I. Coherent Sequences Stevo Todorcevic

Contents

0 Introduction ...... 1 1 Walks in the space of countable ordinals ...... 3 2 The coherence of maximal weights ...... 9 3 Oscillations of traces ...... 21 4 Triangle inequalities ...... 26 5 Conditional weakly null sequences ...... 39 6 An ultraﬁlter from a coherent sequence ...... 44 7 The trace and the square-bracket operation ...... 51 8 A square-bracket operation on a tree ...... 62 9 Special trees and Mahlo cardinals ...... 67 10 The weight function on successor cardinals ...... 73 11 The number of steps ...... 75 12 Square sequences ...... 80 13 The full lower trace of a square sequence ...... 90 14 Special square sequences ...... 96 15 Successors of regular cardinals ...... 98 16 Successors of singular cardinals ...... 109 17 The oscillation mapping ...... 117 18 The square-bracket operation ...... 122 19 Unbounded functions on successors of regular cardinals 129 20 Higher dimensions ...... 138 Bibliography ...... 152 Index ...... 160

0. Introduction

A transﬁnite sequence Cξ ⊆ ξ (ξ < θ) of sets may have a number of ’coherence properties’ and the purpose of this chapter is to study some of them as well as some of their uses. ’Coherence’ usually means that the Cξ’s are cho-

1 2 I. Coherent Sequences

sen in some canonical way, beyond the natural requirement that Cξ is closed and unbounded in ξ for all ξ. For example, choosing a canonical ’fundamental sequence’ of sets Cξ ⊆ ξ for ξ < ε0 relying on the specific properties of the Cantor normal form for ordinals below the first ordinal satisfying the equation x = ωx is a basis for a number of important results in proof theory. In set theory, one is interested in longer sequences as well and usually has a different perspective in applications, so one is naturally led to use some other tools beside the Cantor normal form. It turns out that the sets Cξ can not only be used as ’ladders’ for climbing up in recursive constructions but also as tools for ’walking’ from an ordinal to a smaller one. This notion of a ’walk’ and the corresponding ’distance functions’ constitute the main body of study in this chapter. We show that the resulting ’metric theory of ordinals’ not only provides a unified approach to a number of classical problems in set theory but also has its own intrinsic interest. For example, from this theory one learns that the triangle inequality of an ultrametric e(α, γ) ≤ max{e(α, β), e(β, γ)} has three versions, depending on the natural ordering between the ordinals α, β and γ, that are of a quite different character and are occurring in quite different places and constructions in set theory. For example, the most frequent occurrence is the case α < β < γ when the triangle inequality becomes something that one can call ’transitivity’ of e. Considerably more subtle is the case α < γ < β of this inequality. It is this case of the inequality that captures most of the coherence properties found in this article. Another thing one learns from this theory is the special role of the first uncountable ordinal in this theory. Any natural coherency requirement on the sets Cξ (ξ < θ) that one finds in this theory is satisfiable in the case θ = ω1. The first uncountable cardinal is the only cardinal on which the theory can be carried out without relying on additional axioms of set theory. The first uncountable cardinal is the place where the theory has its deepest applications as well as its most important open problems. This special role can perhaps be explained by the fact that many set-theoretical problems, especially those coming from other fields of mathematics, are usually con- cerned only about the duality between the countable and the uncountable rather than some intricate relationship between two or more uncountable cardinalities. This is of course not to say that an intricate relationship between two or more uncountable cardinalities may not be a profitable detour in the course of solving such a problem. In fact, this is one of the reasons for our attempt to develop the metric theory of ordinals without restricting ourselves only to the realm of countable ordinals. The chapter is organized as a discussion of four basic distance functions on ordinals, ρ, ρ0, ρ1 and ρ2, and the reader may choose to follow the analysis of any of these functions in various contexts. The distance functions will naturally lead us to many other derived objects, most prominent of 1. Walks in the space of countable ordinals 3 which is the ’square-bracket operation’ that gives us a way to transfer the quantifier ’for every unbounded set’ to the quantifier ’for every closed and unbounded set’. This reduction of quantifiers has proven to be quite useful in constructions of various mathematical structures, some of which have been mentioned or reproduced here. I wish to thank Bernhard König,Piotr Koszmider, Justin Moore and Christine Härtlfor help in the preparation of the manuscript.

