claim 1. Then if s � t and We;s contains a member � maxfj; 2eg, thenWe;s+1 \A 6= ; by the construction. Hence if We is in�nite, then We \A 6= ;.

This completes the proof of the theorem.

As we saw in Exercise 4, any isolated member of a c. b. �01 class is com-

putable. Hence, if P is a special c. b. �01 class, then it is perfect and hence has

cardinality 2@0 . It follows that there are 2@0 di�erent degrees of members of P .Several results deal with the comparability of these degrees.

It is easy to see that if F is a computable function from f0; 1gN to f0; 1gN,r 2 f0; 1gN, and P is a �0

1 class such that F (x) = r for all x 2 P , then r iscomputable. (See the exercises.) The next lemma, from Jockusch-Soare [99],improves this observation.

Lemma IV.2.6. Let P be a nonempty c. b. �01 class P and let � : N� NN !

N be a partial computable functional. Suppose that for any n 2 N and anyx1; x2 2 P , �(n; x1) = �(n; x2) whenever they are both de�ned. Then there isa nonempty �0

1 class Q � P such that, for all y 2 Q, if �n�(n; y) is total thenit is computable.


Proof. Let P = [T ] where T is a computable tree and suppose that P and �satisfy the hypothesis. There are two possibilities. Suppose �rst that for somen, �e(n; �) is unde�ned for in�nitely many � 2 T . Then we may let S be thesubtree f� 2 T : �e(n; �) "g and let Q = [S], since y 2 Q then implies that�e(n; y) ". Thus we may suppose that for any given n, �(n; �) # for all but�nitely many � 2 T . Thus �(n; y) # for all y 2 P and by the hypothesis thereis a unique kn such that �(n; y) = kn for all y 2 P . Let r(n) = kn for eachn. It now follows from the remark above that r is computable. Thus in fact�(n; y) = r(n) for all y 2 P and all n.

When F is the identity function, we obtain the following corollary.

Lemma IV.2.7. If P is a nonempty c. b. �01 class with no computable elements,

then any � 2 Ext(T ) has incompatible extensions �1; �2 2 Ext(T ).Theorem IV.2.8. (Jockusch-Soare [99]) For any c. b. �0

1 class P with nocomputable members and any countable set fai : i < !g of noncomputabledegrees, P has continuum many mutually incomparable members x such thatthe degree of x is incomparable with each ai.

Proof. We may assume without loss of generality that P � f0; 1gN and letP = [T ] for some computable tree T � f0; 1g�. For simplicity, we construct theelements of P to be incomparable to a single noncomputable degree a and let zhave degree a; for the general argument simply work on a0; : : : ;an at level n.

We will de�ne a function : f0; 1g� ! Ext(T ) such that �1 � �2 implies (�1) � (�2) and in�nite trees S� � T \ I( (�) such that, for all n

(1) If j�j = n+ 1 and x 2 [S�], then �xn 6= z.

(2) If j� = j� j = n+ 1 and � 6= � , then for any x 2 S� and y 2 S� , �xn 6= y.

Initially, we let F (;) = ;. Now suppose that we have de�ned (�) and S� forj�j = n and let � = �n. By Lemma IV.2.6, there are two possible cases. Firstsuppose that there is a nonempty �0

1 subclass Q of S� such that �yn is eithercomputable or not total for all y 2 Q. Then apply Lemma IV.2.7 to obtainincompatible extensions �0 and �1 of F (�) from Ext(Q), by Lemma IV.2.7 andlet F (�_i) = �i and S�_i = I(�i) \Q for i = 0; 1.

In the second case, by Lemma IV.2.6, F (�) has extensions �0; �1 2 Ext(S�)such that �n(m; �1) 6= �n(m; �2) for some m. Without loss of generality,�n(m; �1) 6= z(m) and thus we �rst extend F (�) to �1 to satisfy condition(1) and then take extensions �0 and �1 as above. In this case, F (�

_i) will be anextension of the provisional value �i for i = 0; 1 after we take care of condition(2). The provisional values for S�_i are S� \ I(�i).

Now we will indicate how to satisfy condition (2). Let �1; �2 have lengthn + 1 with provisional values �1 and �2 for F (�1) and F (�2) and provisionalvalues Q1 and Q2 for S�1 and S�2 with Qi � P \ I(�i). Our goal is to ensurethat �x 6= y for any x 2 S�1 and y 2 S�2 . If �1 came under case 1 above, thenthis is already satis�ed. Otherwise, �ndm and extensions �1; �2 2 S�1 such that


�(m; �1) 6= �(m; �2) as above and �nd an extension � of �2 such that, withoutloss of generality, �(m) 6= �(m; �1). Then the new provisional value of F (�1) is�1, the new provisional value of F (�2) is � ; the new provisional value of S�1 isQ1 \ I(�1) and the new provisional value of S�2 is Q2 \ I(�). This satis�es oneof the �nitely many subconditions of (2).

We repeat this procedure for each pair �1; �2 2 f0; 1gn+1 to obtain �nalprovisional values which will satisfy all of the necessary conditions.

This leads to the following. Two noncomputable reals x; y are said to be aminimal pair if any real computable both in x and in y is itself computable.

Corollary IV.2.9. (Jockusch-Soare [99] Let P � f0; 1gN be a �01 class with no

computable members. Then P contains a minimal pair.

Proof. Let x be any member of P . By Theorem IV.2.8 P contains an element ywhich is incomparable with each of the (countably many) noncomputable realswhich are computable in x.

Theorem IV.2.10. (Jockusch-Soare [98]) There is an in�nite c. b. �01 class

P such that any two di�erent members of P are Turing incomparable.

Proof. We will use a priority argument to de�ne a uniformly computable se-quence fs : f0; 1g� ! f0; 1g� of one-to-one, computable tree homomorphismsand corresponding perfect �0

1 classes Ps = [Ts], where

Ts = f� : (9�)[fs(�) � � ]g:Then the desired (perfect) �0

1 class P is de�ned to be P = \sPs. In particular,each fs satis�es the following for all � 2 f0; 1g�.(1) fs(�

_0) and fs(�_1) are incompatible extensions of fs(�) for all s;

(2) range(fs+1) � range(fs);(3) lims fs(�) = f(�) exists.

It follows that f induces a homeomorphism F from f0; 1gN onto P , de�nedby F (x) = [nf(xdn). It follows that that P is a perfect set, hence uncountableand therefore certainly nonempty.

The construction of P will ensure that for each y 2 P and each partialfunction �e, �e(y) 6= y implies that �e(y) =2 P . To accomplish this, we willensure that for each e and each x 2 f0; 1gN, if z = �e(F (x)) converges andz 6= F (x), then �e(f(xde)) is incompatible with F (x). That is, for each e andeach � 2 f0; 1ge+1, we have the following requirement.

Re;� : (8y)[y 2 I(f(�)) & �e(y) total & �e(y) 6= y ! �e(y) =2 P ]:The priority order on the requirements is lexicographic, �rst on e and then

on �.


Requirement Re;� is said to be satis�ed at stage s if �e(f(�)) =2 Ts. If thisrequirement is satis�ed at stage s and y 2 I(f(�)), then of course requirementRe;� is actually satis�ed, since �e(f(�)) � �e(y).

Initially we set f0(�) = � for all � 2 f0; 1g�.Stage s + 1. Requirement Re;� needs attention at stage s + 1 if it is not

satis�ed at stage s and there exists �0 � � such that �e(fs(�0)) is incompatible

with fs(�0) and extends fs(�

_i) for some � 2 f0; 1ge+1 and some i 2 f0; 1g.If no requirement needs attention at stage s+1, then fs+1 = f�. Otherwise,

let Re;� be the requirement of highest priority which needs attention and let �0,

� and i be given as above. De�ne fs+1 as follows, for any � 2 f0; 1g�.(4) fs+1(�

_�) = fs(�0_�);

(5) If � 6= �, then fs+1(�_�) = fs(�

_(1� i)_�);(6) If � does not extend either � or �, then fs+1(�) = fs(�).

Now suppose that fs+1(�) � y. Then �e(fs+1(�)) = �e(fs(�0)) � fs(�_i) =�

�e(y) and is incompatible with fs+1(�). There are two cases. If � 6= �, thenfs+1(�) = fs(�)

_(1� i) so that fs(�_i) =2 Ts+1. If � = �, then fs(�_i) extends

fs(�) but is incompatible with fs+1(�) and therefore again is not in Ts+1.The functions fs clearly satisfy (1) and (2). To verify (3), we show simul-

taneously that, for each e and �, action is taken on each requirement Re;� atmost �nitely often. Fix e and � and choose s by induction so that for all t � sand all requirements Rd;� of higher priority

(7) fs(�) = ft(�) and

(8) no action is ever taken on requirement Rd;� after stage s.

Then fs(�) = ft(�) for all � � � so that fs(�) � ft(�). Thus once re-quirement Re;� is satis�ed at stage t, it will remain satis�ed. Hence it requiresattention at most one more time after stage s and therefore ft(�) converges.

Finally, suppose that y 2 P , �e(y) is total and �e(y) 6= y. Let y = F (x),� = xdn and � = f(�) � y. Since �e(y) 6= y, there must exist �0 � � suchthat �e(f(�

0)) is incompatible with y. It follows that in fact �e(�) =2 T , sinceotherwise Re;� would require attention at arbitrarily large stages s.

There is a version of this theorem for classes of separating sets.

Theorem IV.2.11. (Jockusch-Soare [98]) There exist disjoint c. e. sets A andB such that A [ B is co�nite, but any two members of Sep(A;B) either have�nite di�erence or are Turing incomparable.

Proof. The disjoint c. e. sets A and B will be de�ned by a priority argument asthe e�ective, increasing unions A = [sAs and B = [Bs. The construction willensure that for each partial function �e and any pair of C;D of separating setsfor A;B, the following requirement is satisifed.

Re: If C 6= D (modulo �nite di�erence), then D 6= �e(C).


As in the construction of a maximal set ([198], p. 188), an increasing sequencefms

i : i < !g will be de�ned at stage s so that for each i, m0i � m1

i � : : :and mi = limsm

si exists. At any stage s, As [ Bs = ! n fms

i : i 2 !g, so thatfmi : i < !g is the complement of A [B.

For �xed e, letDe = fmi : i < eg:

Let C and D be two subsets of De. The subrequirement RC;De associated with

C and D asserts that for any separating sets C1 and D1 for A and B such thatC1 \De = C and D1 \De = D, D1 6= �e(C1).

This requirement is said to be satis�ed at stage s if there exists a string �such that, for any n,

(1) if n 2 De, then n 2 C () �(n) = 1;

(2) if n 2 Dom(�), then n 2 As ! �(n) = 1 and n 2 Bs ! �(n) = 0.

(3) Dom(�) � De [As [Bs(4) �se(m;�) is de�ned and 6= D(m) for some m.

That is, let (1) through (4) hold and suppose that C1 and D1 are separatingsets for A and B such that C1 \De = C and D1 \De = D. Then � � C1 sothat �(m;C) 6= D(m).

RC;De requires attention at stage s if it is not satis�ed at stage s and if thereexists a string � satisfying (1) and (2) and such that

(5) �se(mse; �) is de�ned.

Then we also say that Re requires attention at stage s.The construction proceeds as follows. Initially A0 = B0 = ; and m0

e = e forall e.

Stage s + 1: Choose the requirement Re of highest priority which requiresattention at stage s. (If none exists, let As+1 = As, Bs+1 = Bs and m

s+1i = ms

i

for all i.) Then choose C and D such that RC;De requires attention at stage sand let � be given as above. For m = ms

e, put m 2 As+1 if �se(m;�) = 0 andotherwise put m 2 Bs+1. For m < j�j such that m =2 De and m 6= ms

e, putm 2 As+1 if �(m) = 1 and put m 2 Bs+1 otherwise.

Then the sequence ms+1i is de�ned so that ms+1

i is the ith element not inAs [Bs.

Subrequirement RC;De is now satis�ed at stage s + 1, since �0 satis�es theconditions (1) through (4) where �0 = � except possibly for �(ms

e).It is easy to see that each requirement Re requires attention only �nitely

often and that mse converges for each e. To verify that all requirements are

satis�ed, let C 0 and D0 be separating classes for A and B which have in�nitedi�erence and suppose by way of contradiction that C 0 = �e(D

0). Choose s0 sothat every requirement of priority Re or higher has ceased to require attentionby stage s0. Then for i � e, ms

i = mi for all s � s0. Let C = C 0 \ De andD = D0 \ De. Now at some stage s > s0, �

se(me; D) must converge so that


there is a � � D with �se(me; D) de�ned. But then RC;De requires attention at

stage s, contrary to the assumption on s0. This completes the proof.

The proof of the following theorem is omitted.

Theorem IV.2.12. (Jockusch-Soare [96]) There exist disjoint pairs (A0; B0)and (A1; B1) of disjoint, recursively inseparable c. e. sets such that if C 2S(A0; B0) and D 2 S(A1; B1), then C and D form a minimal pair, that is, anyset computable in both C and D is computable.

Corollary IV.2.13. For any degree a, there is a special c. b. �01 class which

has no members of degree � a.

Proof. Let (Ai; Bi) be given for i = 0; 1 by Theorem IV.2.12. Each classS(Ai; Bi) contains no computable elements. Suppose that S(A0; B0) has a mem-ber of degree �a. Then no member of S(A1; B1) can have degree � a.

Exercises

IV.2.1. Theorem IV.2.3 can be improved to say that for any degree c of a memberof P and any c. e. degree a � c, a = 00. Show this using the full ModulusLemma, which gives a modulus m �T A for C whenever A is a c.e. setsuch that C �T A.

IV.2.2. Show that any nonempty c. b. �01 class P contains members x and y

whose degrees have greatest lower bound 0. If P is special, then this givesa minimal pair of members.

IV.2.3. Show that if F is a computable function from f0; 1gN to f0; 1gN, r 2f0; 1gN, and P is a �0

1 class such that F (x) = r for all x 2 P , then r iscomputable.

IV.3 Measure, Category and Randomness

that is, fq 2 Q : q � rg is a �01 set, since r can be approximated from above

as the limit of a decreasing computable sequence. That is, given a computabletree T � 2<! with P = [T ], let Pn =

SfI(�) : � 2 T \ 2ng. Then m(Pn) is justk=2n, where k = cardf� 2 T \ 2ng and m(P ) = limnm(Pn). The measure isnot necessarily computable, which will follows from the next two results.

The measure of a �01 class of reals will be discussed in detail in Chapter XV.

It is easy to modify the class of diagonally non-computable reals to obtaina �0

1 class which has positive measure but has no computable elements.

Theorem IV.3.1. There is a �01 class P � f0; 1gN with positive measure which

has no computable members.

The proof is left as an exercise.On the other hand, there is a computable basis result.

IV.3. MEASURE, CATEGORY AND RANDOMNESS 107

Theorem IV.3.2. Any �01 class P � f0; 1gN with positive, computable measure

has a computable member.

Proof. Let P have computable measure r > 0. It follows that the measure ofP \ I(�) is computable uniformly in �, since we can approximate the measurefrom below by subtracting the measure of the complement P \ (f0; 1gN � I(�))from r. Thus we can recursively select paths � = (x(0); : : : ; x(n)) of length nsuch that P \ I((x(0); : : : ; x(n))) always has measure � r=2n. The in�nite pathx will be a computable member of P .

Proposition IV.3.3. For any thin �01 class Q � f0; 1gN, �(Q) = 0.

Proof. Let Q be a thin class and let T be a computable tree such that Q =[T ]. Suppose that �(Q) > 0 and choose a rational p > 0 so that �(Q) � p.Uniformly de�ne the class Px to be fy 2 Q : (8n)y � n �lex y � ng. Then letP = fx 2 Q : �(Px) � p. This is a �0

1 class since

x 2 P () (8n)card(f� 2 T : � �lex x � n) � 2np:

P has the property that, for x; y 2 Q, if x 2 P and x �lex y, then y 2 Q. SinceQ is thin, the complement Q � P is also a �0

1 class and �(Q � P ) = p. Butas a closed set Q has a greatest element z. It follows that Q� P = Pz so that�(Pz) = p, which would mean that z 2 P , a contradiction.De�nition IV.3.4. 1. The real x 2 f0; 1gN is said to be random if x does

not belong to any �01 class of measure 0.

2. The real x 2 f0; 1gN is said to be 1-random if for any computable functionf such that �(Pf(n)) > 1� 2�n for all n, x 2 Pf(n) for some n.

It follows from this de�nition that the class of random reals is simply theintersection of all �01 classes of measure 1 and therefore has measure 1 itself.Thus any set of positive measure must contain a random real.

We will also consider degrees of members of �01 classes. For any class P � NN,

D(P ) is the collection of all sets Turing equivalent to a member of P and U(P )is the collection of all sets A such that some member of P is Turing reducibleto A. Note that D(P ) � U(P )

P. Martin-Lof [146] introduced the notion of 1-randomness and showed thatthere is a universal, increasing sequence Pen of �0

1 classes such that �(Pen) >1 � 2�n and such that [nPen is precisedly the class of 1-random reals. Thusthe class of 1-random reals also has measure 1. The degree of a 1-random realis called a 1-random degree. Recall that a function f : ! ! ! is �xed-point-freeif there is no e such that �e = �f(e).

Theorem IV.3.5. (Martin-Lof) There is a computable function g such that ifQn = Pg(n), then

1. Q0 � Q1 � : : :2. For each n, �(Qn) > 1� 2�n


3. [nQn is the set of 1-random reals.

Proof. For each n, we construct a c. e. set Un � f0; 1g� and let Qn = f0; 1gN n[�2UnI(�). First enumerate all c. e. Martin-Lof tests as follows. Let We denotethe eth c. e. subset of f0; 1g� and let We;s � f0; 1gs denote the �nite subset ofWe;s enumerated into We by stage s. Then let We;j;s =

Tji=1Whe;ji;s as long as

�(SfI(�) : � 2 We;j;sg) � 2�(j+1) and otherwise We;j;s = We;j;s�1. Then forWe;j =

SfI(�) : s 2 N; � 2 We;j;sg. the c. e. Martin-Lof tests consist exactlyof the sequences We;0;We;1; : : : for e 2 N.

Now de�neUn = [e2NWe;n+e+1;

and letUn = [fI(�) : � 2 Ung:

Note that �(Un) �Pe2N �(We;n+e+1) �

Pe 2

�(n+e+1) � 2�n. Then fUgn<!is a c. e. Martin-Lof test. Let Qn = f0; 1gN n Un.

If x is 1-random, then x 2 [nQn since it passes all c. e. Martin-Lof tests. Ifx is not 1-random, suppose that it fails the testWe;ne<!

, Then x 2 \n�e+1We;n

and hence x 2 \nUn.Theorem IV.3.6. (Kucera). For any �0

1 class P of positive measure, D(P )contains every 1-random degree.

Proof. Suppose �(P ) > 0. Let W be a c. e. set such that f0; 1gN n P =[�2W I(�). For each n, de�ne a sequence of uniformly c. e. sets Uk by U0 =Wand for each k,

Uk+1 = f�_� : � 2 Uk & � 2Wg:Then for each k, let V k =

S�2Uk I(�). It is easy to see that, for all k, V

k+1 � V kand that �(V k) = �(V )k. Choose m such that �(V k) < 1

2 . Then for each n,�(V nk) < 2�n. Thus \mV m contains no 1-random elements. Suppose now thatA is 1-random. It follows that A =2 V j for some j; let k be the least such. Ifk = 0, then A 2 P . If k > 0, then A 2 V k�1 n V k, so that A = �_B for some� 2 Uk�1 and some B 2 P .

It follows from this proof that for any �01 class P of positive measure and

any random real A, the iterated shift �n(A) 2 P for some n.

Corollary IV.3.7. For each n, D(Qn) equals the set of 1-random degrees.

Proof. By Theorem IV.3.5, every element of Qn is 1-random and by TheoremIV.3.6, D(Qn) contains all 1-random degrees.

This also proves the following result from [99], which signi�cantly improvesTheorem IV.3.1, since now we know that D(Qn) includes the 1-random degreesand thus has measure one.

Corollary IV.3.8. (Jockusch-Soare) There is a c. b. �01 class P with no com-

putable members such that �(D(P )) = 1.

IV.3. MEASURE, CATEGORY AND RANDOMNESS 109

The following is due to Kucera [117].

Theorem IV.3.9. Every 1-random set is bi-e�ectively immune.

Proof. Let Qn be de�ned as in Theorem IV.3.5 above. Following Kucera's proof,we will construct computable functions f and g such that for all k; x 2 N:

A 2 Qn & Wx � A =) card(Wx) � f(n; x);A 2 Qn & Wx � N�A =) card(Wx) � g(n; x):

Now de�ne the uniformly �01 classes Px;n � f0; 1gN by

A 2 Px;m () [A contains the �rst m+ 1 elements of Wx ! card(Wx) < m:

Then �(Px;m) > 1 � 2�m. Now let e = f(n; x) be de�ned so that �e(j) is anindex of Px;j for all j 2 N. It follows from the de�nition of Qe that Qe � Px;e,so that Qk � Px;e. Thus if A 2 Qk & Wx � A, then card(Wx) � e. Theargument for N�A and g is similar.

The next two results are taken from [99].

Theorem IV.3.10. For any closed subset P of NN with no computable member,U(P ) (and hence D(P )) is meager.

Proof. Let P = [T ] where T is not necessarily computable and assume by wayof contradiction that U(P ) is not meager. Then for some e, Se = fA : �Ae 2 Pgis not meager. Now Se is not nowhere dense, so there exists � such that everyextension � of � can be extended to some A 2 Se. Note that Se is not necessarilya closed set. We will use e and � to construct a computable member of P . Byassumption there is an extension A of � in Se so that �Ae 2 P . Thus we can�nd �0 � � such that �e(�0; 0) is de�ned. Proceeding recursively, we can �nd�n+1 � �n such that �e(�n) 2 T and j�e(�n)j > n. It follows that the computablefunction f(n) = �e(�n; n) 2 P . (Although x[n �n is not necessarily an elementof Se.)

This theorem implies the result of Sacks [181] that U(fAg is meager for anynoncomputable set A and the following theorem from [99] will imply the resultfrom [181] that �(U(fAg)) = 0 for noncomputable A.

Theorem IV.3.11. (Jockusch-Soare) For any �01 class P = S[A;B] where A

and B are computably inseparable, �(U(P )) = 0.

Proof. Suppose by way of contradiction that �(U(P )) > 0. Then for some e andsome rationalm, Qe = fC : �Ce 2 Pg has positive measure p withm m=2g:Then each Ci is a c. e. set, since to test whether n 2 Ci, simply �nd k such thatcard(f� 2 f0; 1gk : ��e (n) = ig) > m2k�1. Since �Ce is total for a set of measure


> m, it follows that C0 [ C1 = N. Hence by the reduction principle, there aredisjoint c. e. sets E0 � C0 and E1 � C1 such that E0[E1 = N and therefore E0and E1 are computable. We claim that E1 is a separating set for A and B. Ifn 2 A, then �(fC : �Ce 2 Pg) > m, so that �(fC : �Ce (n) = 1g) > m and hence�(fC : �Ce (n) = 0g) < m=2, so that n =2 C0 and therefore n 2 E1. Similarlyif n 2 B, then n 2 E0. But this contradicts the assumption that A and B arecomputably inseparable.

Exercises

IV.3.1. Construct a �01 class which has positive measurem but has no computable

members. Show that for any real number � > 0, we can make m > 1� �.(Hint: we can ensure that �e =2 P by making �edn =2 T for some largevalue of n. )

IV.3.2. Say that a real r is �01 if fq 2 Q : q < rg is a �01 set. Show that r is �01

(respectively, �01) if and only if r is the limit of a computable, decreasing(resp. increasing) sequence of rationals. Thus r is computable if and onlyif r is both an increasing and a decreasing limit of rationals.

IV.3.3. Show that the class of 1-generic reals has measure 0 and is not a basisfor the family of �0

1 classes of positive measure. (Hint: x is 1-generic if itnever belongs to the boundary of any Pe.)

IV.3.4. Show that 1-generic reals and 1-random reals are also random.

IV.3.5. Show that if a �01 class P contains a random element, then �(P ) > 0.

IV.4 Mathematical Logic: Peano Arithmetic

In this section, we consider further the connection between �01 classes and math-

ematical logic and, in particular, Peano Arithmetic. We begin with some appli-cations of the present chapter together with sections III.III.9.

Theorem IV.4.1. Any axiomatizable theory � has a complete consistent ex-tension of low c. e. degree.

Proof. Let � be an axiomatizable theory and let the �01 class P represent the

family of complete consistent extensions of �, by Theorem III.III.9.1. Then Phas a member of low c. e. degree by Theorem IV.1.4.

On the other hand, we have the following.

Theorem IV.4.2. There is a (propositional) axiomatizable theory � which hasno c. e. complete consistent extension.

Proof. By Theorem IV.2.1, there is a �01 class P with no c. e. member. Now, by

Theorem III.III.9.3, there is an axiomatizable theory � such that P representsthe set of complete consistent extensions of �.

IV.4. MATHEMATICAL LOGIC: PEANO ARITHMETIC 111

Theorem IV.4.3. For any c. e. degree c, there is an axiomatizable theory �such that the c. e. degrees of complete consistent extensions of � are exactly thec. e. degrees above c.

Proof. This is an immediate consequence of Theorems III.III.9.3 and IV.2.4.

Theorem IV.4.4. For any essentially undecidable, axiomatizable theory �,there exists a c. e. degree a such that � has no complete consistent extensionsof degree � a.

Proof. This follows from Theorems III.III.9.1 and IV.2.5.

Theorem IV.4.5. For any essentially undecidable, axiomatizable theory �,there exists two complete consistent extensions, �1 and �2, of � such that anyset computable from �1 and computable from �2 is in fact computable.


Theorem IV.4.6. There is an axiomatizable theory � such that any two com-plete consistent extensions of � are Turing incomparable.


IV.4.1 Peano Arithmetic

The standard model N = (N; S;+; �; <) of arithmetic is fundamental in math-ematics. The language of arithmetic is often de�ned to consist of a one-placefunction symbol S, representing the successor function, and binary functionsymbols + for addition and � for multiplication. The usual linear ordering �may then be de�ned by

x � y () (9z)(x+ z = y);

x < y () x � y & x 6= y:

Each natural number n may be represented by the term Sn0. We will generallyidentify Sn0 with n for simplicity of expression. Peano Arithmetic is a �rst ordertheory for N the consisting of nine axioms needed to de�ne the functions andalso the axiom scheme of induction. The eight axioms of Robinson arithmeticare the following.

S1 Sx = Sy =) x = y

S2 (8x)Sx 6= 0

L1 (8x):x < 0

L2 (8x)(8y)[x < Sy () (x < y _ x = y)]

A1 (8x)x+ 0 = x

A2 (8x)(8y)x+ Sy = S(x+ y)


M1 (8x)x � 0 = 0

M2 (8x)(8y)x � Sy = x � y = y

The axiom scheme of induction provides for each formula � with one freevariable:

IP [�(0) & (8x)(�(x) ! �(Sx))]! (8y)�(y)The fundamental results of G�odel and others [100, 209] connecting Peano

Arithmetic and computability theory is the following.

Theorem IV.4.7. For any c. e. set A, there is a formula � (which representsA) such that, for all m, the following are equivalent:

(1) m 2 A;(2) N j= �(m) and

(3) PA ` �(Sm0):A partial converse to this result is the result of G�odel that any axiomatizable

theory � is computably enumerable (in the sense that the set of G�odel numbersof members of � is a c. e. subset of N). This means that in particular PA itself isa c. e. set. The Incompleteness Theorem of G�odel follows from these results andtells us in particular, that PA can have no axiomatizable, complete consistentextension.

We shall now consider the connection between Peano Arithmetic and �01

classes. It follows from Theorem III.9.1 that the set of complete consistentextensions of PA as well as the set of consistent extensions of PA may be repre-sented as �0

1 classes. It then follows that there is a complete consistent extensionof PA with c. e. degree. Now any axiomatizable extension � of PA is neces-sarily incomplete by G�odel's theorem, hence the family of complete consistentextensions of � will be an uncountable �0

1 class. Certainly not every �01 class

may be represented as the set of complete extensions of such a �. However, theScott Basis Theorem shows that these classes are as complicated as an arbitrary�01 class in a certain sense.

Theorem IV.4.8. (Scott [186]) For any consistent extension � of PA, the setscomputable in � form a basis for the c. b. �0

1 classes.

Proof. Let � be a consistent extension of PA, let P be a c. b. �01 class and let

TP � N� be the tree of nodes which have an extension in P . Then TP is a �01

set and it follows from Theorem IV.4.7 that there is a formula such that, forall strings �,

� =2 TP =) PA ` �(h�i) =) �(h�i 2 �:Now we can de�ne a subtree T of TP which is computable in � by

� 2 T () �(h�i =2 �:The tree T has no dead ends and thus there P contains an in�nite path whichis computable in T and hence computable in �.

IV.4. MATHEMATICAL LOGIC: PEANO ARITHMETIC 113

Tennenbaum [210] observed that no nonstandard model of PA can be com-putable and the following improvement of that result was noted by Jockuschand Soare [98].

Theorem IV.4.9. If M = (N;+M ; �M ) is any nonstandard model of Peanoarithmetic, then the sets computable in M form a basis for the c. b. �0

1 classes.

Proof. We follow the proof of Cohen [54]. Since PA and the set of negationsof sentences in PA form a pair of computabley inseparable c. e. sets, it su�cesby Theorem IV.4.8 to show thatM can compute a separating set for any pairof disjoint c. e. sets A and B. First observe that we can recursively computethe standard numbers inM by starting with a0 and a1 which represent 0 and 1and letting an+1 = an +

M a1 for all n. Similarly we may compute fromM thesequence pn of (�nite) prime numbers of M. Now by the Chinese RemainderTheorem, there must exist an in�nite element a of M such that, in M, a =0(modpn) if and only if a 2 A and a = 1(modpn) if and only if n 2 B. Aseparating set for A and B may now be computed fromM as follows. Given n,compute the unique q and unique r < pn such that a = q �M pn +

M r and putn 2 C if and only if r 6= 1. It is immediate that A � C and B \ C = ;.

We have the following corollary to Theorem IV.2.3.

Corollary IV.4.10. If a is the degree of any consistent extension or nonstan-dard model of Peano Arithmetic and b is a c. e. degree such that a � b, thenb = 00.

It now follows that no consistent extension of PA can have c. e. degree < 00.However, can get degree exactly 00, as announced by Scott and Tennenbaum in[187].

Corollary IV.4.11. There is a complete consistent extension of Peano Arith-metic of degree 00.

Proof. Let P be the c. b. �01 class of all complete consistent extensions of PA.

Then P has a member of c. e. degree b and it follows from Corollary IV.4.10that b = 00.

Here is a nice corollary to Theorem IV.2.8.

Corollary IV.4.12. (Jockusch-Soare [99]) There is a complete consistent ex-tension � of Peano Arithmetic such that every set de�nable in � is either com-putable or not arithmetical.

Proof. By Theorem IV.2.8, there is a theory � whose degree is incomparablewith 0(n) for each n > 0. The corollary now follows from the fact that each setde�nable in � is computable from T .


Chapter V

The Cantor-Bendixson

Derivative

The perfect set theorem states that any closed subset of f0; 1gN is the union ofa perfect closed set (the perfect kernel) and a countable set. The perfect kernelresults from iterating the Cantor Bendixson derivative D�(P ) until a �xed point(an analytic set) is reached. The e�ective version of this theorem is that theperfect kernel of a �0

1 class is a �11 and that the iteration must stop by the �rst

non-computable ordinal !C�K1 .For a countable �0

1 class P , the perfect kernel is the empty set and the iter-ation must stop at some computable ordinal (the C-B rank of P . The Kreiselbasis theorem [116] showed that isolated members of �0

1 classes must be hyper-arithmetic in general, must be computable in 00 if P is bounded, and must becomputable if P is computably bounded. Furthermore, any countable closed setmust have an isolated element.

A �ner analysis was given in [23] for �01 classes in f0; 1gN. The C-B rank of

a real x in a class P is the least ordinal � such that x =2 D�(P ). In particular,if x has C-B rank � + n in P for some limit ordinal � and �nite n, then xis computable in 0�+2n. It follows that every element of a countable class ishyperarithmetic. The C-B rank of a real x is de�ned to be the minimum of theranks of x in any �0

1 class.

V.1 Cantor-Bendixson derivative and rank

The Cantor-Bendixson derivative D(P ) of a compact subset P of NN is the setof nonisolated points of P . Thus a point x 2 P is not in D(P ) if and onlyif there is some open set U containing x which contains no other point of P .Equivalently, x =2 D(P ) if and only if there is some clopen set U such thatU \ P = fxg. Another useful observation is that, for any compact set P , D(P )is empty if and only if P is �nite.

115

116 CHAPTER V. THE CANTOR-BENDIXSON DERIVATIVE

The iterated Cantor-Bendixson derivative D�(P ) of a closed set P is de�nedfor all ordinals � by the following trans�nite induction.

De�nition V.1.1. 1. D0(P ) = P ; D�+1(P ) = D(D�(P )) for any �;

2. D�(P ) =T�<�D

�(P ) for any limit ordinal �.

There is a related derivative which we can de�ne for a tree T � f0; 1g�.De�nition V.1.2. 1. d(T ) = f� 2 T : (9�)[� � � & �_0 2 Ext(T ) & �_1 2

Ext(T )]g.2. d0(T ) = T , d�+1(T ) = d(d�(T )) and d�(T ) = \�<�d�(T ) for limit �.

Then d(T ) �T T (2) and is in fact �02 in T . We iterate this derivative by

Lemma V.1.3. For any countable ordinal � and any tree T � f0; 1g�, D�([T ]) =[d�(T )].

Proof. The proof is by trans�nite induction. The equality is clear for � = 0.The interesting case is when � = 1. Suppose that x 2 [d(T )] and �x n 2 !.

Then xdn 2 d(T ), so that there exists � � xdn such that both �_0 and �_1are in Ext(T ). Thus I(xdn)\ [T ] contains at least two elements. It follows thatthere is some path yn 2 [T ] such that xdn � yn and x 6= yn. Hence x 2 D([T ]).

Next suppose that x 2 D([T ]) and again �x n 2 !. Since x is not isolatedin [T ], there must exist y 2 [T ] such that y 6= x and xdn � y. Let m > n bethe least such that ydm + 1 6= xdm + 1 and let � = xdm. Then xdn � � and�_0 and �_1 are both in Ext(T ), which demonstrates that xdn 2 d(T ). Hencex 2 [d(T )]. v Now suppose that D�([T ]) = [d�(T )]. Then

D�+1([T ]) = D(D�([T ]) = D([d�(T )]) = [d([d�(T ))] = [d�+1(T )]:

For a limit ordinal �,

D�([T ]) = \�<�D�([T ]) = \�<�[d�(T )] = [d�(T )]:

This completes the proof.

The Cantor-Bendixson (CB) rank of a closed set P is the least ordinal �such that D�+1(P ) = D�(P ); then D�(P ) is a perfect closed set, denotedK(P ), called the perfect kernel of P and P n K(P ) is a countable set. ForA 2 P n K(P ), the C-B rank rkP (A) of A in P is the least ordinal � suchthat A 2 D�(P ) n D�+1(P ); the C-B rank rk(P ) is the least � such thatD�(P ) = K(P ). The set A is ranked if there is a �0

1 class P such that A 2P n K(P ), and the C-B rank rk(A) is the least � such that rkP (A) = � forsome �0

1 class P . Kreisel [116] used the Boundedness Principle of Spector [201]to show that rk(P ) � !C�K1 for any �0

1 class P , so that any ranked point A hasrk(A) < !C�K1 .

V.1. CANTOR-BENDIXSON DERIVATIVE AND RANK 117

Theorem V.1.4 (Kreisel). For any �01 class P , K(P ) is a �11 class and

rk(P ) � !C�K1 .

Proof. Fix a computable tree T such that P = [T ] and de�ne the �02 monotoneinductive operator � on f0; 1g� as follows.

� 2 �(A) () � =2 T _ � =2 d(f0; 1g� �A):

It follows that �1 = f0; 1g� � T , �1 = f0; 1g� � d(T ), and so on, so thatCl(�) = f0; 1g� � TK(P ). It follows from Theorem II.II.9.8 that TK(P ) is �

11, so

that K(P ) = [TK(P )] is also �11. It follows from Corollary II.9.13 that rk(P ) �

!1.

We will show in Section VI.VI.7 that there is a �01 class P such that rk(P ) =

!C�K1 and K(P ) is not �11.

For the sake of completeness, let rkP (A) = rk(P ) if A 2 K(P ). This meansthat the rank function will de�ne a prewellordering on P .

A fundamental idea here is that the complexity of an element x of a �01

class P is related to the Cantor-Bendixson rank of x in P . Kreisel [116] �rstnoticed that the Turing degree of a member x of a �0

1 class is related to theCB rank; he used this to show that every member of a countable �0

1 class ishyperarithmetic. In particular, it is easy to see that a real has rank 0 if andonly if it is computable.

We need a series of lemmas from [46]. Note �rst that for two closed sets Pand Q, D(P [Q) = D(P ) [D(Q) but D(P \Q) is in general only a subset ofD(P ) \D(Q).

More generally, we have

Lemma V.1.5. For any closed sets P and Q and any ordinal �:

(a) D�(P [Q) = D�(P ) [D�(Q) and

(b) D�(P \Q) � D�(P ) \D�(Q).

Proof. The proofs are by induction on �. For � = 0, these are trivial.(a) For � = 1, D(P ) and D(Q) are subsets of D(P [Q) since P and Q are

subsets of P [ Q. For the reverse inclusion, suppose that x =2 D(P ) [ D(Q).Then there are clopen sets U and V with U \ P = fxg and V \ Q = fxg.Thus U \ V \ (P [ Q) = fxg, so that x =2 D(P [ Q). Now assume thatD�(P [Q) = D�(P ) [D�(Q). Then using the case � = 1, we have

D�+1(P [Q) = D(D�(P [Q)) = D(D�(P ) [ D�(Q))

= D(D�(P )) [ D(D�(Q)) = D�+1(P ) [ D�+1(Q):

(b) This is left as an exercise.

Note that equality holds in (b) if one of the sets is clopen.


Lemma V.1.6. For any compact subset Q of f0; 1g!, any clopen set K andany ordinal �,

(a) D�(K \Q) = K \D�(Q).

(b) rkK\Q(A) = rkQ(A) for any A.


Lemma V.1.7. For any set A and any computable ordinal �, rk(A) � � if andonly if there is some �0

1 class P such that D�(P ) = fAg.Proof. The \if" direction is immediate from the de�nition of rank. Supposenow that rk(A) � �, so that A 2 D�(Q) nD�+1(Q) for some �0

1 class Q. ThusA is isolated in D�(Q), so that for some clopen K, K \ D�(Q) = fAg. LetP = K \Q. It follows from Lemma V.1.6 that

D�(P ) = D�(K \Q) = K \D�(Q) = fAg.

Lemma V.1.8. (Cenzer-Smith [46])

(a) Let � be a continuous map from f0; 1g! into f0; 1g! and let P;Q becompact sets such that �(P ) = Q. Then for any y 2 Q, rkQ(y) �maxfrkP (x) : x 2 P & �(x) = yg.

(b) For any sets A;B, if A �tt B and B is ranked, then A is ranked andrk(A) � rk(B).

(c) For any sets A;B, if A �tt B and B is ranked, then A is ranked andrk(A) = rk(B).

Proof. Let rkQ(y) = �, so that y 2 D�(Q) n D�+1(Q). It is shown in Lemma1.2 of [46] that D�(�(P )) � �(D�(P )) for any �. Since y 2 D�(Q), it followsthat y = �(x) for some x 2 D�(P ), so that rkQ(y) = � � rkP (x), where it ispossible that x is not ranked in P .

(b) By Theorem II.5.10, if A �tt B, then there is a computable function� : f0; 1g! ! f0; 1g! such that �(B) = A. Now let rk(B) = � and, byLemma 4.2, let P be a �0

1 class such that D�(P ) = fBg, so that rkP (B) =maxfrkP (x) : x 2 Pg, and let Q = �(P ). It follows from (a) that rk(A) �rkQ(A) � rkP (B) = rk(B).

(c) This is immediate from (b).

The following improvement of Lemma 1.4 of [46] is due to J. Owings and C.Laskowski [166]. Recall that �� is the Hessenberg sum of two ordinals, thatA�B is the disjoint union of two sets and that P�Q = fA�B : A 2 P & B 2 Qgfor two classes P and Q of sets.

Theorem V.1.9. (Owings, [166]) For any sets A;B 2 f0; 1gN and any compactP;Q � f0; 1gN, rkP�Q(A�B) = rkP (A)� rkQ(B).

V.1. CANTOR-BENDIXSON DERIVATIVE AND RANK 119

Proof. We �rst show by induction on rkP (A) � rkQ(B) that rkP�Q(x � y) �rkP (A) � rkQ(B). If rkP (A) � rkQ(B) = 0, then A is isolated in P and B isisolated in Q, so that there are open intervals I and J such that P \ I = fAgand Q \ J = fBg. It follows that (P � Q) \ (I � J) = fA � Bg, so thatrkP�Q(A � B) = 0. Now let rkP (A) = � and rkQ(B) = � and suppose theinequality holds for all x; y such that rkP (x)� rkQ(y) < �� . By intersectingwith open intervals, as above, we may assume that D�(P ) = fAg and thatD�(Q) = fBg. It su�ces to show that rkP�Q(x�y) < �� for all x�y 6= A�Bin P � Q. But if x � y 6= A � B, then either x 6= A or y 6= B, so that eitherrkP (x) < � or rkQ(y) < �. In either case, rkP (x) � rkQ(y) < � � �, so thatrkP�Q(x� y) < �+ �.

For the reverse inequality, we prove by induction on �� thatD�(P )�D�(Q) � D��(P �Q).For � � � = 0, this is obvious. We also need the case where � � � = 1.

Suppose without loss of generality that � = 1 and � = 0 and suppose thatA 2 D(P ) and B 2 Q. Then for any interval I � f0; 1g!, there is some A0 6= Ain P \ I. Then for any basic open set I � J 2 f0; 1g! � f0; 1g!, there is anelement A0 �B 6= A�B in (P �Q) \ (I � J). Thus A�B 2 D(P �Q). Thisshows that

D(P )�Q � D(P �Q).Now suppose the inclusion holds for all ordinals �; � with � � � < � � �.

There are two cases.(Case 1) If �� is a limit ordinal, then � and � are both limit ordinals and

�� = supf� � � : � < � & � < �g. Thus

D��(P �Q) = \ <��D (P �Q) = \�<�;�<�D�� (P �Q)= \�<�;�<�D�(P )�D� (Q) = D�(P )�D�(Q):

(Case 2) If �� is a successor, then either � is a successor or � is a successor{without loss of generality say that � = +1, so that �� = ( ��)+1. Then

D�(P )�D�(Q) = D(D (P ))�D�(Q) � D(D (P )�D�(Q))

� D(D ��(P �Q)) = D��(P �Q):This completes the proof.

Thus we have the following corollary.

Theorem V.1.10. For any sets A and B, maxfrk(A); rk(B)g � rk(A�B) �rk(A)� rk(B).Proof. The �rst inequality follows from Lemma V.1.8, since both A and B are�tt A�B. The second inequality follows from Theorem V.1.9.

The basic result for rank 0 is the following.

Lemma V.1.11. For any x 2 f0; 1g!, the following are equivalent:


(a) x is computable;

(b) fxg is a �01 class;

(c) x has Cantor-Bendixson rank 0.

Proof. Suppose �rst that x is computable. Then fxg = [T ], where � 2 T ()(8i < j�j)(�(i) = x(i)). Next suppose that fxg is a �0

1 class. Then the rankof x in fxg is 0, so that the C-B rank of x is 0. Next suppose that x has C-Brank 0 and let P = [T ] be a �0

1 class such that x is isolated in P , where T is acomputable tree. Then for each su�ciently large n, xdn+ 1 is the unique pathof length n which has an extension in P . Thus we may compute xdn + 1 (andtherefore compute x(n)) by searching for the least m such that all strings � 2 Tof length m have the same initial segment �dn.

We remark that Lemma V.1.11 and its proof can be relativized to com-putability in B for any set B.

Exercises

V.1.1. Give a proof of Lemma V.1.5(b).

V.1.2. Give a proof for Lemma V.1.6.

V.1.3. Prove a relativized version of Lemma V.1.11.

V.1.4. Show by induction on ordinals � that for any continuous map � : f0; 1gN !f0; 1gN, and any compact set Q, D�(�(P )) � �(D�(P )).

V.1.5. Show that if rk(B) = rk(A�B), then A is computable in B.

V.1.6. Owings [166] de�ned a Cantor singleton as being the unique noncom-putable element of some �0

1 class. By Theorem III.6.2, every noncom-putable �0

1 retraceable set is a Cantor singleton. Show that any noncom-putable set which is the union of a computable set with a �0

1 retraceableset is also a Cantor singleton.

V.1.7. Show that if B is a Cantor singleton and A �tt B, then A is a Cantorsingleton.

V.1.8. Show that if A�B is a Cantor singleton, then either B is computable orA is computable in B.

V.1.9. A set A is said to be autoreducible if there is a computablee functional Fsuch that, for all n, A(n) = F (n;A n fng). Show that every ranked setA is autoreducible (Owings). (Hint: note that, for any n, A n fng andA [ fng are both �tt A and thus have rank �.)

V.2. BASIS RESULTS 121

V.2 Basis results

In this section, we consider basis results for countable �01 classes. The key

observation here is the following.

Theorem V.2.1. Any countable closed P � f0; 1gN has countable rank and hasan isolated point.

Proof. If P has no isolated points, then P is perfect and hence uncountable.For the rank, observe that for � < rk(P ), D�(P ) nD�+1(P ) is nonempty.

Basis results for countable �01 classes can thus be obtained from the following.

Theorem V.2.2. (Kreisel [116]) Let P be a �01 class.

(a) Any isolated member of P is hyperarithmetic; if P is �nite, then everymember of P is hyperarithmetic.

(b) Suppose that P is bounded. Then any isolated member of P is computablein 00; if P is �nite, then every member of P is computable in 00.

(c) Suppose P is computably bounded. Then any isolated member of P iscomputable; if P is �nite, then every member of P is computable.

Proof. Let x be isolated in P = [T ] and take n large enough so that P\I(xdn) =fxg. Now de�ne the �0

1 class Q to be P \ I(xdn). That is, Q = [S], where S isthe computable tree consisting of all strings in T which are compatible with xdn.It follows that Ext(S) = fxdn : n < !g, so that x is computable in Ext(S).Now consider the three cases.

(a) For an (unbounded) tree S, Ext(S) is �11 by Theorem III.2.13 and since[S] is a singleton, we have

� 2 Ext(S) () (8� 2 !j�j(� 6= � ! � =2 Ext(S)),so that Ext(T ) is also �1

1. It follows that Ext(T ) and hence x are hyper-arithmetic.

(b) For a �nitely branching tree T , Ext(T ) is �02 by Theorem III.2.13. Now

it follows from K�onig's Lemma that for each n, there is some k � n such thatevery sequence in T of length k is an extension of xdn. Thus we have

� 2 Ext(S) () (9k � j�j)(8� 2 !k)(� 2 S ! � � �),so that Ext(S) is also �02. It follows that Ext(S) and hence x are computable

in 00.(c) In this case, Ext(S) is �0

1 by Theorem III.2.13 and is also �01 since� 2 Ext(S) () (9n � j�j)(8� 2 f0; 1; : : : ; f(n)gn)(� 2 S ! � � �),where f is a computable bounding function for S. Thus Ext(S) and x are

both computable.Suppose now that P is �nite. The conclusion in each case follows from the

fact that every member of P will be isolated.

Combining this with the previous result, we have

Theorem V.2.3. (Kreisel) Let P be a countable �01 class.


(a) P has a hyperarithmetic member.

(b) If P is bounded, then P has a member computable in 00.

(c) If P is computably bounded, then P has a computable member.

As a corollary to the proof of Theorem V.2.2, we also have the following.

Theorem V.2.4. Let P = [T ] be a �01 class.

(a) Suppose that T is �nite branching. Then any isolated member of P iscomputable in T 0; if P is �nite, then every member of P is computable inT 0. Thus if P is countable, then P has a member computable in T 0.

(b) Suppose T is computably bounded. Then any isolated member of P iscomputable in T ; if P is �nite, then every member of P is computable inT . Thus if P is countable, then P has a member computable in T .

V.3 Ranked Points and Rank-Faithful Classes

The following notions were introduced by G. Martin [145] and J. Owings [166]. A�01 class is said to be rank-faithful if rkP (x) = rk(x) for all x 2 P . The �0

1 classesconstructed in Theorems III.6.8, V.4.5 and V.4.8 are clearly rank-faithful. A realx is said to be a Cantor Singleton if it is the unique non-computable memberof some �0

1 class. Theorem III.ref2.2 implies that every non-computable �01

retraceable set is a Cantor singleton. G. Martin improved this result by showingthat any noncomputable set which is the union of a computable set with a �0

1

retraceable set is a Cantor singleton. (See the exercises.)

Theorem III.III.6.2 was improved to the following.

Theorem V.3.1. (G. Martin) For each computable ordinal � and every non-zero r.e. degree a, there are c. e. sets A and B of degree a and rank � suchthat A is a Cantor singleton and B belongs to a rank-faithful �0

1 class.

Recall from Theorem III.8.2 that a minimal �01 class is a thin class of rank

one such that all computable elements are isolated and the unique nonisolatedelement is noncomputable. It follows that any minimal �0

1 class is rank-faithful.This can be extended to arbitrary thin classes as follows.

Theorem V.3.2. Any thin �01 class P is rank-faithful.

Proof. Suppose that A 2 P and that rk(A) = �. Let D�(Q) = fAg. ThenA 2 P \Q and P \Q = P \ U for some clopen set U . It follows from LemmaV.1.6 that rkP (A) = rkP\Q(A) � rkQ(A). Equality follows since rkQ(A) isminimal by assumption.

Here are two interesting results from Owings [166].

V.4. RANK AND COMPLEXITY 123

Theorem V.3.3. (Owings) (a) If A� B is a Cantor singleton, then either Bis computable or A is computable in B, so that the degree of A�B is either thedegree of A or the degree of B.

(b) If rk(B) = rk(A�B), then A is computable in B.

Proof. (a) Suppose that A�B is the unique noncomputable element of P andthat B is noncomputable. Let Q = P \(f0; 1g!�fBg and observe that B �tt Cfor every C 2 Q, so that any C 2 Q is noncomputable. Thus A � B is theunique element of Q and is therefore computable in B by the relativized versionof Lemma V.1.11.

(b) Let � = rk(A�B) and let P be a �01 class such that D

�(P ) = fA�Bg.As in (a), let Q = P \ (f0; 1g! � fBg). It follows from Lemma 4.3(b) thatrk(C) � � for all C 2 Q, so that in fact Q = fA � Bg. Then A � B iscomputable in B as above.

Recall that a set A is autoreducible if there is a computable functional Fsuch that, for all n, A(n) = F (n;A n fng).Theorem V.3.4. (Owings) Every ranked set A is autoreducible.

Proof. Let rk(A) = � and let P = [T ] be a �01 class such that D�(P ) = fAg.

Note that for any n, Anfng and A[fng are both �tt A and thus also have rank�. It follows that only one of them can belong to P . Thus, given n and Anfng,we search for the least k such that every � 2 T of length k consistent withA n fng, except possibly at n, has the same value �(n). Then A(n) = �(n).

Exercises

V.3.1. Prove G. Martin's result that any noncomputable set which is the unionof a computable set with a �0

1 retraceable set is a Cantor singleton.

V.3.2. For any sets A;B such that B is a Cantor singleton and A �tt B, showthat A is a Cantor singleton (Owings [166].) Hint: recall the proof ofLemma V.4.3.

V.4 Rank and Complexity

In this section, we examine the connection between the rank and the hyper-arithmetic complexity of a set.

Our �rst theorem will be a generalization of Theorem V.2.2 (c) to arbitraryrank.

First we consider the complexity of the tree resulting from the iterated C-Bderivative.

Lemma V.4.1. For any computable tree T � f0; 1g�, any computable limitordinal � and any �nite n > 0,


(a) dn(T ) �T 0(2n) and is �02n;

(b) d�(T ) �T 0(�+1) and is �0�+1;

(c) d�+n(T ) �T 0(�+2n) and is �0�+2n+1.

Proof. Let T be given and de�ne the inductive operator � on f0; 1g� by

� 2 �(U) () � =2 d(T ) _ � =2 d(f0; 1g� � U):

Then � is �02 by De�nition V.1.2 and it is easy to see that for all ordinals �,

d�(T ) = f0; 1g� � ��:

The result now follows immediately from Theorem III.II.10.9.

For decidable trees, there is a slight improvement.

Lemma V.4.2. For any decidable tree T � f0; 1g�, any computable limit ordi-nal � and any �nite n > 0, dn(T ) �T 0(2n�1) and is �02n�1.

Proof. Since T is decidable, Ext(T ) is a computable set and it follows that d(T )is a �0

1 set. The result now follows from Exercise III.3.

Theorem V.4.3. For any A 2 f0; 1gN, any �01 class P , any �nite n and any

limit ordinal �,

(a) If rkP (A) = n, then A �T 0(2n). Furthermore, if rkP (A) = n for somedecidable P , then A �T 0(2n�1).

(b) If rkP (A) = �+ n, then A �T 0(�+2n+1).

Proof. (a) Let T be a computable binary tree such that [dn(T )] = fAg. Itfollows from Lemma V.4.1 that dn(T ) �T 0(2n) and then by Theorem V.2.4that A �T 0(2n). For a decidable class P = [T ], dn(T ) �T 0(2n�1) by LemmaV.4.2.

(b) Similarly x �T d�+n(T ) �T 0(�+2n).

Theorem V.4.4. For any countable �01 class P � f0; 1gN,

(a) rk(P ) is a computable ordinal;

(b) Every element of P is hyperarithmetic.

Proof. Since P is countable, it follows that the perfect kernel of P is empty. Thusthe inductive de�nition � given in the proof of Lemma V.4.1 has closure f0; 1g�which is a computable set. It follows from the Boundedness Theorem for induc-tive de�nability (Theorem II.II.9.12) that j�j < !1 and therefore rk(P ) < !1. Itnow follows from Theorem V.4.3 that every element of P is hyperarithmetic.


It follows from Theorem V.4.3 that any set of rank one must be computablein 000. We now analyze the rank one sets further. Recall from Theorem III.6.8that every c. e. set is Turing equivalent to a hypersimple r.e. set of rank one.This result has the following improvement.

Theorem V.4.5. (Cenzer-Smith [46] For any noncomputable degree b � 00,there is a hyperimmune set B with degree b of rank one; furthermore, there isa computable tree T with no dead ends such that D([T ]) = fBg.Proof. Let A be a set of degree b. By the limit lemma, there is a computablefunction f such that, for all e,

A(e) = limn f(n; e):Let n(0) be the least n > 0 such that f(n; 0) = A(0) and, for any e, let

n(e+ 1) be the least n > n(e) such that, for all i < e+ 2, f(n; i) = A(i). Thenn(0) < n(1) < � � � is a modulus for the set A, so that B = fn(0); n(1); � � � g hasdegree b.

We de�ne a �01 class P with rkP (B) = 1.

The (possibly �nite) set C = fm(0) < m(1) < � � � g is in P if and only if0 < m(0) and for all e, i, and m:

1. (0 < m < m(0) & C 6= 0)! f(0;m) 6= f(0;m(0));

2. e < i < card(C)! f(m(i); e) = f(m(e); e);

3. (e+1 < card(C) & m(e) < m < m(e+1))! (9j < e+2)(f(m(e+1); j) 6=f(m; j)).

It is clear that P may be de�ned by a tree T without dead ends and in factclosed under extension by 0. Also, P contains all initial subsets of B and in factis closed under initial subsets, so that rkP (B) � 1.

If C = fm(0) < m(1) < � � � g is any in�nite set in P , then it follows from (ii)that f(m(e); e) = f(n(e); e) = A(e) for all e. But it then follows from (i) or, byinduction, from (iii) that m(e) = n(e) for all e, so that C = B.

If C = fm(0) < m(1) < � � � < m(k)g is any �nite set in P , then it followsfrom (i) and (iii) that there are at most two extensions C [ fmg of C in P (onewith f(m; k + 1) = 0 and one with f(m; k + 1) = 1), so that C is isolated in P .

Thus B is the only non-isolated element of P and therefore rkP (B) = 1.To see that B is hyperimmune, suppose by way of contradiction that h were

a computable function with h(e) > n(e) for all e. Then we could de�ne a �01

subclass Q of P by adding the restriction

(iv) (8e)[card(C \ f0; 1; : : : ; h(e)� 1g) > e].

We shall see later that not every �02 degree contains a ranked set. Thus thereis a more complicated result for degrees below 000. The following theorem givesa partial inverse C-B derivative and shows that a large class of degrees containranked points.


Theorem V.4.6. [23] For any real B and any tree S � f0; 1g� such that S iscomputable in B00, there is a tree T computable in B and a homeomorphism �from [S] onto D([T ]) such that for all x 2 [S], x �T �(x) �T x�B0.

Proof. Let F be a function, computable in B, such that, for all � 2 f0; 1g�,�S(�) = lim

p!1limn!1

F (p; n; �):

For each p and �, let Fp(�) = limn!1 F (p; n; �).Let x 2 f0; 1gN be given and assume that x 2 [S]. We de�ne \outer modulus"

values p = pk(x) such that Fp(xdk) = 1 and \inner modulus" values n = nk(x)such that F (pk; n; xdk) = 1.

Let p1 = p1(x) be the least p > 0 such that Fp(xd1) = 1 and let n1 = n1(x)be the least n > p1 such that, for all m � n, F (p;m; xd1) = 1. Now inductivelyde�ne pe+1(x) to the least p > ne such that Fp(xdi) = 1 for all i � e+ 1. Sincex 2 [S], pe+1 exists. Now let ne+1 be the least n > pe+1 such that

(1) for all i � e+ 1 and all m � n, F (pe+1;m; xdi) = 1, and

(2) for all p with ne ne with this property.

Now for x 2 [S], let(3) H(x) = fx(0)_1_0n1_1_0p1_1x(1)_1_0n2_1_0p2_ : : :

It is clear from the de�nition that x is computable from H(x) and that H(x)is computable in 00 �B0.

The tree T is to contain all initial segments of each image H(x) togetherwith certain strings which resemble those initial segments when only �nitelymany values of F are considered. Every element of [T ]�H([S]) will be isolatedin [T ] and each point H(x) will be the limit of a distinct sequence of points from[T ] nH([S]).

Elements of the computable tree T with D([T ]) = fBg will be initial seg-ments of strings � of the form

(4) � = r(0)10n110p11r(1)10n210p2 : : : r(k � 1)10nk10pk1:

The string � as above is said to be consistent if it satis�es the followingconditions:

(5) 0 < p1 < n1 < p2 < n2 < � � � < pk < nk:

(6) For all i, j, k and m, if i � j � k and nj � m � nk, then F (pj ;m; rdi) = 1.

(7) For all j, k and p, if j � k and nj�1 < p < pj , then there is an i � j suchthat, for all m with ni � m � nk, F (p;m; i) 6= F (p;m; rdi) = 0.


If � = �_r or � = �_r_1_0n, then � is consistent if � is consistent. Thestring � = �_r_1_0n_1 is consistent if � is consistent and if conditions (ii)and (iii) hold when nk is replaced by n and the string �_1_0p is consistent if� is consistent and p < n.

It is clear that for all x 2 [S], every initial segment of H(x) is consistent.Now let y 2 f0; 1gN be an extension of � as in (4) above and suppose that everyinitial segment of y is consistent. There are two possibilities. Either y has theform in (3) above or y = �_0!, where � = �_r(k + 1)_ : : :_r(`)_1. In eithercase, the values of nj are unbounded in the initial segments of y. It now followsfrom (5) that:

(8) For all i � j � k, Fpj (rdi) = 1.

It also follows from (v) that:

(9) For all j � k and all p with nj�1 0 for j = 1) such that Fp(rdi) = 1 for all i � j. It follows thatx 2 [S].

A consistent string � with � � � � �_r_1_0nk+1_1_0pk+1 is said to be

exact if each ni is minimal, that is,

(10) For all i; k; n, if i � k and pi < n < ni, thenr(0)10n110p11 : : : 1r(i)10n10pi1 is not consistent.

The desired tree T is de�ned to be the set of all exact strings. It is clearthat for each x 2 [S], H(x) is in [T ]; furthermore, for each k,x(0)_1_0n1(x)_1 : : :_1_x(k)_1_0! belongs to [T ]. Thus H(x) 2 D([T ]) asdesired. It remains to show that all other elements of [T ] are isolated.

First suppose that y 2 [T ] has in�nitely many 1's. Then we saw above thaty has the form (3) where x 2 [S] with each pj minimal. It now follows from(10) that each ni is also minimal so that y = H(x). Thus any element of y of[T ] = H([S]) must have the form

y = r(0)10n110p11r(1)10n210p2 : : : r(k)10! = �_0!

Now suppose that z 6= y and z 2 [T ] \ I(�). Then for some nk and pk,�_1_0nk_1_0pk_1 � z. It follows from (8) and (9) that pk is uniquely de-termined and from (10) that nk is uniquely determined. Thus y is the uniqueelement of [T ] which extends �_1_0nk+1 and hence y is isolated in [T ] as de-sired.

We now have that H is an isomorphism of [S] onto [S] onto D([T ]). Since His computable in B, it must be continuous and since [S] and D([T ]) are compact,it follows that H�1 is also continuous. This completes the proof.


We note that this proof is uniform in S. That is, there is a primitive recursivefunction � such that if e is an index for S as a �0

3-in-B set, then �(e) is an indexfor T as a �0

1-in-B set.

