by Carlos Martinez-Ranero - University of Toronto

Contributions towards a fine structure theory ofAronszajn orderings.

by

Carlos Martinez-Ranero

A thesis submitted in conformity with the requirementsfor the degree of Doctor of PhilosophyGraduate Department of Mathematics

University of Toronto

Copyright c© 2011 by Carlos Martinez-Ranero

Abstract

Contributions towards a fine structure theory of Aronszajn orderings.

Carlos Martinez-Ranero

Doctor of Philosophy

Graduate Department of Mathematics

University of Toronto

2011

The purpose of this thesis is to add to the structure theory of Aronszajn orderings. We

shall focus essentially in four topics. The first topic of discussion is about the relation

between Lipschitz and coherent trees. I will demonstrate that the tree T (ρ0) is coherent

without any extra set theoretic hypothesis. The second topic presents an application

of Todorcevic’s ρ functions to provide some partial answers to an old question of Juhaz

asking whether a standard weakening of Jensen’s diamond principle implies the existence

of a Suslin tree. In the third topic we focus on providing a satisfactory rough classification

result of the class of Aronszajn lines. Our main result is that, assuming PFA, the class of

Aronszajn lines is well-quasi-ordered by embeddability. The last topic is an investigation

of the gap structure of the class of coherent Aronszajn trees. I will show that, assuming

PFA, the class of coherent Aronszajn trees quasi-ordered by embeddability is the unique

saturated linear order of cardinality ℵ2.

ii

Dedication

Dedicated to my suportive wife Marna.

iii

Acknowledgements

First and foremost, I would like to thank Stevo Todorcevic for great supervising of this

work. He has shared with me his insights and unique ways of seeing mathematics. He

has always encouraged and challenged me throughout all this process. I would also like

to thank Justin Moore for his inspiring papers and for sharing his insight on some of the

problems addressed in this thesis. I would like to thank Bohuslav Balcar for many hours

of mathematical discussion. I would like to thank the members of the Toronto Set Theory

seminar for the stimulating environment. I would also wish to thank Michael Hrusak for

showing interest and offering me the encouragement in my mathematical pursuits.

I wish to acknowledge the financial support from the following sources. I was supported

by the Secretaria de Educacion Publica 2005-2009, Conacyt during the period 2005-

2009 and by the University of Toronto Fellowship for the period 2005-2010. I was also

supported by the SGS from 2009-2010. My trips to conferences at which some of these

results where presented were supported from the NSERC grant of Stevo Todorcevic with

supplements from the Association for Symbolic Logic and from the research grant of

Michael Hrusak.

iv

Contents

Introduction. iv

1 Lipschitz trees. 1

1.1 Lipschitz and coherent trees. . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.2 The coherence of T (ρ0). . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

2 Non-special coherent Aronszajn trees. 14

2.1 Star-principles. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

2.2 Star-principles and non-special Aronszajn trees. . . . . . . . . . . . . . . 16

2.3 The star-principles and their relationship with some standard set-theoretic

axioms. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

3 The Aronszajn lines are well-quasi-ordered by embeddability. 24

3.1 Well-quasi-orderings. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

3.2 A-lines. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

3.3 Aronszajn lines and the Weak-Diamond principle. . . . . . . . . . . . . . 33

3.4 Fine structure theory of A-lines under PFA. . . . . . . . . . . . . . . . . 36

4 Gap structure of coherent Aronszajn trees. 45

4.1 Quasi-ordering on trees. . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

4.2 Shift operation on trees. . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

4.3 Comparability of every pair of Lipschitz trees. . . . . . . . . . . . . . . . 54

v

4.4 Coinitiality and cofinality of the class of coherent Aronszajn trees. . . . . 57

4.5 Gaps in (C,). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62

Bibliography 84

Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89

vi

Introduction.

A classical theorem of Cantor [5] states that any linear order X satisfying

(1) X has no first or last element,

(2) X is order complete, and

(3) X is separable in its order topology

is isomorphic to the real line, (R, <).

In an attempt to characterize the order type of the real line Suslin [34], in 1920, asked

whether (3) may be replaced by

(3’) X satisfies the countable chain condition (or ccc), i.e., every family of pairwise

disjoint non-empty intervals is countable.

While Suslin’s problem turn out to be independent of ZFC, this problem has led to a

number of developments in set theory and topology. For example, the complete solution

to Suslin’s problem is responsible for the pioneering work of the fine structure of the con-

structible universe by Jensen (see [7]) and for the invention of forcing axioms by Solovay

and Tennenbaum (see [33]).

In 1935, Kurepa made an important contribution to the subject. Kurepa’s paper [17]

is the first systematic study on trees, where he introduced two fundamental operations

which connect trees and linear orderings. These two operations being the lexicograph-

ical ordering of a tree and the partition tree associated to a linear ordering. Thus, he

vii

demonstrated that the theory of trees is quite close to the theory of linear orderings.

The notion of tree must be interpreted in its order-theoretic sense, i.e., a tree (T,<t) is a

partially ordered set in which the set of predecessors of an element of T are well-ordered

by <T . In the same paper Kurepa proved that there is a Suslin line if only if there is a

Suslin tree (i.e., an uncountable tree in which all chains and antichains are countable).

One of the most important discoveries in the early set theory was made, more or less

simultaneously, by Aronszajn, Kurepa and Jones in the first half of the twentieth century

(see [17]). They each constructed an uncountable tree, now known as an Aronszajn tree,

in which all levels and chains are countable, a phenomenon which is impossible in count-

able trees (see [15]). Contrary to the case of Suslin trees the existence of Aronszajn trees

does not require any additional set-theoretic hypothesis. The study of Aronszajn trees

can be considered as part of Suslin’s program and it has a deep influence in modern set

theory. For example, Todorcevic’s analysis of Aronszajn trees in [35] led to his method

of minimal walks which has seen wide and varied applications (see [37]). The class of

Aronszajn trees is important by itself in the sense that many prominent problems on

combinatorial set theory have its reformulation inside this class.

An immediate corollary of the existence of Aronszajn trees is the existence of a linearly

ordered set, known as Aronszajn lines, whose properties contrast dramatically with those

of the real line. The notion of Aronszajn line came from the notion of Aronszajn tree,

but they certainly have their intrinsic interest. The Aronszajn lines were rediscovered by

Specker in order to solve a question asked by Erdos-Rado [10].

The purpose of this thesis is to provide a satisfactory rough classification result of the

class of Aronszajn lines as well as of the class of Aronszajn trees.

A rough classification result usually depends on a transitive and reflexive binary relation

, i.e., a quasi-order. In this context the quasi-order is usually taken to be isomorphic

embedding, i.e., A B iff there exists an strictly increasing map f : A→ B. One of the

most prominent global conditions, generally considered, as giving a satisfactory rough

viii

classification result is the requirement of being well-quasi-ordered (for more details see

chapter 3).

The first result of this sort is a result of Laver [19] who showed, verifying an old conjec-

ture of Fraisse, that not only the class of countable linear orderings is well-quasi-ordered

but also the class of all σ-scattered linear orderings. Some restriction on linear orderings

in Laver’s result is needed in view of an old results of Dushnik-Miller [9] who proved

that the class of separable linear orderings of size continuum (more precisely, subsets of

the real line) fails badly to be well-quasi-ordered. The idea behind Dushnik-Miller’s con-

struction combined with ideas behind the more recent constructions using the so-called

weak-diamond principle of Devlin and Shelah show that under CH there is basically no

room for extending Laver’s theorem to a larger class of linear orderings. It is for this

reason that one is naturally led to examine this possibility using some alternative to

CH. It turn out that the right alternative to CH is a strong form of the Baire Category

theorem known under the name of Proper Forcing Axiom, PFA. This axiom was first

shown to be relevant in this context by Baumgartner [4]. Most efforts on chapter 3 are

put in the proof of the following:

Theorem 0.0.1 (PFA). The class of Aronszajn lines is well-quasi-ordered by embed-

dability.

The previous theorem gives a surprisingly simple picture of the class of Aronszajn

lines in the presence of PFA which is quite analogous to Laver’s celebrated result on

σ-scattered linear orders.

In chapter 1 we review some of the basic properties of the class of Lipschitz trees. This is

a special class of Aronszajn trees introduced by Todorcevic [36]. It is worth mentioning

that most of the deep results concerning Aronszajn trees as well as Aronszajn lines

involve this class in some way or other. It was shown by Todorcevic that under some

mild hypothesis the Lipschitz trees admit a simple representation as downward closed

subtrees of the tree ω<ω1 which are finite-to-one and coherent. The main result of this

ix

chapter is:

Theorem 0.0.2. The tree T (ρ0) is isomorphic to a coherent tree.

Theorem 0.02 answers a question made by Todorcevic (see [36]).

In chapter 2 we will present some applications of Todorcevic’s ρ functions. We shall give

some partial answers to an old question of Juhaz who asked whether the combinatorial

principle ♣ implies the existence of a Suslin tree. A variant of that question asks whether

♣ implies the existence of a non-special Aronszajn tree. We introduced various weak club

type principles, we called them star principles, in order to construct several non-special

Aronszajn trees. We also explore the relation between the star principles and some well-

known combinatorial principles in set theory such as ♦ and ♦∗.

In chapter 3 we show that the class of Aronszajn lines is well-quasi-ordered. Thus,

we have a satisfactory rough classification result of that class. In chapter 4 we shall

prove an equivalent result for the class of Aronszajn trees. In view of Todorcevic’s

results [36] the class of Aronszajn trees contains infinite strictly decreasing sequences as

well as uncountable antichains. Thus, we can not expect to obtain a satisfactory rough

classification result for the whole class of Aronszajn trees. So we are led to consider some

restricted subclass. The most natural class in this context is the class of Lipschitz trees.

As it has been shown in [36] this class is linearly ordered and cofinal and coinitial in the

class of Aroszan trees. Thus, a rough classification result of that class will provide us

with a big picture of the whole class. In order to provide a rough classification result for

the class of Lipschitz trees, since is not wqo, we need to understand its gap structure.

We shall give a detailed description of the gaps of size at most ω1 under MAω1 . From

our results we infer the following deep result:

Theorem 0.0.3 (PFA). The class of Lipschitz trees is the unique saturated linear order

of cardinality ℵ2.

x

The previous theorem gives us some of the first examples of a saturated structure

under PFA.

Some of the material of this thesis either has been published or will be published in the

coming year. Chapter 2 is essentially a reproduction of [22]. All but section 2 of chapter

3 is reproduced from [23] and the latter section of chapter 4 is reproduced from [24].

xi

Chapter 1

Lipschitz trees.

1.1 Lipschitz and coherent trees.

The notion of a tree in this thesis is to be interpreted in its order-theoretic sense, i.e.,

it is a partially ordered set (T,6T ) with the property that for every node t ∈ T the set

s ∈ T : s <T t is well ordered. Since the trees are well-founded we can define the

height of a node t as the order type of x ∈ T : x <T t, which will be denoted by

ht(t). The αth-level of T is the set Levα(T ) = t ∈ T : ht(t) = α, we may sometimes

use Tα to denote Levα(T ) when there is no risk of confusion. For each A ⊆ ω1 let

T A = t ∈ T : ht(t) ∈ A.

At this point I recall the following definition.

Definition 1.1.1. An Aronszajn tree is a tree of height ω1 in which all levels and chains

are countable.

For the rest of the thesis we shall assume that all trees have height ω1.

Considering trees as a generalization of ordinals, it is natural to say that a tree T is

smaller or equal than a tree S, denoted by T S, if there is a strictly increasing function

f : T → S. Note that if f : T → S is strictly increasing, then g : T → S defined

by g(t) = f(t) ht(t) is also strictly increasing. So, without loss of generality, we may

1

Chapter 1. Lipschitz trees. 2

restrict ourselves to strictly increasing maps that are also level-preserving. It turns out

that there is another way to introduce such maps. Consider the mapping ∆ : T 2 → Ord,

defined by

∆(t, s) = otpx ∈ T : x 6T t and x 6T s.

Definition 1.1.2. A partial map f from a tree T into a tree S is Lipschitz if f is

level-preserving and ∆(t, s) 6 ∆(f(t), f(s)) for all s, t ∈ dom(f).

Remark 1.1.3. The concept of a Lipschitz map was introduced by Todorcevic (see [36])

as a more natural and more general way to introduce strictly increasing maps. Note that

a partial level-preserving map f is Lipschitz if and only if it can be extended to a strictly

increasing map on the downward closure of its domain. So the two notions agree for

downward closed subtrees but being Lipschitz is otherwise more general.

Remark 1.1.4. Let T denote the complete binary tree of height ω+ 1 (26ω) then a map

f : 2ω → 2ω is Lipschitz (in the classical sense) if only if ∆(x, y) 6 ∆(f(x), f(y)) for all

x, y ∈ dom(f), i.e., f is Lipschitz in the sense of definition 1.1.2. We can think definition

1.1.2 as an extension of the classical definition of Lipschitz map to a broader class of

trees. We can view ∆ as a notion of distance where the inequality ∆(x, y) > ∆(x, z)

means that in some sense x is closer to y than to z.

In [36] Todorcevic isolates a special class of trees which have a prolific amount of

Lipschitz self-mappings.

Definition 1.1.5. A Lipschitz tree is any Aronszajn tree T with the property that ev-

ery level-preserving map from an uncountable subset of T into T is Lipschitz on an

uncountable subset of its domain.

Before examining some of the basic properties of Lipschitz trees. It would be helpful

to present a typical example of a Lipschitz tree. This would provide the reader with a

concrete object to test some of the results in this chapter.


Example 1.1.6. Let 〈eα : α < ω1〉 be a sequence of functions (eα : α→ ω for all α < ω1)

such that:

(i) For each α < ω1 the function eα is finite-to-one,

(ii) The sequence 〈eα : α < ω1〉 is coherent, i.e.,

∀α, β < ω1 ξ < minα, β : eα(ξ) 6= eβ(ξ) is finite.

Consider the tree T equal to the downward closure of the sequence 〈eα : α < ω1〉 in

the tree ω<ω1 . Clause (i) implies that the tree does not contain an uncountable branch

and condition (ii) implies that every level is countable. Thus, T is an Aronszajn tree.

Moreover, it can be shown that T does not contain a Suslin subtree, equivalently every

uncountable subset A of T contains an uncountable antichain. This fact together with

condition (ii) imply that T is indeed a Lipschitz tree.

A sequence of functions satisfying (i) and (ii) can be easily constructed by induction (see

[18]). But it turn out that there is a more canonical way to construct such a sequence

by means of a C-sequence, this will be explained in detail in section 2 of this chapter.

Remark 1.1.7. It is worth mentioning that if we restrict ourselves to countably branch-

ing trees of height ω1, there are essentially no Lipschitz trees outside the class of Aronszajn

trees. More precisely, assuming PFA any tree T of height ω1 in which every node has

extensions to all higher levels and satisfies the requirement of definition 1.1.5 it is in fact

an Aronszajn tree.

For the rest of this section we reproduce some results from [37] to be used in the rest

of the thesis. For convenience of the reader we reproduce some of their proofs.

Definition 1.1.8. For every Aronszajn tree T its nth power T ⊗ ... ⊗ T ( n times), is

defined as the set of all n-tuples of elements of T of equal heights equipped with the

coordinatewise ordering.


Lemma 1.1.9. For every Lipschitz tree T , and every positive integer n, and for every

uncountable subset A of the nth power of T . There is an uncountable B ⊆ A such that

∆(ai, bi) = ∆(aj, bj) for all (a1, ..., an) 6= (b1, ..., bn) in B and i, j < n. In particular, this

implies that the nth power of T is a Lipschitz tree.

Proof. Fix i, j < n by applying definition 1.1.5 to the uncountable partial map ai 7→

aj ((a1, ..., an) ∈ A) we can obtain an uncountable A0 ⊆ A such that ∆(ai, bi) 6 ∆(aj, bj).

Applying definition 1.1.5 to the inverse map aj 7→ ai((a1, ..., an) ∈ A0) will provide us with

an uncountable subset A1 ⊆ A0 such that ∆(ai, bi) = ∆(aj, bj). Repeating this procedure

successively for every pair i, j < n, we obtain the conclusion of lemma 1.1.9.

Lemma 1.1.10. Every uncountable subset of a Lipschitz tree T contains an uncount-

able antichain. More generally, every uncountable family A of pairwise disjoint finite

antichains of T contains an uncountable subfamily B such that⋃B is an antichain of

T .

Proof. Let X be given uncountable subset of a Lipschitz tree T . Note that if the down-

ward closure X of X contains an uncountable antichain so does X. Thus, we may assume

that X is already downward closed in T . By shrinking X, if necessary, we may also as-

sume that X does not contain any end-points. For every splitting node x of X we fix two

immediate successors x1 and x2 of x. Let Y denote the set of splitting nodes of X and set

Y0 = y0 : y ∈ Y . Using the fact that T is a Lipschitz tree we can get an uncountable

set Y ′ ⊂ Y0 such that the mapping f(y0) = y1 (y0 ∈ Y0) restricted to Y ′ is Lipschitz.

Note that if x0, y0 were two comparable elements of Y ′, let say x0 6 y0, then

∆(x1, y1) = ht(x) < ht(x) + 1 = ht(x0) = ∆(x0, y0).

Contradicting the fact that f Y ′ is Lipschitz. Therefore, X contain an uncountable

antichain. The second part of the lemma follows from the first and the fact that finite

powers of T are Lipschitz as well (see lemma 1.1.9). This finishes the proof.


Definition 1.1.11. An Aronszajn tree T is irreducible, if T U for every uncountable

downward closed subset U of T .

The following lemma plays a key role in our understanding of the structure of the

Lipschitz trees.

Lemma 1.1.12. Assume MAω1. Every Lipschitz tree is irreducible.

Using the previous lemma we get the following useful corollary.

Corollary 1.1.13. Assuming MAω1 . The following are equivalent for two Lipschitz trees

S and T :

(a) S T ,

(b) There is an uncountable partial Lipschitz level-preserving map f : S → T .

At this point I recall the following definition which encompass the trees that can be

represented as the one in example 1.1.6.

Definition 1.1.14. A coherent tree is a tree which can be represented as a family T of

functions from countable ordinals into ω such that ξ ∈ dom(s) ∩ dom(t) : s(ξ) 6= t(ξ)

is finite for all s, t ∈ T .

It turn out that under some mild assumptions the class of coherent trees exhaust the

class of Lipschitz trees which is the content of the following theorem of Todorcevic.

Theorem 1.1.15 (Todorcevic). Assuming MAω1. Every Lipschitz tree T is isomorphic

to a coherent tree. Conversely, every coherent tree T is Lipschitz.

1.2 The coherence of T (ρ0).

Theorem 1.1.15 tell us that in the context of MAω1 the class of Lipschitz Aronszajn

trees coincide with the class of coherent Aronszajn trees giving us an explanation of why


Lipschitz trees that one can define without appealing to additional axioms of set theory

are almost always coherent. The only Lipschitz tree, which is not given in this way is the

tree T (ρ0) of [37]. In this section of the thesis we shall show without relying on MAω1 or

any other additional set-theoretic principle that T (ρ0) does admit a representation as a

coherent tree.

This will provide an answer to a question of Todorcevic. He asked whether T (ρ0) is

indeed a coherent Aronszajn tree in ZFC. The question appears in print in [36]. In order

to do so we will recall the definition of T (ρ0) as well as some of its basic properties ( see

[38]).

Definition 1.2.1. A sequence 〈Cα : α ∈ ω1〉 is called a C-sequence (or laddersystem)

if satisfies the following properties:

1. Cα+1 = α,

2. Cα is an unbounded subset of α of order-type ω, whenever α is a countable limit

ordinal bigger than 0.

Definition 1.2.2. A step from a countable ordinal β towards a smaller ordinal α is the

minimal point of Cβ that is > α. The cardinality of the set Cβ ∩ α is the weight of the

step.

Definition 1.2.3. A walk (or a minimal walk) from a countable ordinal β to a smaller

ordinal α is the sequence β = β0 > ... > βn = α such that for each i < n, the ordinal

βi+1 is the step from βi towards α.

Analysis of this notion leads to several two-place functions on ω1 that have a rich

structure and many applications see [38]. We shall expose some of these functions in this

and subsequent chapters.


Definition 1.2.4. The full code of the walk is the function ρ0 : [ω1]2 → ω<ω, defined

recursively by

ρ0(α, β) = 〈|Cβ ∩ α|〉_ ρ0(α,min(Cβ \ α)),

with the boundary value ρ0(α, α) = ∅ where the symbol _ refers to the sequence ob-

tained by concatenating the one-term sequence 〈|Cβ ∩ α|〉 with the already known finite

sequence ρ0(α,min(Cβ \ α)) of integers.

For each α < ω1 we consider the fibre map ρ0α : α→ ω<ω defined by ρ0α(ξ) = ρ0(ξ, α).

The tree T (ρ0) is defined as the downward closure in the tree (ω<ω)<ω1 of the set of fibre

maps, in other words:

T (ρ0) = ρ0β α : α 6 β < ω1.

Definition 1.2.5. The full lower trace of the minimal walk is the function F : [ω1]2 →

[ω1]<ω, defined recursively by

F (α, β) = F (α,min(Cβ \ α)) ∪⋃

ξ∈Cβ∩α

F (ξ, α),

with the boundary value F (α, α) = α for all α.

We shall need the following standard facts about this notion.

Lemma 1.2.6. For all α 6 β 6 γ < ω1,

(a) F (α, γ) ⊆ F (α, β) ∪ F (β, γ),

(b) F (α, β) ⊆ F (α, γ) ∪ F (β, γ).

The next lemma is the key for understanding the structure of the tree T (ρ0).

Lemma 1.2.7. For all α 6 β 6 γ,

(a) ρ0(α, β) = ρ0(min(F (β, γ) \ α), β) _ ρ0(α,min(F (β, γ) \ α)),

(b) ρ0(α, γ) = ρ0(min(F (β, γ) \ α), γ) _ ρ0(α,min(F (β, γ) \ α)).


