Fine Hierarchies and Boolean Terms

V.L.Selivanov �

Novosibirsk Pedagogical University

Abstract

We consider �ne hierarchies in recursion theory, descriptive set theory,

logic and complexity theory. The main results state that the sets of values

of di�erent Boolean terms coincide with the levels of suitable �ne hier-

archies. This gives new short descriptions of these hierarchies and shows

that the collections of sets of values of Boolean terms are often almost

well ordered by inclusion. For the sake of completeness we mention also

some earlier results demonstrating the usefulness of the �ne hierarchies.

1 Introduction

Most of the work in hierarchy theory is devoted to hierarchies coarser than Borel(or, in the context of recursion theory, arithmetical). But some attention waspaid also to �ne (i.e. �ner than Borel) hierarchies. In this paper we consideronly �ne hierarchies. Our main results show that the sets of values of di�erentBoolean terms coincide with levels of suitable �ne hierarchies. This seems to beof some interest for two reasons: �rst we get new elegant descriptions of thesehierarchies and secondly we see that the collections of sets of values of Booleanterms are often almost well ordered by inclusion, which is not at all evident. Toget a coherent picture of our subject we mention and discuss also some earliernot broadly known results demonstrating the usefulness of �ne hierarchies. Wehope that this will lead to a better understanding of the place and role of thesehierarchies.

�This work was done at the University of Heidelberg and supported by the Alexander von

Humboldt Foundation.

1

We start in section 2 with a simple result on an abstract version of thedi�erence hierarchy generalizing some earlier results of L.Hay, K.Wagner andG.Wechsung, and the author. In section 3 we extend this to a much deeperresult on a �ne hierarchy in recursion theory introduced and studied by theauthor. This gives probably the clearest "axiomatic" de�nition of this hierar-chy. In section 4 we summarise technical results on an abstract version of thishierarchy. In section 5 we generalize the result from section 3 to this abstracthierarchy. This is useful because it is applicable to some hierarchies naturallyarising outside recursion theory. E.g. in section 7 we discuss such a versionfrom pure logic which leads to an extension of some well{known results on thehierarchy of formulas, in particular of a result of J.Addison. In section 8 weconsider such a version from complexity theory. In section 6 we consider theclassical case of the Borel hierarchy and prove in�nite versions of results fromsections 1|3; modulo a deep result of Louveau on the Borel Wadge classes thepresentation here is quite easy.

2 Di�erence Hierarchy

The di�erence hierarchy was the �rst �ne hierarchy considered in descriptive settheory (F.Hausdor�, K.Kuratowski, M.A.Lavrentiev). It was then intensivelystudied in an abstract setting (J.Addison), in recursion theory (Y.L.Ershov,L.Hay, V.L.Selivanov, etc.) and recently in complexity theory (K.W.Wagner,G.Wechsung, J.Cai, L.Hemachandra, etc.).

Let B = (B;[;\;�; 0; 1) be a Boolean algebra (b.a.) and L be a sublatticeof (B;[;\; 0; 1). Let Dk(k < !) be the set of all elements [i(a2i n a2i+1) whereai 2 L satisfy a0 � a1 � � � � and ak = 0. The sequence fDkgk<! is calledthe di�erence hierarchy over L. Note that the de�nition does not (moduloisomorpism) depend on B. Note also that by the Stone representation theoremwe may without loss of generality think of L as a lattice of sets; we sometimesuse this in our proofs. It is well known (see Hau14 and Ad65) that D0 = f0g,Dk+1 = f�abja 2 Dn; b 2 Lg, Dk[ �Dk � Dk+1 and [k<!Dk is the b.a. generatedby L. Here �Dk = f�xjx 2 Dkg is the dual of the set Dk and ab abbreviates a\ b.

Let T be the set of all �nite Boolean terms with variables vk(k < !) in thelanguage f[;\;�; 0; 1g. Relate to any t 2 T the set t(L) of all values of t whenits variables range over L. We call the sets t(L) classes of the Boolean hierarchy

over L. It is easy to see that any Dk is t(L) for some t (let si = 0 for k � i,si = vi[ : : :[vk�1 for i < k and t be the �nite term equivalent to [(s2i ns2i+1);then Dk = t(L) by the de�nition of Dk). It turns out that any t(L) is also oneof Dk; �Dk.

2.1. Theorem. The classes of the di�erence and the Boolean hierarchies

over L coincide i.e. ft(L)jt 2 Tg = fDk; �Dkjk < !g.Note that particular cases of this simple result are known. E.g. the paper

Hay78 contains almost all relevant computations, and results from Se84 andWe85 immediately imply the particular cases of 2.1 described below.

2.2. Corollary. The classes of the di�erence and the Boolean hierarchies

2

over the lattice L of r.e. sets coincide.

Proof. We have to show that for any t = t(v0; : : : ; vm) the class t(L) isone of Dk; �Dk. Theorem 3 from Se84 states a suÆcient condition for an indexset to be m-complete in one of Dk; �Dk. From this result immediately followsthat any Boolean combination of sets Ai = fxji 2 Wxg, where Wx is the r.e.set with the index x, is m-complete in one of Dk; �Dk. In particular, the setA = t(A0; : : : ; Am) 2 t(L) is m-complete in one of Dk; �Dk, so it suÆces tocheck that A is m-complete in t(L), i.e. any set X = t(X0; : : : ; Xm), Xi 2 L, ism-reducible to A. Let g be a recursive function satisfying Wg(x) = fijx 2 Xig.Then X = t(X0; : : : ; Xm) = t(g�1(A0); : : : ; g

�1(Am)) = g�1(A) completing theproof.

The next assertion follows from a result of K.Wagner and G.Wechsung inWe85.

2.3. Corollary. The classes of the di�erence and the Boolean hierarchies

over the lattice L of NP-sets coincide.

Proof. Let t be a Boolean term as above, let f be the corresponding Booleanfunction and let Kf be the class of sets polynomially btt-reducible to an NP -complete set B when all Boolean functions of the reduction are equal to f . It iseasy to see that Kf = t(L). As a corollary of an indirect proof in We85 it wasshown that Kf is one of Dk; �Dk. This completes the proof.

So Theorem 2.1 was implicit in the literature on the di�erence hierarchy.Nevertheless we write down its proof because it is shorter than the correspond-ing proofs in Hay78 and We85 and because some of our further proofs are itsgeneralizations. The proof is essentially the same as that in Se84.

Proof of 2.1. We have to show that for any term t containing only variablesfrom V = fv0; : : : ; vmg, the class t(L) is one of Dk; �Dk. Relate to any U � V

the termtU = tU (v0; : : : ; vm) = (

\

vi2U

vi) \ (\

vi 62U

�vi):

According to propositional logic t is uniquely representable as tA = [ftU jU 2Ag for a class A of subsets of V . By a 0-chain (1-chain) for t we mean anysequence U0 � � � � � Un of subsets of V such that Ui 2 A $ Ui+1 62 A andU0 62 A (resp. U0 2 A). The number n is called the length of such a chain(U0; : : : ; Un).

Let k be the length of a maximal chain for A. Then A cannot have botha 0-chain (U0; : : : ; Uk) and a 1-chain (V0; : : : ; Vk) because otherwise one of thesequences (;; U0; : : : ; Uk), (;; V0; : : : ; Vk) would be a chain for A of length k +1contradicting the choice of k. So either A has a 0-chain (U0; : : : ; Uk) but doesnot have a 1-chain of length k or A has a 1-chain but does not have a 0-chainof length k.

We claim that t(L) = Dk for the �rst alternative and t(L) = �Dk for thesecond one. It suÆces to check the �rst assertion because the second is its dual.To prove the inclusion Dk � t(L) take any a 2 Dk and represent it in the forma = [i(a2i n a2i+1) where ai 2 L satisfy a0 � a1 � � � � and ak = 0. De�nebU (U � V ) as follows: bU = 0 for U 62 fU0; : : : ; Ukg, bU0 = �a0, bUi+1 = ai n ai+1

3

for i < k. Then elements ci = [fbU jvi 2 Ug belong to L and satisfy bU =tU (c0; : : : ; cm) for any U � V . So a = tA(c0; : : : ; cm) 2 t(L).

It remains to show that t(L) � Dk. Let b 2 t(L), so b = tA(b0; : : : ; bm) forsome bi 2 L. Let ai(i < !) be the union of all tU (b0; : : : ; bm) such that there isa 1-chain (V0; : : : ; Vi) for A with Vi � U . We claim that ai 2 L, a0 � a1 � : : :,ak = 0 and b = a, where a = [i(a2ina2i+1); it would complete the proof becausethen b 2 Dk.

Note that ai is a union of elements of the form c = [ftU (b0; : : : ; bm)jU � Vig.From the de�nition of tU follows that c = \fbj jj 2 Vig, so c; ai 2 L. Inclusionsa0 � a1 � : : : are evident. The set A does not have 1-chains of length k, soak = 0. To prove b � a take any x 2 b. Then x 2 tU (b0; : : : ; bm) for someU 2 A. Then (U) is a 1-chain for A and a fortiori x 2 a0. To prove x 2 a itis now suÆcient to check that if x 2 a2i+1 then x 2 a2i+2. Assume x 2 a2i+1,then x 2 tR(b0; : : : ; bm) for some R � V and some 1-chain (V0; : : : ; V2i+1) forA with V2i+1 � R. But R = U , because elements tU (b0; : : : ; bm) for di�erentU are pairwise disjoint. So (V0; : : : ; V2i+1; U) is a 1-chain for A and a fortiori

x 2 a2i+2. A similar argument shows that �b � �a, so b = a and the theorem isproved.

The next corollary and especially its further generalizations states an in-triguing property of Boolean terms which may be of some independent interestbecause its formulation does not mention hierarchies. Call a collection C of setsalmost well ordered if the partial ordering (C;�) is well founded and almostlinearly ordered (i.e. for all X;Y 2 C either X � Y or Y � �X). Recall thatwith such an ordering C (as well as with any well founded partial ordering) isnaturally associated a unique ordinal known as the length of C; we call it alsothe order type of C.

2.4. Corollary. The collection of classes of the Boolean hierarchy over L

is almost well ordered with the order type � !.The next corollary describes the di�erence hierarchy in terms of the opera-

tion + on subsets of B de�ned by X + Y = fx4yjx 2 X; y 2 Y g, where x4y isthe symmetric di�erence of elements x; y 2 B. This description is obtained inKSW86.

2.5. Corollary. For any k, Dk+1 = L+ � � �+ L (k+1 times).

Proof. It suÆces to show that Dk+1 = t(L), where t = v04� � �4vk. LetA � P (V ), V = fv0; : : : ; vkg, satis�es t = tA. By the evident relation of thesymmetric di�erence to the addition modulo 2 any set U � V is in A i� thecardinality of U is odd. So A has a 0-chain of length k +1 but does not have a1-chain of length k + 1. By the proof of 2.1 t(L) = Dk+1 completing the proof.

We conclude this section with a natural variation on Theorem 2.1. Note thata correct notation for the "elementary conjunction" tU above is tm;U , and thatany equation (or identity) of Boolean terms is equivalent to several equationsof the form tm;U = 0. Let C = f(m;U)jm < !;U � fv0; : : : ; vmgg; then anyset of equations of Boolean terms from vk(k < !) can be identi�ed with someE � C. Now let t(L;E) be the set of all values of terms t = t(v0; : : : ; vm) forvi = ai 2 L satisfying all equations from E. The proof of the next assertion isomitted because it is similar to that of 1.1.

4

2.6. Theorem. For all L and E as above we have the equality

ft(L;E)jt 2 T;E � Cg = fDk; �Dk; Dk \ �Dkjk < !g:

3 Universal Hierarchy

Here we generalize Corollary 2.2 to a hierarchy introduced in Se83. The originalde�nition was in terms of a generalized ternary jump operation iterating whichstarting from empty set one gets the complete sets for levels of the hierarchy.This is natural and useful for some purposes but not suitable for applications andgeneralizations. After acquaintance with Lo83 we gave in Se89 a set{theoreticalde�nition of our hierarchy which we recall and use here.

We need some ordinal arithmetic as described e.g. in KM67. The supremumof the sequence f!; !!; !!

!

; : : :g of ordinals is denoted by "0. Let Ln = �0n+1

be (n + 1)-th level of the arithmetical hierarchy of subsets of ! and �a = ! n abe the complement of a � !. De�ne classes Sn�(n < !) by induction on � < "0as follows (we often shorten a \ b to ab):

3.1. De�nition. (i) Sn0 = f0g; (ii)Sn! = Sn+1 for > 0;

(iii) Sn�+1 = fu0x0 [ u1x1jui 2 Ln; x0 2 Sn� ; x1 2

�Sn� ; u0u1x0 = u0u1x1g;

(iv) SnÆ+! = fu0x0 [ u1x1 [ �u0�u1yjui 2 Ln; x0 2 SnÆ ; x1 2 �SnÆ ; y 2Sn! ; u0u1x0 = u0u1x1g for Æ = ! � Æ0 > 0; > 0:

To see that this de�nition is correct note that every nonzero ordinal � < "0is uniquely representable (see KM67) in the form � = ! 0+ � � �+! k for a �nitesequence 0 � � � � � k of ordinals < � (we call this representation canonical).Applying 3.1 we subsequently get Sn! 0 ; S

n! 0+! 1 ; : : : ; S

n� .