Corollary V.4.7. (a) For any �nite n and any tree S �T 0(2n), there is acomputable tree T and a homeomorphism H from [S] onto Dn([T ]) suchthat, for all x 2 [S], x �T H(x) �T x� 0(2n�1).

(b) For any �nite n and any real x such that 0(2n�1) �T x �T 0(2n) or suchthat 0(2n�2) �T x �T 0(2n�1), there is a computable tree T and a realy �T x with jyjT = jyj = n.

Proof. Part (a) follows easily from Theorem V.4.6 by induction on n. Part (b)now follows from (a) by letting [S] = fxg.

This result can be extended to higher levels of the hyperarithmetic hierarchyas follows. The following is Theorem 42 of [39].

Theorem V.4.8. For any computable ordinal � and any c. b. �02�+1 class Q,

there exists a �01 class P of sets and a homeomorphism from Q onto D�(P )

such that x �T H(x) � x� 02��1 for all x 2 Q.Proof. The proof is by a uniform recursion up to a �xed computable ordinal �with a set of notations as given by Lemma II.8.12. We may assume that Q isactually a class of sets and build a P which is computably bounded. In fact, weonly need to assume that Q = [S], where for all � 2 S and all relevant notationsa, �(i) 6= a for any i.

We will actually de�ne computable functions f and such that �0o(a)�1

(e;a)

is a homeomorphism from Pe;2o(a)+1 onto Do(a)(Pf(e;a)) and such that x �� (e;a)(x) for each x.

The construction will be presented as a trans�nite recursion on o(a), but isactually obtained by the recursion theorem.

We need a series of lemmas. We shall write A = tnBn if A = [nBn and theelements of fBn : n 2 !g are pairwise disjointLemma V.4.9. There is a primitive recursive function � such that, for each� = o(a), Pe;�+3 = [T ], with T = tnU�(e;a;n);�+1.

Proof. From the de�nition, Pe;�+3 = [U ], where U = [Ue;�+3] is uniformly�0�+3. It is then easy to de�ne, as in Proposition III.3.7 a �0�+2 tree S such

that [U ] = [S]. Thus there exists a �0� relation R such that

� 2 S () (9n)(8m)R(m;n; e; �)

As usual, we may assume that :R(m;n; e; �)! :R(m;n+1; e; �) and that(8m)R(m;n; e; �)! (8m)R(m;n+ 1; e; �).

Now de�ne the desired trees by� 2 U�(e;a;hm;ni);�+1 if and only if


1. (8m0)(8i � j�j)R(m0; n; e; �di)2. (8n0 < n)(9m0 < m)(9i � j�j):R(m0; n0; e; �di)3. (8m0 < m)(9n0 < n)(8m00 < m0)(8i � j�j)R(m00; n0; e; �di).

We need a slightly modi�ed notion from [23] of a completely ranked tree.

De�nition V.4.10. A tree T is completely ranked up to � if, for all y 2 [T ]and all notations a with o(a) < �,

1. y 2 Do(a)([T ]) if and only if y(n) = a for in�nitely many n.

2. y =2 Do(a)([T ]) implies that for some n, any extension � 2 T of ydn neverhas �(i) = a for i � n.

T is completely ranked if there exists a recursive ordinal � such that T is com-pletely ranked up to � and, for any � 2 T , any notation a with o(a) � �, andany i, �(i) 6= a.

Let Q be a r. b. strong �02�+1 class of sets. Note that elements of Q are

only allowed to contain notations for ordinals > �. We will de�ne a �01 class P ,

a recursively bounded, tree T , completely ranked up to �+1, such that P = [T ]and a homeomorphism � from Q onto D�(P ) such that x �T �(x) �T �02��1for all x 2 Q. In fact, x is always a subsequence of �(x). Furthermore, anelement y of P will contain in�nitely many notations for ordinals > � only ify = �(x) for some x 2 Q.

For � = 0, just let Q = P and let � be the identity.Successor Case: Suppose that � = o(b) = o(a) + 1 = � + 1 and that

Q = Pe;2�+3 = [T ]. By Lemma V.4.9, we haveT = tnU�(e;a;n);2�+1 = tnTn.Now for x 2 Q, let ni for each i be the unique n such that xdi+ 1 2 Tn and

let�(x) = (bx(0)0n0+1bx(1)0n1+1b : : :):

De�ne the �02�+1 tree U to contain � = (bx(0)0n0+1bx(1)0n1+1 : : : x(k)0n)

and its initial segments provided that xdi+ 1 2 Tni for all i < k.It is easy to see that G = [U ] contains exactly f�(x) : x 2 Qg together with

paths y = (bx(0)0n0+1bx(1)0n1+1bx(k))_0! for all x 2 Q and all k. Each of thelatter will thus be isolated in G and each of the former will have rank 1, so thatQ is homeomorphic to D(G). Furthermore, for x 2 Q, x a subsequence of �(x)and therefore is recursive in �(x), since the values of x are just those followingimmediately after b's. �(x) is recursive in x� 02�+1, since the computation ofthe sequence of witnesses ni may be performed with a �0

2�+1 complete oracle.In addition, for any y 2 G, y 2 D(G) if and only if y has in�nitely many

occurrences of b, and y has no occurrences of any notation � �. Let f be a recur-sive function so that �(i) � f(i) for all � 2 T and all i. Then for any � 2 U andany i, �(i) � maxfb; f(i)g, so that G is recursively bounded. In addition, if y =


(bx(0)0n0+1bx(1)0n1+1bx(k))_0! =2 D(G), then (bx(0)0n0+1bx(1)0n1+1bx(k))_0nk+1

may only be extended in U by 0's.

Now by induction, there is a r. b. �01 class P = [S] which is completely

ranked up to � and a homeomorphism from G onto D�(P ) such that y �T(y) �T y � 02��1. We may assume that y is always a subsequence of (y)and that the values of y follow immediately after the occurrences of a in (y).

It is immediate that is also a homeomorphism from D(G) onto D�(P ).Now let �(x) = �((x)). Then clearly � is a homeomorphism from Q ontoD�(P ).

It remains to show that P is completely ranked up to � + 1. Suppose �rstthat z has in�nitely many occurrences of b. Then, by the construction, z = (y)for some y 2 G such that y has in�nitely many occurrences of b. Thus y 2 D(G),so that z 2 D�(P ). Next suppose that z 2 D�(P ). Then z = (y) for somey 2 D(G). Now y has in�nitely many occurrences of b, so that z must also havein�nitely many occurrences of b.

Finally, suppose that z 2 P has only �nitely many occurrences of b, so thatz =2 D�(P ). Observe that by the construction, the occurrences of b may onlyfollow immediately after occurrences of a. There are two cases.

First, suppose that z =2 D�(P ). Then z has only �nitely many occurrencesof a and by induction, there is some n such that no extension � 2 S of zdn hasany occurrences of a past �(n). It follows from the observation above that bmay not occur in � past �(n) either.

Next, suppose that z 2 D�(P ). Then z = (y) for some y 2 G such thaty has only �nitely many occurrences of b. Thus by the construction, there issome n so that ydn may only be extended by 0's in U . Now choose m so thatzdm includes the subsequence ydn. It follows that no extension of zdm in Smay contain any further occurrences of b. This concludes the proof that P iscompletely ranked up to � + 1.

We note that the construction of P and of are uniform.

Limit Case: Suppose that � = o(b) is a limit ordinal and that Q = Pe;�+1 =[T ]. Let a0; a1; : : : enumerate the set of notations for ordinals less than � andlet �n = o(an). By de�nition, we have Q = \nQn = \n[Tn], where Tn =U�e(an);o(an)+1.

By induction, we have constructed r.b. classes Pn = [Un], completely rankedup to � and homeomorphisms �a from Qn onto D�n+1(Pn). For each x 2 Qand each n, let

�n(x) = (anx(0)�n;0anx(1)�n;1 : : : ;

where each witness �n;i contains no occurrence of an. Now let

�(x) = (bx(0)�0;0bx(1)�1;0bx(2)�1;0 : : : :

Here the sequence of witnesses �i;n is enumerated in order, �rst by the sum i+nand then by the value of n.

It is immediate that x �T �(x) and it follows from the uniformity of theconstruction that �(x) �T x� 0� .


De�ne the �01 tree U to contain all initial segments of

� = (bx(0)�0;0bx(1)�0;1bx(2)�1;0 : : : �n;ibx(k)�) such that

1. For all m; j with m + j < n + i or with m + j = n + i and m � n,amx(0)�m;0amx(1)�m;1 : : : amx(j)�m;j 2 Un and �m;j contains no occur-rence of am.

2. If n 6= 0, then an�1x(0)�n�1;0an�1x(1)�n�1;i : : : an�1x(i+ 1)� 2 Un�13. If n = 0, then ai+1x(0)� 2 Ui+1 and � contains no occurrence of ai+1.

Then P = [U ] clearly contains �(x) for all x 2 Q. It is easy to see that ify 2 P has in�nitely many occurrences of b, then y = �(x) for some x 2 Q. Nowsuppose that y has only �nitely many occurrences of b and let

y = (bx(0)�0;0bx(1)�0;1bx(2)�1;0 : : : �n;ibx(k))_z;

where z has no occurrences of b. Let

� = (bx(0)�0;0bx(1)�0;1bx(2)�1;0 : : : �n;ibx(k)):

This given, we can de�ne a string � as follows. There are two cases.

Case 1 If n 6= 0, thenu = (an�1x(0)�n�1;0an�1x(1)�n�1;i : : : an�1x(i+ 1))_z 2 Pn�1. Let

� = an�1x(0)�n�1;0an�1x(1)�n�1;i : : : an�1x(i+ 1)):

Case 2 If n = 0, then u = ai+1x(0)_z 2 Pi+1. Let� = ai+1x(0):

We will now establish several claims leading to the desired result that P iscompletely ranked and that � is a homeomorphism of Q onto D�(P ). The proofdepends on the de�nition of �. We will give the proofs for Case 1 and leave thesimpler Case 2 to the reader.

We claim �rst that there is some initial segment � of y such that no extensionof � in U has any further occurrences of b. There are two subcases here.

(Subcase a): If z has an occurrence of an�1, then it follows from the de�nitionof U that no further occurrence of b can occur after the �rst an in z has occurredin y. Let � be an initial segment of z containing the �rst an.

(Subcase b): If z has no occurrences of an�1, then since Un�1 is completelyranked, there is some initial segment � of z such that no extension of �_� inUn�1 may contain any further occurrences of an�1. It follows from the de�nitionof U that b can not occur in U past �_�.

Next, we claim that the rank jyjP of y in P equals the rank jujPn�1of u in

Pn�1. We �rst observe that for any z0, if �_z0 2 Un�1, then �_z0 2 U . Thisimplies that jujPn�1

� jyjP . For the other inequality, observe that for any z0

extending �, if �_z0 2 U , then �_z0 2 Un�1.


It now follows that �(x) has rank at least � in P , since it is the limit of thesequence �_n zn, where, for the appropriate value of k,

�n = (bx(0)�0;0bx(1)�0;1bx(2)�1;0 : : : �n;ibx(k)),�n = an�1x(0)�n�1;0an�1x(1)�n�1;i : : : an�1x(i+ 1)), and�n(x) = y = �_n zn.Finally, we show that P is completely ranked up to � + 1. We have already

established that y has rank � � if and only if y has in�nitely many occurrencesof b. Now suppose that jyjP < � and let �, � and z be as above. Then for anya with o(a) � b, jyjP � o(a) if and only if �_z has rank � o(a) in Pn�1, whichis if and only if z has in�nitely many occurrences of a, since Un�1 is completelyranked, and this is if and only if y has in�nitely many occurrences of a. Theprevious discussion already established the other criterion for being completelyranked.

The uniformity of the proof shows that, using the Recursion theorem, wecan actually compute indices for P and for � from an index for Q. We omit thedetails.

Corollary V.4.11. [39] For any computable ordinal � which is either 0 or alimit and any �nite n:

(a) there is a B �T 0(�+2n) with rk(B) = �+ n;(b) for any degree a such that 0(�+2n+1) � a � 0(�+2n+2), there is a B of

degree a with rk(B) = �+ n+ 1.

Proof.

Next we brie y consider sets which cannot be ranked.

Theorem V.4.12. (Cenzer-Smith [46]) For any hyperimmune set A, there isa C �T A which is not ranked.

Proof. Let A = ff(0) < f(1) < � � � g be hyperimmune. Let [T0]; [T1]; : : : enu-merate the �0

1 classes as in Lemma 1.2. We �rst de�ne B �T A so that [Ti]is uncountable whenever B 2 [Ti], which implies that B is not ranked. Thecharacteristic function of B is the limit of a sequence of strings �n of lengthf(n) which is computable in A. Let �0 = 0. Given �n, �n+1 is de�ned in twocases.

(Case 1) There is some i � f(n) and some � of length f(n) such that

1. �n 2 Ti2. � =2 Ti3. �dn = �ndn4. for any j < i, if �n =2 Ti, then � =2 Tj .Then we let i be the least for which there is a � satisfying the conditions

and we let �n+1 be the (lexicographically) least corresponding to that i.

(Case 2) If there is no such i, then �n+1 = �_n 0f(n+1)�f(n).

V.5. COMPUTABLE TREESWITHONEORNO INFINITE BRANCHES133

Let B have characteristic function [n�n. It is clear that B �T A. The proofthat B 2 [Ti] implies [Ti] uncountable is by induction. Suppose true for alli < j, suppose that B 2 [Ti], and choose m large enough that B =2 [Ti] impliesBdm =2 Ti for all i < j. Then for any n > f(m), any extension � =2 Ti of Bdn oflength f(n) will satisfy the conditions of Case 1. Thus the shortest extension �of Bdn not in Ti must have length > f(n). If [Ti] were countable, then every �would have an extension not in Ti, so that we could de�ne a function h(n) to bethe least k such that any string � of length n has an extension of length k whichis not in Ti. Then for n > f(m), h(n) > f(n), contradicting the assumptionthat A is hyperimmune.

It follows that B is unranked. Finally, let C = A � B. Then C �T A andB �tt C, so that C is unranked by Lemma V.1.8.

Of course this implies that every r.e. degree contains an unranked set. Wesay that a is completely unranked if every setA of degree a is unranked. Jockuschand Shore construct in [95] a �02 degree which is completely unranked. Downeyobserved that since sets with the same truth-table degree have the same rankand since all sets in a hyperimmune-free degree have the same truth-table degree(see [163], p. 589), the construction of an unranked set of hyper-immune freedegree will provide a completely unranked degree. This led to the followingimprovement in of the Jockusch-Shore result.

Theorem V.4.13. (Downey [60]) There is a hyperimmune-free degree which iscompletely unranked.

On the other hand, Downey also showed the following, again using a hyper-immune free degree a, so that every A of degree � a is in fact �tt a.Theorem V.4.14. (Downey [60]) There exists a degree a � 000 such that everyset A of degree � a is ranked.

Cenzer and Smith consider in [46] the problem of sets below 00 but with highrank. They showed that for every computable ordinal �, there is a �0

2 set A ofrank �. This was improved by Cholak and Downey in [51].

Theorem V.4.15. (Cholak-Downey [51]) For each computable ordinal �, thereis an r.e. set of rank �.

V.5 Computable Trees with One or No In�nite

Branches

Recall the notions of the height ht(T ) of a well-founded tree T � N� and theheight htT (�) of the nodes of T introduced in Section II.II.10. If T has aunique in�nite branch, then following Clote [53], we let T = supfhtT (�) + 1 :T [s] is well-foundedg. Theorem II.II.10.19 of Clote showed that the hyperarith-metic complexity of the unique in�nite branch is bounded in some sense by the


height of T . In this section we consider the reverse result that every hyper-arithmetic set is reducible to the unique in�nite branch of some computabletree.

We also look at the complexity of the perfect kernel K(Q) of a �01 class in

f0; 1gN.In section III.II.5, we showed that any �0

1 class P � NN can be reduced toa �0

1 class Q � f0; 1gN. In this section, we consider the connection between theheight of a computable tree T such that P = [T ] and the rank of the tree Ssuch that Q = [S].

The following theorem is a variant of the result of Clote [53].

Theorem V.5.1. For any computable ordinal � and any hyperarithmetic indexa 2 H��2, there is a computable tree T with unique in�nite branch x such that T � !� and Ha �T x.Proof. In fact, we will use the recursion theorem to obtain a function suchthat for all hyperarithmetic indices a 2 H��2, T (a is a computable tree withunique in�nite branch xa such that T (a) � !� and Ha � xa. Recall that

Ha = [nN�H�a(n), so that for a 2 H��2, we have

i 2 Ha () (9n)(8k)i 2 H��a(n)(k)(i):

Using Lemmas II.10.6, II.10.7 and II.10.8, we can �nd a function f such that

i =2 Ha () (91n)i 2 Hf(a;n);

where f(a; n) 2 H��2 for some � with � + 1 � �.For the base case of � = 0, we may assume that Ha is uniformly primitive

recursive. Let pi denote the ith prime number. Then for each i, we xa(pi) =h�Ha(i)i and we let xa(j) = 0 for all non-primes j. Let T (a) = fxadn : n 2 Ng.Certainly (T (a)) = 0 for all such a.

Now we may assume that for each a and n, there is a tree Ta;n = T (f(a;n))with unique in�nite branch xa;n and Ta;n � ! � � for some � with � + 1 � �.Furthermore, for every i and n, xa;n(pi) = h�i for some sequence � such that�(0) = �Hf(a;n)(i). Now de�ne ya and xa as follows. For all i and n,

ya(2i+1 � 3n+1) = xa;n(i):

For all i, there are two cases in the de�nition of ya(pi).

Case I: ya(pi) = 0. Then for all k,

ya(pk+1i ) = (least n > ya(p

ki )i 2 Hf(a;n):

Here the sequence of values ya(pk+1i ) provides an in�nite set of witnesses that

i =2 Ha.Case II: ya(pi) = 1. Then

ya(p2i ) = (least n)(8m � n)i =2 Hf(a;m)

V.5. COMPUTABLE TREESWITHONEORNO INFINITE BRANCHES135

and ya(pki ) = 0 for all k > 2. Here ya(p

2i ) provides a witness that i 2 Ha.

Finally

xa(j) = hya(j); ya(j � 1); : : : ; ya(0)i:Observe that Ha may be computed from xa by

i 2 Ha () (xa(i))0 = 1 () :(xa(i))0 = 0:

The computable tree T = T (a) with unique in�nite branch xa is de�ned asfollows. Given a sequence � = (�(0); : : : ; �(m � 1)) which is a candidate formembership in T , we �rst require that there exists a sequence e0; e1; : : : ; em�1such that for each j < m, �(j) has the form hej ; ej�1; : : : ; e0i. Then for j =2i+1 � 3n+1 < m, we require that

(e2i+1�3; e2i+1�9; : : : ; e2i+1�3n+1) 2 Ta;n:

These conditions mean that � provides an answer to whether i 2 Hf(a;n) fora certain set of (i;m; n). For j = pi, ej = 0 indicates the guess that i =2 Ha

and then fepk+1igk is intended to enumerate the in�nite set fn : i 2 Hf(a;n)g.

Similarly ej = 1 indicates the guess that i 2 Ha. Then for j = p2i , ej = n isintended to be the least n such that i =2 Hf(a;m) for all m � n. The string �is in T if and only if the \witnesses" provided by � as to whether i 2 Ha arecon�rmed by the coded subsequences from the trees Ta;n.

It follows that xa is the unique in�nite branch of T . That is, suppose thatx 2 [T ] and has the proper form. Then each of the coded subsequences must bein Ta;n so that each in�nite coded subsequence must be xa;n. Now the codedvalues of ya(p

ki ) must be correct since the witnesses from Ta;n are all correct.

It remains to check that (T ) � ! �! ��. Let � 2 T be a dead end of lengthm in the proper form and let e0; e1; : : : ; em�1 be given as above. There are twocases.Case I: For some n, the coded subsequence � = (e2i+1�3; e2i+1�9; : : : ; e2i+1�3n+1)is a dead end of Ta;n. Then it can be shown by induction that htT (�) �htTa;n(�) � ! � � < ! � � for some � < �.Case II: All coded subsequences from any Ta;n are initial segments of xa;n.Case IIa: ej = ya(j) for all primes j but some witness ej where j = pki isdi�erent from ya(p

ki ). Suppose �rst that ya(i) = 0 is correct and the witnesses

should be n2 = ya(p2i ); n3 = ya(p

3i ); : : : but for j = pki , ej 6= nj . Now let �0 2 T

be any extension of � long enough to predict the value of xa;n(nj). Then � mustbe incorrect as in Case I and thus htT (�) � !�. Thus htT (�) � !��+n < ! ��.A similar argument apples when ya(i) = 1 is correct but the witness ej wherej = p2i is incorrect.Case IIb: For some i and for j = pi, ej 6= ya(j). Suppose �rst that ej = 0but ya(j) = 1. Then there can be only a �nite number K of witnesses in thesequence ya(p

ki ) and any extension of � long enough to predict ya(p

Ki +1) must

code a dead end of Ta;K+1 as in Case IIa. A similar argument applies whenej = 1 but ya(j) = 0.


Corollary V.5.2. For any computable ordinal �, hyperarithmetic index a 2H��2, there is a computable tree T with unique in�nite branch x such that T �!� and x is �0��2 complete.

The next result explores further the relation between trees T in N� with aunique in�nite branch and the natural images �(T ) 2 f0; 1g�, relating the heightof T with the rank of �(T ). Recall from the proof of Theorem III.III.7.3 the def-inition of �(x) = 0x(0)10x(1) : : : mapping NN into f0; 1gN and the correspondingmapping of trees so that

0�(0)10x(1)1 : : : 0x(k�1)10x(k) 2 S = �(T ) () (�(0); : : : ; �(k � 1)) 2 T:

Then a dead end � 2 T of length k corresponds to an isolated point

y� = 0�(0)10x(1)1 : : : 0x(k�1)101 2 [S]:

Theorem V.5.3. Let T � N� be a computable tree with no in�nite paths andlet S = �[T ] be the image of T in f0; 1g� and Q = [S]. Then for any � 2 T ,rkQ(y�) � htT (�).Proof. It is easy to see that Q = fy� : � 2 T . The inequality is proved by induc-tion on � = htT (�). If htT (�) = 0, then � has no immediate extensions in T andhence y� is isolated in Q so that rkQ(y�) = 0. Now let � = (�(0); : : : ; �(k� 1))and suppose that htT (�) = �, so that for any proper extension � 2 T of �,ht(T ) < �. Let y 2 Q be any extension of �� = 0�(0)10x(1)1 : : : 0x(k�1)1 di�er-ent from y�. Then y = y� for some extension � 2 T of �, so that htT (�) < �and hence by induction rk(y) < �. It follows that rkQ(y�) � �.

Exercises

V.5.1. Show that the reverse inequality in Theorem V.5.3 does not hold. In fact,for any computable ordinal �, there is a computable tree T with no in�nitepath and � 2 T such that the rank of y� in �[T ] is 0 but htT (�) = �+ 1.

V.5.2. Combine Theorem V.5.3 and Corollary V.5.2 to show that for any com-putable ordinal �, there is a real x 2 f0; 1gN of degree 0��2 and a �0

1 classQ such that rkQ(x) � ! � �.

V.6 Logical Theories revisited

In this section, we apply the basis results for countable �01 classes to axiomati-

zable theories.

Theorem V.6.1. Let � be an axiomatizable �rst order theory with only count-ably many complete consistent extensions. Then

(a) � has a decidable complete consistent extension.

V.6. LOGICAL THEORIES REVISITED 137

(b) If � has only �nitely many complete consistent extensions, then everycomplete consistent extension is decidable.

(c) Every complete consistent extension of � is hyperarithmetic.

Proof. Let � be an axiomatizable theory with countably many complete con-sistent extensions. By Theorem III.III.9.1, the set of complete consistent ex-tensions of � may be represented as a �0

1 class P . It follows from TheoremV.2.3 that P has a computable member, which represents a decidable completeconsistent extension of �. Similarly, it follows from Theorem V.4.4 that everycomplete consistent extension of � is hyperarithmetic. If � has only �nitelymany complete consistent extensions, then P has only �nitely many elementsand therefore all of the elements of P are computable by Theorem V.2.2 andthus all of the complete consistent extensions of � are decidable.

Given an axiomatizable theory �, let us say that an extension � of � isa �nite extension of � if there is a �nite set F of sentences such that � islogically equivalent to � [ F . Then it is easy to see that a complete consistentextension � of � is a �nite extension if and only if � is isolated in the �0

1 classof complete consistent extensions of �. In particular, any complete consistent�nite extension of � must be decidable. Thus if � is an undecidable completeconsistent extension of �, then � is not a �nite extension. We now want tofocus on complete consistent extensions of rank one.

Theorem V.6.2. Let � be an axiomatizable �rst-order theory which has aunique complete consistent, non-�nite extension �. Then � �T 000 and if �is decidable, then � �T 00.Proof. This follows from Theorem V.4.3.

Here is an existence result.

Theorem V.6.3. (a) For any degree b � 00, there is a decidable theory �with unique complete consistent, non-�nite extension � and � has degreeb.

(b) For any degree b such that 00 � b � 00, there is an axiomatizable theory� with unique complete consistent, non-�nite extension � of degree b.

Proof. (a) Let b � 00. Then by Theorem V.4.5, there is a decidable �01 class P

with unique nonisolated element B of degree b. By Theorem III.III.9.3, thereis a decidable theory � such that P represents the set of complete consistentextensions of � and thus B represents the unique non-�nite complete consistentextension � of �.

(b) This follows from Corollary V.5.2 and Theorem III.III.9.3 as in (a).


Chapter VI

Index Sets

The notions of enumeration and of an index set are fundamental in the studyof the computable functions and computably enumerable sets. The complexity(in the arithemetic hierarchy) of many properties can be measured using indexsets. For example, the index set Inf = fa : Wa is in�niteg is �0

2 complete, sothat from this point of view, the property of being in�nite is �0

2. The chapterbegins with a brief list of such results for c.e. index sets, together with theircomplexity.

We present an enumeration of the �01 classes and then classify several index

sets for �01 classes. In particular, we study index sets for properties related to

cardinality, computable cardinality, measure and category. We then indicatehow these index sets will play an important role in the application of �0

1 classesto various mathematical problems in Part 2.

A set A is said to be an index set (for c. e. sets) if for any a; b, a 2 A and�a = �b imply that b 2 A. We can also de�ne a co-index set to be a set Asuch that for any a; b, (a 2 A & b 2 A & �a = �b) implies that a = b. Thusin particular, ; and ! are index sets. Rice's Theorem ([198], p. 21) states thatthese are the only two computable index sets. We have de�ned the index setsK and K0 in Section II.3. In fact, it is the case that if A is an index set otherthan ;; !, then K �1 A or K �1 �A where K = fa : a 2 Wag. Here are someother examples of index sets which we will employ:

� K1 = fa :Wa 6= ;g;� Fin = fa :Wa is �niteg;� Inf = fa :Wa is in�niteg;� Cof = fa : ! nWa is �niteg;� Coinf = fa : ! nWa is in�niteg;� Rec = fa :Wa is a computable setg;� Tot = fa : �a is totalg;

139

140 CHAPTER VI. INDEX SETS

� Ext = fa : �a is extendible to a total computable functiong;� Ext2 = fa : �a is extendible to a total f0; 1g-valued computable functiong;� Comp = fe :We �T Kg;� U1

1 = fa : (9x)(8n) < xdn >=2Wag.Following Soare [198], p. 66, we de�ne (�mn ;�

mn ) �m (B;C) for a disjoint

pair of sets B and C if for some �mn complete set A, there is a computablefunction f such that, for any a, a 2 A () f(a) 2 B and a =2 A () f(a) 2 C.If B is �mn , C is �mn and (�mn ;�

mn ) �m (B;C), then we will say that the pair

(B;C) is (�mn ;�mn ) complete.

The index sets described above all turn out to be complete for some level ofthe arithmetical hierarchy. Here is a brief list of such complexity results, mosttaken from Soare [198], where the reader can �nd a further discussion of indexsets. We give some details of the proofs as preparation for the work on indexsets for �0

1 classes.

Theorem VI.0.4. (i) K, K0 and K1 are �01 complete sets;

(ii) Tot is a �02 complete set;

(iii) (Fin; Inf) is (�02;�02) complete.

(iv) (Cof;Coinf) is (�03;�03) complete;

(v) Ext, Ext2, and Rec are �03 complete sets;

(vi) Comp is �04 complete;

(vii) U11 is a �11 complete set.

Sketch. (i) It is clear that these are all �01 sets and that K0 is complete. LetW = Dom(�) be any c. e. set and de�ne the partial recursive function �e sothat �e(m; i) = �(m). Let f(m) = S11(e;m) so that �f(m)(i) = �e(m; i) = �(m).Then

m 2W () �(m) # () Wf(m) = !

andm =2W () �(m) # () Wf(m) = ;:

Thus m 2W () f(m) 2Wf(m) () Wf(m) 6= ;.(ii,iii) a 2 Tot () (8m)(9s)m 2 Wa;s and a 2 Fin () (9m)(8n >

m)(8s)n =2Wa;s.For the completeness, let B be a �0

2 set and let R be a computable relatiosuch that

i 2 B () (8m)(9n)R(i;m; n):It is an important observation that we may assume that R(i;m; n)! R(i;m; n+1) and that R(i;m + 1; n) ! R(i;m; n). That is, let R0(i;m; n) () (8m0 �m)(9n0 � n)R(i;m0; n0). Then R0 has the desired properties.

VI.1. INDEX SETS FOR �01 CLASSES 141

Now let �a(i;m) = (�n)R(i;m; n) and let g(i) = S11(a; i), so that �g(i)(m) =(�n)R(i;m; n). If i 2 B, thenWg(i) = !, so that g(i) 2 Tot and also g(i) 2 Inf .If i =2 B, then Wg(i) is �nite, so that g(i) 2 Fin and g(i) =2 Tot.

(iv,v) Let A be a �03 set and let R be a computable relation so that, for alla,

a 2 A () (9m)(8n)(9k)R(a;m; n; k)We will de�ne a primitive recursive function f such that a 2 A if and onlyif Wf(a) is co�nite, which will be if and only if Wf(a) is computable. Let thestandard noncomputable c. e. setK have a computable enumerationK = [sKs.The c. e. set Wf(a) in stages Wf(a);s so that ! nWf(a);s = ffbsa;0 < bsa;1 < : : : g.

Stage 0: Wf(a);s = ;.Stage s + 1: For each m � s such that either m 2 Ks+1 nKs or such that

there is some n � s such that (8n0 � n)(9k � s)R(a;m; n0; k) but :(8n0 �n)(9k < s)R(a;m; n0; k), enumerate bsa;m into Wf(a);s+1.

If a 2 A, then for some m, bsa;m is put into Wf(a);s+1 in�nitely often, sothat lims b

sa;m = 1 and hence Wf(a) is co�nite. If a =2 A, then for every m,

lims bsa;m = ba;m < 1. Thus Wf(a) is coin�nite. Furthermore, K � Wf(a) (so

that Wf(a) is not computable), since m 2 K () m 2 Kba;m and ba;m can beuniformly computed from Wf(a).

This argument can be modi�ed to show that Ext and Ext2 are both �03complete, as follows. Here we want to say that a 2 A if and only if �a isextendible. As before, putm 2Wf(a) at stage s+1 (by de�ning �f(a)(m) = 0) ifthe list of n such that (9k)R(a;m; n; k) becomes longer at stage s. Also, replacethe action associated with the set K with the following. If �m;s(b

sa;m) = j and

(8i < bsa;m)�f(a);s(i) = �m;s(i);

then de�ne �f(a)(bsa;m) = 1 � j, thus putting bsa;m 2 Wf(a). (By the usual

convention, a�b = 0 if a < b.) If a 2 A, then Wf(a) is co�nite as before, so that�f(a) is extendible. If a =2 A, then for each m, �f(a)(ba;m) is either unde�ned ornot equal to �m(ba;m), so that �f(a) is not extendible.

(vi) A proof is given in Soare [198, Ch. XII].(vii) A proof can be found in Hinman [87, p. 84].

Each of the results above can be relativized. That is, letW xe = fn : �xe (n) #g.

Then for example, Finx = fa : W xa is �niteg is �0;x2 complete. In particular,

if we let x = ;(n) denote the n-th jump of the emptyset, then Post's theorem(Theorem II.6.7) implies that a set is �0;xk if and only if it is �0n+k, see Soare

[198]. It follows that, for example, FinK is �03 complete.

VI.1 Index sets for �01 classes

There are several di�erent ways to de�ne index sets to �01 classes. We use here

an approach from [40] based on primitive recursive trees.Let �n denote the string � 2 N� such that h�i = n. Then �0; �1; : : : enu-

merate N�, and furthermore, whenever �i � �j , it must be the case that i < j.


Then a tree T is primitive recursive, computable, etc. if the corresponding setfi : �i 2 Tg is itself primitive recursive, computable, etc..De�nition VI.1.1. Let �e denote the eth primitive recursive function and let� 2 Te () (8� � �)�e(h�i) = 1; let Pe = [Te].

Lemma VI.1.2. (a) For each e, �e is a �01 class;

(b) For each �01 class P , there are in�nitely many e such that P = Pe.

Proof. Part (a) is clear. Part (b) follows from Proposition III.3.1 and the obser-vation that every primitive recursive function has in�nitely many indices.

An enumeration of the strong �0n classes can be given based on the enumer-

ation of the �0n sets. Let Wne be the eth �0n set. To be more precise, Wn

e is thedomain of the function �ne where �ne (m) = �e(m; ;(n)). Then the eth strong�0n+1 class is de�ned as follows, where we identify � 2 N� with h�i as usual for

simplicity of expression.

De�nition VI.1.3. Tn+1e = f� : (8� � �)� 2 Wne g; Pn+1e = [Tn+1e = fx :

(8m)xdm 2Wne g.

Next we look at the complexity of the various notions of boundedness.

Theorem VI.1.4. Let g � 2 be a computable function.

(a) fe : Te is g-boundedg is �01 complete;

(b) fe : Te is almost g-boundedg is �02 complete.

Proof. (a) Let g : N� ! N be an arbitrary function such that g(�) � 2 for all �.The set is �0

1 since Te is g-bounded if and only if

(forall�)(8i)[�_i 2 Te ! i < g(�)]:

For the completeness, we will de�ne a primitive recursive function h such thatTh(e) is g-bounded if and only if e =2 K. Let

� 2 Th(e) () (8t < j�j)[�e;t(e) "! �(t) = 0]:

It follows from the Master Enumeration Theorem II.2.5 and the s-m-n TheoremII.2.7 that h is primitive recursive. If e =2 K, then Th(e) = f0t : t 2 Ng and isclearly g-bounded. If e 2 K and �e;t(e) #, then 0t_i 2 Th(e) for all i, so thatTh(e) is not g-bounded.

(b) This set is �02, since if g = �a, then Te is almost g-bounded if and onlyif

(9k)(8i)(8�)[(j�j � k & �_i 2 Te)! i < �a(�)]:

For the completeness, we de�ne a reduction of Fin as follows. For each e ands, recall that We;s = fi : �e;s(i) #g and that �e;s(i) # implies that i � s. Thuse 2 Fin if and only if fs :We;s+1 nWe;s 6= ;g is �nite. For j�j = s, let

� 2 Th(e) () (8n < s)[We;n+1 nWe;n = ; ! �(n+ 1) < g(�)]:


If e 2 Fin and k satis�es We;k = We, then Th(e) is g-bounded above k. Ife =2 Fin, then for each n such that We;n+1 nWe;n 6= ;, we have 0n_i 2 Th(e) forevery i, so that Th(e) is not almost bounded by g.

Theorem VI.1.5. (a) fe : Te is c. b.g is �03 complete.

(b) fe : Te is almost c. b.g is �03 complete.

Proof. (a) The �rst set is �03, since Te is c. b. if and only if Te is �a-boundedfor some total computable function �a.

For the completeness, we de�ne a reduction f of Rec to our set. This willbe done so that [Tf(e)] is empty if We is �nite and [Tf(e)] has a single element ifWe is in�nite. The primitive recursive tree Tf(e) is de�ned as follows: Put � =(s0; s1; : : : ; sk�1) 2 Tf(e) if and only if s0 < s1 < � � � < sk�1 and there exists asequence m0 < m1 < � � � < mk�1 such that, for each i < k, mi 2We;si nWe;si�1

and mi is the least element of We;sk�1n fm0; : : : ;mi�1g. We observe that if

We is �nite, then Tf(e) is also �nite and therefore recursively bounded. Now�x e and suppose that We is in�nite. Then we may de�ne canonical sequencesn0 < n1 < : : : of elements of We and corresponding stages t0 < t1 < : : : suchthat, for each i, ni 2 We;ti nWe;ti�1 and (t0; t1; : : : ; ti) 2 Tf(e) as follows. Letn0 be the least element of We and, for each k, let nk+1 be the least element ofWe nWe;tk . Then for each k, (t0; : : : ; tk) 2 Tf(e) and nk 2 We;tk . Furthermore,we see by induction on k that

k 2We ! k 2We;tk :

For s = 0, this is because n0 = 0 if 0 2 We. Assuming the statement tobe true for all i < k, we see that if k 2 We, then either k 2 We;tk�1

, orelse nk = k. In either case, we have k 2 We;tk . The key fact here is thatfor any (s0; : : : ; sk) 2 Tf(e), sk � tk. To see this, let (s0; : : : ; sk) 2 Tf(e), let(m0; : : : ;mk) be the associated sequence of elements of We, and suppose byway of contradiction that sk � tk. It follows from the de�nitions of Tf(e) andof t0; : : : ; tk that in fact si = ti and mi = ni for all i � k. Thus Tf(e) has thesequence (t0 + 1; t1 + 1; : : :) as a bounding function.

Suppose now that We is computable. Then the sequence t0 < t1 < : : : isalso computable and thus Tf(e) is computably bounded by. Now suppose thatTf(e) is bounded by some computable function h. Then we must have tk < h(k)for each k. It follows that k 2We () k 2We;h(k), so that We is computable.

(b) This set is �03, since Te is a. c. b. if and only if Te is �a-almost boundedfor some total computable function �a.

For the completeness, use the argument given in (3) above. We may assumethatWe is in�nite, since otherwise the argument goes through trivially. Clearly,if We is computable, then Tf(e) is computably bounded and therefore a. c. b.as well. If Tf(e) is almost bounded by the computable function g, let k be largeenough so that for j�j > k, �i 2 Tf(e) ! i < g(�) and let � = (t0; t1; : : : ; tk).Then we can recursively de�ne a bounding function h(i) � t(i) by letting h(i) =t(i) for i � k and, for each j � k,h(j + 1) = maxfg(h�_(sk+1; : : : ; sj)i) : si � h(i) for each i with k < i � jg:


It follows as above that We is computable.

Theorem VI.1.6. (a) fe : Te is boundedg is �03 complete;

(b) fe : Te is almost boundedg is �04 complete.

Proof. (a) This set is �03, since

Teis bounded () (8�)(9n)(8m > n)(�_m =2 Te):For the completeness, we de�ne a reduction of !nCof as follows. Let �(e;m; s) =(least n > m)(n =2 We;s). This is a primitive recursive de�nition since n 2We;s ! n � s. Then, using the s-m-n Theorem, de�ne the tree Tf(e) by

Tf(e) = f0m : m 2 Ng [ f0m_(s+ 1) : �(e;m; s+ 1) > �(e;m; s)g:Then Tf(e) will be a �nite-branching tree if and only if, for each m, there areonly �nitely many s such that 0m_(s + 1) 2 Tf(e). Now if We is not co�nite,then for each m there is a minimal n > m such that n =2 We. It follows thatlims �(e;m; s) = n, so that �(e;m; s + 1) > �(e;m; s) for only �nitely manys, which will make Tf(e) �nite-branching. On the other hand, if We is co�niteand we choose m so that n 2 We for all m > n, then it is clear that there willbe in�nitely many s such that �(e;m; s + 1) > �(e;m; s), so that 0m will havein�nitely many successors and Tf(e) will not be �nite-branching. Thus we have

e =2 Cof () Tf(e) is bounded:

(b) This set is �04, since

Te is almost bounded () (9k)(8�)(9n)(8m > n)(j�j > k ! �_m =2 Te):For the completeness, �rst modify the proof of part (a) by letting Tg(e) contain0m for eachm together with 0m_(s+1) ifm is the least such that �(e;m; s+1) >�(e;m; s)g This modi�cation ensures that Tg(e) is always almost bounded, sinceonly for the largestm =2We will there be in�nitely many s with 0

m_s+1 2 Tg(e).By the previous argument, Tg(e) will be bounded if and only if e =2 Cof . NowS be an arbitrary �04 set and suppose that a 2 S () (9k)R(a; k); where R is�03. By the usual quanti�er methods, we may assume that R(a; k) implies that

R(a; j) for all j > k. By the argument above, there is a computable functionh such that R(a; k) if and only if Th(a;k) is bounded and such that Th(a;k) isalmost bounded for every a and k. Now simply de�ne

T�(a) = f0n : n < !g [ f(0k1)_� : � 2 Th(a;k)g:If a 2 S, then Th(a;k) is bounded for all but �nitely many k and is almostbounded for the remainder. Thus T�(a) is almost bounded. If a =2 S, then, forevery k, Th(a;k) is not bounded, so that T�(a) is not almost bounded.

Index sets for decidable �01 classes will use the alternate de�nition that

P = [T ] for some computable tree T with no dead ends.


Theorem VI.1.7. (i) fe : Pe is decidable is �02 complete.

(ii) For any recursive g � 2, fe : Pe is decidable and g-bounded is �01 com-

plete.

(iii) For any computable g � 2, fe : Pe is decidable and g-a.b. is D02 complete.

(iv) fe : Pe is decidable and c. b. is �03 complete.

(v) fe : Pe is decidable and almost c. b. is �03 complete.

(vi) fe : Pe is decidable and bounded is �03 complete.

(vii) fe : Pe is decidable and almost bounded is �04 complete.

Proof. (i) This set is �02 since Te has no dead ends if and only if

(8� 2 Te)(9i)(�_i 2 Te):For the completeness, let C be a �0

2 set and R be a computable relation sothat

e 2 C () (8m)(9n)R(e;m; n):Put ; and (m) in Tf(e) for all m and, for any k, put (m;n)_0k 2 Tf(e) if andonly if R(e;m; n). Thus Tf(e) has no dead ends if and only if e 2 C. Note thatTf(e) is g-a.b. for any g.

(ii) This index set is �01, since a g-bounded tree Te has no dead ends if

(8�)(� 2 Te ! (9i � g(�)(� _ i 2 Te):For the completeness, observe that the proof given in Theorem VI.1.4 in factde�nes a tree with no dead ends.

(iii) This index set is D02 since the property of being g-bounded is �02 and

for any tree Ue, Te has no dead ends if and only if

(8� 2 Te)(9i)�_i 2 Te:For the completeness, let A = B \ C, where B is a �02 set and C is a �0

2

set. The tree Tj(e) is constructed in two parts. First, modify the construction ofpart (a) by putting 1_� 2 Tj(e) () � 2 Tf(e). Then Tf(e) \ I((1)) is alwaysg-a.b. and has no dead ends if and only if e 2 C. Then we use the functionh de�ned in Theorem VI.1.6 which has the property that e 2 B if and only ifTh(e) is g-a.b.. Note that Th(e) always has no dead ends. Then put 0_� 2 Tj(e)if and only if 0_� 2 Th(e).

(iv) For this and the remaining cases, the upper bound on the complexityfollows from part (0) above and complexity of the corresponding parts of Theo-rems VI.1.5 and VI.1.6. The completeness of the remaining cases follows from asimple modi�cation of the reductions used to prove the corresponding theoremsabove. That is, one needs only ensure that corresponding trees used in the re-ductions have no dead ends. This is easily accomplished by modifying any givenrecursive tree T to construct a new computable tree T 0 such that (i) 0k 2 T 0 forall k � 0 and (ii) for any n � 1, (�1; : : : ; �n) 2 T i� (�1+1; : : : ; �n+1)_ 0k 2 T 0for all k � 0.


Notions of boundedness for strong �02 classes are considered in the exercises.

Exercises

VI.1.1. Show that fe : T 2e is g-boundedg is �01 complete and

fe : T 2e is computably boundedg is �03 complete.VI.1.2. Show that for any �0

2 function g � 2, fe : T 2e is g-boundedg is �02 complete

and fe : T 2e is highly boundedg is �04 complete. (Hint: relativize fromTheorems VI.1.4.

VI.1.3. Give the details for the proofs of Theorem VI.1.7(4-7).

VI.2 Cardinality

In this section we classify index sets corresponding to cardinality properties of�01 classes.

Theorem VI.2.1. Let g � 2 be a computable function from N� to N.

(a) fe : Pe is g-bounded and nonemptyg is �01 complete;

(b) fe : Pe is g-bounded and emptyg is D01 complete;

(c) (fe : Pe is g-bounded and nonemptyg; fe : Pe is g-bounded and emptyg) is(�0

1;�01) complete.

Proof. We observe �rst that the relation � 2 Ext(Te) has a �01 characterization.

That is,

� 2 Ext(Te) () (8n)(9�)[j� j = n & � � � & � 2 Te];where the quanti�er \(9�)" is bounded by g in the following sense. Let h(0) =g(;) and for each n, let h(n + 1) = maxfg(�) : � 2 f0; 1; : : : ; h(n)gng. Then� 2 Te implies �(n) � h(n) for all n, so that the quanti�er \9�" above may bereplaced by \(9� 2 f0; 1; : : : ; h(n� 1)gn)".

(a,c) Now Pe is nonempty if and only if ; 2 Ext(Te).For the double completeness, de�ne a reduction f for a given �0

1 set A sothat Ph(e) is always g-bounded and is nonempty if and only if e 2 A. Let Rbe a computable relation so that e 2 A () (8n)R(e; n). Then the map maybe de�ned by putting 0n 2 Tf(e) () R(e; n) and putting no other strings inTh(e).

(b) We see that this set is D01 by part (a) and Theorem VI.1.4.

For the completeness, let C = B n A, where A and B are �01 sets and let R

and S be computable relations so that

e 2 A () (8n)R(e; n) and e 2 B () (8n)S(e; n):Then a reduction f of C to our set is given by putting � 2 Tf(e) if and only ifeither

VI.2. CARDINALITY 147

(i) (8i < j�j)[R(e; i) & �(i+ 1) < g(�di))] or(ii) � = (1 + g(;) + n) where not S(e; n) and (8i < n)(S(e; i)).

Clearly Tf(e) is g-bounded if and only if if e 2 B. Similarly Tf(e) is non-empty if and only if e 2 A. Thus Tf(e) is g-bounded and empty if and only ife 2 B �A.Theorem VI.2.2. For any computable g � 2,

(a) fe : Pe is almost g-bounded and nonemptyg is �02 complete;

(b) fe : Pe is almost g-bounded and emptyg is D02 complete.

Proof. It follows from the proof of Theorem VI.1.4 that the relation \Te is g-bounded above k" is �0

1. Now modify the proof of Theorem VI.2.1 to de�neh for � 2 Te with j�j = k by h(�; k) = maxf�(i) : i < kg and for all n,h(�; k + n+ 1) = maxfg(�) : � 2 f0; 1; : : : ; h(k + n)gk+ng. Then for �j = k,

� 2 Ext(Te) () (8n)(9� 2 f0; 1; : : : ; h(�; k + n)gk+n)(� � � & � 2 Te):

Thus the relation \� 2 Te" is �01 when restricted to � of length k when Te is

g-bounded above k. Then for g-bounded Te, Pe is nonempty if and only if thereexists k such that Te is g-bounded above k and there exists � 2 Te such that� 2 Ext(Te). This gives the upper bound on the complexity for both parts.

(a) For the completeness, use the same reduction as in the proof of TheoremVI.1.4.

(b) For the completeness, let A = B \ C, where B is a �02 set and C isa �0

2 set. Suppose that b 2 B () (9m)(8n)R(b;m; n) and c 2 C ()(8m)(9n)S(c;m; n), where R and S are computable. De�ne the function �(b; n)to be the least m < n such that R(b;m; n) for all n0 < n (or �(b; n) = n ifthere is no such m), so that b 2 B if and only if �(b; n) is eventually constant.De�ne the tree Tf(b) recursively as follows. Every string (m) of length 1 is inTf(b). If � 2 Tf(b) is of odd length 2s+1, then �_i 2 Tf(b) if either i < g(�) or�(b; s + 1) > �(b; s). If � 2 Tf(b) is of even length 2s + 2 and �(0) = m, then�_i 2 Tf(e) if i = 0 and either s < m or, for all n � s, :S(b;m; n). Observethat allowing an extension when s < m in the second part of the de�nition ofTf(b) means that we have always have arbitrarily long strings in Tf(b).

Suppose �rst that b 2 B. Then there is some k such that �(b; s+1) = �(b; s)for all s � k. It follows that Tf(b) is g-bounded above k. Next suppose thatb 2 C. Then, for any m, choose nm such that S(b;m; n). It follows that thereis no � of length 2nm + 3 beginning with �(0) = m in Tf(b). It follows thatPf(b) is empty in this case. Thus if b 2 A, then Pf(b) is g almost bounded andis empty. If b =2 B, then, since Tf(b) has arbitrarily long strings, it will not bealmost bounded by g. If b =2 C, then Pf(b) will be nonempty, since for any msuch that :S(b;m; n) for all n, we will have m_0! 2 Pf(b).Theorem VI.2.3. (a) fe : Pe is c. b. and emptyg is �02 complete;


(b) fe : Pe is c. b. and nonemptyg is �03 complete.

Proof. (a) The case of c. b. empty classes is equivalent to bounded empty classesand is treated in Theorem VI.2.10 below.

(b) This set is �03, since Pe is c. b. and nonempty if and only if

(9a)[a 2 Tot & e 2 Pe is �a-bounded and nonempty)]:

For the completeness, modify the reduction f from the proof of Theorem VI.1.5as follows. For any � = (s0; s1; : : : ; sk�1) 2 Tf(e), add �_0k to Tf 0(e) wheneverthere is no s < k such that �_s 2 Tf(e). It is clear that Pf 0(e) will containexactly one element for each e.

Theorem VI.2.4. (a) fe : Pe is bounded and emptyg is �02 complete;

(b) fe : Pe is bounded and nonemptyg is �03 complete.

Proof. (a) The case of bounded empty classes is a special one, since [T ] isbounded and empty if and only if T is �nite, that is, if and only if

(9n)(8�)[� 2 Te !< � >< n]:

For the completeness, de�ne a reduction f of Fin to by letting

Tf(e) = f;g [ f(hn; si) : n 2We;s+1 nWe;sg:

(b) Recall that fe : Te is �nite-branchingg is �03 complete. Now if Te is

�nite-branching, then for any �,

� 2 Ext(Te) () (8i)(9�)[� � � & � 2 Te & j� j � i]:

Thus our set is �03. For the completeness, use the same reduction f as given in

the proof of Theorem VI.1.6, since Pf(e) = f0!g for every e.

Theorem VI.2.5. fe : Pe is a. b. and emptyg and fe : Pe is a. b. and nonemptygare both �04 complete.

Proof. For the \nonempty" case, the set is �04, since Pe is a. b. and nonemptyif and only if

(9k)(9�)[B(k; e) & j�j � k & � 2 Ext(Te)]:For the completeness, use the same reduction as given in Theorem VI.1.6(b).For the \empty" case, the set is �04, since Pe is a. b. and empty if and only

if

(9k)[B(k; e) & (8�)(j�j = k ! � =2 Ext(Te))]:For the completeness, modify the proof of Theorem VI.1.6(b). First de�ne

Tg0(e;k) = fmi : i � m+ kg [ f(mm+k)_s+ 1) : (0m)_s+ 1 2 Tg(e)g:


where g is the function de�ned in Theorem VI.1.6(b). Note that [Tg0(e;k)] isalways empty and that Tg0(e;k) is always almost bounded. Tg0(e;k) is never ac-tually bounded because the empty string has in�nitely many successors (m) foreach m. However, Tg0(e;k) clearly has the following properties.

(i) If e =2 Cof , then every node except ; has �nitely many successors.(ii) If e 2 Cof , then for some m, mm+k has in�nitely many successors.Now let S be any �04 set and suppose that a 2 S () (9k)R(a; k); where

R is �03. By the usual quanti�er methods, we may assume that R(a; k) implies

that R(a; j) for all j > k. Since Cof is �03 complete set, it follows from theabove discussion above that there is a recursive function h0 such that Th0(a;k) isalmost bounded for all a; k, [Th0(a;k)] is empty for all a; k and

(iii) if R(a; k), then every node in Th0(a;k) except ; has �nitely many succes-sors and

(iv) if :R(a; k), then for some m, mm+k has in�nitely many successors.Now de�ne T (a) = f(k)_� : � 2 Th0(a;k)g: [T (a)] is empty since each

[Th0(a;k)] is empty. If a 2 S, then for all but �nitely many k, every node inTh0(a;k) except ; has �nitely many successors, and for the remainder, Th0(a;k) isalmost bounded. Thus T (a) is almost bounded. If a =2 S, then, for every k,Th0(a;k) has a string of length � k with in�nitely many successors, so that T (a)is not almost bounded.

Theorem VI.2.6. (fe : Pe = ;g; fe : Pe 6= ;g) is (�11;�11) complete.

Proof. The upper bounds on the complexity follow from the fact that

Pe 6= ; () (9x)(8n)xdn 2 Te:For the completeness, let A be a �11 set, so that, by the normal form theorem

(see Hinman [87, p. 84]), there is a primitive recursive relation R such that, forall a,

a 2 A () (9x)(8n)R(a; xjn):Then we may de�ne Tf(a) = f� : R(a; �)g by the s-m-n Theorem. Then

a 2 A () Pf(a) 6= ;, as desired.Next we consider index sets for the cardinality of strong �0

2 classes.

Theorem VI.2.7.

(i) (fe : P 2e is g-bounded & emptyg); fe : P 2

e is g-bounded & nonemptyg) is(�02;�

02) complete for any computable g � 2.

(ii) fe : P 2e is c. b. and nonemptyg is �03 complete and fe : P 2

e is c. b. and emptygis �02 complete.

(iii) fe : P 2e is bounded and nonemptyg is �0

3 complete andfe : P 2

e is bounded and emptyg is �02 complete.

(iv) fe : P 2e 6= ;g; fe : P 2

e = ;g) is (�11;�11) complete.


(v) (fe : P 2e is g-bounded and emptyg; fe : P 2

e is g-bounded and non emptyg)is (�02;�

02) complete for any g � 2 which is computable in 00.

(vi) fe : P 2e is highly bounded and nonemptyg is �04 complete and

fe : P 2e is highly bounded and emptyg is is �02 complete.

Proof. The upper bounds on the complexity are routine to check.(i) For the completeness, we de�ne a reduction f such that P2;f(e) is always

a class of sets and such that e 2 Inf if and only if P2;f(e) is nonempty. Simplylet 0n 2 T2;f(e) if and only if there exist a0 < � � � < an�1 each in We.

(ii,iii,iv) In each case, the completeness follows exactly as in Theorems VI.2.3,VI.2.4 .

(v.vi) These are simply relativizations of Theorems VI.2.1 and VI.2.4.

Next we consider �nite cardinality. Results related to almost boundednessare relegated to the exercises.

Theorem VI.2.8. For any positive integer c and any computable function g �2,

(a) (fe : Pe is g-bounded & jPej > cg; fe : Pe is g-bounded & jPej � cg) is(�02;�

02) complete;

(b) fe : Pe is g-bounded and Card(Pe) = c+ 1g is D02 complete;

c) fe : Pe is g-bounded and Card(Pe) = 1) is �02 complete.

Proof. fe : Pe is g-bounded and Card(Pe) > cg is �02, since if Pe is g-bounded,then Card(Pe) > c) if and only if there exist k and incomparable�1; �2; : : : ; �c+1 2 !k such that each �i 2 Ext(Te). For c = 0, this set is in fact�01 by Theorem VI.2.1. These facts imply the upper bounds on the complexity.To prove the �02 completeness for cardinality > c, we de�ne a reduction f of

! n Tot, as follows. For each e, let � = 0m01r0m11r : : : 0mk�11r0mk1t 2 Tfc(e) ifand only if the following conditions are satis�ed.

(i) 1 � r � c and t � r.(ii) for each i < k, if �e;j�j(i) #, then �e;j�j(i) = mi.(iii) if �e;j�j(k) #, then �e;j�j(k) � mk.

Thus if �e is total, then Pf(e) has exactly c elements, 0�e(0)1r0�e(1)1r : : : for

1 � r � c. On the other hand, if �e is not total, then Pfc(e) will be in�nite.Note that the tree Tfc(e) is always g-bounded, since it is a binary tree. Thisreduction shows both the double completeness result as well as the completenessfor cardinality = 1. Note that since Tot is jPi02 complete, it follows that for any�02 set C, there is a reduction hc of C so that card(Phc(e)) = c if e 2 C and

Phc(e) is in�nite otherwise.To prove the D0

2 completeness for cardinal = c+ 1, let A = B \ C where Bis �02 and C is �0

2, let h1 be a reduction of ! � B (as above) so that Ph1(e) isin�nite if e 2 B and card(Pf(e)) = 1 otherwise. Let hc+1 be the reduction of Cdescribed above. Then a reduction � of A to fe : Pe is g-bounded = c+1g maybe given by de�ning T�(e) = Th1(e) � Thc+1(e).


Remark. It follows from Theorem VI.2.8 that fe : card(Pe \ f0; 1g!) > cgis �02 complete, that fe : card(Pe\f0; 1g!) = 1g and fe : card(Pe\f0; 1g!) � cgare both �0

2 complete, and that fe : card(Pe \f0; 1g!) = c+1g is D02 complete.

Theorem VI.2.9. For any positive integer c, fe : Pe is c. b. and Card(Pe) >cg, fe : Pe is c. b. and Card(Pe) � cg, and fe : Pe is c. b. and Card(Pe) =cg, are all �03 complete.

Proof. The g-bounded case above is uniformly �02. Then

Pe is c. b. & jPej > c () (9a)[a 2 Tot & Peis �a-bounded & jPej > c:

A similar argument gives the upper bound for cardinality � c.For the �03 completeness of cardinality > c, recall the modi�ed function g

from the proof of Theorem VI.2.3 such that card(Pg(e)) = 1 for all e and suchthat Pg(e) is c. b. if and only if e 2 Rec. Fix c and let T be a binary tree suchthat [T ] has exactly c + 1 elements. Then let Tk(e) = Tg(e) T . Then k is areduction of Rec to fe : Pe is c. b. and Card(Pe) > cg. This same reduction kalso works for cardinality = c+ 1 and cardinality � c+ 1. Note here that sinceT is binary, Tg(e) T will be r. b. if Tg(e) is r. b. and since [T ] is nonempty,Tg(e) T will be not r. b. if Tg(e) is not computably bounded.

Theorem VI.2.10. For any positive integer c,

(a) fe : Pe is bounded & jPej � cg and fe : Pe is bounded & jPej = 1g areboth �0

3 complete;

(b) fe : Pe is bounded & jPej > cg and fe : Pe is bounded & jPej = c + 1gare both D0

3 complete.

Proof. Let us de�ne here a �03 relation C(c; k; e) such that C(c; k; e) holds i�there is a j � k such that there exist distinct �1; : : : ; �c+1 2 !j \ Te suchthat for all n > j there exists �1; : : : ; �c+1 2 !n \ Te which extend �1; : : : ; �c+1respectively. Note that if Te is bounded above k, then C(c; k; e) implies thatcard(Pe) � c.fe : Pe is bounded and Card(Pe) � cg is �0

3, since

(Pe is bounded & jPej � c) () Pe is bounded & :C(c; 0; e)]:

The upper bounds on the complexity for the other cases follows easily.For the completeness results, let A = B \ C, where B is a �0

3 set and C isa �03 set. It follows from Theorem VI.1.6 that there is a reduction f such thatcard(Pf(e)) = 1 for all e and such that Tf(e) is bounded if and only if e 2 B.This gives the �0

3 completeness in the cases of cardinality � c and cardinality= 1.

Suppose now that e 2 C () (9m)(8n)(9k)R(e;m; n; k), where R is com-putable. We will de�ne, uniformly in e, a computable tree Tg(e) such that Tg(e)is bounded for all e and such that Pg(e) has exactly 2 elements if e 2 C and


exactly one element (0!) otherwise. Let Tg(e) consist of all strings 0m together

with all strings (0m)_(r + 1; k1; k2; : : : kn) such that(i) either m = r = 0 or m > 0 and :R(e;m� 1; r; k) for all k < n;(ii) for all i � n, R(e;m; i; ki) and :R(e;m; i; j) for all j < ki.Clearly Tg(e) is always bounded and has at least one element 0!. There will

be another element (0m)_(r + 1; k1; k2); : : :) when m is the least number suchthat (8n)(9k)R(e;m; n; k). Thus e 2 C if and only if Pg(e) contains exactly twoelements and e =2 C if and only if Pg(e) contains exactly one element. By takinga disjoint union with a �xed set containing exactly c elements, we may obtain arecursive function gc such that e 2 C if and only if Pgc(e) contains exactly c+1elements and e =2 C if and only if Pgc(e) contains exactly c elements.

The reduction of A for cardinality > c is then given by Th(e) = Tf(e)Tgc(e);this also works for the case of cardinality = c+ 1.

Theorem VI.2.11. For any positive integer c,

(a) (fe : Card(Pe) > cg); (fe : Card(Pe) � cg) (�11;�11) complete;

(b) fe : Card(Pe) = cg is �11 complete.