Remark 1.2.8. Given α 6 β, let 0 = ξ0, .., ξn = α denote the increasing enumeration

of F (α, β). For each 0 < i 6 n let σi denote the finite sequence ρ0(ξi, β). It follows from

the previous lemma that ρ0β(η) = σi _ ρ0ξi(η) for all 0 < i 6 n and η ∈ [ξi−1, ξ). This

can be restated as

(∀ 1 < i 6 n) ρ0β [ξi−1, ξi) = (σi _ ρ0ξi) [ξi−1, ξi),

where (σi _ ρ0ξi)(η) = σi _ ρ0ξi(η) for all η < ξi.

In order to show that T (ρ0) is a coherent tree we are going to use an alternative

definition of coherent tree more related to the automorphism structure of such a tree. As

we shall see this is more helpful in the context of T (ρ0). Let us recall that

T t = s ∈ T : s is comparable with t.

Definition 1.2.9. A tree T is called strongly homogeneous if there is a family of auto-

morphisms ht0,t1 : t0, t1 ∈ Tα, α ∈ ω1 with the following properties:

1. ht0,t1 moves T t0 to T t1 , so t0 is mapped to t1. ht0,t1 is the identity in all other parts

of the tree. In particular ht,t is the identity in all T ,

2. (commutativity) hs0,s2(t0) = hs1,s2(hs0,s1(t0)) holds for all s0, s1, s2 ∈ Tα with s0 6

t0,

3. (uniformity) If s0, s1 ∈ Tα with s0 6 t0 and s1 6 hs0,s1(t0) = t1 then ht0,t1 Tt0 =

hs0,s1 Tt0 and

4. (transitivity) If α is a limit ordinal and t0, t1 ∈ Tα then there exists s0, s1 ∈ T α

such that ht(s0) = ht(s1) and hs0,s1(t0) = t1.

Definition 1.2.10. A tree T is uniformly coherent if it is coherent and closed under

finite changes, i.e. T = t ∈ ω<ω1 : t =∗ s for some s ∈ T.1

1Here =∗ means that t is almost equal to s, i.e., ξ ∈ dom(t) ∩ dom(s) : t(ξ) 6= s(ξ) is finite.


We shall prove in the next two lemmas that conditions (1)-(4) of definition 1.2.9

provide us with a precise characterization of uniformly coherent trees.

Lemma 1.2.11. Every uniformly coherent tree is strongly homogeneous.

Proof. Let us assume that T is a uniformly coherent tree, let 〈tα : α < ω1〉 be any

level-sequence, i.e., tα ∈ Tα for all α < ω1. For t0, t1 ∈ Tα, s ∈ T t0 let us define

ht0,t1(s) = t1 ? s.

Where (t1 ? s) α = t1 and (t1 ? s)(ξ) = s(ξ) for α 6 ξ < ht(s). Then T will be strongly

homogeneous via ht0,t1 : t0, t1 ∈ Tα, α ∈ ω1.

Lemma 1.2.12. Every strongly homogeneous tree is isomorphic to a uniformly coherent

tree.

Proof. From now on let T be a strongly homogeneous tree via the family ht0,t1 : t0, t1 ∈

Tα, α ∈ ω1 of automorphisms. The splitting of T does not matter here, but we assume

for simplicity that T is ω-splitting, i.e., every node in T has countable many immediate

successors. We also assume without restriction that T has a root r0. We shall define

π : T → ω<ω1 by induction on the levels Tα (α < ω1) as follows:

If α = 0. Set π(r0) = ∅.

If α is a countable limit ordinal. Given t ∈ Tα set π(t) =⋃s<t π(s).

If α = β+1. Choose an x ∈ Tβ and let f = π(x) and well-order the immediate successors

of x by the enumeration xn : n ∈ ω. Define π(xn) = f ∪ (α, n) and for any other

s ∈ Tα set y = s β and if hy,x(s) = xm, define π(s) = π(r) ∪ (α,m).

Claim 1 π(t)(α) = π(hs0,s1(t))(α) holds whenever ξ 6 α < β, t ∈ Tβ and s0, s1 ∈ Tξ.

We may assume that s0 6 t. Now define t′ = hs0,s1(t). We fix a point x ∈ Tα, and a

fixed enumeration 〈xn : n ∈ ω〉 of the immediate successors of x. Let π(t)(α) = π(t

(α + 1))(α) = m. Since hs0,s1(t α + 1) = t′(α + 1), an application of uniformity yields

htα,t′α(t (α + 1)) = t′ (α + 1).


By commutativity, we can deduce

hx,t′α(htα,x(t (α + 1)) = t′ (α + 1),

so hx,t′α(xm) = t′ (α + 1). But this last equation means in particular that

m = π(t′ (α + 1))(α) = π(t′)(α)

since ht′α,x = h−1x,t′α and so we proved Claim 1.

Claim 2 If t0, t1 ∈ Tδ then the set

α < δ : π(t0)(α) 6= π(t1)(α)

is finite.

By induction on δ. This is obvious for successor steps, so let δ be limit: by transitivity

choose s0, s1 ∈ Tη, η < δ such that hs0,s1(t0) = t1. Claim 1 establishes the following

equation:

α < δ : π(t0)(α) 6= π(t1)(α) = α < η : π(s0)(α) 6= π(s1)(α).

But this last set is finite by induction hypothesis.

Claim 3 π : T → π′′(T ) is an isomorphism.

π is clearly order-preserving. To show that π is one-to-one, let π(s0) = π(s1). We proceed

by induction on β to show that s0, s1 ∈ Tβ implies s0 = s1. We can assume without

restriction that β is a successor ordinal and s0 (β − 1) = s1 (β − 1) = s (else use

induction hypothesis). Then hs,t(s0) = hs,t(s1) holds for any t ∈ Tht(s) by the definition

of π. But hs,t is an automorphism, s0 = s1. All that is left to show is the following Claim:

Claim 4 π′′(T ) is uniformly coherent.

It suffices to show the following: whenever t ∈ Tα, f = π(t) and f =∗ f ′ (f ′ : α → ω)

then f ′ ∈ π′′(T ). But this is clear by uniformity and the fact that the automorphisms are

onto.


In view of the previous lemma it suffices to show that T (ρ0) can be embedded into a

strongly homogeneous tree S. This is where lemma 1.2.7 comes into play. Let 〈Z<ω,_,∅〉

denote the free group of finite words of integers with the convention that 〈n〉_ 〈−n〉 = ∅.

We have a natural action from the group Z<ω into the set (Z<ω)<ω1 defined by:

(σ _ ρ)(ξ) = σ _ ρ(ξ) for all σ ∈ Z<ω, ρ ∈ (Z<ω)<ω1 and ξ ∈ dom(ρ).

Remark 1.2.13. Since it is not feasible to expect that the tree T (ρ0) is strongly homo-

geneous. We are forced to enlarge T (ρ0) to obtain a strongly homogeneous tree. In order

to do so, we are going to add all mappings that can be expressed as a finite union of

restrictions of elements of T (ρ0) shifted by members of the free group Z<ω. We can view

this, to some extent, as the analogous to the process of making a coherent tree closed

under finite changes.

Before giving a precise definition of the strongly homogeneous tree S let us introduce

some notation.

For each n ∈ ω let In be equal to

(ξ0, ..., ξn) ∈ ωn1 : ξ0 = 0 < ξ1 < ... < ξn.

Set S equal to

⋃

16i6n

(σi _ ρ0ξi) [ξi−1, ξi) : σi ∈ Z<ω and (ξ0, ..., ξn) ∈ In for some n ∈ ω.

Note that in view of Remark 1.2.8 we have the inclusion T (ρ0) ⊆ S. So we are left with

the following theorem.

Theorem 1.2.14. The tree S defined above is strongly homogeneous.

Proof. We will define the automorphisms for very pair t0, t1 ∈ Tα by induction on α. We

shall also make sure that the property 4 in definition 1.2.9 is preserved in our inductive

construction. Let α be given and suppose that hso,s1 have been defined for every pair

s0, s1 ∈ Tβ for any β < α.


If α = 0. Let h∅,∅ be the identity function on S,

If α = β + 1. Let t0, t1 ∈ Sα and t ∈ St0 be given. Let s0 = t0 β and s1 = t1 β then

hs0,s1(t0) = (β, σ0) ∪ s1 and t1 = (β, σ1) ∪ s1 for some σ0, σ1 ∈ Z<ω. Define

ht0,t1(t) = s1 ∪ (σ1 _ −σ0 _ hs0,s1(t)) [β, ht(t) + 1).

If α is a countable limit ordinal. Let t0, t1 ∈ Sα be given. If there exist s0, s1 ∈ S α

with ht(s0) = ht(s1) so that hs0,s1(t0) = t1 then define ht0,t1 = hs0,s1 Tt0 . The existence

of s0, s1 is the content of the following Claim.

Claim 1 For every limit ordinal α and t0, t1 ∈ Sα there exist s0, s1 ∈ S α with

ht(s0) = ht(s1) so that hs0,s1(t0) = t1.

We shall proceed by induction on α. By the definition of S there is a ξ < α so that

t0 [ξ, α) = (σ0 _ ρ0α) [ξ, α) and t1 [ξ, α) = (σ1 _ ρ0α) [ξ, α). Since ξ + 1 < α

by induction hypothesis the automorphism ht0(ξ+1),t1(ξ+1) is already defined. Note that

ht0(ξ+1),t1(ξ+1)(t0) = t1 as

ht0(ξ+1),t1(ξ+1)(t0) = t1 ξ ∪ (σ1 _ −σ0 _ σ0 _ ρ0,α) [ξ, α).

This concludes the proof of the claim.

We need to show that the set of automorphism satisfy properties (1)-(4) of definition

1.2.9. Clauses (1) and (3) follow from the construction of the automorphisms and (4) is

the content of Claim 1. So we are left with the following Claim:

Claim 2 The set of automorphism just defined satisfies commutativity.

We proceed by induction on the height of the pair defining the automorphism. The

case t0, t1, t2 ∈ Sα with α limit ordinal follows easily from the induction hypothesis. So

let us assume that α = β+1 and let t ∈ T t0 be given. Then ht0,t2(t) = s2∪ (σ2 _ −σ0 _

hs0,s2(t)) [β, ht(t) + 1) and ht1,t2(ht0,t1(t))) = s2 ∪ (σ2 _ −σ1 _ hs1,s2(hs0,s1(t)))

[β, ht(t) + 1) by induction hypothesis last equation is equal to

s2 ∪ (σ2 _ −σ1 _ σ1 _ −σ0 _ hs0,s2(t)) [β, ht(t) + 1).


Which concludes the proof of the Claim 2.

Chapter 2

Non-special coherent Aronszajn

trees.

In order to show that there is a Suslin tree in Godel’s constructible universe, Jensen

introduced the ♦ principle (see [14]).

Definition 2.0.15. ♦ is statement: There are sets Aα ⊂ α for every α < ω1 so that

(∀A ⊂ ω1) α < ω1 : Aα = A ∩ α is stationary.

The sequence 〈Aα : α < ω1〉 is called a ♦-sequence.

Since then several variations of ♦ has been considered in the literature. One such

variation is the ♣ principle. Let us recall that Λ denote the set of countable limit

ordinals.

Definition 2.0.16. ♣ is the statement: There are sets Aα ⊂ α for α ∈ Λ so that Aα is

cofinal in α for every α ∈ Λ and so that for every uncountable set A ⊂ ω1

α ∈ Λ : Aα ⊂ A is stationary.

The sequence 〈Aα : α < ω1〉 is called a ♣-sequence.

14

Chapter 2. Non-special coherent Aronszajn trees. 15

We recall a well-known question of I. Juhasz asking whether the combinatorial prin-

ciple ♣ implies the existence of Suslin trees. A variant of this question asks whether the

principle ♣ implies the existence of a non-special A-tree. Motivated by this question, we

investigate when a coherent Aronszajn tree associated with the function ρ1 is not special.

To do this, we define principles F0 and F1, and their corresponding weak versions wF0

and wF1. The principlesF0 andF1 are strong enough to construct non-special coherent

Aronszajn trees. All these principles are consistent with MAσ−centered.

2.1 Star-principles.

Definition 2.1.1. The principles F1, wF1, F0 and wF0.

1. F0: There is a C-sequence 〈Sα : α ∈ ω1〉 such that for every ϕ : Λ → ω there are

α, β ∈ Λ such that ϕ(α) = ϕ(β), Sβ ∩ α v Sα and α ∈ Sβ.

2. wF0: There is a C-sequence 〈Sα : α ∈ ω1〉 such that for every ϕ : Λ→ ω there are

α, β ∈ Λ such that ϕ(α) = ϕ(β) and α ∈ Sβ.

3. F1: There is a C-sequence 〈Sα : α ∈ ω1〉 such that for every stationary set S there

are α, β ∈ S such that Sβ ∩ α v Sα and α ∈ Sβ.

4. wF1: There is a C-sequence 〈Sα : α ∈ ω1〉 such that for every stationary set S

there are α, β ∈ S such that α ∈ Sβ.

Following [25] to every C-sequence 〈Cα : α < ω1〉 we associate a function ρ1, defined

as follows:

Definition 2.1.2. The function ρ1 = ρ1(Cα : α < ω1) : [ω1]2 → ω is defined recursively

by

ρ1(α, β) =

max|Cβ ∩ α|, ρ1(α,min(Cβ \ α)) if α < β

0 if α = β

,


Thus, ρ1(α, β) is simply the maximal integer appearing in the sequence ρ0(α, β).

Let us review the main properties of the ρ1 function.

Lemma 2.1.3. For all α < β < ω1 and n < ω,

(a) ξ 6 α : ρ1(ξ, α) 6 n is finite,

(b) ξ 6 α : ρ1(ξ, α) 6= ρ1(ξ, β) is finite.

Let ρ1α : α→ ω be defined by ρ1α(ξ) = ρ1(ξ, α) for every ξ < α. Then it follows from

the previous lemma that the sequence

ρ1α : α→ ω (α < ω1)

of finite-to-one functions is coherent. The corresponding tree

T (ρ1) = ρ1β α : α 6 β < ω1

is a homogeneous Aronszajn tree.

2.2 Star-principles and non-special Aronszajn trees.

The following two theorems show the relevance of the guessing principles F0 and F1.

Theorem 2.2.1. F0 implies that there is a non special coherent Aronszajn tree.

Proof. Let T = T (ρ1) be the coherent Aronszajn tree constructed from a F0-sequence

〈Sα : α < ω1〉 i.e. ρ1 = ρ1(Sα : α < ω1). To prove the theorem it is enough to check

that A = ρ1α : α ∈ Λ ⊆ T is not a countable union of antichains. Given any partion

ϕ : A→ ω of A, we define a new function ϕ : Λ→ ω by ϕ(α) = ϕ(ρ1α) for every α ∈ Λ.

It follows, using F0, that there are α, β ∈ Λ such that ϕ(α) = ϕ(β), Sβ ∩ α v Sα and

α ∈ Sβ. Then let us check that ρ1α ⊆ ρ1β. Let ξk : k 6 n be the increasing enumeration

of Sβ ∩ α. The proof proceeds by cases:


Case 1. If ξ ∈ [0, ξ0] then ρ0(ξ, β) = 〈0〉 _ ρ0(ξ, ξ0). Since Sβ ∩ α v Sα the same

holds for ρ0(ξ, α). Then by the definition of ρ1 we have that ρ1(ξ, α) = ρ1(ξ, β).

Case 2. If ξ ∈ (ξk, ξk+1] then ρ0(ξ, β) = 〈|Sβ ∩ ξ|〉 _ ρ0(ξ,min (Sβ \ ξ)). However,

Sβ ∩ α v Sα implies that xk+1 = min (Sβ \ ξ) = min (Sα \ ξ) and |Sβ ∩ ξ| = |Sα ∩ ξ| so

ρ1(ξ, β) = ρ1(ξ, α).

Case 3.

If ξ ∈ (ξn, α) then ρ0(ξ, β) = 〈n|〉_ ρ0(ξ, α), and ρ0(ξ, α) = 〈|Sα∩ξ|〉_ ρ0(ξ,min (Sα \ ξ)).

However, since Sβ ∩ α v Sα, n 6 |Sα ∩ ξ| so we have that ρ1(ξ, α) = ρ1(ξ, β).

Then ρ1α(ξ) = ρ1β(ξ) holds for every ξ < α. So we are done.

Theorem 2.2.2. F1 implies that there is a coherent Aronszajn tree T which does not

have stationary antichains.

Proof. Let T = T (ρ1) be the coherent Aronszajn tree constructed from a F1-sequence

〈Sα : α < ω1〉 i.e. ρ1 = ρ1(Sα : α < ω1). The result follows using the same argument as

in the previous theorem and the following claim.

Claim. T has a stationary antichain if and only if the set ρ1α : α ∈ ω1 contains one

such antichain.

Let us prove the claim. Let A be a stationary antichain of T , we may assume that

|Tα ∩ A| 6 1 for every α ∈ ω1. For each tα ∈ A there is an Fα ∈ [ω1]ω such that

tα(ξ) = ρ1α(ξ) for every ξ ∈ (α \ Fα). By the pressing down lemma, we can find a

stationary antichain S ⊆ A such that Fα = F for every α ∈ S. Now, set S ′ = ρ1α : tα ∈

S. It follows from the homogeneity of T that S ′ is a stationary antichain contained in

ρ1α : α ∈ ω1, and this finishes the proof.


2.3 The star-principles and their relationship with

some standard set-theoretic axioms.

As we have seen, the principles F0 and F1 are guessing principles which imply the

existence of non-special Aronszajn trees. In order to have a better understanding of

these principles we will compare them with other some well known principles in set

theory, summed up in the following diagram.

♦+

↓

F1 → wF1 8 ♦

↓ ↓

F0 → wF0 → ¬MAω1

↓ ↑

NSAT ♦

Here NSAT is the abbreviation for the statement that there is a non-special Aron-

szajn tree. As the following theorem shows all the principles wF0,F0, wF1 and F1are

relatively consistent with ZFC +MAσ−centered.

Theorem 2.3.1. If V [G] is the generic extension obtained by adding a single Cohen real

then V [G] |=F1.

Proof. From now on assume that c : ω → [ω]<ω is a Cohen-generic real and eα : α →

ω (α < ω1) is a coherent sequence of finite-to-one functions, i.e., ξ < α : eα(ξ) 6= eβ(ξ)

is finite for all α < β ∈ ω1. Let 〈Cα : α < ω1〉 be an arbitrary C-sequence. We use the

Cohen-generic real to obtain a new C-sequence 〈Sα : α < ω1〉 in the following way:

Sα = ξ < α : Cα(n) 6 ξ < Cα(n+ 1), eα(ξ) ∈ c(n) for some n ∈ ω,

where Cα(0) = 0 and Cα(n) is the nth element of Cα for 0 < n < ω. Note that since

e′αs are finite-to-one we have that the order-type of Sα is equal to ω for every α < ω1.


Let us check that Sα is a witness for F1. Assume that A is a stationary subset of ω1.

Note that if A is stationary in V [G], then there is a stationary set A0 ∈ V such that

A0 ⊂ A. So without loss of generality we may assume that A is in the ground model. Fix

p ∈ Fn(ω, [ω]<ω) with dom(p) ∈ ω, use the pressing down lemma to find a stationary set

S ⊂ A such that Sα agrees with Sβ in the initial segment of Sα and Sβ decided by p for

every α, β ∈ S. Pick a β such that ot(β ∩ S) > ω, now choose an α ∈ S in such a way

that Cβ(n0) < α 6 Cβ(n0 + 1) where dom(p) < n0. Let q be defined by

q(n) =

p(n) if n ∈ dom(p)

∅ if dom(p) < n < n0

α if n = n0

then q “Sβ ∩ α @ Sα”.

Corollary 2.3.2. F1 (and hence also F0, wF1 and wF0) are relatively consistent with

MAσ−centered.

Proof. Let V be a model of MA and P a forcing which adds a single Cohen real. By

the previous theorem if G is a P-generic filter then M [G] |= F1 and by the theorem of

Roitman (see [2]) the extension M [G] |= MAσ−centered.

Remark 2.3.3. It should be noted that the fact that a Cohen real r can modify ρ1 to a

coherent ρr1 so that the corresponding tree T (ρr1) has no stationary antichain appear as

lemma 2.2.17 in [38].

The following propositions give us some relationship between ♦ and ♦∗ with our

guessing principles.

Proposition 2.3.4. ♦ implies wF0.

Proof. Let 〈ϕα : α ∈ ω1〉 be a ♦-sequence which guesses elements of ωω1 (i.e. ϕα ∈ ωα).

Define Xα = n : ϕ−1α (n) is cofinal in α for every limit α. For every α ∈ Λ choose


Sα ⊆ α of order type ω such that Sα ∩ ϕ−1(n) is a cofinal in α for every n ∈ Xα. This

is very easy to do. Let us check that the C-sequence 〈Sα : α < ω1〉 has the required

properties. Now, let ϕ : Λ→ ω be given. Set X = n ∈ ω : ϕ−1(n) is cofinal in ω1 and

C = α : ∀n ∈ X (ϕ−1(n) is cofinal in α). It is easy to see that C is a club in ω1. Let be

ξ0 = maxϕ−1(n) : n /∈ X+ 1 and S = α : ϕα = ϕ α. Pick any β ∈ C ∩ S ∩ [ξ0, ω1)

then ϕ(β) = n0 ∈ Xβ. It follows from the properties of Sβ that there is an α ∈ Sβ such

that ϕ(α) = n0.

At this point I recall the following definition.

Definition 2.3.5. ♦∗ is the statement: There are sets Aα ⊂ P(α) for every α < ω1,

such that |Aα| 6 ω and for each A ⊂ ω1, there is a club C so that

∀α ∈ C (A ∩ α ∈ Aα).

The sequence 〈Aα : α ∈ ω1〉 is called a ♦∗-sequence.

Proposition 2.3.6. ♦∗ implies wF1.