The classes Sn for n > 0 play a technical role. By (ii) they are among theclasses S0�. Our hierarchy is f��g�<"0 , where �� = S0�. In Se83 and Se89 wehave shown that the classes �� have m-complete sets, are closed downwardsunder �m and are ordered as a hierarchy, i.e. �� 6� �� and �� [ �� � �� for� < � < "0 (we use standard notation from hierarchy theory, i.e. �� denotesthe dual class for �� and �� = �� \ ��, see e.g. Mo80, Hi78).

Note that by (ii) and (iii) �0n+1 = S0�n+1 , where �0 = 0 and �k+1 = !�k .

So our hierarchy is a re�nement of the arithmetical hierarchy. It is easy to see(Se83, Se89 and section 4 here) that fSnk gk<! is the di�erence hierarchy overLn = �0

n+1, so our hierarchy extends also these di�erence hierarchies. As wewill see below it extends also many other natural hierarchies. This explains thename "universal".

To state an analog of 2.2 for the universal hierarchy we need an extendedversion of the Boolean hierarchy. Let T � be the set of �nite Boolean terms withvariables vnk (k; n < !). We call vnk (k < !) variables of type n and elements ofT � typed Boolean terms. Relate to any t 2 T � the set t(L) of all its values whenthe variables of type n range over Ln = �0

n+1; L denotes the sequence fLng.We call the sets t(L) classes of the typed Boolean hierarchy over L. The analogof 2.2 is now as follows:

3.2. Theorem. The classes of the universal hierarchy coincide with the

classes of the typed Boolean hierarchy, i.e. f��;��j� < "0g = ft(L)jt 2 T �g.

5

Proof. To prove the inclusion from left to right it clearly suÆces to �ndtyped Boolean terms tn� satisfying Sn� = tn�(L). De�ne t

n�(n < !) by induction

on � < "0 as follows:tn0 = 0, tn1 = v, tn! = tn+1 for > 0;tn�+1 = v0�v1t

n�(x) [ �v0v1 �t

n�(y) for limit �; tn�+2 = �tn�+1(x)v,

tnÆ+! = v0�v1tnÆ (x) [ �v0v1 �tnÆ (y) [ �v0�v1t

n! (z) for Æ = ! � Æ0 > 0, > 0:

Here v; v0; v1 are variables of type n and x,y,z are sequences of other di�erentvariables of appropriate types.

The equality Sn� = tn�(L) is checked by induction on � according the de�ni-tion of terms, the three �rst points of the de�nition being trivial. Let � = �+1,so by the induction hypothesis Sn� = tn�(L). Let A 2 tn�(L), so

A = U0 �U1A0 [ �U0U1A1 (1)

for some Ui 2 Ln, A0 2 tn�(L), A1 2 �tn�(L). Let Vi 2 Ln be reducing sets for Ui,

i.e. sets satisfying Vi � Ui; V0V1 = ; and V0 [ V1 = U0 [ U1: Then

A = V0B0 [ V1B1 and V0V1 = ;; (2)

where B0 = U0 �U1A0 and B1 = �U0U1A1. By Se89 B0 2 Sn� and B1 2 �Sn� , soA 2 Sn� .

Conversely, let A 2 Sn� , so A =W0B0 [W1B1 and W0W1B0 =W0W1B1 forsome Wi 2 Ln, B0 2 Sn� and B1 2 �Sn� . Let V0; V1 2 Ln be reducing sets forW0;W1, then we have (2). But then (1) is true for Ui = Vi and Ai = Bi, soA 2 tn�(L).

The case of the ordinal � = Æ+! is considered in the same way. Finally, thecase � = � +2 immediately follows by induction from the following descriptionof Sn�+2 obtained in Se83, Se89: this is the class of intersections of Ln-sets with�Sn�+1-sets.

It remains to prove that any t(L) (t 2 T �) is one of ��;��(� < "0). Firstwe show that the class t(L) is closed downwards under m-reducibility and hasan m-complete set.

Let t = t(v00 ; : : : ; v0n; : : : ; v

n0 ; : : : ; v

nn) contain only the listed variables. Let

X 2 t(L) and Y �m X ; we have to show that Y 2 t(L). Choose �0k+1-sets

Xki with X = t(X0

0 ; : : : ; X0n; : : : ; X

n0 ; : : : ; X

nn ) and a recursive function f with

Y = f�1(X). By induction on terms t we get Y = t(f�1(X00 ); : : : ; f

�1(Xnn )).

By a property of the arithmetical hierarchy f�1(Xki ) 2 �0

k+1, so Y 2 t(L).

It remains to show that t(L) has a complete set. Let Ak be anm-complete setin �0

k+1. Let Bki = fha00; : : : ; a

0n; : : : ; a

n0 ; : : : ; a

nnija

ki 2 A

kg andB = t(B00 ; : : : ; B

nn),

where ha00; : : : ; anni is a recursive coding of all (n+ 1)2-tuples. We have Bk

i �m

Ak, so Bki 2 �0

k+1 and B 2 t(L). We claim that B is t(L)-complete, i.e.every t(L)-set X represented as above is m-reducible to B. Choose recur-sive functions fki m-reducing Xk

i to Ak and de�ne a recursive function f byf(a) = hf00 (a); : : : ; f

0n(a); : : : ; f

n0 (a); : : : ; f

nn (a)i:

Then f�1(B) = t(f�1(B00); : : : ; f

�1(Bnn)) = t(X0

0 ; : : : ; Xnn ) = X: So X �m

B and B is m-complete in t(L).

6

We return to the proof of the theorem. Now it suÆces to prove that the setB is m-complete in one of levels of the universal hierarchy. In Se91 we provedsome suÆcient conditions for this which we now apply. The set B is a Booleancombination of sets Bk

i (i; k < !), so by Theorem 1 in Se91 it suÆces to checkthat the class A = fBk

i ji; k � ng is universal and simple. By lemmas 4 and 5in Se91 it suÆces to check that the singletons Ak = fAkg consisting of �0

k+1-complete sets are universal and simple. But universality of such a singleton justmeans that Ak is �0

k+1-complete. Simplicity of such a singleton by de�nitionmeans the following two conditions, in which Ln is the class of all �0

n+1-setsfrom the Boolean algebra generated by Ak (k is �xed):

a) the set ! is not representable as a nontrivial union of L0-sets;b) for any n, every two disjoint �Ln+1-sets are separable by a Boolean com-

bination of Ln-sets.Both properties are checked in an obvious way completing the proof.3.3. Remark. The reader may be not satis�ed with the short but probably

not very clear proof of the theorem. The reason is that the theorem is a directcorollary of the deep Theorem 1 in Se91. In section 5 we give a complete proofof a generalization of 3.2, and in section 4|proofs of all relevant facts on theuniversal hierarchy.

3.4. Corollary. The classes of the typed Boolean hierarchy are almost well

ordered by inclusion with the corresponding ordinal "0.Theorem 3.2 gives probably the shortest and most natural description of the

classes of the universal hierarchy (e.g. de�nition 3.1 seems at �rst glance ad

hoc). Corollary 3.4 seems interesting in its own right. We do not know its proofnot using the universal hierarchy. So this hierarchy serves as a technical toolfor proving this intriguing and not evident result. The corollary shows that theuniversal hierarchy is very natural and may have some interesting applications.For completeness we would like to cite here two examples of such applications.

The universal hierarchy was invented in Se83 for the classi�cation of somesimple sets not classi�able by means of relativised di�erence hierarchies. Thistopic was continued in Se89, Se91, Se91a and Se92, where we showed that manynaturally arising sets are really m-complete in levels of the universal hierarchy.We cite here only one such result which may be of some interest for pure logic.It generalizes Theorem 2 from Se91 and is proved by the same computations, sowe omit the proof here. Let be a �nite language containing a symbol of arity> 1 or at least two unary functional symbols. Let B be the Tarski{Lindenbaumalgebra of sentences of (B is a b.a.) and � be the numeration of B inducedby the G�odel numeration of -sentences.

3.5. Theorem. For any formula '(v0; : : : ; vn) in the language of Boolean

algebras, the set fhx0; : : : ; xnijB j= '(�x0; : : : ; �xn)g is m-complete in one of the

levels ��;��;��+1 (�; � < "0; � is limit), and all the possibilities are realised.

3.6. Corollary. The structure of m-degrees of sets from 3.5 is almost well

ordered with the corresponding ordinal "0.We conclude this section by explaining the main result from Se83 which gives

another application of the universal hierarchy. For a clearer presentation of themain ideas we are here not completely rigorous and formal. By hierarchy we

7

mean here a sequence fSkgk<! of classes closed downwards and having greatestelements under m-reducibility and such that Sk 6� �Sk and Sk [ �Sk � Sk+1 forall k. We say that a hierarchy fSkg re�nes a hierarchy fTkg in level l + 1, ifTl � [k<!Sk � Tl+1\ �Tl+1. A hierarchy is called discrete if it has no re�nementin any nonzero level.

Relate to any limit ordinal � � "0 the natural increasing sequence of ordinalsf�(�; k)gk<! with supremum � (e.g. �("0; 0) = 0 and �("0; k + 1) = !�("0;k)).LetH be the class of hierarchiesH� = f��(�;k)gk<! for all limit ordinals � � "0.Using a suitable jump operation one can extend any of these hierarchies alongthe Kleene ordinal notation system similarly to the case when the arithmeticalhierarchy is extended to the hyperarithmetical hierarchy (see e.g. Ro67). Thenwe have the following reformulation of the main result from Se83.

3.7. Theorem. The class H has the following properties:

(i) the arithmetical hierarchy belongs to H;(ii) if an H-hierarchy H� is not discrete in level l + 1, then the ordinal

� = �(�; l + 1) is limit, the hierarchy H� re�nes H� in level l + 1 and the

trans�nite extention of H� exhausts ��(�;l+1);(iii) any sequence K0;K1; : : : of H-hierarchies, each of which re�nes the pre-

ceding one in some level, is �nite;(iv) the class of all �nite �-levels of H-hierarchies is exactlyf��j� < "0g.The properties (i)|(iii) show that the class H has natural closure properties

from the point of view of hierarchy theory (one could say that H is closedunder the Suslin{Kleene theorem). So it gives in a sense a complete hierarchicalclassi�cation of the arithmetical sets. The property (iv) means that the universalhierarchy consists exactly of �nite levels ofH-hierarchies. This is the reason whywe call it here universal (the word does not mean that we consider this hierarchyas the best one).

Note that in Se83 we treated also a trans�nite version of 3.6, i.e. we found acomplete hierarchical classi�cation of hyperarithmetical sets. This is very natu-ral for hierarchy theory, but we nevertheless think that the �nite version is moreinteresting for the following reasons: a) classes of the trans�nite �ne hierarchiesare not almost linearly ordered (for the case of the di�erence hierarchy this wasshown in Er68); b) dealing with the trans�nite version is technically extremelycomplicated; c) "natural" sets are usually m-complete in a �nite level of anH-hierarchy; d) the trans�nite version is less "absolute" than the �nite one, e.g.it is not generalizable to the abstract case which we consider in the next twosections.

4 Abstract Universal Hierarchy

Here we summarize some ground facts about an abstract version of the universalhierarchy. Most of them are contained in Se89 and Se91, but the presentationthere is sometimes too short and fragmentary, and these papers are not easilyaccessible in the West. In order to make this paper reasonably self{contained

8

and by the recommendation of the referee we try to give here a full presentation.The material here is not trivial and is somewhat tedious.

Let again (B;[;\;�; 0; 1) be a Boolean algebra. By a base (in B) we meanany sequence fLngn<! of sublattices of (B;[;\; 0; 1) satisfying Ln[ �Ln � Ln+1.Note that De�nition 3.1 is applicable to any base L = fLng and gives us classesSn� � B(n < !; � < "0); we call the sequence fS�g�<"0 , where S� = S0�, theuniversal hierarchy over L. Note also that, as in section 2, this hierarchy doesnot indeed depend on B and we may think of the introduced classes as of classesof sets.

The above de�nition may be reformulated in terms of the operation Bisepon subsets of B de�ned by

Bisep(X;Y0; Y1; Y2) = fx0y0[x1y1[ �x0�x1y2jxi 2 X; yj 2 Yj ; x0x1y0 = x0x1y1g:

Namely, Sn0 = f0g, Sn! = Sn+1 , Sn�+1 = Bisep(Ln; Sn� ;

�Sn� ; Sn0 ), and S

nÆ+! =

Bisep(Ln; SnÆ ;

�SnÆ ; Sn! ). Here and below the ordinals satisfy the same conditions

as in 3.1.The operation Bisep generalizes the corresponding operation from Lo83

which is de�ned (in a slightly di�erent from Lo83 notation) by

bisep(X;Y0; Y1; Y2) = fx0y0 [ x1y1 [ �x0�x1y2jxi 2 X; yj 2 Yj ; x0x1 = ;g:

The reason for this generalization is that the operation bisep used in placeof Bisep gives for many bases not rich enough classes. Let as show that forreducible bases (the case considered in Lo83) we get indeed the same classes.Recall that a class A � B is said to have the reduction property if for all a; b 2 Athere are disjoint a�; b� 2 A (called a reducing pair for a; b) such that a� � a,b� � b and a�[b� = a[b. We call a base L reducible if any Ln has the reductionproperty.