1. Walks in the space of countable ordinals

The space ω1 of countable ordinals is by far the most interesting space considered in this chapter. There are many mathematical problems whose combinatorial essence can be reformulated as problems about ω1, which is in some sense the smallest uncountable structure. What we mean by ’structure’ is ω1 together with system Cα (α < ω1) of fundamental sequences, i.e. a system with the following two properties:

(a) Cα+1 = {α},

(b) Cα is an unbounded subset of α of order-type ω, whenever α is a countable limit ordinal > 0.

Despite its simplicity, this structure can be used to derive virtually all known other structures that have been deﬁned so far on ω1. There is a natural recursive way of picking up the fundamental sequences Cα, a recursion that refers to the Cantor normal form which works well for, say, ordinals 1 < ε0 . For longer fundamental sequences one typically relies on some other principles of recursive deﬁnitions and one typically works with fundamental sequences with as few extra properties as possible. We shall see that the following assumption is what is frequently needed and will therefore be implicitly assumed whenever necessary:

1.1 Deﬁnition. 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, or better to say the order-type of this set, is the weight of the step.

1.2 Deﬁnition. A walk (or a minimal walk ) from a countable ordinal β to a smaller ordinal α is the sequence β = β0 > β1 > . . . > βn = α such that for each i < n, the ordinal βi+1 is the step from βi towards α.

1 One is tempted to believe that the recursion can be stretched all the way up to ω1 and this is probably the way P.S. Alexandroﬀ has found his famous Pressing Down Lemma (see [1] and [2, appendix]). 4 I. Coherent Sequences

Analysis of this notion leads to several two-place functions on ω1 that give a rich structure with many applications. So let us expose some of these functions.

2 <ω 1.3 Deﬁnition. The full code of the walk is the function ρ0 :[ω1] −→ ω deﬁned recursively by

a ρ0(α, β) = h|Cβ ∩ α|i ρ0(α, min(Cβ \ α)),

2 a with the boundary value ρ0(α, α) = ∅ where the symbol refers to the sequence obtained by concatenating the one term sequence h|Cβ ∩ α|i with the already known ﬁnite sequence ρ0(α, min(Cβ \ α) of integers.

Knowing ρ0(α, β) and the ordinal β one can reconstruct two following two traces of the walk from β to α. 1.4 Deﬁnition. The upper trace of the walk from β to α is the set

Tr(α, β) = {ξ : ρ0(ξ, β) v ρ0(α, β)}, where v denotes the ordering of being an initial segment among ﬁnite sequences of ordinals. One also has the recursive deﬁnition of this trace,

Tr(α, β) = {β} ∪ Tr(α, min(Cβ \ α)) with the boundary value Tr(α, α) = {α}. 1.5 Deﬁnition. The lower trace of the walk from β to α is the set

L(α, β) = {λξ(α, β): ρ0(ξ, β) < ρ0(α, β)}, where < is the ordering of being a strict initial segment among the ﬁnite sequences of ordinals and where for an ordinal ξ such that ρ0(ξ, β) < ρ0(α, β),

λξ(α, β) = max{max(Cη ∩ α): ρ0(η, β) v ρ0(ξ, β)}. Note the following recursive deﬁnition of this trace,

L(α, β) = L(α, min(Cβ \ α)) ∪ ({max(Cβ ∩ α)}\ max(L(α, min(Cβ \ α)))) with the boundary value L(α, α) = ∅. The following immediate facts about these notions will be frequently and very often implicitly used.

2 2 Note that while ρ0 operates on the set [ω1] of unordered pairs of countable ordinals, the formal recursive definition of ρ0(α, β) does involves the term ρ0(α, α) and so we have to give it as a boundary value ρ0(α, α) = ∅. The boundary value is always specified when such a function is recursively defined and will typically be either 0 or the empty sequence 2 ∅. This convention will be used even when a function e with domain such as [ω1] is given to us beforehand and we consider its diagonal value e(α, α). 1. Walks in the space of countable ordinals 5

1.6 Lemma. For α ≤ β ≤ γ,

3 a (a) α > L(β, γ) implies that ρ0(α, γ) = ρ0(β, γ) ρ0(α, β) and therefore that Tr(α, γ) = Tr(β, γ) ∪ Tr(α, β), (b) L(α, β) > L(β, γ) implies that L(α, γ) = L(β, γ) ∪ L(α, β). a 1.7 Deﬁnition. The full lower trace of the minimal walk is the function 2 <ω F :[ω1] −→ [ω1] deﬁned recursively by [ F (α, β) = F (α, min(Cβ \ α)) ∪ F (ξ, α),

ξ∈Cβ ∩α with the boundary value F (α, α) = {α} for all α. Thus F (α, β) includes the traces of walks between any two ordinals ≤ α that have ever been referred to during the walk from β to α. 1.8 Lemma. For all α ≤ β ≤ γ, (a) F (α, γ) ⊆ F (α, β) ∪ F (β, γ), (b) F (α, β) ⊆ F (α, γ) ∪ F (β, γ).