Proof. IP (> c) is �11 uniformly in c since e 2 IP (� c) if and only if thereexist distinct x1; : : : ; xc 2 Pe. It then immediately follows that IP (� c) is �1

1

uniformly in c.For IP (= c), we recall from Theorem 1.3 that any countable �0

1 class containsa hyperarithmetic member. Thus we have

e 2 IP (= c) () e 2 IP (� c) & (9x1; : : : ; xc 2 HY P )(xc 2 Pe):

It then follows from the Spector-Gandy Theorem II.10.5 that IP (= c) is �11.

For the completeness, let A be a �11 set and let f be the function fromTheorem VI.2.6 which reduces A to fe : Pe 6= ;g and its complement to fe :Pe = ;g. Let T be a primitive recursive tree such that card([T ]) = c. Then areduction of A to fe : Card(Pe) > cg may be de�ned by Tg(e) = Tf(e) � T andthis simultaneously reduces N n A to fe : Card(Pe) � cg and in fact reducesN nA to fe : Card(Pe) = cg.

Theorem VI.2.12. Let c be a positive integer.

(a) For any computable function g � 2,

(i) (fe : Pe is decidable, g-bounded and Card(Pe) > cg is D01 complete;

(ii) fe : Pe is decidable, g-bounded and Card(Pe) � cg is �01 complete;

(iii) fe : Pe is decidable, g-bounded and Card(Pe) = c + 1g is D01 com-

plete.

(b) fe : Pe is decidable & jPej � cg, fe : Pe is decidable & jPej > cg andfe : Pe is decidable & jPej = cg are all �0

2 complete.


Proof. For a decidable tree Te, Pe has > c members if and only if there Tecontains incomparable �1; : : : ; �c+1. The upper bounds on the complexity nowfollow from Theorem VI.1.7.

(a) (i) For the D01 completeness of this set as well as the set in (iii), let

A = B\C, where B is a �01 set and C is a �01 set. Let R be a computable relation

so that e 2 C () (9n)R(e; n). It follows from the proof of Theorem VI.1.4that there is a computable function g such that if e 2 B, then Pg(e) = f0!gand is 2-bounded and if e =2 B, then Pg(e) is not g-bounded. Let Ph(e) =Pg(e) [ f0n1i+10! : i < c & R(e; n) & (8m < n):R(e;m)g.

(ii) Let Q be a �xed decidable �01 class having exactly c elements. For the

�01 completeness of this set, just let Pg(e) = Ph(e) Q.(iii) For the �0

1 completeness, use the reduction h given in the proof ofTheorem VI.1.4.

(b) For the �02 completeness when cardinality equals 1, modify the reduction

given for the �02 set C in the proof of part (0) of Theorem VI.1.7 by putting ;

in Tf(e) and by putting (n0; n1; : : : ; nk�1) 2 Tf(e) if and only if, for all i < k,ni is the least such that R(e; i; n). Then if e 2 C, Tf(e) will be a decidable treewith exactly one element and if e =2 C, then Tf(e) will be a �nite tree and hencewill not be decidable.

For the �02 completeness in the other two sets, let Pg(e) = Pf(e) Q as in

(ii) above.

There are �ve cases to consider when we examine the possible in�nite car-dinality of a �0

1 class P . P may be �nite, countably in�nite, or uncountable.Negations of two of these adds the notions of in�nite and of countable. Our �rstresult deals with the problem of �nite versus in�nite sets.

Theorem VI.2.13.

(a) For any computable function g � 2,(fe : Pe is g-bounded and in�niteg; fe : Pe is g-bounded and �niteg) is(�0

3;�03) complete.

(b) fe : Pe is c. b. and in�niteg is D03 complete and fe : Pe is c. b. and �niteg

is �03 complete.

(c) (fe : Pe is bounded and in�niteg; fe : Pe is bounded and �niteg) is (�04;�

04)

complete.

(d) (fe : Pe is in�niteg; fe : Pe is �niteg) is (�11;�11) complete.

Proof. In each case, the upper bound on these complexities follows from theuniformity of previous results (Theorems VI.2.8, VI.2.9, VI.2.10 and VI.2.11,since Pe is in�nite if and only if Card(Pe) > c for all c.(a) For the completeness, we de�ne a reduction of Cof to fe : Pe is g-bounded and �nitegwhich simultaneously reduces !nCof to fe : Pe is g-bounded and in�niteg. Let

Tf(e) = f0n : n 2 !g [ f0n10k : n =2We;kg:


Then Tf(e) is always a binary tree and it is easy to see that Pf(e) = f0!g [f0n10! : n =2Weg, so that f(e) 2 IP (g-bounded < @0) () e 2 Cofg.(b) For the �03 completeness in the �nite case, use the same reduction f givenin part (a) above.

For the D03 completeness in the in�nite case, let A = B \C where B is a �03

set and C is a �03 set and let f be the reduction given above applied to ! n C,

so that Tf(a) is always a binary tree and such that Pf(a) is �nite if and only ifa =2 C. It follows from the proof of Theorem VI.2.3 that there is a recursivefunction g such that Pg(a) is always a singleton and is r. b. if and only ifa 2 B. Then the reduction of A to fe : Pe is c. b. and in�niteg is given byTh(a) = Tf(a) Tg(a).

(c) For the double completeness, let A be a �04 set and let C be a �03 set

so that e 2 A () (8m)he;mi 2 C. We may assume that if e =2 A, thenhe;mi 2 C for only �nitely many m. Let the reduction f be given by the proofof Theorem VI.2.13, so that Tf(e;m) is always bounded and Pf(e;m) has oneelement if he;mi =2 C and has two elements if he;mi 2 C. De�ne the reductionh by Th(e) = mTf(e;m). Then Th(e) is always bounded and card(Ph(e) =Qm card(Pf(e;m), so that if e =2 C, then Th(e) is �nite, and, if e 2 C, then Th(e)

is uncountable.(d) For the completeness, just let f be the reduction of Theorem VI.2.6 and

let Th(a) = Tf(a) T , where T is a primitive recursive tree such that [T ] isin�nite.

Remark. It follows from part (a) that fe : Pe \ f0; 1g! is in�niteg is �03

complete and fe : Pe \ f0; 1g! is �niteg is �03 complete.Again we give only two cases for decidable classes.

Theorem VI.2.14.

(a) For any computable function g � 2, (fe : Pe is g-bounded, decidableg; fe :Pe is g-bounded, decidable and �niteg) is (�0

2;�02) complete.

(b) fe : Pe is decidable and in�niteg is �02 complete and

fe : Pe is decidable and �niteg is D02 complete.

Proof. The upper bounds on the complexities of the classes in both parts easilyfollow from the uniformity of the proof of Theorem VI.2.12 as in the previousresult.

For the �02 completeness in the g-bounded in�nite case, we de�ne a reductionf so that Pf(e) is always 2-bounded and decidable and is �nite if and only ifWe is�nite. Let Pf(e) contain 0! and, for each m, contain 0m10! if m = [k; s] wherek 2 We;s+1 n We;s. This also gives the �0

2 completeness in the (unbounded)in�nite case.

For the D02 completeness in the �nite case, let A = B\C where B is �02 and

C is �02. Using the reduction f , it follows that there is a reduction k of B such

that Pk(e) is always 2-bounded and decidable and is �nite if and only if e 2 B.Let h be the reduction from Theorem VI.2.12 so that Ph(e) is decidable and


has cardinality 1 if e 2 C and otherwise Ph(e) is not decidable. Then Pj(e) =Pk(e) Ph(e) de�nes a reduction of A to fe : Pe is �nite and decidableg.

The other three notions of in�nity produce the same level of complexityindependent of the level of boundedness and of the decidability.

Theorem VI.2.15. (fe : Pe is uncountableg; fe : Pe is countableg) is (�11;�11)

complete and fe : Pe is countably in�niteg is �11 complete and the same result

holds for bounded classes, for c. b. classes, and for g-bounded classes, and alsofor decidable classes and for strong Pi02 class, under each possible notion ofboundedness.

Proof. Recall from Theorem VI.1.6 that in each case the underlying set of esuch that Pe is suitably bounded is a �04 set. (For strong �0

2 classes, this alsoholds.) Then the property of being uncountable is �11, since for any tree Te, Pe isuncountable if and only if Pe has a perfect subset, i.e. if and only if there existsan embedding f from f0; 1g<! into Te which preserves the partial order �. Itfollows that the property of being countable is �1

1 and, by Theorem VI.2.13,that the property of being countably in�nite is also �1

1.For the completeness of fe : Pe is uncountableg, we de�ne a reduction of fe :

Pe 6= ;g to fe : Pe is 2-bounded and uncountableg as follows. De�ne the binarytree Tf(e) to consist of all strings 0n0_�_0 0n1_�_1 � � � _ 0nk�1_�k�10

n, where(n0; : : : ; nk�1) 2 Ue and for i < k, �i = (1) or �i = (1; 1). Then for any pathx 2 [Te], Tf(e) will contain uncountably many paths, so that if Pe is nonempty,then Pf(e) will be uncountable. If Pe is empty, then every path in Pf(e) will endin 0!, so that Pf(e) will be countable. Note that f also reduces fe : Pe = ;gto fe : Pe is 2-bounded and countableg. A reduction g of fe : Pe = ;g tofe : Pe is 2-bounded and countably in�niteg is then given by Tg(e) = Tf(e) � T ,where T is some primitive recursive binary tree with [T ] countably in�nite.

It is clear that these reductions work for each of the notions of boundednessand also for decidable classes, since the trees constructed have no dead ends.

Exercises

VI.2.1. A �01class P is said to be intrinsically bounded by g if the tree TP = f� :

P \ � 6= ;g is g-bounded. Show that fe : Pe is c. b.g is �11 complete, and

similarly for all other notions of boundedness.

VI.2.2. Show that fe : Pe is a. c. b. and emptyg and fe : Pe is a. c. b. and nonemptygare both �03 complete. Hint: For the completeness in the \empty" case,modify the reduction f from the proof of Theorem VI.1.6 by letting Tf 0(e)contain strings (n; s0; : : : ; sk�1) such that (s0; : : : ; sk�1) 2 Tf(e) and k < n,so that Tf 0(e) is a. r. b. if and only if Tf(e) is a. c. b. and Pf 0(e) is alwaysempty.

VI.2.3. Let c be a positive integer and g � 2 a computable function. Showthat fe : Pe is g-almost bounded and Card(Pe) > cg is �02 completeand both fe : Pe is g-almost bounded and Card(Pe) � cg and fe :Pe is g-almost bounded and Card(Pe) = cg are D0

2 complete.


VI.2.4. For any positive integer c, show that fe : Pe is a. c. b. and Card(Pe) >cg, fe : Pe is a. c. b. and Card(Pe) � cg, and fe : Pe is a. c. b. and Card(Pe) =cg, are all �03 complete.

VI.2.5. For any positive integer c, show that fe : Pe is a. b. and Card(Pe) > cg,fe : Pe is a. b. and Card(Pe) � cg and fe : Pe is a. b. and Card(Pe) =cg are all �04 complete.

VI.2.6. Show that (�01;�01) �m (fe : Pe is decidable, g-bounded and Card(Pe) >

cg; fe : Pe is decidable, g-bounded and Card(Pe) � cg).VI.2.7. For any positive integer c and any computable function g � 2,

(a) (fe : P 2e is g-bounded and Card(Pe) � cg) is �0

3 complete.

(b) fe : P 2e is g-bounded and Card(P 2

e ) = c+ 1g is D03 complete.

c) fe : P 2e is g-bounded and Card(P 2

e ) = 1) is �03 complete.

(a) (fe : P 2e is g-bounded and �niteg) is �04 complete

VI.3 Computable Cardinality

The computable cardinality of a class P is the cardinality of the set of com-putable members of P . Also, we say that P is computably nonempty if it hasa computable member and computably empty otherwise. In this section, weclassify the various index sets of classes with given computable cardinality con-ditions. The �rst theorem extends the result of Gasarch and Martin [77] thatthe property of being computably nonempty is �03 complete for c. b. �

01 classes.

Theorem VI.3.1. For any computable g � 2,

(a) (fe : Pe is g-bounded and computably nonemptyg; fe : Pe is g-bounded and computably emptyg)is (�03;�

03) complete

(b) fe : Pe is g-bounded, nonempty, and computably emptyg is �03 complete.

Proof. The upper bounds on the complexity follow from Theorem VI.2.1 andthe fact that Pe has a computable member if and only if

(9a)[a 2 Tot & (8n)(�ajn 2 Ue)]:For the completeness of the �rst two sets, we de�ne a reduction of Ext2 byletting Pf(a) equal

fx 2 f0; 1g! : �a � xg = fx : (8m)(8s)(8i)[�a;s(m) = i! x(m) = i]g:For the other completeness, let Q be a nonempty, binary �0

1 class with nocomputable members and let Ph(a) = Pf(a) �Q.Theorem VI.3.2. (a) fe : Pe is c. b. and computably nonemptyg is �03 com-

plete;

VI.3. COMPUTABLE CARDINALITY 157

(b) fe : Pe is c. b. and computably emptyg is D03 complete;

(c) fe : Pe is c. b., nonempty and computably emptyg is D03 complete.

Proof. The upper bounds on the complexity are easily checked.For the completeness in (a) use the reduction from Theorem VI.3.1. For

the completeness of the other two sets, let A = B \ C, where B is �03 and C

is �03. It follow from the proof of Theorem VI.3.1 that there is a computablefunction f such that Pf(a) is always c. b. nonempty and has a computablemember if and only if a =2 B. It follows from the proof of Theorem VI.1.5that there is a computable function g such that Pg(a) is c. b. if and only ifa 2 C. Then a reduction h of A to the sets in (b) and (c) may be given byPh(a) = Pf(a) (Pg(a) [ f0!g):

Proofs of the next two theorems are left for the exercises.

Theorem VI.3.3. (a) fe : Pe is bounded and computably nonemptyg is D03

complete;

(b) fe : Pe is bounded and computably emptyg is �03 complete;

(c) fe : Pe is bounded, nonempty and computably emptyg is �03 complete.

Theorem VI.3.4. (a) fe : Pe is and computably nonemptyg is �03 complete;

(b) fe : Pe is computably emptyg is �03 complete;

(c) fe : Pe is nonempty and computably emptyg is �11 complete.

Theorem VI.3.5. Let c be a positive integer and let g � 2 be a computablefunction.

(a) (fe : Pe is g-bounded and has computable cardinality > cg;fe : Pe is g-bounded and computable cardinality � cg) is (�03;�

03) com-

plete;

(b) fe : Pe is g-bounded, nonempty, and has computable cardinality = cg isD03 complete.

Proof. The upper bounds on the complexity are easily checked.(a) For the completeness, let f be the reduction from Theorem VI.3.1 such

that Tf(a) is always a binary tree and such that Pf(a) has a computable memberif and only if a 2 Ext2. Let Tb be a �xed binary tree such that Pb consists ofexactly c+ 1 computable elements. Then let Ph(a) = Pf(a) Pb.

(b) We begin with a construction for unbounded classes using the �03 com-pleteness of Rec.

De�ne the modulus function �a for the c. e. set Wa by

�a(i) = (least s)[Wa \ f0; 1; : : : ; ig =Wa;s \ f0; 1; : : : ; ig]:


It is easy to see that Wa is computable if and only if �a is computable. Weshall de�ne a tree Tf(a) so that Pf(a) = f�ag and hence Pa has exactly onecomputable element if and only if a 2 Rec.

The tree Tf(a) is de�ned so that a string � of length n is in Tf(a) if and onlyif each of the following three conditions is satis�ed.

1. (8i < n)(i 2Wa;n () i 2Wa;�(i)),

2. �0 > 0! 0 2Wa;�(0) nWa;�(0)�1, and

3. (80 < m < n)[�(m) > �(m� 1)! m 2Wa;�(m) nWa;�(m)�1)].

Now let A = B \ C, where B is �03 and C is �03. It follows from the

completeness of Rec and the above construction that Pf(a) is a singleton foreach a and has a (unique) computable member if and only if a 2 B. Let h bethe reduction from (a) such that Ph(a) has no computable members if a 2 Cand has at least 2 computable members if a =2 C. Let S be a �xed class withexactly c computable members. De�ne the computable function so that

P (a) = S (Pf(a) � Ph(a)):This gives a reduction of A to fe : Pe has exactly c computable membersg.

Finally, let k be the primitive recursive function given in Theorem III.7.7so that for any e, Pk(e) is a �0

1 class of sets such that there is a one-to-onecorrespondence between the members of Pe and the computable members ofPk(e). Then the composite function k( (a)) gives a desired reduction of A to

fe : Pe is g-bounded, nonempty, and has computable cardinality = cg;so that A is D0

3 complete.

The next three theorems essentially follow from the proof of Theorem VI.3.5.Details are left for the exercises.


(a) fe : Pe is c. b. and has computable cardinality > cg is �03 complete;

(b) fe : Pe is c. b. and has computable cardinality � cg is D03 complete;

(c) fe : Pe is c. b. and has computable cardinality = cg is D03 complete.


(a) fe : Pe is bounded and has computable cardinality � cg is �03 complete;

(b) fe : Pe is bounded and has computable cardinality > cg is Dp3 complete;

(c) fe : Pe is bounded and has computable cardinality = cg is D03 complete.


VI.3. COMPUTABLE CARDINALITY 159

(a) (fe : Pe has computable cardinality > cg; fe : Pe has computable cardinality �cg) is (�03;�0

3) complete;

(b) fe : Pe has computable cardinality = cg is D03 complete.

For �nite versus in�nite computable cardinality, all versions of boundednessproduce index sets of the same complexity (excluding almost boundedness).

Theorem VI.3.9. (fe : Pe has �nite computable cardinalityg; fe : Pe has �nite computable cardinalityg)is (�04;�

04) complete and the similar result is true for g-bounded, c. b. and

bounded classes and also for strong �02 classes.

Proof. The upper bounds follow from the uniformity of Theorems VI.3.5, VI.3.6,VI.3.7 and VI.3.8.

For the completeness results, let A be a �04 set, so that for some �

03 relation

R,a 2 A () (8i)R(i; a):

As usual, R may be de�ned so that if a =2 A, then R(i; a) for only �nitely manyvalues of i. By the proof of part (i) of Theorem VI.3.1, there is a computablefunction f so that for each a and i, R(i; a) if and only if Pf(i;a) has a computablemember and Pf(i;a) is a binary class. Now let

T�(a) = f0n : n � 0g [ f0i1_ � : � 2 Uf(i;a):Then it is clear that a 2 A if and only if P�(a) has in�nitely many computablemembers and P�(a) is always a binary class.

Next we consider the problem of whether a �01 class has a member com-

putable in 00, or equivalently whether it has an element in �02. We omit the

almost computably bounded classes, since by Exercise 4.1.3, an a. c. b. �01 class

has a member computable in 00 if and only if it is nonempty.

Theorem VI.3.10. (a) (fe : Pe has a �02 memberg; fe : Pe has no �0

2 memberg)is (�04;�

04) complete and fe : Pe is nonempty but has no �0

2 memberg is�11 complete.

(b) fe : Pe is bounded and has a �02 memberg is �04 complete and

fe : Pe is bounded and has no �02 memberg

and

fe : Pe is bounded and nonempty but has no �02 memberg

are both �04 complete.

Proof. (a) By relativization of Theorem VI.0.4, the set Tot(00) = fe : �00e is totalis a �0

3 complete set and

Pe has a �02 member () (9a)[a 2 Tot(00) & (8n)(�0adn 2 Te)]:


The upper bounds on the complexity follow easily.For the completeness of the �rst two sets, let S be an arbitrary �04 set and

suppose that

a 2 S () (9m)(8n)(9i)(8j)R(a; i; j;m; n);for some recursive relation R. De�ne the reduction f so that

(m; i0; i1; : : :) 2 Pf(a) () (8n)(8j)R(a; in; j;m; n):It is clear that if a =2 A, then Pf(a) is empty and therefore has no membercomputable in 00. On the other hand, if a 2 A, then we may choose m so that(8n)(9i)(8j)R(a; i; j;m; n) and de�ne an in�nite path (m; i0; i1; : : :) 2 Pf(a)which computable in 00 by letting in be the least i such that (8j)R(a; i; j;m; n).

For the �11 completeness result, let A be a complete �11 set and let f be thereduction given in Theorem thm:iue so that Pf(a) is nonempty i� a 2 A. Thenlet g be the computable function such that Pg(a) = Pf(a) �Q, where Q is a �0

1

class with no members computable in 00.(b) The upper bounds on the complexity of the three sets follows as in part

(a). For the completeness of the �rst two sets, let h be the function given inTheorem III.7.7. Then Pe has a (respectively, no) �0

2 member if and only ifPh(e) is bounded and has a (resp. no) �0

2 member.For the �0

4 completeness of the third set, let Q be a nonempty bounded �01

class Q with no �02 members and let Pk(e) = Q� Ph(e).

Finally, we consider the computable and �02 cardinality of strong �

02 classes.

Theorem VI.3.11. (i) For any g � 2 which is computable in 00,

(fe : P 2e is g-bounded and comp. emptyg; fe : P 2

e is g-bounded and comp. nonemptyg)is (�03;�

03) complete and

fe : P 2e is g-bounded, nonempty, and computably emptyg

is �03 complete.

(ii) fe : P 2e is c. b. and computably nonempty is �03 complete and

fe : P 2e is c. b. and computably emptyg

andfe : P 2

e is c. b., nonempty, and computably emptygare D0

3 complete.

(iii) fe : P 2e is bounded and computably nonempty is D0

3 complete and

fe : P 2e is bounded and computably emptyg

andfe : P 2

e is bounded, nonempty, and computably emptygare �0

3 complete.

VI.4. INDEX SETS AND LATTICE PROPERTIES 161

(iv) (fe : P 2e is computably nonemptyg; fe : P 2

e is computably emptyg) is (�03;�03)

complete and fe : P 2e is nonempty but computably emptyg is �0

3 complete.

(v) fe : P 2e is highly bounded and computably nonemptyg is �04 complete,

fe : P 2e is highly bounded and computably emptyg;

and

fe : P 2e is highly bounded, nonempty and computably emptyg

are all �04 complete.

Proof. The upper bounds on the complexity are routine. The completeness ofparts (i){(iv) follow from previous results on �0

1 classes. The completeness inpart (v) follows easily from the �04 completeness of the property of being highlybounded.

Exercises

VI.3.1. Prove Theorem VI.3.3. Hint: Combine the reductions from TheoremsVI.1.6 and VI.3.1.

VI.3.2. Prove Theorem VI.3.4. Hint: Combine the reductions from TheoremsVI.2.6 and VI.3.1.

VI.3.3. Prove Theorems VI.3.6, VI.3.7 and VI.3.8.

VI.3.4. Show that the computable cardinality of a decidable class P is the sameas the cardinality of P , except when P is uncountable but has a count-able in�nite number of computable elements. Then formulate and provedecidable versions of Theorems VI.3.5, VI.3.6, VI.3.7 and VI.3.8.

VI.3.5. Prove Theorem VI.3.11(v).

VI.3.6. Show that (fe : P 2e has a �0

2 member; fe : P 2e has no �0

2 memberg) is(�04;�

04) complete and likewise for g-bounded, c. b. and bounded classes.

VI.3.7. Show that fe : P 2e is bounded and nonempty but has no �0

2 member is�04 complete and likewise for g-bounded and c. b. classes. However, for

unbounded classes the corresponding set is �11 complete.

VI.4 Index Sets and Lattice Properties

In this section, we consider in particular the complexity of the inclusion relationand the lattice operations on the family of �0

1 classes.

Lemma VI.4.1. There are primitive recursive functions i, u, s and psuch that, for all a and b, (a) P i(a;b) = Pa \ Pb; (b) P u(a;b) = Pa [ Pb; (c)P s(a;b) = Pa � Pb; (d) P p(a;b) = Pa Pb


Proof. (a) Here we de�ne i so that T i(a;b) = Ta \ Tb, that is, � i(a;b)(�) =�a(�) � �b(�).

(b) Similarly, T i(a;b) = Ta [ Tb.(c) Here T i(a;b) = f;g[f0_� : � 2 Tag[f1_� : � 2 Tbg, so that � = � u(a;b)

is de�ned by �(;) = 1 and

�(�) =

(�a(�(1); : : : ; �(j�j � 1)); if �(0) = 0

�b(�(1); : : : ; �(j�j � 1)); if �(0) = 1

A formal de�nition of u can now be obtained by the s-m-n Theorem.(d) Here T = T p(a;b) is de�ned to contain � if and only if (�(0); �(2); : : : ) 2

Ta and �(1); �(3); � � � 2 Tb. Details are left to the reader.Next we consider some aspects of the Veri�cation Problem for �0

1 classes,that is, fhi; ji : Pi � Pjg. This problem has been studied for various familiesof !-languages by Klarlund [107], Staiger [205] and others. More generally,Cenzer and Remmel [43] investigated index set problems concerning the size ofthe di�erence of two classes.

Theorem VI.4.2. (Staiger) Let g � 2 be a computable function.

(i) fha; bi : Pa; Pb are g-bounded and Pa � Pbg andfha; bi : Pa; Pb are g-bounded and Pa = Pbg are �0

2 complete.

(ii) fha; bi : Pa; Pb are g-bounded and P 2a � P 2

b g andfha; bi : P 2

a ; P2b are g-bounded and P 2

a = P 2b g are �0

3 complete.

Proof. (i) The �rst set is �02 (and hence also the second set), since

Pa � Pb () (8�)(� 2 Ext(Ta)! � 2 Tb):For the completeness, let b be given so that Pb = f0!g. Let A be a �0

2 set andlet R be a computable relation such that, for all i

i 2 A () (8m)(9n)R(i;m; n)De�ne the tree Tf(i) to contain 0m for all m and also

0m1n 2 Tf(i) () (8j < n):R(i;m; j)Then it is clear that

Pf(i) � Pb () Pf(i) = Pb () i 2 A:(ii) Recall that � 2 T 2e () (8� � �)� 2We), which is a �01 relation. Now letGn = f� : j�j = n & (8m < n)�(m) < g(m)g. Then

� 2 Ext(T 2e ) () (8n)(9� 2 Gn)(� � �& � 2 T 2e );and this relation is �0

2, since the quanti�er (9� 2 Gn) is bounded. It followsas in (i) that the two sets are �0

3. For the completeness, simply relativize theargument from (i). That is, let R now be a �0

1 relation and note that thecorresponding class P 2

f(i) will indeed by a strong �02 class.

VI.4. INDEX SETS AND LATTICE PROPERTIES 163

The complexity of being \almost equal" or \almost a subset" is covered byconsidering di�erences of classes. The following results are taken from [43].

Theorem VI.4.3. For any computable function g and any �nite k � 1:

(i) fha; bi : Pa and Pb are g-bounded and card(Pa�Pb) � kg is �02 complete.

[(ii)] fha; bi : P 2a and P 2

b are g-bounded and card(P 2a � P 2

b ) � kg is �02

complete.

Proof. The completeness follows from Theorem VI.2.8 in (i) and from Exercise7 in (ii). Thus we need only see that index sets have the appropriate complexity.

(i) To see that (i) is �02, we claim that

card(Pa � Pb) � k () (8e)[Pb \ Pe = ; ! card(Pa \ Pe) � k]:Certainly if the condition is false, then card(Pa � Pb) > k. On the other hand,suppose that card(Pa � Pb) > k. Then there are k + 1 elements x0; x1; : : : ; xkin Pa � Pb. For each i, there is a basic open set Ui such that xi 2 Ui andUi \ Pb = ;. Then Pe = U0 [ � � � [ Uk contradicts the condition.

(ii) The argument is similar to (i).

Theorem VI.4.4. For any computable function g � 2:

(i) fha; bi : Pa and Pb are g-bounded and Pa � Pb is �niteg is �03 complete.[(ii)] fha; bi : P 2

a and P 2b are g-bounded and P 2

a �P 2b is �niteg is �0

4 com-plete.

Proof. In each case, the upper bound on the complexity follows from the unifor-mity of Theorem VI.4.3. The completeness follows from Theorem VI.2.13 andExercise 3.7.

Theorem VI.4.5. For any computable function g � 2 and any �nite k � 0:

(i) fha; bi : Pa and Pb are g-bounded and Pa�Pb has � k computable membersgand fha; bi : P 2

a and P 2b are g-bounded and P 2

a�P 2b has � k computable membersg

are �03 complete.

(ii) fha; bi : Pa; Pb are g-bounded and Pa�Pb has < @0 computable membersgand fha; bi : P 2

a ; P2b are g-bounded and P 2

a�P 2b has < @0 computable membersg

are �04 complete.

Proof. (i) The completeness follows from Theorem VI.3.5. For the upper boundson the complexity, we claim that

Pa�Pb has � k computable members () (foralle)[Pb\Pe = ; ! card(Pa\Pe) � k]:The key here is that if there are k + 1 computable elements x0; x1; : : : ; xk inthe di�erence, then fx0; : : : ; xkg is a �0

1 class. Details are left to the reader. Asimilar argument covers strong �0

2 classes.(ii) The upper bounds follow from the uniformity of (i) and completeness

follows from Theorem VI.3.9.


Theorem VI.4.6. fe : Pe � f0; 1gN & Pe is thing and fe : Pe � f0; 1gN & Pe is thingare �0

4 complete sets.

Proof. First observe that Pe is minimal if and only if, for all a, either Pe \Pa is�nite or Pe�Pa is �nite. Thus the property of being minimal is �0

4 by TheoremVI.4.4. Pe is thin if and only if, for every a, there exist �1; : : : ; �k such thatPe \ Pa = Pe \ (I(�1) [ � � � [ I(�k)). Thus the property of being thin is �0

4 byTheorem VI.4.2.

For the completeness, the proof of Theorem III.8.3 may be modi�ed for agiven �0

4 set C to de�ne a reduction f so that Pf(c) is thin and minimal if c 2 Cand otherwise is neither. The modi�cation uses a uniform version of TheoremVI.2.13 that fe : Pe is �niteg is �03 complete to add a new limit point to P ifc =2 C and otherwise to add only isolated points. That is, let c 2 C if and onlyif Pg(c;e) is �nite for all e. Now de�ne the computable function f so that, ateach stage s of the construction of Tf(c), there is a copy of T sf(c;e) below �se butnot below �se+1. If �

se is abandoned, we just extend the �nitely many branches

with 0's. Now if c 2 C, then only �nitely many new points have been addedbelow any �e, so that no new limit point has been added. Then Pf(c) will be aminimal thin class as before. If c =2 C, then for some e, we have attached anin�nite �0

1 class, a copy of Pg(c;e) below �e. Thus there is a second limit pointbelow �e. It follows that Pf(c) is not minimal or thin.

Exercises

VI.4.1. Show that fe : f0; 1gN � Peg is �01 complete.

VI.4.2. Give the details in the proof of Lemma VI.4.1.

VI.4.3. Show that for any cardinal c, fha; bi : card(Pa \ Pb) � cg has the samecomplexity as fa : card(Pa) � cg and similarly for cardinality = c.

VI.4.4. Show that fha; bi : Pa; Pb are g-bounded and Pa \Pb 6= ;g is �01 complete

and fha; bi : P 2a ; P

2b are g-bounded and P 2

a \ P 2b 6= ;g is �0

2 complete.

VI.4.5. Show that fhe; h�ii : I(�) � Peg and fhe; h�ii : Pe is g-bounded and I(�) �Peg are �0

1 complete (for any computable g).

VI.4.6. Show that fhe; h�ii : I(�) � P 2e g and fhe; h�ii : P 2

e is g-bounded and I(�) �Peg are �0

2 complete (for any computable g).

VI.4.7. Show that fha; bi : Pa � Pbg is �11 complete.

VI.4.8. For any computable function g and any �nite k:

(i) fha; bi : Pa and Pb are g-bounded and card(Pa � Pb) = 1g is �02

complete. [(ii)] fha; bi : Pa and Pb are g-bounded and card(Pa �Pb) = k + 1g is D0

2 complete.

VI.4.9. For any computable function g and any �nite k:

VI.5. SEPARATING CLASSES 165

(i) fha; bi : P 2a and P 2

b are g-bounded and card(P 2a � P 2

b ) = 1g is �03

complete. [(ii)] fha; bi : P 2a and P 2

b are g-bounded and card(P 2a �

P 2b ) = k + 1g is D0

3 complete.

VI.4.10. For any computable function g � 2, fha; bi : Pa and Pb are g-bounded and Pa�Pb is countable g and fha; bi : P 2

a and P2b are g-bounded and P 2

a�P 2b is countableg

are �11 complete.

VI.5 Separating Classes

Recall that, for any two sets A andB, the class S(A;B) contains those separatingsets C such that A � C and B \C = ;. When A and B are c. e. sets, S(A;B)is a �0

1 class of sets. There are twp special cases here. The class of supersets ofWe is S(We; ;) and the class of sets disjoint fromWe, is S(;;We). Note that theclass S(A;B) of separating sets has the following property, which we shall referto as being closed under between-ness, that, for any sets X;Y; Z, if X � Y � Zand X;Z 2 P , then Y 2 P .Lemma VI.5.1.

1. For any nonempty �01 class P � f0; 1gN, the following are equivalent.

(a) P is the class of subsets of a �01 set A.

(b) P is the class of subsets of some set A.

(c) P is closed under subsets and under union.

2. For any nonempty �01 class P � f0; 1gN, the following are equivalent.

(a) P is the class of supersets of a �01 set A

(b) P is the class of supersets of some set A

(c) P is closed under supersets and under intersection.

3. For any �01 class P � f0; 1gN, the following are equivalent.

(a) P is the class of separating sets of some pair A;B

(b) P is the class of separating sets of some pair A;B of r. e. sets.

(c) P is closed under union, intersection and between-ness.

Proof. (i) Certainly (a) implies (b) and (b) implies (c). To show that (c) implies(a), suppose that P is closed under subsets and under union and let

A = fn : (9x)[x 2 A & x(n) = 1]g:We claim that P = P(A). First we show by induction that A\f0; 1 : : : ; n�1g 2P for all n. For n = 0, this follows from the subset property and the fact thatA is nonempty. Now suppose that A \ f0; : : : ; n � 1g 2 P . If n =2 A, then


A \ f0; : : : ; ng = A \ f0; : : : ; n � 1g 2 P by assumption. If n 2 A, then byde�nition there is some B 2 P with n 2 B and then, by closure under union,A [ B 2 P and by closure under subset, A \ f0; : : : ; ng 2 P . Finally P is aclosed set and limn!1A \ f0; : : : ; ng = A, so that A 2 P as desired.

Now let P = P(A) = [T ] for some computable tree and recall that Ext(T )is a �0

1 set. Then observe that A may be de�ned by:

n 2 A () () (9� 2 f0; 1gn + 1)[� 2 Ext(T ) & �(n) = 1]:

(ii) This is left as an exercise.(iii) Observe that S = S[A;B] if and only if S is the intersection of the class

of supersets of A with the class of subsets of f0; 1gN �B. Details are left as anexercise.

Lemma VI.5.2. Suppose that P = [T ] where T is a tree with no dead ends.Then

1. P is closed under subsets if and only if for every � � � , if � 2 T , then� 2 T .

2. P is closed under supersets if and only if for every � � � , if � 2 T , then� 2 T .

3. P is closed under union if and only if, for every � and � in T , � [ � 2 T .4. P is closed under intersection if and only if, for every � and � in T ,

� \ � 2 T .Proof. The proof is left as an exercise.

Theorem VI.5.3. 1. Sub = fe : Pe = S(;;Wb) for some bg is �02 com-

plete.

2. Sup = fe : Pe = S(Wa; ;) for some ag is �02 complete.

3. Sep = fe : Pe = S(Wa;Wb) for some a; bg is a �02 complete set.

Proof. (i) By Lemma VI.5.1, e 2 Sub if and only if Pe is closed under subsetsand under intersection. Thus, by Lemma VI.5.2, e 2 Sub if and only if Pe is2-bounded and

(8�; � 2 f0; 1g�)[[(� � � & � 2 Ext(Te))! � 2 Ext(Te)] &[(� 2 Ext(Te) & � 2 Ext(Te))! � \ � 2 Ext(Te)]]:

For the completeness, let A be a �02 set and R a recursive relation so that

a 2 A () (8m)(9n)R(a;m; n):De�ne the �0

1 class Pf(a) as follows:

x 2 Pf(a) () (8m)[x(2m) = x(2m+1) = 0 _ (x(2m) = x(2m+1) = 1 & (8n):R(a;m; n):

VI.5. SEPARATING CLASSES 167

Now if a 2 A, then Pf(a) = f0!g and f(a) 2 Sub. If a =2 A, then choose msuch that (8n):R(a;m; n). Then f2m; 2m + 1g 2 Pf(e), but f2mg =2 Pf(e), sof(e) =2 Sub.

(ii) For any sets B;C,

C 2 S(;; B) () C \B = ; () B � ! n C:Thus e 2 Sub () f(e) 2 Sup, where �f(e)(n) = 1 � �e(n) and similarlye 2 Sup () f(e) 2 Sub. The result now follows from (i).

(iii) It follows easily from Lemma VI.5.1 that Sep is a �02 set and the com-

pleteness follows from part (i).

Theorem VI.5.4. (i) fe 2 Sep : Pe 6= ;g is �02 complete.

(ii) fe 2 Sep : Pe is nonempty but has no computable membersg is �03 com-

plete.

(iii) fe 2 Sep : Pe has a computable memberg is �03 complete.

Proof. (i) This set is �02 by Theorem VI.5.3 and Theorem VI.2.1. The com-

pleteness follows by the proof of part (i) of Theorem VI.5.3.(ii) This set is �0

3 by Theorem VI.5.3 and Theorem VI.3.1. For the com-pleteness, we de�ne a reduction of Rec to Sep. This is done by uniformizing theproof from Odifreddi [163] of Shoen�eld's theorem that every noncomputablec. e. Turing degree contains a recursively inseparable pair of c. e. sets. That is,de�ne computable functions f(e) and g(e) so that

n 2Wf(e) () (9s)(n)1 2We;s+1 nWe;s & �(n)2;s(n) ' 0, andn 2Wg(e) () (9s)(n)1 2We;s+1 nWe;s & �(n)2;s(n) ' 1.Then Wf(e) and Wg(e) are a disjoint pair of c. e. sets with the following two

properties:

(a) Wf(e) and Wg(e) have the same Turing degree as We;

(b) For any separating set D such that Wf(e) � D and Wg(e) \ D = ;, wehave We computable in D.

It follows from (a) that if We is computable, then the pair Wf(e) and Wg(e)

have the computable separating set Wf(e). It follows from (b) that if We isnot computable, then there is no computable separating set for Wf(e);Wg(e).Finally, de�ne the computable function h by letting �h(e)(�) = 1 if and only if

(8i < j�j)[(i 2Wf(e);j�j ! �(i) = 1) & (i 2Wg(e);j�j ! �(i) = 0)]=:

Then we have Ph(e) = S(Wf(e);Wg(e)). It then follows from the discussion abovethat

e 2 Rec () e 2 Sep & Ph(e) has a computable member:

(iii) This follows from the proof of (ii), since Ph(e) is always a nonempty classof separating sets.


For many applications of �01 classes, one demonstrates the di�culty of �nd-

ing a solution to a certain type of computable problem by constructing c. e.sets Wa and Wb and a corresponding separating class Pe = S(Wa;Wb) suchthat the set of solutions to the given problem corresponds to the class Pe. Thuswe want to consider for a given property R of classes, such as the property ofbeing �nite, fe 2 Sep : Pe has property Rg. We note that there is a primi-tive recursive function such that S(Wa;Wb) = P (a;b) for each a and b andconversely, there is a partial computable function � such that for all e 2 Sep,Pe = S(W(�(e))0 ;W(�(e))1).

Theorem VI.5.5.

(i) (fha; bi : S(Wa;Wb) = ;g; fha; bi : S(Wa;Wb) 6= ;g) is (�01;�01) complete.

(ii) For any positive integer c,(fha; bi : card(S(Wa;Wb)) > cg; fha; bi : card(S(Wa;Wb) � cg) is (�02;�0

2)complete, fha; bi : card(S(Wa;Wb)) = c + 1g is D0

2 complete and fha; bi :card(S(Wa;Wb)) = 1g is �0

2 complete.

(iii) (fha; bi : S(Wa;Wb) is �niteg; fha; bi : S(Wa;Wb) is in�niteg) is (�03;�03)

complete.

Proof. In each case, an upper bound on the complexity is given by the reduction noted above together with previous results, Theorems VI.2.1, VI.2.8, VI.2.13and VI.3.1. For the rest of the proof, we set Wb = ;.

(i) Observe that S(We;Wb) is empty if and only if We is empty.(ii) Observe that card(S(We;Wb)) = 2c if and only if card(NnWe) = c. Thus

only powers of 2 need to be considered. Now e 2 Tot () NnWe = ;, which is() Card(S(We;Wb)) = 1 and also () Card(S(We;Wb)) � 1, which givesthe completeness for cardinality > 1 as well. If we let W�(e;c) = fn + c : n 2Weg, then card(N nWe) = c + card(N nWe). Thus card(S(We;Wb) � 1 ()card(W�(e; c); b) � 2c and similarly for SS(> 2c).

Next we show the D02 completeness for cardinality 2c+1. It follows from the

reduction above that, for a given �02 set A, there is a computable function f

such that if a 2 A, then card(N nWa) = c and if a =2 A, then card(N nWa) > c.Let B be a �02 set. We will obtain a reduction g such that if e 2 B, thencard(N nWe) = 0 and if e =2 B, then card(N nWe) = 1. Of course it su�cesto de�ne such a reduction for the �02 complete set Fin, which we do as follows.Given an index e, construct the c. e. set Wg(e) in stages Wg(e);s along with anumber xs which is intended to be the unique member of N nWe, if any. Weassume as usual that at most one element comes into We at any stage s. Theconstruction begins with Wg(e);0 = ; and x0 = 0. At stage s+ 1, there are twocases.

(Case 1) If no element comes intoWe, or if an element x < xs comes intoWe,then we let xs+1 = xs and we put s + 1 2 Wg(e);s+1. In this case, Wg(e);s+1 =f0; 1; : : : ; s+ 1g n fxsg.

VI.6. MEASURE AND CATEGORY 169

(Case 2) If an element x � xs comes into We, then we put xs 2 Wg(e);s+1

and let xs+1 = s+ 1; in this case Wg(e);s+1 = f0; 1; : : : ; sg.If We is �nite, then at some stage, we obtain xs greater than every element

of We, so that Case 1 applies at every later stage t. Thus xt = xs for all t > sand ! n Wg(e) = fxsg. If We is in�nite, then Case 2 applies in�nitely oftenand Wg(e) = !. Finally, we de�ne a reduction of the D0

2 set A \ B by lettingWh(e) =Wf(e) �Wg(e).

(iii) Observe that S(Wa;Wb) is �nite if and only if Wa is co�nite and applyTheorem VI.0.4.

Exercises

VI.5.1. Prove Lemma VI.5.2.

VI.5.2. Give the details in the proof of Lemma VI.5.1.

VI.5.3. Show that that there is a primitive recursive function such that S(Wa;Wb) =P (a;b) for each a and b and there is a partial computable function � suchthat for all e 2 Sep, Pe = S(W(�(e))0 ;W(�(e))1).

VI.5.4. Show that fe 2 Sep : card(Pe) = c+ 1g is D02 complete and similarly for

the Sep versions of cardinality 1, �nite or in�nite from Theorem VI.5.5.

VI.6 Measure and Category

In this section, we consider properties such as being perfect, being meager, andhaving measure > r or � r for some �xed real r.

Recall that a closed set C is perfect if every element of C is a limit pointof C, that is, if D(C) = C. In particular, !!, f0; 1; : : : ; kg! (for any k) and ;are all perfect; some authors exclude the empty set. We can use the method ofTheorem VI.2.15 to classify index sets of perfect classes.

Theorem VI.6.1. (i) For any computable function g � 2,fe : Pe is g-bounded and perfectg and fe : Pe is g-bounded, nonempty and perfectgare �0

3 complete.

(ii) fe : Pe is c. b. and perfectg and fe : Pe is c. b. , nonempty and perfectgare D0

3 complete.

(iii) fe : Pe is bounded and perfectg and fe : Pe is bounded, nonempty and perfectgare �0

4 complete.

(iv) fe : Pe is perfectg and fe : Pe is nonempty and perfectg are �11 complete.

Proof. We �rst observe that if P = [T ] where U is a tree with no dead ends,then P is perfect if and only if d(T ) = T . Thus Pe is perfect if and only if

(�)(8�)[� 2 Ext(Te)! (9�)(9i; j)(� � � & �_i 2 Ext(Te) & �_j 2 Ext(Te))]:


We also observe that there is a �02 relation B � N � N� such that if Te is

�nite branching, then � 2 Ext(Te) () B(e; �). (This is left as an exercise.)

(i) The upper bounds now follow easily from (*) and Theorem VI.2.1.

For the completeness of both index sets, modify the proof of part (i) ofTheorem VI.2.13 by letting Th(e) contain f0n : n 2 !g together with all strings0n1_�1 : : : �k where n =2 We;k and each �i is either (010) or (011). It is theneasy to see that Th(e) is perfect if and only if e =2 Cof .

(ii) It follows from Theorem VI.1.5 and the uniform proof of part (i) thatfe : Pe is c. b. and not perfectg is �03. The upper bounds on the complexitynow follow from (*).

For the completeness of each set, let A = B \ C be a D03 set where B is �03

and C is �03. It follows from the proof of part (iii) of Theorem VI.2.3 that there

is a computable function g0 such that Pg0(e) is always a singleton and is c. b. ifand only if e 2 B.

It follows from our proof of part (i) above that there is a comptable functionh0 such that Ph0(e) is always c. b. and e 2 C if and only if Ph0(e) is nonemptyperfect. For each e, let

P�(e) = Pg0(e) Ph0(e):Then the desired reduction is given by

e 2 A () P�(e) is c. b. and perfect:

and this also works for nonempty perfect.

(iii) The upper bound on the complexity follows from (*) and Theorem VI.2.4as above.

For the completeness results, let A be an arbitrary �04 set and let R be a

computable relation such that for all a,

a 2 A () (8m)(9n)(8j)(9k)R(a;m; n; j; k):

We assume as usual that R(a;m; n; j; k)! R(a;m; n+ 1; j; k). The desiredreduction of A is de�ned as follows. First, for each m, n, and a, let Tf(m;n;a)consist of all strings (k0 + 1; k1 + 1; : : : ; kt + 1), where for each j � t, kj is theleast k such that R(a;m; n; j; k). Then let Tf(a) contain all strings of the form0n together with all strings of the form 0n0 ��_0 0��_1 0� � � � ��r, where for eachm � r, �m 2 Uf(m;nm;a) and nm+1 = j�mj. Each Tf(m;n;a) is �nite-branching,so that Tf(a) is always �nite-branching. 0! 2 Uf(a), so that Pf(a) is alwaysnonempty. Elements of Pf(a), other than 0!, have one of two forms:

(a) 0n0 � �_0 0 � �_1 0 � : : : �_t 0 � x, where for each m � t, �m 2 Tf(m;nm;a)and nm+1 = j�mj and x 2 Pf(m+1;nm+1;a).

(b) 0n0��_0 0��_1 0�: : : , where for eachm, �m 2 Tf(m;nm;a) and nm+1 = j�mj.Suppose that a 2 A. Then for in�nitely many n, there exists xn 2 Pf(0;n;a)

and we have 0n � xn 2 Pf(a). Thus 0! is not isolated. Similarly any string� = 0n0 ��_0 0��_1 0� � � � ��r 2 Ext(Uf(a)), will have in�nitely many extensions0n0 � �_0 0 � �_1 0 � � � � � �_r 0 � xn in Pf(a).


On the other hand, suppose that a =2 A and let M be the least m such that:(9n)(8j)(9k)R(a;m; n; j; k). Then there will be an isolated path 0n0 � �_0 0 ��_1 0 � � � � � �_M�20 � x in Pf(a), where x 2 Pf(M�1;j�M�2j;a).

Thus we have

a 2 A () Pf(a) is bounded, nonempty perfect:

The same reduction applies for bounded, perfect.(iv) First de�ne the �1

1 relation Isol(x; e) which says that x is isolated in Peby

Isol(x; e) () x 2 Pe & (9n)(8y)[(xdn = ydn & x 6= y)! y =2 Pe]:Next recall from Theorem V.2.2 that every isolated point in Pe must be hy-perarithmetic. Thus fe : Pe is perfect is seen to be �11 by the Spector-GandyTheorem II.10.5, since

Pe is perfect () (8HY Px):Isol(x; e):It follows from Theorem VI.2.6 that fe : Pe is nonempty perfectg is also �11.

For the completeness in the nonempty perfect case, let f be the reductiongiven in Theorem VI.2.6 so that, for an arbitrary �11 set A, Pf(a) is nonemptyif and only if a 2 A, and let Tg(a) = Tf(a) f0; 1g<!. Then Pg(a) is nonemptyperfect if a 2 A and is empty otherwise. For the other case, let g be as aboveand let

Th(a) = f0n_(�(0) + 1; : : : ; �(k � 1) + 1) : n 2 ! & � 2 Tg(a)g:Thus

Ph(a) = f0!g [ f0n_(x(0) + 1; x(1) + 1; : : :) : n 2 N & x 2 Pg(a)g:For a 2 A, Ph(a) is clearly a perfect set, and for a =2 A, Ph(a) = f0!g.

Next we consider the notions of category. We begin with a few de�nitions.A set K � NN is said to be dense in another set M if M � Cl(K). For a closedset K, K is dense in M if and only if M � K. K is said to be nowhere densein !! if there is no string � such that K is dense in the interval I(�). Similarly,K � f0; 1g! is nowhere dense if there is no � 2 f0; 1g<! such that K is dense inI(�)\f0; 1g!. Thus a closed set K is nowhere dense if and only if it includes nointerval. Note that a nonempty open set can never be nowhere dense. A set issaid to be meager or �rst category if it is the countable union of nowhere densesets. A meager set includes no interval, by the Baire Category Theorem, andthus a closed meager set is itself nowhere dense. Thus a closed set K is meagerif and only if it includes no interval. A set is said to be non-meager or secondcategory if it is not meager. Thus a closed set K is second category if and onlyif it includes an interval. Note that a nonempty open set always contains aninterval and thus is always non-meager. Finally, a set is said to be comeager ifit is the complement of a meager set. It follows that a closed set K is comeagerif and only if K = !! (or f0; 1g!).


Theorem VI.6.2.

(i) For all � 2 !<!, fe : I(�) � Peg is �01 complete and for all � 2 f0; 1g<!,

fe : I(�) \ f0; 1g! � Pe \ f0; 1g!g is �01 complete.

(ii) fe : Pe is meagerg and fe : Pe is meager in f0; 1gNg are both �02 complete.

Proof. (i) is left as an exercise.(ii) The upper bound on the complexity follows from the fact that Pe is

non-meager if and only if I(�) � Pe for some �.For the completeness, let A be a �02 set and let R be a computable relation

so that

a 2 A () (9m)(8n)R(m;n; a):Then a reduction of A to fe : Pe is meagerg given by

Tf(a) = f0m : m 2 !g [ f(0m)_1_� : � 2 f0; 1g<! & (8n < j� j)R(m;n; a)g:

It follows that, for example, fe : Pe 6= NNg is �01 complete and fe :Pe is non-meagerg is �02 complete. Also note thatfe : Pe is co-meagerg = fe : NN = Peg and is �0

1 complete.Next we consider the complexity of index sets associated with measure. Re-

call that the measure on f0; 1gN is de�ned by setting �(I(�) = 2�j�j and the mea-sure on NN is de�ned (with �(NN) = 1) by setting the measure of fx : x(m) = ngto be 2�n�1, so that I(�) has measure 2�(m0+m1+��+mk�1+k). Recall from sec-tion II.II.4 that a real number r is said to be �0

1 (respectively, �01, etc.) iffq 2 Q : q < rg is a �0

1 (resp. �01, etc.) set. We note that the ordered ring Q of

rationals is a computable structure and can be coded into N for computabilitypurposes.

Lemma VI.6.3.

(a) For any �01 class P , �(P ) is a �0

1 real number.

Proof. Let T be a computable tree such that P = [T ], let �0; �1; : : : be a com-putable enumeration of N� � T . For each n 2 N, let Kn = NN � Si�n I(�i).The result now follows from the fact that �(P ) is the decreasing limit of thecomputable sequence h�(Km)im2! of dyadic rationals.

It is an exercise to show that �(P ) need not be computable.

Theorem VI.6.4. (i) For any �01 real r 2 (0; 1], (fe : �(Pe) < rg; fe :�(Pe) � rg) is (�01;�0

1) complete if r is not computable, then fe : �(Pe) �r) is �01 complete.

(ii) For any �01 real r < 1, (fe : �(Pe) > rg; fe : �(Pe) � rg) is (�02;�

02)

complete and fe : �(Pe) = r) is �02 complete. If r is �0

1 complete, then(fe : �(Pe) < rg; fe : �(Pe) � rg) (�02;�0

2) complete.


(iii) For any �01 real r 2 (0; 1], (fe : �(Pe\f0; 1gN) < rg; fe : �(Pe\f0; 1gN) �rg) is (�01;�0

1) complete if r is not computable, then fe : �(Pe\f0; 1gN) �r) is �01 complete.

(iv) For any �01 real r < 1, (fe : �(Pe\f0; 1gN) > rg; fe : �(Pe\f0; 1gN) � rg)

is (�02;�02) complete and fe : �(Pe \ f0; 1gN) = r) is �0

2 complete. If r is�01 complete, then (fe : �(Pe \ f0; 1gN) < rg; fe : �(Pe \ f0; 1gN) � rg) is

(�02;�02) complete.

Proof. (i) Let �0; �1; : : : enumerate N� and let

Pe;n = NN �

[fI(�i) : i < n & �i =2 Teg:

Then the function �(Pe;n) is computable and we have for any rational q:

�(Pe) � q () (8n)�(Pe;n) � q:

If r is �01 and not rational, then

�(Pe) � r () �(Pe) > r () (8q 2 Q)[q < r ! �(Pe;n) � q:

For the completeness, let A be a �01 set and R a computable relation such that

a 2 A () (8n)R(n; a):

The necessary reduction f of A is de�ned so that Pf(a)) = f0; 1gN when a 2 Aand Pf(a) = ; if a =2 A. Just let The reduction f is de�ned by Tf(a) = f� :(8n < j�j)R(n; a)g.

(ii) Let r be a �01 real. Then we have

�(Pe) � r () (8q 2 Q)(q � �(Pe)! q � r)

and similarly

�(Pe) � r () (8q 2 Q)(q � r ! q � �(Pe)):

It follows that fe : �(Pe) � rg, fe : �(Pe) � rg, and fe : �(Pe) = rg are all �02

sets. Next we show the completeness of the latter two sets. Let B be a �01 set

so that r =Pi2B 2�i�1 and let PB = f0!g [Si2B I(0i1), so that �(PB) = r.

Since r 6= 1, we may assume that B is co-in�nite. Let A be a �02 set and R a

computable relation so that

a 2 A () (8m)(9n)R(m;n; a):

Here we assume as usual that if a =2 A, then (9n)R(m;n; a) for only �nitelymany m. Now de�ne the reduction g by

Pg(a) = f0!g [[fI(0m1) : i 2 B or (8n):R(m;n; a)g:


If a 2 A, then clearly Pg(a) = PB so that �(Pg(a)) = r. If a =2 A, then Pg(a)includes PB together with co�nitely many intervals I((m)), so that �(Pg(a)) > r.

For the completeness of measure � r when r is �01 complete, let B be a �0

1

set such that �(PB) = r. Let A be a �02 set and, by the completeness, let f be

a computable functions such that, for any a,

a 2 A () (8m)f(a;m) =2 B:

De�ne the uniformly �01 set Ca = B n ff(a;m) : m 2 Ng, so that for any a, we

have a 2 A () C(a) = B and otherwise, Ca is a proper subset of B. Thende�ne

Pg(a) = f0!g [[fI((0n1)) : n 2 C(a)g:

If a 2 A, then Pg(a) = PB , so that �(Pg(a)) = r and if a =2 A, then Pg(a) is asubset of PB � I(0n1) for some n 2 B and thus �(Pg(a)) < r.

Parts (iii) and (iv) follow immediately.

Exercises

VI.6.1. De�ne a �02 relation B � N� N� such that if Te is �nite branching, then

� 2 Ext(Te) () B(e; �).

VI.6.2. For any computable function g � 2, show that fe : Pe is g-a.b. and perfectgand fe : Pe is g-a.b., nonempty and perfectg are �0

3 complete.

VI.6.3. fe : Pe is a.c.b. and perfectg and fe : Pe is a.c.b., nonempty and perfectgare D0

3 complete.

VI.6.4. fe : Pe is a.b. and perfectg and fe : Pe is a.b., nonempty and perfectg areD04 complete.

VI.6.5. Prove part (i) of Theorem VI.6.2.

VI.6.6. De�ne a �01 class P such that �(P ) is a �0

1 complete real.

VI.7 Derivatives

In this section we consider the uniform (arithmetic) complexity of D�(Pe) andthe complexity of various cardinality properties ofD�(Pe). These problems were�rst studied in the context of Polish spaces by Kuratowski, see [120], where theCantor-Bendixson derivative is viewed as a mapping from the space of compactsubsets of f0; 1g! to itself. Kuratowski showed that the derivative is a Borelmap of class exactly two. In particular, he showed that the family D�1(f;g) of�nite closed sets is a universal �0

2 class and posed the problem of determiningthe exact Borel class of the iterated operator D�. Cenzer and Mauldin showedin [30, 31] and that the iterated operator Dn is of Borel class exactly 2n for�nite n and that for any limit ordinal � and any �nite n, D�+n is of Borel class

VI.7. DERIVATIVES 175

exactly � + 2n + 1. In particular it is shown that for any �, the family T� ofclosed sets K such that D�(K) = ; is a universal �0

2� set. Lempp [127] givesan e�ective version of this result.

We �rst observe that the basic results on the cardinality of �01 classes can

be relativized. For any �xed set X, let PXe enumerate the binary classes whichare �0

1 in X. That is, let PXe = [TXe ], where

TXe = f� : (8� � �)(he; �i =2WXe g:

Theorem VI.7.1. For any set X,

1. (fe : PXe is emptyg; fe : PXe is nonemptyg) is (�01X ;�01X) complete,

2. fe : card(PXe ) = 1g is �02X

complete.

3. For any positive integer c, (fe : card(PXe ) > cg; fe : card(PXe ) � cg) is(�02

X;�0

2X) complete and fe : card(PXe ) = c+ 1g is D0

2X

complete.

4. (fe : PXe is �niteg; fe : PXe is in�niteg) is (�03X ;�03X) complete.

Proof. This follows from the proofs of Theorems VI.2.1, VI.2.8 and VI.2.13.

The strong �0n classes were de�ned in Section III.III.3. Here we need a

uniform de�nition of the strong �0� classes for any computable ordinal �. Let

Te;� = f� : (8� � �)(he; �i =2 O�g:and let

Pe;� = [Te;�]:

Then a closed set P is said to be �0� if it equals Pe;� for some index e. It follows

that P is �0�+1 if and only if P is �0

1 in O�. Furthermore, for any ordinal �, P

is a strong �0� class if and only if TP is a �0

� set. (See the exercises.)The following result from [39] is now immediate for successor ordinals.

Theorem VI.7.2. For any computable ordinal �,

1. (fe : Pe;�+1 is emptyg; fe : Pe;�+1 is nonemptyg) is (�0�+1;�0�+1) com-

plete,

2. fe : card(Pe;�+1) = 1g is �0�+2 complete.

3. For any positive integer c, (fe : card(Pe;�+1) > cg; fe : card(Pe;�+1) � cg)is (�0�+2;�

0�+2 complete and fe : card(Pe;�+1) = c+1g is D0

�+2 complete.

4. (fe : Pe;�+1is �niteg; fe : Pe;�+1is in�niteg) is (�0�+3;�0�+3) complete.

We need a uniform version of Lemma V.V.4.1.

Lemma VI.7.3. [[39]] For any computable limit ordinal � and any �nite n > 0,

(a) fhe; �i : � 2 dn(Te \ f0; 1g�)g is �02n;


(b) fhe; �i : � 2 d�(Te \ f0; 1g�)g is �0�+1;

(c) fhe; �i : � 2 d�+n(Te \ f0; 1g�)g is �0�+2n.

By the uniform proof of Theorem V.V.4.8, we have the following.

Theorem VI.7.4. There is a primitive recursive function � such that, for anycomputable ordinal �, if Q is the �0

2�+1 class with index e, then P�(e) is theindex of a �0

1 class P of sets such that there is a homeomorphism H from Qonto D�(P ) with x �T H(x) � x� 02��1 for all x 2 Q.

Theorem VI.7.5. For any recursive ordinal �,

1. (fe : D�(Pe) is emptyg; fe : D�(Pe) is nonemptyg) is (�02�+1;�02�+1)

complete.

2. fe : card(D�(Pe)) = 1)g is �02�+2 complete.

3. For any positive integer c, (fe : card(D�(Pe)) � c); fe : card(D�(Pe)) >cg) is (�02�+2;�0

2�+2) complete and fe : card(D�(Pe)) = c + 1) is D02�+2

complete.

4. (fe : D�(Pe) is �niteg; fe : D�(Pe) is in�niteg) is (�02�+3;�02�+3) com-

plete.

Proof. The upper bound on the complexity follows from Lemma VI.7.3 andTheorem VI.7.2. That is, for example, �x � = � + n, where � is a limit andn > 0. Then D�(Pe) = [d�+n(Te)] and it follows from Lemma VI.7.3 that thisequals Pf(e);�+2n+1] for some computable function f . Since �+2n+1 = 2�+1,the complexity follows from Theorem VI.7.2.

The completeness follows from Theorems VI.7.2 and VI.7.5. That is, for ex-ample, Pe;�+1 is �nite if and only if D

�(Pf(e)) is �nite, and fe : Pe;�+1 is �nitegis �0�+3 complete, therefore fe : D�(Pf(e)) is �niteg is also �0�+3 complete.