Proof. Let 〈Aα : α ∈ ω1〉 be a ♦∗-sequence. For each α, let Sα ⊂ α be a sequence of

order-type ω such that Sα ∩ A 6= ∅ for every A ∈ Aα (this can be done by an easy

induction). Let us verify that 〈Sα : α ∈ ω1〉 is a witness for wF1. Given a stationary

set S, there is a club C such that ∀ α ∈ C (S ∩ α ∈ Aα). Pick any β ∈ (C ∩ S) then

Sβ ∩ (S ∩ β) 6= ∅, now choose α ∈ Sβ ∩ (S ∩ β). Then α, β ∈ S and α ∈ Sβ. So we are

done.

We do not know if in the previous propositions we can replace the weak versions for

the stronger ones. However, we have some limitations as the following theorem shows.

Theorem 2.3.7. ♦ does not implies wF1.

To prove the theorem we need the following lemmas.


Lemma 2.3.8. For every C-sequence 〈Sα : α ∈ ω1〉 there is an α such that for every

β > α, γ : (Sγ \ α) ∩ β = ∅ is stationary.

Proof. Suppose that this is not the case. Then for every α there is a β(α) and a club Cα

such that (Sγ \α)∩ β(α) 6= ∅, whenever γ ∈ Cα. Pick α0 ∈ ω1 and define αn+1 = β(αn).

Let ξ ∈⋂n∈ω Cαn be greater than α = supαn : n ∈ ω . Since Sξ intersects each interval

[αn, an+1), α is an accumulation point of Sξ, so the order-type of Sξ is greater than ω,

which is a contradiction.

The following lemma is a well known fact.

Lemma 2.3.9. 1. Countable support iteration of σ-closed forcings is σ-closed,

2. Every σ-closed forcing preserves ♦.

Proof of theorem 2.3.7. For every C-sequence C = 〈Cα : α ∈ ω1〉, define the notion of

forcing PC where

PC = p ∈ 2<ω1 : ∀α ∈ p−1(1), Cα ∩ p−1(1) = ∅ and p αC ≡ 0

Here αC is the α in the previous lemma which correspond to the C-sequence C, and the

order is by extension.

Claim 1. PC is a σ-closed forcing.

Let pn be a decreasing sequence of conditions in PC and set pω =⋃n∈ω pn. Obviously,

pω ∈ 2<ω1 and pω αC ≡ 0. Suppose that there are α, β ∈ p−1ω (1) such that α ∈ Cβ i.e.

Cβ ∩ p−1ω (1) 6= ∅, then there are n,m ∈ ω such that α ∈ dom(pn) and β ∈ dom(pm) but

this implies that α, β ∈ p−1m+n(1) and Cβ ∩ p−1

m+n(1) 6= ∅ wich is a contradiction.

Claim 2. PC force that C is not a wF1-sequence.

Let fG be the PC-generic function and S = p−1(1). To see that C is not a witness for

wF1 in M [G] it suffices to prove that S is stationary in M [G]. Let C be a name for a

club and p a condition such that p “C is a club”. Let ξ be a bound for all the places of


S fixed by p, ξ > αC. Then by the previous lemma, we can find a sequence M0 ⊆M1 ⊆

... ⊆ Mn ⊆ ... of countable elementary submodels of H(c) such that (Lδn \ αC) ∩ ξ = ∅

where δn = Mn ∩ ω1, and p, 〈Lα : α ∈ ω1〉 ∈ Mo. Set Mω =⋃n∈ωMn and δ = Mω ∩ ω1.

We will construct a sequence pn of conditions such that pn+1 6 pn, pn “δn ∈ C”,

p−1n (1) ∩ Lδ = ∅ and pn ∈Mn by recursion as follow: let ξ0 = max(Lδ ∩ δn), and extend

p to a condition q = p ∪ (α, 0) : α ∈ [dom(p), ξ0]. Since M0[G] |= “C is a club” there

is a η0 ∈ ω1 ∩M0 and a r ∈ P ∩M0 such that r “η0 ∈ C”. Set po ∈ M0 (p0 6 r, q).

For the inductive step assume that we have constructed pk for k 6 n with the required

properties, let ξ be a bound for all the places fixed by pn since pn ∈ Mn it follows that

we may assume without loss of generality that ξ < δn now set ξn+1 = max(L∩δn+1) since

ξ < dn the function q = pn∪(α, 0) : α ∈ [dom(pn), ξn+1] is a well defined condition using

that q “C is a club” there is a ηn+1 ∈ δn+1 and a condition r such that r “ηn+1 ∈ C”

then set pn+1 ∈Mn+1 with pn+1 6 q, r. Finally, pick a q ∈Mω such that qω 6 pn for every

n ∈ ω, and let pω = q ∪ (α, 0) : α ∈ [dom(q), δ], pω is well defined since dom(q) ∈ Mω.

Then supn∈ωηn = δ and pω “ηn ∈ C and p−1ω (1)∩Lδ = ∅. Since C is a name for a club

pω “δ ∈ S ∩ C. So S is stationary and Claim 2 holds.

Assume V |= V = L and construct a countable support iteration P = 〈Pα, Qα :

α < ω2〉 so that Pα “Qα = PC for some C-sequence C”. By a standard book-keeping

argument one can make sure that all C-sequences in the intermediate models are listed.

Let G be a P-generic. Since every C-sequence C in M [G] has a Pα-name for some α < ω2,

and in some stage α < β < ω2 we have that Qβ = PC. Then Qβ is adding a stationary

set S witnessing that C is not a wF1-sequence in M [Gβ]. Since the countable support

iteration of σ-closed forcing is proper, S remains stationary through all the itereation.

So M [G] |= ¬wF1 and by the lemma 2.3.9. we have that M [G] |= ♦ so we have that

M [G] |= ¬wF1 ∧ ♦.

Finally we show that none of the principles is consistent with Martin’s Axiom.


Theorem 2.3.10. MA(ω1) implies ¬wF0.

Proof. Let 〈Cα : α < ω1〉 be a C-sequence. Define

P = p : A→ ω : A ∈ [Λ]<ω, (∀α < β)(ϕ(α) = ϕ(β) → α /∈ Cβ)

ordered by inverse inclusion. It is easy to see that, if fG is the generic function, then fG

is defined on Λ and forces that 〈Lα : α < ω1〉 is not a wF0-sequence, to assure both we

need to meet on ω1 many dense sets. The theorem follows from the following claim.

Claim P is a c.c.c. forcing.

Suppose that pα : α ∈ ω1 is an antichain. By a standard ∆-system type argument,

we can assume that their domains form a ∆-system with root r, such that there is a

N ∈ ω with |dom(pα)| = N for each α ∈ ω1 and all the functions agree on r. Moreover,

we can assume that dom(pα) ∩ dom(pβ) = ∅ for every α, β ∈ ω1, and

max(dom(pα)) < min(dom(pβ)) if α < β.

Now, set dom(pω·N+1) = ξ1, ..., ξN. Since pω·N+1 is incompatible with pα for every

α < ω ·N + 1,

(N⋃i=1

Lξi) ∩ dom(pα) 6= ∅ for every α < ω ·N + 1.

Then by the pigeon hole principle there is a i such that ot(Lξi) > ω + 1. But this

contradicts the fact that 〈Cα : α < ω1〉 is a C-sequence, so we are done.

Chapter 3

The Aronszajn lines are

well-quasi-ordered by embeddability.

A rough classification result for a given class K of mathematical structures usually de-

pends on a reflexive and transitive relation , i.e., a quasi-ordering where for A and B

in K the relation A B means that in some sense A is simpler than B. The strength

of the rough classification result depends not only on how fine the corresponding equiv-

alence relation is (A ≡ B iff A B and B A) but also on the information about the

quasi-ordering (K,) it gives. One of the most prominent global conditions, generally

considered, as giving a satisfactory rough classification result is the requirement of being

well-quasi-ordered. Recall that a class (K,) is well-quasi-ordered (or wqo) if for every

infinite sequence An(n ∈ ω) of elements of K there exists n < m such that An Am. The

sense of strength of such a rough classification result comes from the fact that whenever

(K,) is well-quasi-ordered then the complete invariants of the equivalence relation ≡

on K are only slightly more complicated than the ordinals.

In this chapter we are interested in proving such rough classification result for a class

K of linear orderings. Recall that in this context the quasi-ordering is usually taken

to be isomorphic embedding, i.e., A B iff there is a strictly increasing map f : A→ B.

24

Chapter 3. The Aronszajn lines are well-quasi-ordered by embeddability.25

The first result of this sort is a result of Laver [19] who showed, verifying an old conjecture

of Fraisse [11], that not only the class of countable linear orderings is well-quasi-ordered

but also the class of all σ-scattered linear orderings. Some restrictions on the linear

orderings in Laver’s result are needed in view of a classical result of Dushnik-Miller [9]

who proved that the class of separable linear orderings of size continuum (more precisely,

suborders of the real line) fails badly to be well-quasi-ordered. The idea behind Dushnik-

Miller’s construction combined with ideas behind the more recent constructions using

the so-called weak-diamond principle of Devlin and Shelah show that under CH there is

basically no room for extending Laver’s theorem to a larger class of linear orderings. It is

for this reason that one is naturally led to examine this possibility using some alternative

to CH such as the Proper Forcing Axiom, PFA.

However even assuming PFA the situation is not so simple as there are natural restrictions

that do not depend on CH or any other additional axioms. For example, in [4], Baum-

gartner showed that the class Bℵ1 of non σ-scattered linear orderings A of size ℵ1 with

the property that every uncountable subset of A contains an uncountable well-ordered

set (an isomorphic copy of ω1) is not well-quasi-ordered. Thus one is led to consider only

restricted classes of uncountable linear orderings. One such class is the class Rℵ1 of car-

dinality ℵ1 suborders of the real line. These were completely classified by Baumgartner

[3].

Theorem 3.0.11. [3] (PFA) Any two ℵ1-dense suborders of R are order-isomorphic. In

particular, any two elements of Rℵ1 are equivalent.

Another class which arises naturally in this context is the class of Aronszajn lines,

i.e., the linear orderings which do not contain uncountable separable suborders or ω1 or

ω∗1. The following theorem is one of the main result of the present thesis.

Theorem 3.0.12. (PFA) The class A of Aronszajn lines is well-quasi-ordered by embed-

dability.


As we mention above there is a very close relationship between Aronszajn trees and

Aronszajn lines, any lexicographical ordering of an Aronszajn tree is an Aronszajn line,

and conversely, any binary partition tree of an Aronszajn line is an Aronszajn tree (see

[35]). It is for this reason surprising that there is a discrepancy between Aronszajn trees

and Aronszajn lines when it comes to the wqo-theory. Our Theorem 3.0.12 is in contrast

with a result of Todorcevic from [36] where it is proved that the class of Aronszajn trees

is not wqo under embeddability. The question of whether A is well-quasi-ordered appears

in print in the survey article by Moore [25].

Besides the ideas of Nash-Williams and Laver from the wqo-theory, while proving

theorem 3.0.12, we shall also rely on some ideas behind the deep results obtained by

Todorcevic [36] and Moore [26].

In 1970 Countryman [6] made a brief but important contribution to the subject by asking

whether there is an uncountable linear order C whose square is the union of countably

many chains. Such an order is called Countryman (or C-line). It can be seen that every

C-line is Aronszajn and that if C is a C-line and C∗ denotes its reverse, then no uncount-

able linear order can be embedded into both C and C∗. Shelah [31] proved that such

orders exists in ZFC and Todorcevic [39] produced a number of concrete representations

of C-lines. Moore [27] proved, solving a longstanding conjecture of Shelah, the following

deep result.

Theorem 3.0.13. [27](PFA) The class of A-lines contains a two element basis consisting

of C and C∗ where C is any Countryman line.

Furthermore, in [26] Moore proved that, assuming PFA, there is a universal A-line

ηC .

Theorem 3.0.14. [26] (PFA) Every A-line is isomorphic to a suborder of ηC.

Moreover ηC can be easily described in terms of a fix C-line C. Let D = C∗ + 1 + C

then ηC consist of all elements of Dω which are eventually zero ordered lexicographically.


It is worth mentioning that in view of the following result of Abraham and Shelah [1]

some extra assumptions, such as PFA, are needed in our main result.

Theorem 3.0.15. [1] (2ℵ0 < 2ℵ1) There is a collection F of pairwise incomparable

A-lines of cardinality 2ℵ1 .

Furthermore, the previous theorem can be easily modified, by using the tree T (ρ1) (

see [36]), to obtain a family F of pairwise incomparable C-lines of cardinality 2ℵ1 .

3.1 Well-quasi-orderings.

It will be helpful to fix some notation and review some basic facts about the theory of

well-quasi-orderings. First, recall that a quasi-order is a structure of the form (Q,)

where is a transitive and reflexive binary relation.

Definition 3.1.1. A quasi-order (Q,) is a well-quasi-order (or wqo) if satisfies any of

the following two equivalent conditions:

1. For any function f : ω → Q there exists i < j such that f(i) f(j).

2. Any strictly decreasing sequence of members of Q is finite, and every antichain of

members of Q is finite.

The equivalence between both definitions is an immediate consequence of Ramsey’s

theorem.

Example 3.1.2. The most typical example of a well-quasi-ordering is the class ON of

ordinal numbers with its usual order.

A more interesting example is giving by the class C of countable linear orders quasi-

ordered by embeddability. This is a non-trivial fact and it is due to Laver [19].

We will define some useful operations on quasi-orderings.


Definition 3.1.3. Given a quasi-ordering (A,≺) we define its reverse A∗ = (A,≺∗) by

a ≺∗ b iff b ≺ a.

Definition 3.1.4. Let A,B be quasi-orderings.

1. We define A × B as the lexicographical ordering on the cartesian product, i.e.,

(a1, b1) ≺ (a2, b2) if (a1 ≺A a2) or (a1 = a2 and b1 ≺B b2).

2. Define A+B as the quasi-ordering on (A× 0) ∪ (B × 1) given by a ≺A+B b if

(π1(a) ≺ π1(b)) or (π1(a) = π1(b) and π0(a) ≺ π0(b)).

3. Let I be a quasi-ordering and let Ai (i ∈ I) be a collection of quasi-orderings. We

define the sum∑

i∈I Ai to be the quasi-ordering (C,≺C) where C =⋃i∈I Ai × i

and x ≺C y if (π1(x) ≺I π1(y)) or (π1(x) = π1(y) and π0(x) ≺Aπ1(x)π0(y)).

In order to make the induction hypothesis in our main theorem go through we will

need a generalization of a quasi-ordering called Q-type. Intuitively a Q-type is a quasi-

ordered set whose points are labeled by members of Q.

Definition 3.1.5. If Q quasi-ordering, a Q-type is a pair (A, f) where A is a linearly

ordered set and f is a function from A into Q.

We quasi-order the class of Q-types by the following embeddability relation:

(A1, f1) (A2, f2) if there is a strictly increasing function f : A1 → A2

such that f1(x) f2(f(x)) for all x ∈ A1.

One way to motivated this definition is as follows: Given two linear orderings A,B

formed by sums of linear orderings, say,

A =∑x∈X

Ax and B =∑y∈Y

By.

We can associate to A,B a natural Q-type structure given by (A, x 7→ Ax) and (B, y 7→

By), respectively. Observe that if (A, x 7→ Ax) (B, y 7→ By) as Q-types then A embeds

into B.


Definition 3.1.6. If M is a class of linear orders and Q is a quasi-order let QM =

(A, f) : A ∈M, f : A→ Q

In order to prove our result we will need a generalization of the theory of well-quasi-

orderings called better-quasi-orderings (bqo), a concept introduced by Nash-Williams

[29].

Remember that [ω]ω represents the set of infinite subsets of natural numbers. We will

consider [ω]ω as a topological space with the Ellentuck topology, which is topologized

by the basic open sets of the form

[s, A] = X ∈ [ω]ω : s @ X ⊂ A

for A ∈ [ω]ω and s ∈ [ω]<ω.

We are now in position to state the concept of better-quasi-order.

Definition 3.1.7. Let Q be a quasi-order. Q is a better-quasi-ordering (or bqo) if for

every Borel map f : [ω]ω → Q there exists X ∈ [ω]ω such that f(X) f(X \min(X)).

Even though the concept of bqo might appear unintuitive at first it is more natural

in the sense that the property of being bqo is preserved for almost all operations on

quasi-orders contrary to the case of well-quasi-orders.

It is worth mentioning that our definition of better-quasi-ordering is not the original one

given by Nash-Williams[29]. Our definition comes from [38].

It is easy to see that every better-quasi-ordered set is well-quasi-ordered but the reverse

implication does not necessarily holds as seen in the next example:

Example 3.1.8 (Rado). Let Q denote the set of all pairs (i, j) so that i 6 j quasi-

ordered by (i, j) ≺ (k, l) if either i = k and j < l or j < k. It can be easily seen that Q

is an example of a well-quasi-order which is not better-quasi-order. Moreover, the class

Qω is not well-quasi-ordered by embeddability as a Q-type.

Using the concept of better-quasi-order we can state the main theorem of Laver.


Theorem 3.1.9. [19] Let S denote the class of σ-scattered linear orderings, and let Q

be a better-quasi-order. Then QS is better-quasi-ordered by embeddability as a Q-type. In

particular QC is bqo where C denotes the class of countable linear orders.

Observe that in view of example 3.1.8 the requirement of being better-quasi-ordered

in Laver’s theorem is essential.

3.2 A-lines.

In this section we shall collect some standard facts about the class of A-lines which be

helpful in the rest of the chapter.

Definition 3.2.1. Let A denote the class of Aronszajn lines.

We are ready to define an operation which connects A-trees and A-lines. So let

T be an A-tree. For every α < ω1, we fix a linear ordering 6α of Tα. Then the

lexicographical ordering l of T induced by6α: α < ω1 is defined by t l s iff.

(i) t 6T s or,

(ii) t, s are incomparable and t∆(s,t) 6∆(t,s) s∆(t,s).

The following are standard facts about A-lines and A-trees (see [35]).

Fact 3.2.2. Every lexicographical ordering of an Aronszajn tree is an Aronszajn line.

Let us now define an inverse operation which connects A-lines and A-trees. This

operation is called a process of atomization of a A-line (A,6A), and it is a process of

inductive construction of families Tα, α ∈ ON of non-empty convex subsets of A such

that:

(i) If α = 0, then Tα = A,


(ii) If α = β + 1, then for each non-trivial interval I ∈ T β there exists disjoint I0, I1 ∈

T a such I0 ∪ I1 = I, and

Tα = I0, I1 : I ∈ T β and |I| > 2.

(iii) If α is a limit ordinal, then

Tα = ⋂

b : b ⊂⋃β<α

T β, b ∩ T β 6= ∅ for all β < α, and⋂

b 6= ∅.

It is clear that for some α, Tα = ∅, hence we may define

ht(T ) = minα : Tα = ∅ and T =⋃

α<ht(T )

Tα.

Then (T,⊇) is a tree and Tα is the α-th level of T for all α < ht(T ). Any tree which is

a result of an atomization process of A is called a partition tree of A.

Fact 3.2.3. Every partition tree of an Aronszajn line is an Aronszajn tree.

As we can see from Facts 3.2.2 and 3.2.3 there is a strong duality relation between the

Aronszajn lines and the Aronszajn trees. At this point we recall the following definitions.

Definition 3.2.4. A Suslin tree is a tree T such that |T | = ℵ1 and every chain and

every antichain of T has countable cardinality.

Definition 3.2.5. A Suslin line is a non-separable linear ordering A with the countable

chain condition (or ccc), i.e., every family of pairwise disjoint non-empty open intervals

of A is countable,

Fact 3.2.6. (i) Every lexicographical ordering of a Suslin tree is a Suslin line. Moreover,

every Suslin line A is isomorphic to a lexicographical ordering of a Suslin tree.

(ii) Every partition tree of a Suslin line is a Suslin tree.


Definition 3.2.7. A linear ordering A is ℵ1-dense if A has cardinality ℵ1, if it has no

end-points, and if between any two elements of A there are exactly ℵ1 elements of A.

Definition 3.2.8. An uncountable linear ordering C is a Countryman line (C-line, in

short) if its lexicographical square is a countable union of chains. C2 is quasi-ordered by

(a1, b1) (a2, b2) if a1 a2 and b1 b2.

The C-lines play a prominent role in the structure theory of the class of A-lines, under

PFA, as they constitute the building blocks of the class of A-lines. This will be explained

in detail in section 4.

Fact 3.2.9. If C is Countryman, then C does not contain a Suslin suborder.

Proof. Observe that if C is Countryman then C remains Countryman in any forcing

extension which preserves ℵ1. It suffices to show that any Suslin line A fails to be

Countryman in some ccc forcing extension. Let A be a Suslin line. Using fact 3.2.6 we

can find a Suslin tree (T,6T ) so that A is isomorphic to a lexicographical ordering of T .

Forcing with the ccc poset (T,>T ) we add a copy of ω1 to A which implies that A is not

Countryman in V [G].

For the rest of the paper fix an ℵ1-dense Countryman line which we denote by C. For

example, to be specific, we fix an ℵ1-dense subordering C of C(ρ0) (see [37], p.25 )

Fact 3.2.10. [37](MAω1) C is equivalent to any uncountable suborder A of C.

Fact 3.2.11. (MAω1) C is equivalent to C × C.

Proof. It should be clear that C C ×C. So let us show that C ×C C. By fact 3.2.9

C is not Suslin so we can find a family Iα (α ∈ ω1) of pairwise disjoint non-empty open

intervals of C. For each α < ω1, we fix an element xα ∈ Iα and set X = xα : α < ω1. By

fact 3.2.10 and the since C is ℵ1-dense we can find a strictly increasing map fα : C → Iα


for all α < ω1 and a strictly increasing map f : C → X. Define a map F : C× → C by

F (x, y) = fα(y) where f(x) = xα.

It is easy to check that F is a strictly increasing map.

The following result of Moore [26] is a generalization of the existence of a two element

basis for the class of Aronszajn lines.

Theorem 3.2.12. [26](PFA) If A is an A-line, then either A is equivalent to ηC or else

A contain an interval equivalent to C or C∗.

3.3 Aronszajn lines and the Weak-Diamond princi-

ple.