Proof. The proof is by induction on γ. Let F (α, γ1) of F (α, γ) where γ1 = min(Cγ \ α). To prove (a) consider ﬁrst the case γ1 < β. By the inductive hypothesis,

F (α, γ1) ⊆ F (α, β) ∪ F (γ1, β) ⊆ F (α, β) ∪ F (β, γ), since γ1 ∈ Cγ ∩ β, so F (γ1, β) ⊆ F (β, γ). If γ1 ≥ β then γ1 = min(Cγ \ β) and again F (β, γ1) ⊆ F (β, γ). By the inductive hypothesis

F (α, γ1) ⊆ F (α, β) ∪ F (β, γ1) ⊆ F (α, β) ∪ F (β, γ).

Let now us consider F (ξ, α) for some ξ ∈ Cγ ∩ α. Using the inductive hypothesis we get

F (ξ, α) ⊆ F (α, β) ∪ F (ξ, β) ⊆ F (α, β) ∪ F (β, γ), since ξ ∈ Cγ ∩ β and F (ξ, β) is by deﬁnition included in F (β, γ). To prove (b) consider ﬁrst the case γ1 < β. By the inductive hypothesis:

F (α, β) ⊆ F (α, γ1) ∪ F (γ1, β).

3For two sets of ordinals F and G, by F < G we denote the fact that every ordinal from F is smaller than every ordinal from G. When G = {α}, we write F < α rather than F < {α}. 6 I. Coherent Sequences

Since γ1 ∈ Cγ ∩ β, the set F (γ1, β) is included in F (β, γ). Similarly by the deﬁnition of F (α, γ), the set F (α, γ1) is included in F (α, γ). Suppose now that γ1 ≥ β. Note that in this case γ1 = min(Cγ ∩ β), so F (β, γ1) is included in F (β, γ). Using the inductive hypothesis for α ≤ β ≤ γ1 we have

F (α, β) ⊆ F (α, γ1) ∪ F (β, γ1) ⊆ F (α, γ) ∪ F (β, γ). a 1.9 Lemma. For all α ≤ β ≤ γ,

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

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

Proof. The proof is again by induction on γ. Let α1 = min(F (β, γ)\α) and γ1 = min(Cγ \ α). Let ξ1 be the minimal ξ ∈ (Cγ ∩ β) ∪ {min(Cγ \ β)} such that α1 ∈ F (ξ, β) (or F (β, ξ)). Applying the inductive hypothesis to α ≤ ξ1 ≤ β (or α ≤ β ≤ ξ1) and noting that α1 = min(F (ξ1, β) \ α) (or α1 = min(F (β, ξ1) \ α)) we get

a ρ0(α, β) = ρ0(α1, β) ρ0(α, α1), (I.1) the part (a) of Lemma 1.9. Note that depending on whether α ≤ γ1 ≤ β or α ≤ β ≤ γ1 we have that F (γ1, β) ⊆ F (β, γ) or F (β, γ1) ⊆ F (β, γ), and therefore α2 := min(F (γ1, β) \ α) ≥ α1 or α2 := min(F (β, γ1) \ α) ≥ α1. Applying the inductive hypothesis to α ≤ γ1 ≤ β (or α ≤ β ≤ γ1) we get:

a ρ0(α, β) = ρ0(α2, β) ρ0(α, α2), (I.2)

a ρ0(α, γ1) = ρ0(α2, γ1) ρ0(α, α2). (I.3)

Comparing (I.1) and (I.2) we see that ρ0(α, α1) is a tail of ρ0(α, α2), so (I.3) can be rewritten as

a a a ρ0(α, γ1) = ρ0(α2, γ1) ρ0(α1, α2) ρ0(α, α1) = ρ0(α1, γ1) ρ0(α, α1). (I.4)

a a Finally since ρ0(α, γ) = h|Cγ ∩ α|i ρ0(α, γ1) = ρ0(γ1, γ) ρ0(α, γ1) holds, the equation (I.4) gives us the desired conclusion (b) of Lemma 1.9. a 1.10 Deﬁnition. Recall that right lexicographical ordering on ω<ω denoted by

1.11 Lemma. The cartesian square of the total ordering

p(α, β) = p(γ, δ) = p, and α

We must show that β ≤c δ. Let

ξαβ = min(F (α, β) \ ∆(α, γ)), and

ξγδ = min(F (γ, δ) \ ∆(α, γ)).