4.1. Proposition. Classes of the universal hierarchy over any reducible

base coincide with the corresponding classes obtained by using the operation

bisep in place of Bisep.

Proof. It clearly suÆces to show that if X � B has the reduction propertythen Bisep(X;Y0; Y1; Y2) = bisep(X;Y0; Y1; Y2). The inclusion from right to leftis trivial. Conversely, let y 2 Bisep(X;Y0; Y1; Y2), so y = x0y0 [ x1y1 [ �x0�x1y2and x0x1y0 = x0x1y1 for some xi 2 X; yj 2 Yj . Let (x

�0; x

�1) be a reducing pair

for (x0; x1), then y� = x�0y0[x

�1y1[ �x

�0�x

�1y2 2 bisep(X;Y0; Y1; Y2). But it is easy

to see that y� = y. This completes the proof.Next we want to give an alternative technically important description of the

universal hierarchy. First some notation and terminology. Let !̂! (!̂2) be theset of all �nite strings of numbers (resp. of numbers < 2). By � � � we denotethat a string � is an initial segment of a string � . By j�j we denote the lengthof a string �, by �� (�k)|the concatenation of strings �; � (respectively of astring � and a number k). For k < 2 let �k = 1 � k, and let �� be de�ned by�; = ; and �k = ���k. By 0m we denote the string of m zeros.

By a tree (in B) we mean a sequence fa�g�2!̂2 of elements of B almost allof which are zeros; the elements ~a� = a� n (a�0 [ a�1) are called components of

9

the tree. Such a tree is called: normal, if a; = 1; monotone, if it is normal anda� � a�k; reduced, if it is monotone and a�0a�1 = 0; a �-tree for a base L (where� 2!̂ !), if a� = 0 for j�j > j�j and a�k 2 L�(j�j) for j�j < j�j. We say that anelement a 2 B is de�ned by a tree fa�g, if a � a0 [ a1, aa�0 � a�00 [ a�01 and�aa�1 � a�10 [ a�11.

Note that a tree fa�g de�nes an element a i� the tree fb�g, de�ned byb; = 1 and b�k = a�k, de�nes a (indeed a does not depend on a;). Below wefor simplicity sometimes identify fa�g with the normal tree fb�g. Let us statesome properties of the introduced notions.

4.2. Properties. (i) A normal tree fa�g de�nes an element a i� a = [�~a�1and �a = ~a; [ ([�~a�0).

(ii) A normal tree fa�g de�nes some element i� the components ~a;, ~a�0 are

disjoint from the components ~a�1.(iii) Any reduced tree has pairwise disjoint components and de�nes some

element.

(iv) If fa�g de�nes a, then fa�xg de�nes ax.

(v) If a monotone tree fa�g de�nes a, then fa�0�g de�nes aa�0 and fa�1��gde�nes �aa�1 for any � 2!̂ 2.

(vi) If fa�g de�nes ax, fb�g de�nes ay and a� � x, b� � y, then fa� [ b�gde�nes a(x [ y).

(vii) Let fa�g de�nes a and let b; = 1, b1 = a1[x and b� = a��x for � 6= ;; 1.Then fb�g de�nes a [ x.

(viii) Let fb�g be constructed from fa�g and from a number k as follows:

b�i = a�i [ ([fa� : �i � �; j� j = k +1g) for j�j < k, b�i�j = a�ia�i�j for j�j = k.

If fa�g de�nes a, then fb�g also de�nes a.

(ix) Let a is de�ned by a monotone �-tree fa�g. De�ne a string � and a tree

fb�g by �(k) = �(k+1) and b; = 1; b�k = a0�k

[a1�k

. Then fb�g is a monotone

�-tree and de�nes an element b satisfying a = w�b, where w = a00 [ a01 [ a1.(x) If an element is de�ned by a �-tree, then it is de�ned by a monotone

�-tree.

Proof. (i) Let fa�g de�nes a. Then aa; � a0 [ a1. Intersecting this with�a0�a1 we get a~a; = 0. The same argument shows that a~a�0 = 0 and �a~a�1 = 0.So ~a;; ~a�0 � �a and ~a�1 � a. But it is clear that [�~a� = 1, so we have thedesired equalities a = [�~a�1 and �a = ~a; [ ([�~a�0). The converse computationstates the converse implication.

(ii) and (iii) immedately follow from (i).(iv)|(ix) follow from the de�nitions by evident computations, so let us check

as an example only (ix). It is clear that fb�g is a monotone �-tree satisfying~b�k � ~a0�k [ ~a1�k . By (ii) fb�g de�nes an element b, so it remains to checkthe equality a = w�b, i.e. x 2 a i� x 2 w�b for any x 2 1. Assume �rst thatx 2 b0 [ b1 = [fa� : j�j = 2g. Then x 2 ~a�k and x 2 ~b�n for some � 6= ;; �and k; n < 2. But ~b�n � ~a0�n [ ~a1�n, so x 2 ~am�n for some m < 2. Note thatk = �n (otherwise x 2 ~a�k~am���k contradicting the disjointness of ~a�k and ~am���k).So x 2 a i� x 2 �b. But x 2 w, so x 2 a i� x 2 w�b as desired.

Now assume that x 62 b0 [ b1, then x is in one of the disjoint components ~a;,~a0, ~a1. If x 2 ~a; [ ~a0, then x 62 a [ w. If x 2 ~a1, then x 2 aw and x 2 ~b; � �b.

10

In any case x 2 a i� x 2 w�b.(x) Let fa�g be a �-tree de�ning a. We may think that j�j > 1 because

otherwise the tree fa�g is itself monotone. Let fki : i < j�jg be a permutationof elements 0; : : : ; j�j�1 satisfying �(k0) � �(k1) � � � �. Let fb�g be constructedfrom fa�g and k0 as in (viii). Then fb�g is a �-tree de�ning a and satisfyingb� � b�i � b� for � � �i � � and j�j = k0. Let fc�g be constructed similarlyfrom fb�g and k1. Then fc�g is a �-tree de�ning a and satisfying c� � c�i � c�for � � �i � � , j�j = k0, and c� � c�i � c� for � � �i � � , j�j = k1. Repeatingthe construction j�j times we get a monotone �-tree de�ning a. This completesthe proof of the properties.

Now we can describe the universal hierarchy in terms of trees. Let M�

(R�) be the set of all elements de�ned by monotone �-trees (respectively byreduced �-trees) over a base L. De�ne strings �n�(n < !) by induction on �

as follows: �n0 = ;; �n�+1 = n�n�; �n! = �n+1 for > 0, �nÆ+! = �n! n�

nÆ for

Æ = ! � Æ0 > 0; > 0: Let �� = �n�.4.3. Theorem. For any base we have S� = M�� . For any reducible base

we have S� = R�� .Proof. First let us prove the following auxiliary assertion: if strings �; �

and a number n are such that n < �(i) for i < j�j and n � �(j) for j < j�j,then Bisep(Ln;M� ; �M� ;M�) =M�, where � = �n�.

Let a 2 Bisep(Ln;M� ; �M� ;M�), so a = x0a0 [ x1a1 [ �x0�x1a2 and x0x1a0 =x0x1a1 for some xi 2 Ln; a0 2 M� ; a1 2 �M� ; a2 2 M�. The elements a0; �a1and a2 are de�ned respectively by some monotone �-trees fa�g; fb�g and bya monotone �-tree fc�g. By (iv) the �-trees fa�x0g; fb�x1g and the �-treefc� �x0�x1g de�ne respectively the elements ax0 = x0a0; �ax1 = x1�a1 and a�x0�x1 =�x0�x1a2. De�ne a monotone �-tree fd�g as follows:

d; = 1; d0l = c0l �x0�x1 [ x0 [ x1 for 0 < l � j�j; d� = c��x0�x1 for j�j �j�j; � 6� 0j�j; d0j�jk = xk ; d0j�j0�k = a�kx0, d0j�j1�k = b

�kx1; d� = 0 for

j�j > j�j; � 6� 0j�j.By cases it is not diÆcult to check that fd�g de�nes a, i.e. a 2M�.Conversely, let a be de�ned by a monotone �-tree fd�g. By (v) �-trees fd�0�g

and fd�1��g de�ne respectively the elements ad�0 and �ad�1 for all � of length j�j.Let xk = [fd�k : j�j = j�jg for k < 2. By (vi) ax0 and �ax1 are de�nedrespectively by �-trees fa�g and fb�g, where a� = [fd�0� : j�j = j�jg andb� = [fd�1�� : j�j = j�jg. So a0 = ax0 2 M� and a1 = �ax1 2 �M� . By (iv) the�-tree fd��x0�x1g de�nes a2 = a�x0�x1, so a2 2M�. But a = x0a0 [ x1a1 [ �x0�x1a2and x0x1a0 = x0x1a1, so a 2 Bisep(Ln;M� ; �M� ;M�) completing the proof ofthe auxiliary assertion.

To prove the �rst assertion of the theorem it suÆces to check by inductionon � that Sn� =M�n

�for all n. For � = 0; ! this is evident. For � = � + 1 we

have by the inductive hypothesis and by the auxiliary assertion

Sn� = Bisep(Ln; Sn� ;

�Sn� ; Sn0 ) = Bisep(Ln;M�n

�

; �M�n�

;M;) =Wn�n�

=M�n�:

The case � = Æ + ! is considered in the same way.

11

For a reducible base we note that the proof of the auxiliary assertion worksif we replace Bisep by bisep and M by R. Taking into account Proposition 4.1we see that the proof from the preceding paragraph works also for a reduciblebase. This completes the proof of the theorem.

Now we state some technical properties of the classes of the universal hier-archy. In these properties � denotes any limit ordinal < "0 and other ordinalssatisfy (if other conditions are not formulated explicitly) conditions from Def-inition 3.1. We use the following notation: X � Y = fxyjx 2 X; y 2 Y g,X + Y = fx4yjx 2 X; y 2 Y g and ~X = X \ �X , where x4y is the symmetricdi�erence of x; y.

4.4. Properties. (i) We have �Sn�+1 = Bisep(Ln; Sn� ;

�Sn� ;�Sn0 ) and

�SnÆ+! =

Bisep(Ln; SnÆ ;

�SnÆ ;�Sn! ).

(ii) The class Sn� (respectively �Sn1+�; Sn� ;

�Sn�) is closed under intersection

with any element from Ln (respectively from ~Ln; ~Ln+1; ~Ln+1).(iii) Sn�+2 =

�Sn�+1 � Ln and fSnk gk<! is the di�erence hierarchy over Ln.

(iv) If x0; x1 2 Ln (respectively ~Ln, Ln) and ax0; ax1 2 Sn� (respectively �Sn�,

�Sn� ), then a(x0 [ x1) 2 Sn� (respectively �Sn�,

�Sn� ).(v) SnÆ+! (respectively �SnÆ+! ) is the class of all elements a such that ax0 2

SnÆ , ax1 2�SnÆ and a�x0�x1 2 S

n! (respectively a�x0�x1 2 �Sn! ) for some x0; x1 2 Ln.

(vi) Bisep(Ln; SnÆ+! ; S

nÆ+! ; S

n! ) = Sn! (here Æ is any ordinal of the form

! � Æ0).(vii) Sn� [

�Sn� � Sn� for � < � < "0.

(viii) a 2 SnÆ+! �2 i� ax 2 �SnÆ+! and a�x 2 Sn! for some x 2 Ln (here Æ is

any ordinal of the form ! � Æ0).(ix) Sn! (k+1) = Sn! + Snk .

(x) Let b; c be elements of the b.a. generated by Ln, let b [ c = 1 and

ab; ac 2 Sn! [�Sn! . Then a 2 S

n! �k for some k < !.

(xi) If fLng and fL0ng are bases in Boolean algebras B and B0 respectively

and g : B ! B0 is a homomorphism satisfying g(Ln) � L0n for all n < !, then

g(S�) � S0� for all � < "0.

Proof. (i) The condition y = x0y0 [ x1y1 [ �x0�x1y2 ^ x0x1y0 = x0x1y1 isclearly equivalent to the condition

8t 2 x0(t 2 y $ t 2 y0) ^ 8t 2 x1(t 2 y $ t 2 y1) ^ 8t 2 �x0�x1(t 2 y $ t 2 y2):

Taking into account that Bisep(X;Y0; Y1; Y2) = Bisep(X;Y1; Y0; Y2) we imme-diately get the desired equalities.

(ii) Let a 2 Sn� and x 2 Ln; we have to show that ax 2 Sn� . By 4.3 a isde�ned by a monotone �n�-tree fa�g. By 4.2.(iv) fa�xg de�nes ax. By de�nitionof the string �n� any its component is � n, so fa�xg is a �

n�-tree. By 4.3 ax 2 S

n� .

Now we check that �Sn1+� �~Ln � �Sn1+�. It suÆces to prove the dual assertion

that a [ x 2 Sn1+� for a 2 Sn1+� and x 2 ~Ln. By 4.3 a is de�ned by a monotone�n1+�-tree fa�g. Let fb�g be the tree from 4.2.(vii), so fb�g de�nes a [ x. Butfb�g is a �

n1+�-tree, because 1 + � > 0. So a [ x 2 Sn1+�.