In this section we shall use the weak-diamond principle to construct and antichain of

A-lines of size 2ℵ1 .

Lemma 3.3.1. There exist an scheme Ts : s ∈ 2ω1 such that Ts ⊂ (ωω)<|s|, for s of limit

height the |s| level of Ts_0 and Ts_1 are disjoint and for all f ∈ 2ω1⋃α∈ω1

Tfα is an

A-tree.

Proof. In order to do that we will be using the tree T (ρ0) of [37]. Let Cα, C′α (α ∈ ω1)

be C-sequences such that Cδ ∩ C ′δ = ∅ for every limit ordinal δ.

Claim Let δ be a limit ordinal then T (ρ0(〈Cα : α < ω1〉))δ and T (ρ0(〈C ′α : α < ω1〉))δ

are disjoint.

Let ρ0α δ and ρ′0β δ elements of δth level of T (ρ0(〈Cα : α < ω1〉)) and T (ρ0(〈C ′α : α <

ω1〉)), respectively. Let ξ0 = min(Cδ \ ((Cα ∪C ′β)∩ δ) and ξ1 = min(C ′δ \ ((Cα ∪C ′β)∩ δ).

ρ0(ξ0, α) = ρ0(δ, α) _ 〈|Cδ ∩ ξ0|〉 & ρ′0(ξ0, β) = ρ′0(δ, β) _ ρ′0(ξ0, δ)


Length of ρ0(ξ0, δ) is at least two since ξ0 ∈ (Cδ \ C ′δ). Hence ρ′(δ, β) ⊆ ρ0(δ, α), by

a similar argument using ξ1 we can conclude that ρ0(δ, α) = ρ′0(δ, β). It follows that

ρ0(ξ0, α) 6= ρ′0(ξ0, β) which finish the proof of the claim.

The rest of the construction follows from a straight forward recursive definition.

The following is a well-known fact.

Fact 3.3.2. If (A,6) is an A-line and T and S are trees associated with A, then

α : Tα = Sα contains a club.

The previous fact give us a hint of how construct an antichain of A-trees. It suffices

to provide a family of A-tree that are pairwise non club embeddable into each other, i.e.,

if T, S are trees in such a family and C is a club then T C is not embeddable into S.

It is worth mentioning that under PFA any two A-trees are club-isomorphic so this can

not be done in ZFC. In order to do this we will use the so called weak-diamond principle

of Devlin and Shelah that will be denote by WDP for short in this paper.

Let us first review some basic definitions and results about WDP

The Weak Diamond Principle (WDP): Suppose |A| 6 2ℵ0 then

(∀F : 2<ℵ1 → 2)(∃g : ω1 → 2)(∀f : ω1 → A) α : g(α) = F (f α) is stationary.

It is proved in Devlin-Shelah [8] that 2ℵ0 < 2ℵ1 implies WDP and the converse is trivial,

so in fact WDP is a reformulation of 2ℵ0 < 2ℵ1 .

Furthermore, WDP is self-improving to the following extent (see [3]):

Lemma 3.3.3. Suppose |A| 6 2ℵ0 and for all α < 2ℵ0 , Fα : 2<ω1 → 2. Then (∃g : ω1 →

2)(∀f : ω1 → A) and (∀α < 2ℵ0)α : g(α) = Fα(f α) is stationary.


Let X ⊂ ω1. We say that WDP (X) holds provided that if A has size at most

continuum, and F : 2<ω1 → 2, then

(∃g : X → 2)(∀f : ω1 → A) α ∈ X : g(α) = F (f α) is stationary.

In [8], Devlin and Shelah show that X ⊂ ω1 : WDP (X) fails is a non-trivial

normal ideal. It follows immediately from an old argument of Ulam that there is a

partition 〈Xα : α < ω1〉 of ω1 such that WDP (Xα) holds for all α. Using this fact and

same techniques of [1] we can now prove.

Theorem 3.3.4. If 2ℵ0 < 2ℵ1 then there is a family of 2ℵ1 pairwise incomparable C-lines.

Proof. Let Ts : s ∈ 2<ω1 be an scheme as in Lemma 3.3.2 Note that since each tree in

our construction is of the form T (ρ0) for some C-sequence, we have that every A-line is

actually Countryman( see Lemma 2.1.11 in [37]).

Let W = (s, t, h) : s ∈ 2α+1, t ∈ 2α(α ∈ ω1) h is a partial mapping from Ts into Tt.

Let (s1, t1, h1) 6 (s2, t2, h2) iff s1 @ s2, t1 @ t2 and h2 (Ts1 × Tt1) = h1. Via a coding

device we can see that WDP applies to W .

Let 〈Xα,i : α ∈ ω1, i ∈ 2〉 be a partition of ω1 so that WDP (Xα,i) holds for all α ∈ ω1

and i ∈ 2. We must define F : W → 2:

First suppose (s, t, h) ∈ W with s ∈ 2α+1, t ∈ 2α and furthermore α ∈ Xβ,i, s(β) 6= i and

t(β) = i.

Case 1 : If α is a limit ordinal and there is a closed unbounded c ⊂ α such that h is an

isomorphism of Ts c into Tt and ht(h(x)) : x ∈ Ts c is cofinal in α. In that case

h determines a mapping carrying each x ∈ (Ts)α to a branch B(x) = z ∈ Tt : (∃y <Ts

x)z @ h(x) of length α. Since Ts_0α and Ts_1α are disjoint, there exists i ∈ 2 such that

for some x ∈ (Ts)α, there is no element of Ts_iα which extends every element of B(x).

Let F (s, t, h) = i.

Case 2. Otherwise let F (s, t, h) be arbitrary.

Now for each (α, i) there exists gαi : ω1 → 2 which satisfies WDP (Xαi) with respect to


every branch through W . Let G be the set of all functions g : ω1 → 2 such that

(∗)(∀α < ω1) if α ∈ Xβi and g(β) = i, then g(α) = gβi(α).

Note that if α ∈ Xβi, and g(β) 6= i then (∗) impose no requirement on g(α). Hence

|G| = 2ℵ0 .

Claim The trees Tg : g ∈ G are pairwise non-club embeddable into each other.

Let f, g ∈ G be given. Suppose that Tf is club embeddable into Tg since f 6= g there

is (β, i) such that g(β) = i and f(β) 6= i Let C be a club and let h : (Tf ) C → Tg be an

isomorphism. Let B = (f α + 1, g α, h∩ ((Tf ) (α+ 1)×Tgα) : α ∈ ω1. Then B is

an uncountable branch trough W . If D = α ∈ C : (f α + 1, g α, h∩ ((Tf ) (α+ 1)×

Tgα) satisfies case 1 then D is a club and since gβi satisfies WDP (Xβi) it follows from

the definition of gβi that h can not be an isomorphism, which is a contradiction.

3.4 Fine structure theory of A-lines under PFA.

Observe that if A is an Aronszajn line which contain ηC , then A is equivalent to ηC . This

leads us to the following definition.

Definition 3.4.1. An A-line A is fragmented if ηC A.

If we have any hope to show that the class A is wqo we need to introduce a notion of

rank. Theorem 3.2.12 give us a hint about how to associate a rank to each fragmented

Aronszajn line, i.e., roughly speaking the rank corresponds to how many applications of

a derivative operation are necessary in order to trivialize it. This will be explained in

detail below.

Given A ∈ AF consider the following relation:

x ∼ y iff [x, y] ∑i∈I

Ai

where I Q and the linear orderings Ai (i ∈ I) belong to C,C∗. It is clear that ∼ is

an equivalence relation and that each equivalence class is convex. For each x ∈ A let [x]


denote the equivalence class of x, i.e., [x] = y ∈ A : x ∼ y.

We have a natural map associated to this equivalence relation given by:

c : A→ A1

here A1 = [x] : x ∈ A ordered by [x] < [y] if x < y. We call A1 a condensation of

A. Thus a condensation map is a map from A into a partition of A in convex intervals.

Since every partition of a fragmented A-line into convex intervals is itself a fragmented

A-line we can iterate this process as follows:

Definition 3.4.2. For every ordinal α and any A-line A we will construct a con-

densation map cα : A→ Aα recursively as follows:

1. For α = 0 let c0 = Id and let A0 = A

2. For α = β + 1 set cβ+1(x) = y : c(cβ(x)) = c(cβ(y)) and Aβ+1 = (Aβ)1,

3. For a non zero limit ordinal α, let cα(x) =⋃cβ(x) : β < α here Aα = cα(x) :

x ∈ A.

We have the following:

Theorem 3.4.3. Let A ∈ AF be given then there is an ordinal α < ω2 such that cβ(x) =

cα(x) for all x ∈ A and β > α. The least such α will be called the C-rank of A.

Proof. Note that cα(x) ⊆ cβ(x) for α 6 β. Since A has size ℵ1 this process must stop for

some ordinal less than ω2.

Note that the condensation of an A-line is itself an A-line. If α is equal to the C-rank

of A then Aα is either 1 or does not contain a non-trivial interval embeddable in either

C or C∗. In the later case we have, by virtue of theorem 3.2.12, that ηC A. We have

the following theorem which is an analogue of the well known result of Hausdorff about

(countable) scattered linear orders.

For every α < ω2, recursively define the classes Aα as follows:


Definition 3.4.4. Suppose α < ω2 is given and that Aβ has been defined for all β < α.

1. For α = 0, let A0 denote the class of Countryman lines.

2. For a non zero ordinal α, let Aα be the class of all linear orderings which are

equivalent to the ones of the form

∑i∈I

Ai,

where I C or I C∗ and linear orderings Ai (i ∈ I) with Ai in⋃ξ<αAξ.

Theorem 3.4.5. (PFA) The class AF of fragmented A-lines allows a decomposition as

AF =⋃ξ<ω2

Aξ.

Proof. We will prove by induction on α that any A-line A of C-rank α is in⋃ξ<α+2Aξ.

Suppose α is given and that any A-line of C-rank β < α is in⋃ξ<β+2Aξ. Let A be an

A-line of C-rank α.

If α = 0, then C has cardinality one which is impossible.

If α = β + 1, then cα(x) = A. Since Aβ = cβ(x) : x ∈ A is Aronszajn, it has both

countable cofinality and countable coinitiality. So let xn (n ∈ Z) be such that xn < xm

for n < m and cofinal and coinitial in Aβ. We are focusing in the case where A does

not have a first or last element since the argument applies with routine modifications

to the degenerate cases. Thus, [cβ(xn), cβ(xn+1)) (n ∈ Z) it is a partition of Aβ. It is

sufficient to show that each interval [xn, xn+1) is in⋃ξ<α+2Aξ. For each n ∈ Z, we fix a

set Xn ⊂ [xn, xn+1) so that |Xn ∩ cβ(x)| = 1 for all xn 6 x < xn+1, i.e., Xn is a set of

representatives of the interval [cβ(xn), cβ(xn+1)). We have that

[xn, xn+1) =⋃x∈Xn

cβ(x).

Since cβ(x) has C-rank β for all x ∈ A and cβ(xn) ∼ cβ(xn+1) for all n ∈ Z. We obtain

that Xn ∑

i∈I Ai where I C or I C∗ and Ai is Countryman for each i ∈ I, and


cβ(x) is in⋃ξ<β+2Aξ for all x ∈ A. Thus, we infer that [xn, xn+1) ∈

⋃ξ<α+2Aξ.

If α is a non zero limit ordinal, then A =⋃β<α c

β(x) for some (any) x ∈ A. First note

that if cof(α) = ω1 and αξ (ξ ∈ ω1) is strictly increasing and cofinal in α then picking an

element xξ ∈ (cαξ+1(x)∩ [x,∞)) \ cαξ(x) or xξ ∈ (cαξ+1(x)∩ (−∞, x])) \ cαξ(x) we obtain

a copy of ω1 or ω∗1, respectively. Therefore, cof(α) = ω. Let αn be strictly increasing

and cofinal in α. Then

A =⋃n∈ω

[(cαn+1(x) ∩ [x,∞)) \ cαn(x)] ∪⋃n∈ω

[(cαn+1(x) ∩ [x,∞)) \ cαn(x)].

Since the C-rank of the intervals [(cαn+1(x)∩ [x,∞))\ cαn(x) and [(cαn+1(x)∩ (−∞, x]))\

cαn(x) is αn we infer that A ∈⋃ξ<α+2Aξ.

Note that if α < β, then Aα ⊆ Aβ and we have a natural rank

Definition 3.4.6. Given A ∈ AF let rank(A) = minα : A ∈ Aα.

Note that A B implies rank(A) 6 rank(B). We are now ready to prove an

important structural result about the class AF of fragmented Aronszajn lines.

Lemma 3.4.7. (MAω1) For every ordinal α < ω2 there exist two incomparable A-lines

D+α and D−α of rank α such that :

1. C ×D+α ≡ D+

α , C∗ ×D−α ≡ D−α ,

2. D−α C∗ ×D+α , D+

α C ×D−α and

3. For every A ∈ Aα either A ≡ D+α or A ≡ D−α or else both A ≺ D+

α and A ≺ D−α

holds.

Proof. The proof is by induction on α. Suppose that α is given and that D+β and D−β

satisfying clauses (1)-(3) has been defined for all β < α.

If α = 0, then let D+0 = C and D−0 = C∗. Clause (1) follows from fact 3.2.11, clause (2)

is trivial and clause (3) follows from fact 3.2.10.

If α = β+ 1, then let D+α = C×D−β and D−α = C∗×D+

β . (1) follows from fact 3.2.11, for


(2) note that D−α C∗ ×D+α is equivalent to C∗ ×D+

β C∗ × (C ×D−β ) which follows

from the induction hypothesis D+β C∗ ×D−β .

In order to prove (3), let A ∈ Aα be given. We may assume that A =∑

x∈C Ax where

rank(Ax) 6 β for all x ∈ C. We will show that either A ≡ D+α or both A D+

α and

A D−α . Let X = x : Ax ≡ D−β (note X 6= ∅ otherwise A D+β which has rank < α).

If X is uncountable, then by fact 3.2.10 we have that X ≡ C. Using the embedding of

C into X we obtain D+α A, since A D+

α it follows that they are equivalent.

So suppose D+α A and hence X is countable.

Consider the following relation on C \X:

x ∼ y iff [x, y] ∩X = ∅.

It is easy to see that ∼ is an equivalence relation with convex classes.

Since C does not contain uncountable real types we have

|[a] : a ∈ C \X| = ω.

By the induction hypothesis we have that for each a ∈ C\X we can write Ba =∑

x∈[a] Ax

where Ax D+β for all x ∈ [a].

Since [a] has countable cofinality and coinitiality, it follows Ba D+β . Let I be a set such

that |I ∩ [a]| = 1 for all a ∈ C. Therefore A =∑

i∈(I∪X) Xi where Xi = Bi for i ∈ I and

Xi = Ai for i ∈ X. Hence X is a countable sum linear orders which are embeddable into

either D+β or D−β and therefore A is a countable sum of linear orders which embed into

both D+α and D−α .

If α is a non zero limit ordinal. We first note that properties (1)-(3) imply that every

A ∈ Aβ+1 \ Aβ must contains a copy of both D+β and D−β for all β < α. Thus, by clause

(3) again we infer that A contains every line of smaller rank. Observe that if A has rank

α then for each β < α there exists an A-line embedded into A with rank bigger than β.

Thus, we have the following useful property rank(A) < rank(B) 6 α implies A B.

Fix a strictly increasing sequence (αj) converging to α (we use the convention that j ∈ ω


or j ∈ ω1 depending on whether cof(α) = ω or cof(α) = ω1, respectively). Let

D+α =

∑x∈C

Ax, D−α =∑x∈C∗

Ax

where Ax = D+αj

for some j. Moreover, for all j the set x : Ax = D+αj is dense in both

C and C∗. By fact 3.2.9 C is not Suslin, so let I = Iα : α < ω1 be an uncountable

family of pairwise disjoint nonempty intervals of C. We order I by

I < J iff (∀x ∈ I)(∀y ∈ J) x <C y.

Since I is isomorphic to an uncountable subset of C, we have by fact 3.2.10 that C I.

This give us an embedding of C ×D+α into D+

α , i.e., (1) holds. Part (2) should be clear

from the definition of D+α and D−α .

We shall prove that (3) holds. Let A ∈ Aα be given, we may assume that A =∑

x∈C Ax

where rank(Ax) 6 α, for all x ∈ C. Define the following relation on C:

a ∼ b iff suprank(Ax) : x ∈ [a, b] < α.

It is easy to see that ∼ is an equivalence relation with convex equivalence classes. Since

[a] has countable coinitiality and cofinality it follows that the ordering

A′a =∑x∈[a]

Ax

has rank 6 α. Let C ′ ⊂ C be such that |C ′ ∩ [a]| = 1 for all a ∈ C. Then we also have

the following equality

A =∑a∈C′

A′a.

The point with this new representation is that if a, b ∈ C ′, a < b and (a, b) 6= ∅ then

(∀j)(∃x ∈ (a, b)) rank(Ax) > αj.

If this was not the case then we would get that a ∼ b which is impossible.

Observe that if C ′ is countable then A is a countable sum of linear orders which embed


into both D+α and D−α . So it suffices to show that if C ′ is uncountable then A ≡ D+

α . By

going to a subset of C ′, we may assume that C ′ is ℵ1-dense, using fact 3.2.9 we can find

an uncountable family

Iξ := (aξ, bξ) : ξ < ω1

of pairwise disjoint nonempty intervals of C ′. As before we order X = Iξ : ξ < ω1 by

Iξ ≺ Iη iff (∀x ∈ Iξ)(∀y ∈ Iη) x <C y.

By fact 3.2.10 there is a strictly increasing map F : C → X. We will use F to construct

a map f : C → C ′ with the property that f(x) ∈ F (x). Note that this guarantees that

f is strictly increasing. Given x ∈ C find a ∈ F (x) such that rank(Ax) < rank(A′a).

Thus, f provides an embedding of D+α into A, by a similar argument we can obtain the

reverse embedding. Therefore A is equivalent to D+α which concludes the proof of the

lemma.

Theorem 3.4.8. (PFA) The class of Aronszajn lines is bqo under embeddability.

Proof. We will prove by induction on α that Aα is bqo. Suppose that α is given and that

Aβ is bqo for all β < α.

If α = 0, then the result follows from theorem 3.0.13 and fact 3.2.10.

If α = β + 1, then let

f : [ω]ω → Aα

be a given Borel map. Consider the following partition

[ω]ω = X1 ∪X2 ∪X3

where:

X1 = A ∈ [ω]ω : f(A) ≡ D+α ,

X2 = A ∈ [ω]ω : f(A) ≡ D−α ,

X3 = A ∈ [ω]ω : f(A) ≺ D+α ∧ f(A) ≺ D−α .


By the Galvin-Prikry theorem (see [12]) there is X ∈ [ω]ω such that f”[X]ω ⊂ Xi for

some i ∈ 1, 2, 3. If i = 1, 2 then

f(X) ≡ f(X \ min(X))

and the result holds. If i = 3 then f maps X into X = Aα \B ∈ Aα : B ≡ D+α ∨ B ≡

D−α . By the previous lemma every element of X is a countable sum of linear orders

which are embeddable into either D+β or D−β . Thus, X can be identified with the set

(Aβ)C of Q-types. By theorem 3.1.9 X is bqo and therefore there exists an Y ∈ [X]ω so

that

f(Y ) f(Y \ minY ).

If α is a non zero limit ordinal. Then it follows from lemma 3.4.7 that

Aα = A : A ≡ D+α ∪ A : A ≡ D−α ∪ (

⋃ξ<α

Aξ)C

by theorem 3.1.9 and Galvin-Prikry theorem it is enough to show that⋃ξ<αAξ is bqo.

Let f : [ω]ω →⋃ξ<αAξ be a given Borel map. Consider the following partition of [ω]ω:

X1 = A ∈ [ω]ω : rank(f(A)) > rank(f(A \ min(A)),

X2 = A ∈ [ω]ω : rank(f(A)) < rank(f(A \ min(A)),

X3 = A ∈ [ω]ω : rank(f(A)) = rank(f(A \ min(A)).

Again by the Galvin-Prikry theorem there is an X ∈ [ω]ω such that f”[X]ω ⊂ Xi for

some i ∈ 1, 2, 3. The case i = 1 is impossible as it would give us a strictly decreasing

sequence of ordinals. Since the case i = 2 gives us the desired conclusion, we may assume

i = 3. By going to an infinite subset X ′ of X we can assume that rank(f(Y )) is constant

for all Y ∈ [X ′]ω then the conclusion follows from the induction hypothesis.

From lemma 3.4.7 and theorem 3.4.8 we can obtain the following corollary reminiscent

of the Cantor normal form for the ordinals.


Definition 3.4.9. We say that an A-line is additively indecomposable if A = A1 + A2

implies A A1 or A A2.

Fact 3.4.10. Note that if A is additively indecomposable and A ≡ B then B is additively

indecomposable.

Before stating the corollary let us recall the next result of Laver ( see theorem 4.7

[19]).

Lemma 3.4.11. If Q is bqo, and L ∈ QS for some S ∈ S then L is a finite sum of

elements of H(Q).

Corollary 3.4.12. Every A-line is a finite sum of additively indecomposable ones.

Proof. It is clear that a universal Aronszajn line is indecomposable as the sum of frag-

mented A-lines is fragmented and hence not universal. So it suffices to show that the

class of fragmented A-lines satisfies this property.

This will be done by induction on the rank of a fragmented A-line. The result is clear for

A0. To treat the case α > 0, we first note that D+α and D−α are additively indecomposable

for the same reason that the universal line is additively is indecomposable. Moreover note

that a line A of rank ≤ α which is not bi-embeddable with one of the two lines D+α and

D−α must belong to [⋃ξ<αAξ]S , where C denotes the class of countable linear orderings.

As we have proved that⋃ξ<αAξ is a bqo the conclusion now follows from the previous

lemma.

Chapter 4

Gap structure of coherent Aronszajn

trees.

4.1 Quasi-ordering on trees.

In this chapter we add to the study of the class of Aronszajn trees relative to the following

quasi-ordering relation.