Note that F (α, β)∩∆(α, γ) = F (γ, δ)∩∆(α, γ) so ξαβ and ξγδ correspond to each other in the isomorphism of the (α, β) and (γ, δ) structures. It follows that: ρ0(ξαβ, α) = ρ0(ξγδ, γ)(= tα,γ ), (I.6)

ρ0(ξαβ, β) = ρ0(ξγδ, δ)(= tβ,δ). (I.7) Applying Lemma 1.9 we get:

a ρ0(∆(α, γ), α) = tαγ ρ0(∆(α, γ), ξαβ), (I.8)

a ρ0(∆(α, γ), γ) = tαγ ρ0(∆(α, γ), ξγδ). (I.9)

It follows that ρ0(∆(α, γ), ξαβ) 6= ρ0(∆(α, γ), ξγδ). Applying Lemma 1.9 for β and δ and the ordinal ∆(α, γ) we get:

a ρ0(∆(α, γ), β) = tβδ ρ0(∆(α, γ), ξαβ), (I.10) a ρ0(∆(α, γ), δ) = tβδ ρ0(∆(α, γ), ξγδ). (I.11)

It follows that ρ0(∆(α, γ), β) 6= ρ0(∆(α, γ), δ). This shows ∆(α, γ) ≥ ∆(β, δ). A symmetrical argument shows the other inequality ∆(β, δ) ≥ ∆(α, γ). It follows that ∆(α, γ) = ∆(β, δ)(= ξ¯). ¯ ¯ Our assumption is that ρ0(ξ, α)

1.12 Notation. Well-ordered sets of rationals. The set ω<ω ordered by the right lexicographical ordering

1.13 Lemma. ρ0(α, γ)

The sequence ρ0α (α < ω1) of members of σQr naturally determines the subtree T (ρ0) = {(ρ0)β ¹ α : α ≤ β < ω1}.

Note that for a ﬁxed α, the restriction ρ0β ¹ α is determined by the way ρ0β acts on the ﬁnite set F (α, β). This is the content of Lemma 1.9. Hence all levels of T (ρ0) are countable, and therefore T (ρ0) is a particular example of an Aronszajn tree, a tree of height ω1 with all levels and chains countable. We shall now see that T (ρ0) is in fact a special Aronszajn tree, i.e. one that admits a decomposition into countably many antichains or, equivalently, admits a strictly increasing map into the rationals.

1.14 Lemma. {ξ < β : ρ0(ξ, β) = ρ0(ξ, γ)} is a closed subset of β whenever β < γ. Proof. Let α < β be a given accumulation point of this set and let β = β0 > . . . > βn = α and γ = γ0 > . . . > γm = α be the traces of the walks β → α and γ → α, respectively. Using the assumption (c) of the fundamental sequences and the fact that α > 0 is a limit ordinal we can ﬁnd an ordinal ξ < α such that ρ0(ξ, β) = ρ0(ξ, γ) and

ξ > max(Cβi ∩ α) and ξ > max(Cγj ∩ α) for all i < n and j < m. (I.13) Note that this in particular means that the walk α → ξ is a common tail of the walks β → ξ and γ → ξ. Subtracting ρ0(ξ, α) from ρ0(ξ, β) we get ρ0(α, β) and subtracting ρ0(ξ, α) from ρ0(ξ, γ) we get ρ0(α, γ). It follows that ρ0(α, β) = ρ0(α, γ). a

It follows that T (ρ0) does not branch at limit levels. From this we can conclude that T (ρ0) is a special subtree of σQ since this is easily seen to be so for any subtree of σQ which is finitely branching at limit nodes. 1.15 Definition. Identifying the power set of Q with the particular copy 2Q of the Cantor set, define for every countable ordinal α,

Q Gα = {x ∈ 2 : x end-extends no ρ0β ¹ α for β ≥ α}. 2. The coherence of maximal weights 9

1.16 Lemma. Gα (α < ω1) is an increasing sequence of proper Gδ-subsets of the Cantor set whose union is equal to the Cantor set. a

1.17 Lemma. The set X = {ρ0β : β < ω1} considered as a subset of the Cantor set 2Q has universal measure zero.

Q Proof. Let µ be a given nonatomic Borel measure on 2 . For t ∈ T (ρ0), set

Q Pt = {x ∈ 2 : x end-extends t}.