The remaining inclusions Sn� �~Ln+1 � Sn� and �Sn� �

~Ln+1 � �Sn� are provedby induction on �. For � = ! they follow from the two preceding paragraphs

12

because Sn� = Sn+1 and > 0. Let � = Æ + ! , a 2 Sn� and x 2 ~Ln+1; we have

to prove that ax 2 Sn� (the case of �Sn� is considered similarly using (i)). We havea = x0a0 [ x1a1 [ �x0�x1a2 and x0x1a0 = x0x1a1 for some a0 2 S

nÆ , a1 2

�SnÆ anda2 2 S

n! . Then ax = x0(a0x)[x1(a1x)[�x0�x1(a2x) and x0x1(a0x) = x0x1(a1x).

By the inductive hypothesis a0x 2 SnÆ , a1x 2

�SnÆ and a2x 2 Sn! , so ax 2 S

n� .

(iii) Note that �Sn� � Sn�+1 (for any a 2�Sn� we have a = 0�a [ 1a 2 Sn�+1). So

by (ii) �Sn�+1 � Ln � Sn�+2 � Ln = Sn�+2. Conversely, let a 2 Sn�+2, so by 4.3 a is

de�ned by a monotone �n�+2-tree fa�g. By 4.2.(ix) a = w�b for w = a00[a01[a1and for an element b de�ned by a monotone �n�+1-tree (so b 2 Sn�+1). But

�n�+2 = nn�n�, so w 2 Ln and a 2 �Sn�+1 � Ln.By de�nition we have Sn0 = f0g and Sn1 = Ln, so by the preceding paragraph

Snk+1 = �Snk � Ln for any k < !. In section 2 we noted that for the di�erence

hierarchy fDkgk<! over Ln we also have D0 = f0g and Dk+1 = �Dk � Ln. SoSnk = Dk by induction on k < !.

(iv) Let ax0; ax1 2 Sn� , then they are de�ned by some �n�-trees fa�g andfb�g respectively. We may assume that a� � x0 and b� � x1 (otherwise replacea� by a�x0 and use 4.2.(iv)). By 4.2.(vi) the �

n�-tree fa�[b�g de�nes a(x0[x1),

so a(x0 [ x1) 2 Sn� .

Now let xi 2 ~Ln and axi 2 �Sn�. Then �a[�xi 2 Sn�, so by (ii) �axi = (�a[�xi)xi 2

Sn� . By the preceding paragraph �a(x0[x1) 2 Sn� , so a[�x0�x1 2

�Sn� . Again by (ii)a(x0 [x1) = (a[ �x0�x1)(x0 [x1) 2 �Sn� (we may assume � > 0 because otherwisethe assertion is trivial). The last assertion is checked in the same way.

(v) Let a 2 SnÆ+! , so a = x0a0 [ x1a1 [ �x0�x1 for suitable ai; xi as usual.

Then by (ii) ax0 = x0a0 2 SnÆ , ax1 = x1a1 2 �SnÆ and a�x0�x1 = �x0�x1a2 2 S

n! as

desired. Conversely, let we have the last relations. Then for a0 = ax0, a1 = ax1and a2 = a�x0�x1 we have a = x0a0 [ x1a1 [ �x0�x1 and x0x1a0 = x0x1a1, soa 2 SnÆ+! . For

�SnÆ+! the argument is the same.(vi) Let a 2 Bisep(Ln; S

nÆ+! ; S

nÆ+! ; S

n! ), then as in (v) for some x0; x1 2

Ln we have axi 2 SnÆ+! and a�x0�x1 2 Sn! . For Æ = 0 we deduce from (iv)that a 2 Sn! as desired. Now let Æ 6= 0. For x = x0 [ x1 we have a�x 2 Sn! and ax 2 SnÆ+! by (iv). By (v) there are u0; u1 2 Ln satisfying axu0 2 SnÆ ,

axu1 2 �SnÆ and ax�u0�u1 2 Sn! . We have

a = ax [ a�x = axu0 [ axu1 [ ax�u0�u1 [ a�x = a(xu0) [ a(xu1) [ a(x�u0�u1 [ �x)

and by (iv) a(x�u0�u1 [ �x) 2 Sn! . But x�u0�u1 [ �x = xu0xu1, so for the elementsvi = xui we have av0 2 S

nÆ , av1 2

�SnÆ and a�v0�v1 2 Sn! . By (v) a 2 S

nÆ+! .

(vii) Note that if sets Yi � B are nonempty then Yi � Bisep(Ln; Y0; Y1; Y2)(e.g. for yi 2 Yi we have y0 = 1y0 [ 0y1 [ �1�0y2 2 Bisep(Ln; Y0; Y1; Y2)). Thisremark together with the de�nition of Sn� shows that 0; 1 2 Sn� for � > 0. In

the proof of (iii) we noted that �Sn� � Sn�+1; a similar argument shows that

Sn� [�Sn� � Sn�+1 and S

nÆ [

�SnÆ � SnÆ+! .The needed inclusion is checked by induction on �. For � = 0 the inclusion

follows from the preceding paragraph because Sn0 = f0g. So assume � > 0 andconsider the canonical representations � = !�0 + � � �+!�k , � > �0 � : : : � �k,

13

and � = !�0 + � � � + !�l , � > �0 � : : : � �l. It is known (see e.g. KM67)that � < � i� (�0; : : : ; �k) is lexicographically less than (�0; : : : ; �l). So either(�0; : : : ; �k) is a proper initial segment of (�0; : : : ; �l) or there is a numberi � k; l such that �i < �i and �j = �j for j < i. For the �rst alternativethe inclusion Sn� [

�Sn� � Sn� follows from the preceding paragraph, so assumethe second alternative. We have Sn

!�0+���+!�i� Sn� , so we may assume that

i = l. From the inductive hypothesis we easily get Sn!�l [�Sn!�l � Sn

!�l, because

Sn!�l

= Sn+1�l. Let Æ = !�0 + � � � + !�l�1 for l > 0 and Æ = 0 otherwise. Then

SnÆ [�SnÆ � Sn� , so by (vi) and (i)

SnÆ+!�l [�SnÆ+!�l � Bisep(Ln; S

nÆ ;

�SnÆ ; Sn!�l

) � Bisep(Ln; Sn� ; S

n� ; S

n!�l

) = Sn� :

In the case l = k the proof is �nished, otherwise just repeate the last argumentk � l times.

(viii) First assume Æ = 0. Let a�x 2 Sn! and ax 2 �Sn! for some x 2 Ln.For x0 = 0 and x1 = x we have ax0 2 Sn! , ax1 2

�Sn! and a�x0�x1 2 Sn! , soa 2 Sn! �2 by (v). Conversely, if a 2 Sn! �2, then there are x0; x1 2 Ln with thelast properties. By (ii) we have ax0�x1 2 S

n! , so by (iv) a�x1 2 S

n! and we can

take x = x1.For the case Æ > 0 the �rst part is proved in the same way as for Æ = 0. Now

let a 2 SnÆ+! �2, so for some x0; x1 2 Ln we have ax0 2 SnÆ+! , ax1 2�SnÆ+!

and a�x0�x1 2 Sn! . For the element ax0 there are y0; y1 2 Ln with the properties

described in (v); then for y = y0 [ y1 we have ax0y 2 �SnÆ+! and ax0�y 2 Sn! .

Let x = x1 [ x0y, then ax 2 �SnÆ+! by (iv) and a�x = a(�x1�x0 [ �x1x0�y) 2 Sn! by

(ii) and (iv).(ix) is proved by induction on k the case k = 0 being trivial. Let k = l + 1

and assume �rst a 2 Sn! + Snk , so a = b4c for some b 2 Sn! and c 2 Snk .By (iii) c = �c1x for some c1 2 Snl and x 2 Ln, so a = b4(�c1x). We havea�x = b�x 2 Sn! and ax = (b4�c1)x. By the induction hypothesis b4�c1 2 �Sn! �k,so ax 2 �Sn! �k and by (viii) a 2 Sn

! (k+1). Conversely, let a 2 Sn! (k+1), then by

(viii) ax 2 �Sn! �k and a�x 2 Sn! for some x 2 Ln. By the induction hypothesisax = b14�c1 for some b1 2 Sn! and c1 2 Snl . Let b = a�x [ b1x and c = �c1x,then b 2 Sn! and c 2 Snk as above. We have (b4c)x = (b14�c1)x = ax and(b4c)�x = a�x, so a = b4c 2 Sn! + Snk .

(x) The elements b; c are in the subalgebra generated by �nitely many el-ements d0; : : : ; dm 2 Ln. As in section 2 let tU (U � V ) be the "elementaryconjunctions" from variables V = fv0; : : : ; vmg. Let < be a linear ordering onP (V ) extending the relation �, so P (V ) = fU0 < � � � < Uk�1g, k = 2m+1. LetAk � � � � � A0 be all subsets of P (V ) closed upwards under <, so e.g. Ak = ;,Ak�1 = fUk�1g and Ak�2

�Ak�1 = fUk�2g. This shows that the elements b; care unions of several elements of the form ai�ai+1, where ai = tAi

(d0; : : : ; dm).We clearly have a0 � � � � � ak and ak = 0. Note that the sets Ai are closedupwards under �, so by the proof of 2.1 ai 2 Ln for i � k. Let d0 = a�a0,d1 = aa0�a1; : : : ; dk = aak�1�ak. Any ai�ai+1 is contained in one of b; c, so by (ii)di 2 Sn! [

�Sn! for all i. We may assume (possibly doubling some ai and so

14

increasing k) that d2i 2 Sn! and d2i+1 2 �Sn! . Let d = d0 [ �d1 [ d2 [ �d3 [ � � �and e = [i(a2i�a2i+1), then e 2 Snk and by (iv) d 2 Sn! . But a = d4e, soa 2 Sn

! (k+1) by (ix).

(xi) Let a 2 S�, then by 4.3 a is de�ned by a ��-tree fa�g. Then g(a)is de�ned by the ��-tree fg(a�)g. So g(a) 2 S� completing the proof of theproperties.

Now we answer a natural question on on the classes from 4.3. We knowthat any class S� is among classes M� (� 2

!̂ !). It turns out that the converseinclusion is also true.

4.5. Theorem. For any base L, the collections fM�j� 2!̂ !g and fS�j� <

"0g coincide.

Proof. It suÆces to show by induction on j�j that if m � �(i) for i < j�j,then M� = Sm� for some � < "0. For � = ; take � = 0. Otherwise represent� as in the proof of 4.3, � = �n�, where n is the least member of the string�. It suÆces to check that M� = Sn� for some � (for m < n this will followfrom 3.1.(ii)). By the induction hypothesis M� = Sn+1 and M� = SnÆ for

some ; Æ < "0. We have M� = Bisep(Ln;M� ; �M� ;M�), so it suÆces to checkthat M� = Bisep(Ln; S

nÆ ;

�SnÆ ; Sn+1 ) coincides with SnÆ+! (then we could take

� = Æ + ! ). For = 0 this is true by De�nition 3.1. For > 0 we haveSn+1 = Sn! , but we cannot immediately apply De�nition 3.1, because theordinal Æ may not satisfy the condition Æ = ! � Æ0 > 0. Assume �rst thatÆ < ! , so Æ + ! = ! . Then we have Sn! �M� by the proof of 4.4.(vii), andM� � Bisep(Ln; S

n! ; S

n! ; S

n! ) = Sn! by 4.4.(vii) and 4.4.(vi). Now let Æ �

and Æ = ! 0 + � � �+ ! k be the canonical representation, then 0 � . Let i bethe greatest number with i � , and let Æ1 = ! 0+ � � �+! i , Æ2 = Æ�Æ1. ThenÆ2 < ! , so Æ+! = Æ1+! and Æ1 is a nonzero ordinal of the form ! � Æ0. By4.4.(vi) and 4.4.(vii) for � = Æ + ! we have

Sn� � Bisep(Ln; SnÆ1; �SnÆ1 ; S

n! ) �M� � Bisep(Ln; S

n� ; S

n�; S

n! ) = Sn�

completing the proof of the theorem.Next we show that the universal hierarchy over some bases is perfect, which

by de�nition from Ad65 means that for any � < "0 the class ~S� = S� \ �S�coincides with the class

S�<�(S� [

�S�). We call a base L interpolable (the termis justi�ed by a relation to a version of the Craig Interpolation Theorem) if forall n < ! any two disjoint elements a; b 2 �Ln+1 are separable by an element cof the b.a. (Ln) generated by Ln (i.e. a � c � �b). We call a base L perfect if itis interpolable and the element 1 is join irreducible in the lattice L0.

4.6. Theorem. The abstract universal hierarchy over any perfect base is

perfect.

Proof. First we prove that the universal hierarchy over an interpolable baseis perfect in all limit levels, namely we check by induction on the limit ordinal� < "0 that ~Sn� =

S�<�(S

n� [

�Sn� ). Let �rst � = !, then Sn� = Sn+11 = Ln+1.

So we have to prove that a 2 [k<!Snk for a 2 ~Ln+1. Let a; �a 2 �Ln+1, then the

elements a; �a are separable by an element c 2 (Ln). But then a = b 2 [k<!Snk

by 4.4.(iii) and by a property of the di�erence hierarchy over Ln.