Definition 4.1.1. For two Aronszajn trees T and S, by T S we denote the fact that

there is a strictly increasing map f : T → S. Let T ≺ S whenever T S and S 6 T

and let S ≡ T whenever T S and S T . When T ≡ S, we will say that the trees S

and T are equivalent.

The main goal of this chapter is to provide a satisfactory rough classification result

for the class of Aronszajn trees. As we saw in the previous chapter one such notion is

the requirement of being well-quasi-ordered.

Todorcevic results show that the structure (A,) is quite big, so we can not expect

some satisfactory rough classification result for the whole class A of Aronszajn trees.

Therefore, we shall restrict ourselves to some subclass where we can expect such a rough

classification result. The most natural class in this context would be the class C of all

45

Chapter 4. Gap structure of coherent Aronszajn trees. 46

coherent Aronszajn trees. As it has been shown in [37] the class (C,) is a very important

linearly ordered subclass of A-trees which is cofinal and coinitial in (A,≺). Thus, a rough

classification result for the class C will provide us with a big picture of the whole class

A of A-trees. On the other hand, since every element of the linear order C is compatible

with every Aronszajn tree it follows that any rough classification result of a given subclass

K of Aronszajn trees should mention the class C in some way.

In order to provide a classification result for the linear order (C,≺), since it is not w.q.o,

we need to understand its gap structure. We shall use MAω1 to force a tree filling any

gap of cardinality at most ω1. In other words, we show that there are no (κ, λ∗)-gaps

for λ, κ ∈ 1, ω, ω1. From our results about gaps we shall infer that (C,) is universal

for all the linear orders of cardinality ℵ2. Moreover, we shall prove that assuming PFA

(C/ ∼,) is the unique saturated linear order of cardinality ℵ2, where T ∼ S means that

T is equivalent to the nth successor of S or S is equivalent to the nth successor of T , for

some n. We will make this more precise in following sections.

4.2 Shift operation on trees.

Todorcevic [38] used a shift operation to construct an infinite strictly decreasing sequence.

This operation will play an important role in our study of gaps. For this reason, we shall

reproduce some of the results of [38] that will be used in the rest of the section. For

convenience of the reader we shall reproduce also some of the proofs.

In this and the following sections we assume that trees T are represented in such a

way that its elements on a given level α are simply functions from α into ω. Let us

recall that a tree is uniformly coherent if it contains all finite changes of its elements (see

definition 2.2.10). More precisely, if there is a k 6 ω such that rang(t) ⊂ k for all t ∈ T,

then we can define the uniform-k-closure T ∗ of T as the set of all s : α → k for which

we can find t ∈ Tα such that ξ < α : s(ξ) 6= t(ξ) is finite. Thus, a tree is uniformly


coherent, if it is equal to its uniform closure, but we shall use this word to even cover the

case when the tree is equal to its k-closure for k not necessarily equal to ω.

Notation 4.2.1. Let Λ denote the set of all countable limit ordinals, and for a positive

integer n, let Λ + n = λ+ n : λ ∈ Λ.

Definition 4.2.2. For an integer m and a tree T , we let T (m) be its mth-shift, the

downward closure of t(m) : t ∈ T Λ, where for a limit node t of T , we let t(m) be the

function with the same domain λ as t defined by,

t(m)(ξ) = t(ξ −m),

when x−m exits; otherwise (i.e. when m is positive and the largest limit ordinal is less

than or equal to ξ is less than m steps away), we let t(m)(ξ) = 0.

Remark 4.2.3. Note that a positive shift T (m) of any Lipschitz tree T is Lipschitz and

that the map t 7→ t(m) is a strictly increasing map from T into T (m). It follows that

T T (m) for all m > 0. Note also that for non-negative integers m and n,

T (m+n) = (T (m))(n).

Lemma 4.2.4. If T is a Lipschitz tree, then T (m) ≺ T (n) for every pair of non-negative

integers m and n such that m < n.

Proof. Consider two m,n non-negative integers with m < n and let f : T (m) → T (n) be

any given level-preserving map. For each ordinal δ < ω1, pick a representative tδ from

δth level of T and let sδ ∈ Tδ be such that

f(t(n)δ ) = s

(m)δ .

By using lemma 1.1.9, we can find an uncountable set Γ ⊂ Λ such that ∆(tγ, tδ) =

∆(sγ, sδ) for all γ, δ ∈ Γ, γ 6= δ. By lemma 1.1.10, we can choose γ 6= δ in Γ such that

tγ is incomparable to tδ and sγ incomparable to sδ. Note that this implies

α = ∆(tγ, tδ) = ∆(sγ, sδ) < minγ, δ.


Then

∆(t(n)γ , t

(n)δ ) = α + n > α +m = ∆(s(m)

γ , s(m)δ ).

Which is a contradiction.

The following lemma shows that in the class of Lipschitz trees, T (1) is a minimal tree

above T . We shall see later on that T (1) is indeed the immediate successor of T .

Lemma 4.2.5. Assume MAω1. For every pair S and T of Lipschitz trees, S ≺ T implies

S(1) T .

Proof. Let sδ ∈ Sδ(δ < ω1) and tδ ∈ Tδ(δ < ω1) be representatives of the δth level of S

and T , respectively . In the next section (see lemmas 4.3.1 and 4.3.2) we show that there

is an uncountable set Γ ⊂ Λ so that ∆(sγ, sδ) < ∆(tγ, tδ) for all γ 6= δ in Γ. As we have

shown in the proof of the previous lemma, we get the equation

∆(s(1)γ , s

(1)δ ) = ∆(sγ, sδ) + 1 6 ∆(tγ, tδ)

for all γ 6= δ in Γ. By applying lemma 4.3.1 again, we infer that S(1) T .

It follows that under MAω1 for every Lipschitz tree T , the chain (T (n) : n ∈ ω) of

positive shifts is really an ω-chain, i.e., its convex closure inside the class of Lipschitz

trees is isomorphic to ω as an ordered set. It is worth mentioning that this result can be

extended to the whole class of Aronszajn trees by assuming PFA. . The case of negative

shifts is more subtle, although we shall see that they do behave as expected.

Definition 4.2.6. A tree T is orthogonal to a set of ordinals Γ, if there is an uncountable

subset X of T such that ∆(X) ∩ Γ = ∅, where ∆(X) = ∆(x, y) : x, y ∈ X, x 6= y.

Lemma 4.2.7. Suppose that n < m 6 0 and that T is a Lipschitz tree which is orthogonal

to Λ + k for all 0 6 k 6 |n|. Then T (m) and T (n) are also Lipschitz and T (m) T (n).


Proof. Let us first show that T (m) T (n). For each 0 6 k 6 |n|, we can find an

uncountable Xk ⊆ T such that ∆(Xk) ∩ (Λ + k) = ∅. For each δ ∈ Λ and k 6 |n|, pick

tδ(k) ∈ Xk so that ht(tδ(k)) > δ. Shrinking Xk, we may assume that the map δ 7→ tδ(k) is

one-to-one for k 6 |n|. By lemma 1.1.10 there is an uncountable Γ ⊆ Λ so that tγ(k) γ

is incomparable with tδ(k) δ for all γ, δ ∈ Σ, γ 6= δ and for all 0 6 k 6 |n|. Applying

lemma 1.1.9 to 〈tδ(k) δ : 0 6 k 6 |n|〉δ∈Γ we get an uncountable set Γ0 ⊆ Γ such that

for all γ 6= δ in Γ and 0 6 j, k 6 |n|, we have

∆(tγ(j) γ, tδ(j) δ) = ∆(tγ(k) γ, tδ(k) δ).

Let tδ = tδ(0) δ for δ ∈ Γ0. This gives us an uncountable level-sequence with the

property that ∆(tγ, tδ) /∈ Λ + k for all γ 6= δ in Γ0 and 0 6 k 6 |n|. In other words,

X = tδ : δ ∈ Γ0 is an uncountable subset with the property that ∆(X) ∩ (Λ + k) = ∅

for all 0 6 k 6 |n|.

Given any level-preserving map f : T (m) → T (n) and for each γ ∈ Γ0 find sγ ∈ Tγ such

that

f(t(m)γ ) = s(n)

γ .

Applying lemma 1.1.9 once again, we find an uncountable Γ1 ⊆ Γ0 such that ∆(sγ, sδ) =

∆(tγ, tδ) whenever γ 6= δ are chosen from Γ1. Applying lemma 1.1.10 we can find γ 6= δ

in Γ1 so that tγ and tδ are incomparable and sγ and sδ are incomparable as well. Note

that this implies that α = ∆(sγ, sδ) = ∆(tγ, tδ) < minγ, δ. Let λ be the maximal limit

ordinal below α. Then α > λ+ |n|, so the ordinals α+m and α+n are well-defined and

greater than λ. Note that

∆(t(m)γ , t

(m)δ ) = α +m > α + n = ∆(s(n)

γ , s(nδ ).

This shows that f is not a Lipschitz map. Let us now show that T (n) is a Lipschitz

tree. Let f : T (n) → T (n) be any uncountable partial level-preserving map. It should

be clear that a similar argument to the one in the first part of the proof will provide


us with an uncountable X ⊂ dom(f) so that X can be enumerated as a level-sequence

tδ(δ ∈ Γ) and ∆(X) ∩ (Λ + k) = ∅ for all 0 6 k 6 |n|. Moreover, we can assume that

Y = f(tδ) = sδ : δ ∈ Γ is so that ∆(Y ) ∩ (Λ + k) = ∅ for all 0 6 k 6 |n|. For each

γ ∈ Γ we choose representatives t′γ ∈ Tγ and s′γ ∈ Tγ such that t′(n)γ = tγ and s

′(n)γ = sγ.

Applying lemmas 1.1.9 and 1.1.10 we get an uncountable Γ0 ⊂ Γ so that t′γ(γ ∈ Γ0) and

s′γ(γ ∈ Γ0) are both antichains and

∆(t′γ, t′δ) = ∆(s′γ, s

′δ) /∈

⋃06k6|n|

(Λ + k) for all γ 6= δ in Γ0.

It follows that

∆(tγ, tδ) = ∆(t′(n)γ , t

′(n)δ ) = ∆(t′γ, t

′δ) + n = ∆(s′γ, s

′δ) + n = ∆(s′(n)

γ , s′(n)δ ) = ∆(sγ, sδ).

This finishes the proof.

The following result summarizes what has been shown so far about the shift operation

on the class of Lipschitz trees.

Theorem 4.2.8 (Todorcevic). Assuming MAω1. For every Lipschitz tree T which is

orthogonal to Λ + k for all k > 0, the shifts T (n) : n ∈ Z form a family of Lipschitz

trees with the following properties:

1. T (m+n) ≡ (T (m))(n),

2. T (n) ≺ T (m) if only if n < m,

3. there is no Lipschitz tree S such that T (n) ≺ S ≺ T (n+1) for all n ∈ Z.

Lemma 4.2.9. There exists a coherent Lipschitz tree that is orthogonal to Λ + k for

every non-negative integer k.

Proof. Let 〈Cα : α < ω1〉 be a C-sequence with the additional property that for all

α ∈ Λ \ 0, if ξ ∈ Cα and k = |Cα ∩ ξ|, then ξ = λ + k + 1, where λ is the maximal

limit ordinal 6 ξ. Following [37] we define for every β < ω1, a function tβ : β → 2 by


letting tβ(α) = 1 if only if ρ1(α, β)(ξ) = ρ1(α, β), where ξ is the last step of the walk.

The following facts are easily established by induction on α < β < ω1:

1. (∀β) α < β : tβ(α) 6= 0 ∩ (Λ + k) is finite for all k < ω,

2. tα =∗ tβ α, whenever α < β < ω1,

3. there is no t : ω1 → 2 such that t α =∗ tα for all α < ω1.

It follows that if we let T be the uniform 2-closure of the collection tβ α : α 6 β < ω1,

we get a tree satisfying the conclusion of the lemma.

Lemma 4.2.10. There is a uniform coherent Lipschitz tree T so that the corresponding

shifts T (n) : n ∈ Z are all coherent and Lipschitz and moreover T (m) ≺ T (n) holds

whenever m < n.

Proof. By Lemmas 4.2.4, 4.2.7 and 4.2.8 it remains to show that if T is the tree of lemma

4.2.9, then T (m) T (n) whenever m < n 6 0. Let

S = t ∈ T : t(ξ) = 0 for all ξ ∈ dom(t) ∩⋃k<|m|

Λ + k.

Then S is a downward closed subset of T with the property that S(m) = T (m). So it

suffices to observe that t(m) 7→ t(n) (t ∈ S Λ) is a partial Lipschitz map from S(m) into

S(n) ⊆ T (n).

We shall introduce a generalization of shifting maps and in order to do so we need

some definitions.

Definition 4.2.11. Recall the definition of the distance function ∆ : T 2 → ω1 on a tree

of height ω1,

∆(x, y) = otpz ∈ T : z 6 x and z 6 y.

For X ⊆ T , let

∆(X) = ∆(x, y) : x, y ∈ X, x 6= y


We use this notation to describe a family of subsets of ω1 as follows:

U(T ) = Γ ⊆ ω1 : ∆(X) ⊆ Γ for some uncountable X ⊆ T.

Lemma 4.2.12. The family U(T ) is an uniform filter on ω1 for every Lipschitz tree T .

Proof. Given to uncountable subsets X and Y of T , we need to find an uncountable

subset Z of T such that

∆(Z) ⊆ ∆(X) ∩∆(Y ).

For each γ ∈ Λ, pick xγ ∈ X and yγ ∈ Y such that ht(xγ) > γ and ht(yγ) > γ. Shrinking

X and Y , we can find an uncountable Σ ⊂ Λ so that the sets xγ γ : γ ∈ Σ and

yγ γ : γ ∈ Σ are both antichains. For γ ∈ Σ, let x′γ = xγ γ. It follows that the

sets X ′ = x′γ ∈ Tγ and Y ′ = y′γ : γ ∈ Γ satisfy ∆(X ′) ⊂ ∆(X) and ∆(Y ′) ⊂ ∆(Y ).

Apply lemma 1.1.9 to the subset (x′γ, y′γ) : γ ∈ Γ of T ⊗ T and obtain an uncountable

set Σ ⊆ Γ such that

∆(x′γ, x′δ) = ∆(y′γ, y

′δ) for all γ, δ ∈ Σ, γ 6= δ.

So we can take Z to be any of the sets x′δ : δ ∈ Σ or y′δ : δ ∈ Σ. This finishes the

proof.

Definition 4.2.13. Suppose that g is a partial map from ω1 into ω1. Then the g-shift

of T , denoted by T (g), is the downward closure of t(g) : t ∈ T Ω, where Ω = δ < ω1 :

g”δ ⊆ δ and t(g) is defined by

t(g)(ξ) = t(g(ξ))

if ξ ∈ dom(g); otherwise t(g)(ξ) = 0.

The shifts considered previously where associated to maps of the form γ(ξ) = ξ −m.

Lemma 4.2.14. If g is a partial strictly increasing map on ω1 and if rang(g) ∈ U(T )

for some Lipschitz tree T , then the g-shift T (g) is also a Lipchitz tree.


Proof. Let f : T (g) → T (g) be a given uncountable partial leveling-preserving map.

Choose an uncountable X ⊂ dom(f) such that ∆(X) ⊆ rang(g). We may assume

that for some fixed Γ ⊆ ω1, the set X is enumerated as a level-sequence of the form

xγ(γ ∈ Γ). For each γ ∈ Γ find α(γ) ∈ Ω, tα(γ) ∈ Tα(δ) such that t(g)α(γ) γ = xγ and

β(γ) ∈ Ω, sβ(γ) ∈ Tβ(γ) such that s(g)β(γ) γ = f(xγ).

By restricting one of the nodes we may assume that α(γ) = β(γ). We can define an

uncountable partial level-preserving map from T into T by

f(tα(γ)) = sα(γ).

Since T is a Lipschitz tree we can find an uncountable subset Γ0 of Γ so that

f tα(γ) : γ ∈ Γ0

is a Lipschitz map. Note that for every γ, δ ∈ Γ0

∆(xγ, xδ) = g−1(∆(tα(γ), tα(δ)) 6 g−1(∆(sα(γ), sα(δ)) = ∆(f(xγ), f(xδ)).

This shows that f restricted to the uncountable set Γ0 is a Lipschitz map.

Lemma 4.2.15. Suppose that T is a Lipschitz tree and g is a strictly increasing partial

map on ω1 such that rang(g) ∈ U(T ). If g is regresive (i.e. g(ξ) < ξ for all ξ ∈ dom(g)),

then T (g) T . On the other hand, if g is expansive (i.e. g(ξ) > ξ for all ξ ∈ dom(g)),

then T T (g).

Proof. We shall prove that g regressive implies T (g) T (the other case is similar).

Consider a level-preserving map f : T (g) → T . For each γ ∈ Ω, pick a representative

tγ ∈ Tγ and let sγ ∈ Tγ be such that

f(t(g)γ ) = sγ.

Find an uncountable subset Γ0 of Γ so that ∆(tγ : γ ∈ Γ0) ⊂ rang(g). Moreover, by

lemma 1.1.9 we can assume that for all γ, δ ∈ Γ0

∆(tγ, tδ) = ∆(sγ, sδ).


Choose γ 6= δ ∈ Γ0 such that

α = ∆(tγ, tδ) = ∆(sγ, sδ)

is smaller than both γ and δ (i.e. tγ is incomparable to tδ and sγ is incomparable to sδ).

Then

∆(t(g)γ , t(g)δ ) = g−1(α) > α = ∆(sγ, sδ).

This shows that f is not a Lipschitz map, finishing the proof.

Remark 4.2.16. Note that if g is a strictly increasing regressive partial map on ω1 such

that rang(g) = ω1, then T < T (g) holds for every Lipschitz tree T . This observation

can be used to construct both strictly increasing and strictly decreasing ω1-sequences of

Lipschitz trees.

4.3 Comparability of every pair of Lipschitz trees.

The purpose of this section is to analyze the comparability of Lipschitz trees under MAω1 .

Again the results are taken from [36] and reproduced here since we will need them for

the rest of the thesis. For the time being, fix a pair S and T of Lipschitz trees, and for

every δ < ω1, fix representatives sδ and tδ from the δth level of S and T respectively.

This give us two mappings

∆s : [ω1]2 → ω1 and ∆t : [ω1]2 → ω1

defined as follows:

∆(α, β) = ∆(sα, sb) and ∆t(α, β) = ∆(tα, tβ).

For Γ ⊆ ω1, we let

∆s(Γ) = ∆s(α, β) : α, β ∈ Γ, α 6= β and ∆t(Γ) = ∆t(α, β) : α, β ∈ Γ, α 6= β.


Lemma 4.3.1. Assume MAω1. The following are equivalent for every pair of Lipschitz

trees T and S:

(a) S T ,

(b) there is an uncountable Γ ⊆ ω1 such that ∆s(α, β) 6 ∆t(α, β) for all α, β ∈ Γ,

(c) for every uncountable Σ ⊂ ω1 there is an uncountable Γ ⊆ Σ such that ∆s(α, β) 6

∆t(α, β) for all α, β ∈ Γ, α 6= β.

Proof. To deduce (b) from (a), suppose we are given a strictly increasing level-preserving

map f : S → T . Applying lemmas 1.1.9 and 1.1.10 to the sequence 〈(tδ, f(tδ)) : δ ∈ ω1〉

we obtain an uncountable Γ ⊆ ω1 such that for all γ 6= δ:

1. tγ and tδ are incomparable,

2. f(sγ) and f(sδ) are incomparable,

3. ∆(tγ, tδ) = ∆(f(sγ), f(sδ)).

Clearly, this Γ satisfies (b). Similarly one shows that (a) implies (c). Note that the clause

(b) simply says that he map sδ 7→ tδ (δ ∈ Γ) is Lipschitz and it therefore extends to a

strictly increasing map from the downward closure S0 of the set sδ : δ ∈ Γ in S. By

lemma 1.1.12, S S0 T . This shows that (b), and therefore the stronger (c), implies

(a). This finishes the proof.

The next lemma give us a convenient reformulation of the inequality T S.

Lemma 4.3.2. The following are equivalent for every pair S and T of Lipschitz trees:

(a) T S,

(b) there is an uncountable Γ ⊆ ω1 such that ∆s(α, β) < ∆(α, β) for all α, β ∈ Γ, α 6=

β,


(c) for every uncountable Σ ⊆ ω1 there is an uncountable subset Γ ⊆ Σ such that

∆s(α, β) < ∆(α, β) for all α, β ∈ Γ, α 6= β.

Proof. To see (a) implies (b), let P be the poset of all finite p ⊂ ω1 such that

1. ∆t(α, β) 6 ∆s(α, β) for all α, β ∈ p, α 6= β, ordered by reverse inclusion.

If P would satisfy the countable chain condition, an application of MAω1 would give

us an uncountable set Γ ⊆ ω1 such that ∆t(α, β) 6 ∆s(α, β) for all α, β ∈ Γ, α 6= β

which by the previous lemma would give us T S, contradicting (a). So let

〈pδ : δ ∈ ω1〉 be a given sequence of pairwise incompatible members of P . We may

assume that min(pδ) > δ for all δ ∈ ω1. For δ ∈ ω1, let

aδ = sξ δ : ξ ∈ pδ and bδ = tξ δ : ξ ∈ pδ.

For δ ∈ Λ, let h(δ) be the maximum of all ordinals that have the form ∆(x, y) +

1, x, y ∈ aδ, x 6= y, and ∆(x, y) + 1, x, y ∈ bδ, x 6= y. Find an stationary Ω ⊂ Λ

such that h is constant on Ω and let ξ be the constant value. Shrinking Ω, we

may assume that all aδ(δ ∈ Ω) are of some fixed size m, and that all bδ(δ ∈ Ω) are

of some fixed sized n. Let aδ(i)(i < m) and bδ(j)(j < n) be fixed enumerations.