Lempp used di�erent methods in [127] to prove parts (i) and (iv). He gaveweaker versions of parts (ii) and (iii), showing that

(�02�+1;�02�+1) � (I

(�)P (empty); I

(�)P (= 1)).

We now consider the complexity of the perfect kernel K(Pe). It follows fromTheorem VI.2.15 that fe : K(Pe) = ;g is �1

1 complete. It follows from TheoremVI.7.5 that, for every computable ordinal �, there exists e such that D�(Pe) isnonempty but D�+1(Pe) = ;. This gives us the following.

Theorem VI.7.6. There is a �01 class P � f0; 1gN such that

(i) rk(P ) = !C�K1 .

(ii) TK(P ) is �11 complete, that is, f� : K(P ) \ I(�) 6= ;g is �11 complete.

VI.8. INDEX SETS FOR LOGICAL THEORIES 177

Proof. Let

P =[e

f0e1x : x 2 Peg:

Then

K(Pe) = f0! [[e

f0e1x : x 2 K(Pe)g

and, for each �,

D�(P ) = f0! [[e

f0e1x : x 2 D�(Pe)g:

We know that rk(P ) � !C�K1 by Theorem V.V.1.4 and it now follows fromTheorem VI.7.5 that rk(P ) = !C�K1 . It also follows from TheoremV.V.1.4 thatK(P ) is a �11 class, so that TK(P ) is a �11 set. For the completeness, observethat K(Pe) 6= ; if and only if Pe is uncountable and that fe : Pe is uncountablegis �11 complete by Theorem VI.2.15. Then we have

K(Pe) 6= ; () K(P ) \ I(0e1) 6= ;;

which shows that K(P ) is �11 complete.

Exercises

VI.7.1. Give a careful proof of Lemma VI.7.3.

VI.7.2. Show that, for any ordinal �, P is a strong �0� class if and only if TP is a

�0� set.

VI.8 Index Sets for Logical Theories

In this section we de�ne index sets for (propositional) logical theories and con-sider the complexity of properties associated with the consistency and complete-ness of such theories.

Let Sent denote the set of sentences f 0; 1; : : : g of the propositional lan-guage with variables fA0; A1; : : : g, enumerated �rst by length and then lexico-graphically. The e'th axiomatizable theory �e � Sent may be de�ned as theset of consequences of f i : i 2Weg. The following lemma is left as an exercise.

Lemma VI.8.1. fi : i 2 �eg is a c. e. set and in fact there is a computablefunction f such that fi : i 2 �eg =Wf(e).

As in section III.III.9, a �01 class P � f0; 1gN represents a class G of subsets

of Sent if, for any x 2 TN , x 2 P if and only if fAi : x(i) = 1g 2 G.The next result now follows easily from the uniformity of the proof of The-

orem III.9.1.


Lemma VI.8.2. There is a primitive recursive function f such that, for all e,Pf(e) represents the set of complete consistent extensions of �e. Furthermore,if �e is a decidable theory, then Pf(e) is a decidable �0

1 class, that is, f� :Pf(e) \ I(�) 6= ;g is a computable set.

Note here that when �e is decidable, we do not necessarily have Tf(e) to bea tree without dead ends; there simply exists a tree T without dead ends suchthat Pf(e) = [T ].

On the other hand, Theorem III.9.3 may be uniformized as follows.

Lemma VI.8.3. There is a primitive recursive function g such that, for all e,Pe represents the set of complete consistent extensions of �g(e). Furthermore, ifPe is a decidable �0

1 class, then �g(e) is a decidable theory.

Note again that when Pe is decidable (which is true whenever Te has no deadends), then �g(e) is a decidable theory but it not necessarily true that Wg(e) isa computable set.

We can now apply the index set result sets of this chapter to obtain somecomplexity results for axiomatizable theories.

Theorem VI.8.4. 1. fe : �e is consistentg is �01 complete.

2. fe : �e is consistent and completeg is �02 complete.

3. fe : �e is essentially undecidableg is �03 complete.

Proof. (1) Using the function f from Lemma VI.8.2, �e is consistent if andonly if Pf(e) is nonempty, and this is a �0

1 condition by Theorem VI.2.1. Forthe completeness, Pe is nonempty if and only if �g(e) is consistent, where g isthe function from Lemma VI.8.3. The completeness now follows from TheoremVI.2.1.

(2) �e is consistent and complete if and only if it has a unique completeconsistent extension, that is, if and only if card(Pf(e)) = 1, which is a �0

2

condition by Theorem VI.2.8. The completeness follows from Theorem VI.2.8since card(Pe) = 1 if and only if �g(e) is consistent and complete.

(3) �e is essentially undecidable if and only if it has no computable completeconsistent extension, that is, if and only if Pf(e) has no computable element,which is a �0

3 complete condition by Theorem VI.3.1. The completeness followsfrom Theorem VI.3.1 since Pe has no computable element if and only if �g(e) isessentially undecidable.

We can also classify the index sets of theories with a given number of com-plete consistent extensions (and similarly for computable complete consistent ex-tensions). The next theorem follows from Theorems VI.2.8, VI.2.13 and VI.2.15as above. Let us abbreviate \computable consistent extensions" by CCEs.

Theorem VI.8.5. Let c > 0 be �nite.

1. (fe : �e has > c CCEsg; fe : �e has � c CCEsg is (�02;�02) complete.

VI.8. INDEX SETS FOR LOGICAL THEORIES 179

2. fe : �e has exactly c CCEsg is D02 complete.

3. (fe : �e has �nitely many CCEsg; fe : �e has in�nitely many CCEsg is(�03;�

03) complete.

4. fe : �e has exactly @0 CCEsg is �11 complete.

5. (fe : �e has uncountably many CCEsg; fe : �e has countably many CCEsgis (�11;�

11) complete.

For a given number of computable complete consistent extensions, we applyTheorems VI.3.5 and VI.3.9.

Theorem VI.8.6. Let c > 0 be �nite.

1. (fe : �e has > c computable CCEsg; fe : �e has � c computable CCEsg is(�03;�

03) complete.

2. fe : �e has exactly c computable CCEsg is D03 complete.

3. (fe : �e has < @0 computable CCEsg; fe : �e has � @0 computable CCEsg)is (�04;�

04) complete.

Finally, we consider thin Martin{Pour-El (MPE) theories.

Theorem VI.8.7. fe : �e is MPEg is a �04 complete set.

Proof. Let f be the function from Lemma VI.8.2 so that Pf(e) represents the setof complete consistent extensions of �e. Then �e is MPE if and only if Pf(e)is thin, and this is a �0

4 condition by Theorem VI.4.6. For the completeness,let g be the function from Lemma VI.8.3 such that Pe represents the classof complete consistent extensions of �g(e). Then Pe is thin if and only if �e isMartin{Pour-El and the �0

4 completeness now follows from Theorem VI.4.6.

.Exercises

VI.8.1. Prove Lemma VI.8.1



VI.8.4. Let f be the function from Lemma VI.8.1. Show that fe :Wf(e) =Weg is�02 complete. That is, the property of being a logical theory is �

02 complete

for sets of sentences.


Chapter VII

Reverse Mathematics

There is a close connection between �01 classes and certain subsystems of second

order arithmetic which are used in the so-called Reverse Mathematics developedby Friedman and Simpson (see [192]). In particular, the system WKL0 (WeakKonig's Lemma) corresponds roughly to the statement that every in�nite treein f0; 1g� has an in�nite. The system ACA0 arithmetic comprehension) corre-sponds to the statement that every in�nite, �nitely branching tree has an in�nitepath. Thus the representation theorems from Part B may be used to show thatcertain standard in�nite combinatorial theorems are logically equivalent, overthe base theory RCA0, to either WKL0 or to ACA0.

For example, consider the completeness theorem for (countable) proposi-tional logic. Given a consistent theory �, the set of complete consistent exten-sions of � can be viewed as the in�nite paths through a certain in�nite binarytree and thus Weak Konig's Lemma can be used to prove that a complete con-sistent extension exists. On the other hand, given an arbitrary in�nite treeT � f0; 1g�, we showed that there exists a consistent theory � such that Trepresents the class of complete consistent extensions of �. The completenesstheorem tells us that a complete consistent extension exists and therefore T pos-sesses an in�nite path. This gives an (informal) proof of Weak Konig's Lemmafrom the completeness theorem and demonstrates that the two are logicallyequivalent.

We also present in this chapter the reverse mathematics of propositionallogic. Later, in Part B, we will we will consider the proof-theoretic strenth oftheorems from various areas of algebra, analysis and combinatorics. This willincludee the Cantor-Schroder-Bernstein Theorem and related theorems aboutsymmetric marriages in a highly computable society which are equivalent, var-iously, to Weak Konig's Lemma or to Arithmetic Comprehension. We alsoexamine several results on in�nite partially ordered sets, including Dilworth'stheorem that any poset of width n can be covered by n chains.

181

182 CHAPTER VII. REVERSE MATHEMATICS

VII.1 Subsystems of Second Order Arithmetic

In this section, we discuss the language of second order arithmetic, models ofsecond order arithmetic and the basic axiom system for second order arithmeticas well as certain subsystems closely related to �0

1 classes. These are RCA0,WKL0 and ACA0. For details, see Simpson's [192].

A second order structure includes both objects and sets of objects. Thus amodel of second order arithmetic includes a model of �rst order arithmetic, witha set of objects intended as natural numbers together with the usual operationsof addition and subtraction, as well as a collection of sets of numbers and themembership relation (n 2 X) between the objects and the sets.

The language L2 of second order arithmetic thus includes the usual languageof �rst order arithmetic, with constant symbols 0 and 1, intended to denote thecorresponding natural numbers, with binary function symbols + and �, intendedto denote the addition and multiplication functions on the natural, and with arelation symbol < intended to denote the ordering of the natural numbers, aswell as the usual equality symbol = from predicate logic.

There is also a relation symbol 2 which denotes the membership relation.There are two sorts of variables intended to range over numbers and oversets. Number variables i; j; k;m; n; : : : are intended to range over the set N =f0; 1; : : : g of natural numbers and set variablesX;Y; Z; : : : are intended to rangeover subsets of N.

Terms are de�ned as in �rst order arithmetic to compose the smallest setof strings containing the two constant symbols and all number variables andclosed under t = t1 + t2 and t = t1 � t2. Atomic formulas are t1 = t2, t1 < t2and t1 2 X, where t1 and t2 are terms and X is a set variable. Formulascompose the smallest set containing all atomic formulas and closed under thepropositional connectives, number quanti�ers (8n) and 9n) and also under setquanti�ers (8X) and (9X).

A model for the language L2 has the form

M = hM;SM ;+M ; �M ; 0M ; 1M ; <M i;

where M is the universe of M, SM is a set of subsets of M , +M and �M arebinary operations on M , 0M and 1M are distinguished elements of M and <Mis a binary relation on M .

The intended model for L2 is hN;P(N);+; �; 0; 1; <i.An !-model M is a model of L2 with universe N and with the standard

operations + and �, constants 0 and 1, and binary relation <, but with SMmerely a subset of P(N). In this case, we simply identify M with the familyS = SM . In addition to the intended model, we will be interested in thefollowing.

(1) REC is the !-model with S the set of recursive sets of natural numbers.

(2) ARITH is the !-model with S the set of arithmetical sets of natural num-bers.

VII.1. SUBSYSTEMS OF SECOND ORDER ARITHMETIC 183

These de�nitions can be relativized to RECB and ARITHB for any �xedB � N.

The axioms of second order arithmetic include the eight axioms of RobinsonArithmetic (see Section IV.IV.4).

For the induction axiom, we can now discuss sets rather than formulas, sowe have

IS (8X)[(0 2 X ^ (8n)(n 2 X ! n+ 1 2 X))! (8n)n 2 X]

To ensure that some sets exist, we have a Comprehension Axiom for eachformula �(n) of L2:

C (9X)(8n)[n 2 X () �(n)].

Here we allow number and set parameters in the formula �. These lasttwo axioms imply the induction scheme IP of Peano Arithmetic and in fact astronger, full second order induction scheme where the formula � in IP may beany formula of L2. Note also that any !-model also satis�es full second orderinduction.

These axioms compose the formal system Z2 of second order arithmetic.By a subsystem of second order arithmetic, we mean a theory included in Z2,generally obtained by weakening the axioms of induction and comprehension.

VII.1.1 Recursive Comprehension

The fundamental system RCA0 consists of the basic axioms of Robinson Arith-metic together with �01 induction and �0

1 comprehension. Some de�nitions arerequired to de�ne the hierarchy of formulas.

If t is a numerical term not containing n and � is any formula of L2, thenthe bounded quanti�ers 8n < t) and (9n < t) are de�ned by

(8n < t)� � (8n)(n < t! �) and

(9n < t)� � (9n)(n < t & �)

A formula � of L2 is said to be a bounded quanti�er (or �00) formula if all ofits quanti�ers are bounded. � is said to be �01 if it is of the form (9m) where is a bounded quanti�er formula and � is said to be �0

1 if it is of the form (8m)�where � is a bounded quanti�er formula. More generally, � is �0k if it is of theform (9n1)(8n2) : : : nk� where � is a bounded quanti�er formula, and simlarlyfor the �0

k formulas.

De�nition VII.1.1. A �0k (respectively �0k) induction scheme has the form

[�(0) & (8n)(�(n)! �(n+ 1))] =) (8n)�(n);

where � is any �0k (resp. �0k) formula of L2. Here � may have other number

and set variables.


De�nition VII.1.2. 1. A �0k (respectively �0k) comprehension scheme has

the form(9X)(8n)(n 2 X () �(n));

where � is any �0k (resp. �0k) formula of L2.

2. A bounded �0k comprehension scheme has the form (8n)(9X)(8i)(i 2X () (i < n & �(n)).

3. A �0k comprehension scheme has the form

(8n)(�(n) () (n)) =) (9X)(8n)(n 2 X () �(n));

where � is a �0k formula and is a �0k formula.

As above, � and may have other number and set variables.

The !-models of RCA0 may be characterized as follows.S is an !-model of RCA0 if and only if

S 6= ;;A 2 S and B 2 S imply A�B 2 S;A 2 S and B �T A imply B 2 S.It follows that RCA0 has a minimum !-model,

REC = fA 2 P(N) : A is recursivegSimpson [192] outlines the development of ordinary mathematics within

RCA0. In particular the coding function hn1; : : : nki are de�nable in RCA0 andG�odel numbering of propositional and also �rst-order logic may be done there.Functions may be de�ned by primitive recursion and also by minimization.

Here are some other important results from [192].

Lemma VII.1.3. The following are provable in RCA0.

1. For any in�nite set X � N, there exists a strictly increasing function� : N! N such that X is the range of �.

2. Let �(n) be a �01 formula in which X and f do not occur freely. Theneither there exists a �nite set X such that (8n)(n 2 X () �(n)), orthere exists a one-to-one function f : N ! N such that (8n)(�(n) ()(9m)(f(m) = n)).

Proof. (1) Let �(0) be the least n 2 X and for each k, let �(k + 1) be the leastk > �(n) such that k 2 X.

(2) Suppose no �nite set X exists as stated. Let � be �00 such that �(n) ()(9j)�(j; n) and let Y = fhj; ni : �(j; n) & (8i < j):�(i; n)g. Then Y is in�nite,so by part (1), there is a function � which enumerates Y in increasing order.Let f(m) = �2(�(m)), where �2 is the projection of hx; yi onto y.


Theorem VII.1.4. RCA0 proves bounded �01 comprehension.

Corollary VII.1.5. RCA0 proves �01 induction.

Here is the basic results for trees.

Theorem VII.1.6. The following is provable in RCA0. If T � f0; 1g� is atree with no dead ends, then T has an in�nite path.

Proof. The leftmost path through T may be de�ned by primitive recursion.

On the other hand, K�onig's Lemma is not provable in RCA0, since RECwill contain a computable tree with no computable in�nite path and thereforeno path in REC.

VII.1.2 Weak K�onig's Lemma

In this section we consider the stronger system WKL0 and its relation to �01

classes.

De�nition VII.1.7. 1. Weak K�onig's Lemma is the statement that everyin�nite subtree of f0; 1g� has an in�nite path.

2. WKL0 is the subsystem of Z2 consisting of RCA0 plus Weak K�onig'sLemma.

It is clear that REC is not a model of WKL0 so that WKL0 is a properextension of RCA0. The formal systemWKL0 was �rst introduced by Friedman[71]. !-models ofWKL0 are sometimes known as Scott systems in the literature,referring to [186]. The development of ordinary mathematics inWKL0 is carriedout in great detail by Simpson in [192].

The following equivalent forms of Weak K�onig's Lemma are frequently usedin the applications.

Theorem VII.1.8. The following are equivalent are equivalent over RCA0

1. WKL0, i.e. every in�nite tree T � f0; 1g<N has an in�nite path.

2. (�01 separation) Let �i(n), i = 0; 1 be �01 formulas in which X is does notoccur freely. If :9n(�0(n) ^ �1(n)), then

9X8n((�0(n)! n 2 X) ^ (�1(n)! n =2 X)):

3. If f; g : N! N are one-to-one with (8m;n)f(m) 6= g(n), then there existsa set X such that, for all m, f(m) 2 X ^ g(m) =2 X.

Proof. It is clear that (2) implies (1).(1) =) (2). Assume (1) and let T � NN and g be given as stated

and de�ne T � � f0; 1g� as follows. For any � 2 T with j� j = n, let �� =0�(0)10�(1) : : : 0�(n�1. Then de�ne T by �0

1 Comprehension so that � 2 T � ifand only if � � �� for some � 2 T with j� j � g(0) + g(1) + : : : g(j�j).


Then T � is an in�nite subtree of f0; 1g� and therefore possesses an in�nitepath f�. Now de�ne an in�nite path f 2 [T ] by primitive recursion so thatf(0) = the least k < g(0) such that f�(k) = 0 and for each n, f(n + 1) is theleast k < g(n+ 1) such that f�(k + g(n)) = 1:



2. (Bounded Konig's Lemma) If T � NN is an in�nite tree and there is afunction g such that for all � 2 T and all m < j� j, �(m) < g(m), then Thas an in�nite path.

Proof. (1) =) (2). Assume (1) and let �0; �1 be given as stated and let �0; �1be bounded quanti�er formulas so that �i(n) () (9m)�i(m;n). Now de�neT � f0; 1g� by �0

1 Comprehension so that

� 2 T () (8i < 2)(8m;n < j�j)[�i(m;n) =) �(n) 6= i]:

T is an in�nite tree and therefore has an in�nite path X by Weak K�onig'sLemma, which will satisfy the conclusion of (3).

(2) =) (1). Let T � f0; 1g� be an in�nite tree. De�ne the �01 formulas �iso that

�i(�) () (9n)(9� 2 f0; 1gn)[�_(i)_� 2 T & (8� 2 f0; 1gn):(�_(1�i)_� 2 T )]:

Then �0; �i satsify the hypothesis of (3), so there exists a set X such that forall �, �0(�)! � 2 X and �1(�)! � =2 X. We can now de�ne an in�nite paththrough T as follows. Let �0 = ; and for each k, let �k+1 = �k

_0 if �k 2 Xand otherwise �k+1 = �k

_1. Then f = [k�k belongs to [T ].(2) =) (3). Assume (2) and let f and g be given as stated. Let �0(n) ()

(9m)f(m) = n and �1(n) () (9m)g(m) = n. The hypothesis of (2) issatis�ed by assumption and therefore there exists X as in the conclusion of (3),which will also satisfy the conclusion of (3).

(3) =) (2). Assume (3) and let �i be given as stated. Apply LemmaVII.1.3 to obtain two cases. First there may exist �nite sets Xi = fn : �i(n)g.If this holds for i = 0, let X = X0 and if this holds for i = 1, let X =N � X1. If neither set exists, then there are one-to-one functions fi such that�i(n) () (9m)(fi(m) = n). It follows from the hypothesis of (3) that(8m;n)f(m) 6= g(n). Hence by the conclusion of (2), we obtain a set X suchthat, for allm, f(m) 2 X and g(m) =2 X. This setX then satis�es the conclusionof (3).

Scott [186] characterized the countable !-models of WKL0 as those M �P(N) such that there exists a complete extension � of Peano Arithmetic suchthat M is the family of subsets of N which are representable in �.


VII.1.3 Arithmetic Comprehension

In this section we consider the system ACA0 and its relation to �01 classes. A

formula is said to be arithmetical if it is �0k for some k.

De�nition VII.1.10. 1. Arithmetical Comprehension The arithmeti-cal comprehension scheme is (9X)[n 2 X () �(n)] where � is anarithmetical formula of L2 in which X does not occur freely.

2. ACA0 is the subsystem of Z2 whose axioms are arithmetical comprehen-sion, full induction and the basic axioms of Robinson arithemtic.

The !-models of ACA0 may be characterized as follows.

S is an !-model of ACA0 if and only if

S 6= ;;

A 2 S and B 2 S imply A�B 2 S;

A 2 S and B �T A imply B 2 S.

A 2 S implies A0 2 S.

It follows that ARITH is the minimum !-model for ACA0.

Theorem VII.1.11. The following are equivalent over RCA0.

1. ACA0.

2. �01 comprehension.

3. If f : N! N is an injection, then the range of f is a set.

Proof. The implications (1) Implies (2) and (2) =) (3) are trivial. Theimplication (2) =) (3) follows easily from Lemma VII.1.3. For the impli-cation (1) =) (2) we prove by induction that �0k comprehension implies�0k+1 comprehension. Let �(n) () (9j) (n; j) where is �0

k. By �0kcomprehension, let Y = f(n; j) : : (n; j)g. Then by �01 comprehension letX = fn : (9j)(n; j) =2 Y g.

Theorem VII.1.12. The following are equivalent over RCA0.

1. ACA0.

2. (K�onig's Lemma) If T is an in�nite, �nitely branching tree, then there isan in�nite path through T .

3. K�onig's Lemma restricted to trees T such that each � 2 T has at most twoimmediate successors in T .


Proof. (1) =) (2). Let T be an in�nite, �nite-branching tree. By arithmeticcomprehension, there is a subtree T � of T consisting of all � 2 T such that �has in�nitely many extensions in T . Since T is �nite branching, every � 2 T �has at least one immediate successor in T �. Clearly ; 2 T � and for each n, wemay de�ne g(k) to be the least n such that (g(0); : : : ; g(k � 1); n) 2 T �. Theng 2 [T ] as desired.

(2) =) (3) is immediate, so it remains to prove (3) =) (1). Let f : N! Nbe one-to-one. Use �00 comprehension to de�ne the tree T by � 2 T if and onlyif

(8m;n < j� j)[f(m) = n () �(n) = m+ 1]

and(8n < jtj)[�(n) > 0 =) f(�(n)� 1) = n]:

Each � 2 T has at most two possible immediate successors, �_0 and �_(m+1)where f(m) = j�j. T is in�nite by the following argument. Fix k and de�neY by bounded �01 comprehension to be fn < k : (9m)f(m) = ng. Now let�(n) = 0 if n =2 Y and �(n) = m + 1 if n 2 Y ^ f(m) = n for n < k. Thenj�j = k and � 2 T . Hence by (3), there exists g 2 [T ]. By �0

1 comprehensionlet X = fn : g(n) > 0g. Then X is the range of f as desired.

VII.2 Mathematical Logic

In this section, we consider the connection between logical theories, in�nite treesand subsystems of second order arithmetic.

A weak form of the completeness theorem can be proved even for �rst-orderlogic. Here is the propositional version.

Theorem VII.2.1. [[192]] The following is provable in RCA0. If Gamma isa consistent propositional theory, then there exists a countable model M for Xsuch that M ` � for all � 2 �.Proof. Recall from Theorem III.III.9.3 that, for each �nite sequence � = (�(0); : : : ; �(n�1)), we de�ned P� = C0^C1^� � �^Cn�1, where Ci = Ai if �(i) = 1 and Ci = :Aiif �(i) = 0. Given the theory �, de�ne the tree T without dead ends by

� 2 T () :P� =2 �g:

Since � is a theory, � 2 T if and only if P� is consistent with �. T has nodead ends since � is consistent. That is, if j�j = n and P� is consistent with �,then either P� ^ Ai is consistent with � or P� ^ :Ai is consistent with �. Itfollows from Theorem VII.1.6 that T has an in�nite path X and the model Mis de�ned by letting M ` Ai if X(i) = 1 and M ` :Ai if X(i) = 0.

Using Weak K�onig's Lemma, we can prove the completeness theorem andthe reverse is also true. (See [192].) We note �rst that the representationtheorem III.III.9.1 for propositional logic can be proved in RCA0, that is, for

VII.2. MATHEMATICAL LOGIC 189

any countable set � of sentences, there exists a tree T � f0; 1g� such that [T ]represents the set of complete consistent extensions of �. Likewise the reverserepresentation theorem III.III.9.3 can be proved in RCA0. That is, given thetree T , de�ne the set �(T ) to consist of all P� ! An such that � 2 T and�_0 =2 T and all P� ! :An such that � 2 T and �_1 =2 T , where j�j = n. Thenthere is a one-to-one correspondce between the complete consistent extensionsof �(T ) and the in�nite paths through T .



2. Lindenbaum's Lemma: every countable consistent set of sentences has acomplete consistent extension.

3. The completeness theorem for propositional logic with countably manyvariables.

4. The compactness theorem for propositional logic with countable many vari-ables.

Proof. (1) =) (2) follows from the representation theorem as discussed above.That is, given a consistent set �, we can build in RCA0 an in�nite tree repre-senting the set of complete consistent extensions of � and then use Weak K�onig'sLemma to �nd an in�nite path X through T and hence a complete consistentextension of �.

(2) =) (3). For propositional logic, this is immediate. Just let � be acomplete consistent extension of � and let M(Ai) = 1 if and only if Ai 2 �.

(3) =) (4). Suppose that every �nite subset of � is satis�able. Then � isconsistent and hence has a model M by (3) and is therefore satis�able.

(4) =) (1). Assume (4) and let T � f0; 1g� be an in�nite tree. Let �(T )be constructed as above. Then �(T ) is �nitely satis�able and hence has a model(and therefore a complete consistent extension) by (4). But this implies that Thas an in�nite path.

Exercises

VII.2.1. Show that a set of natural numbers is c. e. if and only if it is de�nableby a �01 formula over the standard model of arithmetic and therefore iscomputable if and only if it is �0

1 de�nable.

VII.2.2. Show that S � P(N) is a model of RCA0 if and only if

(i) S 6= ;;A 2 S and B 2 S imply A�B 2 S;A 2 S and B �T A imply B 2 S.


Chapter VIII

Complexity Theory

In this chapter, we examine the notions of computable trees and e�ectivelyclosed sets in a resource-bounded setting. We consider the complexity of themembers of e�ectively closed sets as in Chapter IV from this point of view. Weshow that for any �0

1 class P � NN, there is a polynomial time tree T suchthat P = [T ]. Resource-bounded variations on the notions of boundedness fortrees and classes are de�ned, such as locally p-time, highly p-time, and p-timebounded are de�ned and basis and antibasis results given. For example, everylocally p-time tree possesses an in�nite path which is computable in doubleexponential time and also, there is a p-time bounded �0

1 class with a uniqueelement which is not p-time computable.

We also look at the representation problem and show that for essentiallyall of the problems from Part 2, polynomial time presented problems su�ceto represent all �0

1 classes. This is based on the result that any computablerelational structure is computably isomorphic to a polynomial time structure.

Let � be a (usually �nite) alphabet. Then �� denotes the set of �nite stringsof letters from � and �! denotes the set of in�nite sequences. In particular, eachnatural number n may be represented in unary form by the string tal(n) = 1n

if n > 0 and tal(n) = 0 if n = 0 and in (reverse) binary form by the stringbin(n) = i0 � � � ik, where n = i0 + i1 � 2 + � � �+ ik � 2k.

We let Tal(!) = ftal(n) : n 2 !g and Bin(!) = fbin(n) : n 2 !g. Bothsets are included in f0; 1g�. The tally and binary representation of the naturalnumbers will be essential for our study of feasible structures, problems andsolutions. The main reason is due to the fact the feasibility of an algorithm isusually measured in terms of the computation time as a function of the length ofthe input to the algorithm. Note that since the tally representation of a numberis of exponential length in comparison to the binary representation, it followsthat a function which is polynomial time computable in the tally representationof the natural numbers is not necessarily polynomial time computable in thebinary representation of the natural numbers. Indeed, we can only concludethat such a function is exponential time computable in the binary representation.

191

192 CHAPTER VIII. COMPLEXITY THEORY

Thus it is essential that a de�nite representation be given for a feasible structure.A related reason is that two feasible sets need not be feasibly isomorphic. Inparticular, Tal(!) and Bin(!) are not p-time isomorphic. Thus we may have ap-time structure, say a graph, with universe Tal(!), which is not isomorphic toa p-time structure with universe Bin(!).

Our basic computation model is the standard multitape Turing machine ofHopcroft and Ullman [89]; see also Papadimitriou [167]. Note that there are dif-ferent heads on each tape and that the heads are allowed to move independently.This implies that a string � can be copied in linear time. An oracle machine isa multitape Turing machine M with a distinguished work tape, a query tape,and three distinguished states QUERY, YES, and NO. At some step of a com-putation on an input string �, M may transfer into the state QUERY. In stateQUERY,M transfers into the state YES if the string currently appearing on thequery tape is in an oracle set A. Otherwise, M transfers into the state NO. Ineither case, the query tape is instantly erased. The set of strings accepted byMrelative to the oracle set A is L(M;A) = f�j there is an accepting computationof M on input � when the oracle set is Ag. If A = ;, we write L(M) instead ofL(M; ;).

Let t(n) be a function on natural numbers. A Turing machine M is saidto be t(n) time bounded if each computation of M on inputs of length n wheren � 2 requires at most t(n) steps. A function f(x) on strings is said to be inDTIME(t) if there is a t(n)-time bounded deterministic Turing machine Mwhich computes f(x). For a function f of several variables, we let the lengthof (x1; : : : ; xn) be jx1j+ � � �+ jxnj. A set of strings or a relation on strings is inDTIME(t) if its characteristic function is in DTIME(t). A Turing machineM is said to be t(n) space bounded if each computation of M on inputs oflength n where n � 2 the work space required to carry out the computation isbounded by t(n) . A function f(x) on strings is said to be in DSPACE(t) ifthere is a t(n)-space bounded deterministic Turing machine M which computesf(x). For a function f of several variables, we let the length of (x1; : : : ; xn) bejx1j + � � � + jxnj. A set of strings or a relation on strings is in DSPACE(t) ifits characteristic function is in DSPACE(t).

We let

LOGTIME =Sc�1DTIME(c � log2(n)),

LOG =Sc�1DSPACE(c � log2(n)),

LIN =Sc>0DTIME(cn),

P =Si2!DTIME(ni),

PSPACE =Si2!DSPACE(n

i),

DEXT =Sc�0DTIME(2c�n),

193

EXPSPACE =Sc�0DSPACE(2

c�n),

DOUBEXT =Sc�0DTIME(22

c�n

),

DOUBEXPSPACE =Sc�0DSPACE(2

2c�n),

EXPTIME =Sc�0DTIME(2n

c

),

DOUBEXPTIME =Sc�0DTIME(22

nc

), and in general,

DEX(S) =St(n)2S DTIME(2t(n))g.

We say that a function f(x) is polynomial time if f(x) 2 P , is exponentialtime if f(x) 2 DEXT , and is double exponential time if f(x) 2 DDOUBEXT .

A function f(x) on strings is said to be in NTIME(t) if there is a t(n)-timebounded nondeterministic Turing machine M which computes f(x). A set ofstrings or a relation on strings is in NTIME(t) if its characteristic function isin NTIME(t). We let

NLOG =Sc�1NSPACE(C � log2(n)),

NP =Si2! NTIME(ni),

NEXT =Sc�0fNTIME(2c�n)g,

NEXPTIME =Sc�0NTIME(2n

c

),

A function f is said to be non-deterministic polynomial time (NP) if there isa �nite alphabet �, a polynomial p, a p-time relation R and a p-time functiong such that, for any � and � ,

f(�) = � () (9� 2 �p(�))[R(�; �) & g(�; �) = � ]:

Similar de�nitions apply to other complexity classes.

We �x enumerations fPigi2N and fNigi2N of the polynomial time boundeddeterministic oracle Turing machines and the polynomial time bounded non-deterministic oracle Turing machines respectively. We may assume that pi(n) =max(2; n)i is a strict upper bound on the length of any computation by Pi orNi with any oracle X on inputs of length n. PXi and NX

i denote the oracleTuring machine using oracle X.

For A; B � ��, we shall write A �Pm B if there is a polynomial-time functionf such that for all x 2 ��, x 2 A i� f(x) 2 B. We shall write A �PT B if Ais polynomial time Turing reducible to B. For r equal to m or T , we writeA �Pr B if A �Pr B and B �Pr A and we write A jPr B if not A �Pr B and notB �Pr A.


VIII.1 Complexity of Trees

We think of a computable tree T as a set of �nite sequences (n0; : : : ; nk�1) ofnatural numbers and of an in�nite path (n0; n1; : : : ) through T as a functionfrom the natural numbers into the natural numbers which maps i to ni. Weshall de�ne two natural representations of T which will be useful for the studyof the complexity of trees and paths through trees. First we de�ne the binaryrepresentation of T , bin(T ), as the set of �nite strings f(bin(n0); : : : ; bin(nk�1)) :(n0; : : : ; nk�1) 2 Tg. We also de�ne the tally representation of T , tal(T ), to bethe set of strings f(tal(n0); : : : ; tal(nk�1) : (n0; : : : ; nk�1) 2 Tg. The stringsin bin(T ) and tal(T ) are over the �nite alphabet f0; 1;0 ;0 ;0 (0;0 )0g which hassymbols for the comma and the left and right parentheses. We say that Tis p-time in binary if bin(T ) is a polynomial time subset of ��. Similarly wesay T is p-time in tally if tal(T ) is p-time subset of ��. Since bin(n) can becomputed in polynomial time from tal(n), it follows that if bin(T ) is p-time,then tal(T ) is also p-time. Given an in�nite path x = (n0; n1; : : : ) through T ,the binary representation of x is the function bin(x) from Tal(!) to Bin(!)de�ned by bin(x)(tal(i)) = bin(ni). The tally representation of x, tal(x), issimilarly de�ned by tal(x)(tal(i)) = tal(ni). Then we say that x is a polynomialtime path in binary if the function bin(x) is the restriction of p-time functionfrom f0; 1g� to f0; 1g�; we say that x is p-time in tally if tal(x) is the restrictionof a p-time function from f0; 1g� to f0; 1g�. It is clear that if x is p-time intally, then x is also p-time in binary, since bin(x)(tal(i)) can be computed fromtal(x)(tal(i)) for each i. The reason for using Tal(!) for the domain of bin(x) isthe following. For any path x, x is computable if and only if the initial segmentfunction x is computable, where x(i) = (n0; : : : ; ni�1). We want to have asimilar result for p-time paths, and this would be impossible if x had to mapbin(i), which has length roughly log2(i), to a string which must have length atleast i. Similar de�nitions can be given for other notions of complexity, such asexponential time, non-deterministic polynomial time (NP), etc.

Recall that a tree T � !<! is highly computable if there is a recursivefunction f such that, for any node � 2 T , f(�) is the number of immediatesuccessors �_i of � in T . Now given the number of successors of a node, wecan search through all the possible immediate successors and �nd the largestone. Thus we can �nd a computable function g such that g(�) is the largesti such that if � = (�0; : : : ; �n) 2 T , then (�0; : : : ; �n; i) is in T . Finally, wecan also compute recursively the sequence h(�) = (i1; : : : ; id) which lists alli such that (�0; : : : ; �n; i) is in T in increasing order. It is clear that f iscomputable if and only if g is computable and if and only if h is computable.The situation is di�erent for polynomial time complexity. Consider �rst thebinary representation of T so that we identify a node � 2 T with a sequence ofnumbers in Bin(!). It is not hard to see that if h is p-time, then both f and gare p-time. However, these are the only relations which are guaranteed to holdbetween the three functions. To see this, consider the following three examples.

Example VIII.1.1. De�ne the sequence x0; x1; : : : of natural numbers by let-

VIII.1. COMPLEXITY OF TREES 195

ting x0 = 1 and, for each n, xn+1 = 2xn and let T = f(x0; : : : ; xi�1) : i 2 !g.Then the tree T is p-time and f is p-time, since f(�) = 1 for all � 2 T . How-ever, the function g cannot be p-time since, for if � = (x0; : : : ; xn), then in thebinary representation j�j � 3jxnj whereas jxn+1j = 2jxnj.

Example VIII.1.2. De�ne the tree T1 computably by putting ; 2 T1 and,for any � = (x0; : : : ; xn�1) 2 T1, putting �

_i 2 T1 if and only if i � 1 +x0 + � � � + xn�1. T1 is clearly p- time and the function g is also p-time sinceg((x0; : : : ; xn�1)) = 1 + x0 + � � � + xn�1. However, the function h which liststhe immediate successors of any node is not p-time because, for any n, if � =(1; 2; 4; : : : ; 2n), then h(�) = (0; 1; : : : ; 2n+1), so that in the binary representationjh(�)j > 2n+1, whereas j�j = (n+ 2)(n+ 3)=2.

Example VIII.1.3. For this example, we will appeal to the intractability ofthe well-known P = NP conjecture. That is, we will de�ne a p-time tree T2 forwhich the function g is p-time and such that if the associated function f were p-time, then the P = NP conjecture would be true. The tree T2 will be de�ned sothat � = (n0; n1; : : : ; n2k+1) 2 T2 if and only if, for each i � k, bin(n2i) codes agraph on i vertices and bin(n2i+1) either codes a Hamiltonian path on the graphcoded by bin(n2i) or is a string of 1's of the appropriate length. Now a graphGi on i vertices v1; : : : ; vi is determined by a set of unordered pairs (vr; vs) ofvertices (the edges of the graph). There are

�i2

�= i(i � 1)=2 possible edges in

Gi and these may be lexicographically ordered so that a sequence e1; e2; : : : ; e(i2)

codes the graph Gi where, for all t � �i2�, et = 1 if Gi has the t'th edge andet = 0 otherwise. Of course the (reverse) binary representation bin(ni) mustend with a 1, so the graph Gi will actually be coded by the string (e0; : : : ; e(i2)

; 1).

Observe that this code for Gi will always be a string of length 1 +�i2

�and that

any binary number bin(n) of length 1+�i2

�will code a graph on i vertices. Now

a Hamiltonian path on Gi is a permutation (vr0 ; vr1 ; : : : ; vri) of the vertices suchthat there is an edge joining vrt with vrt+1 for all t < i. Such a path will be codedby the binary sequence 0r010r11 : : : 0ri1, which will always be a binary numberof length

�i+12

�and binary number bin(n) of length

�i+12

�will code a possible

Hamiltonian path on a graph of i vertices if and only if bin(n) has exactly i 1's.It is easy to see that there is a p-time algorithm which will decide, given twobinary numbers bin(n) and bin(m), whether bin(n) has length

�i2

�+ 1 for some

i < jbin(n)j and therefore codes a graph G on i vertices and whether bin(m) codesa Hamiltonian path on that graph. The tree T2 can now be de�ned by putting� = (bin(n0); : : : ; bin(n2k+1) 2 T2 if and only if, for each i < k, bin(n2i) codes agraph Gi on i vertices and bin(n2i+1) either codes a Hamiltonian path on Gi orequals tal(

�i+12

�). It follows from the discussion above that T2 is a p-time tree.

Now the function g for this tree is p-time since, for any � = (�0; : : : ; �t) 2 T2,we have g(�) = 2(

i+12 ) � 1 if t = 2i and g(�) = 21+(

i

2) � 1 if t = 2i � 1.(In each case, g(�) is just a string of 1s of the right length.) On the otherhand, the function f associated with the tree T2 has the property that for any� = (�0; : : : ; �2i) 2 T2, f(�) = 1 if and only if the graph Gi coded by �2i hasno Hamiltonian path. Now suppose that f were p-time and let bin(n) be a code


for a �nite graph on k vertices. For all i < k, let n2i = 2(i+12 ) � 1 and let

n2i+1 = 21+(i+12 ) � 1. Finally, let � = (bin(n0); : : : ; bin(n2k�1); bin(n)). Then

the sequence � can be computed from bin(n0) in polynomial time and G has aHamiltonian path if and only if f(�) > 1. It follows that if f were p-time, thenthe Hamiltonian path problem would be p-time. But it is well-known that theHamiltonian path problem is NP-complete. (See Garey and Johnson [6] for anexplanation of NP-completeness and the P = NP problem.) Thus we havedemonstrated that if the function f associated with the tree T2 were p-time, thenP = NP would true.

Now the situation is slightly di�erent for tally representation of T wherewe identify a � 2 T with a sequence of numbers in Tal(!). Once again it iseasy to see that if h is p-time, then f and g are p-time. Moreover, example (1)above will still show that f may be p-time without g being p-time. Howeverin this case, if T is p-time in tally and g is p-time, then h is also p-time. Tosee this, suppose � = (�0; : : : ; �n). Note that to �nd h(�), we need only checkwhether (�0; : : : ; �n; i) 2 T for i � g(�). Now in the tally representation,j(�0; : : : ; �n; 0)j < j(�0; : : : ; �n; 1)j < : : : < j(�0; : : : ; �n; g(�))j. Then if it takesq(j�j) steps to check whether � 2 tal(T ) for each � = (tal(�0); : : : ; tal(�n)),than it will take approximately

j(�0;:::;�n;g(�))jXi=0

q(i) � q(j(�0; : : : ; �n; g(�))j)2

steps to check whether (�0; : : : ; �n; i) 2 T for i � g(�). Thus it is easy to seethat we can �nd h(�) in polynomial time in the tally representation of �.

We say that a tree T is locally p-time in binary (respectively in tally) if allthree of the functions de�ned above are p-time in binary (resp. tally). In thecase that T is not itself p-time, then we will say that T is locally p-time if eachof the functions is the restriction to T of a function which is p-time (in binaryor tally).

Next we will show that if T is locally p-time, then T is also p-time. Thesame argument works for both binary and tally. Let Q be either bin(T ) ortal(T ) and suppose that the function h associated with Q is p-time. Given asequence � = (�0; : : : ; �k), here is the procedure for testing whether � 2 tal(T ).Begin by computing h(;) = (tau1; : : : ; taud) and checking whether �0 = �i forsome i � d. Then, for j < k in turn, compute h(�0; : : : ; �j) and check to seethat �j+1 is in this list. Suppose that h(�) may be calculated in time p(j� j),where p is some polynomial, then since each (�0; : : : ; �j) is a substring of �, wesee that we can do each of the h computations in time no greater than p(j�j).To read the resulting list of possible successors of (�0; : : : ; �j) and compare eachone with �j+1 can then be done in time at most (c � 1)p(j�j) for some �xedconstant c. Thus each step of the procedure takes time at most cp(j�j). Nowthere are k such steps and k � j�j, so that the entire procedure takes time atmost cj�jp(j�j), which is again a polynomial function of j�j.


The functions f , g and h describe the behavior of the tree at a particularnode. Now sometimes we need to have a global bound as well. Note that fora computably bounded tree, there is a computable function p such that, for allnatural numbers k and all � = (n0; : : : ; nk) 2 T , nk � p(k). We will say thata tree T is p-time bounded in binary if there is a p-time function p such that,for all natural numbers k and all � = (n0; : : : ; nk) 2 T , jbin(nk)j � p(1k). Atree T is p-time bounded in tally if there is a p-time function p such that forany � = (n0; : : : ; nk) 2 T , we always have nk = jtal(nk)j � p(1k). Since we cancompute bin(n) from tal(n) in polynomial time, it follows that any tree whichis p-time bounded in tally is also p-time bounded in binary. Note that any treeT � f0; 1g<! is p-time bounded, so that a tree may be p-time bounded withoutbeing p-time. One additional observation is worth making at this point. If T isp-time bounded in tally, then there will also be a p-time function q such that,for any � = (n0; : : : ; nk) 2 T , jtal(�)j � q(1k). To see this, note that � consistsof the strings tal(ni) for i � k, separated by commas and with parentheses atthe beginning and end. Thus

j� j = 2 + k + jn0j+ � � �+ jnkj � 2 + k + p(10) + � � �+ p(1k):

Thus we can de�ne a p-time bound q(1k) = 2+ k+ p(10)+ � � �+ p(1k), which isclearly p-time computable. The same observation holds for p-time bounded inbinary.

Now suppose that T is p-time bounded in tally and that Tal(T ) is p-time.This implies that there are at most p(1k) possible choices for tal(nk), that is, thestrings 1e for e < p(1k). To compute h(�), where � = (tal(n0); : : : ; tal(nk�1)),we simply use the p-time algorithm for membership in tal(T ) to test whether� � tal(i) 2 tal(T ) for all i � p(1k) and compile the list (tal(i1); : : : ; tal(id)) =h(�) of all tal(i) such that � � tal(i) 2 T . This shows that the function h isp-time in tally. It then follows by the discussion above that g and f are alsop-time in tally. Hence in the tally representation, any p-time bounded, p-timetree is also locally p-time.

Let us say that a tree T is highly p-time in binary if T is p-time, locallyp-time and also p-time bounded in binary. Similarly, T is highly p-time in tallyif T is p-time, locally p-time and also p-time bounded in tally. Then we haveshown that in tally, p-time plus p-time bounded implies highly p-time. On theother hand, we have also seen that these notions are distinct for the binaryrepresentation.

Similar de�nitions can be given for other notions of complexity. Our nexttheorem shows that any �0

1-class can be realized as the set of in�nite pathsthrough a p-time tree.

Theorem VIII.1.4. Let T be a computable tree. Then there is a polynomialtime tree P such that [T ] = [P ]. Furthermore, if T is computably bounded, thenP is also computably bounded and if T is p-time bounded, then P is also p-timebounded.

Proof. The same argument works for the binary and for the tally representation.We will give the binary argument for the �rst part and the tally argument for


the second part, since these are the stronger results. Let � be a computablefunction from !<! into f0; 1g such that � 2 bin(T ) () �(�) = 1. Let �s

denote the partial computable function which results by computing � for exactlys steps on any input and let T s be the s'th approximation to T , given by

� 2 T s () �s(bin(�)) = 1 or is unde�ned :

Thus T 0 � T 1 � � � � and, for any �, � 2 T () (8s)(� 2 T s).Now de�ne the p-time tree P by letting

� 2 P () (8� � �)� 2 T jbin(�)j:Note that P is a p-time tree in binary since to compute whether � 2 T jbin(�)j

requires jbin(�)j steps for all � so that to compute whether � 2 P requiresroughlyjbin(�)j(jbin(�)j+ 1) steps.

It follows from the de�nition of P that T � P , so that [T ] � [P ]. Nowsuppose that x =2 [T ]. Then there is some initial segment � = x � n which is notin T. This means that, for some s, � =2 T s. Since the sequence T s is decreasing,we may assume that s > n. Now let � = x � s, so that jbin(�)j � s. It followsfrom the de�nition of P that � =2 P . This implies that x =2 [P ]. Thus [T ] = [P ].

Now suppose that T is computably bounded in tally and let p be the com-putable function which computes, for each k, an upper bound p(1k) (in tally)for the possible value of nk for any node � = (n0; : : : ; nk) 2 T .

Suppose �rst that p is actually p-time. Then we can recursively de�ne a treeQ such that T � Q � P by putting � = (n0; : : : ; nk) 2 Q if and only if � 2 Pand, for all i � k, ni � p(1i). It is clear that [Q] = [T ] and that Q is p-timesince P and p are p-time.

Finally, suppose only that p is recursive and let ps be the usual result ofcomputing p for s steps. Once again we can de�ne a highly recursive tree Qsuch that T � Q � P by putting � = (n0; : : : ; nk) 2 Q if and only if � 2 P and,for all i � k, either pk(1i) is unde�ned or ni � pk(1i). Then again it is easy tocheck that Q is p-time in binary and that [Q] = [T ].

Next we would like to consider conditions which might force the tree Tto have a p-time (exponential time, etc.) path. Recall that a �0

1 class P isdecidable if P = [T ] for a computable tree with no dead ends (or with Ext(T )computable) and that a decidable �0

1 class always has a computable member.Recall also that any �0

1 singleton is necessarily computable. Next we show thatthe obvious p-time analogues of these results fail for p-time decidable trees.

Theorem VIII.1.5. For any computable x 2 f0; 1g!, there is a tree T whichis polynomial time in binary and in tally and such that [T ] = fxg.

Proof. This follows from Theorem VIII.1.4, since for x 2 f0; 1g!, the tree T =f(x(0); : : : ; x(n� 1) : n < !g is computable


Theorem VIII.1.5 shows that even if a polynomial time bounded p-time treehas a unique in�nite path �, � may not be polynomial time. However thereare some natural conditions which we can put on T which will ensure that insuch situations we can at least get double exponential time paths or winningstrategies and in some cases actually guarantee the existence of polynomial timepaths or winning strategies.

Theorem VIII.1.6. (a) Let Ext(T ) be a locally p-time tree in tally (respec-tively binary) and let [T ] be nonempty. Then [T ] contains an in�nite pathwhich is double exponential time computable in tally (resp. binary). Fur-thermore, if Ext(T ) is locally p-time in tally (resp. binary) and [T] is�nite, then every element of [T ] is computable in double exponential timein tally (resp. binary).

(b) Let Ext(T ) be a locally p-time tree in tally (respectively binary) and let[T ] be nonempty. Moreover, assume that there is a linear time functionh such that for all � = (n0; : : : ; nk) 2 T , h(b(�)) lists all b(n) such that(n0; : : : ; nk; n) 2 T where b() = tal() if T is p-time in tally and b() = bin()if T is p-time in binary. Then [T ] contains an in�nite path which isexponential time computable in tally (resp. binary). Furthermore, if [T ]is �nite, then every element of [T ] is computable in exponential time intally (resp. binary)

(c) If Ext(T ) is a highly p-time tree in tally (resp. binary) and [T ] is nonempty,then [T ] contains an in�nite path which is p-time time in tally (resp. bi-nary). Furthermore, if [T ] is �nite, then every element of [T ] is p-time intally (resp. binary).

(d) If Ext(T ) is a p-time bounded, p-time tree in binary and [T ] is nonempty,then [T ] contains an in�nite path which is EXPTIME in binary. Fur-thermore, if [T ] is �nite, then every element of [T ] is NP in binary.

Proof. To simplify the discussion, we will assume in all cases that T = Ext(T ),that is, that T has no dead ends. Thus the conditions set out for Ext(T ) willbecome the conditions for T .

(a) We give the proof for the binary representation of T . The proof for thetally representation is exactly the same except for replacing bin(:::) with tal(:::)at appropriate locations throughout. Let h be the p-time function such that forall � = (n0; : : : ; nk) 2 T , h(bin(�)) lists all bin(n) such that (n0; : : : ; nk; n) 2 T .Then we can recursively de�ne the p-time path x through T by letting x(k) bethe number n such that bin(n) is the �rst entry of h(bin(x � k)). It remains tobe checked that the computation of bin(x(n)) from 1n can be done in doubleexponential time. Let c be a number such that h(�) can be computed from �in time bounded by j� jc�1 for all � 2 Bin(Ext(T )) with j� j � 2. For each k, let�k = (bin(x(0); : : : ; bin(x(k � 1)). Then �0 = ; and, for each k > 0, �k+1 canbe computed from �k in time bounded by j�kjc. (Just start the computationof h(�k), stop it as soon as you have the �rst element � = bin(x(k)) in the listand then append � to the end of �k). Thus in particular j�k+1j � j�kjc for all


k > 0. Now choose c large enough so that j�1j � 2c. It then follows by induction

that, for all k, j�kj � 2ck

. It follows that the computation of �k+1 from �k can

be done in time bounded by (2ck

)c�1 � 2ck+1

. Thus the entire computation of�n = (bin(x(0)); : : : ; bin(x(n))) from 1n takes time bounded by

�k<n2ck+1 < 2c

n+1 < 2(c+1)n

;

which shows that x is computable in double exponential time in binary.(b) Now suppose that [T ] is �nite and let x 2 [T ]. By resticting T to the

extensions of x � n for su�ciently large n, we may assume that x is the uniquein�nite path through T . The result now follows immediately from the �rst partabove.

(c) The proof is essentially the same as the proof of (a). Again we shall onlygive the proof in the case that T is p-time in binary. The point is that if h islinear time then it follows that for all k � 1, �k+1 can be computed from �kin time c � j�kj for some �xed constant c. If we pick c so that c � j�0j, then itis easy to prove by induction that j�kj � ck+1 for all k � 0. Thus the entirecomputation of �n = (bin(x(0)); : : : ; bin(x(n))) from 1n takes time bounded by

�k<nck+2 < (n+ 1)cn+1 < c2n+2;

which shows that x is computable in exponential time in binary.Now if [T ] is �nite and x 2 [T ], then again by resticting T to the extensions

of x � n for su�ciently large n, we may assume that x is the unique in�nite paththrough T . Then by the above argument if follows that x 2 DEXT .

(d) The proof will be a minor modi�cation of the proof of (a) above. Againthe proofs are the same for tally and for binary so we will just give a binaryversion. Let q be a p-time function such that for any � = (bin(n0); : : : ; bin(nk) 2bin(T ), j� j � q(1k). Since j1kj = k, it follows that for some constant b and allk > 1, we have j�kj � kb. Let c be a number such that h(�) can be computedfrom � in time bounded by j� jc�1 for all � 2 Bin(Ext(T )) with j� j > 1. Thenthe computation of �k+1 from �k can be done in time bounded by j�kjc � kbc

for all k > 1. Now let a be large enough so that a > bc and also large enoughso that �0 and �1 can both be computed in time bounded by a. Then the entirecomputation of �n = (bin(x(0)); : : : ; bin(x(n))) from 1n takes time bounded by

a+ 2a + 3a + � � �+ (n� 1)a � �k<nak � n2a;which shows that x is computable in polynomial time in binary.If we assume further that [T ] is �nite, then the same argument as given in (a)

and (b) above shows that every element of [T ] is polynomial time computable.(d) As in (c), we may assume that if � = (bin(n0); : : : ; nk) 2 T and k > 1,

then j� j � kb so that in particular nk � kb. In this case, we are not assumingthat T is locally p-time, so that we need a di�erent algorithm for producing anin�nite path x in [T ]. We will de�ne x(k) recursively by making x(k) be the leastnumber n such that (x(0); : : : ; x(k � 1); n) 2 T . This means that we may haveto check whether (x(0); : : : ; x(k� 1); x) 2 T for all x with jbin(x)j � kb. This is


where the binary representation di�ers from the p-time representation, because

there will now be 2kb

di�erent strings to check. Each check will require time atmost (kb)c, so that the computation of bin(x(k)) from (bin(x(0)); : : : ; bin(x(k�1)) will require time less than 2k

bc+b

for k > 1. Now let a be large enough so thata � bc+b and also large enough so that bin(x(0)) and bin(x(1)) can be computedin time � a. Then the entire computation of �n = (bin(x(0)); : : : ; bin(x(n)))from 1n takes time bounded by

a+ 22a

+ 23a

+ � � �+ 2n�1)a � �k<na2

k � 2n2a

;

which shows that x 2 EXPTIME.If we assume further that [T ] is �nite, then the same argument as given in

(a) above shows that every element of [T ] is EXPTIME. However, it is easyto show that the in�nite paths through T are actually NP computable.

As above, we may assume that T has no dead ends and has a unique in�nitepath x. Thus for any k, x(0); x(1); : : : ; x(k)) is the unique �nite path in T withk + 1 entries. Furthermore, since T is p-time bounded, we know as above thatj(bin(x(0)); : : : ; bin(x(k)))j � kb for some �xed b. Thus to compute x(k) non-deterministically, we simply guess a string � = (bin(n0); : : : ; bin(nk)) of length� kb and then use the p-time algorithm for T to test whether � 2 T . When theanswer is yes, we read the value of x(k) from the end of �. Since there is onlyone possible correct guess for �, this procedure will compute x(k).

Next we shall give two examples to show that the bounds given in parts (a)and (b) of Theorem VIII.1.6 can not be improved. Consider the following.

Example VIII.1.7. A locally p-time tree T with a unique in�nite path x suchthat T = EXT (T ) and x is double exponential time.

Let x(n) = 22n

for all n and let the tree T consist of all initial segments ofx. Then (n0; : : : ; nk) 2 T if and only if n0 = 1 and, for all i < k, ni+1 = n2i .It is clear that both tal(T ) and bin(T ) are p-time. Furthermore, for any � =(n0; : : : ; nk) 2 T , we have h(�) = n2k, so that T is locally p-time in both binaryand tally.

Example VIII.1.8. A locally p-time tree T with a unique in�nite path x suchthat T = EXT (T ), there is a linear time function h such that for all � =(n0; : : : ; nk) 2 T , h(b(�)) lists all b(n) such that (n0; : : : ; nk; n) 2 T whereb() = tal() if T is p-time in tally and b() = bin() if T is p-time in binary, andx is exponential time.

Let x(n) = 2n for all n and let the tree T consist of all initial segments of x.Then (n0; : : : ; nk) 2 T if and only if n0 = 1 and, for all i < k, ni+1 = 2ni. Itis clear that both tal(T ) and bin(T ) are p-time and that the function h is lineartime in both cases, so that T is locally p-time in both binary and tally.

For �01 classes in f0; 1gN, boundedness conditions are not needed and tally

and binary representations are identical. Here are the basis results for classesof various complexity. These will be applied later to logical theories and othermathematical examples.


Theorem VIII.1.9. Let P = [T ] be a �01 class in f0; 1gN, where T has no dead

ends.

(a) If T is computable in time nlog(n)O(1), then P has a member which iscomputable in time nlog(n)O(1).

(b) If T is LIN , then P has a member which is computable in time O(n2).

(c) If T is PTIME, then P has a PTIME member.

(d) If T is DEXT , then P has a DEXT member.

(e) If T is EXPTIME, then P has an EXPTIME member.

Proof. Part (c) follows from Theorem VIII.1.6. We give the proof of (d) andleave the others as an exercise 2.

As in the proof of Theorem VIII.1.6, we compute the desired path recursivelyso that x(k) = 1 if (x(0); : : : ; x(k � 1); 0) 2 T and x(k) = 1 otherwise. Byassumption, there exists c so that this last step in the computation of x(k)takes time � 2c(k+1) and thus the total computation requires time

� 2c + 22c + � � �+ 2c(k+1) � 2c(k+2) = 22c � 2ck;

plus a little time for bookkeeping.

Theorem VIII.1.10. Let P = [T ] be a �01 class in f0; 1gN, where T has no

dead ends.

(a) If T is LINSPACE, then P has a member which is computable in LINSPACE.

(b) If T is PSPACE, then P has a PSPACE member.

(c) If T is EXPSPACE, then P has an EXPSPACE member.

Proof. We just sketch the proof of (a) and leave the others to the reader. Tocompute x(k), we need to have stored x(0); : : : ; x(k � 1), whoch requires spacek and then test whether x(0); : : : ; x(k � 1); 0) 2 T , which requires additionalspace ck, where c is �xed. This additional space may be reused, so that we canstay in LINSPACE.

xxxxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxxxxxExercises

VIII.1.1. Show that there is a highly computable tree T with no dead ends suchthat there is not highly p-time tree wihout dead ends such that [T ] = [S].

VIII.1.2. Show that if T is decidable in time nlog(n)O(1), then T has an in�nitepath which is computable in time nlog(n)O(1), if T is decidable in linearspace, then T has a in�nite path which is computable in time O(n2) andif T is EXPTIME, then [T ] has an EXPTIME member.

VIII.2. COMPLEXITY OF STRUCTURES 203

VIII.2 Complexity of Structures

Complexity theoretic or feasible model theory is the study of resource-boundedstructures and isomorphisms and their relation to computable structures andcomputable isomorphisms. This subject has been developed during the 1990'sby Cenzer, Nerode, Remmel and others. See the survey article [38] for an intro-duction. Complexity theoretic model theory is concerned with in�nite modelswhose universe, functions, and relations are in some well known complexityclass such as polynomial time, exponential time, polynomial space, etc. Byfar, the complexity class that has received the most attention is polynomialtime. One immediate di�erence between computable model theory and com-plexity theoretic model theory is that it is not the case that all polynomial timestructures are polynomial time equivalent. For example, there is no polynomialisomorphism f with a polynomial time inverse f�1 which maps the binary rep-resentation of the natural numbers Bin(N) = f0g [ f1gf0; 1g� onto the tallyrepresentation of the natural numbers Tal(N) = f1g�. This is in contrast withcomputable model theory where all in�nite computable sets are computablyisomorphic so that one usually only considers computable structures whose uni-verse is the set of natural numbers N.

There are two basic types of questions which have been studied in polynomialtime model theory. First, there is the basic existence problem, i.e. whether agiven in�nite computable structure A is isomorphic or computably isomorphicto a polynomial time model. That is, when we are given a class of structures Csuch as a linear orderings, Abelian groups, etc., the following natural questionsarise.

(1) Is every computable structure in C isomorphic to a polynomial time struc-ture?

(2) Is every computable structure in C computably isomorphic to a polynomialtime structure?

For example, the authors showed in [34] that every computable relational struc-ture is computably isomorphic to a polynomial time model and that the standardmodel of arithmetic (!;+;�; �; <; 2x) with addition, subtraction, multiplication,order and the 1-place exponential function is isomorphic to a polynomial timemodel. The fundamental e�ective completeness theorem says that any decidabletheory has a decidable model. It follows that any decidable relational theory hasa polynomial time model. These results are examples of answers to questions(1) and (2) above. However, one can consider more re�ned existence questions.For example, we can ask whether a given computable structure A is isomorphicor computably isomorphic to a polynomial time model with a standard universesuch as the binary representation of the natural numbers, Bin(N), or the tallyrepresentation of the natural numbers, Tal(N). That is, when we are given aclass of structures C, we can ask the following questions.