15

Now let � = ! +1 = ! � ! for some > 0, so Sn� = Sn+1 +1 and we have to

show that ~Sn+1 +1 � [k<!Sn! �k. Let a; �a 2 S

n+1 +1 , so

a = u0a0 [ u1a1; u0u1a0 = u0u1a1; �a = v0b0 [ v1b1; v0v1b0 = v0v1b1 (3)

for some ui; vi 2 Ln+1, a0; b0 2 Sn+1 and a1; b1 2 �Sn+1 . Then u0[u1[v0[v1 = 1

and a fortiori �u0�u1�v0�v1 = 0. From the separability condition follows that forsome ci; di 2 (Ln) we have �ui � �ci �vi � �di and �c0�c1 �d0 �d1 = 0, so ci � ui, di � viand c0 [ c1 [ d0 [ d1 = 1. Then a = ac0 [ ac1 [ ad0 [ ad1 and by 4.4.(ii)ac0 = au0c0 = a0c0 2 Sn+1 . A similar argument shows that �ad0 2 Sn+1 and

ac1; �ad1 2 �Sn+1 . From �ad0 2 Sn+1 we get ad0 = (a[ �d0)d0 2 �Sn+1 and similarly

ad1 2 Sn+1 .

We have proved that all aci; adi belong to Sn+1 [ �Sn+1 = Sn! [�Sn! . Let

e0; : : : ; em be a �nite sequence of pairwise disjoint elements from (Ln) whosejoin is 1 and every of which is contained in one of ci; di. Then a = ae0[� � �[aemand aej 2 S

n! [

�Sn! for all j � m. By 4.4.(x) a 2 Sn! �k for some k < !.If � = !�1 for a limit ordinal �1, then the assertion is true by the induction

hypothesis.Finally, let � = Æ+! and a; �a 2 Sn� ; we have to show that a 2 Sn� for some

� < �. Let ui; vi 2 Ln satisfy au0; �av0 2 SnÆ , au1; �av1 2

�SnÆ , and a�u0�u1; �a�v0�v1 2Sn! . As above we have av0 2 �SnÆ and av1 2 SnÆ , so by 4.4.(iv) aw0 2 SnÆ ,aw1 2 �SnÆ and a �w0 �w 2 Sn! , where w0 = u0[v1 and w1 = u1[v0. Replacing a by

�a, we similarly get �a �w0 �w 2 Sn! , so a �w0 �w 2 �Sn! . We checked that a �w0 �w 2 ~Sn! ,so by the induction hypothesis a �w0 �w 2 Sn� for some � < ! . Of course we maythink that � = ! 1 � k for some 1 < ; k < !. From 4.4.(viii) follows thata 2 SnÆ+� completing the proof of the perfectness in the limit levels.

It remaines to show that ~S�+1 � S� [ �S� for every �. Let a; �a 2 S�+1,so we have (3) for some ui; vi 2 L0; a0; b0 2 S� and a1; b1 2 �S�. We haveu0 [ u1 [ v0 [ v1 = 1, so by the join irreducibility condition one of ui; vi is 1.Assume e.g. that u0 = 1. Then a = au0 = u0a0 = a0 2 S� [ �S� completing theproof of the theorem.

We conclude this section by a result on a structural property of the classes ofthe universal hierarchy. Recall that a class A � B is said to have the separationproperty if any disjoint elements a; b 2 A are separable by an element c 2 ~A =A\ �A. It is well known that if A has the reduction property then the dual class�A has the separation property. It is also known that many levels of naturalhierarchies have one of these properties. Let us show that in some cases theclasses of the abstract universal hierarchy have the separation property.

4.7. Theorem. For any reducible base, every class �S� has the separation

property.

Proof. By induction on � we check that the class �Sn� has the separationproperty. Cases � = 0 and � = ! > 1 are trivial. Let � = Æ + ! anda; b 2 �Sn� ; ab = 0. By 4.1 there are ui; vi 2 Ln such that

au0; bv0 2 SnÆ ; au1; bv1 2

�SnÆ ; a�u0�u1; b�v0�v1 2�Sn! ; u0u1 = v0v1 = 0:

Let u�; v� 2 Ln be a reducing pair for u = u0 [ u1 and v = v0 [ v1 and letu�i = u�ui, v

�i = v�vi. By the induction hypothesis there is an element c 2 ~Sn!

16

separating a�u from b�v. We claim that the element d = au�0[au�1[

�bv�0[�bv�1[c�u�v

is in ~Sn� and separates a from b. We have au�0 = au0u�0 and �bv�1 = bv1v

�1 , so

by 4.4.(ii) au�0;�bv�1 2 SnÆ . For d0 = au�0 [

�bv�1 we have d0u�0 = au�0 2 SnÆ and

d0v�1 =

�bv�1 2 SnÆ , so d0 2 S

nÆ by 4.4.(iv). Similarly we have d1 = au�1[

�bv�0 2�SnÆ .

We have also d = w0d0 [ w1d1 [ �w0 �w1c, where w0 = u�0 [ v�1 and w1 = u�1 [ v

�0 .

By 4.4.(i) d 2 ~Sn� . It remains to show that a � d � �b, and this is done by aneasy set{theoretic argument which we do not write down.

Now let � = � + 1 and again a; b 2 �Sn� ; ab = 0. Then

a = u0a0 [ u1a1 [ �u0�u1; b = v0b0 [ v1b1 [ �v0�v1; u0u1 = v0v1 = 0

for some ui; vi 2 Ln; a0; b0 2 Sn� and a1; b1 2 �Sn� . We have �a [ �b = 1, �a =

u0�a0 [ u1�a1 and �b = v0�b0 [ v1�b1; so u0 [ u1 [ v0 [ v1 = 1. De�ne elementsu; v; u�; v�; u�i ; v

�i as above. The elements u�0; u

�1; v

�0 ; v

�1 are pairwise disjoint,

belong to ~Ln and their union is 1. So by 4.4.(iv) c = a0u�0 [ a1u

�1 [

�bv�0 [�bv�1 is

in ~Sn� . Again an easy set{theoretic argument shows that a � c � �b completingthe proof.

From a result in VW76 follows that the above result cannot be strengthendto the assertion that every S� has the reduction property.

5 Abstract Hierarchy and Boolean Terms

Here we prove an abstract version of 3.2 applicable to some natural hierarchies,and discuss some relevant results. The next theorem is probably the main resultof this paper. By classes of the typed Boolean hierarchy over a base L we meansets t(L) = ft(ank )ja

nk 2 Lng of all values of t = t(vnk ) 2 T � when varibles

vnk (k < !) of type n range over Ln.5.1. Theorem. For any reducible base L, the classes of the universal and of

the typed Boolean hierarchies over L coincide, i.e. fS�; �S�j� < "0g = ft(L)jt 2T �g:

Proof. The inclusion from left to right is checked just as in the proof ofTheorem 3.2, so we have only to prove that any t(L) is one of S�; �S�. Let m < !

be such that the term t contains only variables from V = fvnk jn; k � mg. Notethat any set U � V is uniquely representable in the form U = U0 [ � � � [ Um,where Un is the set of all variables of type n from U . De�ne preorderings�n (n < !) on the class P (V ) of all subsets of V as follows: R �n U i� Ri = U i

for i < n and Rn � Un. Let Ln be the class of all subsets of P (V ) closedupwards under �n, and L = fLngn<!. Note that for n > m the relation �n

coincides with equality and a fortiori Ln = P (P (V )). Let us prove �rst fourtechnical assertions.

a) L is a perfect base and the universal hierarchy fS�g over L is perfect.We need to prove only the �rst assertion because the second is a corollary

of 4.6. The class Ln is clearly closed under union and intersection and contains; and P (V ). The inclusion Ln [ �Ln � Ln+1 follows from the evident remarkthat R �n+1 U implies R �n U and U �n R. To check that P (V ) is notrepresentable as a nontrivial union of L0-sets assume that P (V ) = X [ Y for

17

some X;Y 2 L0. Then ; is in one of X;Y . But ; �0 U for every U � V , so oneof X;Y is P (V ).

It remains to show that any two disjoint sets X;Y 2 �Ln+1 are separable bya set from (Ln). Note that for all R 2 X;U 2 Y there is i � n with Ri 6= U i

(suppose not: R0 [ � � � [Rn = U0 [ � � � [Un = T for some R 2 X;U 2 Y ; thenT �n+1 R;U and both X;Y are downward closed under �n+1, so T 2 X \ Ycontradicting the disjointness of X;Y ). Now let Z be the class of all P � V suchthat R0 [ � � � [ Rn = P 0 [ � � � [ Pn for some R 2 X . Then Z clearly separatesX from Y . For any �xed R, the class fP jR0 [ � � � [Rn = P 0 [ � � � [ Png is thedi�erence of Ln-classes fP jR

0 [ � � � [ Rn � Pg and fP jR0 [ � � � [ Rn � Pg, soZ 2 (Ln) completing the proof of a).

b) For all ank 2 Ln, the function X 7! tX(ank ) is a homomorphism of b.a.

P (P (V )) in b.a. B and tX(ank ) 2 Lp for X 2 Lp.

The �rst assertion is clear, because tX = [ftU jU 2 Xg and from the def-inition of tU we immediately get tX[Y (a

nk ) = tX(a

nk ) [ tY (a

nk ) and similarly

for \;�; 0; 1. Now let X 2 Lp. Then X is a union of classes of the formY = fR � V jU �p Rg for U 2 X . So it suÆces to show that tY (a

nk ) 2 Lp.

By de�nition of �p, Y is the di�erence of classes Z = fRjU0 [ � � � [ Up � Rgand T = fRjU0 [ � � � [ Up�1 � Rg. But as we noted in the proof of 2.1,tZ(a

nk ) = \fank jv

nk 2 U0 [ � � � [ Upg 2 Lp. We have also T 2 Lp�1 (or T = ;

for p = 0), so by induction on p we may assume that tT (ank ) 2 Lp�1 � �Lp. So

tY (ank ) = tZ(a

nk ) n tT (a

nk ) 2 Lp completing the proof of b).

Let A � P (V ) and � 2!̂ !. By a �-aternating tree for A we mean asequence fU� : � 2!̂ 2; j�j � j�jg of subsets of V such that U; 62 A andU�0 62 A;U�1 2 A;U� ��(j�j) U�k for j�j < j�j and k < 2. Let �� be the stringfrom theorem 4.3.

c) For any A � P (V ); A 62 �S�, there is a ��-alternating tree for A.

Let us prove the contraposition: if there is no ��-alternating tree for A, thenA 2 �S�. By 4.3 it suÆcies to prove that for all � 2

!̂ ! if there is no �-alternatingtree for A, then A 2 �M�. This is proved by induction on j�j. Let j�j = 0, then�-alternating tree for A has the form fU;g, for U; 62 A. There is no such a tree,so A = P (V ) 2 �M�.

Let j�j > 0, then � = �n� for some �; � 2!̂ !; n 2 !, with 8i < j�j(n < �(i))and 8i < j�j(n � �(i)). Let B0 (respectively B1) be the set of all U � V suchthat there is no �-alternating tree fV"g for �A (respectively for A) satisfyingU �n V;. It is clear that B0; B1 2 Ln. Note that there are no �-alternating

trees for AB0 and �AB1. (Suppose e.g. that fV"g is a �-alternating tree for AB0.Then V; 2 AB0. But V; �n V" and B0 2 Ln, so V" 2 B0 for all " of length� j�j. So fV"g is a �-alternating tree for �A. By de�nition of B0 then V; 62 B0,a contradiction.)

Now let us check that there is no �-alternating tree fWÆg for A [ B0 [ B1.Suppose the contrary, then W; 2 �B0

�B1�A. By the choice of � and n we have

W; �n+1 WÆ , so W; �n WÆ . Then WÆ 2 �B0�B1 for all Æ of length � j�j and a

fortiori fWÆg is a �-alternating tree for A. For every Æ of length j�j we haveWÆ 2 �B0

�B1, so by the de�nition of B0; B1 there are �-alternating trees fP Æ" g

18

for A and fQÆ"g for

�A such that WÆ �n PÆ; and WÆ �n Q

Æ;. Let U� = W� for

� of length � j�j and UÆ0" = P Æ" , UÆ1" = QÆ

" for Æ of length j�j and " of length� j�j. Then fU�g is a �-alternating tree for A, a contradiction.

We have proved that there are no �-alternating trees for A[B0 [B1 and �-

alternating trees for AB0 and �AB1. By the induction hypothesis A[B0 [B1 2�M� and AB0; �AB1 2 �M� . Then B0; B1 2 Ln, A0 = AB0 2 �M� , A1 = �AB1 2M� , A2 = �A �B0

�B1 2M� and �A = B0A0[B1A1[ �B0�B1A2, B0B1A0 = B0B1A1.

By the proof of 4.3 �A 2M� completing the proof of c).d) Let fU� : j�j � j�jg be a �-alternating tree (for some A) and fb�g be

a monotone �-tree over L (see 4.2). Then for any n; k < ! the element ank =

[f~b�jvnk 2 U�g belongs to Ln.

It suÆces to show that if vnk 2 U�, then ~b� � c � ank for some c 2 Ln. Let�0 be the least string (with respect to �) such that �0 � � and �(j� j) > n

for all � satisfying �0 � � � �. Then vnk 2 U�0 (for �0 = � this is trivial,otherwise U�0 �n+1 U� and vnk 2 U�, so v

nk 2 U�0 by the de�nition of �n+1).