Applying lemmas 1.1.9 and 1.1.10, we can find an uncountable Σ ⊆ Γ such that for

all γ 6= δ in Σ:

2. aγ(i) ξ = aδ(i) ξ for all i < m,

3. bγ(i) ξ = bδ(i) ξ for all i < n,

4. aγ(i) and aδ(j) are incomparable for all i, j < m,

5. bγ(i) and bδ(j) are incomparable for all i, j < n,

6. ∆(aγ(i), aδ(i)) = ∆(aγ(j), aδ(j)) for all i, j < m,

7. ∆(bγ(i), bδ(i)) = ∆(bγ(j), bδ(j)) for all i, j < n. Consider γ 6= δ in Σ. Then pγ ∪ pδ

fails to satisfy condition (1), i.e. there exist ξ ∈ pγ and η ∈ pδ such that ∆(sξ, sη) <


∆(tξ, tη). Let sξ γ = aδ(j), sη δ = aδ(j), tξ γ = bγ(k) and tη δ = bδ(l). Using

(2)-(4) and the fact that pγ and pδ satisfy (1), we conclude that i = j and k = l

and therefore, by (6) and (7), we have the following:

8. ∆(aγ(0), aδ(0)) = ∆(aγ(i), aδ(i)) < ∆(bγ(k), bδ(k)) = ∆(bγ(0), bδ(0)). Of course,

we may assume that the enumeration of aδ and bδ are given in a way such that

if ξ(δ) = min(pδ), then sξ(δ) δ = aδ(0) and tξ(δ) = bδ(0) for all δ in Σ. Let

Γ = ξ(δ) : δ ∈ Σ. Then Γ satisfies clause (b), finishing thus the proof that (a)

implies (b). Similarly, one proves that (a) in fact implies (c). The implication from

(b) to (a) follows from the previous lemma. This finishes the proof.

Finally, we are in a situation to state the main result of this section.

Theorem 4.3.3 (Todorcevic). Assume MAω1. Every two Lipschitz trees are comparable.

Proof. Suppose we are given a pair S and T of Lipschitz trees such that T S. By

lemma 4.3.2 , there is an uncountable Γ ⊂ ω1 such that ∆s(α, β) < ∆t(α, β) for all

α, β ∈ Γ, α 6= β. By lemma 4.3.1, we conclude that T S. This completes the

proof.

4.4 Coinitiality and cofinality of the class of coherent

Aronszajn trees.

In this section we reproduce some of the results of [37] which show that the chain C of

coherent trees is cofinal as well as coinitial in (A,).

Lemma 4.4.1. For every A-tree S, there is a Lipschitz tree T such that S T .

Proof. Let P be the set of all finite partial functions p from S × ω1 into ω such that

1. ξ < ht(x) for all (x, ξ) ∈ dom(p),


2. p(x, ξ) = p(y, ξ) for all (x, ξ), (y, ξ) ∈ dom(p) with ξ < ∆(x, y). We let p extends q

if p extends q as a function and

3. p(x, ξ) = p(y, ξ) for all x, y ∈ dom0(q) and ξ < ht(x), ht(y) such that ξ /∈ dom1(q),

4. p(x, ξ) 6= p(x, η) for all (x, η) ∈ dom(q) and (x, ξ) ∈ dom(p) \ dom(q). A simple

∆-system argument (contained in the proof of lemmas 4.3.2 above) shows that P

satisfies the countable chain condition, so an application of MAω1 give us a map g

from S × ω1 into ω so that its fibres gx(ξ) = g(x, ξ) are total maps from ht(x) into

ω for all x ∈ S and such that

5. gx : ht(x)→ ω is a finite-to-one map for all x ∈ S,

6. ∆(x, y) 6 ∆(gx, gy) for all x, y ∈ S,

7. ξ : gx(ξ) 6= gy(ξ) is finite for all x, y ∈ S. It follows that the downwards closure

T of gx : x ∈ S is a coherent A-tree and that x 7→ gx is a Lipschitz map from S

into T . This finishes the proof.

Lemma 4.4.2. Assume MAω1. For every A-tree S, there is a Lipschitz tree T such that

T S.

Proof. Let P be the set of all finite partial functions p from S × ω1 into ω such that


2. for every pair x and y of incomparable nodes from dom0(p), there is ξ 6 ∆(x, y)

with (x, ξ), (y, ξ) ∈ dom(p) and p(x, ξ) 6= p(y, ξ). We let p extends q if p extends q

as a function and

3. p(x, ξ) = p(y, ξ) for all x, y ∈ dom0(q) and ξ < ht(x), ht(y) such that ξ /∈ dom1(q),

4. p(x, ξ) 6= p(x, η) for all (x, η) ∈ dom(q) and (x, ξ) ∈ dom(p) \ dom(q). To prove

that P satisfies the countable chain condition, we start with an uncountable subset


Ξ of P and perform a ∆-system argument combined with lemmas 1.1.9 and 1.1.10

to obtain two conditions p and q in Ξ such that for some ξ < α < β, v0, ..., vn ∈

Sξ, s0, ..., sn ∈ Sα and t0, ..., tn ∈ Sβ we have

5. every node of dom0(p) is either of height less than ξ or it extends some si(i 6 n),

6. every node of dom0(q) is either of height less than ξ or it extends some ti(i 6 n),

7. dom1(p) ⊆ ξ ∪ (α, β) and dom1(q) ⊂ ξ ∪ (β, ω1),

8. vi 6= vj for i 6= j 6 n,

9. si and ti extend vi but are incomparable for all i 6 n,

10. p and q are isomorphic conditions via an isomorphism that is the identity on ξ,

vi(i 6 n) and maps si to ti for all i 6 n. We claim that such p and q can be amal-

gamated into a condition r of P that extends them both. Let ξ = min∆(si, ti) :

i 6 n. Then ξ 6 ξ < α. Let k = max(rang(p)) = max(rang(q)). Let dom(r) be

equal the union of dom(p), dom(q) and the following two sets:

D = (x, ξ) : x ∈ dom0(p), ht(x) > α,

E = (y, ξ) : y ∈ dom0(q), ht(y) > β.

Define r by giving it constant value k+1 on D and constant value k+2 on E. Note

that r satisfies condition (1) and (2) as well as conditions (3) and (4) for extending

both p and q.

Applying MAω1 to P give us a partial map g : S × ω1 → ω so that if gx(ξ) = g(x, ξ)

then

11. gx is a finite-to-one map from ht(x) into ω for all x ∈ S,

12. ∆(gx, gy) 6 ∆(x, y) for all x, y ∈ S,


13. ξ : gx(ξ) 6= gy(ξ) is finite for all x, y ∈ S. It follows that the downward closure T

of gx : x ∈ S is a Lipschitz tree and that gx 7→ x is a partial Lipschitz map from

T into S, witnessing the relation T S. This finishes the proof.

Theorem 4.4.3 (Todorcevic). Assume MAω1. There is no maximal Aronszajn tree.

Proof. Given an Aronszajn tree S by lemma 4.4.1 we find a Lipschitz tree T such that

S T . By lemma 4.2.4, T ≺ T (1), so in particular S ≺ T (1).

Lemma 4.4.4. Assume PFA. For every Lipschitz tree T there is a Lipschitz tree S such

that S ≺ T .

Proof. Fix a level-sequence tα ∈ Tα(α ∈ ω1) in a given Lipschitz tree T . Let ∆t : [ω1]2 →

ω1 be the corresponding distance function ∆t(α, β) = ∆(tα, tβ). For Γ ⊆ A, let

∆t(Γ) = ∆t(α, β) : α, β ∈ Γ, α 6= β.

Let P be the poset of all pairs p = (fp,Γp) such that

1. Γp is a finite subset of ω1,

2. fp is a finite partial strictly increasing map from ω1 into ω1 which can be extended

to a total increasing and continuous map f : ω1 → ω1 so that rang(f) is disjoint

from ∆t(Γp) and separates the points of ∆t(Γp) (i.e. between every two members

of ∆t(Γp), there is a member of rang(f)).

We order P by coordinatewise inclusion. To show that P is proper, consider a

countable elementary submodel M of some large enough structure of the form

〈H(θ),∈〉 such that M contains P , T and the level-sequence 〈tα : α ∈ ω1〉. For a

given p ∈ P ∩M let q = 〈fp ∪ 〈δ, δ〉,Γp〉, where δ = M ∩ ω1. We claim that q is

an M -generic condition of P . To show this, consider a dense-open subset D of P

such that D ∈ M and an extension r of q. We need to show that r is compatible

with a member of D ∩M . Extending r, we may assume r ∈ D. Let vi(i 6 n) be a


one-to-one enumeration of tα δ : α ∈ Γp \ δ. Let p = r M. Then p ∈ P ∩M

and so we can find an extension f : ω1 → ω1 of fp satisfying (2) for p such that

f ∈ M. Let ξ ∈ (max(Γp, δ) be a fixed point of f . Find a copy r of r in D ∩M

such that if we let d and vi(i 6 n) be its versions of δ and vi(i 6 n), then

3. vi ξ = vi ξ for all i 6 n,

4. vi and vi are incomparable for all i 6 n,

5. ∆(v0, v0) = ... = ∆(vn, vn). It is clear that we can combine the function f with the

normal functions witnessing (2) for r, r and obtain a strictly increasing continuous

function f : ω1 → ω1 which fixes ξ and witnesses (2) simultaneously for r, r and

moreover, the ordinal ∆(v0, v0) is not in its range. Since

∆t(Γr ∪ Γr) = ∆(Γr) ∪∆(Γr) ∪ ∆(v0, v0),

this shows that 〈fr ∪ fr,Γr ∪ Γr〉 is a member of P witnessing the compatibility of

r.

Applying PFA to P gives us an uncountable Γ ⊆ ω1 and a closed unbounded set

C ⊆ ω1 such that C ∩∆t(Γ) = ∅ and C separates the points of ∆t(Γ). For δ < ω1,

let δ+ be the minimal point of C above δ. Define

C0 = δ ∈ C : (δ, δ+) ∩∆t(Γ) 6= ∅.

Note that for δ ∈ C, there is only one point of ∆t(Γ) in the interval (δ, δ+). Call

this point g(δ). This defines a strictly increasing map g from C0 onto ∆t(Γ). So

in particular, rang(g) ∈ U(T ). Let S = T (g). From previous results on shifting

transformation we conclude that S is a Lipschitz tree. Since clearly g(δ) > δ for all

δ0 ∈ C0, we conclude that T S. This finishes the proof.


4.5 Gaps in (C,).

In this section we shall give a full description of the gap structure of (C,). Our first

lemma concerns with the structure of countable gaps.

Lemma 4.5.1. Assume MAω1 . For every sequence

T 00 ≺ T 0

1 ≺ ... ≺ T 0n ≺ ... ≺ T 1

n ≺ ... ≺ T 11 ≺ T 1

0 (n ∈ ω),

of coherent Aronszajn trees there is a coherent Aronszajn tree T such that T 0n ≺ T ≺ T 1

n

for all n ∈ ω.

Proof. We may assume without loss of generality that T in∩T jm = ∅ for all (n, i) 6= (m, j)

and n,m ∈ ω, i, j ∈ 2. First of all we shall define some auxiliary notions to help us with

the construction. It follows from our assumptions on the sequence that there is a well

defined index function

χ :⋃

(n,i)∈ω×2

T in → ω × 2

defined by

χ(x) = (n, i) if and only if x ∈ T in.

Define a lexicographic ordering on ω × 2 by

(n, i) <lex (m, j) if only if i < j or i = j = 0 and n < m or i = j = 1 and n > m.

Now construct a subset X of⋃

(n,i)∈ω×2 Tin with the following properties:

(i) X ∩ T in is uncountable for all n ∈ ω, i ∈ 2,

(ii) ∀α < ω1 |X ∩ Levα(T in)| 6 1 for all n ∈ ω, i ∈ 2,

(iii) ∀α ∈ ω1 |(n, i) ∈ ω × 2 : X ∩ Levα(T in) 6= ∅| 6 1 and

(iv) (∀n < ω) (∀i < 2) (∀x, y ∈ X ∩ T in) ∆(x, y) > ω.


We shall use MAω1 to produce a tree T witnessing

∀k < ω, T 0k ≺ T ≺ T 1

k .

Let P = P(T in : n ∈ ω, i < 2) be the poset of all partial finite mappings p : X × ω1 → ω

with the following properties:


2. ∀x, y ∈ dom0(p) and ∀ξ < ht(x), ht(y) [(x, ξ) ∈ dom(p) if and only if (y, ξ) ∈

dom(p)],

3. For all x, y ∈ dom0(p),

(a) If χ(x) = χ(y) = (n, 0) for some n, then

[(x, ξ), (y, ξ) ∈ dom(p), ξ < ∆(x, y) implies p(x, ξ) = p(y, ξ)].

(b) If χ(x) = χ(y) = (n, 1) for some n, then there exist

ξ < ∆(x, y) such that p(x, ξ) 6= p(y, ξ).

We let p extend q if p extend q as a function and

4. p(x, ξ) = p(y, ξ) for all x, y ∈ dom0(q) and ξ < minht(x), ht(y), ξ /∈ dom1(q),

5. p(x, ξ) 6= p(x, η) for all (x, η) ∈ dom(q) and (x, ξ) ∈ dom(p) \ dom(q).

It follows that if P satisfies the countable chain condition then an application of

MAω1 to the dense sets defined in Claim 1 and Claim 2 give us a map f : X × ω1 →

ω so that it is fibers fx : ht(x) → ω are total maps. Conditions (4) and (5) imply

that 〈fx : x ∈ X〉 is a sequence of coherent finite-to-one maps. Let

T = fx ξ : x ∈ X, ξ 6 ht(x).

Then T is a coherent Aronszajn tree.

Claim 1 For all x0 ∈ X, the set Dx0 = p ∈ P : x0 ∈ dom0(p) is dense open.


Proof. Let p ∈ P be given with the property that x0 /∈ dom(p). We shall proceed

by cases:

Case 1: If χ(x) = (k, 1) for some k. Let n = max(rang(p)).

Subcase a: If there is a x ∈ (dom0(p) \ x0) such that χ(x) = χ(x0). Set

ξ = min∆(x0, x) : x ∈ (dom0(p) \ x0), χ(x0) = χ(x).

By (iv) ξ > ω so we can find ξ0 < ξ, ξ0 /∈ dom1(p). Let q be a map with domain

equal to

dom(p)∪(x, ξ0) : x ∈ dom0(p), ξ0 < ht(x)∪(x0, ξ0)∪(x0, ξ) : ξ ∈ dom1(p), ξ < ht(x0).

Define q by q dom(p) = p, q(x, ξ0) = n + 2 for x ∈ dom0(p) and q(x0, ξ) =

q(x0, ξ0) = n+ 1 for all ξ ∈ dom1(p), ξ < ht(x0). Let us show that q is a condition

in the partial order P . It should be clear from our definition of q that clauses (1)

and (2) holds. Consider x, y ∈ dom0(q) if χ(x) = χ(y) and π1(χ(x)) = 0. Let ξ <

∆(x, y), ξ ∈ dom0(q) be given then either ξ ∈ dom0(p) implying q(x, ξ) = q(y, ξ)

since (x, ξ), (y, ξ) ∈ dom(p) or ξ = ξ0 which implies q(x, ξ0) = n + 2 = q(y, ξ0).

Thus, clause 3(a) holds. To verify clause 3(b), let x, y ∈ dom0(q) be given so that

χ(x) = χ(y) and π1(χ(x)) = 1. If x, y ∈ dom0(p) then using the fact that p is a

condition we can find η < ∆(x, y) in dom1(p) so that p(x, η) 6= p(y, η), otherwise if

x0 = x then q(x0, ξ0) = n+ 1 6= n+ 2 = q(y, ξ0). It follows that q is a condition in

Dx0 . Let us verify that q extends p. By definition q extends p as a function. For

all x, y ∈ dom0(p) we have q(x, ξ0) = n+ 2 = q(y, ξ0), where ξ0 is the only element

in dom1(q) \ dom1(p). Moreover, q(x, ξ0) = n + 2 > q(x, ξ) for all (x, ξ) ∈ dom(p).

Therefore q is a condition in Dx0 extending p.

Subcase b: If there is no x ∈ dom0(p) so that χ(x0) = χ(x). Then let q be the

function with domain

dom(p) ∪ (x0, ξ) : ξ ∈ dom1(p), ξ < ht(x0)


defined by q dom(p) = p and q(x0, ξ) = n + 1 for all ξ ∈ dom1(p), ξ < ht(x0). It

should be clear in this case that q is a condition in Dx0 extending p.

Case 2: If χ(x) = (k, 0) for some k. Let n = max(rang(p)).

Subcase a: If there is x ∈ dom0(p) \ x0 such that χ(x0) = χ(x). Pick x′0 ∈

dom0(p) \ x0 such that χ(x′0) = χ(x0) and

∆(x0, x′0) = max∆(x0, x) : x ∈ dom0(p), χ(x0) = χ(x), x0 6= x.

Let q be map with domain

dom(p) ∪ (x0, ξ) : ξ ∈ dom1(p), ξ < ht(x0).

Define q by q dom(p) = p and q(x0, ξ) = p(x′0, ξ) for ξ ∈ dom1(p) ∩ ∆(x0, x′0)

and q(x0, ξ) = n + 1 for ξ ∈ dom1(p) ∩ [∆(x0, x′0), ht(x0)). Let us show that q is

a condition. First note that q satisfies clauses (1), (2) and 3(b) by definition. In

order to verify clause 3(a) let x, y ∈ dom0(q) be given so that χ(x) = χ(y) and

π1(χ(x)) = 0. Let ξ < ∆(x, y), ξ ∈ dom1(q) be given then either x, y ∈ dom0(p)

so q(x, ξ) = q(y, ξ) since (x, ξ), (y, ξ) ∈ p and p is a condition or x = x0 which

would imply ξ < ∆(x0, y) = ∆(x′0, y) so q(y, ξ) = q(x′0, ξ) = q(x0, ξ). Thus, q is

indeed a condition. Moreover, q extends p as clauses (5) and (6) are vacuously true.


Subcase b: If there is no x ∈ dom0(p) so that χ(x0) = χ(x). Then let q be the

condition with domain


be defined by q dom(p) = p and q(x0, ξ) = n + 1 for all ξ ∈ dom1(p), ξ < ht(x0).

It should be clear in that q is a condition in Dx0 extending p.

Claim 2 For all (x0, ξ0) ∈ X × ω1 with ξ0 < ht(x0), the set D(x0,ξ0) is open-dense

in P .


Proof. Let p ∈ P be given. By Claim 1 we can find a condition p′ extending p so

that x0 ∈ dom0(p′). Let n = max(rang(p′)). Set q to be the function with domain

dom(p′) ∪ (x, ξ0) : x ∈ dom0(p′), ξ0 < ht(x)

be defined by q dom(p′) = p′ and q(x, ξ0) = n+ 1 for all x ∈ dom0(p′), such that

ξ0 < ht(x). Let us verify that q is a condition. It should be clear that clauses (1) and

(2) holds. Let x, y ∈ dom0(q) be given so that χ(x) = χ(y) and π1(χ(x)) = 1 since

x, y ∈ dom0(p′) and p is a condition we can find η ∈ dom1(p′) such that η < ∆(x, y)

and p(x, η) 6= p(y, η). On the other hand, given x, y ∈ dom0(q), ξ ∈ dom1(q) such

that χ(x) = χ(y), π1(χ(x)) = 0 and ξ < ∆(x, y) then either ξ ∈ dom1(p′) which

would imply q(x, ξ) = q(y, ξ) since (x, ξ), (y, ξ) ∈ dom(p′) and p′ is a condition or

ξ = ξ0 in that case q(x, ξ0) = n + 1 = q(y, ξ0). We are now left to prove that q

extends p′. First note that q(x, ξ0) = n+ 1 = q(y, ξ0) for all x, y ∈ dom0(q) so that

ξ0 < ht(x), ht(y), where ξ0 is the only element of dom1(q) \ dom1(p′). Moreover,

q(x, ξ0) = n+ 1 > q(x, ξ) for all (x, ξ) ∈ dom(p′). Thus, D(x0,ξ0) is open dense in P .

Claim 3: T 0n ≺ T ≺ T 1

n for all n < ω.

Proof. Set X0n = x ∈ X : χ(x) = (n, 0). Consider the map ϕn : X0

n → T given by

ϕn(x) = fx it follows from 3(a) we infer that ϕn is a Lipzschitz map for all n ∈ ω.

Using lemma 1.1.13 we obtain the inequality T 0n T for all n ∈ ω. The strict

inequality follows from T 0n ≺ T 0

n+1 T .

Let Yn = fx : χ(x) = (n, 1). Define the map ψn : Yn → T 1n as follows ψn(fx) = x.

Then by 3(b) we infer ψn is a Lipszchitz map. Using lemma 1.1.13, we conclude

that T T 1n for all n ∈ ω.We are left with proving the following claim.

Claim 4 The partially ordered set P satisfies the countable chain condition.

Let 〈pδ : δ ∈ ω1〉 be an uncountable sequence of elements of P . Using Claim 1 we can

assume that for all δ ∈ ω1 there exists x, y ∈ dom0(pδ) such that π1(χ(x)) = 0 and

π1(χ(y)) = 1. By the ∆-system lemma and by a counting argument, we may assume


that 〈dom(pδ) : δ ∈ ω1〉 form a ∆-system with root D and pδ D = pδ′ D for all

δ, δ′ ∈ ω1. Moreover, we may assume that the image mapping δ 7→ χ′′(dom0(pδ)) is

constant.

For δ < ω1, let

aδ = x δ : x ∈ dom0(pδ), ht(x) > δ and bδ = x ∈ dom0(pδ) : ht(x) < δ.

Moreover, for δ < α, we let

h(δ) = max(∆(x, y) : x, y ∈ aδ, x 6= y, χ(x) = χ(y)∪(dom1(pδ)∩δ)∪ht(x) : x ∈ bδ)+1.