(3) Is every computable structure in C isomorphic to a polynomial time struc-ture with universe Bin(N) or Tal(N)?


(4) Is every computable structure in C computably isomorphic to a polynomialtime structure with universe Bin(N) or Tal(N)?

It is often the case that when one attempts to answer questions of type (3)and (4) that the contrasts between computable model theory and complexitytheoretic model theory become more apparent. For example, Grigorie� [80]proved that every computable linear order is isomorphic to a Ptime linear orderwhich has universe Bin(N). However Grigorie�'s result can not be improvedto the result that every computable linear order is computably isomorphic to aPtime linear order over Bin(N). For example, Cenzer and Remmel [34] provedthat for any in�nite polynomial time set A � f0; 1g�, there exists a computablecopy of the linear order ! + !� which is not computably isomorphic to anypolynomial time linear order which has universe A. Here ! + !� is the orderobtained by taking a copy of ! = f0; 1; 2; : : :g under the usual ordering followedby a copy of the negative integers under the usual ordering.

The general problem of determining which computable models are isomor-phic or computably isomorphic to feasible models has been studied by the au-thors in [34], [35], and [37]. For example, it was shown in [35] that any com-putable torsion Abelian group G is isomorphic to a polynomial time group Aand that if the orders of the elements of G are bounded, then A may be takento have a standard universe, i.e. either Bin(N) or Tal(N). It was also shown in[35] that there exists a computable torsion Abelian group which is not isomor-phic, much less computably isomorphic, to any polynomial time (or even anyprimitive recursive) group with a standard universe. Feasible linear orderingswere studied by Grigorie� [80], by Cenzer and Remmel [34], and by Remmel[175, 176]. Feasible vector spaces were studied by Nerode and Remmel in [160]and [161]. Feasible Boolean algebras were studied by Cenzer and Remmel in[34] and by Nerode and Remmel in [159]. Feasible permutation structures andfeasible Abelian groups were studied by Cenzer and Remmel in [35] and [37].By a permutation structure A = (A; f), we mean a set A together with a unaryfunction f which maps A one-to-one and onto A.

General conditions were given in [38] which allow the construction of modelswith a standard universe such as Tal(N) or Bin(N) and these conditions wereapplied to graphs and to equivalence structures. For example, it was shownthat any computable graph with all but �nitely many vertices of �nite degreeis computably isomorphic to a polynomial time graph with standard universe.On the other hand, a computable graph was constructed with every vertexhaving either �nite degree or �nite co-degree (i.e. joined to all but �nitely manyvertices) which is not computably isomorphic to any polynomial time graph witha standard universe. An equivalence structure A = (A;RA) consists of a set Atogether with an equivalence relation. It was also shown that any computableequivalence structure is computably isomorphic to a polynomial time structurewith a standard universe.

In this section, we want to consider the connection between computablestructures and resource-bounded structures and the corresponding connectionbetween computable trees and resource-bounded trees as developed in section

VIII.2. COMPLEXITY OF STRUCTURES 205

VIII.1.A relational structure is simply a structure which has no functions. We

will present an improved version of the theorem (�rst due to Grigorie� [80])from [34] that every computable relational structure is computably isomorphicto a polynomial time structure. This theorem will be our primary tool in theanalysis of computable combinatorial structures. It is important to note thatthe polynomial time structure provided will have for its universe a polynomial-time set possibly di�erent from f1g� or f0; 1g�. An example is constructed in[34] which shows that the theorem fails if any �xed polynomial time set A isspeci�ed in advance as the universe of the structure. The improved versionof the theorem presented here applies to structures with two distinct types ofobjects, the �rst type being the normal universe of the structure, and withfunctions which map the �rst type into the second type. The type of examplethat we have in mind is a function from the vertices of a graph into the naturalnumbers which computes the degree of a vertex. The universe of the graph isnow expanded by adding a p-time set which represents the natural numbersand the degree function now becomes part of the structure. Naturally, the newobjects are not vertices and therefore are not joined to any other objects byedges.

Theorem VIII.2.1. Let

C = (C;A;B; fRCi gi2S ; ffCi gi2T ; );

be a computable structure such that

(i) A and B are disjoint subsets of C with C = A [ B and B is a polynomialtime set.

(ii) there is a computable isomorphism from Bin(!) onto a subset of Bin(!)nBwith a p-time inverse.

(iii) for each i 2 T , fi maps C into B.

(iv) for each i 2 S, the relation Ri is independent of B, that is, for any(x1; : : : ; xn) 2 Cn, where n = s(i), any j � n such that xi 2 B, andany b 2 B, RCi (x1; : : : ; xn) if and only if RCi (x1; : : : ; xj�1; b; xj+1; : : : ; xn).

(v) for each i 2 T , the function fi is independent of B, that is, for any(x1; : : : ; xn) 2 Cn, where n = t(i), any j � n such that xi 2 B, andany b 2 B, fCi (x1; : : : ; xn) = fCi (x1; : : : ; xj�1; b; xj+1; : : : ; xn).

Then there is a computable isomorphism � of C onto a p-time structureM suchthat �(b) = b for all b 2 B.Proof. The idea of the proof is that we will replace each element x of A by astring y which codes x and is long enough to allow us to compute whether x 2 Ain time jyj and also to compute the relations and functions on A in time jyj forall inputs which are less than or equal to x. These new strings may accidentally


be in the set B, which must be kept disjoint from AM. This is the reason for thep-time mapping which takes an arbitrary string to one which is not in B. Let be a p-time map from Bin(!) into Bin(!) nB such that �1 is also p-time.We can assume that A is an in�nite set, since, if A is �nite, then C is p-timeitself. Let �0; �1; : : : be an e�ective enumeration of A in the usual order. Let b0be the shortest element of B. For any x 2 A, we let �(x) denote the number ofsteps needed to run the following algorithm.

First start to list �0; �1; : : : until we �nd an s such that �s = x. Nextfor each i � s such that i 2 S [ T , list all sequences (x1; : : : ; xn)from fb0; �0; : : : ; �sgn for n = s(i) or t(i) and then, for i 2 S,compute whether Ri(x1; : : : ; xn) holds and, for i 2 T , computefCi (x1; : : : ; xn).

Observe that the algorithm is completely uniform in x because our de�nitionof computable structure ensures that there is a computable relation R such thatR(i; hx1; : : : ; xt(i)i) () Ri(x1; : : : ; xt(i)) and a computable function f suchthat f(i; hx1; : : : ; xt(i)i) = fi(x1; : : : ; xt(i)) Note that in order to obtain the list�0; : : : ; �s, we have to test whether a 2 A for all a � x. We then de�ne astructure

M = (M; fRMi gi2S ; ffMi gi2T )as follows. For each a 2 A, let �(a) = (a_0_1�(a)) and, for each b 2 B, welet �(b) = b. It is clear that � is a computable isomorphism from C onto asubset M of Bin(!), that �(B) = B and that �(A) is disjoint from B. Thestructure M is the image of C under the isomorphism �. This means thatAM = f�(a) : a 2 Ag, BM = B, and M = AM [ BM. For each i 2 S and(x1; : : : ; xn) 2 C, RMi is de�ned by

RMi (�(x1; : : : ; �(xn)) () RAi (x1; : : : ; xn);

where s(i) = n. For each i 2 T , fMi is de�ned by

fMi (�(x1); : : : ; �(xn)) = �(fAi (x1; : : : ; xn));

where t(i) = n:It is clear that the function � is a computable isomorphism from A ontoM.

It remains to be seen thatM is a polynomial time structure, that is, that M isa polynomial time set and that each relation RM and function fM is p-time.

We show that M is p-time as follows. It clearly su�ces to show that AM isp-time, since BM = B is p-time. The procedure for testing whether an inputy is in AM is to compute �1(y), check to make sure that it has a 0 in it, andthen determine x and n such that �1(y) = x_0_1n. Then we simply run thealgorithm outlined above to input x for n steps. Then y 2 AM if and only ifthe algorithm terminates in exactly n steps and gives the answer that x 2 A.

We show that the function fMi is p-time as follows. Fix i and let f = fi,let n = t(i) and let c be the maximum amount of time required to computefC(x1; : : : ; xn) when fx1; : : : ; xng � fb0; �0; �1; : : : ; �i�1g. Now given input

VIII.3. PROPOSITIONAL LOGIC 207

(y1; : : : ; yn), where each yi 2 M , the procedure for computing fM(y1; : : : ; yn)is the following. First replace every xi 2 B with x0i = b0 and let x0i = xifor xi 2 A. Then compute fC(x01; : : : ; x

0n). We claim that this computation

takes time at most c + maxfjyj j : 1 � j � ng. There are two cases of thisclaim to consider. First, if fx01; : : : ; x0ng is a subset of fb0; �0; : : : ; �i�1g, then,by the de�nition of c, the computation takes at most c steps. On the otherhand, if at least one of the x0j = xj = �s for some s � i, then by the de�nitionof �, the computation takes less than �(xj) steps for some j; but of course�(xj) < jyj j � maxfjyj j : 1 � j � ng.

The argument for the relations is similar. This completes the proof of The-orem VIII.2.1.

For an example, let (N; R; 0; f) be a computable structure where R is abinary relation de�ning a tree with root 0 on the set of even numbers and f isan injection mapping the even numbers onto the odd numbers so that f(0) = 1and, for each n, R(f�1(2n + 1); f�1(2n + 3)). That is, f de�nes an in�nitecomputable path through T . (We assume that R(m;n) implies that both mand n are even.) Then the theorem provides a polynomial time tree with apolynomial time in�nite path starting from the root.

VIII.3 Propositional Logic

In this section, we shall consider the complexity of theories in propositionallogic and of the corresponding �0

1 class of complete consistent extensions of thetheory.

It is �rst necessary to de�ne the length j�j of a formula �. Suppose that theunderlying set of propositional letters in our propositional language is fA0; A1; : : :g.In the standard or binary representation of a sentence �, the numeral i ina propositional letter Ai is written in binary representation bin(i) so thatthe length jAij in binary is 1 + jbin(i)j. That is, jbin(Ai)j = r + 2 when2r � i < 2r+1. In the tally representation, the numeral i is written as 1i

so that jtal(Ai)j = i + 1. A complete consistent theory � is represented by asubset of !, S(�) = fi : Ai 2 �g, or, equivalently, by the characteristic functionin f0; 1g! of S(�). The set of all complete consistent extensions of a consis-tent set � of sentences is denoted as CC(�). We shall let a �nite sequence� 2 f0; 1gn represent the sentence B(�) = B0 ^ B1 ^ : : : Bn, where Bi = Ai if�(i) = 1 and Bi = :Ai if �(i) = 0.

We note that there is a lower limit on the complexity of non-trival propo-sitional theories. To be more precise, the set SAT of consistent, or satis�able,sentences is the classic NP complete set. Now a sentence � is valid if and onlyif :� is not satis�able. Thus the smallest theory, the set of valid sentencesis Co-NP complete. On the other hand, any complete propositional theory isdetermined by its underlying set of literals. That is, let V = fA0; A1; : : : g bea set of propositional variables, S be any subset of V and �(S) be the conse-quences of fAi : i 2 Sg [ f:Ai : i =2 Sg. Then S is computable from �(S) in


constant time. On the other hand, given any sentence � containing variablesAi1 ; : : : ; Aik , we can decide whether � 2 �(S) by �rst making each At true ifit is in S and false if not, and then evaluating �. That is, � 2 �(S) if andonly if the value of � is true. Thus �(S) is computable from S in linear timeand linear space. Thus there are complete propositional theories in any of thestandard complexity classes such as linear time, linear space, polynomial time,polynomial space, etc.

Lemma VIII.3.1. tal(B(�)) has length O(n2) and may be computed in timeO(n2) and Bin(B(�)) has length O(n � log n) and may be computed in timeO(n log n).

Proof. The sentence B(�) contains the atoms A0; A1; : : : ; An�1, n� 1 conjunc-tion symbols ^ and between 0 and n negation symbols :. The total length ofthe atoms in tally is

2 + 2 + 3 + 4 + � � �+ n =n2

2+n

2+ 1;

so thatn2

2+3n

2� jtal(B(�)j � n2

2+5n

2

In binary, suppose �rst that n = 2k+1 � 1. Then the total length of theatoms is

2 � 2 +kXj=1

(j + 1)2j�1

so that the total length of the atoms is strictly between (k + 1)2k�1 + 1 and(k + 1)2k and the length of bin(B(�)) is between (k + 5)2k�1 and (k + 5)2k.Now suppose that k � log(n) � k + 1, so that 2k � n < 2k+1. It follows that(5+ log(n))n=2 � (k+5)2k�1 � jbin(B(�))j � (k+5)2k � (5+ log(n)) � 2n.

A set � of sentences is said to be P -decidable in binary (in tally) if thereis a polynomial time Turing machine which given as input the binary (tally)representation of a formula �, computes 1 if � ` � and computes 0 otherwise.We say that � is weakly P -decidable in binary (in tally) if there is a polynomialtime Turing machine which given as input the binary (tally) representationof a conjunction � of literals, computes 1 if � 2 SAT (�) and computes 0otherwise. One can de�ne the notion of � being (weakly) C-decidable in binaryor tally for any complexity class C in a similar manner. A theory � is said tobe P -axiomatizable if it possesses a polynomial time set � of axioms such that� = Cn(�). Again similar de�nitions apply to other notions of complexity.

Recall that the tree T represents CC(�), the set of complete consistentextensions of �, if the set [T ] of in�nite paths through T equals the family ofsets S � fA0; A1; : : :g such that �(S) is a complete consistent extension of �.The canonical tree T which represents CC(�) is given by � 2 T () B(�) 2SAT (�).


Theorem VIII.3.2. Let � be a propositional theory.

(a) If � is weakly DTIME(n log(n)O(1)) decidable in binary, then CC(�)may be represented as the set of paths through a tree in DTIME(n log(n)O(1)).

(b) If � is weakly P -decidable (respectively PSPACE decidable) in eitherbinary or tally, then CC(�) may be represented as the set of paths througha P -tree (resp. PSPACE-tree).

(c) If � is weakly DEXT -decidable (respectively, EXPSPACE-decidable) intally or binary, then CC(�) may be represented as the set of paths through

an EXPTIME-tree (resp.Sk2!DSPACE(2

nk)-tree).

Proof. In each case, we shall let T be the canonical tree which represents CC(�).That is, � 2 T () B(�) 2 SAT (�).

(a) Suppose that � is weakly DTIME(n log(n)O(1)) decidable in binary.By Lemma VIII.3.1, we can compute bin(B(�)) from � in time O(n log n), sothat T is in DTIME(n log(n)O(1)).

(b) It easily follows from Lemma VIII.3.1 that we can compute bin(B(�))and tal(B(�)) in polynomial time and space from �. Thus if � is weakly P -decidable (weakly PSPACE-decidable), then T is a P -tree (PSPACE-tree).

(c)If � is weakly DEXT -decidable in tally (EXPSACE-decidable), it will

require on the order of 2j�j2

time (space) to determine if B(�) 2 SAT (�) sothat T is an EXPTIME-tree (

Sk2!DSPACE(2

nk)-tree). Similarly if � isweakly DEXT -decidable in binary (EXPSACE-decidable), it will require onthe order of 2j�jlog(j�j) time (space) to determine if B(�) 2 SAT (�) so that

again T is an EXPTIME-tree (Sk2!DSPACE(2

nk)-tree).

Thus we have the following corollary.

Corollary VIII.3.3. Let � be a propositional theory.

(a) If � is weakly P -decidable (respectively PSPACE decidable) in tally orbinary, then � has a P -decidable (resp. PSPACE-decidable) completeconsistent extension in tally

(b) If � is weakly DEXT -decidable (respectively, EXPSPACE-decidable)in tally or binary, then � has a complete consistent extension which isEXPTIME decidable (resp. EXPSPACE decidable) in tally.

Proof. (a) Let � be weakly PTIME decidable in tally or binary. By TheoremVIII.3.2, CC(�) may be represented as the set of paths through a P -decidabletree T . It follows from Theorem VIII.1.9 that T has an in�nite PTIME pathx 2 f0; 1gN. The complete consistent extension � corresponding to x has ax-ioms Ai for x(i) = 1 and :Ai for x(i) = 0. Thus given an arbitrary sentence�(A0; : : : ; An) of length geqn, we can �rst compute x(0); : : : ; x(n)) in polyno-mial time and then use this to substitute true and false for the occurrences ofthe propositional variables in � to compute the value of �. This can certainlybe done in polynomial time in tally. The proof for PSPACE is similar.

(b) The proof is similar to (a).


For the binary representation, note that jAnj is of order logn and hence itmay take exponential time to decide An from a polynomial time x 2 f0; 1gN.Thus we have

Corollary VIII.3.4. Let � be a propositional theory.

(a) If � is weakly P -decidable (respectively PSPACE decidable) in eitherbinary or tally, then � has a DEXT -decidable (resp. EXPSPACE-decidable) complete consistent extension in binary.

(b) If � is weakly EXPTIME-decidable (respectively, EXPSPACE-decidable)in tally or binary, then � has a complete consistent extension which isDOUBEXT decidable (resp. DOUBEXPSPACE decidable) in binary.

Since any PTIME decidable theory is certainly weakly PTIME decidableand likewise for other comlexity classes, these results hold with the \weakly"removed from the hypothesis.

It was shown in [44] that this di�erence in the complexity of the completeconsistent extension between the tally and binary representations is necessary.That is,

(1) There is a propositional theory which is NP -decidable in binary but hasno P -decidable complete consistent extension in binary.

(2) There is a propositional theory which is DEXT -decidable in binary buthas no EXPTIME-decidable complete consistent extension in binary.

There are no nice basis results for axiomatiable theories. The correspondingrepresentation results for axiomatizable theories do not require any restrictionon the complexity of the set of axioms. In fact, our next result strengthensTheorem 4.1 of [36] which showed that any �0

1 class may be represented as theset of paths through a polynomial time tree.

A computable function f is said to be time constructible if and only if thereis a Turing machine which on every input of size n halts in exactly f(n) steps.In particular, the functions logk2 (n) are time constructible for k � 1 wherewe de�ne logk2 (n) by induction as log1(n) = log(n) and for k > 1, logk2 (n) =log2(log

k�12 (n)).

It was shown in [44] that for any time constructible function f which isnondecreasing and unbounded and any axiomatizable propositional theory �, �has a DTIME(O(f)) set of axioms and may be represented as the set of pathsthrough a DTIME(O(f))-tree. Note that this is not necessarily a decidabletree.

The results for decidable trees are somewhat surprising. Let us �rst give afew de�nitions. Recall that SAT is the set of satis�able, or consistent, proposi-tional sentences and is the standard NP -complete set.

Theorem VIII.3.5. The following are equivalent:

(a) P = NP ;


(b) Every P -decidable tree represents the set of complete consistent extensionsof some theory which is P -decidable in tally.

(b) ! (a). Let T = f0; 1g� and suppose that � is a theory which is P -decidablein tally such that f0; 1g! = [T ] represents the set of complete consistent exten-sions of �. Then it is easy to see that SAT (�) = SAT . But this means that

� 2 SAT () :[� ` :�]: (VIII.1)

Since � is P -decidable in tally, (VIII.1) would imply that SAT is polynomialtime and hence P = NP .

[(a) ! (b)] Next suppose that P = NP and let T be a P -decidable tree.Let �(A0; : : : ; An) be a propositional formula whose propositional letters are aa subset of fA0; : : : ; Ang and which contains An. The canonical theory � suchthat [T ] represents the set of complete consistent extensions of � is de�ned by

� ` �(A0; : : : ; An) () (8� 2 T \ f0; 1gn)(B(�) ` �(A0; : : : ; An)):

We will show that SAT (�) is NP and hence in P by our assumption. Intally, n � jB(�)j � 2n2, so that for each � 2 f0; 1gn, we can test whetherB(�) ` :�(A0; : : : ; An) in polynomial time in the length of �(A0; : : : ; An). Thuswe can test �(A0; : : : ; An) 2 SAT (�) in the usual NP fashion, by guessing astring � of length n and checking that � 2 T and B(�) implies �(A0; : : : ; An).Thus SAT (�) is in P .

The corresponding result does not follow relative to the binary representationof theories. That is, the direction [(ii)! (i)] still holds, since the SAT problemis NP -complete in either tally or binary. However, the argument given forthe reverse direction only shows that SAT (�) is DTIME(2O(1))-decidable inbinary. This is due to the fact that a short formula � with a high numberedvariable, such as a propositional variable A2n�1 requires us to check whetherB(�) ` �(A0; : : : ; An) for j�j = 2n � 1 which would require time of order 2n

since jB(�)j � 2n. Thus since jA2n�1j = n + 1, such a check would requireexponential time in j�j.


Part B

Applications of �01Classes

213

215

E�ectively closed sets arise naturally in the study of computable mathemat-ics. In many problems associated with mathematical structures, such as theproblem of �nding a 4-coloring of a planar graph, the family of solutions maybe viewed as a closed set under some natural topology. Thus for a computablestructure, the set of solutions may be viewed as a �0

1 class. For another exam-ple, the set of zeroes of a continuous real function on the unit interval is a closedset and the set of zeroes of a computable real function will be a �0

1 class.We will say that the �0

1 class P represents the set of solutions to a givenproblem if there is a one-to-one degree preserving correspondence between theelements of the class P and the solutions to the problem. It will be importantwhether the class P is bounded, or computably bounded. For example, the setof 4-colorings of a given computable graph G may be represented as a subclass off0; 1; 2; 3gN and is therefore computably bounded. Then we may apply the basisresults surveyed in Chapter IV, for example, Theorem III.2.15, and concludethat if G has a 4-coloring, then it has a 4-coloring of c.e. degree.

Now a fundamental observation is that computable problems, such as thegraph-coloring problem, often do not have computable solutions. The resultsfrom Chapter IV on special �0

1 classes strengthen in various ways the basic resultthat a computably bounded �0

1 class may not have any computable members. Inorder to be able to transfer these results to results about the degrees of solutionsto a given computable mathematical problem of a given type, one must establishthat every computably bounded �0

1 class represents the set of solutions to somecomputable problem of that type. For example, Remmel [174] showed that, upto a permutation of the colors, every computably bounded �0

1 class representsthe set of 3-colorings of some computable graph. It then follows from TheoremIV.2.4 that for any c.e. degree c, there is a computable, 3-colorable graph Gsuch that every 3-coloring of G has degree � c.

It is very important to make the representation e�ective. For each typeof mathematical problem, we shall establish a natural enumeration of the com-putable (and sometimes the c.e.) problems and then use the e�ective correspon-dence between solution sets and �0

1 classes to transfer results on index sets fromChapter VI. For example, we will show that the set of indices of computablegraphs which have a computable 3-coloring is a �03 complete set.

Many important problems in the history of �01 classes come from mathemat-

ical logic. The problem associated with a logical theory � is to �nd a completeconsistent extension of �. For any e�ectively presented language L, the set ofsentences of L may be e�ectively listed as f�n : n 2 Ng. Then � is said to bedecidable if fn : �n 2 �g is a computable set and � is said to be axiomatizableif fn : �n 2 �g is a c.e. set.

Shoen�eld [189] showed in 1960 that the set of complete consistent extensionsof a axiomatizable �rst theory can always be represented by a �0

1 class. Theclassical undecidability theorem of Turing and Church may be viewed as showingthat the �0

1 class of complete consistent extensions of Peano Arithmetic has nocomputable element. The complete consistent extensions of a decidable theory� may be represented by a decidable �0

1 class.Ehrenfeucht [69] showed in 1961 that, conversely, every computable bounded

216

�01 class P represents the set of complete consistent extensions of some axiom-

atizable theory �. In particular, � may be a theory of propositional logic or atheory for the language with a single, binary relation symbol. If P is decidable,then we may take � to be a decidable theory.

The chapters below include proofs of the theorems cited above, togetherwith applications of results on members of �0

1 classes and on index sets for �01

classes.

Chapter IX

Algebra

Three types of computable and computably enumerable algebraic structuresare considered: Boolean algebras, abelian groups, and commutative rings withunity. The associated problems are to �nd proper, prime and maximal idealsof rings and Boolean algebras and to �nd proper and maximal subgroups ofabelian groups.

The set of prime ideals of a c. e. Boolean algebra or commutative ring withunity may always be represented by a c. b. �0

1 class, and the set of maximalideals of a computable Boolean algebra may be represented by a decidable c. b.�01 class. The set of maximal ideals of a c. e. commutative ring with unity or

of a c. e. Boolean algebra may always be represented by a �02 class.

For the reverse direction, any c. b. �01 class P may be represented by the

set of maximal ideals of a c. e. Boolean algebra B and if P is decidable, thenB may be taken to be computable. Finally, any �0

1 class of separating sets maybe represented as the set of prime ideals of some c. e. commutative ring withunity [72].

A recursively presented group, ring, or �eld consists of a recursive subset Uof !, the universe of the structure, together with appropriate partial recursivefunctions over U for addition, subtraction, multiplication and/or division func-tions as required. Unless, explicitly stated otherwise, we will assume that all ourstructures are countably in�nite so that there is no loss in generality in assum-ing that the underlying universe is !. A c. e. ring is the quotient of a recursivering modulo an r:e: ideal, a c. e. group is the quotient of a computable groupmodulo a c. e. normal subgroup and a c. e. Boolean Algebra is the quotient ofa recursive Boolean Algebra modulo a c. e. ideal.

We will show that the set of prime ideals of c. e. commutative ring with unityand the set of prime ideals of a c. e. Boolean algebra can always be representedby a c. b. �0

1 class. We will show that the set of maximal ideals of a c. e.commutative ring with unity and the set of maximal subgroups of a c. e. groupcan always be represented by a �0

2 class. We shall also show that the set ofall ideals or the set of all maximal ideals of a recursive Boolean algebra can berepresented as the set of paths through a recursive tree with no dead ends.

217

218 CHAPTER IX. ALGEBRA

Reversing such results, we will show that any c. b. bounded �01 class can be

strongly represented by the set of maximal ideals of an c. e. Boolean algebra.We show that the set of paths through any recursive tree T with no dead endscan be represented as the set of maximal ideals of a recursive Boolean algebra.We shall also show that the set of separating set S(A;B) of a pair of c. e. setscan be represented by the set of prime ideals or the set of maximal ideals of ac. e. commutative ring with identity.

We refer the reader to Downey [61] for a general survey of computable alge-bra.

Some de�nitions are needed. Recall that a subset H of an Abelian groupG = (G;+G;�G; 0G) is a subgroup if it satis�es the following conditions:

(i) 0G 2 H.

(ii) a 2 H and b 2 H implies a�G b 2 H.

H is a maximal subgroup if, in addition, there is no subgroup J of G such thatH � J � G.

A subset I of a commutative ring with unity R = (R;+R;�R; �R; 0R; 1R)is an ideal I is a subgroup of R = (R;+R; 0R) and it satis�es the followingadditional conditions:

(iii) a 2 I and r 2 R implies a �R b 2 I.(iv) 1R =2 I.I is a prime ideal if, in addition,

(v) a �B b 2 I implies a 2 I or b 2 I.I is a maximal ideal if, in addition, there is no ideal J such that I � J . It iseasy to see that any maximal ideal is prime, but the converse is not always true.

The classical results that every proper subgroup of a group has an extensionto a maximal (and therefore proper) subgroup and that every ideal in a ring hasan extension to a maximal (and therefore prime) ideal follow easily from Zorn'sLemma. In particular, if the commutative ring R with unity is not a �eld, thenR has, for each non-unit a a proper ideal Ra = fra : r 2 Rg and therefore hasa maximal ideal.

Any Boolean algebra (B;_B ;^B ;:B ; 0B ; 1B) may be viewed as a commuta-tive ring with unity where a �B b = a^ b and a+ b = (a^B :Bb)_B (:Ba^B b).In a Boolean ring any prime ideal is maximal, so it follows from the Booleanalgebra results that, for any �0

1 class P , there is a c. e. commutative ring withunity such thatMax(R) = Prime(R) is represented by P . However, there turnsout to be a signi�cant di�erence between Boolean rings and rings in general.The proof that any recursive Boolean ring has a recursive maximal ideal cannotbe extended to arbitrary rings and in fact, a recursive ring need not have arecursive maximal ideal. This naturally led to the conjecture that any �0

1 classcould be represented as the set of prime ideals of some commutative ring. Byconsidering rings of polynomials, Friedman-Simpson-Smith obtained in [73] thepartial result that any �0

1 class of separating sets can be represented as the setof prime ideals of some recursive commutative ring with unity.

IX.1. BOOLEAN ALGEBRAS 219

IX.1 Boolean algebras

The Stone Representation Theorem implies that every Boolean algebra is iso-morphic to the Boolean algebra of clopen sets of a topological space (indeedof a Boolean space). If the Boolean algebra is countable, the proof shows thatit is isomorphic to the Boolean algebra RC(P ) of relatively clopen sets of aclosed class P contained in f0; 1gN, and of course RC(P ) is countable for everyclosed class P contained in f0; 1gN. In this section we point out e�ectivizedversions of this correspondence and use them to transfer some of our results on�01 classes to results on computable and c. e. Boolean algebras. In particular, we

give an e�ective version of the Stone Representation Theorem, that every com-putable (c. e. ) Boolean algebra is isomorphic to the set of its prime ideals. Wedetermine the meaning of thinness and of the Cantor-Bendixson derivative inthe setting of Boolean algebras. We also look at the connection between com-putable Boolean algebras and theories of propositional calculus, in particularwith Martin{Pour-El theories. Finally, we interpret the results of the previoussections on �0

1 classes for computable and c. e. Boolean algebras. Here the �01

class represents the set of prime ideals of a c. e. Boolean algebra.Some of the results are known as part of the folklore of the subject. For

more on computable Boolean algebras, see Remmel [172].A computable Boolean algebra B is given by a model (N;�;:;_;^) where �

is a computable binary relation, : is a computable unary operation, and _ and^ are computable binary functions satisfying the usual properties of a Booleanalgebra. In particular, there is a �-least element 0 and a �-greatest element 1,and we assume that 0 2 N names the last and 1 2 N names the greatest. Wenote that the complement :a may be computed by searching for the elementb 2 B such that a^b = 0 and a_b = 1, and thus we do not need to assume thatit is computable. The partial ordering � may be de�ned (and in fact computed)from the two binary operations in that a � b () a _ b = b () a ^ b = a.(See the exercises.) We will also use the operation a+ b = (a ^ :b) _ (b ^ :a),which will be computable for any computable Boolean algebra.

An element a of a Boolean algebra A is an atom if there does not exists b 2 Asuch that 0 < b < a. A is said to be atomless if it has no atoms. Alternatively,we may say that A is dense if the ordering � is dense, that is, whenever a < bin A, then there exists c 2 A such that a < c < b. A is said to be atomic if forevery b 2 A, there exists an atom a 2 A such that a � b.

The fundamental computable atomless Boolean algebra Q may be thoughtof as the family of clopen subsets of f0; 1g!. Each clopen set has a uniquerepresentation as a �nite union of disjoint intervals I(�1) [ � � � [ I(�k), whereeach �i has the same length and k is taken to be as small as possible. Thenthe join (_) and meet (^) operations are clearly computable, as well as thecomplement operation and the partial ordering relation on Q.

Alternatively, we may consider the fundamental Boolean algebra Q(!) to bethe Lindenbaum algebra of propositional calculus over an in�nite set fA0; A1; : : : gof propositional variables. Here two propositions p and q are equal in Q(!) ifthey have the same truth table, so that this is a computable equivalence relation.


A c. e. Boolean algebra is given by a model (N;�;_;^) such that � is ac.e. relation which is a pre-ordering, _;^ are total computable binary functions,and the quotient structure B = (N;�;_;^)= � is a Boolean algebra (wheren � m () n � m & m � n). We can suppose that 0 2 N names the leastand 1 2 N the greatest element of B. Note here that � is preserved under theoperations.

A subset I of a Boolean algebra B is said to be an ideal if for all a; b 2 B,(i) If a 2 I and b 2 I, then a _ b 2 I;(ii) If b 2 I and a �B b, then a 2 I.An ideal I is proper if 1 =2 I and is prime if, for all a; b: (iii) a _ b 2 I ! a 2 Ior b 2 I.

An ideal I is principal if, for some b 2 B, I = I(b) = fa 2 B : a � bg.Finally, an ideal I is maximal if I is proper and there is no proper ideal J

with I � J .For any ideal I, the equivalence relation �I is de�ned by

a �I b () a+ b 2 I:It is clear that �I is c. e. if I is c. e. and is computable if I is computable.

Conversely, given an operation-preserving equivalence relation � on B, thecorresponding ideal I may be de�ned as fa : a � 0g. Then I will be computable(c. e. ) if � is computable (c. e. ).

The dual notion of an ideal is a �lter. A subset M of a Boolean algebra Bis a �lter if it is closed uner ^ and is closed upwards. It is easy to see that Mis a �lter if and only if the set Md = f:b : b 2Mg is an ideal; similarly for anyideal I, we may de�ne the dual �lter Id = f:b : b 2 Ig. Downey [59] developsthe theory of c. e. Boolean algebras from the point of view of c. e. �lters.

Let us de�ne a computable quotient Boolean algebra to be the quotientB= �B , where B = (B;�B ;:B ;^B ;_B) is a computable structure such thatB � !, such that �B is an equivalence relation on B, such that the unary op-eration :B and the two binary operations _B and ^B preserve the equivalenceclasses, and hence the set of equivalence classes forms a Boolean algebra.

Lemma IX.1.1. Any computable quotient Boolean algebra B is isomorphic toa computable Boolean algebra A.Proof. De�ne the universe A of A by

A = fb 2 B : (8a < b):(a �B b)g.For any b 2 B, let (b) be the least a such that a �B b. Then de�ne the

operations on A by:A(a) = (:B(a)),a _A b = (a _B b), anda ^A b = (a ^B b).It is clear that the set A together with these operations forms a Boolean

algebra which is isomorphic to the Boolean algebra on the equivalence classesof B and that the set A and each of the Boolean operations is computable.


For any Boolean algebra B with universe B = !, let P (B) be the class ofmaximal ideals B. It is easy to see that P (B) is a closed subclass of 2!, wherean ideal J is represented as by its characteristic function.

Theorem IX.1.2. If A is a c. e. quotient Boolean algebra, then P (A) is a �01

class and if A is a computable Boolean algebra, then P (A) is a decidable �01

class.

Proof. Suppose that A = B= � is a c. e. quotient Boolean algebra. We canrepresent the class P (A) of prime ideals on A as follows.

x 2 P (A) ()(1) (8a)(8b)[a � b! x(a) = x(b)] and(2) (8a)(8b)[x(a) = x(b) = 1! x(a _B b) = 1] and(3) (8a)(8b)[x(a) = 1! x(a ^B b) = 1] and(4) (8a)[x(a) = 1 () x(:Ba) = 0].This clearly de�nes a �0

1 class. Observe that either x(0B) = 1 or x(1B) = 1

by (4) and hence x(0B) = 1 by (3), so that x(1B) = 0. Thus any x 2 P (A)represents a proper prime ideal. If B is actually a computable Boolean algebra,then we can omit clause (1) and de�ne a computable tree T with no dead endssuch that P (B) = [T ], as follows. T is de�ned to be the set of �nite sequencesx = (x(0); : : : ; x(n� 1)) which satisfy the following, where lh(x) = n.

(2)' (8a < n)(8b < n)[(x(a) = x(b) = 1a _B b < n)! x(a _B b) = 1] and(3)' (8a < n)(8b < n)[(x(a) = 1a ^B b < n)! x(a ^B b) = 1] and(4)' (8a < n)(foralli < 2)[(x(a) = i:Ba < n)! x(:Ba) = 1� i].(5)0k (8a1 < a2 < � � � < ak < n)(x(a1) = � � � = x(ak) = 0 ! a1 ^B a2 � � � ^B

ak 6= 1B).Clause (5) is needed to establish the �nite intersection property for fa < n :

x(a) = 0g which will ensure that any � 2 T can be extended to a prime idealin P (A). This then implies that T has no dead ends.

We can now apply our general results about �01 classes to Boolean algebras.

The following is a consequence of Theorems III.2.15 and IV.1.4.

Theorem IX.1.3. (i) For any c. e. Boolean algebra B, B has a prime idealJ of low c. e. degree (so that J is computable in 00).

(ii) For any computable Boolean algebra B, B has a computable prime ideal.

Theorem IX.1.4. For any c. e. Boolean algebra B with no computable primeideal, there exists a c. e. degree a such that B has no prime ideals of degree � a.

Proof. This follows from Theorem IV.thm:nla.

Theorem IX.1.5. For any For any c. e. Boolean algebra B with no computableprime ideals, there exists two prime ideals, I and J , of B such that any setcomputable from I and computable from J is in fact computable.

Proof. This follows from Theorem IV.IV.2.12.


The following theorem is a corollary of Theorems V.V.2.3,V.4.4 and V.2.2.

Theorem IX.1.6. Let B be a c. e. Boolean algebra only countably many primeideals. Then

(a) B has a computable prime ideal.

(b) If B has only �nitely many prime ideals, then every prime ideal is com-putable.

(c) Every prime ideal of B is hyperarithmetic.

Now we will brie y consider the notion of rank for ideals. It is an exercisebelow 5 that for any �0

1 class P , an element U of RC(P ) is an atom if and onlyif U \ P is a singleton.

Proposition IX.1.7. For any prime ideal J of a c. e. Boolean algebra B, J isisolated in P (B) if and only if J is principal.

Proof. Suppose that J is a principal and prime ideal. Then for some b, we haveJ = fa : b � ag. It follows that J is isolated in the interval I(J � b+ 1).

Suppose that J is isolated in the interval I(�) where lh(�) = n. Let b� =b0 ^ b1 ^ : : : bn�1 where bi = i if x(i) = 1 and bi = :i if x(i) = 0. The isolationmeans that J is the only prime ideal of B that contains b�. It follows that J isgenerated by b�.

Note that a principal ideal is prime if and only if it is generated by an atom.

The following is a corollary of Theorem V.V.4.3.

Theorem IX.1.8. Let B be a c. e. Boolean algebra which has a unique non-principal prime ideal J . Then J �T 000 and if B is computable, then J �T 00.

Next we consider the reverse direction of the correspondence between �01

classes and c. e. quotient Boolean algebras.

For an arbitrary �01 class P , let RC(P ) be the Boolean algebra of relatively

clopen subsets of P , that is, fU \ P : U 2 Qg under the standard set opera-tions. Let B(P ) denote the c. e. Boolean algebra resulting from Q by taking theequivalence relation

U �P V () U \ P = V \ P:

(Note that the corresponding ideal I = fU 2 Q(f0; 1gN) : U \ P = ;g.)The notion of a perfect closed set corresponds to the notion of an atomless

Boolean algebra in the following sense.

Proposition IX.1.9. For any closed set P , P is perfect if and only if RC(P )is atomless.


Proof. Assume �rst that P is perfect and let U \ P 6= ;. Then U \ P containsat least two elements and hence U can be partitioned into two clopen sets U1and U2 such that U1 \ P and U2 \ P are distinct nonempty sets in RC(P ). Itfollows that RC(P ) is atomless.

Next assume that P is not perfect and let x be isolated in P . This meansthat there is a clopen set U such that U \ P = fxg. Clearly U \ P is an atomin RC(P ).

Theorem IX.1.10. Let P � f0; 1gN be a �01 class. Then the quotient algebra

B(P ) is isomorphic to the Boolean algebra RC(P ), the equivalence relation �Pis computably enumerable and hence RC(P ) is a c. e. Boolean algebra. Fur-thermore, if P is decidable, then �P is computable and B(P is a computableBoolean algebra.

Proof. The isomorphism mapping U=equivP to U \ P is clearly a computableisomorphism from B(P ) to RC(P ).

To see that �P is a c. e. relation, let P = [T ] where T is a computable treeand suppose that U = I(�1) [ : : : I(�k) and V = I(�1) [ : : : I(�m). Then

U \ P � V \ P () (8i � k)I(�i) \ P � V \ P:But for any �, we have

I(�)\P � V \P () (9n)(8�)[(lh(�) = n&� � � & � 2 T )! (9i � m)(�i � �)]:Finally, �P is c. e. , sinceU �P V () [U \ P ) � V \ PV \ P � U \ P ].If P is decidable, then T has no dead ends, so we can take k to be the

maximum of flh(�i) : i � mg, so that �P will be computable and hence eachoperation of B(P ) also computable.Theorem IX.1.11. (a) For any �0

1 class P � f0; 1gN, P is computablyhomeomorphic to the set of prime ideals of RC(P ).

(b) For any Boolean algebra B with universe B = !, RC(P (B)) is isomorphicto B. For a c. e. Boolean algebra, this isomorphism is e�ective.

Proof. (a) Map the element x of P to the prime ideal J(x) = fU \ P : x =2 Ug.If x 6= y, then there must be a clopen set P such that x 2 U and y =2 U , so thatthe map is injective. Given a prime ideal J of RC(P ), we claim that there mustbe a unique element xJ of P which belongs to every U \ P in J . Every �nitesubset of J has nonempty intersection since ; =2 J , hence by compactness

TJ

is nonempty. Let x be an element ofTJ , suppose that y 6= x and let U be a

clopen set such that x 2 U and y =2 U . Then U \ P 2 J but y =2 U and hencey =2 T J . Thus T J is a singleton. We leave it to the exercises to show that xJmay be computed e�ectively from J .

(b) Let B be a Boolean algebra with universe B = !. The isomorphism fromB toRC(P (B)) is given by mapping the element b to fJ : J is a prime ideal of B & b =2Jg, that is, to P (B)\U(b), where U(b) is the clopen set de�ned by x 2 U(b) ()x(b) = 0.


We now have the following corollaries.

Theorem IX.1.12. For any c. e. degree c, there is a c. e. Boolean algebra Bsuch that the c. e. degrees of prime ideals of B are exactly the c. e. degrees abovec.

Proof. This is an immediate consequence of Theorems IV.IV.2.4 and IX.1.10.

Theorem IX.1.13. There is a c. e. Boolean algebra B such that any two primeideals of � are Turing incomparable.

Proof. This follows from Theorem IV.IV.2.10.

Next we consider the translation of the notion of a thin �01 class to the

corresponding notion for Boolean algebras. Let us say that a c. e. Booleanalgebra is thin if every c. e. ideal of B is principal. Downey [59] de�ned a c. e.�lter M in Q to be superthick if, for every c. e. �lter W such that M �W � Q,there exists b 2 Q such that W = hM; bi. Here a 2 hM; bi if and only if thereexists x 2M such that b ^ x � a.Lemma IX.1.14. Let A be the c. e. Boolean algebra de�ned as the quotient ofQ modulo the ideal I. Then the following are equivalent.

(i) The �01 class P (A) is thin.

(ii) The �lter Id is superthick.

(iii) B is thin.

Proof. We will show that (i) and (iii) are equivalent and leave the rest as anexercise. We may assume by Theorem IX.1.10 that P � f0; 1gN is a �0

1 class,that I = fV 2 Q : V \P = ;g is a c. e. ideal in Q and that A = f[U ] : U 2 Qg,where [U ] is the equivalence class in Q of U under the equivalence relationde�ned by U �P V () U \ P = V \ P . Then we have, for any x 2 f0; 1gN,

x 2 P () (8U 2 I)x =2 U:Suppose �rst that P is thin and let J be a c. e. ideal in A. De�ne the �0

1

class Q � P by

x 2 Q () (8V 2 Q)([V ] 2 J =) x =2 V ):By assumption, Q = P \ U for some U 2 Q and it follows that J = f[V ] :

[V ] � [U ]g and is principal.For the converse, suppose that A is thin and let Q � P be a �0

1 class. ThenJ = f[V ] : V \ Q = ;g is a c. e. ideal in A. By assumption, there exists Usuch that, for all V 2 Q, [V ] 2 J () [V ] � [U ]. It is then easy to see thatQ = P \ U .


Here is an existence result concerning �01 classes of rank one. This follows

from Theorem V.V.4.5 and Corollary V.V.5.2.

Theorem IX.1.15. (a) For any degree b � 00, there is a computable Booleanalgebra B with a unique non-principal prime ideal J such that J has degreeb.

(b) For any degree b such that 00 � b � 00, there is a c. e. Boolean algebra Bwith unique non-principal prime ideal J of degree b.

Exercises

IX.1.1. Show how to compute the partial ordering �B of a Boolean algebra Bfrom the _B and ^B operations.

IX.1.2. Show that the � relation in a c. e. Boolean algebra is preserved under theoperations. That is, if a � b then :a � :b and similarly for the binaryoperations.

IX.1.3. Show how to carefully de�ne the Boolean algebra of clopen sets to see thatit is in fact computable.

IX.1.4. Show how to compute xJ in the proof of Theorem IX.1.11. (Hint: for anyclopen U , exactly one of U and f0; 1gN � U belongs to J ; to �nd x(0),check which one of I((0)) and I((1)) belongs to J .)

IX.1.5. Show that for any closed set P , U is an atom in the Boolean algebraRC(P ) if and only if U \ P is a singleton.

IX.1.6. Complete the proof of Lemma IX.1.14.


Chapter X

Computer Science

Non-monotonic logics arose in attempts to formalize several notions of \common-sense" reasoning. These systems include the default logic of Reiter [171] andthe stable semantics of general logic programs [78] due to Gelfond and Lifschitz.Classical logic is monotonic in that a deduction from a set of premises remainsvalid for any larger set of premises. Minsky [153] suggested that there is an-other form of reasoning which is not monotonic. That is, common sense andeven scienti�c reasoning forces one to make assumptions in the absence of com-plete information. Thus new information may naturally lead to the rejection ofprevious beliefs. The set of stable models of a logic program is in some sense anon-monotonic generalization of the set of complete consistent extensions of aset of premises. Marek, Nerode and Remmel [137] showed that di�erent versionsof a logic program may be used to represent c. b., bounded and unbounded �0

1

classes.Another area of theoretical computer science where �0

1 classes have appli-cation is the study of !-languages, that is, sets of in�nite words. Here an!-language is the set of in�nite words which are accepted, in some fashion, bya program. In particular, a �0

1 class may be viewed as the set of in�nite wordswhich are accepted by a deterministic automata M in the sense that an in�nitesequence x = (x(0); x(1); : : : is accepted by M if M is always in an acceptingstate after reading each initial segment (x(0); : : : ; x(k)) of x. L. Staiger andK. Wagner [207, 204, 205, 206] have examined several other widely studied no-tions of acceptance which produce di�erent classes. The relation between thesenotions and �0

1 classes is developed in [43].

X.1 Non-monotonic Logic

227

228 CHAPTER X. COMPUTER SCIENCE

Chapter XI

Graphs

There are several combinatorial problems associated with computable graphs.These include the graph coloring problem, the problems of Hamiltonian andEuler circuits, the vertex partition problem, and various matching or marriageproblems. In each case, the set of solutions to any such problem may be repre-sented by a �0

1 class. To obtain a bounded �01 class, it is sometimes necessary

to assume that each vertex of the computable graph has �nite degree and toobtain a c. b. �0

1 class, it is sometimes necessary to assume that the graph ishighly computable, that is, the set of vertices joined to vertex v can be computedfrom v.

For the reverse direction, there are a variety of results. In each case, the setof solutions can represent an arbitrary �0

1 class of separating sets. For the graph-coloring problem, Remmel [174] showed that the 3-coloring problem for highlycomputable graphs can represent an arbitrary c. b. �0

1 class. Manaster andRosenstein [133] showed that the set of surjective marriages in a symmeticallyhighly computable society can likewise represent an arbitrary c. b. �0

1 class. Onthe other hand, Remmel [173] showed that this last result does not hold wheneach person knows at most two other people; this problem is related to thethe Schroder-Bernstein theorem, where one tries to construct an isomorphismbetween two sets given injections in each direction.

For each section, we begin by giving a list of the problems and the requiredde�nitions together with some of the history of each problem. Next we explain(in varying detail) how to prove that the set of solutions to any such problem canbe represented by a computably bounded �0

1 class. Then we apply the results ofChapters IV and V to obtain corollaries which apply to the set of solutions of anysuch problem. Conversely we also consider for each problem, whether the set ofsolutions to such a problem can represent any c:b: �0

1 class. In each case, we showthat the set of solutions to such a problem can represent the class of separatingsets of any two disjoint c:e: sets. Then we apply the results of Chapters IV andV to obtain corollaries which give the existence of \pathological" problems ofeach type. Next we consider index sets for such problems using the methods ofChapter VI. Then we examine the reverse mathematics of such problems as in

229

230 CHAPTER XI. GRAPHS

Chapter VII. Finally, we look at complexity-theoretic versions of some of theproblems.

XI.1 Matching problems

A computable society S = (B;G;K) consists of disjoint computable sets B, theset of boys, and G, the set of girls, and a computable binary relation K � B�G.HereK(b; g) means b knows g. The solutions in this case are the set ofmarriages,or matchings, that is, 1:1 maps f : B ! G such that K(b; f(b)) holds for allb. For any subset B0 of B, let K(B0) = fg : (9b 2 B0)K(b; g)g. Marshall Hall[82] extended the classical Philip Hall Theorem to in�nite societies and provedthat, for any countable society S = (B;G;K), if every boy knows only �nitelymany girls and, for any �nite subset B0 � B, jB0j � jK(B0)j, then there isa marriage for S. We say that a computable society S = (B;G;K) is highlycomputable if there is a partial computable function k : B ! ! such that, foreach b 2 B, k(b) equals the cardinality of K(b). We say that S is symmetricallyhighly computable if there is also a partial computable function k such that, foreach g 2 G, k(g) is the cardinality of the set of boys which know g.

The problems which we consider are:

(i) The general problem of �nding a marriage in a highly computable societyS,

(ii) the surjective matching problem, that is, �nding a marriage f : B ! Gwhich is both 1:1 and onto in a symmetrically highly computable societyS, and

(iii) the surjective matching problem, where each person knows at most twoother people in a symmetrically highly computable society S.

Problems (i) and (ii) were analyzed by Manaster and Rosenstein in [133, 134],who showed that the set of marriages in case (i) and (ii) is always a c:b: �0

1

class, but does not always contain a computable element. Moreover, Manaster,Rosenstein showed that in case (ii), the set of surjective marriages can representan arbitrary c:b: �0

1 class. We note that problem (iii) contains a computableversion of Banach's strengthening of the Schroder-Bernstein theorem, whichwas shown to be none�ective by Remmel [173]. That is, suppose we take 1:1computable functions with computable ranges f : B ! G and g : G ! Bwhere B and G are computable sets. Then we can form a highly computablesociety S = (B;G;K), where K(x; y) holds if and only if f(x) = y or g(y) = x.For such a society S, the only surjective marriages h arise from some partitionB = B1[B2, where h = fdB1[g�1dB2, and the existence of such marriages areguaranteed by Banach's result. (See [173] for details.) It was shown by Remmelin [174] that the set of surjective marriages in case (iii) cannot represent anarbitrary c:b: �0

1 class in contrast to the Manaster-Rosenstein result for case(ii).

XI.1. MATCHING PROBLEMS 231

In each case, the set of solutions to such a problem can be represented by a�01 class [133, 134].

Theorem XI.1.1. For any computable instance of each of the three matchingproblems described above, the set of solutions can be represented by a �0

1 class.If the given graph is highly computable, then the class is computably bounded.

Proof. We may assume that B is the set of even numbers and G is the set ofodd numbers. In (1), a marriage is simply a 1:1 map g : B ! G such that(b; g(b)) 2 K for all b 2 B. We can represent g by a map xg : N ! N byletting xg(i) = 2g(2i) + 1. Thus the �0

1 class P � NN which represents the setof solutions is given by

x 2 P () (8i)(2i; 2x(i) + 1) 2 K & (8i; k)(x(i) = x(k)! i = k):

If S is highly computable, then given i, we can compute the �nite set Gi = fj :(2i; 2j + 1) 2 Kg. Since xg(i) 2 Gi, this shows that P is computably bounded.

For problems (ii) and (iii), the solution is a pair of functions, one from Binto G and one from G into B, which are inverses of each other. This matchingcan be represented by a single function from N to N and the set of solutions willagain be a �0

1 class, and will be c. b. if S is highly computable.

We can derive a number of immediate corollaries to Theorem XI.1.1.

Theorem XI.1.2. For each highly computable society S and matching problemof type (i), (ii), or (iii), the following hold.

(a) If S has a solution, then S has a solution in some c. e. degree.

(b) If S has a solution, then S has solutions s1 and s2 such that any functioncomputable in both s1 and s2 is recursive.

(c) If S has a solution but only has countably many solutions, then S has acomputable solution.

(d) If S has only �nitely many solutions, then each solution is computable.

(e) If S has a solution but has no computable solution, then for any count-able sequence of nonzero degrees faig, S has continuum many solutions swhich are mutually Turing incomparable and such that the degree of s isincomparable with each ai.

Next we consider the reverse direction of this correspondence. That is, givenan arbitrary c. b. �0

1 class P , is there a matching problem of a given type suchthat P represents the set of matchings?

Theorem XI.1.3. [[133]] The problem of �nding a surjective marriage in acomputable society can represent an arbitrary bounded �0

1 class and the problemof �nding a surjective marriage in a symmetrically highly computable societycan represent an arbitrary c. b. �0

1 class.


Proof. Let P be the set of in�nite paths through a computable tree T . Let B =f2 < � >: � 2 T f;gg and G = f2 < � > +1 : � 2 Tg. K consists of all pairs(2 < � >; 2 < � > +1) for � 2 T as well as the pairs 2 < � >; 2 < � � n > +1)for any � 2 T with length n + 1. An in�nite path x 2 T corresponds to thematching which assigns girl 2 < x � n > +1 to boy 2 < x � n+ 1 > and assignsgirl 2 < � > +1 to boy 2 < � > if � is not an initial segment of x. It is clearthat if T is highly computable, then K will also be highly computable.

Theorem XI.1.4. The following problems can represent the c. b. �01 class of

separating sets for any pair of disjoint in�nite c. e. sets.

(i) The problem of �nding a marriage in a highly recursive society.

(ii) The problem of �nding a surjective marriage in a symmetrically highlyrecursive society where each person knows at most two other people.

Proof. (i) For each i 2 !, we will specify a boy bi and two girls g0;i and g1;i sothat bi knows both g0;i and g1;i and no other. Our highly computable societyS = (B;G;K) will be such that G = fg0;i; g1;i : i 2 !g and B = R[fbi : i 2 !g,where R = frs : (As [ Bs) � (As�1 [ Bs�1) 6= ;g is some in�nite set of boysheld in reserve. A marriage f for S will code a set Cf by specifying that i 2 Cfif and only if f(bi) = g1;i. We then determine who the boys in R know in stagesin such a way that

(a) if i 2 A, then one boy in R knows g1;i and no others and no boy in R knowsg0;i;

(b) if i 2 B, then one boy in R knows g0;i and no others and no boy in R knowsg1;i;

(c) if i =2 A [B, then no boy in R knows g0;i or g1;i.

Then if i enters A [ B at stage s, we put rs 2 B and we put (rs; g1;i) inK if i 2 A and (rs; g0;i) in K if i 2 B. It is clear that this de�nes a highlycomputable society S and that there is a one-to-one degree-preserving corre-spondence between the marriages f for S and the separating sets C of A andB, given by mapping f to Cf .

(ii) Fix a pair A andB of in�nite disjoint c. e. sets and recursive enumerationsfAsgs2! and fBsgs2! such that, for all s, As; Bs � f0; 1; : : : ; sg and there is atmost one element of A [B which comes into A [B at stage s.

We �rst partition ! into a computable sequence (G0; B0; G1; B1; : : :) of in-�nite computable sets . For any �xed i, let g0i < g1i < : : : and b0i < b1i < : : :list the elements of Gi and Bi in increasing order. Our symmetrically highlycomputable society S = (B;G;K) will be thought of as a bipartite graph withB = [iBi and G = [iGi. The idea is to construct a connected component of Swith vertex set Gi [ Bi for each i. We construct the i-th component in stages,so that at stage s, we determine the edges out of gki and b

ki for k � 2s. We begin

as if we are going to construct the two-way in�nite chain in which b0i is joined

XI.2. GRAPH-COLORING PROBLEMS 233

to g0i and g1i and such that, for each n > 0, b2ni is joined to g2n�2i and g2ni and

b2n�1i is joined to g2n�1i and g2n+1i . See Figure XI.1Observe that there are exactly two possible surjective marriages f for such

a component depending on whether f(b0i ) = g0i or f(b0i ) = g1i . A marriagef : B ! G for S will code a separating set Cf for A and B by letting i 2 Cf ifand only if f(b0i ) = g1i . Then it is easy to see that all we need to do to ensurethat each marriage f of S corresponds to a separating set Cf for A and B isto construct the i-th component so that it is a one-way chain starting in Bi ifi 2 A, a one-way chain starting in Gi if i 2 B, and the full two-way in�nitechain if i =2 As [ Bs. Thus we build the chain until we see that i 2 A [ B atsome stage s. That is, at each stage t, we add bki and gki for k 2 f2t; 2t + 1gas pictured in Figure XI.1. Then if i 2 Bs omit b2ni and g2ni from the chain forall n � s so that the chain will be a one-way in�nite starting a girl g2s�2i . Ifi 2 As, then add b2si and we omit g2si plus all boys and girls of the form b2ni andg2ni for n > s from the chain so that the chain will be a one-way in�nite chainstarting at b2si .

We note that we can consider this example as a computable version of prob-lem (ii) by simply directing the edges of the graph down the left hand side ofthe graph and up the right hand side of the graph. That is, we can de�ne thefunction f : B ! G by saying that f(b�) = g� is there is a directed edge fromb� to g� in some component and de�ne the function g : G ! B by saying thatg(g�) = b� if there is a directed edge from g� to b� in some component.

Given these representation results, we have the usual corollaries.

Theorem XI.1.5. (a) For each one of the three matching problems,

(1) There is a computable society S which has a matching but has no com-putable matching.

(2) There is a computable society S such that that any two distinct matchingsare Turing incomparable.

(3) If a is a Turing degree and 0 <T a �T 00, then there is a computablesociety s which has a matching of degree a but has no computable matching.

(b) For the surjective matching problem, the following also hold.

(4) There is a computable society S such that if a is the degree of any matchingand b is a c. e. degree with a �T b, then b = 00.

(5) If c is any c. e. degree, then there exists a computable society S such thatthe set of c: e: degrees which contain matchings equals the set of c. e.degrees �T c.

XI.2 Graph-coloring problems

A countable in�nite graph G = (V;E) consists of a subset V of the naturalnumbers called vertices together with a symmetric subset E of V � V , called


b

b

b

b

b

b

g

g

g

g

g

g

b

..

.

..

.

0

0 1

12

23

4

4 5

56i

i

3i

i

i

i

i

i

i

i

i

i

i

Figure XI.1: Generic component of the symmetric society

XI.2. GRAPH-COLORING PROBLEMS 235

the edges. G is said to be computable if the sets V and E are computable. Wesay that vertices u; v are joined by an edge (u; v). The degree of a vertex uof G is the cardinality of the set of vertices joined to u. A k-coloring of thegraph G is a map g from V into f1; 2; : : : ; kg such that g(u) 6= g(v) whenever(u; v) 2 E. The k-coloring problem for a graph G is to determine whether G hasany k-colorings. The set of solutions to this problem is the set of k-colorings ofG. We make the convention that, unless stated otherwise, the graphs we shalldiscuss are assumed to be connected, have no loops or multiple edges, and havethe property that each vertex v of G is of �nite degree.

The graph coloring problem has been studied in combinatorics for over acentury. Two classical results for �nite graphs are Brooks' Theorem [19] thatevery graph with all vertices of degree � k and with no k + 1-cliques is k-colorable and the Four Color Theorem of Haaken and Appel [3] that everyplanar graph is 4-colorable. These results are easily extended to in�nite graphsby a compactness argument. A natural question is whether such results canbe e�ectivized. The answer to this question is yes for Brooks' Theorem, thatis, Schmerl showed in [183] that every computable graph with all vertices ofdegree � k and with no k+1-cliques has a computable k-coloring. On the otherhand, the Four Color Theorem cannot be e�ectivized. Bean constructed in [12]a 3-colorable, computable, planar connected graph which has no computablek-coloring for any k.

A computable graph G = (V;E) is said to be highly computable if there isa partial computable function f : V ! ! such that, for each v 2 V , f(v) isthe degree of v. Highly computable graphs are of interest for several reasons.One reason is the result of Bean [12] that any highly computable k-colorablegraph has a computable 2k-coloring, in contrast to the result cited above forarbitrary computable graphs. This result was improved by Schmerl [182] from2k to 2k � 1, who also showed that 2k � 1 is the best possible result. It followsfrom the work of Bean and Schmerl that every highly computable planar graphhas a computable 6-coloring. This result was improved by Carstens [20] from 6to 5, but the highly computable four color problem remains open.

Bean showed in [12] that the set of k-colorings of a highly computable graphis always a computably bounded �0

1 class. (See Exercise 1.)Conversely, Remmel [174] showed that every c. b. �0

1 class can actually bestrongly represented by a highly computable k-coloring problem.

The problem of feasible graphs and colorings has been studied by Cenzerand Remmel in [37].

Exercises

XI.2.1. Show that for any highly computable graph G = (V;E) and any �nite k,the set of k-colorings of G may be represented by a computably bounded�01 class P � f0; 1; : : : ; k � 1gN.

XI.2.2. Show that if G is a planar graph, then the set of 5-colorings of G alwayshas cardinality 2@0 and hence not every c.b. �0

1 class may be representedas the set of 5-colorings of a planar graph. (Hint: every planar graph Gis 4-colorable, by the theorem of Appel and Haken [3].)