By minimality of �0 we have �0 = ; or �(j�0j � 1) � n; in any case b�0 2 Ln.Let m be the least number i with j�j � i < j�j and �(i) < n, or m = j�j,

if there is no such an i. For any string � of length m + 1 we have eitherb� 2 L�(m) or b� = 0 (for m = j�j); in any case �b� 2 Ln. So the element

c = b�0 n [fb� : � � �0; j� j = m+ 1g belongs to Ln. It is clear also that ~b� � c,so it remains to check that c � ank . Note that v

nk 2 U� for all � � �0 of length

� m (because vnk 2 U�0 �n U�); so ~b� � ank for all such �. But c is the union of

all such components ~b�, so c � ank completing the proof of d).Let us return to the proof of the theorem and show that any t(L) is one of

S�; �S�. Let A � P (V ) satisfy t = tA. Before a) we noted that Lm+1 = P (P (V )),so A 2 Lm+1 = Sm+1

1 and by 3.1.(ii) A 2 S� [ �S� for some � < "0. Choosethe least such �. By a) A is in one of the classes S� n �S�; �S� n S�. It suÆcesto show that for the �rst alternative t(L) = S�, because then for the secondalternative we dually have t(L) = �S�. So assume A 2 S� n �S�. By b) and4.4.(xi) t(ank ) 2 S� for ank 2 Ln, i.e. t(L) � S�.

It remains to check the inverse inclusion. Let b 2 S�, then b 2 R�� by4.3. Let fb�g be a reduced ��-tree de�ning b. We have to show that b 2 t(L),i.e. b = t(ank ) for some a

nk 2 Ln. Let ank = [f~b�jv

nk 2 U�g. By d) ank 2 Ln,

so t(ank ) 2 t(L). So it suÆces to check that b = tA(ank ) (then b 2 t(L) and

S� � t(L)).Note that if x 2 ~b� then vnk 2 U� $ x 2 ank (implication from left to

right is by the de�nition of ank ; conversely, if x 2 ank , then x 2 ~b� for some

� with vnk 2 U� ; but by 4.2.(iii) the components ~b� are pairwise disjoint, so� = � and vnk 2 U�). From the de�nition of the terms tU� in the proof of

Theorem 2.1 it follows that x 2 tU� (ank ) for x 2

~b� , i.e. ~b� � tU� (ank ). But then

b = [�~b�1 � [�tU�1(ank ) � tA(a

nk ) and

�b = ~b; [ ([�~b�) � tU;(ank ) [ ([�tU�0(a

nk )) � t �A(a

nk ) = �tA(a

nk ):

So b = tA(ank ) completing the proof of the theorem.

19

Note that 5.1 generalizes 3.2 in the same way as 2.1 generalizes 2.2. Theorem5.1 is an analog of 2.1. There is also a natural analog of 2.6. For some questionsit is interesting to classify sets t(L;E) for t 2 T � and a �xed set E of equalitiesof the typed Boolean terms. This leads to subtle generalizations of 5.1, some ofwhich are implicit in Se91. We formulate an example of this. The proof is thesame as that of 5.1. Recall from section 2 that the set E is in a natural wayidenti�ed with a subset of P (V ). Let Ln be the class of all subsets A � P (V )nEclosed upwards under �n. Let LE = fLng be the base formed by the latticesLn.

5.2. Corollary. If the base LE is perfect and the base L is reducible, then

any class t(L;E) coincides with one of the classes of the universal hierarchy

over L.

As in section 2 we immediately get5.3. Corollary. The classes of the typed Boolean hierarchy over any re-

ducible base are almost well ordered with the corresponding ordinal � "0.Another immediate corollary of the proof of 5.1 is5.4. Corollary. Inclusions between classes of hierarchies in 5.1 are ef-

fective, i.e. there are algorithms computing from a given level of the universal

hierarchy the corresponding level of typed Boolean hierarchy and vice versa.

The next corollary states a natural closure property of the universal hierar-chy. Relate to any (untyped) Boolean term s = s(v0; : : : ; vn�1) 2 T an n-aryoperation Fs on subsets of b.a. B as follows:

Fs(A0; : : : ; An�1) = fs(a0; : : : ; an�1)jai 2 Aig:

E.g. for s = v0 \ v1 we have Fs(A0; A1) = fa0 \ a1jai 2 Aig.5.5. Corollary. The collection of classes of the universal hierarchy over

any reducible base L is closed under these operations and the structure formed

by this collection and these operations is recursive.

Proof. By 5.4 it suÆces to prove that the structure of the classes of typedBoolean hierarchy over L with the operations Fs(s 2 T ) is recursive, i.e. froma given s = s(v0; : : : ; vn�1) 2 T and t0; : : : ; tn�1 2 T � we can compute t 2 T �

with t(L) = Fs(t0(L); : : : ; tn�1(L)). First compute an m < ! such that allvariables of the terms ti(i < n) are contained in V = fvlkjk; l � mg. Thencompute pairwise disjoint sets Vi(i < n) of di�erent typed variables "lookinglike V " (i.e. any Vi is of the form fwl

kjl; k � mg for some variables wlk of type

l). Let t0i be the term obtained by replacing variables vlk(l; k � m) in ti by wlk,

and t = s(t00; : : : ; t0n�1) be the term obtained by replacing variables vi(i < n) in

s by terms t0i. Then t clearly has the desired properties completing the proof.Theorem 5.1 generalizes 3.2 and is applicable to some other cases, e.g. for

the arithmetical and the Borel hierarchies of subsets of the Baire space. Letus state a result generalizing all these examples. Let B be a �-algebra (i.e.a b.a. closed under countable unions and intersections) and let (M ;�) be anumbered subalgebra of the b.a. B (i.e. M is a subalgebra of B and � is afunction from ! onto M modulo which Boolean operations are recursive). LetS0n+1 = S0n+1(M ;�) be the set of all elements [i0 \i1 � � �in �fhi0; : : : ; ini, where

20

the symbols [;\; : : : alternate and f is a recursive function. The sequencefS0n+1gn<! is called the e�ective Borel hierarchy generated by (M ;�) insideB. It is an abstract version of arithmetical hierarchy and was considered e.g.in Ro67 and Se82. By fS0n+1gn<! we denote the �nite (none�ective) Borelhierarchy over subalgebra M of B (it may be de�ned just as above but witharbitrary � : ! !M and f : ! ! !).

5.6. Theorem. The e�ective and none�ective Borel hierarchies are re-

ducible bases, so Theorem 5.1 is applicable to both of them.

Proof. The assertion that both sequences are bases is known and almostevident. Let us prove reducibility for the e�ective case, the none�ective beingsimilar. First we show the reduction property for S01 , and indeed we state thefollowing slightly more general assertion: for any uniform sequence faigi<! ofelements of S01 (i.e. sequence of the form ai = [k<!�fi(k) for a recursivesequence ffig of recursive functions) there is a uniform sequence fa�i gi<! ofelements of S01 such that the a�i are pairwise disjoint, a

�i � ai and [ia

�i = [iai.

Let bhi;ki = �fi(k), b�0 = b0 and b�n+1 = bn+1 n (b

�0 [ � � � [ b

�n). It is clear that

the sequence fb�i g is uniform, elements b�i are pairwise disjoint, b�i � bi and[ib

�i = [ibi. Then the elements a�i = [kb

�hi;ki have the desired properties.

The class S0n+1 for any n has a natural enumeration induced by the standardenumeration of r.e. sets. Let M 0 be the b.a. generated by S0n+1 and �

0 be itsenumeration induced by the enumeration of S0n+1 and by the G�odel enumerationof Boolean terms. Then (M 0;�0) is a numbered subalgebra of B and it is easyto see that S0n+2 = S01(M

0;�0) (see Se82). So the above argument yields alsothe reduction property for S0n+2 completing the proof.

Theorem 5.1 treats bases of very special kind. Is it true for an arbitrary base?If not then what classes of the universal hierarchy over an arbitrary base areclasses of the typed Boolean hierarchy over this base? The next result is a partialanswer to these questions. Let � be the least set of ordinals such that 0 2 �and if � 2 � then !� �k 2 � for k < !. It is easy to see that any level Snk (k < !)of the di�erence hierarchy over Ln(n < !) is among S�(� 2 �). But the lastsequence contains also some other classes because the ordinal corresponding to(fSnk jn; k < !g;�) is !2 and the ordinal type of (�;�) is !!.

5.7. Theorem. (i) There is a base L over which the collections fS�; �S�j� <"0g and ft(L)jt 2 T

�g are incomparable by inclusion.

(ii) Any class S�(� 2 �) over any base L is a class of the typed Boolean

hierarchy over L.

Proof. (i) Let R0 and R1 be recursive sets for which all sets R0R1, R0�R1,

�R0R1, �R0�R1 are in�nite. Let L0 = f;; R0R1; R0; R1; R0 [ R1; !g, L1 = �0

1,Ln+2 = P (!), P and Q are r.e. nonrecursive subsets of R0

�R1 and �R0R1 re-spectively, t0 = u0�u1v0 [ �u0u1�v1, t1 = t0 [ u0u1 and t2 = t0 [ �u0�u1, where thevariables ui are of type 0 and vi are of type 1. Let Xi be the value of ti whenuk is interpreted as Rk, v0 as P and v1 - as Q. An easy computation shows thatXi 2 ti(L) n (tj(L) [ tk(L)) for all pairwise di�erent i; j; k < 3. So for the baseL = fLng the structure (ft(L)jt 2 Tg;�) is not almost linearly ordered and a

fortiori some members of t(L) are di�erent from all levels of the corresponding

21

universal hierarchy.It remains to �nd a class of the universal hierarchy over L which is not a

class of the typed Boolean hierarchy. We will show that S!+1 is such a class, i.e.S!+1 6= t(L) for any t 2 T �. Fix t = tA and consider �rst the case A 62 S!+1 (weuse notation from the proof of 5.1). We have �A 62 �S!+1, so there is a sequencefU� : � 2

!̂ 2; j�j � j�!+1jg for �A as described in the assertion c) from the proofof 5.1. De�ne elements b�(� 2

!̂ 2) as follows:b; = !; b0 = R0; b1 = R1; b01 = P; b10 = Q; b� = ; for other �.We have �!+1 = �0!+1 = 0�0! = 0�11 = 01 (see the de�nition before 4.3), so

fb�g is a �!+1-sequence. Let ank = [f~b�jv

nk 2 U�g, so a

nk 2 Ln as in the proof

of 5.1, because the corresponding proof in Se91 does not use the additionalproperty b�0b�1 = 0 which is not true in this case. So a = tA(a

nk ) 2 t(L). Note

that�R0

�R1; R0�R1

�P;Q � a and �R0R1�Q;P � �a: (4)

This follows from the argument at the end of the proof of 5.1, because any ofthe listed sets is contained in a unique ~b� and is disjoint from ~b� for � 6= �.It suÆces to prove that a 62 S!+1, because this implies the desired inequalityS!+1 6= t(L). Suppose the contrary, so for some ui 2 L0; c0 2 L1; c1 2 �L1 wehave

a = u0c0 [ u1c1 and u0u1c0 = u0u1c1: (5)

By (4) �R0�R1 � a � u0 [ u1. But u0 [ u1 2 L0, so u0 [ u1 = ! by the

de�nition of L0. Again from the structure of the lattice L0 we get ui = !

for some i < 2. Let e.g. u0 = !, the second case being similar. We havea = u0a = u0c0 [ u0u1c1 = u0c0 = c0, so a 2 L1 = �0

1. From (4) followsR0

�R1a = R0�R1

�P , so P is recursive, a contradiction.It remains to consider the case A 2 S!+1. LetXi 2 L0; A0 2 L1 and A1 2 �L1

satisfyA = X0A0 [X1A1 and X0X1A0 = X0X1A1: (6)

From L0 [ �L0 � L1 we get X0X1A0 2 L1 and X0X1A1 2 �L1, so the setY = X0X1A0 belongs to L1 \ �L1. By the assertion a) in the proof of 5.1Y 2 (L0).

Consider the set a = P [ �R0R1�Q [ T , where P;Q are as above and T is a

nontrivial recursive subset of R0R1. For the sets ui = Ri 2 L0, c0 = P [T 2 L1and c1 = �R0

�Q [ T 2 �L1 we have (5), so a 2 S!+1. It suÆces to show thata 62 tA(L), because then S!+1 6= tA(L) as desired. Suppose the contrary: a =tA(a

nk ) for some a

nk 2 Ln. By the assertion b) from the proof of 5.1 we deduce

from (6) that xi 2 L0, a0 2 L1, a1 2 �L1, a = x0a0 [x1a1 and x0x1a0 = x0x1a1,where x0 = tX0

(ank ) and so on. Sets x0; x1 are incomparable by inclusion (ife.g. x0 � x1, then a = ax0 = x0a0 2 L1, so a is r.e.|a contradiction). By thestructure of the lattice L0 this implies fx0; x1g = fR0; R1g, so let e.g. xi = Ri.Then y = R0R1a0 = R0R1a = T , so T 2 (L0). But again by the de�nitionof L0 the b.a. (L0) contains only trivial subsets of R0R1. This contradictioncompletes the proof of (i).