By the pressing down lemma there is a stationary set Γ of countable limit ordinals

on which the mapping h, and the mapping

δ → dom(pδ) ∩ (X δ)× δ

as well as the mapping δ 7→ bδ are constant. Let ξ, F and B be the constant values,

respectively. Shrinking Γ, we may assume that all aδ (δ ∈ Γ) are of some fixed size

n. Moreover, we may assume that the function δ 7→ aδ ξ has a constant value

a, and that the conditions pδ(δ ∈ Γ) generate isomorphic structures over ξ, F and

a. Thus, we want in particular the isomorphism between the pδ (δ ∈ Γ) to respect

a fixed enumeration aδ(i)(i < n) of aδ. Moreover, the isomorphism between the

conditions pγ and pδ is the identity on ξ, and aδ(i) ξ (i < n) and maps nodes

extending aγ(i) into nodes extending aδ(i) for all i < n. Using lemmas 1.1.9 and

1.1.10 we find an uncountable subset Σ ⊂ Γ such that for ξ < γ < δ in Σ:

(6) aγ(i) ξ = aδ(i) ξ for all i < n,

(7) every node of dom0(pγ) is either of height less or equal to ξ or it extends some

aγ(i)(i < n),

(8) every node of dom0(pδ) is either of height less or equal to ξ or it extends some

aδ(i)(i < n),


(9) dom1(pγ) ⊂ ξ ∪ (γ, δ) and dom1(pδ) ⊂ ξ ∪ (δ, ω1),

(10) aγ(i) and aδ(j) are incomparable for all i, j < n, whenever χ(aγ(i) = χ(aδ(j)),

(11) ∆(aγ(i), aδ(i)) = ∆(aγ(j), aδ(j))) for all i, j < n, whenever χ(aγ(i)) = χ(aδ(i)) =

χ(aγ(j)) = χ(aδ(j)) and

(12) ∆(aγ(i), aδ(i)) < ∆(aγ(j), aδ(j))), whenever χ(aγ(i)) = χ(aδ(i)) <lex χ(aγ(j)) =

χ(aδ(j)).

We claim that if ξ < γ < δ are in Σ, then pγ and pδ are compatible. Let

m = max[(range(pγ) ∪ (range(pδ)].

Let (k, 0) be the maximum in the <lex order of the set χ(aδ(i)) : i < n, π1(χ(aδ)) = 0

and let ξ0 = ∆(aγ(i0), aδ(i0)) for some (every) i0 such that χ(aδ(i0)) = (k, 0). We define

p ∈ P by letting its domain be

dom(pγ) ∪ dom(pδ) ∪ (x, ξ0) : x ∈ dom0(pγ) ∪ dom0(pδ), ht(x) > ξ0,

and letting p(x, ξ0) = m + 1 for x ∈ dom0(pγ), ht(x) > ξ0, and p(x, ξ0) = m + 2 for

x ∈ dom0(pδ), ht(x) > ξ0. and let p (dom(pγ) ∪ dom(pδ)) = pγ ∪ pδ.

Note that p is a function extending pγ, pδ and it clearly satisfy (1) and (2). Note moreover

that dom1(q) = dom1(pγ) ∪ dom1(pδ) ∪ ξ0 and that dom0(q) = dom0(pγ) ∪ dom0(pδ).

Let us show that p satisfies clause (3) of being a condition. In order to do so we shall

proceed by cases:

Case 1: Consider x ∈ dom0(pγ) and y ∈ dom0(pδ) with χ(x) = χ(y) = (N, 0) for some

N . Let ξ < ∆(x, y) be given such that (x, ξ), (y, ξ) ∈ dom(q).

Subcase a: If x γ = aγ(i) and y δ = aδ(i) for some i < n.

Then since (N, 0) 6lex (k, 0) and so by (11) and (12) and the definition of ξ0 we get that

ξ < ∆(x, y) = ∆(aγ(i), aδ(i)) 6 ξ0.


It follows that ξ ∈ (dom1(pγ)∩dom1(pδ)) ⊂ ξ. Let y′ in dom0(pγ) be the copy of y relative

to the isomorphism between pγ and pδ. Then y′ extends aγ(i) and pδ(y, ξ) = pγ(y′, ξ).

Since x also extends aγ(i) we have ξ < γ 6 ∆(x, y′) and χ(y′) = (N, 0) = χ(x). So by

3(a) for the condition pγ we conclude that pγ(x, ξ) = pγ(y′, ξ) = pδ(y, ξ) as required.

Subcase b: If x γ = aγ(i) and y γ = aδ(j) with i 6= j < n.

Let y′ be the copy of y in dom0(pγ) relative to the isomorphism between the conditions

pγ and pδ since y′ extends aγ(j) we get that ∆(x, y) = ∆(aγ(i), aγ(j)) = ∆(x, y′), so in

particular, ξ < ξ. By 3(a) of pγ we infer pγ(x, ξ) = pγ(y′, ξ) = pδ(y, ξ).

Subcase c: If x ∈ B or y ∈ B, where B is the constant value of the mapping δ 7→ bδ. Then

either x, y ∈ dom0(pγ) or x, y ∈ dom0(pδ). Let us focus on the case x, y ∈ dom0(pγ), as

the other one is similar. This implies that ξ < ξ, so in particular (y, ξ) is a fixed point

of the isomorphism between pγ and pγ so we get pγ(y, ξ) = pδ(y, ξ) applying 3(a) for

condition pγ we obtain pγ(x, ξ) = pγ(y, ξ). Finishing the proof.

Case 2: Consider the case x ∈ dom0(pγ), y ∈ dom0(pδ) and χ(x) = χ(y) = (N, 1) for

some N .

Subcase a: If x γ = aγ(i), y δ = aδ(i) for some i < n.

Since (k, 0) <lex (N, 1) it follows from (12) and the definition of ξ0 that ∆(x, y) =

∆(aγ(i), aδ(i)) > ξ0. Moreover p(x, ξ0) = m+ 1 6= m+ 2 = p(y, ξ0) so ξ0 is a witness for

clause 3(b).

Subcase b: If x γ = aγ(i), y δ = aδ(j) with i 6= j < n.

Then ∆(x, y) = ∆(aγ(i), aγ(j)) = ∆(aδ(i), aδ(j)) which is less than ξ. Let x′ and y′

be the copies of x and y relative to the isomorphism between pγ and pδ, respectively.

Note that x′ extends aδ(i) and y′ extends aγ(j). Since pγ is a condition there exists

η < ∆(x, x′) = ∆(y, y′) such that pγ(x, η) 6= pγ(x′, η). Since the isomorphism between

pγ and pδ is the identity for elements less than ξ we get that η ∈ dom1(pδ). Moreover,

pγ(x, η) = pδ(x′, η) and pγ(y

′, η) = pδ(y, η) which implies that pγ(x, η) 6= pδ(y, η)so η is a

witness for 3(b).


Subcase c: If x ∈ B or y ∈ B, where B is the constant value of the mapping δ 7→ bδ. Then

either x, y ∈ dom0(pγ) or x, y ∈ dom0(pδ). Since both cases are similar let us assume that

x, y ∈ dom0(pγ). Let y′ be the copy of y relative to the isomorphism between pγ and pδ.

By 3(b) for the condition pγ we can find η < ∆(x, y) so that pγ(x, η) 6= pγ(y′, η) = pδ(y, η).

Finishing the proof.

We are left with proving that p extends both pγ and pδ. Given x, y ∈ dom0(pγ) or

x, y ∈ dom0(pδ) we have that p(x, ξ0) = m + 1 = p(y, ξ0) or p(x, ξ0) = m + 2 = p(y, ξ0),

respectively, where ξ0 is the only new member of dom1(p)\dom1(pγ) and the only element

of dom1(pδ). On the other hand, given x ∈ dom0(pγ) or x ∈ dom0(pδ) we have that

p(x, ξ0) = m + 1 > p(x, η) for all (x, η) ∈ dom(pγ) or p(x, ξ0) = m + 2 > p(x, η) for all

(x, η) ∈ dom(pγ) and all (x, η) ∈ dom(pδ). Thus, p satisfies clauses (4) and (5). This

finishes the proof.

Remark 4.5.2. Note that the above argument shows that for every sequence

T 00 ≺ T 0

1 ≺ ... ≺ T 0ξ ≺ ... ≺ T 1

ξ ≺ ... ≺ T 11 ≺ T 1

0 (ξ < γ)

of coherent Aronszajn trees of some countable limit length γ, the poset

P(T iξ : ξ < γ, i < 2)

satisfies the countable chain condition.

Lemma 4.5.3. Assume MAω1. For every sequence

T 00 ≺ T 0

1 ≺ ... ≺ T 0ξ ≺ ... ≺ T 1

ξ ≺ ... ≺ T 11 ≺ T 1

0 (ξ < ω1)

of coherent Aronszajn trees there is a coherent Aronszajn tree T such that T 0ξ ≺ T ≺ T 1

ξ

for all ξ < ω1.

Proof. We may assume without loss of generality that T iξ ∩ T jη = ∅ for (η, j) 6= (ξ, i) ∈

ω1 × 2. Construct a subset X of⋃δ∈ω1i∈2 T

iδ with the following properties:


(i) X ∩ T iδ is uncountable for all δ < ω1 and i ∈ 2,

(ii) ∀α < ω1 |X ∩ Levα(T iδ)| 6 1 for all δ < ω1 and i ∈ 2,

(iii) ∀α < ω1|(δ, i) ∈ ω1 × 2 : X ∩ Levα(T iδ) 6= ∅| 6 1 and

(iv) ∆(X ∩ T iδ) > ω for all δ < ω1, i < 2.

It follows from our assumptions on the elements of the sequence that there is an index

function

χ :⋃

(δ,i)∈ω1×2

T iδ → ω1 × 2

such that χ(x) = (δ, i) if x ∈ T iδ . We shall use MAω1 to produce a tree T witnessing

∀δ < ω1, T0δ T T 1

δ .

Let P(T iξ : ξ < ω1, i < 2) be the poset of all partial finite mappings p : X × ω1 → ω

with the following properties:


2. ∀x, y ∈ dom0(p)

∀ξ < ht(x), ht(y) [(x, ξ) ∈ dom(p) if and only if (y, ξ) ∈ dom(p)],

3. For all x, y ∈ dom0(p),

(a) If χ(x) = χ(y) = (δ, 0) then

(x, ξ), (y, ξ) ∈ dom(p), ξ < ∆(x, y) implies p(x, ξ) = p(y, ξ).

(b) If χ(x) = χ(y) = (δ, 1) then there exist

ξ < ∆(x, y) such that p(x, ξ) 6= p(y, ξ).



4. p(x, ξ) = p(y, ξ) for all x, y ∈ dom0(q) and ξ < minht(x), ht(y) such that ξ /∈

dom1(q),


Note that if P satisfies the countable chain condition then an application of MAω1

and an analogous arguments to those in Claim1 and Claim 2 of lemma 4.5.1 will provide

us with a tree T filling the gap. Therefore, the following claim is all we need to complete

the proof.

Claim The poset P satisfies the countable chain condition.

Proof of the Claim: Let 〈pγ : γ ∈ ω1〉 be given uncountable sequence of conditions. By

the ∆-system Lemma and a counting argument we may assume that 〈dom(pγ) : γ ∈ ω1〉

form a ∆-system with root D and moreover pγ D = pγ′ D for all γ, γ′ ∈ ω1. Let

Aγ = χ′′[dom0(pγ)]

By applying ∆-system lemma to the sequence 〈Aγ : γ ∈ ω1〉 we get an uncountable set

Γ ⊆ ω1 and a finite set A such that 〈Aγ : γ ∈ ω1〉 forms a ∆-system with root A. We

consider two cases:

Case 1: If Aγ = A for all γ ∈ Γ. Let α = maxδ : (∃i < 2)(δ, i) ∈ A+ω.Then 〈pγ : γ ∈ Γ〉

is an uncountable sequence in the suborder P(T iξ : ξ < α, i < 2) of P . In view of remark

4.5.2 the suborder P(T iξ : ξ < α, i < 2) satisfies the countable chain condition. So there

is an r ∈ P(T iξ : ξ < α, i < 2) and γ, δ ∈ Γ such that r 6 pγ and r 6 pδ. It should be

clear that r extends both pγ and pδ as conditions in the larger partial order P .

Case 2: If Aγ \ A 6= ∅ for all γ ∈ Γ. Let us consider the reduction 〈pAγ : γ ∈ Γ〉 of the

sequence 〈pγ : γ ∈ Γ〉 to the suborder P(T iξ : ξ < α, i < 2), where pAγ = pγ (x, ξ) ∈

dom(pγ) : χ(x) ∈ A and α = maxδ : (∃i < 2)(δ, i) ∈ A + ω. By remark 4.5.2 there

are γ, δ ∈ Γ and r ∈ P(T iξ : ξ < α, i < 2) such that r 6 pAγ , pAδ . We can amalgamate r

with pγ, pδ to a new condition q of P(T iξ : ξ < ω1, i < 2) as follows. First of all note

that r ∪ pγ ∪ pδ is a function. Let n = max[rang(r) ∪ rang(pγ) ∪ rang(pδ]. Let q be the


function with domain

(x, ξ) : x ∈ dom0(r ∪ pγ ∪ pδ), ξ ∈ dom1(r ∪ pγ ∪ pδ) ξ < ht(x).

Defined by q dom(r) = r, q dom(pγ) = pγ, q dom(pδ) = pδ and q(x, ξ) = n + 1 for

all (x, ξ) ∈ dom(q) \ (dom(r) ∪ dom(pγ) ∪ dom(pδ). Since condition 3 is only meaningful

for pairs x, y ∈ dom0(q) so that χ(x) = χ(y). It follows from our construction that

this only applies to pairs x, y so that either x, y ∈ dom0(pγ) or x, y ∈ dom0(pδ) or

χ(x) = χ(y) ∈ A. In the later case clause 3 holds since x, y ∈ dom0(r) and r is a

condition of P(T iξ : ξ < α, i < 2). Thus, q is a condition. Let us now show that q extend

both pγ and pδ. Consider x, y ∈ dom(pγ)orx, y ∈ dom0(pδ) it follows from the definition

of q that q(x, ξ) = n + 1 = q(y, ξ) for all ξ /∈ dom1(pγ) or ξ /∈ dom1(pδ). Moreover,

q(x, ξ) = n + 1 > q(x, η) for all ξ /∈ dom1(pγ) and η ∈ dom1(pγ) or ξ /∈ dom1(pδ) and

η ∈ dom1(pδ).This finishes the proof.

Note that a similar argument establishes the following two lemmas.


T 00 ≺ T 0

1 ≺ ... ≺ T 0n ≺ ... ≺ T 1

ξ ≺ ... ≺ T 11 ≺ T 1

0 (n < ω, ξ < ω1)

of coherent Aronszajn trees there is a coherent Aronszajn tree T such that T 0n ≺ T ≺ T 1

ξ

for all n < ω, ξ < ω1.


T 00 ≺ T 0

1 ≺ ... ≺ T 0ξ ≺ ... ≺ T 1

n ≺ ... ≺ T 11 ≺ T 1

0 (n < ω, ξ < ω1)

of coherent Aronszajn trees there is a coherent Aronszajn tree T such that T 0ξ ≺ T ≺ T 1

n

for all n < ω, ξ < ω1.

Definition 4.5.6. Let ∼ be the equivalence relationship on C determined as follows:

T ∼ S iff (∃n ∈ Z) T (n) ≡ S.


Note that this equivalence relation is giving us a dense linear ordering (C/ ∼,). We

are now in position to state and prove the main theorem of the chapter.

Theorem 4.5.7. Assume MAω1. There are no gaps of type (κ, λ∗)-gaps on (C/ ∼,)

for λ, κ ∈ 1, ω, ω1.

Proof. By lemmas 4.5.1, 4.5.3, 4.5.4 and 4.5.5 we are left formally to consider the cases of

(1, ω∗), (1, ω∗1), (ω, 1) and (ω1, 1) gaps in (C/ ∼,). The first two cases are quite similar

so we consider the case of (1, ω∗1)-gaps.

Consider a gap

T ≺ ... ≺ Tξ ≺ ... ≺ T1 ≺ T0 (ξ < ω1)

of coherent trees. Note that since there are no coherent trees between T and its shift

T (1) we conclude that T (1) ≺ Tξ for all ξ < ω1. Repeating it for other finite shifts T (n) of

T we get that

T = T (0) ≺ T (1) ≺ ... ≺ T (n) ≺ ... ≺ Tξ ≺ ... ≺ T1 ≺ T0 (n ∈ ω, ξ < η ∈ ω1).

It follows from lemma 4.5.4 that there is a tree T such that

T (n) ≺ T ≺ Tξ for all (n < ω, ξ < ω1).

Thus, T fills the (1, ω∗1)-gap.

So we are left to show that the quotient ordering does not have (ω, 1) and (ω1, 1) gaps.

Since the two cases are similar let us consider, the case of (ω1, 1)-gaps. So consider a gap

T0 ≺ T1 ≺ ... ≺ Tξ ≺ ... ≺ T (ξ < ω1).

Note that if T (−n) is defined for all n < ω then we would be done by lemma 4.5.5 Let N be

the maximal integer so that T (−N) is defined. Consider the poset P(T (−N), Tξ (ξ < ω1)) as

we have seen above this forcing notion satisfies the countable chain condition and provide

us with a coherent tree T such that Tξ T ≺ T (−N) for all ξ < ω1 the relation T ≺ T (−N)

follows from the Clause 3(b) of the definition of the forcing and lemma 4.3.2.


We finish this chapter with a complete description of the quotient structure under the

stronger forcing axiom PFA. In order to do so we have to analyze the existence of upper

and lower bounds for strictly increasing sequences of length ω1 and strictly decreasing

sequences of length ω1, respectively.

Lemma 4.5.8. MAω1 . For every sequence of coherent Aronszajn trees

T0 ≺ T1 ≺ ... ≺ Tξ ≺ ... (ξ < ω1)

there is a coherent Aronszajn tree T such that Tξ ≺ T for all ξ < ω1.

Proof. Let xα(α < ω1) be an enumeration of the tree T0. Consider the tree

T ′ = xα ∪ t [htT0(xα), htTα(t)) : α < ω1, t ∈ Tα.

It should be clear that T ′ is an Aronszajn tree and that lemma 2.1.7 implies Tξ ≺ T ′ for

all ξ < ω1. By lemma 4.4.1 we can find a coherent Aronszajn tree T such that T ′ T .

this finishes the proof.

The next two lemmas together with lemma 4.4.4 above show that under PFA the

coinitiality of (C/ ∼,≺) is bigger or equal to ω2. This answers question 9.9 of [25]

attributed to Todorcevic.

Lemma 4.5.9. MAω1 . For every sequence of coherent Aronszajn trees

... ≺ Tn ≺ ... ≺ T1 ≺ T0 (n < ω)

there is a coherent Aronszajn tree T such that T ≺ Tn for all n < ω.

Proof. We would like to use the naturally defined poset P(Tn : n < ω) but unfortunately

in view of lemma 4.4.4 it is unlikely that P(Tn : n < ω) satisfies the countable chain

condition. Therefore we are going to modify the poset by weakening the clause 3(b) in

the definition of the forcing. Consider a strictly decreasing sequence

... ≺ Tn ≺ ... ≺ T1 ≺ T0 (n < ω)


We may assume without loss of generality that Tn ∩ Tm = ∅ for all n 6= m < ω. As

before we define some auxiliary notions to help us with the construction. It follows from

our assumptions on the sequence that there is a well defined index function

χ :⋃n<ω

Tn → ω

defined by

χ(x) = n if and only if x ∈ Tn.

Now construct a subset X of⋃n<ω Tn with the following properties:

(i) X ∩ Tn is uncountable for all n ∈ ω,

(ii) ∀α < ω1 |X ∩ Levα(Tn)| 6 1 for all n ∈ ω and

(iii) ∀α ∈ ω1 |n ∈ ω : X ∩ Levα(Tn) 6= ∅| 6 1.

We shall use MAω1 to produce a tree T witnessing

∀k < ω, T ≺ Tk.

Let P = P(Tn : n ∈ ω) be the poset of all partial finite mappings p : X × ω1 → ω with

the following properties:


2. ∀x, y ∈ dom0(p) and ∀ξ < ht(x), ht(y) (x, ξ) ∈ dom(p) if and only if (y, ξ) ∈ dom(p),

3. For all x, y ∈ dom0(p) (χ(x) = χ(y) = n) implies that there exist

ξ 6 ∆(x, y) such that p(x, ξ) 6= p(y, ξ).

We let p extend q if p extends q as a function and

4. p(x, ξ) = p(y, ξ) for all x, y ∈ dom0(q) and ξ < minht(x), ht(y), ξ /∈ dom1(q),



It follows that if P satisfies the countable chain condition then an application

of MAω1 to the dense sets defined in Claim 1 and Claim 2 will give us a map

f : X × ω1 → ω so that it is fibers fx : ht(x) → ω are total maps. Conditions (4)

and (5) imply that 〈fx : x ∈ X〉 is a sequence of coherent finite-to-one maps. Let

T = fx ξ : x ∈ X, ξ 6 ht(x).

Then T is a coherent Aronszajn tree.

Claim 1 For all x0 ∈ X, the set Dx0 = p ∈ P : x0 ∈ dom0(p) is dense open.

Proof. Let p ∈ P be given with the property that x0 /∈ dom(p). We shall proceed

by cases:

Case 1: If there is a x ∈ (dom0(p) \ x0) such that χ(x) = χ(x0). Set

ξ0 = min∆(x0, x) : x ∈ (dom0(p) \ x0), χ(x0) = χ(x).

Let q be a map with domain equal to

dom(p)∪(x, ξ0) : x ∈ dom0(p), ξ0 < ht(x)∪(x0, ξ0)∪(x0, ξ) : ξ ∈ dom1(p), ξ < ht(x0).

Define q by q dom(p) = p, q(x, ξ0) = n + 2 for x ∈ dom0(p) and q(x0, ξ) =

q(x0, ξ0) = n + 1 for all ξ ∈ dom1(p), ξ < ht(x0), where n = max(rang(p)). Let

us show that q is a condition in the partial order P . It should be clear from our

definition of q that clauses (1) and (2) holds. Consider x, y ∈ dom0(q) so that

χ(x) = χ(y). If x, y ∈ dom0(p) then using the fact that p is a condition we can

find η < ∆(x, y) in dom1(p) so that p(x, η) 6= p(y, η), otherwise if x0 = x then

q(x0, ξ0) = n + 1 6= n + 2 = q(y, ξ0). It follows that q is a condition in Dx0 .