Q Note that each Pt is a perfect subset of 2 and therefore is µ-measurable. Let S = {t ∈ T (ρ0): µ(Pt) > 0}. Then S is a downward closed subtree of σQ with no uncountable antichains. By an old result of Kurepa (see [74]), no Souslin tree admits a strictly increasing map into the reals (as for example σQ does). It follows that S must be countable and so we are done. a

2. The coherence of maximal weights

2.1 Deﬁnition. The maximal weight of the walk is the two-place function 2 ρ1 :[ω1] −→ ω deﬁned recursively by

ρ1(α, β) = max{|Cβ ∩ α|, ρ1(α, min(Cβ \ α))},

4 with the boundary value ρ1(α, α) = 0 . Thus ρ1(α, β) is simply the maximal integer appearing in the sequence ρ0(α, β).

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

(a) {ξ ≤ α : ρ1(ξ, α) ≤ n} is ﬁnite,

(b) {ξ ≤ α : ρ1(ξ, α) 6= ρ1(ξ, β)} is ﬁnite. Proof. The proof is by induction. To prove (a) it suﬃces to show that for every n < ω and every A ⊆ α of order-type ω there is ξ ∈ A such that ρ1(ξ, α) > n. Let η = sup(A). If η = α one chooses arbitrary ξ ∈ A with the property that |Cα ∩ ξ| > n, so let us consider the case η < α. Let α1 = min(Cα \ η). By the inductive hypothesis there is ξ ∈ A such that:

ξ > max(Cα ∩ η), (I.14)

ρ1(ξ, α1) > n. (I.15)

4 This is another use of the convention ρ1(α, α) = 0 in order to facilitate the recursive deﬁnition. 10 I. Coherent Sequences

a Note that ρ0(ξ, α) = h|Cα ∩ η|i ρ0(ξ, α1), and therefore

ρ1(ξ, α) ≥ ρ1(ξ, α1) > n.

To prove (b) we show by induction that for every A ⊆ α of order-type ω there exists ξ ∈ A such that ρ1(ξ, α) = ρ1(ξ, β). Let η = sup(A) and let β1 = min(Cβ \ η). Let n = |Cβ ∩ η| and let

B = {ξ ∈ A : ξ > max(Cβ ∩ η) and ρ1(ξ, β1) > n}.

Then B is inﬁnite, so by the induction hypothesis we can ﬁnd ξ ∈ B such that ρ1(ξ, α) = ρ1(ξ, β1). Then

ρ1(ξ, β) = max{n, ρ1(ξ, β1)} = ρ1(ξ, β1), so we are done. a

Note the following unboundedness property of ρ1 which immediately follows from the previous lemma and which will be elaborated on in latter sections of this Chapter.

2.3 Corollary. For every pair A and B uncountable families of pairwise disjoint ﬁnite subsets of ω1 end every positive integer n there exist uncountable subfamilies A0 ⊆ A and B0 ⊆ B such that for every pair a ∈ A0 and 5 b ∈ B0 such that a < b we have ρ1(α, β) > n for all α ∈ a and β ∈ b. a

2.4 Corollary. The tree

∗ 6 T (ρ1) = {t : α −→ ω : α < ω1, t = ρ1α} is a coherent, homogeneous, and R-embeddable Aronszajn tree. a

The following is an interesting property of T (ρ1) or any other coherent tree of ﬁnite-to one mappings from countable ordinals into ω.

2.5 Theorem. The cartesian square of any lexicographically ordered coherent tree of ﬁnite-to-one mappings is the union of countably many chains.

Proof. Let T (a) be a given tree given by a coherent sequence aα : α −→ ω (α < ω1) of ﬁnite-to-one mappings. Let

5For sets of ordinals a and b we write a < b whenever α < β for all α ∈ a and β ∈ b. 6 ∗ Here, ρ1α : α → ω are deﬁned from ρ1 in the usual way, ρ1α(ξ) = ρ1(ξ, α), and = denotes the fact that the functions agree on all but ﬁnitely many arguments. 2. The coherence of maximal weights 11 and l(t) are levels as well as domains of the mappings s and t, respectively. Given such such pair (s, t), let

D(s, t) = {ξ : ξ = l(s) or ξ < l(s) and s(ξ) 6= t(ξ)}, and let n(s, t) be the maximal value s or t takes on D(s, t). Let

F (s, t) = {ξ : ξ = l(s) or ξ < l(s) and s(ξ) ≤ n(s, t) or t(ξ) ≤ n(s, t)}.