XI.3 The Hamiltonian circuit problem

Let G = (V;E) be a countably in�nite graph. Two vertices u; v of G areadjacent if (u; v) 2 E and two edges (u1; v1) and (u2; v2) are adjacent if eitherv1 = u2 or u1 = v2. A one-way (respectively two-way) Hamiltonian circuit (orHamiltonian path) for G is a one-to-one correspondence f between the naturalnumbers ! (resp. the integers Z) and V such that consecutive vertices areadjacent, i.e. (f(i); f(i + 1)) 2 E for all i. The dual concepts are the one-way(respectively two-way) Euler path, which is a one-to-one correspondence betweenthe natural numbers ! (resp. the integers Z) and E such that consecutive edgesare adjacent. For each of these four notions, let us also de�ne the associatednotion of being such a path for a subgraph. That is, a one-way Hamiltoniansub-path for G will be a one-to-one embedding of the natural numbers intoV such that consecutive vertices are adjacent. The other three de�nitions aresimilar.

In each case, the problem here is whether a given graph has such a path.We will focus on the sub-path problems.

Theorem XI.3.1. For each of the following problems, the set of solutions canbe represented as a �0

1 class. In cases (a) and (b), the class is bounded if theeach vertex has �nite degree and is c. b. if the graph is highly computable.

(a) The one-way Hamiltonian (Euler) sub-paths starting from a �xed vertexin a recursive graph.

(b) The two-way Hamiltonian (Euler) sub-paths through a �xed vertex in arecursive graph.

(c) The one-way Hamiltonian (Euler) paths starting from a �xed vertex in arecursive graph.

(d) The two-way Euler paths through a �xed vertex in a recursive graph.

Proof. (a) Let the computable graph G = (V;E) with �xed vertex v0 be given.Then a one-way Hamiltonian (Euler) sub-path is a function f from ! into Vwith f(0) = v0 such that (f(n); f(n + 1) 2 E for all n and such that, for theHamiltonian path, m 6= n implies that f(m) 6= f(n) and, for the Euler path,m 6= n implies that the edges (f(m); f(m+1)) and (f(n); f(n+1)) are di�erent.In each case, this clearly de�nes a �0

1 class P . If each vertex v has �nite degree,then there is a function g such that all vertices joined to vertex v are � g(v).It follows that we can compute a bound h(m) for the possible value of f(m)by letting h(0) = v0 and in general h(m + 1) = supfg(v) : v � h(m)g. Thisshows that P is bounded. If G is highly computablee, then the function g maybe taken to be computable, so that P is computably bounded.

(b) Again let the computable graph G = (V;E) with �xed vertex v0 be given.Then a two-way Hamiltonian (Euler) sub-path

: : : ; �(�1); �(0) = v0; �(1); : : :

XI.3. THE HAMILTONIAN CIRCUIT PROBLEM 237

can be represented as a function f from ! into V with f(0) = v0 such that(v0; f(1)) 2 E, such that (f(n); f(n + 2)) 2 E for all n and such that, for theHamiltonian path, the function f is one-to-one, and, for the Euler path, no edgeoccurs twice in the list

: : : ; (f(3); f(1)); (f(1); f(0)); ((f(0); f(2)); (f(2); f(4)); : : : :

It follows as in (a) that the class P of two-way Hamiltonian (Euler) sub-pathsis a �0

1 class, is bounded if each vertex of G has �nite degree, and is c. b. if Gis highly computable.

(c) We �rst give the proof for one-way Hamiltonian paths. Recall that V = !and represent a one-way Hamiltonian path

� = (�(0) = v0; �(1); �(2); : : :)

by a function f such that f(2n) = �(n) and f(2v + 1) = n such that v = �(n).This is clearly a one-to-one degree-preserving correspondence between the one-way Hamiltonian paths ofG and the �0

1 class P . Then the �01 class P of solutions

is the set of functions f such that f(0) = v0, such that (f(2n); f(2n+2)) 2 E forall n, and such that, for all v and n, f(2n) = v if and only if f(2v+1) = n. Forthe one-way Euler paths �, we take f(2n) = �(n) and let f(2[u; v] + 1) = n+ 1such that �(n) = u and �(n+1) = v if (u; v) 2 E and otherwise f(2[u; v]+1) = 0.In either case, the assumption that G is highly computable does not necessarilyimply that P is even bounded.

(d) Represent a two-way Hamiltonian path by a function f so that the pathis given by : : : ; f(4); f(1); f(0) = v0; f(3); f(6); : : : and such that f(3v + 2) = nsuch that n 6= 2 mod 3 and f(n) = v. Represent a two-way Euler path �again by a function f so that � = : : : ; f(4); f(1); f(0) = v0; f(3); f(6); : : : andnow such that f(3[u; v] + 2) = n such that n 6= 2 mod 3 and f(n) = u andf(n+ 3) = v.

It follows that if each vertex of G has �nite degree and G has a one-wayor two-way Hamiltonian (Euler) sub-path, then it has such a sub-path whichis computable in 000. In the cases (c) and (d) of the Hamiltonian and Eulerpaths, we can only conclude, even for a highly computable graph G, that G hasa solution recursive in some �11 set. We leave the other usual corollaries for thereader.

Bean [13] showed that if G is highly computable and has an Euler path,then G will actually have a computable Euler path. This is not the case forHamiltonian paths, by the following reasoning. If every highly computablegraph G with a Hamiltonian path had a hyperarithmetic Hamiltonian path,then the set of highly computable graphs with Hamiltonian paths would be �1

1,by the Spector-Gandy theorem II.10.5. However, Harel [84] showed that theproblem of the existence of (one-way or two-way) Hamiltonian paths in a highlycomputable graph is �11-complete and therefore not �

11. It follows that the set of

Hamiltonian paths of a highly computable graph is not always a c. b. �01 class.

That is, if it were always a c. b. class, then by Theorem every highly computable


graph with a Hamiltonian path would have a Hamiltonian path computable in00 and hence a hyperarithmetic Hamiltonian path. This applies to one-way andtwo-way paths.

For the reverse direction, we have the following result of Bean [13].

Theorem XI.3.2. For any c. b. �01 class P , there is a highly computable planar

graph G and a one-to-one computable isomorphism between P and the set ofHamiltonian paths for G.

Proof. Let P be the set of in�nite paths through the highly computable tree T .G is constructed in stages, beginning with vertices 0 and 0out and edge (0; 0out).At stage n, let �0; �1; : : : ; �m be the nodes of T at level n and introduce a circuitof 3(m+ 1) vertice in G given by

(�out0 ; �in0 ; �0;1; �out1 ; �1;2; : : : ; �

outm ; �inm ; �m;0:�

out0 ):

For every node �i at level n� 1 and every successor �j at level n of �i, also addan edge (�outi ; �inj ) to G. (For the two-way circuit, also add a vertex vn andedge (vn�1; vn), where v0 = 0. It is clear that G is a highly computable planargraph. The desired correspondence between P and the Hamiltonian paths of Gis given as follows. The node �j of T follows the node �i on the in�nite paththrough T if and only if the vertex �inj immediately follows the vertex �outj onthe Hamiltonian path.

It follows that there is a highly computable graph G which has Hamiltonianpaths but has no computable Hamiltonian paths. Other corollaries are left tothe reader.

This problem, posed by S. Ulam, is to show that for each partition of thevertex set V of a graph G = (V;E) into sets of uniformly bounded cardinality,there is at least one set of the partition which is adjacent to m (or more) othersets of the partition. Here we say that two sets S1 and S2 are adjacent if thereexist vertices v1 2 S1 and v2 2 S2 such that (v1; v2) 2 E. The partition numberm of a graph G is the least numberm for which the statement is true. The vertexpartition problem was studied by Cenzer and E. Howorka [27], who computedthe vertex partition numbers of various well-known graphs, including the m-regular trees Tm and the planar mosaic graphs M3, M4 and M6. The tree Tmmay be viewed as f1; 2; : : : ;mg�. The graphs M3, M4 and M6 may be viewedas tilings of the plane by regular hexagons, squares and equilateral triangles. Ineach case, the partition number of the graph turns out to be the degree of thegraph. In this situation, the �0

1 class arises from the dual problem. That is,given the graph G and numbers k and m, to �nd a k-partition P of the graphsuch that no set has m neighbors. Here a k-partition is a partition of V intosets of cardinality � k. The solution to such a problem may be represented asa function f from V �V into f0; 1g which is to be the characteristic function ofthe equivalence relation with equivalence classes being the sets of the partition.

XI.3. THE HAMILTONIAN CIRCUIT PROBLEM 239

Theorem XI.3.3. For any highly computable graph G = (V;E), and any �nitek and m, the set C of k-partitions of V such that no set in the partition isadjacent to m other sets may be represented by a �0

1 class in f0; 1gN.Proof. Let G = (V;E) be a highly recursive graph and let k;m be positiveintegers. Let C be the set of k-partitions of V such that no set in the partitionis adjacent to m other sets. As indicated above, we may represent a partition bythe characteristic function f of the corresponding equivalence relation. Let usassume that V = ! for simplicity and let C be the class of all such functions forwhich there is no set in the partition represented by f which has m neighbors.Now a function f 2 f0; 1g! will be in the class C if it satis�es the followingconditions:

(i) (8u)[f(u; u) = 1].

(ii) (8u; v)[f(u; v) = f(v; u)].

(iii) (8u; v; w)[f(u; v) = f(v; w) = 1! f(u;w) = 1].

(iv) (8u1; u2; : : : ; uk+1)(9i; j � k + 1)[f(ui; uj) = 0].

(v) (8u1; v1; u2; v2; : : : ; um; vm)[((8i; j � m)[f(ui; uj) = 1] &(8i � m)[E(ui; vi)])! (9i; j � m)[f(vi; vj) = 1]].

The �rst three clauses are the requirement that f is the characteristic func-tion of an equivalence relation. The fourth clause is the requirement that eachset in the corresponding partition has cardinality � k and the �nal clause is therequirement that no set in the partition is adjacent to m other sets.


Chapter XII

Orderings

There are several problems associated with partially ordered sets (posets) andalso with computable linear orderings and ordered structures.

Problems to be considered include the decomposition of a poset into chainsand into antichains, as well as the problem of expressing a partial ordering asthe intersection of �nitely many linear orderings. For each computable instanceof these problems, the set of solutions can be represented as a c. b. �0

1 classand can represent an arbitrary �0

1 class of separating sets.For a computable linear ordering A, we consider the problem of �nding

suborderings of type ! or !�, the problem of �nding an !-successivity or !�-successivity, and the problem of �nding a self-embedding of A.

Finally, we consider the problem of �nding an ordering of a computableAbelian group or formally real �eld. As usual, the set of orderings can alwaysbe represented by a c. b. �0

1 class and Metakides and Nerode [151] showedthat any c. b. class. On the other hand, Solomon [199] showed that not everyc. b. �0

1 class can be represented as the set of orderings of a computable abeliangroup.

XII.1 Partial orderings

In this section we consider three problems associated with partially ordered sets(posets). Two of these are the dual problems of covering a poset with chains orwith antichains. The third problem is the dimension problem, that is, expressinga poset as the intersection of linear orderings.

We �rst describe the problems and show that the solution set to a computableproblem always forms a c. b. �0

1 class, and then apply the results of Part Oneto obtain corollaries which apply to the set of solutions of any such problem.We also consider for each problem, whether, conversely, the set of solutions tosuch a problem can represent any c. b. �0

1 class. For each problem, we showthat the set of solutions to such a problem can represent the class of separatingsets of any two disjoint c. e. sets and we apply the results of Part One to ob-

241

242 CHAPTER XII. ORDERINGS

tain corollaries which give the existence of \pathological" problems of each type.

Decomposition problems for posets

Here we start with a computable poset A = (A;�A), which consists of a com-putable subset A of N and a computable partial ordering �A. The width ofA is the maximum cardinality of an antichain in A and the height of A is themaximum cardinality of a chain in A.

(a) The �rst decomposition theorem we consider is Dilworth's theorem [58],which states that any poset A of width n can be covered by n chains. Theproblem here is to �nd such a covering of A by n chains and the set of solutionscorresponds to the various coverings of A by n chains. The e�ective version ofDilworth's theorem has been analyzed by Kierstead in [102], where he showedthat every computable poset A of width n can be covered by (5n � 1)=4 com-putable chains, while for each n � 2, there are computable posets of width nwhich cannot be covered by 4(n� 1) chains. See Kierstead's article [101] in thisvolume for details and more results.

Thus, the set of solutions of this problem for a computable poset A can berepresented as the set of maps f : A! f1; 2; : : : ; ng such that f�1(fig) = fx 2A : f(x) = ig is a chain for each i, which is clearly a c. b. �0

1 class.(b) There is a natural dual to Dilworth's theorem which says that every

poset of height n can be covered by n antichains. The problem again is to�nd such a covering. The e�ective version of the latter theorem was analyzedby Schmerl, who showed that every computable poset of height n can be cov-ered by (n2 + n)=2 computable antichains while for each n � 2, there is acomputable poset of height n which cannot be covered by (n2 + n)=2� 1 com-putable antichains. Furthermore, Szem�eredi and Trotter showed that there existcomputabl partial orders of height n and computable dimension 2 which stillcannot be covered by (n2 + n)=2 � 1 computable antichains. These results arereported by Kierstead in [102].

(2) Dimension of posets problem The poset A = (A;R) is de�ned to be n-dimensional if there are n linear orderings of A, (A;L1); : : : ; (A;Ln), such thatR = L1\ � � �\Ln. The notion of the dimensionality of posets is due to Dushnikand Miller, who showed in [68] that a countable poset (A;R) is n-dimensionalif and only if it can be embedded as a subordering in the product ordering Qn,where Q is the set of rational numbers under the usual ordering. A (computable)poset (A;R) has (computable) dimension equal to d, for d �nite, if there are d(computable) linear orderings (A;L1); : : : ; (A;Ld) such that R = L1 \ � � � \ Ld,but there are not d�1 (computable) linear orderings (A;L01); : : : ; (A;L0d�1) suchthat R = L01 \ � � � \ L0d�1. Kierstead, McNulty and Trotter have analyzed in[105], the computable dimension of computable posets and have shown that ingeneral, the computable dimension of a poset is not equal to its computabledimension.

Given a countable poset (A;R) with A � !, we can code a set of d linearorderings of A, (A;L1); : : : ; (A;Ld) as follows. Let a0 < a1 < � � � be an increas-

XII.1. PARTIAL ORDERINGS 243

ing enumeration of A. Then given d linear orderings of fa0; : : : ; an�1g, thereclearly are (n + 1)d ways to extend the d linear orderings to d linear orderingson fa0; : : : ; ang. One can �x some e�ective enumeration of these extensions foreach n, so that it then becomes possible to code each d-tuple of linear order-ings by a function f : A ! ! where f(an) � (n + 1)d � 1 for all n. Thus theset of solutions for the n-dimensionality problem of a computable poset (A;R)can be represented as the set of all f : A ! ! such that f codes an n-tuple,(A;L1); : : : ; (A;Ln), of linear orderings on A such that R = L1 \ � � �Ln, whichis a c. b. �0

1 class.We state the �rst theorem and leave the details of the representation to the

reader.

Theorem XII.1.1. For each speci�c computably presented instance of one ofthe poset problems P listed above, the set of solutions can be represented as ac. b. �0

1 class.

As usual, we can now derive a number of immediate corollaries from theresults of Part One. We state only a few of these and leave the rest to thereader. For example, the following is true.

Theorem XII.1.2. (a) If a computable poset A has a covering by n chains,then A can be covered by n chains C1; : : : Cn such that C1 � � � � � Cn hasc. e. degree.

(b) If A = (A;R) is a computable poset such that the family of setsf(A;L1); (A;L2); : : : ; (A;Ln)g of n linear orderings such thatR = L1 \ L2 \ � � � \ Ln is countably in�nite, then A has computabledimension � n.

(c) If a computable poset A has a covering by n antichains, but has no coveringby n computable antichains, then for any countable sequence of nonzerodegrees faig, A has a continuum of coverings fA1; A2; : : : ; Ang by n an-tichains, which are pairwise Turing incomparable and such that the degreeof fA1; A2; : : : ; Ang is incomparable with each ai.

Next we consider the reverse direction of this correspondence.

Theorem XII.1.3. Each of the three problems described above can stronglyrepresent the c. b. �0

1 class of separating sets for any pair of disjoint in�nitec. e. sets.

Proof. Fix a pair A and B of in�nite disjoint c. e. sets and computable enu-merations fAsgs2! and fBsgs2! such that, for all s, As; Bs � f0; 1; : : : ; sg andthere is at most one element of A [B which comes into A [B at stage s.

(1) The problem of covering a computable poset of width k by k chains.First consider the case k = 2. We begin with the poset D0 consisting of twoone-way chains fai;j : i = 0; 1 ^ j 2 !g and fbi;j : i = 0; 1 ^ j 2 !g where wehave ai;j � ai;k whenever j < k and a0;j � a1;j as well, and similarly for the


bi;j . The two chains are linked by having a0;j � b1;j and similarly b0;j � a1;j .Let us call the posets fa0;i; a1;i; b0;i; b1;ig the i-th block of the poset D0. Thei-th block of D0 is pictured in Figure XII.1(A).

Our �nal poset D = (D;�D) will consist of the poset D0 together with anin�nite computable set E whose relations to the elements of D0 and amongthemselves is to be speci�ed in stages. Now it is clear that a decomposition ofthis poset, up to renaming the chains, is completely determined by the choice,for each i, of either

(a) putting a0;i and a1;i in one chain and b0;i and b1;i in the other, or

(b) putting a0;i and b1;i in one chain and a1;i and b0;i in the other.

Thus we can think of a chain decomposition f : D ! f1; 2g as coding up aset Cf , where i 2 Cf if and only if we use choice (b) for the i-th component,that is, if and only if f(a0;i) = f(b1;i). Now the idea is to de�ne the relationsbetween the remaining computable set E so that we introduce an element e inthe i-th component between a0;i and a1;i if i 2 B, see Figure XII.1(B). Thiswill force e, a0;i and a1;i to be in the same chain. We introduce an elementf in the i-th component between b0;i and a1;i if i 2 A, see Figure XII.1(C).This will force f , b0;i and a1;i to be in the same chain. Finally we have no newelement in the i-th component if i =2 A [ B. It is not di�cult to see that thiscan be accomplished so as to ensure that D is a computable poset of width 2and that such actions will ensure that the correspondence f ! Cf will be aone-to-one degree-preserving correspondence between the decompositions of Dinto two chains and the separating sets of A [ B. We leave the details to thereader. For the case where k > 2, one simply adds to the poset described a setof k � 2 computable in�nite one-way chains, all of whose elements are incom-parable with D and so that elements from di�erent chains are also incomparable.

(2) The problem of covering a computable poset of width k by k antichains.Again we shall initially consider the case k = 2. The poset D = (D;�D) willconsist of two parts. The �rst part of the poset will consist of a computableantichain c0; c1; : : :, and the second part will consist of two antichains a0; a1; : : :and b0; b1; : : : where a0 � b0 and, for each i, ai+1 � bi and ai+1 � bi+1, seeFigure XII.1.

We will complete the partial ordering on D by specifying the relations be-tween the two parts in stages. Clearly, up to renaming the antichains, thereis a unique decomposition of the second part of the poset into two antichains.We think of a decomposition of D into two antichains as coding up a set Cf byspecifying i 2 Cf if and only if f assigns ci to the same antichain as the a's.Then, for each i, we de�ne ci to be greater than as if i 2 As+1 nAs and incom-parable to as otherwise, and de�ne ci to be less than bs if i 2 Bs+1 n Bs andincomparable to bs otherwise. It is then easy to check that D is a computableposet of height two and that, up to renaming the antichains, the correspondencef ! Cf is a one-to-one degree preserving correspondence between decomposi-tions of P into two antichains and separating sets of A[B. For the case where


a

a

a

a

a

b

b

b

b

be

f

b

a

i not in A or B

i in B

i in A

(A)

(C)

(B)

1,i

0,i

1,i

0,i

1,i

0,i

1,i

0,i

1,i

0,i

1,i

0,i

Figure XII.1: Blocks for width 2 poset


. . .

. . .

a a a a aa

b b b b bb0

1 20 43 5

1 3 4 52

Figure XII.2: Height 2 poset


k > 2, one simply adds to the poset described a set of k� 2 computable in�niteantichains, all of whose elements are comparable with every element of D andso that elements from di�erent antichains are also comparable.

(3)The problem of expressing a computable poset P = (P;� P ) of dimension das the intersection of d linear orderings.

We consider the case oftwo dimensional partial orderings. First we partitionN into two in�nite computable sets C = fc0 < c1 < � � � g and D = fd0 < d1 <� � � g. For each i, we let Ci = fc5i; c5i+1; c5i+2; c5i+3; c5i+4g. We shall de�nea computable partial ordering <P on ! in stages. Given any two sets E andF, E <P F will denote that, for any e 2 E and f 2 F , e <P f . We start byde�ning <P so that C0 <P C1 <P C2 <P � � � . This means that if <1 and <2 aretwo linear orderings such that <1 \ <2=<P , then the only di�erence between<1 and <2 on C is how <1 and <2 order the elements within the blocks Ci.For each block Ci, <P is de�ned so that we have the Hasse diagram in FigureXII.1(A).

It is then easy to check that, up to a permutation of the indices of the linearorderings <1 and <2, there are precisely two ways to de�ne <1 and <2 on Ciso that <1 \ <2 equals <P restricted to Ai, namely,

(I) c5i <1 c5i+1 <1 c5i+2 <1 c5i+3 <1 c5i+4 andc5i+2 <2 c5i+4 <2 c5i+3 <2 c5i <2 c5i+1, or

(II) c5i <1 c5i+1 <1 c5i+2 <1 c5i+4 <1 c5i+3 andc5i+2 <2 c5i+3 <2 c5i+4 <2 c5i <2 c5i+1.

Note that the di�erence between (I) and (II) is that in the ordering wherethe elements c5i; c5i+1 precede the elements c5i+2; c5i+3; c5i+4, we have c5i+3preceding c5i+4 in (I), while in (II) c5i+4 precedes c5i+3.

We can thus use a pair of linear orderings <1 and <2 such that <1 \ <2=<Pis de�ned within the blocks Ci to code a set S(<1; <2) � ! by declaring i 2 Sif and only if <1 and <2 are of type (I) on Ci.

The key to our ability to code up a tree of separating sets for a pair of disjointc. e. sets A and B is the following. If we add an element d to the Hasse diagramas pictured in Figure XII.1(B), then only linear orderings <1 and <2 of type (I)can be extended to Ci [ fdg so that <1 \ <2=<P and if we add an element dto the Hasse diagram as pictured in Figure XII.1(C), then only linear orderings<1 and <2 of type (II) can be extended to Ci [ fdg so that <1 \ <2=<P .

That is, it is easy to check that, up to a permutation of indices there is onlyone way to de�ne linear orderings <1 and <2 on Ci [ fdg so that <1 \ <2=<Pif <P has the Hasse diagram as pictured in Figure XII.1(B), namely

(I 0): c5i <1 d <1 c5i+1 <1 c5i+2 <1 c5i+3 <1 c5i+4 andc5i+2 <2 c5i+4 <2 d <2 c5i+3 <2 c5i <2 c5i+1.

Similarly, up to a permutation of indices, there is only one way to de�ne linearorderings <1 and <2 on Ci [ fdg so that <1 \ <2=<P if <P has the Hassediagram as pictured in Figure 5 (C), namely


c

c

c

c

c

d

d

i in A

i in B

A or Bi in neither

(A)

(B)

(C)

5i+2

c5i+1

c

c5i+3

5i

c

c

c

c

c

c

c5i+4

5i+1

5i

5i+3 5i+4

5i+2

5i+1

5i

5i+3 5i+4

5i+2

Figure XII.3: The i-th block of the dimension 2 poset

XII.2. LINEAR ORDERINGS 249

(II 0): c5i <1 d <1 c5i+1 <1 c5i+2 <1 c5i+4 <1 c5i+3 andc5i+2 <2 c5i+3 <2 d <2 c5i+4 <2 c5i <2 c5i+1.

Now to complete our de�nition of <P on !, we proceed in stages as follows.

Stage 0 If i 2 A0, let Ci�1 <P fd0g <P Ci+1, and de�ne <P on Ci [ fd0gso that we have a Hasse diagram as in Figure XII.1(B). If i 2 B0, let Ci�1 <Pfd0g 0. Assume we have de�ned <P on C [ fd0; : : : ; ds�1g so that for

all j < s, Ci�1 <P fdjg <P Ci+1 if i 2 (Aj [Bj) n (Aj�1 [Bj�1) and fdjg <PC [ fd0; : : : ; dj�1g otherwise. Then if i 2 As nAs�1, let Ci�1 <P fdsg <P Ci+1and de�ne <P on Ci [ fdsg so that we have a Hasse diagram as pictured inFigure XII.1(B). If i 2 Bs nBs�1, let Ci�1 <P fdsg <P Ci+1 and de�ne <P onCi [ fbsg so that we have a Hasse diagram as pictured in Figure XII.1(C). If(As [ Bs) n (As�1 [ Bs�1) = ;, de�ne fdsg <P C [ fd0; : : : ; ds�1g. Again thisde�nes <P on all of C [ fd0; : : : ; dsg by transitivity.

This completes the proof of Theorem XII.1.3.

As usual, there are a number of immediate corollaries and we state only afew.

Theorem XII.1.4. (a) There is a computable poset of width k which has nocovering by k chains.

(b) There is a computable poset A of height k such that any two distinctcoverings of A by k antichains are Turing incomparable, where distinctmeans not obtainable from the other by a permutation of the antichains incombination with the shifting of a �nite number of elements.

(c) If a is a Turing degree and 0 <T a �T 00, then there is a computableposet A = (A;R) of dimension d, but not of computable dimension d suchthat there exists a set f(A;L1); : : : ; (A;Ld)g of degree a of linear orderingssuch that R = L1 \ � � � \ Ld.

XII.2 Linear orderings

There are three problems discussed in this subsection related to a given com-putable linear ordering A = (A;�A).

(1) The problem of �nding a subordering of A of type ! or of type !�.

(2) The problem of �nding an !-successivity or an !�-successivity in A.(3) The problem of �nd a self-embedding of A.


(1) Suborderings of type ! or !�

A standard classical result is that any in�nite linear ordering has a suborder-ing ff(0); f(1); : : :g of order type either ! or !� (the order type of the negativeintegers). Tennenbaum and independently Denisov showed that there is an in�-nite computable linear ordering of order type ! + !� which has no computablyenumerable subordering of either type (see Rosenstein [179] or Downey [62]).The suborderings of type ! (respectively !�) are simply the functions f : ! ! Asuch that f(n) �A f(n+1) (resp. f(n+1) �A f(n)) for all n. Thus in each casethe set of solutions to the problem of �nding such a subordering is a �0

1 class,but is clearly not bounded. For example, if A is the standard ordering (!;�),then the class of suborderings of A of type ! is just the class of all increasingsequences of natural numbers, which is homeomorphic to !! and not even com-pact. We observe that the class of suborderings of type ! is always a perfectset, since for any such subordering f and any n, there is another subordering oftype ! given by (f(0); f(1); : : : ; f(n); f(n+ 2); f(n+ 4); : : :).

Theorem XII.2.1. For any computable linear ordering A = (A;�A), the classof suborderings of A of type ! (respectively, of type !�) is a perfect �0

1 class.

Thus all we can say is that if a computable linear ordering has a suborder-ing of type ! (respectively, type !�), then it has such a subordering which iscomputable in some �11 set. It was shown by Manaster that any computablelinear ordering has a �0

1 subordering of type ! or of type !� (see Downey [62]).

(2) SuccessivitiesAn element b of A is said to be the successor of an element a if a <A b

and there is no c such that a <A c <A b; in such a case, we write b = SA(a).We say that a subordering f of type ! in A is an !-successivity if f(n + 1)is the successor of f(n) in the linear ordering for each n, and similarly de�nean !� � successivity. Then the family P of !-successivities is a �0

1 class andlikewise the family of !�-successivities.

Observe that the class of !-successivities of the standard ordering on !consists of all sequences (n; n + 1; n + 2; : : :) and is thus a countable set inwhich all elements are isolated. As for the suborderings above, this class is notnecessarily compact.

In general, there is at most one !-successivity f for each starting elementf(0) = a, so that every member of the class P of !-successivities is isolated;a class with this property is said to be scattered. Clearly P is also countable.Furthermore, we can de�ne a bounded computable tree T with P = [T ] by(a0; a1; : : : ; an) 2 T if and only if

(8i < n)[ai <A ai+1 & (8m < ai+1):(ai <A m <A ai+1)]:

A similar argument applies for !�-successivities.

Theorem XII.2.2. For any computable linear ordering A = (A;�A), the classof !-successivities (respectively !�-successivities) of A is a scattered, bounded�01 class.

XII.3. ORDERED ALGEBRAIC STRUCTURES 251

As an immediate application, we have the following.

Corollary XII.2.3. Every !-successivity (respectively !�-successivity) of acomputable linear ordering A is computable in 00.

This of course may also be proven directly from the de�nition of a successiv-ity. It follows from the result of Tennenbaum and Denisov that there is a com-putable linear ordering of type !+ !� which has no computable !-successivity.

(3) Self-embeddingsAnother classical result is due to Dushnik and Miller [68], who showed that

an in�nite countable linear ordering always has a non-trivial self-embedding.Hay, Manaster and Rosenstein [86] constructed a computable linear ordering oftype ! with no non-trivial computable self-embedding. A map f : A ! A is aself-embedding of A if, for all a and b, f(a) �A f(b) if and only if a � b. Thefamily of self-embeddings of a computable linear ordering is again seen to be a�01 class. For the standard ordering on !, it is clear that a self-embedding is the

same thing as a subordering of type !. Thus the class of self-embeddings neednot be compact.

Now A always has a computable self-embedding, namely the identity func-tion. If A has a non-trivial self-embedding, then we can �x an element a andconsider the �0

1 class of self-embeddings f such that f(a) 6= a. It follows asusual that A at least has a non-trivial self-embedding which is computable insome �11 set.

Theorem XII.2.4. For any computable linear ordering A = (A;�A), the classof self-embeddings of A is a �0

1 class.

Theorem XII.2.5. For any computable linear ordering A = (A;�A), if A hasa non-trivial self-embedding, then A has a self-embedding computable in a �11set.

Downey and Lempp [67] showed that the proof-theoretical stregnth of theDushnik-Miller theorem is ACA, which implies that every computable linearordering has a self-embedding which is computable in 00.

XII.3 Ordered algebraic structures

In this section, we consider two problems:

(1) The problem of �nding an ordering of an Abelian group.

(2) The problem of �nding an ordering of a formally real �eld.

In each case, the set of solutions to a given e�ective problem can always berepresented by a r:b: �0

1 class and in case (2), any r:b: �01 class can be represented

by such a set.


In this section, we will assume that a computably presented group, ring,or �eld is given by computable addition, subtraction, multiplication and divi-sion functions on the set !, as appropriate. A c. e. ring is the quotient of acomputable ring modulo a c. e. ideal and a c. e. group is the quotient of a com-putable group modulo a c. e. normal subgroup. An ordering will be representedby the cone of positive elements.

A formally real �eld is a �eld F such that no sum of (non-zero) squaresequals zero. A �eld (F;+F ; �F ) is said to be ordered by the relation � providedthat � is a linear ordering such that for all a; b; c 2 F ,(i) a � b! a+F c � b+F c.(ii) (0 � a & 0 � b)! 0 � a �F b.An ordering for a commutative group (G;+G; 0G) is de�ned similarly except inthis case the ordering � need only satisfy condition (i).

The set C = C� = fa 2 F : 0 � ag clearly satis�es the following for anya; b 2 F :(i) a; b 2 C ! a+F b 2 C.(ii) a; b 2 C ! a �F b 2 C.(iii) (a 2 C & 0F �F a 2 C) () a = 0F .

(iv) a 2 C _ 0F �F a 2 C.A subset C of F satisfying (i) to (iv) is said to be a positive cone of F . Thusany linear ordering of F de�nes a positive cone and conversely any positive coneC of F de�nes a linear ordering by

a � b () b�F a 2 C:Thus we will identify the set of linear orderings of a �eld F with the set ofpositive cones of F .

For a commutative group (G;+G; 0G), a cone C need only satisfy (i), (iii)and (iv).

The classical result of Artin-Schreier [5] is that any formally real �eld can beordered. Craven showed in [55] that any closed subset C of the Cantor space canbe represented as the set of orderings of some formally real �eld F . Metakidesand Nerode [150] made this proof e�ective by showing that if C is a �0

1 class,then F may be taken to be a computable �eld. Downey and Kurtz observedthat the �eld F may have additional orderings which are compatible with thegroup structure although not compatible with the �eld structure. The classicalresult for groups is due to Levi [128], who showed that an Abelian group can beordered if and only if it is torsion-free. Downey and Kurtz constructed in [66] acomputable group isomorphic to �!Z which has no computable ordering.

Theorem XII.3.1. For each speci�c c. e. instance of the problems (1) and (2)listed above, the set of solutions can be represented as a c. b. �0

1 class.

XII.3. ORDERED ALGEBRAIC STRUCTURES 253

Proof. For computable structures, this is immediate from the discussion above.For a c. e. structure, say, F = R=I, observe that a positive cone C on R=Icorresponds to a subset C 0 of R satisfying clauses (i), (ii) and (iv) along withthe following modi�ed version of clause (iii).

(iii) (a 2 C 0 & 0F �F a 2 C 0) () a 2 I.We leave it to the reader to translate these four clauses into a de�nition of a

computable tree T such that [T ] represents the set of positive cones on F . Theproof for ordered groups is similar.

We can as usual derive a number of immediate corollaries from the resultsof Part One. For example,

Theorem XII.3.2. (a) Any c. e. presented group which has an ordering hasan ordering of c. e. degree.

(b) If the set of orderings of the c. e. presented group G is countably in�niteand nonempty, then G has a computable ordering.

(c) If the c. e. presented �eld F has only �nitely many orderings, then everyordering of F is computable.

Next we turn to the other direction of our correspondence, that is, repre-senting an arbitrary �0

1 class by the set of solutions to certain of these problems.The problem of orderings of formally real �elds was solved by Metakides andNerode in [151].

Theorem XII.3.3. Any c. b. �01 class P can be represented by the set of or-

derings of a formally real �eld.

Proof. Let the computable tree T � f0; 1g<! be given so that P = [T ]. Theconstruction begins with the underlying ring R = Q[xi : i 2 !] (the ring ofpolynomials with rational coe�cients in in�nitely many variables). We de�ne acomputable maximal ideal of R such that the set of orderings of the �eld R=Irepresents [T ]. We sketch a proof is which is somewhat di�erent from that in[151].

The �rst step of our construction is to adjoin to Q the radicalsppi, where pi

is the i-th prime. That is, we put x2i �pi into I for each i. Thus we have initiallya continuum of possible orderings on Q[

ppi : i < !], where to each � 2 f0; 1g!

there corresponds the orderingR(�) determined by taking xi > 0 if �(i) = 0 andxi < 0 if �(i) = 1. Now for any � =2 T , we use an auxiliary variable y� to elim-inate the ordering corresponding to � in the following manner. We uniformlyand e�ectively de�ne, for � of length n, a polynomial f�(x0; : : : ; xn�1) such thatfor (e0; : : : ; en�1) 2 f0; 1gn, f�((�1)e0

p2; (�1)e1p3; � � � ; (�1)en�1

ppn�1) < 0

if and only if (e0; : : : ; en�1) = �. Then we add to I the polynomial y2� =f�(x0; : : : ; xn�1), thus adjoining to our �eld a square root for f�(x0; : : : ; xn�1).It follows that if � � �, then the ordering R(�) is not compatible with the�eld, since it forces a negative element to have a square root. The function f� is


de�ned to be f�(x0; : : : ; xn�1) = c�� (�1)�(0)x0�� (�1)�(n�1)xn�1, wherec� is the least integer c such that

p2 +p3 � � �+ppn�1 > c.

(For example, let � = (0; 1). Then we want f�(p2;�p3) < 0,

f�(p2;p3) > 0, f�(�

p2;�p3) > 0, and f�(�

p2;p3) > 0. We compute

that 3 <p2 +p3 < 4 and de�ne f�(x0; x1) = 3� x0 + x1.)

Finally, to prevent any additional orderings from arising due to the newroots in the �eld, we add a sequence of roots yi;j to the �eld such that yi;0 = yiand y2i;j+1 = yi;j . Thus each yi and each yi;j is forced to be positive.

This representation theorem has, as usual, a number of immediate corollariesof which we state only a few.

Theorem XII.3.4. (a) There is a computable formally real �eld which hasno computable ordering.

(b) There is a computable formally real �eld which has continuum many or-derings and such that any two distinct orderings are Turing incomparable.

(c) There is a computable formally real �eld F such that if a is the degree ofany ordering of F and b is a r:e: degree with a �T b, then b =T 0

0.

(d) There is a computable formally real �eld F which has a unique non-computable ordering �0, such that this ordering �0 has degree 00, andsuch that for any other ordering � of F , there is some �nite subset Aof F such that for any ordering �0 of R, if � agrees with � on A, then�=�0.

D. R. Solomon [199] showed that the analogue of the Metakides-Nerode the-orem fails for Abelian groups, that is, every abelian group has either two or hasin�nitely many orderings and therefore not every �0

1 class may be representedas the set of orderings of a computable abelian group.

Chapter XIII

In�nite Games

The set of winning strategies for an e�ective, closed f0; 1g-game of perfect in-formation was shown in [36] to strongly represent any c. b. �0

1 class. We willconsider more general closed games here.

For any subset C of NN, the in�nite game G(C) of perfect information isde�ned as follows. Two players, I and II, alternately play an in�nite sequencez = (x(0); y(0); x(1); y(1); : : :) and player II wins this play if z 2 C. A strategyfor Player II is a (partial) function � from !<! into !. For any play x =(x(0); x(1); : : :) of the game by Player I, the play �(x) of the game when �is applied to x is given by (x(0); y(0); x(1); : : :), where, for each n, y(n) =�((x(0); y(0); : : : ; y(n�1); x(n)). The strategy � is said to be a winning strategyfor Player II in the game G(C) if, for any play x of the game by Player I,�(x) 2 C. The notion of a strategy and a winning strategy for Player I issimilarly de�ned. The game G(C) is said to be determined if one of the twoplayers has a winning strategy. Gale and Stewart showed in [75] that the gameG(C) is determined if C is either closed or open. For a closed set C, we haveC = [T ] for some tree T , and we will sometimes refer to G(C) as G(T ). Wesay that G(T ) is a computably presented Gale-Stewart game if T is a recursivetree and that G(T ) is bounded (respectively, computably bounded) if the set [T ]is bounded (resp. c. b. ).

As pointed out in [36], strategies need to be coded to avoid always having aperfect set of winning strategies.

Let �0; �1; : : : e�ectively enumerate the nonempty elements of !<! in in-creasing order where we order the sequences by the sum of the sequence plusthe length and then lexicographically. Thus �0 = (0); �1 = (00); �2 = (1); : : :.For each � 2 !<!, let n(�) be the unique n such that � = �n. Then an arbi-trary sequence z = (z(0); z(1); : : :) 2 !! codes a strategy �z for Player II inthe following manner. For any play x = (x(0); x(1); : : :) of Player I, the strat-egy �z produces the following response y = (y(0); y(1); : : :) by Player II. First,y(0) = z(n((x(0))) and for any k, y(k + 1) = z(n), where �n = (x(0); : : : ; x(k)),that is, �z((x(0); y(0); : : : ; y(k � 1); x(k)) = z(n). Thus z(0) = �z((0)), z(1) =�z(0;�z(0); 0), z(2) = �z((1)), and so on. It is clear that the result �z(x) of

255

256 CHAPTER XIII. INFINITE GAMES

applying this strategy to a play x = (x(0); x(1); : : :) of the game by Player I canbe computed from x and z by a computable functional. For a �nite sequencezdn = (z(0); : : : ; z(n�1)), �zdn is a partial strategy which, applied to any partialplay xdm + 1 = (x(0); : : : ; x(m)) of Player I with n(xdm) < n, gives a partialresponse �zdn((x(0); y(0); : : : ; y(m � 1); x(m))) = y(m) where for all r � m,y(r) = z(kr) if n((x(0); : : : ; x(r)) = kr.

Now, for any tree T � !<!, let WS(T ) be the set of codes

z = (z(0); z(1); : : :) 2 !!

for winning strategies of Player II in the game G(T ).

Theorem XIII.0.5. For any computable tree T :

(a) WS(T ) is a �01 class.

(b) If T is �nitely branching, then WS(T ) is bounded.

(c) If T is highly computable, then WS(T ) is computably bounded.

Proof. We will de�ne a computable tree Q such that WS(T ) = [Q], as follows.First ; is in Q and then for any � = (z(0); : : : ; z(n�1)), � 2 Q if and only if, forall sequences � = (x(0); : : : ; x(r�1)) where n(�) < n, the result of applying thepartial strategy �� coded by � to the partial play � is in T . It follows from thediscussion above that there is a computable function g such that, for each n, thevalue z(n) of a coded strategy gives the play y(g(n)) of player II at step g(n).If T is �nitely branching, then there are only �nitely many possible choices fory(g(n)) which allow player II to win the game, so that only �nitely many valuesare possible for z(n). This makes WS(T ) bounded. If T is highly computabl,then we can actually compute a list of these possible values from g(n). ThusWS(T ) will be computably bounded.

As usual, we can derive a number of immediate corollaries. We state thefollowing and leave the rest to the reader.

Theorem XIII.0.6. Let T be a computable tree such that player II has a win-ning strategy for the Gale-Stewart game G(T ).

(a) There is a winning strategy which is computable in some �11 set and, ifthere are only �nitely many winning strategies, then each winning strategyis hyperarithmetic.

(b) If T is �nitely branching, then there is a winning strategy which is com-putable in 000.

(c) If T is highly computable, then there is a winning strategy of c. e. degreeand, if there are only countably many winning strategies, then there is acomputable winning strategy.

(d) If T is highly computable and there is no computable winning strategy, thenthere is a continuum of pairwise Turing incomparable winning strategies.

257

Next we consider the set of winning strategies for Player I (who is trying toget the play into the open set). Let WS0(T ) be the set of codes for winningstrategies of Player I. Note that for a Gale-Stewart game G(C), the set ofwinning strategies of Player I is in general not a closed set or an open set.

Theorem XIII.0.7. For any computable tree T :

(a) WS0(T ) is a �11 class.

(b) If T is �nitely branching, then WS(T ) is an open set.

(c) If T is highly computable, then WS(T ) is a �01 class.

Proof. We describe the class of actual strategies � and leave it to the readerto translate this into the coded strategies as in the proof of Theorem XIII.0.5.In general, � is a winning strategy for Player I if and only if, for all plays y ofPlayer II, the result �(y) of the game when Player I uses the strategy � is notin the set [T ], that is,

(8y)(9n)[(x(0); y(0); x(1); y(1); : : : ; x(n); y(n)) =2 T ]

where x(i+ 1) = �((x(0); y(0); : : : ; x(i); y(i))) for all i.If T is �nitely branching, let f(n) give an upper bound for the possible values

of �(n) for any � 2 T . Then we can use K�onig's In�nity Lemma as usual toexpress this in the form

(�) (9n)(8(y(0); y(1); : : : ; y(n)))[(x(0); y(0); x(1); y(1); : : : ; x(n); y(n)) =2 T ];

where each y(i) � f(2i), so that the (8) quanti�er is bounded, which shows thatWS0(T ) is an open set.

Finally, if T is highly computable, then we may take the function f tobe computable, so that the characterization (*) above makes WS0(T ) a �01class.

Theorem XIII.0.8. Let T be a computable tree such that Player I has a win-ning strategy for the Gale-Stewart game G(T ).

(a) There is a �12 winning strategy and, if there are only �nitely many winning

strategies, then each winning strategy is �12.

(b) If T is �nitely branching, then there is a computable winning strategy.

Proof. (a) This follows from the theorem that �12 is a basis for �1

1, which is acorollary of the Novikov-Kondo-Addison Uniformization Theorem (see Hinman[87], pp. 196-198) for details).

(b) Since WS0(T ) is open and nonempty, there must be an interval of codedwinning strategies, which of course will contain a computable strategy.

Now we consider the reverse direction of the correspondences given in The-orems XIII.0.5 and XIII.0.7.


Theorem XIII.0.9. For any computable tree Q, there is a computable treeT and an e�ective one-to-one degree preserving correspondence between the �0

1

class [Q] of in�nite paths through Q and the class WS(T ) of winning strategiesfor the e�ectively closed game G(T ). If Q is �nitely branching (respectivelyhighly computable), then T may be taken to be �nitely branching (resp. highlycomputable).

Proof. Let the computable tree Q be given. Our basic idea is that each path� = (�(0); �(1); : : :) 2 [Q] should correspond to a strategy �� which acts asfollows. Given any partial play, ((x(0); : : : ; x(m)) of Player I, �� will respondwith

��((x(0); y(0)); : : : ; y(m� 1); x(m))) = y(m)

where y(m) = 0 if x(i) > 0 for any i � m and y(m) = �(m) if x(i) = 0 forall i � m. Thus whenever Player I plays a value x(i) > 0, then ever after ��

will respond with a 0 and if Player I plays all 0's, then �� will respond byreproducing the path �. It is easy to see that when we code the strategy ��

via a sequence z = (z(0); z(1); : : :) that z will have the same Turing degree as�. Thus the correspondence �! �� will be an e�ective 1:1 degree preservingcorrespondence. Thus all we need to do is recursively de�ne a computable treeT � !<! so that WS(T ) = fz : z is a code of �� for some � 2 Qg. We beginwith sequences (a; b) of length 2 by putting (a; b) 2 T if and only if, either a > 0and b = 0 or a = 0 and (b) 2 Q. (This ensures that if Player I starts with anx > 0, then any winning strategy � for Player II must respond with a 0, whereasif Player I starts with a 0, then Player II must respond by starting a sequencein Q. Similar remarks will apply to the subsequent nodes we put in T .) Then,for each n and each � = (x(0); y(0); : : : ; x(n); y(n)) 2 T , do the following:

(1) If x(k) > 0 for some k � n, then put �_a_0 2 T and leave �_a_b out ofT for all a and for all b > 0.

(2) If x(k) = 0 for all k � n, then put �_a_b 2 T if and only if, either a > 0and b = 0 or a = 0 and (y(0); : : : ; y(n); b) 2 Q.

It easily follows from the de�nition of T that for any � = (�(0); �(1); : : :) 2[Q], �� is a winning strategy for Player II for the game G(T ). Now supposethat � is a winning strategy for Player II for G(T ). Then we can de�ne a� = (�(0); �(1); : : :) 2 [Q] such that � = �� by recursion as follows. For eachn, let �(n) = �((0; �(0); 0; �(1); : : : ; 0; �(n � 1); 0)). It is easy to see from ourde�nition of T that � 2 Q and that � = ��. Thus the correspondence �! ��

is our desired e�ective 1:1 degree preserving correspondence between [Q] andWS(T ).

Suppose now that Q is �nitely branching (respectively, highly computable).Let f(�) be an upper bound on fs : �_s 2 Qg; if Q is highly computablee, thenf is computable. Now given a partial code � = (z(0); : : : ; z(n � 1)) 2 WS(T )for a strategy for the game G(T ), we will indicate how to compute an upperbound g(�) for ft : �_t 2 WS(T )g. First compute the n-th �nite sequence

259

�n = (�(0); : : : ; �(k � 1)) in the enumeration described above, and use � tocompute the partial play � = (�(0); y(0); : : : ; �(k � 2); y(k � 2); �(k � 1)){thiscan be done since for any i < k, �di appears before � in the enumeration. Nowthere are two cases in the computation of g(�). If �(k � 1) > 0, then g(�) = 0and if �(k � 1) = 0, then g(�) = f(�). Thus WS(T ) is �nitely branchingand if Q is highly computable, then g is recursive so that WS(T ) is highlycomputable.

As usual, there are a number of immediate corollaries and we state only afew. Note that all of the examples below are games in which player II (who istrying to force the play into the closed set) has the winning strategy.

Corollary XIII.0.10. (a) There is a computably presented Gale-Stewart gamesuch that Player II has a winning strategy but has no hyperarithmetic win-ning strategy.

(b) There is a computably presented, bounded Gale-Stewart game G(C) suchthat Player II has a winning strategy and for any winning strategy � with00 <T � �T 000, there is a �02 set A such that 00 <T A �T �.

(c) For any c. e. degree c, there is a computably presented, computably boundedGale-Stewart game G(C) such that Player II has a winning strategy andthe set of c. e. degrees which contain winning strategies for G(C) equalsthe set of c. e. degrees �T c.

Next we consider the reverse direction for games in which Player I has a win-ning strategy. Here the bounded games all have computable winning strategiesand nothing more can be said. For the unbounded games, the reverse direc-tion demonstrates the connection between �1

1 classes and the game quanti�erof Moschovakis. Recall that the �0

1 class with index e is the set [Te] of in�nitepaths through the e-th primitive recursive tree Te. A theorem of Moschovakisstates that the set of indexes e such that Player I has a winning strategy forthe game G(Te) is a universal �1

1 set. See [155] for a discussion of the gamequanti�er and this theorem.

Note that every winning strategy for Player I is a limit point of the set ofwinning strategies for Player I, since once the play of the game has gotten intothe open set, Player I may play anything at all from that point on. Thus wecannot hope to represent even every �0

1 class with a one-to-one correspondence.

Theorem XIII.0.11. For any �11 class Q � !!, there is a recursively presented

Gale-Stewart game G(C) and a recursive function F such that y 2 V ()F (y) 2WS0(C).

Proof. Suppose that y 2 Q () (8x)(9n)R(xdn; ydn). De�ne the closed set Cto be f(x; y) : (8n):R(xdn; ydn)g. For each y 2 !!, let F (y) code the strategywhich simply plays y in response to any play x of Player I. Then it is clear thatF (y) codes a winning strategy if and only if y 2 Q.


Theorem XIII.0.12. (a) There is a computably presented Gale-Stewart gameG(C) such that the set WS0(T ) of winning strategies for Player I is not�11.

(b) There is a computably presented Gale-Stewart game G(C) for which PlayerI has a winning strategy but has no hyperarithmetic winning strategy.

Proof. (a) This is immediate from Theorem XIII.0.11.(b) Let Q = fzg be a �1

1 singleton such that z is not hyperarithmetic andlet the game G(C) and the recursive function F be given by Theorem XIII.0.11.Then it is clear that Player I has a unique winning strategy which consists ofplaying z(n) at his n-th turn, and that this strategy has the same degree asz.

Chapter XIV

The Rado Selection

Principle

In this section, we summarize the results of Jockusch, Lewis and Remmelfrom [93]. A Rado Family consists of collection of �nite subsets fAi : i 2Ig of A = [i2IAi and a collection of �nite partial functions f�F :2 AF :F is a �nite subset of Ig such that for each �nite subset F of I, �F (i) 2 Ai forall i 2 F . The Rado selection problem is to �nd a choice function f : I ! Asuch that for any �nite subset F of I, there is a �nite extension E � Fsuch that f(i) = �E(i) for all i 2 F . We call such a choice function aRado selector. Rado proved in [170] that any such family has a Rado selec-tor. A �nite set F = fx1 < : : : < xng of natural numbers may be coded byk = 2x1 + 2x2 + � � � + 2xn . In this case, we write F = Dk. We let 0 code theempty set. Then a family fAi : i < !g of �nite sets may be coded by a functionf such that Ai = Df(i) for each i. Similarly a family of �nite partial choicefunctions �F may be coded by a single function g such that g(i) = j if and onlyif Dj = f2x+13y+1 : x 2 Di & �Di(x) = yg. A Rado family together with thecoding described above is an e�ective Rado family A = I = ! and if the codingfunctions f and g are both computable.

Given an e�ective Rado family F as above, let Ch(F) be the set of functionsh : ! ! ! such that

(i) h(i) 2 Ai for each i and

(ii) for each �nite F � !, there is a �nite extension E such that �E(i) = h(i)for all i 2 F .

The following is Theorem 3 of [93].

Theorem XIV.0.13. For any e�ective Rado family F , there is a boundedstrong �0

2 class P and an e�ective, degree-preserving correspondence between Pand Ch(F).

261

262 CHAPTER XIV. THE RADO SELECTION PRINCIPLE

Proof. We can de�ne a tree T which is computable in 00 such that [T ] = Ch(F)as follows. A �nite path (y0; y1; : : : ; yn) is in T if and only if

(i) yi 2 Ai for all i � n and(ii) there exists a �nite set M such that f0; : : : ; ng �M and �M (i) = yi for

all i � n.Applying Theorems III.2.15, and V.2.2, we obtain the following.

Corollary XIV.0.14. Let F be an e�ective Rado family. Then

(a) F has a Rado selector of �02 degree.

(b) If F has only countably many Rado selectors, then F has a Rado selectorwhich is computable in 00.

The following is Theorem 2 of [93].

Theorem XIV.0.15. For any nonempty bounded strong �02 class P , there

exists an e�ective Rado family F and an e�ective, degree-preserving correspon-dence between P and Ch(F).

We can now prove the following.

Corollary XIV.0.16. (i) There is an e�ective Rado family such that, forany degree a of a Rado selector for F and any �02 degree b �T a, b = 000.

(ii) There is an e�ective Rado family such that, for any two degrees a, b ofRado selectors for F , a 6� b _ 00.

(iii) There is an e�ective Rado family such that, for any degree a �T 000 of aRado selector for F , there is a �02 degree b with 00 �T b �T a.

Chapter XV

Analysis

Computable functions on real numbers are just e�ectively continuous functionsand �0

1 classes of reals are just e�ectively closed sets. Computable real functionsmay be represented by computable functions on natural numbers, by enumer-ating a countable basis of rational intervals. E�ectively closed, compact setsof reals may be represented by �0

1 classes in f0; 1gN. Weihrauch [217] has pro-vided a comprehensive foundation for computability theory on various spaces,including the space of compact sets and the space of continuous real functions.

The basic example of a �01 class of reals is the sete of zeros of a computable

function. Nerode and Huang [158] showed that any �01 class in f0; 1gN may

be represented as the set of zeroes of a computable real function and Ko [112]showed that this can be done by polynomial time computable functions. Thisleads easily to the set of points where extreme values occur and to the set of�xed points of a computable function. E�ective real dynamical systems havebeen studied by Cenzer [22], [113] and more recently by Cenzer, Dashti, King,Toska and Wyman [24, ?] and by S. Simpson.

Index sets for e�ectively closed sets of reals were studied by Cenzer andRemmel in [41, 42, 43].

Results from Chapter VI are used to obtain the complexity of index setsrelated to the cardinality, computable cardinality, measure and category of ef-fectively closed sets of reals. Index sets for computable real functions are de�nedand lead to complexity results for index sets corresponding to the zeroes, ex-trema, and �xed points of such functions.

Brattka and Weihrauch [16, 217] identify three di�erent types of \e�ectivelyclosed" sets in Euclidean space <n. These are determined from an enumerationIm of the basic open sets (or intervals) and considering whether the set of msuch that Im (or In meetsK (or not) is a computably enumerable set. Of course,there are four possible notions here, and these can be re�ned further by askingwhether the c.e. sets are in fact computable. These notions are developedin [42] and applied to the graphs of computable functions. In particular, weexamine the question of whether a function with an e�ectively closed graph is\necessarily" continuous.

263

264 CHAPTER XV. ANALYSIS

Polynomial time and NP versions of e�ectively closed sets are studied and aversion of the \P=NP" problem is given. Here the choice of the basic open setsis crucial.

The fundamental problems for which the solution sets may be representedas �0

1 classes are the following.

(1) Zeroes of continuous functionsThe classical problem here is to �nd a zero for a continuous function. The

intermediate value theorem can be used to show the existence of a zero for acontinuous function which is negative at one point and positive at another point.The e�ective version of this theorem also holds, that is, any computable func-tion on the reals which is negative at one point and positive at another pointhas a computable zero, which can be computed by repeatedly splitting the in-terval between the two initial points. (See Pour-El{Richards [169] for a proof.)However, Lacombe [124, 125] showed that there are computable functions whichhave zeroes but have no computable zeros. We will give the improvement of thisresult due to Nerode and Huang [158] by showing that every �0

1 class is the setof zeroes of some computable function.

(2) The Extreme Value TheoremThe classical result here is that any function which is continuous on a com-

pact set takes on a maximum and a minimum on that set. The problem here isto �nd a point where the maximum or minimum is attained. Lacombe showedthat the extreme values of a computable function on [0; 1] are themselves com-putable and also constructed a computable function F on [0; 1] which does notattain its maximum at any computable point. We will present the result ofNerode and Huang [158] that any �0

1 class may be represented as the set ofpoints where some e�ectively continuous function attains its maximum.

(3) Fixed points of continuous functionsThe problem here is to �nd a �xed point for a given continuous function. A

simple application of the intermediate value theorem shows that any continuousfunction F on [0,1] has a �xed point. It is well known that if F is e�ectivelycontinuous, then F will have a computable �xed point. The Brouwer FixedPoint Theorem says that a continuous function on [0; 1] � [0; 1] will also havea �xed point, but Orevkov [164] showed that there need not be a computable�xed point. J. Miller [152] de�ned the notion of a �xable set as a �0

1 classQ � [0; 1] � [0; 1] for which there exists a computably continuous function Fsuch that Q = fz : F (z) = zg. He gave a beautiful result which characterizesthe �xable sets. Results for other spaces are di�erent. On the real line, thecontinuous function F (x) = x + 1 has no �xed point. On !!, the functionF ((x(0); x(1); : : :) = (1 + x(0); 1 + x(1); : : :) has no �xed point. On the Cantorspace the function F (x(0); x(1); : : :) = (1�x(0); 1�x(1); : : :) has no �xed point.

(4) Dynamical systems

265

We will give a few results on e�ective real dynamical systems from Cenzer[22] and from Ko [113]. We shall view a dynamical system as determined by acontinuous function F on a spaceX. The associated problem is to determine thebehavior of the sequence x; F (x); F (F (x)); : : : for a given x. In particular, wewant to �nd those points x for which this sequence is bounded or converges tosome �nite number and those x for which the sequence is unbounded or divergesto in�nity whereX is either the real line or the Baire space. If F is a polynomial,then it is always possible to compute a bound c such that fF (n)(x) : n < !gis bounded if and only if jF (n)(x)j < c for all n. In fact, we can take c largeenough so that F (x) > x + 1 for all x > c, so that limn!1F

(n)(x) = 1 forall x > c. In this situation, we say that 1 is an attracting point for F . Thenfx : jF (n)(x)j � c for all ng is called the Julia set of F . (See Blum, Shub andSmale [15].) It is then easy to see that the Julia set of any continuous functionmust be a compact set and we will show that for a computably continuousfunction, the Julia set is a �0

1 class. The �rst problem for dynamical systems isto �nd a member of the Julia set.

A point x is said to be a periodic point of a continuous function F if F (n)(x) =x for some �nite n. The basin of attraction B(x) of x is de�ned to be fu :limnF

(n)(u) = xg. The periodic point x is said to be attracting if there is someopen neighborhood U about x such that U � B(x). The basin of attractionof in�nity may also be de�ned as fu : limnF

(n)(u) = 1g. Thus the basinof attraction is an open set. We will show that for a computably continuousfunction, the complement of a basin of attraction is a �0

1 class. If 1 is anattracting periodic point of a function F on f0; 1g! or [0; 1], then we will referto the complement of B(1) as the Julia set of F . The problem here is to �nd apoint not in the basin of attraction.

Before turning to the problems mentioned above, we give a brief introductionto computable analysis, including the problem of characterizing the computableimage of the interval and the related concept of a real as a Dedekind cut ofrationals, which was studied by Soare in [196, 197].

A basic principle of computable analysis is that a computable function onthe real numbers is an e�ectively continuous function and a �0

1 class is ane�ectively closed set. We will consider the real line <, as well as three subspaces:the space of irrationals, which is homeomorphic to the Baire space !! and twocompact subspaces, the interval [0,1] and the Cantor space, which is computablyhomeomorphic to f0; 1g!. Since < is computably homeomorphic to the openinterval (0,1) via the order-preserving map ex

1+ex , we will frequently identify <with (0; 1) and treat it as a subset of [0; 1].

Let D be the set of dyadic rationals in [0; 1]. Then [0; 1] has a basis of openintervals (a; b),[0; c) or (d; 1] where a; b; c; d 2 D. Thus an open subset of [0,1] isa countable union

U = [n(an; bn)[[n[0; cn)

[[n(dn; 1]

of dyadic intervals. The open set U is said to be e�ectively open, or �01, if thesequences an, bn, cn and dn are computable. Then a closed set C is said to be


e�ectively closed, or �01, if it is the complement of an e�ectively open set.

Any x 2 f0; 1gN represents a real rx =Pn x(n)=2

n 2 [0; 1]. In addition,for any � 2 f0; 1g�, �_0! represents the dyadic rational q� =

Pi<n �(i)=2

i.Some di�culty arises from the fact that q� has another representation, �d(n�1)_0_1! (assuming that � ends in a 1). Each dyadic rational is of coursecomputable, so that we may unambiguously say that r is a computable real ifr = rx for some computable sequence x 2 f0; 1g!. Then a subset P of f0; 1g!represents a subset of [0; 1] if and only if, for all x; y such that rx = ry, we havex 2 P if and only if y 2 P . For any � 2 f0; 1g<! of length n, the members ofI(�) represent the members of the real closed interval [q�; q� + 2�n], which wedenote by U(�). More generally, if r < s are computable reals, then the interval[r; s] is a �0

1 class, since, if r = rx and s = ry, then

rz 2 [r; s] () (8n)[qxdn � 2�n � qzdn � qydn + 2�n]:

Lemma XV.0.17. (a) The following are equivalent for any subset K of [0; 1].

(1) K is a �01 class

(2) K is closed and fhp; ri 2 D2 : K \ [p; r] = ;g is a c. e. set.

(3) K is represented by a �01 class P � f0; 1g!.