Let us verify that q extends p. By definition q extend p as a function. For all

x, y ∈ dom0(p) we have q(x, ξ0) = n + 2 = q(y, ξ0), where ξ0 is the only element

in dom1(q) \ dom1(p). Moreover, q(x, ξ0) = n + 2 > q(x, ξ) for all (x, ξ) ∈ dom(p).



Case 2: If there is no x ∈ dom0(p) so that χ(x0) = χ(x). Then let q be the function

with domain


defined by q dom(p) = p and q(x0, ξ) = n + 1 for all ξ ∈ dom1(p), ξ < ht(x0),

where n = max(rang(p)). It should be clear in this case that q is a condition in

Dx0 extending p.

Claim 2 For all (x0, ξ0) ∈ X × ω1 with ξ0 < ht(x0), the set D(x0,ξ0) is open-dense

in P .

Proof. Let p ∈ P be given. By Claim 1 we can find a condition p′ extending p so

that x0 ∈ dom0(p′). Let n = max(rang(p′)). Set q to be the function with domain

dom(p′) ∪ (x, ξ0) : x ∈ dom0(p′), ξ0 < ht(x)

be defined by q dom(p′) = p′ and q(x, ξ0) = n + 1 for all x ∈ dom0(p′), such

that ξ0 < ht(x). Let us verify that q is a condition. It should be clear that

clauses (1) and (2) holds. Let x, y ∈ dom0(q) be given so that χ(x) = χ(y) since

x, y ∈ dom0(p′) and p is a condition we can find η ∈ dom1(p′) such that η < ∆(x, y)

and p(x, η) 6= p(y, η). We are now left to prove that q extends p′. First note that

q(x, ξ0) = n+ 1 = q(y, ξ0) for all x, y ∈ dom0(q) so that ξ0 < ht(x), ht(y), where ξ0

is the only element of dom1(q) \ dom1(p′). Moreover, q(x, ξ0) = n+ 1 > q(x, ξ) for

all (x, ξ) ∈ dom(p′). Thus, D(x0,ξ0) is open dense in P .

Claim 3: T ≺ Tn for all n < ω.

Proof. Let Xn = fx : χ(x) = n. Define the map ϕn : Xn → Tn as follows

ϕn(fx) = x. Then by 3 we infer ϕn is a Lipszchitz map. Using lemma 1.1.13, we

conclude that T Tn for all n ∈ ω. The strict inequality follows from T Tn+1 ≺

Tn for all n < ω.


We are left with proving the following claim.

Claim 4 The partially ordered set P satisfies the countable chain condition.

Proof. Let 〈pδ : δ ∈ ω1〉 be an uncountable sequence of elements of P . By the ∆-

system Lemma and by a counting argument, we may assume that 〈dom(pδ) : δ ∈ ω1〉

form a ∆-system with root D and pδ D = pδ′ D for all δ, δ′ ∈ ω1. Moreover, we

may assume that the image mapping δ 7→ χ′′(dom0(pδ)) is constant.

For δ < ω1, let

aδ = x δ : x ∈ dom0(pδ), ht(x) > δ and bδ = x ∈ dom0(pδ) : ht(x) < δ.

Moreover, for δ < α, we let

h(δ) = max(∆(x, y) : x, y ∈ aδ, x 6= y, χ(x) = χ(y)∪(dom1(pδ)∩δ)∪ht(x) : x ∈ bδ)+1.

By the pressing down lemma there is a stationary set Γ of countable limit ordinals

on which the mapping h, and the mapping

δ → dom(pδ) ∩ (X δ)× δ

as well as the mapping δ 7→ bδ are constant. Let ξ, F and B be the constant values,

respectively. Shrinking Γ, we may assume that all aδ (δ ∈ Γ) are of some fixed size

n. Moreover, we may assume that the function δ 7→ aδ ξ has a constant value

a, and that the conditions pδ(δ ∈ Γ) generate isomorphic structures over ξ, F and

a. Thus, we want in particular the isomorphism between the pδ (δ ∈ Γ) to respect

a fixed enumeration aδ(i)(i < n) of aδ. Moreover, the isomorphism between the

conditions pγ and pδ is the identity on (X ξ)× ξ, and aδ(i) ξ (i < n) and maps

nodes extending aγ(i) into nodes extending aδ(i) for all i < n. Using lemmas 1.1.9

and 1.1.10 we find an uncountable subset Σ ⊂ Γ such that for ξ < γ < δ in Σ:

(6) aγ(i) ξ = aδ(i) ξ for all i < n,


(7) every node of dom0(pγ) is either of height less or equal to ξ or it extends some

aγ(i)(i < n),

(8) every node of dom0(pδ) is either of height less or equal to ξ or it extends some

aδ(i)(i < n),

(9) dom1(pγ) ⊂ ξ ∪ (γ, δ) and dom1(pδ) ⊂ ξ ∪ (δ, ω1),

(10) aγ(i) and aδ(j) are incomparable for all i, j < n, whenever χ(aγ(i) = χ(aδ(j)),

(11) ∆(aγ(i), aδ(i)) = ∆(aγ(j), aδ(j))) for all i, j < n, whenever χ(aγ(i)) = χ(aδ(i)) =

χ(aγ(j)) = χ(aδ(j)) and

(12) ∆(aγ(i), aδ(i)) < ∆(aγ(j), aδ(j))), whenever χ(aγ(i)) = χ(aδ(i)) > χ(aγ(j)) =

χ(aδ(j)).

We claim that if ξ < γ < δ are in Σ, then pγ and pδ are compatible. Let

m = max[(range(pγ) ∪ (range(pδ)].

Let k be the maximum of the set χ(aδ(i)) : i < n and let ξ0 = ∆(aγ(i0), aδ(i0)) for

some (every) i0 such that χ(aδ(i0)) = k. We define p ∈ P by letting its domain be

dom(pγ) ∪ dom(pδ) ∪ (x, ξ0) : x ∈ dom0(pγ) ∪ dom0(pδ), ht(x) > ξ0,

and letting p(x, ξ0) = m + 1 for x ∈ dom0(pγ), ht(x) > ξ0, and p(x, ξ0) = m + 2 for

x ∈ dom0(pδ), ht(x) > ξ0. and let p (dom(pγ) ∪ dom(pδ)) = pγ ∪ pδ.

Note that p is a function extending pγ, pδ and it clearly satisfy (1) and (2). Note moreover

that dom1(q) = dom1(pγ) ∪ dom1(pδ) ∪ ξ0 and that dom0(q) = dom0(pγ) ∪ dom0(pδ).

Let us show that p satisfies clause (3) of being a condition. In order to do so we shall

proceed by cases: Consider the case x ∈ dom0(pγ), y ∈ dom0(pδ) and χ(x) = χ(y) = N

for some N .

Case 1: If x γ = aγ(i), y δ = aδ(i) for some i < n.


Since k > N it follows from (12) and the definition of ξ0 that ∆(x, y) = ∆(aγ(i), aδ(i)) >

ξ0. Moreover p(x, ξ0) = m+ 1 6= m+ 2 = p(y, ξ0) so ξ0 is a witness for clause (3).

Case 2: If x γ = aγ(i), y δ = aδ(j) with i 6= j < n.

Then ∆(x, y) = ∆(aγ(i), aγ(j)) = ∆(aδ(i), aδ(j)) which is less than ξ. Let x′ and y′

be the copies of x and y relative to the isomorphism between pγ and pδ, respectively.

Note that x′ extends aδ(i) and y′ extends aγ(j). Since pγ is a condition there exists

η < ∆(x, x′) = ∆(y, y′) such that pγ(x, η) 6= pγ(x′, η). Since the isomorphism between

pγ and pδ is the identity for elements less than ξ we get that η ∈ dom1(pδ). Moreover,

pγ(x, η) = pδ(x′, η) and pγ(y

′, η) = pδ(y, η) which implies that pγ(x, η) 6= pδ(y, η) so η is

a witness for (3).

Case 3: If x ∈ B or y ∈ B, where B is the constant value of the mapping δ 7→ bδ. Then

either x, y ∈ dom0(pγ) or x, y ∈ dom0(pδ). Since both cases are similar let us assume that

x, y ∈ dom0(pγ). Let y′ be the copy of y relative to the isomorphism between pγ and pδ.

By (3) for the condition pγ we can find η < ∆(x, y) so that pγ(x, η) 6= pγ(y′, η) = pδ(y, η).

Finishing the proof.

We are left with proving that p extends both pγ and pδ. Given x, y ∈ dom0(pγ) or

x, y ∈ dom0(pδ) we have that p(x, ξ0) = m + 1 = p(y, ξ0) or p(x, ξ0) = m + 2 = p(y, ξ0),

respectively, where ξ0 is the only new member of dom1(p)\dom1(pγ) and the only element

of dom1(pδ). On the other hand, given x ∈ dom0(pγ) or x ∈ dom0(pδ) we have that

p(x, ξ0) = m + 1 > p(x, η) for all (x, η) ∈ dom(pγ) or p(x, ξ0) = m + 2 > p(x, η) for all

(x, η) ∈ dom(pγ) and all (x, η) ∈ dom(pδ). Thus, p satisfies clauses (4) and (5). This

finishes the proof.

As before we make the following remark about the proof of Claim 4.

Remark 4.5.10. Note that the above argument shows that for every sequence

... ≺ Tξ ≺ ... ≺ T1 ≺ T0 (ξ < γ)

of coherent Aronszajn trees of some countable limit length γ, the poset P(Tξ : ξ < γ)


satisfies the countable chain condition.

Lemma 4.5.11. MAω1 . For every decreasing sequence of coherent Aronszajn trees

... ≺ Tξ ≺ ... ≺ T1 ≺ T0 (ξ < ω1)

there is a coherent Aronszajn tree T such that T ≺ Tξ for all ξ < ω1.

Proof. We may assume without loss of generality that Tξ ∩ Tη = ∅ for η 6= ξ ∈ ω1. As

before we shall define some auxiliary notions. Construct a subset X of⋃δ∈ω1

Tδ with the

following properties:

(i) X ∩ Tδ is uncountable for all δ < ω1,

(ii) ∀α < ω1 |X ∩ Levα(Tδ)| 6 1 for all δ < ω1 and

(iii) ∀α < ω1|δ ∈ ω1 : X ∩ Levα(Tδ) 6= ∅| 6 1.

It follows from our assumptions on the elements of the sequence that there is an index

function

χ :⋃δ∈ω1

→ Tδ

such that χ(x) = δ if x ∈ Tδ. We shall use MAω1 to produce a tree T witnessing

∀δ < ω1, T ≺ Tδ.

Let P(Tξ : ξ < ω1) be the poset of all partial finite mappings p : X × ω1 → ω with the

following properties:


2. ∀x, y ∈ dom0(p)

∀ξ < ht(x), ht(y) (x, ξ) ∈ dom(p) if and only if (y, ξ) ∈ dom(p),


3. For all x, y ∈ dom0(p), χ(x) = χ(y) = δ implies that there exist

ξ 6 ∆(x, y) such that p(x, ξ) 6= p(y, ξ).


4. p(x, ξ) = p(y, ξ) for all x, y ∈ dom0(q) and ξ < minht(x), ht(y) such that ξ /∈

dom1(q),


Note that if P satisfies the countable chain condition then an application of MAω1

and an analogous arguments to those in Claim1 and Claim 2 of lemma 4.5.8 will provide

us with a tree T satisfying ∀δ < ω1, T ≺ Tδ. Therefore, the following claim is all we

need to complete the proof.

Claim The poset P satisfies the countable chain condition.

Proof of the Claim: Let 〈pγ : γ ∈ ω1〉 be given uncountable sequence of conditions. By

the ∆-system Lemma and a counting argument we may assume that 〈dom(pγ) : γ ∈ ω1〉

form a ∆-system with root D and moreover pγ D = pγ′ D for all γ, γ′ ∈ ω1. Let

Aγ = χ′′[dom0(pγ)]

By applying ∆-system lemma to the sequence 〈Aγ : γ ∈ ω1〉 we get an uncountable set

Γ ⊆ ω1 and a finite set A such that 〈Aγ : γ ∈ ω1〉 forms a ∆-system with root A. We

consider two cases:

Case 1 If Aγ = A for all γ ∈ Γ. Let α = max(A) + ω.Then 〈pγ : γ ∈ Γ〉 is an

uncountable sequence in the suborder P(Tξ : ξ < α) of P . In view of remark 4.5.10

the suborder P(Tξ : ξ < α) satisfies the countable chain condition. So there is an

r ∈ P(T iξ : ξ < α, i < 2) and γ, δ ∈ Γ such that r 6 pγ and r 6 pδ. It should be clear

that r extends both pγ and pδ as conditions in the larger partial order P .

Case 2 If Aγ \ A 6= ∅ for all γ ∈ Γ. Let us consider the reduction 〈pAγ : γ ∈ Γ〉 of

the sequence 〈pγ : γ ∈ Γ〉 to the suborder P(T iξ : ξ < α), where pAγ = pγ (x, ξ) ∈


dom(pγ) : χ(x) ∈ A and α = max(A) + ω. By remark 4.5.10 there are γ, δ ∈ Γ and

r ∈ P(T iξ : ξ < α) such that r 6 pAγ , pAδ . We can amalgamate r with pγ, pδ to a new

condition q of P(T iξ : ξ < ω1) as follows. First of all note that r ∪ pγ ∪ pδ is a function.

Let n = max[rang(r) ∪ rang(pγ) ∪ rang(pδ)]. Let q be the function with domain

(x, ξ) : x ∈ dom0(r ∪ pγ ∪ pδ), ξ ∈ dom1(r ∪ pγ ∪ pδ) ξ < ht(x).

Defined by q dom(r) = r, q dom(pγ) = pγ, q dom(pδ) = pδ and q(x, ξ) = n + 1 for

all (x, ξ) ∈ dom(q) \ (dom(r) ∪ dom(pγ) ∪ dom(pδ). Since condition 3 is only meaningful

for pairs x, y ∈ dom0(q) so that χ(x) = χ(y). It follows from our construction that

this only applies to pairs x, y so that either x, y ∈ dom0(pγ) or x, y ∈ dom0(pδ) or

χ(x) = χ(y) ∈ A. In the later case clause 3 holds since x, y ∈ dom0(r) and r is a

condition of P(T iξ : ξ < α). Thus, q is a condition. Let us now show that q extend both

pγ and pδ. Consider x, y ∈ dom(pγ) or x, y ∈ dom0(pδ) it follows from the definition

of q that q(x, ξ) = n + 1 = q(y, ξ) for all ξ /∈ dom1(pγ) or ξ /∈ dom1(pδ). Moreover,

q(x, ξ) = n + 1 > q(x, η) for all ξ /∈ dom1(pγ) and η ∈ dom1(pγ) or ξ /∈ dom1(pδ) and

η ∈ dom1(pδ).This finishes the proof.

Theorem 4.5.12. Assuming PFA. The ordering 〈C/ ∼,〉 is the unique ω2-saturated

linear order of cardinality ω2.

Proof. The result is the content of theorem 4.5.7 as well as lemmas 4.5.8, 4.5.9 and 4.5.11

and the well-known fact that |C| = ω2 under PFA.

Bibliography

[1] U. Abraham and S. Shelah. Isomorphism Types of Aronszajn Trees. Israel J. Math.,

50 (1985), 75-113.

[2] T. Bartoszynski and H. Judah, Set Theory on the Structure of the Real Line, A K

Peters, 1995.

[3] J. Baumgartner. Order Types of Real Numbers and Other Uncountable Orderings

Ordered Sets. 239-277 (NATO Advanced Study Institute Series ,1981).

[4] J. Baumgartner. All ω1-dense Sets of Reals can be Isomorphic. Fund. Math., 79 (1973),

101-176.

[5] G. Cantor. Beitrage Zur Begrundung Der Transfiniten Mengenihre. Math. Ann., 46

(1895) 481-512.

[6] R.S. Countryman, Spaces having a σ-monotone base, Preprint 1970.

[7] K. J. Devlin. Constructibility. Perspectives in Mathematical Logic. Springer-Verlag,

Berlin, 1984.

[8] K. J. Devlin and S. Shelah. A Weak Version of ♦ which follows from 2ℵ0 < 2ℵ1. Israel

J. Math., 29 (1978), 239-247.

[9] B. Dushnik and E. W. Miller. Concerning similarity transformation of linearly ordered

sets. Bull. Amer. Math. Soc., 46(1940) 321-326.

85

Bibliography 86

[10] P. Erdos and R. Rado. A partition calculus in set theory. Bull. Amer. Math. Soc.,

62, 427-489.

[11] R. Fraisse. Sur la comparaison des types d’ordres. C. R. Acad. Sci. Paris, 226: 1330,

1948.

[12] F. Galvinand K. Prikry. Borel sets and Ramsey’s theorem. J. Symbolic Logic 38

(1973), 193–198.

[13] F. Hausdorff. Grundzuge einer Theorie der geordneten Mengen. Math. Ann., 65

(1908), 435-505.

[14] R. B. Jensen. The fine structure of the constructible hierarchy. Ann. Math. Logic, 4

(1972), 229-308.

[15] D. Konig. Uber eine Schlussweisse aus dem Endlichen ins Undenliche. Acta. Litt.

Acad. Sci. Hung., Zesged, 3, 121-130.

[16] B. Konig. Local coherence. Ann. of Pure and App. Logic, 124 (2003), 107-139.

[17] D. Kurepa.Ensembles Ordonnes et Ramifies. Publ. Math. Univ. Belgrade, 4 (1935),

1-138.

[18] K. Kunen. Set theory. Studies in Logic and the Foundations of Mathematics, 102.

North-Holland (1983).

[19] R. Laver.On Fraisse’s Order Type Conjecture. Anns. of Math., (2) 93, (1971) 89-111.

[20] R. Laver. An order type decomposition theorem. Anns. of Math., (2) 98, (1973),

96-119.

[21] R. Laver, Better-quasi-orderings and a class of trees. Studies in foundations and

combinatorics pp. 31-48, Adv. in Math. Suppl. Stud., Academic-Press, New York-

London, 1978.

Bibliography 87

[22] M. Hrusak and C. Martinez-Ranero. Some remarks on non-special coherent Aron-

szajn trees. Act. Univ. Carolin. Math. Phys., (46) 2005.

[23] C. Martinez-Ranero. Aronszjan Lines are Well-Quasi-Ordered, submitted to Funda-

menta Mathematicae, 2010.

[24] C. Martinez-Ranero and S. Todorcevic. Gap structure of coherent Aronszajn trees.

submitted to Mathematical Research Letters, 2010.

[25] J. Moore. Structural analysis of Aronszajn trees. Logic Colloquium 2005, 85–106,

Lect. Notes Log., 28, Assoc. Symbol. Logic, Urbana, IL, 2008

[26] J. Moore. A universal Aronszajn line. Math. Res. Lett. 16 (2009), no. 1, 121-131.

[27] J. Moore. A Five Element Basis for Uncountable Linear Orderings. Ann. Math., 19

(2006), 717-736.

[28] C. St. J. A. Nash-Williams. On well-quasi-ordering finite trees. Proc. Cambridge

Philos. Soc., 59 (1963), 833-835.

[29] C. St. J. A. Nash-Williams. On well-quasi-ordering Infinite Trees. Proc. Cambridge

Philos. Soc., 61 (1965), 697-720.

[30] C. St. J. A. Nash-Williams. On well-quasi-ordering transfinite sequences. Proc. Cam-

bridge Philos. Soc., 61(1965) 33-39.

[31] S. Shelah. Decomposing Uncountable Squares to Countably Many Chains. J. Combi-

natorial Theory Ser. A., 21(1): 110-114, 1976.

[32] W. Sierpinski. Sur les Types d’ordre des Ensembles Lineares. Fun. Math., 37 (1950),

253-264.

[33] R. Solovay and S. Tennenbaum. Iterated Cohen extensions and Souslins problem.

Ann. of Math., 94:201245, 1971.

Bibliography 88

[34] M. Suslin. Probleme 3. Fund. Math., 1 (1920), 223.

[35] S. Todorcevic. Trees and Linearly Ordered Sets, in Handbook of Set Theoretic Topol-

ogy, 235-293 (North-Holland, Amsterdam, 1984).

[36] S. Todorcevic. Lipschitz Maps on Trees. Journal of the Inst. Jussieu 6(3), (2007),

527-566.

[37] S. Todorcevic. Walks on Ordinals and Their Characteristics. Progress in Mathemat-

ics 263, (2007) Birkhuser Verlag AG.

[38] S. Todorcevi. Ramsey Methods in Analysis. Advanced Course in Mathematics- CRM

Barcelona (2005), Birkhauser Basel.

[39] S. Todorcevic. Partitioning Pairs of Countable Ordinals. Acta Math., 159(3-4): 261-

294, 1987.

Index

Λ, 14

ℵ1-dense, 31

F, 15

♣, 14

♦, 14

♦∗, 20

ηC , 26

ρ0, 6

ρ1, 15

Aronszajn line, 25

Aronszajn tree, vii, 1

atomization process, 30

better-quasi-order, 29

C-rank, 37

C-sequence, 6

coherent tree, 5

condensation, 37

Countryman line, 32

fragmented , 36

height, 1

indecomposable, 43

irreducible tree, 4

level, 1

level-preserving, 2

lexicographical ordering, 30

Lipschitz map, 2

Lipschitz tree, 2

lower trace, 7

minimal walk, 6

partition tree, 30

Q-type, 28

rank, 39

reverse ordering, 27

scheme, 33

shift, 47

step, 6

strongly homogeneous, 8

Suslin line, 31

Suslin tree, 31

uniformly coherent, 8

weak-diamond-principle, 34

89

INDEX 90

well-quasi-order, 27

Documents

by Carlos Martinez-Ranero - University of Toronto