22

(ii) It suÆces to prove that for any nonzero � 2 � and any n < ! there isa �nite sequence of numbers n0 � � � � � nk with nk = n, Sn� = Ln0 + � � �+ Lnkand, conversely, for any sequence n0 � � � � � nk there is a nonzero � 2 �with Snk� = Ln0 + � � � + Lnk . Here + is the operation from section 2. Onedirection is by induction on �. For � = k + 1 < ! we have Sn� = Ln + � � �+ Ln(k + 1 times) by 2.5. Now let � = !�1(k + 1) for some nonzero ordinal �1 2 �.By the induction hypothesis there is a sequence n0 � � � � � ni with ni > n

and Sn+1�1= Ln0 + � � � + Lni . By 4.4.(iii), 4.4.(ix) and 2.5 we have Sn� =

Sn+1�1+Snk = (Ln0 + � � �+Lni)+(Ln+ � � �+Ln) (k members in the second sum)

as desired. The opposite direction is proved by induction on the cardinality ofthe set fn0; : : : ; nkg using a similar computation. This completes the proof ofthe theorem.

5.8. Remark. By the proof of (ii) we have fS�j� 2 �; � 6= 0g = fLn0 +� � �+ Lnk jk; ni < !g, so the fragment fS�g�2� has an especially easy and nat-ural description. We guess (but cannot yet prove) that the assertion 5.7.(ii) isoptimal, i.e. for any � 2 "0 n� there is a base over which S� is not a class of thetyped Boolean hierarchy. It seems plausible that the argument from the proofof 5.7.(i) can be generalized to prove this.

The next result concerns the structure L� = (ft(L)jt 2 T �g;�) over arbi-trary base L. By the proof of 5.7.(i) it need not be almost linearly ordered. Itseems plausible that it is always well founded. The last result of this sectionshows that this question is in a sense reducible to a similar question for a veryspecial base. Let G be the free b.a. with generators vnk (n; k < !) satisfyingequalities vnk = vn+13k and �vnk = vn+13k+1. Let Fn be the subalgebra of (G;[;\; 0; 1)generated by elements vnk (k < !). Simplifying notation we identify elements ofG with the corresponding terms; then elements of Fn are lattice terms in thevariables of type n.

The sequence F = fFng is clearly a base. We call this base free becausefor any base L with countable components there is a homomorphism f from G

into B satisfying f(Fn) = Ln. (By the way, if we enrich the language by binaryfunctional symbols rn0 ; r

n1 and add equalities stating that these functions give a

reducing pair for variables of any type n, then we get a reducible base with asimilar "freedom" property.) Another universality property of the free base isthe following

5.9. Proposition. For any base L, the structure L� is a homomorphic

image of the structure F �.Proof. It suÆces to show that for all s; t 2 T � the inclusion s(F ) � t(F )

implies the inclusion s(L) � t(L), because then t(F ) 7! t(L) is the desiredhomomorphism from F � onto L�. So assume that s(F ) � t(F ) and a 2 s(L);we have to show that a 2 t(L). Let ank 2 Ln satisfy a = s(ank ). De�ne b

nk 2 Ln

by b0k = a0k, bn+13k = bnk , b

n+13k+1 = �bnk and bn+13k+2 = an+1k . Let f be the unique

homomorphism from G into B satisfying f(vnk ) = bnk . Then f(Fn) � Ln andf(wn

k ) = ank , where w0k = v0k and wn+1

k = vn+13k+2. We have wnk 2 Fn, so s(w

nk ) 2

s(F ) and a fortiori s(wnk ) 2 t(F ). Let tnk 2 Fn satisfy s(wn

k ) = t(tnk ). As wenoted before the formulation of the theorem we may assume that tnk are lattice

23

terms in variables of type n, tnk = tnk (vni ) (we apologise for the not completely

rigorous notation). Let cnk = tnk (bni ) be the value of this term when the vni are

interpreted as bni . Then cnk 2 Ln and a fortiori t(cnk ) 2 t(L). But

a = s(f(wnk )) = f(s(wn

k )) = f(t(tnk (vni ))) = t(tnk (b

ni )) = t(cnk );

so a 2 t(L) completing the proof.The proposition implies that the well{foundedness of L� for any base L would

follow if there were no sequence fxkgk<! of elements of F � such that xk 6� xlfor k < l, but unfortunately we were unable to prove this.

6 Wadge Hierarchy

Let !1 be the �rst uncountable ordinal and f�01+�g�<!1 be the (none�ective)

Borel hierarchy of subsets of the Baire space I =! !. By Theorem 2.1 the classesof the Boolean hierarchy over �0

1 coincide with the classes of the di�erencehierarchy over �0

1. By Theorem 5.1 the classes of the typed Boolean hierarchyover the base f�0

1+kgk<! coincide with the classes of the universal hierarchyover this base. Here we generalize these results to in�nitary Boolean terms.

De�ne !1-terms by induction: constants 0; 1 and variables vk(k < !) are!1-terms; if ti(i < !) are !1-terms, then so are the expressions �t0, t0[t1, t0\t1,[i<!ti and \i<!ti. The notion of typed !1-term is the same but as variables wetake v�k (� < !1; k < !). Denote by T1 (by T

�1 ) the set of all (typed) !1-terms.

For t = t(vk) 2 T1 let t(�01) be the set of all values of t when each vk(k < !)

is interpreted as some Ak 2 �01. For t = t(v�k ) 2 T �

1 and L = f�01+�g�<!1

let t(L) be the set of all values t(A�k ) of t when each v�k (� < !1; k < !) is

interpreted as some A�k 2 �0

1+�.To formulate in�nitary analogs of 2.1 and 5.1 we need an analog for the

corresponding hierarchies. This analog is fairly general and is de�ned as follows.By a Wadge class we mean any class of the form fBjB �W Ag, where B �W A

i� B = F�1(A) for some continuous function F : I ! I . Such a Wadge class iscalled Borel if A is Borel.

The next theorem shows that the in�nitary analogs of 2.1 and 5.1 coincideand gives an exact sense to the intuitive assertion from Se89 that the univer-sal hierarchy from section 3 is the �nite e�ective version of the Borel Wadgehierarchy.

6.1. Theorem. For any � � P (I), the following conditions are equivalent:

(i) � is a nonselfdual Borel Wadge class;

(ii) � = t(�01) for some t 2 T1;

(iii) � = t(L) for some t 2 T �1 :

Proof. It suÆces to show that (ii) is equivalent to (iii), (ii) implies (i) and(i) implies (iii). The implication (ii)!(iii) is trivial because we may consider!1-terms as typed !1-terms containing only variables of type 0.

To check the implication (iii)!(ii) �rst note that by the de�nition of theBorel hierarchy for any � < !1 there is an !1-term s�(vm) satisfying �

01+� =

s�(�01). Choose in�nite pairwise disjoint sets R�

k (� < !1; k < !) of natural

24

numbers and relate to any t 2 T �1 the term t� 2 T1 obtained by replacing in

t variables v�k by terms s�(vg�k(m)), where g

�k is the function satisfying R�

k =

fg�k (0) < g�k (1) < � � �g. An evident induction on t shows that t(L) = t�(�01)

which proves our implication.To prove the implication (ii)!(i) it suÆces to check that for any t 2 T1

the class t(�01) is closed downwards under �W and has a universal set (it is

well-known that then t(�01) has a �W -greatest element and is nonselfdual). let

A 2 t(�01), i.e. A = t(Ak) for some Ak 2 �0

1, and B �W A, i.e. B = F�1(A)for some continuous F . By induction on t we have F�1(t(Ak)) = t(F�1(Ak)).But F�1(Ak) 2 �0

1, so B 2 t(�01).

It remains to �nd a universal set in t(�01). Let U be universal in �0

1, i.e.U 2 �0

1 and every �01-set is fgjhf; gi 2 Ug for some f 2 I (here h = hf; gi is

the code of pair (f; g) de�ned by h(2n) = f(n) and h(2n + 1) = g(n)). LetXk = fhf; gijhfk; gi 2 Ug, where fk is de�ned by fk(x) = fhk; xi, and letW = t(Xk). Sets Xk are �0

1, so W 2 t(�01). Now let Y be an arbitrary set

from t(�01), then Y = t(Yk) for some Yk 2 �0

1. Let hk be functions satisfyingYk = fgjhhk; gi 2 Ug and h = �khk be the function de�ned by hhk; xi = hk(x).Then Y = fgjhh; gi 2Wg and W is universal in t(�0

1).The last (and most diÆcult) implication (i)!(iii) follows from a deep result

in Lo83. To avoid a long citation and cumbersome notation we give only asketch of proof with the corresponding references. By VW76 and St80 for anynonselfdual Borel Wadge class � exactly one of classes �; �� has the separationproperty. So it suÆces to prove that any nonselfdual Borel Wadge class �without the separation property has the form t(L) for some t 2 T �

1 . FromTheorem 1.9 in Lo83 follows that � is �u for some u 2 D as described inDe�nition 1.2 in Lo83. Considering in turn points of this inductive de�nition in away similar to that from the proof of 3.1 here one gets the desired representation�u = t(L). This completes the proof of the theorem.

By a remarkable result of W.Wadge and D.Martin (see VW76) the structureof the Borel Wadge classes is almost well ordered by inclusion. So we have thefollowing analog of corollaries 2.4 and 5.3.

6.2. Corollary. The structures (ft(�01)jt 2 T1g;�) and (ft(L)jt 2 T �

1 g;�)are almost well ordered.

The next result is a little generalization of 6.2 because any t(�01) coincides

with s(�01) for some s 2 T1. Here�

01 = �0

1\�01 is of course the class of clopen

subsets of I .6.3. Corollary. The structure (ft(�0

1)jt 2 T1g;�) is almost well ordered.

Proof. It suÆces to prove that for any t = t(vk) 2 T1, the class t(�01)

is a Wadge class because then we can use the mentioned result of Wadge andMartin. Downward closedeness of t(�0

1) is checked just as in the proof of 6.1.It remaines to �nd a W -complete set in t(�0

1). Let C be a W -complete set in�0

1, and Ak = ff jfk 2 Cg, where fk(x) = fhk; xi. We claim that the set A =t(Ak) 2 t(�

01) is W -complete in t(�0

1). Let X 2 t(�01), so X = t(Xk) for some

Xk 2 �01. Let Fk (k < !) be continuous functions satisfying Xk = F�1

k (C);then F (f) = �kF (fk) is a continuous function satisfying Xk = F�1(Ak) for all

25

k. But then X = t(Xk) = t(F�1(Ak)) = F�1(t(Ak)) = F�1(A). So X �W A

completing the proof.Let us relate !1-terms to set{theoretic operations which were very popular

in classical descriptive set theory (and became less popular after the interven-tion of logic in this �eld). People were most interested in so called positiveanalytic or Æs-operations introduced independently by A.N.Kolmogorov andF.Hausdor� and intensively studied by L.V.Kantorovich and E.M.Livenson andby A.A.Liapunov (for references and a modern presentation see Hi78). Theseoperations are connected with coarse hierarchies but not so well connected with�ne hierarchies. Here we need more general operations studied e.g. in Och55.

Relate to any A �! 2 an in�nitary term

tA = tA(vk) =[

f2A

tf ; where tf = (\

f(k)=1

vk) \ (\

f(k)=0

�vk):

Note that tf is an in�nitary analog of the "elementary conjunctions" tU (tosee this just identify U � V with its characteristic function) and tA|of thecorresponding terms in the proofs of 2.1 and 5.1. We call two in�nitary Booleanterms equivalent if they de�ne the same in�nitary operation in any complete b.a.The next result relates !1-terms to a natural class of set{theoretic operations.The evident proof is omitted (one direction is by induction on !1-terms, theother|by induction on � satisfying A 2 �0

1+�).6.4. Theorem. Any !1-term is equivalent to tA for some Borel A �! 2

and vice versa.

We conclude this section with an abstract version of 6.3. As in section 5let B be a �-algebra and M be a subalgebra of the b.a. B (M need not be a�-algebra). Corollary 6.3 is a particular case of the next assertion for B = P (I)and M =�0

1.6.5. Theorem. The structure (ft(M)jt 2 T1g;�) is almost well ordered.

Proof. By 6.4 it suÆces to prove that structure (ftA(M)jA �! 2; A is Borelg;�) is almost well ordered. By the cited result of W.Wadge and D.Martin thestructure of Borel subsets of !2 with the relation �W is well founded and for anyof its elements either A �W B or B �W

�A. Taking into account that t �A(M) isthe dual class of tA(M) we see that it suÆces to deduce tA(M) � tB(M) fromA �W B.

So let F be a continuous function on the Cantor space !2 satisfying A =F�1(B). From the de�nition of the topology on !2 (basic open sets are ff j� �fg for � 2!̂ 2, where � � f means that � is the initial segment f [n] of thesequence f 2 I of length n = j�j) it follows that there are a strictly increasingfunction u 2! ! and sets C� �

!̂ 2(� 2!̂ 2) with the following properties:(i) sets C� are pairwise disjoint;(ii) [fC� : j�j = ng = f� 2!̂ 2 : j� j = u(n)g;(iii) � [u(j�j)] 2 C� for � 2 C�k;(iv) g = F (f) is the unique function satisfying f [u(n)] = g[n] for all n < !.We have to show that tA(M) � tB(M). Let a 2 tA(M), so a = [f2Atf (ak)

26

for some ak 2M(k < !). Let bn = [ft� (ak)j9�(j�j = n ^ � 2 C�1)g; where

t� (vk) = (\

�(k)=1

vk) \ (\

�(k)=0

�vk)

is the �nite version of the term tf . The elements bn are in M , so b = tB(bk)is in tB(M) and it remains to show that a = b, i.e. [f2Atf (ak) = [g2Btg(bk).This follows from the inclusion tf (ak) � tf(F )(bk), because A = F�1(B), tf (ak)are pairwise disjoint and [f2!2tf (ak) = 1.