(b) K may be represented by a computable binary tree with no dead ends ifand only if fhp; ri 2 D2 : K \ [p; r] = ;g is computable.

Proof. (a) We show that both (1) and (3) are equivalent to (2). Suppose �rstthat K is a �0

1 class and let

[0; 1] nK = [n(an; bn)[[n[0; cn)

[[n(dn; 1]:

Then

K \ [p; r] = ; () (9n)[p; r] � [m<n(am; bm)[[m<n[0; cn)

[[m<n(dn; 1]:

Suppose next that A = fhp; ri : K \ [p; r] = ;g is an c. e. set. Then K is a�01 class since

[0; 1] nK =[f(p; r) : hp; ri 2 Ag:

Furthermore, K is represented by [T ] where the �01 tree T is de�ned as

follows. Given � of length n, let

� 2 T () [q�; q� + 2�n] 6� [f(p; r) : hp; ri 2 Ang:

Here we replace q + 2�n with 1 if q = 1.)Finally suppose that K = frx : x 2 Pg for some �0

1 class P = [T ] � f0; 1g!.Then for any �,

K \ [q�; q� + 2�n] = ; () � =2 Ext(T )

267

Since any dyadic interval [p; r] may be decomposed into a �nite union of intervalsof the form [q�; q� + 2�n], it follows that fhp; ri : K \ [p; r] = ;g is an c. e. set.

(b) This follows from the observation that, if K is represented by [T ], then

� 2 Ext(T ) () K \ [q�; q� + 2�j�j] 6= ;:

An arbitrary �01 class P � f0; 1g! can be represented by a �0

1 subclass of[0,1] by the following lemma.

Lemma XV.0.18. For any �01 class P � f0; 1g!, there is a �0

1 subclass Q �f0; 1g! which represents a subset of [0; 1]nD which is computably homeomorphicto P.

Proof. Let the computable homeomorphism � be de�ned by

�(x(0); x(1); : : :) = (1; 0; x(0); 1; 0; x(1); : : :)

and let Q = �[P ]. Q represents a subset of [0,1] since every element of Q hasboth in�nitely many \1"s and in�nitely many \0"s.

We can characterize those intervals which are �01 classes using the notion

of the Dedekind cut L(r) = fq 2 D : q � rg of a real number r. Soareshowed in [196, 197] that if x 2 f0; 1g! is the characteristic function of a �0

1 set(respectively a �01 set), then L(rx) is a �0

1 set (resp. a �01 set) and that theseimplications are not reversible.

The set !<! and the space !! may be linearly ordered by the lexicographicordering �L, where x <L y if, for some n, x(n) < y(n) and x(i) = y(i) for alli < n. This ordering is computable on !<! and thus is �0

1 on !!, since

x �L y () (8n)xdn � ydn:

We now de�ne the interval [x; y] = fz : x �L z �L yg and also [x;1] = fz :x �L zg. Then we let L(x) = f� 2 !<! : �_0! � xg. These notions may alsobe restricted to f0; 1g! and f0; 1g<!. Observe that for non-dyadic rationals rxand ry, rx < ry if and only if x <L y.

Lemma XV.0.19. (a) For any x < y in either [0; 1], f0; 1g!, or !!, theinterval [x; y] is a �0

1 class if and only if L(x) is a �01 set and L(y) is a�01 set.

(b) In either [0; 1], f0; 1g!, or !!: L(x) is a computable set (respectivelycomputable in A) if and only if x is computable (resp. in A)

(c) For any x 2 f0; 1g!, if x is the characteristic function of a �0;A1 (respec-

tively �0;A1 ) set, then L(x) is a �0;A1 (resp. �0;A

1 ) set.


Proof. (a) First consider the case, where x < y and x; y 2 !!. We claim that[x; y] is a �0

1 class if and only if both [x;1] and [0; y] are �01 classes. The

if direction follows from the fact that [x; y] = [x;1] \ [0; y]. For the otherdirection, choose � 2 !! and n such that xdn �L � <L ydn and observe that[x;1] = [x; y] [ [�_1!;1] and [0; y] = [0; �_0!] [ [x; y].

Thus we need only show that [x;1] is a �01 class i� L(x) is a �01 set and

that [0; y] is a �01 class i� L(x) is a �

01 set. Suppose that [x;1] = [T ] for some

computable tree T . Then

� 2 L(x) () �_0! =2 [T ] () (9n)�_0n =2 T

and hence L(x) is �01 set. Vice versa, suppose that L(x) is a �01 set. Then wehave

z 2 [x;1] () (8m)(zdm =2 L(x))so that [x;1] is a �0

1 class.Similarly, if [0; y] = [T ] for some computable tree, then

� 2 L(y) () (8n)(�_0n 2 T )

so that L(y) is a �01 set. Vice versa, if L(y) is �

01 set, then

z 2 [0; y] () (8n)(zdn 2 L(y))

so that [0; y] is a �01 class.

For x; y 2 f0; 1g!, the argument is similar, except that [x;1] is replaced by[x; 1!].

For rx; ry 2 [0; 1], the problem reduces to the previous case of f0; 1g!, aslong as we take x to end in 0! whenever rx 2 D and y to end in 1! wheneverry 2 D, so that q� 2 [rx; ry] () � 2 [x; y].

(b) We give the argument for !!. L(x) is computable in x, since � 2L(x) () � �L xdj�j. Also, x is computable in L(x), since for each n, x(n+1)is the least a such that xdn_a 2 L(x) & xdn_a+ 1 =2 L(x).

(c) Now suppose that x is the characteristic function of a �01 set, i.e. x

is the characteristic function of ! n A where A is an r.e. set. Then let As

for s � 0 be some e�ective enumeration of A. Thus x is the decreasing limitof a sequence (x0; x1; : : :) where xs is the characteristic function of As. Then� 2 L(x) () (8n)(� �L xndj�j). Similarly, if x is the characteristic functionof a �01 set A then x is the increasing limit of the sequence (xn). Hence � 2L(x) () (9n)(� �L xndj�j).

It follows from part (b) that L(x) is �02 if and only if x is �0

2, and that if xis �0

2 (respectively, �02), then L(x) is �

02 (resp. �

02.)

Theorem XV.0.20. (a) Let x 2 !!. If x is the maximum element of a c. b.�01 class, L(x) is a �0

1 set. If L(x) is a �01 set and, in addition, x is not

hyperimmune, i.e. there is a computable function f such that x(e) � f(e)for all e, then x is the maximum element of some r:b: �0

1 class. If x is the

269

minimum element of some r:b: �01 class, then L(x) is a �01 set. If L(x)

is �01 set and, in addition, x is not hyperimmune, then x is the minimumelement of some r:b: �0

1 class.

(b) For any x in [0; 1], x is the maximum element of some �01 class if and

only if L(x) is �01 and x is the minimum element of some �0

1 class if andonly if L(x) is �01.

(c) For any x 2 !! or [0; 1], x is the maximum (respectively, minimum)element of a �0

1 class represented by a tree with no dead ends if and onlyif x is computable.

(d) For any x 2 !!, if x is the maximum element of a bounded �01 class, then

L(x) is a �02 set and if x is the minimum element of a bounded �0

1 class,then L(x) is a �02 set.

(e) For any x 2 !!, if x is the maximum element of a �01 class, then L(x) is

a �11 set and if x is the minimum element of a �01 class, then L(x) is a

�11 set.

Proof. We just give proofs for the maximum element versions.(a) Suppose that L(x) is a �0

1 set and there is a computable function fsuch that x(e) � f(e) for all e. Then x is the maximum element of the �0

1

interval [0; x] by Lemma XV.0.19. Hence x is the maximal element of the r:b:�01 class [0; x]\ [T ] where T is the computable tree such that � 2 T () (8i �j�j)(�(i) � f(i)).

Now let x be the maximum element of a r:b: �01 class P = [T ]. Then � 2 L(x)

if and only if

(9y)[y 2 P & � �L y] () (9� 2 !j�j)[� 2 Ext(T ) & � �L � ]:

Since T is r:b:, the search for � is bounded and, since Ext(T ) is a �01 set, L(x)

is a �01 set.

If T has no dead ends, then Ext(T ) is computable, so that L(x) is com-putable. This completes the proof of part (a) as well as part (c).

Part (b) now follows from Lemma XV.0.17.(c) Any computable x is the maximum element of the r:b: class fxg. The

maximum element of a r:b: �01 class P = [T ] is computed by letting x(n) be the

largest i such that (xdn� 1)_i 2 T .Parts (d) and (e) follow from the characterization of L(x) given above, since

Ext(T ) is always �11 and is �02 if P is bounded.

XV.0.1 Computable continuous functions

Next we turn to the de�nition of computably continuous functions. For functionson !! or f0; 1g!, a computable function y = F (x) is given by an oracle Turingmachine which uses input x as an oracle to compute the values y(n) and iscontinuous since each value y(n) depends on only �nitely many values of x.


Lemma XV.0.21. A function F : !! ! !! (respectively, F : f0; 1g! !f0; 1g!) is computably continuous if and only if there is computable functionf : !<! ! !<! (resp. f : f0; 1g<! ! f0; 1g<!) such that

(1) for all � � � , f(�) � f(�),(2) for all x 2 !!, limn!1 jf(xdn)j =1, and

(3) for all x 2 !!, limn!1 f(xdn) = F (x).

Proof. Given such a representation f for F , clearly we can compute y(n) fory = F (x) from x by computing f(xdk) for su�ciently large k.

Given a computable function F , de�ne the representation f as follows. Oninput � of length n, compute the values of �(i) where � = f(�) for each i < nby applying the algorithm for F for n steps, using oracle �. The length of � willbe the least k < n such that �(k) does not converge in n steps.

In general, a function F on the Baire space is continuous if and only if it hasa representation f as above. Thus F is continuous if and only if it is computablein some parameter x 2 !!.

The de�nition of computably continuous real functions is more di�cult.

De�nition XV.0.22. A function F : [0; 1] ! [0; 1] is computable (or com-putably continuous if there is a uniformly computable sequence of functions fn :D ! D such that, for any x 2 f0; 1g!, F (rx) = limi fi(qxdi) and a computablefunction � : ! ! ! such that, for all natural numbers m;n; k and all dyadicrationals q; r, if jq � rj < 2��(k) and m;n > �(k), then jfm(q)� fn(r)j < 2�k.

This de�nition is easily seen to be equivalent to other standard de�nitions,such as those given by Lacombe [124]. See Pour-El{Richards [169] for somehistory.

Note for any computable real function, F (x) is computable real for anycomputable real x.

Functions of several variables are treated similarly, thus a uniformly com-putable sequence of functions ffngn2! and a computable function � representa continuous function F : [0; 1]2 ! [0; 1] if lim fi(qxdi; qydi) = F (x; y) for any

reals x; y and if jfm(q1; q2)� fn(r1; r2)j < 2�k whenever m;n > �(k) and bothjq1� r1j; jq2� r2j < 2��(k). For example, the standard distance function jx� yjmay be represented by taking fn(q; r) = jq � rj for all n and �(k) = k + 1.

We say a function F : f0; 1g! ! f0; 1g! represents a real functionG providedy = F (x) whenever ry = G(rx).

Lemma XV.0.23. If F is a continuous (respectively computably continuous)map on f0; 1g! such that F (x) = F (y) whenever rx = ry, then F represents acontinuous (respectively computably continuous) map on [0; 1].

Proof. Given the representation function f for F , let fi(q�) = qf(�) for all i andlet �(k) be the least n such that jf(�)j > k for all � 2 f0; 1gn.

271

We remark that not every computably continuous real function may be rep-resented by a computable function on f0; 1g!; the distance function jx� rj forany �xed rational r 2 (0; 1) is a counterexample. For example, suppose thatr = 1

6 and G(x) = jx � 16 j. Now suppose that F : f0; 1g! ! f0; 1g! represents

G and that f : f0; 1g<! ! f0; 1g<! represents F . Now G( 23 ) =12 which has two

representations x1 = 1_0! and x0 = 0_1!. Let x2 = (10)! so that x3 repre-sents 2

3 . Then either F (x2) = x1 or F (x2) = x0. Suppose �rst that F (x2) = x0.Then for some n, 0 � f((10)n). But then (10)n

_1! is a number greater than

23 so that 1 � f((10)n1k) for some k which is a contradiction. Similarly ifF (x2) = x1, then for some n, 1 � f((01)n). But then (10)n

_0! is a number

less than 23 so that 0 � f((10)n0k) for some k which is again a contradiction.

A computable metric on the Baire space is de�ned by �(x; y) = 1=2n = 0n1!,where n is the least such that x(n) 6= y(n), and �(x; y) = 0 = 0! if x = y.

The graph of a function F : X ! X is de�ned as usual to be gr(F ) =f(x; F (x)) : x 2 Xg. For X = !!, we can view the graph as a subset of X byassociating the pair (x; y) with the element z = x y, where z(2n) = x(n) andz(2n+ 1) = y(n). For any class P and any x 2 X, let �x(P ) = fy : x y 2 Pg.For a function F from [0; 1] to [0; 1], the graph may be represented by a subsetof f0; 1g!, namely fx y : f(rx) = ryg.

A classical result says that a function on the interval is continuous if andonly if the graph is closed. We give the e�ective version here.

Theorem XV.0.24. (a) The graph of a computably continuous function on!! is a �0

1 class.

(b) Let X be either f0; 1g! or [0; 1]. Then a function F : X ! X is com-putably continuous if and only if the graph of F is a �0

1 class. Furthermore,the graph of any computably continuous function may be represented by atree with no dead ends.

Proof. (a) Suppose �rst that F : !! ! !! is computably continuous and isrepresented by f : !<! ! !<!. De�ne the computable tree T with [T ] = gr(F )by putting � � 2 T if and only if � is consistent with f(�).

(b) Given a computable F : f0; 1g! ! f0; 1g!, de�ne the computable treeT with gr(F ) = [T ] as in (a). Then Ext(T ) is �01, and therefore computable,by the following easily veri�ed claim.

CLAIM: � � 2 Ext(T ) () (9�0 � �)� � f(�0):Given a computable tree T so that gr(F ) = [T ], de�ne the computable

representing function f by letting f(�) be the common part of f� : � � 2 Tg.Next suppose that F is a computably continuous function on [0; 1] and let

the computable sequence fi of dyadic rational functions and the computablemodulus function � be given as in De�nition XV.0.22. We can assume that�(k) > k for all k. In this case, we can de�ne our desired computable tree Twith gr(F ) = [T ] to be the set of pairs � � of length 2n � 1 or 2n such thatjfn(q�)� q� j � 21�k for all k such that j�j � �(k). Again it is easy to see thatEXT (T ) is computable.


Suppose now that gr(F ) is a �01 class and, by Lemma XV.0.17, let T �

f0; 1g<! be a computable tree so that [T ] represents gr(F ). T may not be thegraph of a function, since each dyadic real has two representations. Howeverany two representations of length n di�er by 2�n. Thus, for any i and any � oflength n, we let fi(q�) = q� for the lexicographically least � of length j�j suchthat � � 2 T , and let �(k) be the least n such that, for all � of length n andany �1; �2 with � �1 and � �2 both in T , �q(�1; �2) < 2�k.

We next examine the complexity of the image of a �01 class under a com-

putablly continuous function. The classical results are that the image of anycompact set under a continuous function is compact and that the image of aclosed set is an analytic set.

Theorem XV.0.25. Let F be a computably continuous function on a �01 sub-

class P = [T ] of !! or [0; 1] and let F [P ] = fF (x) : x 2 Pg. Then(a) F [P ] is a �11 class,

(b) if P is bounded, then F [P ] is a strong �02 class, and

(c) if P is computably bounded, then F [P ] is a computably bounded �01 class

and, furthermore, if there is a computable tree T with no dead ends suchthat P = [T ], then there is a computable tree S with no dead ends suchthat F [P ] = [S].

Proof. (a) This part follows immediately from the fact that y 2 F [P ] ()(9x)(x 2 P & hx; yi 2 gr(F )).

(b) Suppose that T is a �nitely branching, computable tree and let S be acomputable tree such that gr(F ) = [S]. Then it follows from K�onig's Lemmathat F [P ] = [R], for the �nitely branching �01 tree R de�ned by

� 2 R () (9�)[� 2 T and � � 2 S]:(c) Now suppose that T is computably bounded and let F be represented by

the computable function f : !<! ! !<!. Then it is easy to see the de�nitionabove in (b) becomes computable.

To �nd a bound for the possible value of �(n) for � 2 R, compute the leastmsuch that jf(�)j > n for all � 2 T of length m. Then we compute the maximumvalue h(r) of f(�(n)) for all � 2 T of length n. Thus R is highly computable.

ly continuous map F (x) = r + (s� r)x.Suppose thatK is the image of the computably continuous map F . It follows

from the Intermediate Value Theorem that K = [r; s] where the reals r and s arethe maximum and minimum elements of P . It follows from Theorem XV.0.25that K may be represented by a tree with no dead ends and then from TheoremXV.0.20 that r and s are computable.

Corollary XV.0.26. Let F be a computably continuous function on !!, f0; 1g!,or [0; 1]. Then the maximum and minimum values of F on P are computablereals (if they exist).

273

Theorem XV.0.27. Each of the following sets is a �01 class for any computably

continuous function F : X ! X, where the space X may be f0; 1g!, [0; 1], !!or <. In case (3), the class always has a compuable member when X = [0; 1].In case (4), the class is always bounded when X = <.

(a) The set of points x where F (x) = x0 for any �xed computable x0.

(b) The set of points where F attains its maximum (minimum).

(c) The set of �xed points of F .

(d) The Julia set of F where X = !! or <.(e) The complement of the basin of attraction of a computable periodic point.

Proof. (a) This is immediate from Theorem XV.0.24.(b) It follows from Corollary XV.0.26 that the maximum and minimum are

computable if they exist. The result now follows from part (b).(c) This is easily reduced to part (b). For <, x is a �xed point of F if and

only if F is a zero of G(x) = F (x)� x. For [0; 1], take G(x) = jF (x)� xj. Forf0; 1g! or !!, de�ne z = G(x) by z(n) = jF (x)(n)� x(n)j.

A computable �xed point r may be found for a computably continuous func-tion on X = [0; 1] by the standard procedure. If F has a dyadic �xed point,then there is nothing to do. If not, then repeatedly split the interval in two andchoose the subinterval with F (x) < x on one end and F (x) > x on the other.Then r is the unique element in the intersection of these intervals.

(d) This is immediate from the characterization of the Julia set as fx :(8n)jFn(x)j � cg for a �xed computable point c. Note that in !!, fx : x �L x0gis not a bounded �0

1 class in our sense of being the paths through a �nitebranching tree.

(e) Given an attracting point c for F , there is some computable interval(a; b) � B(x) containing c. Then the complement of the basin of attraction maybe characterized as

fx : (8n)(Fn(x) � a _ Fn(x) � b)g

As usual, we give a few immediate corollaries from the results of Part One.

Theorem XV.0.28. Let F : X ! X be a computably continuous map, whereX is either f0; 1g!, [0; 1] or <.(a) If F attains a maximum M , then there are two points x1 and x2 with

F (x1) = F (x2) =M such that any function computable in both x1 and x2is computable.

(b) If F has only countably many zeroes, then F has a computable zero.


(c) If F has only �nitely many �xed points, then every �xed point of F iscomputable.

(d) If the Julia set of F : < ! < has no computable member, then it containsa continuum of pairwise Turing incomparable elements.

(e) If the basin of attraction B(x0) of a computable �xed point x0 of F is notall of X, then there is a point x of c. e. degree which is not in B(x0).

Proof. We just note that in each case, a function de�ned on < may be restrictedto a �nite interval and thus be treated as a map on the interval. For example,if F has a zero, take a computable interval [a; b] on F has a zero and let [c; d]be the image of [a; b] under F . Then F may be composed with maps between[0; 1] and the two intervals to obtain a map G : [0; 1]! [0; 1] so that the set ofzeroes of G is homeomorphic to a subset of the set of zeroes of F .

Theorem XV.0.29. Let F : !! ! !! be a computably continuous map.

(a) If F attains a maximum M , then F (x) = M for a point x which is com-putable in some �11 set.

(b) If F has only countably many zeroes, then F has a hyperarithmetic zero.

(c) If F has only �nitely many �xed points, then every zero of F is hyper-arithmetic.

Next we give the collection of converses to Theorem XV.0.28. The �rst threeparts are due to Nerode and Huang [158] and may also be found in Ko [112].

Next we give the collection of converses to Theorem XV.0.28. The �rst threeparts are due to Nerode and Huang [158] and may also be found in Ko [112].

Theorem XV.0.30. Let P be a �01 subclass of the space X, either f0; 1g!,

[0; 1], !! or <.(1) There is a computably continuous function F such that P is the set of

zeroes of F .

(2) There is a computably continuous function F with maximum valueM suchthat P = fx : F (x) =Mg.

(3) (a) If X is either f0; 1g!, !! or <, then there is a computably continuousfunction F such that P is the set of �xed points of F .

(b) If X is [0; 1] and P has a computable member, then there is a com-putably continuous function F such that P is the set of �xed pointsof F .

(4) If P is bounded and has both a computable maximum and a computableminimum element, then there is a computably continuous function suchthat

275

(a) P is the complement of the basin of attraction of a computable peri-odic point, where X = [0; 1] or <.

(b) P is the Julia set of F where X = <Proof. (1) First suppose P � !! and let T be a computable tree such thatP = [T ]. De�ne the computable function F by

F (x) =

�0!; for x 2 P0n_1_0!; if n is the least with xdn =2 T :

If P represents a subset of [0; 1], then the function F is modi�ed when xrepresents a dyadic, so that F (�_1_0!) = F (�_0_1!) for all �. Thus whenrx = ry is dyadic, we let

F (x) = F (y) =

�0!; for x 2 P0n_1_0!; if n is the least with xdn =2 T and ydn =2 T :

For a subset P of <, let Q be the image of P under the isomorphism G with(0; 1) together with the point 0, if P has no lower bound and the point 1, if Phas no upper bound. Then let H be the computably continuous map with setQ of zeroes. It follows that P is the set of zeroes of H �G.

(2) Let F be the function de�ned in the proof of (1) and observe that 0 isthe minimum value of F in each case. For the maximum argument on [0; 1] or<, just take G(x) = 1 � F (x). For the maximum argument on !!, note thatthe range of F is a subset of f0; 1g! and take G(x)(n) = 1� F (x)(n).

(3) (a) Let F be given by (1) so that P is the set of zeroes of F . Now letG(x) = F (x) + x for the real line, and, for !! or f0; 1g!, let G(x)(n) = x(n) ifF (x)(n) = 0 and G(x)(n) = 1� x(n), if F (x)(n) 6= 0.

(b) Let x0 be a computable member of the �01 class P and let F be the

function given by (1) so that x 2 P if and only if F (x) = 0. De�ne G(x) to bex+ (x0 � x)F (x).

(4) (a) Let P be a �01 proper subclass of [0; 1]. Then there is some computable

element x0 =2 P . Let F be the computable function given by part (1) such thatF (x) = 0 for x 2 P and F (x) > 0 for x =2 P . Let P1 = P \ [0; x0] andP2 = P \ [x0; 1]. Let M1 be the maximal element of P1 and let M2 be theminimal element of P2, so that both M1 and M2 are computable. Now de�nethe function G by cases.

G(x) =M1 + F (x)(x0 �M1), for x �M1;G(x) = x+ (x�M1)(x0 � x), for M1 � x � x0;G(x) = x� (M2 � x)(x� x0), for x0 � x �M2;G(x) =M2 � F (x)(M2 � x0), for M2 � x.

Then x0, M1 and M2 are all �xed points of G. We claim that P is the comple-ment of the basin of attraction of x0. The following inequalities are immediatefrom the above de�nition.

M1 � G(x) � x0, for x < M1;x < G(x) < x0, for M1 < x < x0;x0 < G(x) < x, for x0 < x < M2;


x0 � G(x) < M2, for M2 < x.

First we show that the basin of attraction of x0 for G includes [M1;M2].Given M1 < x < x0, we see that x < G(x) < x0. It follows that Gn(x) is anincreasing sequence with limit L such that G(L) = L and M1 < L � x0. Thuswe must have L = x0. A similar argument works for x0 < x < M1.

Next suppose that x =2 P and either x < M1 or x > M2. Then eitherG(x) 2 [M1;M2], so that x is in the basin of attraction of G.

Now suppose that x 2 P , so that either x 2 P0 or x 2 P1. For x 2 P0,we have F (x) = 0 and x � M1, so that G(x) = M1 and thus Gn(x) = M1 forall n > 0. Thus x is not in the basin of attraction of G. Similarly for x 2 P1,Gn(x) =M2 for all n > 0, so that x is not in the basin of attraction of G.

For X = <, just identify X with a subclass of (0; 1) as in (1) above.

(b) Let P be a bounded �01 class of reals with a computable minimal element

m and a computable maximal element M and let the computably continuousfunction F be given by (1) so that F (x) = 0 for all x 2 K and F (x) > 0 for allx =2 K. Now de�ne the function F in the following cases.

G(x) = m+M � x, for x � m.

G(x) =M + F (x), for m � x �M .

G(x) = 2x�M for x �M .

Since any countable �01 subset of [0; 1] and any �0

1 subset which may berepresented by a tree with no dead ends has a computable member, we have thefollowing immediate corollary.

Corollary XV.0.31. (a) If the nonempty �01 subclass K of [0,1] may be rep-

resented by a tree with no dead ends, then K is the set of �xed points ofsome computably continuous function from [0,1] into [0,1].

(b) Any countable, nonempty �01 subclass K is the set of �xed points of some

computable function from [0,1] into [0,1].

As usual, we have a number of immediate corollaries, of which we state onlya few.

Theorem XV.0.32. Let X be f0; 1g!, !!, <, or [0; 1].

(a) For any r:e: degree c, there is a computably continuous function F on Xsuch that the set of c. e. degrees which contain zeroes of F equals the setof c. e. degrees �T c.

(b) There is a computably continuous function F on X which has a �xed pointand such that any two distinct �xed points are Turing incomparable if Xis f0; 1g!, !!, <. There is a computably continuous function F on [0; 1]which has a unique computable �xed point and uncountable many non-computable �xed points and such that any two distinct non-computable�xed points are Turing incomparable.

XV.1. SYMBOLIC DYNAMICS 277

(c) There is a computably continuous function F which has a maximum Mon X, such that there is a unique non-comptable point x0 where M isattained and x0 is also the unique accumulation point of the set where Mis attained.

(d) There is a computably continuous function on < with attracting point atin�nity such that every computable point is attracted to in�nity but notevery point is attracted to in�nity.

(e) There is a computably continuous function on [0; 1] with a attracting pointat in�nity such that every computable point is attracted to in�nity but notevery point is attracted to in�nity.

Proof. Note that in part (b) when X = [0; 1], we may add a single computablepoint to the �0

1 class so that it can represent the set of �xed points.

Theorem XV.0.33. (a) There is a computably continuous function on !!

which has a zero but has no hyperarithmetic zero.

(b) There is a computably continuous function on !! which attains a maxi-mum M such that F (x) 6=M for any hyperarithmetic point x.

(c) There is a computably continuous function on !! which has a �xed pointbut has no hyperarithmetic �xed point.

Ko [113] improved part (4) of Theorem 10.15 by showing that if the �01 class

P has either a p-time maximum element or a p-time minimum element, thenthere is a p-time computable function f with Julia set P . Furthermore, Koshows in [113] that there is such a set P which has a non-computable Hausdor�dimension, which implies that there is a p-time computable function f such thatthe Julia set of f has non-computable Hausdor� dimension.

XV.1 Symbolic Dynamics

In this section, we examine computable dynamical systems and symbolic dy-namics associated with computable functions on the Cantor space f0; 1gN andthe unit interval [0; 1].

Computable real dynamical systems have been studied by Cenzer [?], wherethe the Julia set of a computably continuous real function is shown to be a �0

1

class and Ko [113], who examined fractal dimensions and Julia sets. Computablecomplex dynamical systems have recently been investigated by Braverman andCook [?] and Braverman and Yampolsky [?], who showed that there is a complexnumber c such that the Julia set corresponding to the function f(z) = z2 + c isnot decidable.

In particular, the computability of a closed set K in a computable metricspace (X; d) may be de�ned in terms of the distance function dK , where dK(x)is the in�mum of fd(x; y) : y 2 Kg. K is a �0

1 class if and only if dK is


upper semi-computable and K is a decidable (or computable) �01 class if dK is

computable.xxxput this in earlierzzzFor any �nite k, the shift function on f0; 1; : : : ; kg is de�ned by �(x) = y,

where y(n) = x(n+ 1). A closed set Q � f0; 1; : : : ; kg is said to be a subshift ifit is closed under the shift function. We will refer to a �0

1 class which is also asubshift as a subsimilar �0

1 class.Fix a �nite alphabet �, let F : �N ! �N be a computable function and let

a partition fU0; U1; : : : ; Ukg of �N into clopen sets be given. The itinerary of apoint x 2 �N is the sequence It(x) 2 f0; 1; : : : ; kgN where

It(x)(n) = i () Fn(x) 2 Ui:Now let IT [F ] = fIt(x) : x 2 �Ng. We show that IT [F ] is a decidable subsimilar�01 class and that, for any decidable subsimilar �0

1 class Q � �N, there exists acomputable F such that Q = IT [F ].

XV.1.1 Undecidable subshifts

In this section, we construct a subsimilar �01 class with no computable element.

We will give the construction in f0; 1gN, but it can be generalized to �N forany �nite �. Now every decidable �0

1 class has a computable element (in fact,the leftmost path is computable). Hence we have an undecidable subsimilar �0

1

class.Let us say that a string v is a factor of a string w if there exist w1 and w2

such that w = w1_v_w2. For any set S of strings, we may de�ne a closed set

PS , where x 2 PS if and only if, for all n and all w 2 S, w is not a factor ofxdn. If the set PS is nonempty, then S is said to be avoidable. For this section,we restrict ourselves to � = f0; 1gLemma XV.1.1. Given any sequence x0; x1; : : : of elements of f0; 1gN, thereis a nonempty subshift containing no xi.

Proof. De�ne the sequence l0; l1; : : : by l0 = 3 and, for n > 0,

ln = 3(2n(n+3)

2 ):

This will imply that ln+1 = 2n+2ln. Now let wn = xnd2ln for each n and de�nesubshift P to consist of all x which do not contain any wn as a factor. Clearlyxi =2 P for all i. It remains to show that P is nonempty, that is, fwn : n 2 Ngis avoidable.

It is important to notice that given any word w of length 2k, it has at mostk + 1 distinct factors of length k. Since there are 2k words of length k, for klarge enough so that 2k > k+1, there are words of length k that do not appearas a factor of w. With this in mind, we construct recursively two sequences ofwords < An >n2N and < Bn >n2N such that, for all n:


1. jAnj = jBnj = ln.

2. A0 and B0 are not factors of w0; this is possible since jw0j = 6 so w0 hasat most 4 distinct factors of length 3.

3. An+1 and Bn+1 are taken from fAn; Bng�, have An as a pre�x, and havelength m = 2n+2 = ln+1=ln. This is possible since there are 2m�1 suchwords, but there are at most ln+1 + 1 factors of length ln+1 in wn+1 and2m�1 = ln+1 + 1 + 2.

Now let x = limnAn. This exists since each An � An+1. We claim thatx 2 P . Suppose by way of contradiction that some wn is a factor of x. Wecan view x as an in�nite concatenation of blocks length ln, where each blockis either An or Bn. Since wn has length 2 ln, it must completely contain oneof the blocks, which would imply that either An or Bn is a factor of wn. Thiscontradiction shows that x 2 P .

We need to improve this lemma in two ways. First, we may have only asubset of words wk of length lnk . Second, we need an e�ective version.

Theorem XV.1.2. There is a recursive sequence of natural numbers l0; l1; : : :such that if for any subsequence < lnk >k2N and any set S = fvk : k 2 Ngof words such that jvkj = lnk , S is avoidable. Furthermore, if � is a partialcomputable function such that �(nk) = vk, then there is a nonempty subsimilar�01 class P such that no element of P contains any factor vk.

Proof. For the �rst part, simply let wnk = vk and choose arbitrary words wi oflength li for i =2 fnk : k 2 Ng and apply the lemma.

For the second part, we have

x 2 P () (8n)(8k)[vk is not a factor of xdn]In more detail, notice that vk is not a factor of xdn if and only if, for all v, if�s(k) = v, then v is not a factor of xdn.Theorem XV.1.3. There is a nonempty subsimilar �0

1 class P with no com-putable element.

Proof. Let the sequence< ln > be given as in Lemma XV.1.1. Let Let �0; �1; :::; �e; : : :be an enumeration of partial computable functions. Now de�ne the partial re-cursive function � by

�(k) =

(�kdlk; if �k(i) # for all i < 2lk;

unde�ned; otherwise.

By Theorem XV.1.2, there is a nonempty subsimilar �01 class P such that no

element of P has any word �(k) as a factor. Now let y be any computableelement of f0; 1gN. Then y = �k for some k such that �k is a total function.Thus �(k) = �kdk is de�ned and is not a factor of any x 2 P and hence certainly�k =2 P .


XV.1.2 Symbolic Dynamics of Computable Functions

Fix a �nite alphabet � = f0; 1; : : : ; kg, let F : �N ! �N be a computablefunction and let a partition fU0; U1; : : : ; Ukg of �N into clopen sets be given.The itinerary of a point x 2 SN is the sequence It(x) 2 f0; 1; : : : ; kgN where

It(x)(n) = i () Fn(x) 2 Ui:

Now let IT [F ] = fIt(x) : x 2 �Ng. We observe that IT [F ] is a subshift. Thatis, suppose y = It(x) 2 IT [F ]. Then �(y) = It(F (x)), so that �(y) 2 IT [F ] aswell. The function It is continuous and hence IT [F ] is a closed set, as seen bythe proof of the following lemma.

Lemma XV.1.4. The function from �N ! f0; 1; : : : ; kgN mapping x to I(x) iscomputable.

Proof. Given clopen sets U0; : : : ; Uk, there exists a �nite j and a �nite subsetW of f0; 1gj such that each Ui is a �nite union of intervals I[w] for some set ofw 2W . Thus one can determine from ydj the unique i for which y 2 Ui. Givenx 2 �N, let y = I(x). To compute y(n), it su�ces to �nd the �rst j values ofFn(x), which can be computed uniformly from x and n.

Theorem XV.1.5. Fix a computable function F : �N to �N, let a partitionfU0; U1; : : : ; Ukg of �N into clopen sets be given and let I(x) denote the itineraryof x under F . Then

(a) For any computable x 2 �N, I(x) is computable.

(b) The set IT [F ] of itineraries is a decidable, subsimilar �01 class.

Proof. Part (a) follows from the well-known result that computable functionsmap computable points to computable points and (b) follows from the fact thatthe image of a decidable �0

1 class under a computable function is a decidable�01 class. See [?, ?].

Next we prove the converse. Note that F 0(x) = x for all x 2 �N andtherefore IT [F ] meets every Ui. Note that if Q is a subshift and Q does notmeet I[i], then Q � f0; 1; : : : ; i� 1; i+ 1; : : : ; kg.Theorem XV.1.6. Let � = f0; 1; : : : ; kg be a �nite alphabet and let Q � �N

be a decidable, subsimilar �01 class which meets I[i] for all i. Then there exists

a partition fU0; : : : ; Ukg and a computable F : �N ! �N such that Q = IT [F ].

Proof. We will use the partition given by Ui = I[i]. Since Q is decidable, wecan de�ne a function G : �N ! Q such that G(x) = x for all x 2 Q. LetQ = [T ] where T is a computable tree without dead ends. The approximatingfunction g for G is de�ned as follows. For any w 2 f0; 1; : : : ; kgn, �nd thelongest initial segment v such that v 2 T and let g(v) be the lexicographicallyleast (or leftmost) extension of v which is in T \ f0; 1; : : : ; kgn; this exists sinceT has no dead ends. Now let F (x) = �(G(x)). We claim that �F = Q.


For any x 2 Q, we have F (x) = �(x) and �(x) 2 Q, since Q is a subshift.Hence Fn(x) = �n(x), so that Fn+1(x)(0) = x(n + 1) and belongs to the setUx(n). Thus the itinerary I(x) = x. This shows that Q � IT [F ].

Next consider any x 2 �N. We will show by induction that Fn(x) =�n(G(x)). For n = 1, this is the de�nition. Then

Fn+1(x) = �(G(Fn(x))) = �(G(�n(G(x))));

by induction. But G(x) 2 Q, so that �n(G(x)) 2 Q by subsimilarity andtherefore G(�n(G(x))) = �n(G(x)) and �nally Fn+1(x) = �n+1(G(x)), asdesired. It follows that for n > 0, It(x)(n) = G(x)(n). But for n = 0,the assumption that Q meets I[x(0)] implies that G(x)(0) = x(0) and henceIt(x)(0) = x(0) = G(x)(0) as well. Therefore It(x) 2 Q as desired.


Chapter XVI

Feasible versions of

combinatorial problems

The main goal in this chapter is to apply the results of Chapter VIII to themathematical problems discussed above in Chapters IX to XV

We observe that any feasible structure is computable, therefore the set ofsolutions to a feasible problem is also the set of solutions to a computableproblem. Thus results such as Theorems ??, XI.1.1 and XII.1.1 have feasibleversions. The reverse direction is more interesting.

We consider computable representation theorems such as Theorems ??, XI.1.3and XII.1.3, and corollary results such as Theorems ??, XI.1.5 and XII.1.4.

These representation theorems showed that the set of solutions to a com-putable problem of various sorts can represent either every c. b. �0

1 class orat least every �0

1 class of separating sets. In this section we obtain better re-sults, in most cases, by improving \recursive" to \polynomial-time". Now anin�nite computable problem may be assumed to have universe !, since any twoin�nite computable sets are recursively isomorphic. (Here the universe of agraph-coloring problem, for example, is the set the vertices.) However, it is nottrue that any two polynomial-time sets are polynomial-time isomorphic. (Forexample, it is clear that there is no p-time map from Tal(!) onto Bin(!).)Thus a polynomial-time structure with some p-time set for its universe may notbe computably isomorphic to any p-time structure with universe !. For exam-ple, a p-time Abelian groups with all elements of �nite order is constructed byCenzer and Remmel in [35] which is not even isomorphic to any p-time groupwith standard universe either (Tal(!) or Bin(!)). For many of the problemsconsidered above, we will show that any computable problem can be reduced�rst to a p-time problem and then to a p-time problem with standard universe.

We illustrate the general strategy with the graph coloring problem. Recallfrom Section XI.XI.2 that, for k � 3, the set of k-colorings of a recursive graphcan be represented by a c. b. �0

1 class and conversely can represent an arbi-trary c. b. �0

1 class. Let G = (V;E) be a computable graph. Then the set of

283

284CHAPTER XVI. FEASIBLE VERSIONS OF COMBINATORIAL PROBLEMS

k-colorings of G can be represented as the �01 class [T ] of in�nite paths through

a computable tree T . Now Theorem VIII.1.4 constructs for us a p-time tree Psuch that [T ] = [P ]. Then the converse representation creates from P a graphwhose k-colorings are in an e�ective degree-preserving �nite-to-one correspon-dence with the in�nite paths through P . Furthermore, inspection of the prooffrom [174] shows that this graph will actually be polynomial time, since P ispolynomial time. This shows that the k-colorings of any computable graph canalways be placed in an e�ective degree-preserving correspondence with the k-colorings of some p-time graph, and, therefore, that the k-colorings of a p-timegraph can strongly represent any c. b. �0

1 class.However, there is no natural correspondence between the recursive graph and

the p-time graph constructed in this manner. We can do better using TheoremVIII.2.1.

Theorem XVI.0.7. For each computable instance P of any of the followingproblems, there is a p-time instance Q of the problem which is computably iso-morphic to P . Furthermore, except in cases (13) and (14), if P has a computablesolution, then we can take Q to have a p-time solution.

(1) Finding a k-coloring for a k-colorable highly computable graph, for anyk � 3.

(2) Finding a marriage in a highly computable society.

(3) Finding a surjective marriage in a symmetrically highly computable soci-ety.

(4) Finding a surjective marriage in a symmetrically highly computable societywhere each person knows at most two other people.

(5) Finding a k-partition of a highly computable graph such that no set in thepartition is adjacent to m other sets, for m > 2.

(6) Finding a one-way (or two-way) Hamiltonian (or Euler) path starting froma �xed vertex for a highly computable graph.

(7) Covering a computablee poset of width k by k chains, for any k � 2.

(8) Covering a computable poset of height k by k antichains, for any k � 2.

(9) Expressing a computable partial ordering on a set as the intersection of dlinear orderings on the set.

(10) Finding a subordering of type ! (or of type !�) of a computable ordering.

(11) Finding an !-successivity (or an !�-successivity) in a computable linearordering.

(12) Finding a non-trivial self-embedding of a computable linear ordering.

(13) Finding a winning strategy for an e�ectively closed binary game.

285

(14) Finding a prime ideal of a recursive Boolean algebra.

Proof. For problems (1) through (12), this follows immediately from TheoremVIII.2.1, since each of these problems can be viewed as a relational structureand the given solution can be viewed as a function mapping to a �xed range.In the dimension of posets problem, we can interpret the solution as a �nite setof relations. For problem (13), Theorem 4.4 of [36] shows that any computablegame may viewed as a p-time game in that the set of in�nite paths which arewinning for Player I will be the set of in�nite paths through a p-time tree. Forproblem (14), Theorem 2.6 of [34] shows that any computable Boolean algebrais computably isomorphic to a p-time Boolean algebra.

We note that a computable game with a computable winning strategy isnot necessarily isomorphic to a p-time game with a p-time winning strategy,since by Theorem 4.5 of [36], there is a computable game with unique winningstrategy, which is computable but not p-time.

Corollary XVI.0.8. For each recursive instance P of any of the problemslisted in Theorem XVI.0.7, there is a p-time instance Q of the problem suchthat the �0

1 class of solutions to P is computably homeomorphic to the �01 class

of solutions to Q.

Proof. In each case, it is easy to see that the computable isomorphism betweenP and Q gives rise to a computable homeomorphism between the �0

1 classes ofsolutions.

For example, we consider the coloring problem. Recall that the �01 class of

k-colorings on a computable graph G = (V;E) (where V may be assumed toequal !) is the set [T ] of in�nite paths through the computable k-ary tree T ,where a �nite sequence (�(0); : : : ; �(n�1)) 2 f1; 2; : : : ; kgn is in T if and only if�(i) 6= �(j) for all (i; j) 2 E. Now suppose that f is a computable isomorphismmappingG to the computable graphG0 = (V 0; E0), so that V 0 = ff(0); f(1); : : :gand (f(i); f(j)) 2 E0 if and only if (i; j) 2 E. Then we can de�ne the tree k+1-ary T 0 by having (�(0); : : : ; �(n� 1)) 2 f0; 1; : : : ; kg in T 0 if and only if

(1) �(v) = 0 () v =2 V 0;

(2) �(u) 6= �(v) whenever (u; v) 2 E0.

Then [T 0] represents in a reasonable way the set of legal k-colorings on G0 andwe have a natural homeomorphism from [T ] to [T 0] de�ned by H(x)(f(i)) = x(i)and H(x)(v) = 0 if v =2 V 0.

We can now represent �01 classes as the set of solutions to p-time problems

of the types listed above. We list only some of the results.

Corollary XVI.0.9. For each of the problems (1) through (9), and (13) listedin Theorem XVI.0.7,


(a) The problem of �nding a computable solution to a p-time problem canstrongly represent the c. b. �0

1 class of separating sets for any pair ofdisjoint in�nite c. e. sets.

(b) There is a p-time instance of the problem with no computable solution.

(c) If a is a Turing degree and 0 <T a <T 00, then there is a p-time instance P

of the problem such that P has a solution of degree a but has no computablesolution.

For problems (1), (3), (6) and (13), we have also:

(d) The problem of �nding a computable solution to a p-time problem canstrongly represent an arbitrary c. b. �0

1 class.

(e) There exists a p-time instance P of the problem such that

(i) P has a unique non-computable solution y which is also the uniquelimit solution and has degree 00 and such that any other solution iscomputable;

(ii) if R is any computable sub-problem of P and z is any computablesolution of R, then either (i) there are only �nitely many solutions ofP which extend z, or (ii) all but �nitely many solutions of P extendz.

(iii) if x is any computable solution of P , then there is some �nite sub-problem F of P such that any solution of P which agrees with x onF must equal x.

We have seen that, by changing the names of the vertices, we can transform acomputable graph into a p-time graph. However, we would prefer for a countablyin�nite graph to have the set V of vertices equal to some standard universesuch that the tally or binary representation of the set of natural numbers. Thiswould, for instance, allow us to de�ne the homeomorphism of Corollary XVI.0.8without worrying about the set of non-vertices. The p-time graph constructedby Theorem VIII.1.4 will have a rather sparse set of vertices and this appearsto be an essential part of the theorem. We will next indicate how to �ll out thep-time structure given by Theorem XVI.0.7 to a structure with universe Bin(!)such that there is a degree-preserving correspondence, which is one-to-one (upto a �nite permutation), between the �0

1 classes of solutions of the associatedproblems. For example, in the coloring problem, we add vertices whose colorswill be determined, up to a permutation, by the coloring of the vertices of Q.

Theorem XVI.0.10. For each computable instance P of the combinatorialproblems listed below, there is a p-time instance Q with universe Bin(!) and adegree-preserving correspondence between the solutions of P and the solutionsof Q.


287



(4) Finding a surjective marriage in a symmetrically highly computable societywhere each person knows at most two other people.

(5) Finding a k-partition of a highly computable graph such that no set in thepartition is adjacent to m other sets.

(6) Finding a (one-way or two-way) Hamiltonian or Euler path for a highlycomputable graph.

(7) Covering a computable poset of width k by k chains, for any k � 2.

(8) Covering a computable poset of height k by k antichains, for any k � 2.

(9) Expressing a computable partial ordering on a set as the intersection of dlinear orderings on the set.


Proof. In each case, we may assume by Corollary XVI.0.8 that we start with ap-time instance of the problem which is a relational structure B with some uni-verse B � Bin(!). Now it follows from Lemma 2.3 of [35] that B is computablyisomorphic to a p-time structure A with universe A � Tal(!). Then Lemma 2.6of [35] says that the disjoint union A�Bin(!) is p-time isomorphic to Bin(!),where X�Y = fh0; xi : x 2 Xg[fh1; yi : y 2 Y g. Then we will create a p-timestructure C with universe A � Bin(!) which has a copy of A together with acopy of Bin(!), where the relations will be de�ned on Bin(!) and between Aand Bin(!) so as to determine the degree-preserving correspondence betweenthe solutions of A and those of the extension C. Since the universe C of C isp-time isomorphic to Bin(!), it follows from Lemma 2.2 of [35] that C is com-putably isomorphic to a p-time structure with universe Bin(!). Then we willlet Q be the problem associated with this structure. It follows that there willbe a degree preserving correspondence between the set of solutions of Q and theset of solutions of the original problem P . In each case, we will assume that ouroriginal structure is p-time and has for its universe a p-time subset A of Tal(!)and that there is a p-time list of Bin(!)nA. These assumptions are justi�ed bythe above discussion. In each case, the correspondence will be one-to-one unlessotherwise indicated.


This is Theorem 2.1 of [37]. Here the correspondence is one-to-one, up to a�nite permutation of the colors on the new vertices.



Let S = (B;G;K) be a p-time society. Then we will directly extend S to ahighly recursive p-time society S0 = (B0; G0;K 0) where B0 = G0 = Bin(!). LetBin(!)nB = fb0; b1; : : :g and let Bin(!)nG = fg0; g1; : : :g be p-time lists of thenew boys and girls in the society S0. Then K 0 is de�ned by putting (bi; gi) 2 K 0

for all i. It is clear that any marriage f on S has a unique extension f 0 to S0

de�ned by letting f 0(bi) = gi for all i. It follows that f and f 0 have the samedegree.


The extension is the same as in (2). It is clear that f 0 will be onto if andonly if f is onto.

(4) Finding a surjective marriage in a symmetrically highly computable soci-ety where each person knows at most two other people.

The extension is again the same as in (2). It is clear that if each person inS knows at most two other people, then each person in the extension S0 alsoknows at most two other people.

(5) Finding a k-partition of a highly computable graph such that no set inthe partition is adjacent to m other sets, with m > 2.

Let the p-time graph G = (V;E) be given. We de�ne a p-time graph G1 =(V1; E1) to be a regular m � 1-ary tree of complete k-graphs. That is, de�nethe regular (m� 1)-ary tree Tm�1 to consist of a root node ; together with theset fbin(0); bin(1); : : : ; bin(m� 1)g � fbin(1); bin(2); : : : ; bin(m� 2)g�, where ;has m � 1 successors (bin(i); ;) for i < m and (bin(i); �) has m � 2 successors(bin(i); �_bin(j)) for j < m� 1. Then we let

V1 = fbin(1); bin(2); : : : ; bin(k)g � Tm�1;and we let ((bin(i); �); (bin(j); �)) 2 E1, where � = (�(0); : : : ; �(s � 1)) and� = (�(0); : : : ; �(t� 1)), provided that � = � or either � is a successor of � or �is a successor of � . It is clear that if the graph is computably partitioned intothe complete k-graphs corresponding to the nodes of Tm�1, then each set in thepartition is adjacent to at most m� 1 other sets. We see also that each node ofTm�1 has m� 1 neighbors, so that any two distinct nodes have at least 2m� 4other neighbors. Now let fAi : i < !g be a k-partition of G1. Suppose thatsome Ai contains vertices u and v corresponding to di�erent nodes of the Tm�1.Then u and v taken together have at least 2(k� 1)+ (2m� 4)k = (2m� 2)k� 2other adjacent vertices in G1. Since k � 2 of these could belong to Ai, we seethat Ai has at least (2m� 3)k adjacent vertices. Since each set in the partition

289

has at most k vertices, it follows that Ai is adjacent to at least 2m � 3 sets inthe partition. Thus since m > 2, we have 2m� 3 > m� 1 so that the set Ai isadjacent to too many sets.

Now let G0 = G � G1. It is clear that for any k-partition of G, there is apartition of G0 of the same degree which is given by adding the recursive parti-tion of G1 into the nodes of the tree as de�ned above. We claim that these arethe only possible partitions. That is, suppose fBj : j < !g is a k-partition ofG0. It su�ces to show that for any u 2 V1 and any j, if u 2 Bj , then the entirenode to which u belongs must be included in Bj . Suppose that this is false.It follows from the argument above that Bj may not contain an element of adi�erent node of T . Thus the set Bj has all k(m� 1) vertices from the adjacentnodes as neighbors as well as at least one vertex from the node of u. But thisclearly implies that at least m sets of the partition must be adjacent to Bj .

(6) Finding a one-way (or two-way) Hamiltonian Euler path starting from a�xed vertex for a highly recursive graph.

Let the p-time graph G = (V;E) be given with V = fv0 < v1 < � � � g a sub-set of Tal(!). Let each edge (tal(m); tal(n)) of V with tal(m) < tal(n) becoded as 0n+11m+1 in Bin(!). Let b0; b1; : : : enumerate the codes of edges inincreasing order and let bi = 0ni+11mi+1 for each i. Now let V 0 = V � Bin(!)and let E0 be de�ned by joining h1; bii to h0; tal(mi)i, joining h0; tal(ni)i toh1; bi+1 � 1i, joining h1; bi to h1; b + 1i whenever b + 1 6= bi for any i, andjoining h1; b0 � 1i to h0; tal(m0)i. Note that other than the initial sequenceof edges connecting h1; 0i to h1; b0 � 1i and then to h0; tal(m0)i, this has thee�ect of replacing an edge (m;n) 2 E where m < n and bi = 1m+10n+1 by asequence of edges (h0;mi; h1; bii); (h1; bii; h1; bi+1i); : : : ; (h1; bi+1�2i; h1; bi+1�1i); (h1; bi+1 � 1i; h0; ni). See Figure XVI.

Thus to test whether h1; bi and h1; ci are joined by an edge, where b < c,we simply check that c = b + 1 and that, if c = 0n+11m+1 with m < n, then(tal(m); tal(n)) =2 E. To determine whether h0; vi and h1; ci are joined by anedge, we �rst check that v = tal(m) 2 V and that either (i) c = 0s+11r+1 or(ii) c = 0s+11r+1 � 1 for some edge (tal(r); tal(s)) in G with r < s. Finally, incase (i), we check that r = m and, in case (ii), we either check that m = m0

and that c+ 1 = b0 or else we compute the largest code 0q+11p+1 of an edge of

G less than c+ 1 and check that m = q. If everything checks, then there is anedge and otherwise there is not. Thus G0 is a p-time graph.

Now suppose that f is a one-way Euler path on G starting from f(0) = v0 =tal(m0). Then we can de�ne a corresponding Euler path on G0 starting fromh1; bin(0)i by beginning with the sequenceh1; bin(0)i; h1; bin(1)i; : : : ; h1; b0 � 1i; h0; v0i and then replacing in turn eachedge (f(i); f(i + 1)) which joins tal(m) to tal(n) with m < n, either by thesequence h0; f(i)i; h1; bin(b)i; h1; bin(b + 1)i; : : : ; h1; bin(c � 1)i; h0; f(i + 1)i, iff(i) = tal(m) < f(i+ 1), or by the sequence h0; f(i)i; h1; bin(c� 1)i; h1; bin(c�2)i; : : : ; h1; bin(b+ 1)i; h1; bin(b)i; h0; f(i+ 1)i if f(i) = tal(n) > f(i+ 1), whereb = 0n+11m+1 and c is the least code greater than b for an edge of G. It is clear


m < n

b = 1 0m+1 n+1

i

<1,b > <1,b +1> <1,b -2> <1,b -1>i i i+1 i+1

m n

. . .

Figure XVI.1: Replacement for edge (m;n)

291

that this one-way Euler path is computable in f .Conversely, let g be a one-way Euler path in G0 starting from h1; bin(0)i. It

is clear that the path must proceed throughh1; bin(1)i; h1; bin(2)i; : : : ; h1; b0 � 1i and then to h0; v0i. Now let f : ! ! V bede�ned by letting h0; f(i)i be the i-th vertex of the form h0; xi in the path g.Then f is a one-way Euler path for G and it follows from the construction thatg is the corresponding path as de�ned above, since there is only one way in G0

to go from h0; f(i)i to h0; f(i+ 1)i.For two-way Euler paths, modify the construction by eliminating the �nite

initial sequence h1; bin(0)i; h1; bin(1); : : : ; h1; bin(m0)�1i of vertices of G0 alongwith the edges through those vertices. Then the remaining vertex set is stillp-time isomorphic to Bin(!) and the argument goes through as above.

The Hamiltonian paths require a di�erent construction. We assume withoutloss of generality that the vertices of G include 0 and that all are multiples of4 (in binary) and let 4m0; 4m1; : : : enumerate the vertices of G in increasingorder. De�ne the graph G0 to have vertex set V 0 = Bin(!) with edges de�nedas follows. For each i, there will be two sequences of edges joining the set ofbinary numbers from 4mi + 1 up to 4mi+1, as follows:

(i) 4mi+1; 4mi+1 � 4; 4mi+1 � 8; : : : ; 4mi + 4; 4mi + 2; 4mi + 6; : : : ; 4mi+1 � 2,

(ii) 4mi+1; 4mi+1 � 3; 4mi+1 � 7; : : : ; 4mi + 1; 4mi + 3; : : : ; 4mi+1 � 1.

These are the vertices associated with 4mi+1. In addition, for each edge joining4mi and 4mj in G with mi;mj 6= 0, there are edges joining 4mi�1 with 4mj�2and joining 4mi� 2 with 4mj � 1. For an edge in G joining 4mi with 0, there isan edge joining 4mi � 1 with 0. The procedure for determining whether thereis an edge joining a and b is the following. First look for the largest m and nsuch that 4m 2 V and 4m < a and 4n 2 V and 4n < b. In the special casethat a = 0, a and b are joined if and only if b+1 is joined to 0 as vertices in G.Otherwise there are several cases. First suppose that m = n; then a are b arejoined if and only if, either they di�er by exactly 4 or fa; bg = f4m+1; 4m+3gor fa; bg = f4m + 2; 4m + 4g. Next suppose that m 6= n. Then a and b arejoined if and only if either a + 1 and b + 2 are joined as vertices in G or a + 2and b+ 1 are joined as vertices in G. Thus G0 is a p-time graph.

Now let f be a one-way Hamiltonian path on G starting from v0 = 0 andsuppose that f(i) = 4mri . Then there is a corresponding Hamiltonian path g inG0 obtained by replacing the edge from v0 to 4mr1 with the sequence of edgesjoining v0 to 4mr1 � 1 and then on to 4mr1 � 3 and 4mr1 as described above,and for i > 0, replacing each edge (f(i); f(i + 1)) with the sequence of edges�rst joining 4mri to 4mri � 4 and then on to 4mri � 2 as described above, thenjoining 4mri � 2 to 4mri+1 � 1, and closing with the sequence joining 4mi+1� 1to 4mi+1� 3 and then 4mi+1. Thus for each i > 0, the even vertices associatedwith f(i) are joined to the odd vertices associated with f(i+ 1).

Conversely, let g be a one-way Hamiltonian path in G0 starting from v0 = 0and de�ne f(i) = 4mri so that 0; 4mr1 ; : : : lists the members of G in order ofappearance in the path g. It follows from the construction that f is a one-way


Hamiltonian path for G0 starting from v0 and that g is the corresponding pathas de�ned above.

For the two-way Hamiltonian paths, the construction is modi�ed by addingan edge joining v0 with 4mi � 2 for each edge joining v0 with 4mi in G. Thenfor any two way Hamiltonian path f in G, there will be two corresponding two-way paths in G0, one in which the even vertices associated with f(i) are joinedto the odd vertices associated with f(i + 1) and one in which the odd verticesassociated with f(i) are joined with the even vertices of f(i + 1). Thus thecorrespondence here is two-to-one.

(7) The problem of covering a computable poset of width k by k chains, forany k � 2.

Let P = (P;�P ) be a p-time poset where P � Tal(!). Then de�ne a p-time poset R = (R;�R) where R = P � (fbin(1); : : : ; bin(k)g � Bin(!)) andh0; pi �R h0; qi i� p �P q, h0; pi �R h1; ni for all p and n, andh1;mi �R h1; ni i� m = hbin(i); bin(r)i and n = hbin(i); bin(s)i where r � s.Then it is clear that for any covering f of P by k chains induces a covering f 0

of covering of R by k chains where

(i) f 0(h0; pi) = f(p) for all p 2 P and

(ii) f 0(h1; hbin(i); nii) = f 0(h1; hbin(i);mii) for all i,m and n.

Thus the covering is determined by the value of f 0 on the �nitely many newpoints h1; hbin(1); 0ii; : : : ; h1; hbin(1)0ii. This shows that f 0 has the same degreeas f and that f 0 is unique up to a permutation of the names of chains. Then Ris p-time isomorphic to p-time linear ordering S whose universe is Bin(!).

(8) The problem of covering a computable poset of height k by k antichains, forany k � 2.

This is the dual of problem (7). The partial order is now de�ned by mak-ing h1; hbin(i); nii � h1; hbin(j);mii () (i < j & m = n).

(9) The dimension of posets problem.

Let P = (P;�P ) be a poset and let Bin(!) n P = fvi : i < !g. The par-tial order �0 is de�ned on Bin(!) by making p �0 vi for all p 2 P and all i andmaking vi �0 vj if and only if vi � vj (where < is the usual ordering on Bin(!).


Given a p-time linear ordering L1 = (A;<1) on a p-time set A = fa0 < a1 <� � � g, we may assume a0 = 0 and that each ai = bin(4mi). Now de�ne thep-time ordering L2 = (Bin(!); <2) by replacing each point a = bin(4mi) with

293

a block B(a):

bin(4mi + 1) < bin(4mi + 5) < � � � < bin(4mi+1 � 3)

< bin(4mi+1 � 1) < bin(4mi+1 � 5) < � � � < bin(4mi + 3)

< bin(4mi) < bin(4mi + 4) < bin(4mi + 8) < � � � < bin(4mi+1 � 4)

< bin(4mi+1 � 2) < bin(4mi+1 � 6) < � � � < bin(4mi + 2)

That is, we use the elements between 4mi and 4mi+1 which are equivalent to 1mod 4 to form a chain between 4mi+1 and 4mi+1�1, then we use the elementsbetween 4mi and 4mi+1 which are equivalent to 3 mod 4 in reverse order toform a chain between 4mi+1 � 1 and 4mi, etc.

Now suppose that f is an !-successivity in L1. Then we can recursivelyobtain an !-successivity g in L2 by replacing each point f(i) with the blockB(f(i)). Conversely, given an !-successivity g in L2, the !-successivity f ofL1 may be de�ned by making f(i) the i-th binary number in the successivityg which is divisible by 4 and it then follows that g is the successivity obtainedfrom f as above. The argument for !�-successivities is similar.

We remark that, for the three-coloring problem, it is possible to improvethis result by having the 3-colorings of the the original computable graph berestrictions of the 3-colorings of the p-time graph to the original recursive vertexset.

Theorem XVI.0.10 can now be applied to obtain improved versions of Corol-lary XVI.0.9. We list only a few here.

Corollary XVI.0.11. (a) There exists a p-time graph G with universe Bin(!)which has a unique non-computable Hamiltonian path �, where � has de-gree 00 and such that any other Hamiltonian path is the unique extensionto G of a Hamiltonian path on some �nite subgraph F of G.

(b) There is a p-time partial ordering with universe Bin(!) of width k whichhas no computable covering by k chains.

(c) For any x �T 00, there is a p-time linear ordering A with universe Bin(!)such that there is !-successivity (respectively !�-successivity) of A of de-gree x and every !-successivity (respectively !�-successivity) of A is eithercomputable or has the same Turing degree as x.

Theorem XVI.0.10 and Corollary XVI.0.11 demonstrate that the problem of�nding solutions to feasible problems is just as di�cult as the problem of �ndingsolutions to recursive problems. Therefore more conditions will have to be puton a problem than just feasibility if our goal is to guarantee the existence of afeasible solution, or even the existence of a recursive solution. There are manypossible approaches to this goal, some of which were explored in [37] for thegraph-coloring problem.

Finally, we consider the problem of �nding a prime ideal of a recursiveBoolean algebra, or more generally, of a recursive ring.


Theorem XVI.0.12. For any recursive Boolean algebra B, there is a p-timecommutative ring R with unity, having universe Bin(!), and a one-to-onedegree-preserving map between the class of prime ideals of R and the class ofprime ideals of B.Proof. By Theorem XVI.0.7, we may assume that B is a p-time Boolean alge-bra, and thus a Boolean ring. Now de�ne the ring R = B � Q. Q is chosenhere because it has no (proper) prime ideals. The ring Q of rationals may berepresented as a p-time ring with universe Bin(!) and it follows from Lemmas2.2 and 2.6 of [35] that R is p-time isomorphic to a ring with universe Bin(!).For any prime ideal I of B, it is easy to check that I �Q is a prime ideal of Rand that these are the only prime ideals of R.Corollary XVI.0.13. (i) For any degree a <T 00, there exists a recursive

commutative ring R with a prime ideal I of degree a such that I is theunique non-recursive prime ideal of B and such that any other prime idealof B is �nitely generated.

(ii) There is a computable commutative ring with unity, R, which has a uniquenon-computable prime ideal I, such that any other prime ideal of R is�nitely generated, and such that for any c. e. ideal J of R, either thereare only �nitely many prime ideals of R extending J or else all but �nitelymany of the prime ideals of R extend J .

Part C

Advanced Topics and

Current Research Areas

295

Chapter XVII

The Lattice of �01classes

The inclusion lattice E� of �01 classes has an interesting algebraic structure, in

some ways analogous to the dual of the lattice E of c.e. sets. Recent work hasfocused on comparing and contrasting the two lattices. Important issues includethe de�nability and complexity of various properties, automorphisms of the lat-tice and orbits under automorphisms, and the analysis of certain substructuresof the lattice.

Here is an example. Given two �01classes P � Q, the interval [P;Q] = fR :

P � R � Qg of P and in particular [;; Q] is an initial segment of E�. A �01class

P is said to be thin if [;; Q] is a Boolean algebra. P is perfect if every elementof P is a limit point. Cholak et al. [50] have shown that the family of allperfect thin classes is in certain ways analogous to the hyper-hypersimple c.e.sets. That is, any two perfect thin classes are automorphic in E�, the family ofperfect thin classes is de�nable in E� and the degrees of perfect thin classes areexactly the c.e. array noncomputable degrees. (Here the degree of P = [T ] isthe degree of the set of nodes of T which have an extension in P .)

An in�nite �01 class P is minimal if every �0

1 subclass of P is either �niteor co�nite in P . This is of course dual to the notion of a maximal c.e. set.For any lattice L, let L� be the quotient lattice of L modulo �nite di�erence.Then P is minimal if and only if [0; P ]� is the trivial Boolean algebra. Cenzer,Downey, Jockusch and Shore [25] �rst constructed a minimal thin class. Cenzerand Nies [32] characterized the order types of the �nite intervals of E�� as �nitedistributive lattices with the dual reduction property. Furthermore, for eachsuch lattice L, the theory of L is decidable. In particular, this means that thereare intervals (in fact, initial segments) of order type n for any �nite ordinal n.This contrasts with the classic result that �nite intervals of E� are all Booleanalgebras. However, it is shown in [32] that for any decidable �0

1 class P , if[0; P ]� is �nite, then it must be a Boolean algebra. Finally, if P is decidable and[0; P ] is not a Boolean algebra, then the theory of [0; P ] interprets the theory ofarithmetic and is therefore undecidable.

End intervals [P; TN ] were studied in [33] where it was shown that there

297

298 CHAPTER XVII. THE LATTICE OF �01 CLASSES

are exactly two possible isomorphism types of end intervals (where P is eitherclopen or not).

In his thesis (and continued in joint work with Cenzer), Riazati [177, 45]studied minimal extensions of �0

1 classes (an analogue of maximal subsets). Heproved an analogue of the Owings Splitting Theorem and used it to prove thatdecidable minimal extensions are not possible.

Lawton [126] introduced the notion of minor superclasses of �01classes, as

an analogue of major subsets of c.e. sets and gave a characterization of the �01

classes which have strong minor superclasses.It is easy to see that the family of �nite classes is invariant under auto-

morphism. Cenzer and Nies [33] showed that the property of being �nite isde�nable in the lattice and in general, the family of countable �0

1 classes ofrank � is de�nable if and only if � < !.

XVII.1 Countable thin classes

Theorem XVII.1.1. (C-D-J-S) For any computable ordinal �, there is a thin�01 class P� with Cantor-Bendixson rank �. Furthermore, we may take P� as

the set of paths through a computable tree with no dead ends.

Proof. We �rst sketch the proof for � = 1. We construct a sequence �e 2f0; 1g<! such that �_e 1 � �e+1 for all e, a set A = [e�e and a �0

1 class P = [T ]such that

(1) D(P ) = fAg, and(2) for any e, if A 2 [Te] then P \ I(�e) � [Te].These conditions imply that A is non-computable, since if A were com-

putable, then fAg = [Te] for some e, so that by (2), P \ I(�e) = fAg, contra-dicting (1).

These conditions imply that A is non-computable, since if A were com-putable, then fAg = [Te] for some e, so that by (2), P \ I(�e) = fAg, contra-dicting (1).

These conditions also imply that P is minimal (and therefore thin by Theo-rem 5.2). To see this, suppose that [Te] � P . If A =2 [Te], then [Te] has no limitpoint and is therefore �nite. If A 2 [Te], then P n [Te] � P n I(�e) by (2) has nolimit point and is �nite.

The construction is in stages, so that at stage s we have a tree T s and strings�se . At stage s+1, we simply look for e � s such that some � � �se is in T s n Teand let �s+1e = � for the least such e and � . For i < e, let �s+1i = �si and for�s+1e+i = 0�_1i. We leave the details to the reader.

The general construction for a computable ordinal � is accomplished usinga computable system of notations for � and a uniformly computable familyof trees of rank up to � as in the proof of Theorem V.4.8. The details areomitted.

Next we consider the possible degrees of members of thin �01 classes.

XVII.1. COUNTABLE THIN CLASSES 299

Theorem XVII.1.2. (C-D-J-S) There is a �01 set A of degree 00 and a mini-

mal, thin �01 class P such that D(P ) = fAg.

Proof. Let B = 00 be the union of uniformly computable sets Bs. Let T0; T1; : : :be an e�ective enumeration of the primitive recursive trees on 2<!. We willde�ne a �0

1 retraceable set A = fa0 < a1 < � � � g and a corresponding �01 class

P = I(A) of initial subsets of A, by Theorem III.6.2, such that(1) For any e, e 2 B () e 2 Bae .(2) For any �0

1 class Pe = [Te], if A 2 Pe, then An 2 Pe for all n � e.By property (1), 00 is recursive in A, so that, since A is �0

1, A has degree 00.It then follows from (2) as in the proof of Theorem XVII.1.1 that P is minimaland thin.

The sequence a0 < a1 < � � � is de�ned by �01 recursion in the style of

Theorem III.6.5 by making an the least a which satis�es the following:(i) For all m < n, am < a.(ii) n 2 B ! n 2 Ba.(iii) For all m < n, either < a0; : : : ; an�1; a >=2 Tm or(8x)(< a0; : : : ; an�1; x >2 Tm).(iv) For all x < a, either

(a) x � xn�1 or(b) n 2 Ba nBx or(c) for some m < n, < x0; x1; : : : ; xn�1; x >2 Tm& < x0; x1; : : : ; xi; a >=2 Tm.

The details are left to the reader.

The following result is Theorem 2.13 of [25] (p. 102).

Theorem XVII.1.3. (C-D-J-S) Let T be a recursive tree and P a �01 class

such that P = [T ]. Then for any set A 2 P ,(a) If P � P(A), then A �T Ext(T ).(b) If A is a �0

1 set and P is thin, then A �T Ext(T )(c) If T has no dead ends and A is either r. e. or co-r. e., then A is recursive.

Proof. (a) To test whether n 2 A, simply see if there is a � 2 Ext(T ) of lengthn+ 1 such that �(n) = 1.

(b) Note that P(A) is a �01 class, so that Q = P \ P(A) is a �0

1 subclass ofP and is nonempty since A 2 Q. Since P is thin, we must have Q = P \ U forsome clopen U = I(�0) [ � � � [ I(�k). If we now de�ne

TQ = f� 2 T : �iscompatiblewith�i; forsomei � kg,then it is clear that Q = [TQ] and thatExt(Tq) = f� 2 Ext(T ) : �iscompatiblewith�i; forsomei � kg,so that Ext(TQ) is recursive in Ext(T ). Now Q � P(A), so that by (a) we

have A �T Ext(TQ) �T Ext(T ).(c) It is immediate from (b) that if A is �0

1 set, then A is recursive. If A is anr.e. set, then ! nA is �0

1 and belongs to the thin �01 class f! nX : X 2 Pg.


Theorem XVII.1.4. (C-D-J-S) Let T be a recursive tree such that P = [T ] isa thin �0

1 class and let A 2 P . Then(a) A0 �T A� 000 (so that it is not possible that A �T 000.)

(b) If T has no dead ends, then A0 �T A � 00 (so that it is not possible thatA �T 00.)

Proof. (a) Let P = [T ] be thin and suppose A 2 P . For each e, let Qe =fC : �Ce (e) "g. Then Qe is a �0

1 class, so there is a clopen set U(e) such thatP \ Qe = P \ U(e). Thus if �Ae (e) ", then there is some � = Adn such that �forces �Ae (e) ", that is, such that, for any B 2 P , if � � B then �Be (e) ". Nowde�ne the �0

2 relation R(e; �) which says that � forces �Be (e) ", byR(e; �) () (8� � �)[(� 2 T&��e (e) #)! � =2 Ext(T )]:Then we can compute from A together with 000, whether e 2 A0 by searching

for the least n such that, for � = xdn, either ��e (e) #, in which case e 2 A0, orR(e; �), in which case e =2 A0.

(b) 0bserve that if Ext(T ) is recursive, then the relation R de�ned abovewill be recursive in 00.

It follows from (b) and Theorem 4.2 (b) above that if A has rank one in athin �0

1 class P = [T ], where T has no dead ends, then A has low degree a, thatis, a0 = 00.

Part (a) of this theorem is best possible in the sense that, as shown inTheorem 2.18 of [25], there is a minimal thin �0

1 class P and a set A such thatD(P ) = fAg and A� 00 �T 000.

We conclude this section by stating without proof several further resultsfrom [25].

Theorem XVII.1.5. (C-D-J-S) Between any two distinct r. e. degrees b < c,there is a degree a, a set A of degree a and a minimal, thin �0

1 class P withD(P ) = fAg.

There is a family of c. e. degrees which contain members of thin �01 classes.

In particular, it follows from Theorem 4.9 of Downey-Jockusch-Stob [65] thatall array non-computable ( a.n.c.) degrees and hence all non� low2 degreescontain members of thin �0

1 classes.

Theorem 5.8 tells us that no set of degree 000 can even belong to a thin �01

class. Two further results give lower degrees which also contain no members ofthin classes.

Theorem XVII.1.6. (C-D-J-S) (a) There is an r.e. degree a such that no setB of degree a belongs to any thin �0

1 class.

(b) There is a minimal degree a < 00 such that no set A of degree a is amember of any thin �0

1 class.

In contrast, we have the following improvement of Theorem V.4.14.

XVII.2. INITIAL SEGMENTS OF THE LATTICE 301

Theorem XVII.1.7. (C-D-J-S) There is a non-recursive set A �T 000 suchthat every non-recursive set B �T A is a rank 1 member of a minimal, thin �0

1

class.

Finally, there is another connection with maximal c. e. sets.

Theorem XVII.1.8. (C-D-J-S) There is a maximal set A which is not a mem-ber of any thin �0

1 class.

XVII.2 Initial Segments of the Lattice

XVII.3 Global Properties of the Lattice

XVII.4 Almost complemented classes

XVII.5 Perfect thin classes

The proof of the Low Basis Theorem IV.1.4 shows that every nonempty c.b. �01

class contains a member of c. e. degree a such that a is low, that is, a � 00 =a0 = 000. The method of Theorem III.8.1 can be used to construct a nonempty�01 class with no computable members and no members of high degree, where

the c. e. degree a is high if a0 = 000.

Theorem XVII.5.1. There exists a perfect, thin, c.b. �01 class P with no

computable members such that if a is the degree of a member of P , then a0 �a� 00.

Proof. Let P be the �01 class constructed in Theorem III.8.1 and let f be the

function de�ned therein. Then f is the limit of a uniformly computable sequenceof functions and is therefore computable in 00 by the Limit Lemma. Now letUe = f� : ��e (e) "g. It follows from the s-m-n theorem that there is a computablefunction � such that Ue = T�(e) for each e. Then for any element A of P andany e, we have

e 2 A0 () (9� � A)� 2 Ue () xdf(2e+ 2) 2 T�(e).This shows that A0 is computable in A� 00.xxxThere will be results here from the paper [50]


Chapter XVIII

Degrees of Di�culty

The Medvedev lattice was introduced in [149] to classify problems according totheir degree of di�culty. A mass problem is a subset of NN and is thought ofas representing the set of solutions to some problem. For example, the problemof separability of sets A and B is S(A;B) = ff : i 2 A ! f(i) = 0 & i 2 B !f(i) = 1g. The coloring problem for a given countably in�nite graph G may begiven as a set of functions each mapping ! into f1; 2; 3; 4g. A mass problem issaid to be solvable if it contains a computable function. See the survey paperby Sorbi [200] for more background.

In this chapter we study the Medvedev and Muchnik degrees of nonempty �01

classes. Each of these partial orderings is in fact a distributive lattice with topelement, which can be viewed as the degree of the set of completions of Peanoarithemtic, and bottom element, which is the degree of any set containing acomputable member.

XVIII.1 Reducibility

P is Medvedev reducible to Q (P �M Q) if there is a partial computable func-tional � which is de�ned for all X 2 Q and maps Q into P . Thus any solutionof Q may be used to compute a solution of P , so we say that P has a lower(Medvedev) degree of di�culty than Q. There is also a nonuniform notion,Muchnik reducibility, given by P �w Q if every member X of Q computes amember of P , that is, Y �T X for some Y 2 P . As usual, P �M Q means thatboth P �M Q and Q �M P , P <M Q means P �M Q but not Q �M P , andthe Medvedev degree dgM (P ) of P is the class of all sets Q such that P �M Q.Similar notations applies to Muchnik reducibility. Observe that P �M Q im-plies P �w Q, so that the Medvedev degree of P is a subset of the Muchnikdegree of P . Let PM denote the lattice of Medvedev degrees of �0

1 classes andlet Pw denote the lattice of Muchnik (or weak) degrees.

We will focus primarily on the Medvedev degrees.

303

304 CHAPTER XVIII. DEGREES OF DIFFICULTY

First we show that only total functionals are needed for Medvedev reducibil-ity of �0

1 classes.

Lemma XVIII.1.1. For any �01 subclasses P and Q of !!, if P �M Q, then

there exists a total computable functional F : NN ! NN such that F [Q] � P .Proof. Given that P �M Q, there is a partial computable functional � whichmaps Q into P . This means that there is a partial computable function �mapping �nite sequences to �nite sequences such that �(X) = [n�(Xdn) andwith the property that � � � implies �(�) � �(�). Now Q may be expressed asthe set of in�nite paths through some computable tree T . Then we can extendthe mapping � from Q to a total mapping F representing function f de�nedrecursively as follows. Let f(;) = ;. Then for any �nite sequence � and any n,de�ne f(�_n) in two cases. If �_n 2 T , let f(�_n) = �(�_n), which must bede�ned. If �_n =2 T , let f(�_n) = f(�)_0.

Simpson [194] has observed that Medvedev reducibility can be viewed as theanalog for degrees of di�culty of the truth-table degrees for the Turing degrees,whereas Muchnik reducibility is the analog of Turing reducibility.

Proposition XVIII.1.2. If P �M Q, then for any x 2 Q, there is a y 2 Psuch that y �tt x.Proof. Suppose P �M Q for �0

1 classes P and Q. Then by Lemma XVIII.1.1,there is a total computable � which maps Q into P . Then for any element x ofQ, it follows from Lemma II.5.10 that �(x) �tt x.

The meet and join operations of the Medvedev lattice turn out to be thestandard sum P � Q and product P Q de�ned earlier. We summarize heresome basic facts about these meet and join operations.

Proposition XVIII.1.3. For any �01 classes P , Q and R,

(i) P �Q �M Q� P and P Q �M Q P (so that also P �Q �w Q� Pand P Q �w Q P

(ii) The Medvedev (Muchnik) degree of P � Q is the meet, or greatest lowerbound, of the Medvedev (resp. Muchnik) degrees of P and Q;

(iii) The Medvedev degree of P Q is the join, or least upper bound, of theMedvedev degrees of P and Q

(iv) P(Q�R) �M (PQ)�(PR) and P(Q�R) �M (PQ)�(PR).(v) If P �M Q, then, for any R, (P R)�Q �M P (Q�R); if P �w Q,

then, for any R, (P R)�Q �w P (Q�R)Proof. (i) is obvious since these sets are in fact computably homeomorphic.

(ii) P � Q �M P via the map �(x) = 0_x and similarly P � Q �M Q.Suppose now that R � P via � and R � Q via . Then R � P � Q via themap taking 0_x to �(x) and 1_x to (x).

XVIII.1. REDUCIBILITY 305

(iii) P �M P Q via the map �(x) = (x(0); x(2); : : : ) and similarly Q �MPQ. Suppose now that P �M R via � and Q �M R via . Then PQ �M Rvia the map taking x to h�(x);(x)i.

(iv) To see that P � (QR) �M (P �Q) (P �R), we de�ne computablefunctionals in each direction. First de�ne � : P � (QR)! (P �Q) (P �R)by

�(0_X) = h0_X; 0_Xiand

�(1_hY;Zi) = h1_Y; 1_Zi:Then de�ne : (P � Q) (P � R) ! P � (Q R) as follows. Given Z =hV;W i 2 (P �Q) (P �R), there are three cases.

If V = 0_X; let (Z) = V ;

if V = 1_Y and W = 0_X; let (Z) =W ;

if V = 1_Y and W = 1_Z; let (Z) = 1_hY;Zi:The other equivalence of (iv) is left as an exercise.(v) Since P �M Q, we have P �Q �M P and P Q �M Q. Then

(P R)�Q �M (P �Q) (R�Q) �M P (Q�R):The same argument works for the Muchnik degrees.

Corollary XVIII.1.4. Both PM and Pw are distributive lattices.

Next we observe that PM has both a least and a greatest element. The leastelement 0 consists of all classes P which contain a computable element. To seethis, just let X0 be a computable element of P and de�ne F (X) = X0 for anyX. Then F maps any class Q into P , so that P �M Q. In particular, the classesf0; 1g! and f0!g are both in 0. Sorbi points out in [200] that this means thatthe solvable Medvedev degree is de�nable in the lattice (as the least element).

Proposition XVIII.1.5. PM has a greatest element.

Proof. Since there is an enumeration fPege2! of the �01 classes, it seems nat-

ural to take the product of these classes as the univeral set and hence the topMedvedev degree. There is one hitch in that the empty set is not included inPM . To enumerate the nonempty �0

1 classes, �rst recall the usual enumerationfTege2N of the primitive recursive trees in f0; 1g� and let

� 2 T+e () [� 2 Te _ (8m � j�j)(�dm =2 Te ! (8� 2 f0; 1gm)� =2 Te)]:Now if Pe = ; and m is the least such that Te \ f0; 1gm+1 = ;, then it is clearthat P+

e = [T+e ] =SfI(�) : � 2 f0; 1gm \ Teg and is still a �0

1 class. If Pe 6= ;,P+e = Pe. We claim that

Qe P

+e is the greatest element of PM and hence also of

Pw. That is, for any nonempty �01 class Pe, the projection map �e takes

Qe P

+e

to P+e .


XVIII.2 Completeness

Let B denote the computable Boolean algebra of clopen sets in f0; 1gN; recallthat these are �nite unions of intervals I(�). Note that B is computably iso-morphic to the Boolean algebra of propositional logic over an in�nite set ofvariables{see Section IX.1 of Chapter IX. In particular, let bn 2 B denotefx : x(n) = 1g.

De�nition XVIII.2.1. Let P be a nonempty subset of f0; 1gN. A splittingfunction for P is a computable function g : N ! B such that, for all e, ifPe � P and Pe 6= ;, then Pe \ g(e) and Pe � g(e) are both nonempty. P is saidto be productive if it has a splitting function.

Clearly a productive �01 class can have no subset that is a singleton and hence

can have no computable member. It follows in particular that P is nowheredense.

We observe that the class DNC2 of diagonally non-computable functions inf0; 1gN is productive. This will be shown in the next section.

Theorem XVIII.2.2. (Simpson) For any productive �01 class P � f0; 1gN and

any nonempty �01 class Q � f0; 1gN, there is a computable functional � from P

onto Q. Thus any productive class is Medvedev complete.

Proof. Let P and Q be �01 subsets of f0; 1gN and suppose that P is productive.

We will de�ne a computable monomorphism f : B ! B such that, for allb 2 B, Q \ b = emptyset () P \ f(b) = ;. Then the computable map� : P ! Q is de�ned by letting Y = �(X) be the unique element of Q suchthat X 2 f(Y dn) 6= ; for all n. The approximating function for � is somethinglike the inverse of f . For an arbitrary element Y 2 Q, Q \ I(Y dn) 6= ; for all nand hence P \ I(f(Y dn)) 6= ; for all n, so that \n[P \ I(f(Y dn))] 6= ; and anyelement of this set will map to Y . This shows that � will map X onto Y .

It clearly su�ces to de�ne f for intervals I(�) and for ease of notation wewill just write f(�) for f(I(�)). Then f(�) = ��1(I(�)) under the function �de�ned above. That is, if Y = �(X), then X 2 f(�) () � � Y .

The function f is de�ned recursively beginning with f(;) = ;.For the recursive step, suppose that f(�) = a is given so that Q \ I(�) =

; () P \ a = ; and that a � I(�) for some � with j� j � j�j. We will showhow to compute f(�_0) = a0 and f(�

_1) = a1 such that each ai is includedin some I(�i) with j�ij > j� j and such that

�_i 2 TQ () P \ ai 6= ;:

Since P is productive, it is nowhere dense so we can partition a non-triviallyinto b0 [ b1 [ b2 so that P \ b0 = P \ b1 = ; and hence P \ a = P \ b2.

XVIII.2. COMPLETENESS 307

By the Recursion Theorem, we can compute e 2 N such that

Pe =

8>>><>>>:P \ b2 if �_0 2 TQ and �_1 2 TQ;P \ b2 \ g(e) if �_0 =2 TQ and �_1 2 TQ;P \ b2 � g(e) if �_0 2 TQ and �_1 =2 TQ;; if �_0 =2 TQ and �_1 =2 TQ:

Now let a0 = b0 [ (b2 \ g(e)) and a1 = f0; 1gN� a0 = b1 [ (b2� g(e)). We claimthe following:

1. Q \ I(�_0) = ; () P \ a0 = ;;2. Q \ I(�_1) = ; () P � a1 = ;:There are several cases to check. Suppose �rst that � =2 TQ. Then by

assumption P \ a = ;, so that we have P \ a0 = P \ a1 = ; = Q \ I(�_0) =Q \ I(�_1). Now suppose that � 2 TQ, so that Q \ I(�) and P \ a are bothnonempty. There are three cases. First suppose that both �_0 and �_1 are inTQ. ThenQ\I(�_0) andQ\I(�_1) are both nonempty, so that Pe = P\a 6= ;.It follows that P \ a0 = P \ b2 \ g(e) = Pe \ g(e) 6= ; (since g is a splittingfunction for P ) and similarly P \ a1 6= ;. Next suppose that �_0 =2 TQ but�_1 2 TQ, so that Pe = P \ b2 \ g(e) = P \ a0. Then Pe � g(e) = ; and, sinceg is a splitting function for P , it follows that Pe = ;, and hence

P \ a1 = (P \ b1) [ (P \ b2 \ g(e)) = ;:The remaining case where �_0 2 TQ and �_1 =2 TQ is similar.

This result can be improved for two productive classes.

Theorem XVIII.2.3. (Simpson) Any two productive �01 classes P;Q � f0; 1gN

are computably homeomorphic.

Proof. We simply add a back-and-forth argument to the proof of TheoremXVIII.2.3 to make the monomorphism onto. That is, at stage n, we will have a�nite isomrphism fn : Bn ' B0n where each of Bn and B0n are �nite subalgebrasincluding b0; : : : ; bn such that P \ a = ; if and only if Q \ fn(a) = ;. We startas above with f((0)) = a0 and f((1)) = a1 as above so that P \ ai = ; ()Q \ I((i)) = ;. Now use the splitting function for P to obtain bij � I(i) fori; j 2 f0; 1g so that ai \ (j) \Q = ; () bij \ P = ; and let B0 be generatedby fbij : i; j 2 f0; 1gg and B1 be generated by fI((0)); I((1)); a0; a1g. Note thatf�1(I(e)) = be0 [ be1. We leave the details to the reader.

Question XVIII.2.4. Suppose in general that P and Q are �01 subsets of

f0; 1gN and that there there exist mappings � from P onto Q and from Qonto P . Does it follows that P and Q are computably homeomorphic?

Lemma XVIII.2.5. (Simpson) Let P and Q be nonempty subsets of f0; 1gN.If P �M Q and P is productive, then Q is productive.


Proof. Since P �M Q, there is a computable function � : Q ! P . De�nef : B ! B by f(b) = ��1(b). Let g : N! B be a splitting function for P . By thes-m-n Theorem, let h : N! N be primitive recursive such that Ph(e) = �(Pe\Q)for all e. We claim that the composition f � g � h is a splitting function for Q.To see this, suppose that Pe � Q and Pe 6= ;. Then Ph(e) = �(Pe) � P andPh(e) 6= ;, so that Ph(e) \ g � h(e) 6= ; and Ph(e) � g � h(e) 6= ; and thereforePe \ f � g � h(e) 6= ;.Corollary XVIII.2.6. Let P � f0; 1gN be a nonempty �0

1 class. Then P isproductive if and only if P is Medvedev complete.

Proof. By Theorem XVIII.2.2, if P is productive, then P is Medvedev complete.Lemma XVIII.2.5 implies that any complete �0

1 class is productive.

Simpson and Slaman (unpublished) have shown the following.

Theorem XVIII.2.7. Every nonzero degree in Pw contains in�nitely manyMedvedev degrees from PM .

Proof. xxxx

Exercises

XVIII.2.1. Prove parts (ii) and (iii) of Proposition XVIII.1.3 for the Muchnik degrees.

XVIII.2.2. Show that P (Q�R) �M (P Q)� (P R).XVIII.2.3. Show that the map taking the Medvedev degree of P to the Muchnik

degree of P is a lattice homomorphism of PM onto Pw.

XVIII.3 Separating Classes

Here is a general result which will provide a large class of Medvedev complete �01

classes. A pair of disjoint c. e. sets A and B are said to be e�ectively inseparableif there is a computable function � such that, for any x and y, if A � Wx

and B � Wy and Wx \Wy = ;, then �(x; y) =2 Wx [Wy. For example, it iswell known that the set PA = A1 of theorems of Peano Arithmetic and the setB1 of negations of theorems of Peano Arithmetic are e�ectively inseparable{seeOdifreddi [163], p 356. The following lemma will then imply that S(A1; B1) isMedvedev complete.

Proposition XVIII.3.1. If A are B are e�ectively inseparable c.e. sets, thenS(A;B) is a productive �0

1 class.

Proof. Let P = S(A;B) where A and B are e�ectively inseparable c. e. setsand and let � be given as above. De�ne Wf(e) = fn : (8X 2 Pe)n 2 Xg andWh(e) = fn : (8X 2 Pe)n =2 Xg. To see that these are indeed c. e. sets, notethat Wf(e) has an alternate de�nition, that is,

XVIII.3. SEPARATING CLASSES 309

n 2Wf(e) () (8� 2 f0; 1gn+1)(� 2 Te =) �(n) = 1);

Clearly Wf(e) \Wh(e) = ;, and if Pe � P , then A � Wf(e) and B � Wh(e).Thus �(f(e); h(e)) = n =2Wf(e) [Wh(e). Hence there exist X and Y in Pe suchthat n 2 X and n =2 Y . The splitting function for P can thus be de�ned byg(e) = fX : �(f(e); h(e)) 2 Xg.

Proposition XVIII.3.2. The �01 class DNC2 is the separating class of a pair

of e�ectively inseparable c. e. sets.

Proof. Let A = fe : �e(e) = 0g and B = fe : �e(e) = 0g. Then S(A;B) =DNC2. Now suppose that A �Wx and B �Wy and suppose that Wx \Wy =;. We will show how to compute �(x; y) = e such that e =2 Wx [ Wy. Let (e; x; y) = 1, if e 2Wx and = 0, if e 2Wy. That is, (e; x; y) searches for theleast s such that e 2 Wx;s [Wy;s and then outputs 1 if e 2 Wx;s and outputs0 if e 2 Wy;s �Wx;s. Let �(x; y) = e so that �e(i) = (e; x; y). We claim thate =2 Wx [Wy. To see this, suppose that e 2 Wx, so that by de�nition of ; �and e, �e(e) = 1. Then e 2 B, which implies that e 2 Wy and contradictsWx \Wy = ;. The argument when e 2Wy is similar.

Corollary XVIII.3.3. DNC2 is Medvedev complete.

Proof. By Propositions XVIII.3.2 and XVIII.3.1, DNC2 is productive and henceby Theorem XVIII.2.2, it is Medvedev complete.

The family of c.e. separating classes are closed under join, since

S(A;B) S(C;D) = S(hA;Ci; hB;Di):However, there is no non-trivial meet for c.e. separating classes, as shown

by the following.

Lemma XVIII.3.4. For any �01 class P and any clopen sets G, and H, if

P \G �M P \H, then P \G �M P \ (G [H).

Proof. First, P \ (G [H) �M P \ G via the identity map. Fix a computablefunctional � : P \H ! P \G and de�ne : P \ (G [H)! P \G by

(X) :=

(X; if X 2 G;�(X); otherwise.

Note that is computable since clopen sets are simply �nite unions of intervals.

Lemma XVIII.3.5. For any c.e. separating class P and any clopen set G, ifP \G 6= ;, then P \G �M P .


Proof. By Lemma XVIII.3.4, it su�ces to prove this for intervals, and weproceed by induction on the length n of �. If n = 0, then I(�) = 2!, soP \ I(�) = P . Assume as induction hypothesis that P \ I(�) �M P for some� of length n, and suppose that P \ I(�_(e)) 6= ;. If P \ I(�_(1 � e)) = ;,then P \ I(�_(e)) = P . Otherwise, P \ I(�_(e)) �M P \ I(�_(1� e)) via thecomputable maps X 7! X [ f0g and X 7! X=f0g. Then by Lemma XVIII.3.4again,

P \ I(�_(e)) �M P \ �I(�_(e)) [ I(�_(1� e))� = P:

Proposition XVIII.3.6. For any �01 classes P and Q and any c.e. separating

class R, if P �Q �M R, then either P �M R or Q �M R.

Proof. Fix a computable functional � : R! P �Q and set G := fX : �(X) 2I((0))g. G is clopen as the continuous inverse image of an interval. P �M R\Gvia the map X 7! �

k 7! �(X)(k+1)�. If R\G 6= ;, then by Lemma XVIII.3.5

R \ G �M R, so P �M R. Otherwise R n G 6= ; and we have similarlyQ �M R.

This suggests that we should consider the sublattice of PM generated bythe family of c.e. separating degrees. This turns out to have a simple directcharacterization.

De�nition XVIII.3.7. For any tree T � f0; 1g<! and any �01 class P �

f0; 1g!,(i) T is homogeneous i� (8�; � 2 T )(8i < 2),

j�j = j� j =) (�_i 2 T () �_i 2 T );

(ii) T is almost homogeneous i� 9n(8�; � 2 T )(8i < 2),

n � j�j = j� j ^ � � n = � � n =) (�_i 2 T () �_i 2 T );The least such n is called the modulus of T ;

(iii) P is (almost) homogeneous i� TP is (almost) homogeneous; a Medvedevdegree is (almost) homogeneous i� it contains an (almost) homogeneousclass; AH denotes the family of almost homogeneous degrees.

Proposition XVIII.3.8. For any �01 class P ,

P is homogeneous () P is a c.e. separating class:

Proof. If P = S(A;B) for c.e. sets A and B, then

TP = f� : (8i < j�j)��(i) = 0 ^ i =2 A) _ (�(i) = 1 ^ i =2 B)�g:This is clearly a homogeneous tree. Conversely, if TP is homogeneous, thenP = S(A;B) for

A = fn : 0n_0 =2 TP g and B = fn : 0n_1 =2 TP g:

XVIII.3. SEPARATING CLASSES 311

Corollary XVIII.3.9. For any �01 class P , if P is almost homogeneous with

modulus n, then P is the disjoint union of 2n c.e. separating classes.

Proof. Given P 2 AH with modulus n, for each sequence � of length n, letP [�] := fX 2 P : � � Xg. Each P [�] is homogeneous, so is a c.e. separatingclass, and clearly P is the disjoint union of the P [�].

Proposition XVIII.3.10. For any �01 classes P and Q, if P and Q are almost

homogeneous, then also P �Q and P Q are almost homogeneous.

Proof. If P and Q are almost homogeneous with moduli m and n, respectively,then easily P � Q is almost homogeneous with modulus maxfm;ng + 1 andP Q is almost homogeneous with modulus 2maxfm;ng.Theorem XVIII.3.11. AH is the smallest sublattice of PM which includesthe family of c.e. separating degrees.

Proof. By the preceding two propositions, AH is a sublattice of PM whichincludes the family of c.e. separating degrees. Let L be any other such lattice;we prove by induction that for all n,

P is almost homogeneous with modulus n =) dgM (P ) 2 L:For n = 0 this is true by Proposition XVIII.3.8, so assume as induction hypoth-esis that it holds for n and that P is almost homogeneous with modulus n+ 1.Then if for i < 2 we set Pi := fX : (i)X 2 Pg, Pi is almost homogeneous withmodulus n, so dgM (Pi) 2 L and clearly P = P0 � P1 so also dgM (P ) 2 L.

Of particular interest are the generalizations

DNCk = fX 2 kN : (8n)X(n) 6= �n(n)g:The next theorem is due to Jockusch [92]. We will use the following lemma.

Lemma XVIII.3.12. (Cenzer-Hinman) For any l < k, any s > 0 and anyfunction F : ks ! l, there exists j < l and a tree T � k�s such that

(i) for all � 2 T , there exist i0 6= i1 < k such that �_it 2 T for i = 0; 1;

(ii) for all � 2 T \ ks, F (�) = j.

Proof. The proof is by induction on s. For s = 1, this is just the pigeonholeprinciple. Now given F : ks+1 ! l, de�ne G : ks ! l by

G(�) = (least j < l)[(9i0 < i1)F (�_i0) = F (�_i1) = j];

such a j must exist for each � by considering the map F� : k ! l de�ned byF� (i) = F (�_i).

Now by induction there exists j < l and a tree TG � k�s satisfying (i) and(ii) above with respect to G. Let

T = TG [ f�_i : � 2 TG \ ks & F (�_i) = jg:It is easy to check that T satis�es conditions (i) and (ii) with respect to F .


Theorem XVIII.3.13. For all n > 1, DNCk+1 <M DNCk.

Proof. Note here that in general �e : N ! N and we let �ke(n) = maxfk �1; �e(n)g to get the eth function in kN. DNCk+1 �M DNCk by the the map�(X) = X. Now suppose by way of contradiction that DNCk �M DNCk+1and let � : DNCk+1 ! DNCk. We will show how to use � to compute anelement Y of DNCk, which is the contradiction.

Given n, we can compute Y (n) as follows. First compute a level s suchthat �(�; n) # for all � 2 (k + 1)s and consider the map F : (k + 1)s ! kde�ned by F (�) = �(�; n). By Lemma XVIII.3.12, there exits jn < k and atree T � (k + 1)�s such that F (�) = jn for all � 2 T \ (k + 1)s and such thatany � 2 T \ (k + 1)<s has at least two extensions in T . We now show thatthere is in fact some � 2 T \ (k + 1)s such that I(�) \DNCk+1 6= ; and hencesome X 2 DNCk+1 such that �(X)(n) = �(�; n) = jn. Since �(X) 2 DNCk,it will follow that �n(n) 6= jn and hence we can compute Y 2 DNCk by takingY (n) = jn for each n.

The path � = (e0; e1; : : : ; es�1 2 T\(k+1)s exists by the following (althoughwe cannot directly compute it). For t = 0, we have (i0) 6= (i1) in T andat least one of these does not equal �0(0); let e0 be the least such. Given� = (e0; : : : ; et�1) 2 T , again there exist i0 < i1 such that both �_i0 and �

_i0are in T and again at least on of i0; i1 is not equal to �t(t) so we can choose e0to be the least such.

Exercises

XVIII.3.1. Say that c. e. sets A and B are weakly e�ectively inseparable if there isa computable function F , mapping !2 into the family of �nite sets ofnatural numbers, such that, for any x and y, if A �Wx and B �Wy andWx \Wy = ;, then F (x; y) contains at least one element which is not inWx [Wy. Show that if S(A;B) is productive, then A and B are weaklye�ectively inseparable.

XVIII.3.2. Recall from Section III.III.9 the class CC(T ) of complete consistent ex-tensions of a c. e. propositional theory T . Show that for any c. e. theoryU , CC(U) is Medvedev complete if and only if, for every c. e. theory T ,there exists a computable function � : CC(U)! CC(T ).

XVIII.3.3. Let the eth c. e. theory Te = f i : i 2 Weg, where i : i 2 Ng enumeratesthe sentences of propositional logic. Let us say that a theory T is e�ectivelyincompletable if there exists a computable mapping � : N! Sent such thatfor all a, if T � Ta and Ta is consistent, then both Wa [ f�(a)g and Wa [f:�(a)gare consistent. Show that, for any c. e. theory U , U is e�ectivelyincompletable if and only if CC(U) is productive and hence CC(U) isMedvedev complete if and only if CC(U) is e�ectively incompletable.

XVIII.3.4. For a pair A;B of disjoint c. e. sets, let S(A;B) represent the logicaltheory U(A;B) with axioms fAi : i 2 Ag [ f:Ai : i 2 Bg. Show thatfor any pair A;B of e�ectively inseparable c. e. sets, U(A;B) is e�ectively

XVIII.4. MEASURE 313

incompletable. Hint: There exist computable functions f and g such thatWf(e) = fi : Ai 2 Tegand Wg(e) = fi : :Ai 2 Teg.

XVIII.3.5. Let U be an e�ectively incompletable c. e. theory. Then for any c. e.theory T , there exists a computable mapping � : Sent3 ! Sent such thatif both T [ f�g and U [ f g are consistent, then(a) T [ f�; �g is consistent � U [ f ; �(�; ; �)g is consistent;(b) T [ f�;:�g is consistent � U [ f ;:�(�; ; �)g is consistent.Hint: Use the Recursion Theorem to compute an index a from �; ; �such that Ta equals Con(U [ f ; �ag) if T [ f�; �g is consistent, equalsCon(U [ f ;:�ag) if T [ f�;:�g is consistent, and equals Con(U [ f g)otherwise.

XVIII.3.6. Show directly that any e�ectively incompletable c. e. theory U is Medvedevcomplete. Hint: given any c. e. theory T , recursively de�ne a mapping :f0; 1g� ! Sent taking � to � by (;) = p0_:p0 and for any � of length n, (�_0) = � ^ :�(q�; �; An) and (�_1) = � ^�(q�; �; An), whereq� denotes the conjunction over i < n of fAi : �(i) = 1g[f:Ai : �(i) = 0g.Then T [ fq� is consistent if and only if U [ f (�)g is consistent and forX 2 CC(U), we may de�ne �(X) 2 CC(T ) to be the unique Y such thatX 2 CC(U [ f (Y dn)g for all n.

XVIII.4 Measure

In this section we give the result of Cenzer and Hinman [26] that there is noMedvedev complete �0

1 class of positive measure and indeed that no class ofpositive measure can be �M any separating class. The following lemma is dueto Simpson [194].

Lemma XVIII.4.1. (Simpson) Let fFngn2! be a sequence of nonempty �nitesubsets of N of bounded cardinality and let S =

Qn Fn. Let P � f0; 1gN have

positive measure and let Q � NN be nonempty.

1. If S �M P Q, then S �M Q.

2. If S �w P Q, then S �w Q.Proof. (1) The proof is similar to that of Theorem IV.3.11. Suppose thatcard(Fn) < k for all n. Let U and V be clopen so that �(V � P ) < �(P )and �(V � U) are both < �(P )=4k and therefore �(U � P ) < �(U)=K. It isimportant here that �(U) is rational. Let � be a computable function such that�(x� y) 2 S for all x 2 P and y 2 Q. Given y 2 Q and n 2 N, we can computem = (y)(n) such that �(fx 2 U : �(x� y)(n) = mg) > �(U)=k and thereforem 2 Fn. Thus maps Q onto S and hence S �M Q.


(2) Fix y 2 Q and note that S �w P fyg. Now for each x 2 P , thereis a function �e(x) such that �e(x)(x; y) 2 S. By countable additivity of �,there is a single function � and a subset Pg of P of positive measure such thatS �M Pgfyg Now by lemma XVIII.4.1, S �M fgg. It follows that S �s Q.

Note that in particular these lemmas apply to any separating class S.

Theorem XVIII.4.2. (Simpson) Let P;Q be nonempty �01 subsets of f0; 1gN

and suppose that P has positive measure. If Q <M 1, then P Q <M 1 andsimilarly if Q <w 1, then P Q <w 1.

Proof. This follows from Lemma XVIII.4.1, since there is a Medvedev completeseparating class by Corollary XVIII.3.3.

Corollary XVIII.4.3. If P � f0; 1gN is a �01 class with positive measure, then

P <w 1.

Proof. This follows from Theorem XVIII.4.2 by letting Q = f0; 1gN.

Exercises

XVIII.4.1. Show that for any �01 classes P and Q, if P has positive measure and

DNCk �M P Q, then DNCk �M Q. Conclude that P DNCk >MP DNCk+1 for all k.

XVIII.5 Randomness

Recall the notion of 1-randomness from Section 4.3. Let R be the class of all1-random reals in f0; 1gN. It follows from Theorem 4.IV.3.5 that there is a �0

1

class P with positive measure such that P � R.Theorem XVIII.5.1. (Simpson) For any �0

1 class P � R with positive mea-sure, P �w R.Proof. Since P � R, it follows that R �w P . On the other hand, Theorem4.IV.3.6 tells us that P has elements of every 1-random degree, so for anyx 2 R, there exists y 2 P with y �T x and hence P �w R.Corollary XVIII.5.2. If a �0

1 class P contains a random real, then P �wR.

Corollary XVIII.5.3. (Simpson) The Muchnik degree of R can be character-ized as the unique largest Muchnik degree of any �0

1 class P � f0; 1gN such that�(P ) > 0.

Proof. By Theorem XVIII.5.1, there is a �01 class P � f0; 1gN with positive

measure and with P �w R. Now let P be any �01 class P of positive measure.

It follows from Theorem 4.IV.3.6 that P �w R.Exercises

XVIII.5.1.

XVIII.6. THIN CLASSES 315

XVIII.6 Thin Classes

In this section, we prove the following theorem of Simpson.

Theorem XVIII.6.1. If Q � f0; 1gN is a nonempty perfect thin �01 class

and R � f0; 1gN is the set of all Martin-L�of random reals, then Q and R andMuchnik incomparable.

The theorem is proved using a sequence of lemmas.

Lemma XVIII.6.2. Let Q � f0; 1gN be nonempty thin �01 class, let x be

Martin-L�of random, and let y 2 Q be almost computable. Then x is not Turingreducible to y.

Proof. Suppose by way of contradiction that x �T y. Then by Theorem II.5.13x �tt y. Now by Theorem II.5.10, there is a total computable functional � :f0; 1gN ! f0; 1gN mapping y to x. Now by Lemma III.8.4, the image �[Q] isa thin �0

1 class and hence has measure zero by Theorem ?? But x 2 �[Q] isMartin-L�of random and hence �(�[Q]) > 0 by Exercise 5.

Since every nonempty �01 class contains an almost computable member by

Theorem ??, it follows that R is not Muchnik reducible to Q. The followinglemma is due to Demuth [57].

Lemma XVIII.6.3. Let x 2 f0; 1gN be Martin-L�of random and let y �tt x benoncomputable. Then there exists z �T y such that z is Martin-L�of random.

Lemma XVIII.6.4. If Q � f0; 1gN is a nonempty perfect thin �01 class, let

y 2 Q and let x be Martin-L�of random and almost computable. Then y is notTuring reducible to x.

Proof. Suppose by way of contradiction that y 2 Q and y �T x. Then y �tt xsince x is almost computable. It follows from Lemma XVIII.6.3 that there is arandom z �T y. But then z �T x, contradicting Lemma XVIII.6.2.

To complete the proof of Theorem ??, let x be random and almost com-putable and let Q be a nonempty perfect thin �0

1 class. It follows from LemmaXVIII.6.4 that no member of Q is computable from x.

Corollary XVIII.6.5. There is a �01 separating class Q and a �0

1 class Q0 such

that Q < wQ0, and furthermore, for any separating class P , if P is Muchnikreducible to Q0, then P is Muchnik reducible to Q.

Proof. Let Q be a perfect thin �01 class which is separating III.8.1, let R be a

�01 class of randoms and let Q0 = Q R. It follows from Theorem ?? that Q0

is not Muchnik reducible to Q. The furthermore remark follows from Lemma??.


Chapter XIX

Random Closed Sets

The study of algorithmic randomness has been of great interest in recent years.The basic problem is to quantify the randomness of a single real number; herewe will extend this problem to the randomness of the set of paths through a�nitely-branching tree. Early in the last century, von Mises [214] suggested thata random real should obey reasonable statistical tests, such as having a roughlyequal number of zeroes and ones of the �rst n bits, in the limit. Thus a randomreal would be stochastic in modern parlance. If one considers only computabletests, then there are countably many and one can construct a real satisfying alltests.

An early approach to randomness was through betting. E�ective betting ona random sequence should not allow one's capital to grow unboundedly. Thebetting strategies used are constructive martingales, introduced by Ville [213]and implicit in the work of Levy [130], which represent fair double-or-nothinggambling.

Martin-L�of [146] observed that stochastic properties could be viewed as spe-cial kinds of meaure zero sets and de�ned a random real as one which avoidscertain e�ectively presented measure 0 sets. That is, a real x 2 f0; 1gN isMartin-L�of random if for any e�ective sequence S1; S2; : : : of c.e. open setswith �(Sn) � 2�n, x =2 \nSn.

At the same time Kolmogorov [114] de�ned a notion of randomness for �nitestrings based on the concept of incompressibility. For in�nite words, the strongernotion of pre�x-free complexity developed by Levin [129], G�acs [74] and Chaitin[47] is needed. Schnorr later proved [185] that the notions of constructive mar-tingale randomness, Martin-L�of randomness, and pre�x-free randomness areequivalent. In this chapter, we will consider algorithmic randomness on thespace C of nonempty closed subsets P of f0; 1gN.

The betting approach to randomness is formalized as follows:

De�nition XIX.0.6 (Ville [213]). 1. A martingale is a function d : n<N !

317

318 CHAPTER XIX. RANDOM CLOSED SETS

[0;1) such that for all � 2 n<N,

d(�) =1

n

n�1Xi=0

d(�_i):

2. A martingale d succeeds on X 2 nN if

lim supm!1

d(Xdm) =1:

That is, the betting strategy results in an unbounded amount of moneymade on the binary string X.

3. The success set of d is the set S1[d] of all sequences on which d succeeds.

That is, a martingale on 2<N is the representation of a fair double-or-nothingbetting strategy. When working on 3<N the strategy is triple-or-nothing.

De�nition XIX.0.7. A martingale d is constructive (e�ective, c.e.) if it is

lower semi-computable; that is, if there is a computable function d : n<N�N!Q such that

1. for all � and t, d(�; t) � d(�; t+ 1) < d(�), and

2. for all �, limt!1 d(�; t) = d(�).

In other words, d(w) is approximated from below by rationals uniformly inw. A sequence in 2N is considered random in this setting if no constructivemartingale succeeds on it.

Martin-L�of randomness for reals, as de�ned above, is extended to closedsets by giving an e�ective homeomorphism with the space f0; 1; 2gN and simplycarrying over the notion of randomness from that space.

Pre�x-free randomness for reals is de�ned as follows. Let M be a pre�x-freefunction with domain � f0; 1g�; that is, if � v � are strings in the domain ofM ,then � must equal � . For any �nite string � , let KM (�) = minfj�j;1 :M(�) =�g. There is a universal pre�x-free function U such that, for any pre�x-free M ,there is a constant c such that for all �

KU (�) � KM (�) + c:

We let K(�) = KU (�). Then x is called pre�x-free random if there is a constantc such that K(xdn) � n� c for all n. This means that the initial segments of xare not compressible.

For a tree T , we want to consider the compressibility of Tn = T \ f0; 1gn.This has a natural representation of length 2n since there are 2n possible nodesof length n. We will show that any tree TP can be compressed, that is, K(Tn) �2n � c is impossible for a tree with no dead ends.

XIX.1. MARTIN-L �OF RANDOMNESS OF CLOSED SETS 319

XIX.1 Martin-L�of Randomness of Closed Sets

In this section, we de�ne a measure on the space C of nonempty closed subsetsof f0; 1gN and use this to de�ne the notion of randomness for closed sets. Wethen obtain several properties of random closed sets.

An e�ective one-to-one correspondence between the space C and the space3N is de�ned as follows. Let a closed set Q be given and let T = TQ be the treewithout dead ends such that Q = [T ].

Then de�ne the code x = xQ 2 f0; 1; 2gN for Q as follows. Let ; =�0; �1; �2; : : : enumerate the elements of T in order, �rst by length and thenlexicographically. We now de�ne x = xQ = xT by recursion as follows. For eachn, x(n) = 2 if �_n 0 and �_n 1 are both in T , x(n) = 1 if �_n 0 =2 T and �_n 1 2 Tand x(n) = 0 if �_n 0 2 T and �_n 1 =2 T .

Now de�ne the measure �� on C by

��(X ) = �(fxQ : Q 2 Xg):

Informally this means that given � 2 TQ, there is probability 13 that both

�_0 2 TQ and �_1 2 TQ and, for i = 0; 1, there is probability 13 that only

�_i 2 TQ. In particular, this means that Q\ I(�) 6= ; implies that for i = 0; 1,Q \ I(�_i) 6= ; with probability 2

3 .

Let us comment brie y on why some other natural representations were re-jected. Suppose �rst that we simply enumerate all strings in f0; 1g� as �0; �1; : : :and then represent T by its characteristic function so that xT (n) = 1 () �n 2T . Then in general a code x might not represent a tree. That is, once we have(01) =2 T we cannot later decide that (011) 2 T . Suppose then that we allowthe empty closed set by using codes x 2 f0; 1; 2; 3g� and modify our originalde�nition as follows. Let x(n) = i have the same de�nition as above for i � 2but let x(n) = 3 mean that neither �_n 0 nor �_1 is in T . Informally, thiswould mean that for i = 0; 1, � 2 T implies that �_i 2 T with probability 1

2 .The advantage here is that we can now represent all trees. But this is also adisadvantage, since for a given closed set P , there are many di�erent trees Twith P = [T ]. The second problem with this approach is that we would have[T ] = ; with positive probability. We brie y return to this subject in Section 6.

Now we will say that a closed set Q is (Martin-L�of) random if the code xQ isMartin-L�of random. Let Qx denote the unique closed set Q such that xQ = x.Since random reals exist, it follows that random closed sets exists. Furthermore,there are �0

2 random reals, so we have the following.

Theorem XIX.1.1. There exists a random closed set Q such that TQ is �02.

Note that if TQ is �02, then Q must contain �0

2 elements (Theorem XV.1.6).Since there exist strong �0

2 classes with no �02 elements, there are strong �0

2

classes Q such that TQ is not �02.

The following lemma will be needed throughout.


Lemma XIX.1.2. For any Q � 2N which is either closed or open, ��(fP :P � Qg) � �(Q).Proof. Let PC(Q) denote fP : P � Qg. We �rst prove the result for (nonempty)clopen sets U by the following induction. Suppose U = [�2SI(�), whereS � f0; 1gn. For n = 1, either �(U) = 1 = ��(PC(U)) or �(U) = 1

2 and��(PC(Q)) = 1

3 . For the induction step, let Si = f� : i_� 2 Sg, let Ui =[�2SiI(�), let mi = �(Ui) and let vi = ��(PC(Ui)), for i = 0; 1. Then consid-ering the three cases in which S includes both initial branches or just one, wecalculate that

��(PC(U)) = 1

3(v0 + v1 + v0v1):

Thus by induction we have

��(PC(U)) � 1

3(m0 +m1 +m0m1):

Now2m0m1 � m2

0 +m21 � m0 +m1;

and therefore

��(PC(U)) � 1

3(m0 +m1 +m0m1) � 1

2(m0 +m1) = �(U):

For a closed set Q, let Q = \nUn, with Un+1 � Un for all n. Then P � Q ifand only if P � Un for all n. Thus

PC(Q) = \nPC(Un);

so that��(PC(Q)) = lim

n!1��(PC(Un)) � lim

n!1�(Un) = �(Q):

Finally, for an open set Q, let Q =Sn Un be the union of an increasing sequence

of clopen sets. Then, by compactness,

PC(Q) = [nPC(Un);

so that��(PC(Q)) = lim

n!1��(PC(Un)) � lim

n!1�(Un) = �(Q):

This completes the proof of the lemma.

Next we consider the intersection of a random closed set with an intervalI(�) and the disjoint union of random closed sets.

Let us call the coding of a closed set Q by the nodes of its representative treewith no dead ends the canonical code of Q. We wish now to introduce a secondmethod of coding, the ghost code. A ghost code of Q is an in�nite ternary stringwhose bits correspond to all nodes of 2<N in lexicographical order. The bitscorresponding to the nodes of Q's tree (the \canonical nodes") hold the same


values as the corresponding bits in the canonical code; the remaining \ghostnodes" may hold any values. Ghost codes are non-unique, and every closed sethas a non-random ghost code (if the closed set itself is random take the codewith ghost nodes all equal to zero).

We de�ne randomness for closed sets in the world of ghost codes as possessionof a random code. This method of coding is more convenient for some purposes;for example, we will use it to show that if Q0; Q1 are closed sets and Q =f0_x : x 2 Q0g [ f1_x : x 2 Q1g, Q is random if and only if the Qi arerandom relative to each other. With canonical coding it is straightforward toshow relative randomness of the half trees is su�cient for randomness of the fulltree, but not its necessity.

However, the utility of the ghost codes rests entirely on the following corre-spondence.

Theorem XIX.1.3. The canonical code of a closed set Q � 2N is random ifand only if Q has some random ghost code.

Proof. (() Suppose the canonical code of Q is nonrandom. Then there is a c.e.martingale m that succeeds on it. From any initial segment � of a ghost codeg for Q, the subsequence � of exactly the canonical nodes of � is computable.Therefore it is computable whether the bit of g after � is canonical or ghost.From m, de�ne the martingale m0 which bets as follows:

m0(�_i) =

(m(�_i)m(�) m0(�) next bit is a canonical node

m0(�) next bit is a ghost node.

That is, m0 holds its money on ghost nodes and bets proportionally to m (infact, identically) on canonical nodes. It is clear that m0 succeeds on the ghostcode g and thus g is nonrandom.

()) Now suppose the canonical code r for Q is random, and let q be an in�niteternary string that is random relative to r (so therefore r is also random relativeto q). We claim the ghost code g obtained by using the bits of r as the canonicalnodes and the bits of q in their original order as the ghost nodes is random. Itis clear that g is a ghost code for Q.

Suppose m is a c.e. martingale that bets on g. We de�ne two martingalesfrom m, mr and mq, that bet on the bits of r and q respectively, with oracle qand r respectively, according to m's actions on the corresponding bits of g, andshow that if they do not succeed, m does not succeed. As q and r are relativelyrandom, mr and mq cannot succeed, and so g will be random.

We de�ne mr only; for mq swap the roles of r and q, and of canonical andghost nodes. From a string � compute the unique initial segment � of a ghostcode such that

1. � is the substring of canonical nodes of � ,

2. the node to follow � is a canonical node, and


3. the ghost nodes of � are an initial segment of the oracle q.

Then mr(�_i) = m(�_i)

m(�) mr(�).

Since each of r; q is random relative to the other, neither mr nor mq cansucceed. That is, there is some b such that mr(�) � b for all � � r and likewisemq(�) � b for all � � q. We claim that the values of m on initial segments of gare bounded by b2.

At each node of the code r the martingale mr will multiply the capital heldbefore the node by some constant factor. Let fcrngn2N be the sequence of level-nmultiplicative factors for mr. Assuming an initial capital of 1, the values mr

achieves on initial segments of r are

Yk=0

crk

for 0 � ` <1. By assumption, every product of this form is less than or equalto b. Substituting the corresponding de�nition for mq, fcqngn2N, gives the sameresult.

The original martingale m behaves on the subequences r and q exactly asthe martingales mr and mq do, by construction of mr and mq. Therefore it hasthe same collection of multiplicative factors. Again assuming initial capital 1,the values it achieves are thus products of the form

Yk=0

crk

!�0@ `0Yk=0

cqk

1A :

As each of the subproducts is bounded by b, the entire product is boundedby b2 and m does not succeed on g. Since m was arbitrary, no c.e. martingalesucceeds on g and thus Q has a random ghost code.

Note that this theorem relativizes so we can assert that the canonical codeis, say, A-random for some A if and only if it has an A-random ghost code.

The primary purpose of the ghost codes is to remove the dependence onthe particular closed set under discussion when interpreting bits of the code asnodes of the tree. This is especially useful when subdividing the tree, as in thefollowing de�nition.

De�nition XIX.1.4. The tree join of closed sets P0 and P1 is the closed setQ = f0_x : x 2 P0g[f1_x : x 2 P1g. Given ghost codes r0; r1 for the Pi, theirtree join r0 � r1 is the code for Q with the corresponding ghost node values.

This is distinguished from what we will call the recursion-theoretic join r0�r1, where elements of r0 and r1 alternate.

We wish to relate the recursion-theoretic join and the tree join. First recallvan Lambalgen's theorem.

Theorem XIX.1.5 (van Lambalgen [212]). The following are equivalent.


1. A�B is n-random.

2. A is n-random and B is n-A-random (or vice-versa).

3. A is n-B-random and B is n-A-random.

Lemma XIX.1.6. Given two ghost codes r0; r1, the tree join r0� r1 is randomif and only if the recursion theoretic join r0 � r1 is random.

Proof. We show a Martin-L�of test that contains one version of the join may betransformed into a test that contains the other version, and therefore if one isnonrandom, both are nonrandom. To simplify the proof we ignore the initialbit of the tree join, as it has no matching bit in the recursion-theoretic join.

It is clear that initial segments of the two versions of join may be transformedinto each other via a computable algorithm that is independent of the value ofthe bits, provided the strings are of length `n := 2n+1� 1 (i.e., the �nal node isthe end of a level of the tree). For strings that terminate in the middle of levels,a single initial segment � becomes a �nite collection of strings �j , one for eachextension of � to a string of length `n for the least n such that j�j � `n. Therewill be 2`n�j�j such strings and hence the measure of the intervals around themwill total 2�`n � 2`n�j�j = 2�j�j, equal to the measure of the interval around �.

Therefore, let fUi : i 2 Ng be a Martin-L�of test failed by r0 � r1. When� enters Ui for some i, enumerate the �nite collection of �j as described above

into a new set Ui. The collection fUi : i 2 Ng is clearly a Martin-L�of test, as itis enumerated simultaneously with the fUig and �(Ui) = �(Ui) for all i 2 N. Ifr0 � r1 extends � 2 Ui, r0 � r1 will extend one of the corresponding �j in Ui.

Therefore since r0�r1 fails fUig, r0�r1 will fail fUig. Symmetrically, if r0�r1is nonrandom we may construct a Martin-L�of test that r0 � r1 also fails.

We now obtain the following corollary of Theorems XIX.1.3 and XIX.1.5,and Lemma XIX.1.6.

Corollary XIX.1.7. Suppose Pi, i = 0; 1, are closed sets with canonical codesri and let P be the tree join of P0; P1. Then P is random if and only if r0 � r1is random.

Proof. (() Suppose that r0 � r1 is random. Then by Theorem XIX.1.5, theri are mutually relatively random. By Theorem XIX.1.3, there are ghost codesgi for the Pi that are also mutually relatively random. Again by XIX.1.5, therecursion-theoretic join g0 � g1 is random; then by Theorem XIX.1.6 the treejoin g0 � g1 is also random, and hence P possesses a random ghost code and israndom.()) Suppose now that P is random, and therefore possesses a random ghostcode g. The code g may be thought of as a tree join g0 � g1, which is thereforerandom, and so by Theorem XIX.1.6, g0 � g1 is random. By Theorem XIX.1.5,the individual codes g0; g1 are therefore mutually relatively random, and so byTheorem XIX.1.3 the canonical codes r0; r1 for the half trees are as well. Thusagain by XIX.1.5, r0 � r1 is random.


XIX.2 Members of Random Closed Sets

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