So it remains to check that tf (ak) � tg(bk) for any f 2! 2, where g = F (f).

Take any x 2 tf (ak) and prove that x 2 tg(bk). We have tg(bk) = \ft�(bk)j� �gg, so it suÆces to prove that x 2 t�(bk) for any � � g. This is by inductionon j�j. For � = ; we have t;(bk) = 1, so x 2 t�(bk). Let j�j = n; x 2 t�(bk)and i = g(n); we have to show that x 2 t�i(bk). By the property (iv) above� = f [u(n+1)] 2 C�i. We also have x 2 t� (ak), because tf (ak) = \ft�(ak)j� �fg: By de�nition of bn and properties of C� and t� we have: t� (ak) � bn fori = 1 and t� (ak) � �bn for i = 0. So x 2 bn for i = 1 and x 2 �bn for i = 0. Butt�i(bk) = t�(bk)\ bn for i = 1 and t�i(bk) = t�(bk)\�bn for i = 0. So x 2 t�i(bk)completing the proof of the theorem.

7 Hierarchy of Formulas

The aim of the two last sections is to discuss �ne hierarchies naturally arisingoutside the traditional sphere of hierarchy theory. In this section we brie yreview and discuss such a hierarchy in logic (all cited results are from Se91a),and in the next section|some hierarchies in complexity theory.

To any �rst-order theory T in a language we can naturally associate a baseL(T ) = f�0

k+1(T )gk<!, where �0k+1(T ) is the set of -sentences equivalent in

the theory T to a �0k+1-sentence (more exactly, �

0k+1(T ) is the set of equivalence

classes corresponding to these sentences). By the construction from section 5we have the corresponding universal hierarchy fS�(T )g�<"0 over L(T ) which wecall the universal hierarchy of sentences over T . It has the following remarkableproperty which is mainly due to a �ne version of the Craig interpolation theorem:

7.1. Theorem. If T is axiomatizable by �02-sentences and has a model

embeddable in any model of T, then the universal hierarchy over T is perfect.

This shows that our hierarchy of sentences is in a sense a "�nest" hierarchy.Note that we can also consider a hierarchy of formulas (not only sentences)

over T . For this hierarchy we have the following weaker version of 7.1:7.2. Theorem. The universal hierarchy of formulas fS�g over any �0

2-axiomatizable theory T is perfect in all limit levels, i.e. S� \ �S� = [�<�S� for

any limit ordinal � < "0.This generalizes the property of the predicate logic that any �0

n+2-formulais equivalent to a �nite Boolean combination of �0

n+1-formulas.The next result gives a natural model{theoretic characterisation of the classes

S�(T ). For -structureA denote by A the language consisting of all -symbols

27

and a new constant symbol ca for every a 2 A. If A is a substructure of B(A � B), then let BA denotes the A-structure coinciding with B on andinterpreting ca as a. Let A �n B (n < !) denote that every A-sentence from�0n+1 (i.e. equivalent to a �

0n+1-sentence and to a �

0n+1-sentence) has the same

value in AA and BA.Now let � 2!̂ ! and ' be a -sentence. By �-tree for ' in T we mean any

sequence fA� : � 2!̂ 2; j�j � j�jg of T -models such that A� ��(j�j) A�k forj�j < j�j and k < 2, ' is true in A;;A�0, and ' is false in A�1 for j�j < j�j.Let �� be the string de�ned in 4.3.

7.3. Theorem. The class S�(T ) coincides with the class of all -sentencesnot having a ��-tree in T .

An immediate corollary of 7.3 is the following description of the classesDk(T ) (k < !) of di�erence hierarchy over �0

1(T ) which was implicit in Ad65:' 2 Dk(T ) i� there is no sequence A0 � � � � � Ak of T -models such that ' istrue in A2i and is false in A2i+1.

Now let fS�g be the universal hierarchy of unary formulas of primitive re-cursive arithmetic. It turns out to be closely related to the hierarchy f��g fromsection 3:

7.4. Theorem. For any � < "0, the class �� is the class of sets de�nable

in the standard model by a formula from S�.These results may be used to give a partial explanation of the known phe-

nomenon that subsets of ! with "simple" de�nitions are usually m-complete ina �nite level of a suitable hierarchy (this phenomenon was discussed in Ro67).Let T be primitive recursive arithmetic with induction over quanti�er{free for-mulas. Let A � ! be de�ned by a formula '(v) of primitive recursive arithmeticin the standard model. We call this de�nition simple if from A 2 �� it followsthat ' 2 S� (i.e. if the fact A 2 �� is provable in a weak theory T ). Then by7.2 A cannot be in �� n[�<��� for limit �. In other words A is in one of classes�� n �� , �� n �� , ��+1 n (�� [ ��) for some � < "0. Note that this follows(somewhat unexpectedly) from a version of the Craig Interpolation Theorem.

Moreover, if we restrict ourselves to formulas ' satisfying the additionalproperty that ' 62 S� implies ' 6� for any 2 S� e�ectively on (e.g. if' 6� for any such is provable in T intuitionistically) then one can show thatA is really m-complete in one of levels ��, �� ,��+1 (� < "0) of the universalhierarchy (for details see Se91a).

We do not know if the classes of the typed Boolean hierarchy of formulasare useful for some purposes or not.

8 Re�nements of Polynomial Hierarchy

We conclude the paper by a brief discussion of �ne hierarchies naturally arisingin complexity theory. First note that here we have to consider sets of wordsover a �nite alphabet in place of sets of natural numbers in the recursive case.Without loss of generality we restrict ourselves to subsets of !̂2. Let L =f�p

n+1gn<! be the polynomial hierarchy of subsets of !̂2. It is well known (see

28

e.g. BDG88) that L is a base, so the de�nitions of sections 4 and 5 give us twonatural re�nements of the polynomial hierarchy|the typed Boolean hierarchyft(L)gt2T� and the universal hierarchy fS�g�<"0 over L.

By the same argument as that in the proof of 3.2 (just replace recursive-ness by polynomial computability) we get that any class t(L) is closed down-wards under polynomial m-reducibility and has a polynomially m-complete set.Moreover, if �p

k+1-complete sets Ak in this proof code "natural" combinatorial

problems then so does also the set B. For this reason we believe that the typedBoolean hierarchy may be useful for classi�cation of some "natural" combina-torial problems. The main lack of this "hierarchy" is that we can say nothingabout the structure (ft(L)jt 2 T �g;�), e.g. we do not know if it is well foundedor almost linearly ordered.

The universal hierarchy fS�g�<"0 over L is in this respect better: the col-lection of its classes is almost well ordered. From 4.4.(xi) it is easy to deducethat any class of the universal hierarchy over L is downward closed under poly-nomial m-reducibility. But this hierarchy also has a disadvantage|we cannotprove that any of its classes has a polynomially m-complete set.

In section 5 we de�ned an "absolute" common part of both hierarchies |fS�g�2�. This fragment has the good properties of both hierarchies and containsclasses Snk (k < !) of the di�erence hierarchy over �p

n+1 for any n < !. By Re-mark 5.8 this fragment has an especially nice description in terms of the classesof the polynomial hierarchy. This fragment has also another good property: wecan prove that if the polynomial hierarchy does not collapse (i.e. �p

n+1 6= �pn+1

for all n < !), then the fragment also does not collapse (i.e. S� 6= �S� for all� 2 �). The proof is a generalization of the proof of the corresponding result ofJ.Kadin for the di�erence hierarchy over NP; we hope to publish it somewhereelse. We know nothing about generalizations of this result for the classes of ourhierarchies outside this fragment.

We can prove also no inclusion between ft(L)jt 2 T �g and fS�; �S�j� < "0g.By 5.1 these collections would coincide if the base L were reducible. Unfortu-nately, it is not even known if there is an oracle modulo which the polynomialhierarchy does not collapse and for any n one of �p

n+1, �pn+1 has the reduction

property (for n = 0 this was asked already in BG84). In the opposite directionK.Ambos-Spies (private communication) constructed a recursive oracle mod-ulo which �p

1 6= �p1 and none of these classes has the reduction (or even the

separation) property.So the main problems on our hierarchies in the context of complexity theory

are open and seem to be diÆcult. We hope they are of some interest to this�eld.

9 Conclusion

We see that �ne hierarchies arise naturally in di�erent areas. For the case ofBorel sets the situation is ideal and admits a complete description even forthe in�nitary version. In the recursive case the situation is very good for the

29

�nite levels of the �ne hierarchies but is not so good for the in�nite levels. Thesituation with hierarchies in complexity theory is very complicated even for the�nite levels. Note that the recursive case uses techniques which are relevant todescriptive set theory as well as to complexity theory. One could say that therecursive world is a bridge between the Borel Paradise in descriptive set theoryand the Cook{Karp Hell in complexity theory.

In conclusion I would like to thank the unknown referee for the careful read-ing and for improving of my English.

References

[Ad65] J.Addison. The method of alternating chains. In: The theory of models,Amsterdam, North Holland, 1965, p.1|16.

[BDG88] J.L.Balc�azar, J.D�iaz, and Gabarr�o. Structural Complexity I, volume11 of EATCS Monographs on Theoretical Computer Science. Springer-Verlag, 1988.

[BG84] A.Blass and Y.Gurevich. Equivalence relations, invariants, and normalforms. In Logic and Machines: Decision Problems and Complexity, v.171of Lecture Notes in Computer Science, p.24|42. Springer-Verlag, 1986.

[Er68] Y.L.Ershov. On a hierarchy of sets II. Algebra i Logika, v.7, No 4 (1968),p.15|47 (English translation: Algebra i Logika, v.7 (1968), p.212|232).

[Hau14] F.Hausdor�. Grundz�uge der Mengenlehre. Leipzig, 1914.

[Hay78] L.Hay. Convex subsets of n2 and bounded truth-table reducibility. Dis-crete mathematics, v.21 (1978), p.31-46.

[Hi78] P.Hinman. Recursion-theoretic hierarchies. Berlin, Springer, 1978.

[KM67] K.Kuratowski and A.Mostowski. Set Theory. North Holland, 1967.

[KSW86] J.K�obler, U.Sh�oning and K.W.Wagner: The di�erence and truth-table hierarchies for NP. Dep. of Informatics, Koblenz. Preprint 7 (1986).

[Lo83] A.Louveau. Some results in the Wadge hierarchy of Borel sets. Lec.Notes in Math., No 1019 (1983), p.28|55.

[Mo80] Y.N.Moschovakis. Descriptive set theory. Amsterdam, North Holland,1980.

[Och55] Y.S.Ochan. Theory of operations on sets. Uspekchy Mat. Nauk, N 3(65) (1955), p.71|128 (in Russian).

[Ro67] H.Rogers jr., Theory of recursive functions and e�ective computability.

New York, McGraw-Hill, 1967.

30

[Se82] V.L.Selivanov. On index sets in the Kleene-Mostowski hierarchy. TrudiMat. Inst. SO AN SSSR, N 2 (1982), p.135|158 (in Russian).

[Se83] V.L.Selivanov. Hierarchies of hyperarithmetical sets and functions. Al-gebra i Logika, 22, No 6 (1983), p.666|692 (English translation: Algebraand Logic, 22 (1983), p.473|491).

[Se84] V.L.Selivanov. Index sets in hyperarithmetical hierarchy. Sib. Mat.

Zhurn., 25, No 3 (1984), p.164|181 (English translation: Sib. Math.

J., 25 (1984), p.474|488).

[Se89] V.L.Selivanov. Fine hierarchies of arithmetical sets and de�nable indexsets. Trudi Mat. Inst. SO AN SSSR, v.12 (1989), p.165|185 (in Rus-sian).

[Se91] V.L.Selivanov. Fine hierarchies and de�nable index sets.Algebra i logika,30, No 6 (1991), p.705|725 (English translation: Algebra and logic, 30(1991), p.463|475).

[Se91a] V.L.Selivanov. Fine hierarchy of formulas. Algebra i Logika, v. 30, No5 (1991), p.568|582 (English translation: Algebra and Logic, v. 30(1991), p.368|379).

[Se92] V.L.Selivanov. Computing degrees of de�nable classes of sentences. Con-temporary math., 131, part 3 (1992), p.657|666.

[St80] J.Steel. Determinateness and the separation property. J. Symbol. Logic,

45 (1980), p.143|146.

[We85] G.Wechsung. On the Boolean closure of NP. In Proceedings of the 1985

Int. Conf. on Fundamentals of Computation theory, v.199 of LectureNotes in Computer Science, p.485|493. Springer-Verlag, 1985.

[VW76] Van Wesep. Wadge degrees and descriptive set theory. Lec. Notes in

Math., No 689 (1976), p. 151|170.

Current address: Victor Selivanov, Mathematisches Institut der Universit�atHeidelberg, Im Neuenheimer Feld 294, 69120 Heidelberg, Germany.

e-mail vseliv@math.uni-heidelberg.de

31