Note that D(s, t) ⊆ F (s, t). Finally, to the pair (s, t) we associate a hereditarily ﬁnite set p(s, t) which codes the behavior of the restrictions of s and t on the set F (s, t). Suppose we are given two pairs (s, t) and (u, v) such that p(s, t) = p(u, v) and such that s

F (s, t) ∩ α = F (u, v) ∩ α is a common initial part of F (s, t) and F (u, v). Since s and u agree below α, we conclude that t and v must also agree below α. If α∈ / F (s, t) ∪ F (u, v), then t(α) = s(α) < u(α) = v(α), so t

t(α) ≤ n(s, t) = n(u, v) < v(α), so t

F (s, t) = F ∪ {l(s)} and F (u, v) = F ∪ {l(u)}

If l(s) < l(t) then t(l(s)) ≤ n(s, t) = n(u, v) < v(l(s)). Since t and v agree bellow l(s) we conclude that t

2.6 Deﬁnition. Consider the following extension of T (ρ1):

Te(ρ1) = {t : α −→ ω : α < ω1 and t ¹ ξ ∈ T (ρ1) for all ξ < α}.

If we order Te(ρ1) by the right lexicographical ordering

[tahmi, tahm + 1ia~0], where t ∈ T (ρ1) and m < ω. Removing the right-hand points from all the jumps we get a linearly ordered continuum which we denote by Ae(ρ1).

2.7 Lemma. Ae(ρ1) is a homogeneous nonreversible ordered continuum that can be represented as the union of an increasing ω1-sequence of Cantor sets. Proof. For a countable limit ordinal δ > 0, set

Aeδ(ρ1) = {t ∈ Ae(ρ1): l(t) ≤ δ}.

Then each Aeδ(ρ1) is a closed subset of Ae(ρ1) with the level Tδ(ρ1) being a countable order-dense subset. One easily concludes from this that Aeδ(ρ1)(δ limit < ω1) is the required decomposition of Ae(ρ1). To show that Ae(ρ1) is homogeneous, consider two pairs x0 < x1 and y0 < y1 of non-endpoints of Ae(ρ1). Choose a countable limit ordinal δ strictly above the lengths of x0, x1, y0 and y1. So we have a Cantor set Aeδ(ρ1), an order-dense subset Tδ(ρ1) of it and two pairs of points x0 < x1 and y0 < y1 in Aeδ(ρ1) \ Tδ(ρ1).

By Cantor’s theorem there is an order isomorphism σ : Aeδ(ρ1) −→ Aeδ(ρ1) such that: 00 σ Tδ(ρ1) = Tδ(ρ1), (I.16)

σ(xi) = yi for i < 2. (I.17)

Extend σ to the rest of Ae(ρ1) by the formula

σ(t) = σ(t ¹ δ)at ¹ [δ, l(t)).

It is easily checked that the restriction is the required order-isomorphism. Let us now prove that there is no order-reversing bijection

π : Ae(ρ1) −→ Ae(ρ1).

Since T (ρ1) with the lexicographical ordering does not have a countable 00 order-dense subset, the image π T (ρ1) must have sequences of length bigger 2. The coherence of maximal weights 13

than any given countable ordinal. So for every limit ordinal δ we can ﬁx tδ in T (ρ1) of length ≥ δ such that π(tδ) also has length ≥ δ. Let

f(δ) = max{ξ < δ : tδ(ξ) 6= π(tδ)(ξ)} + 1. By the Pressing Down Lemma ﬁnd an uncountable set Γ of countable limit ¯ ordinals, a countable ordinal ξ and s, t ∈ Tξ¯(ρ1) such that for all δ ∈ Γ: f(δ) = ξ,¯ (I.18) ¯ ¯ tδ ¹ ξ = s and π(tδ) ¹ ξ = t. (I.19) Moreover we may assume that

{tδ ¹ δ : δ ∈ Γ} and {π(tδ) ¹ δ : δ ∈ Γ} are antichains of the tree T (ρ1). Pick γ 6= δ in Γ and suppose for deﬁnite- ness that tγ

It follows that ∆(π(tγ ), π(tδ)) = ∆(tγ , tδ)(= ξ) and that

π(tγ )(ξ) = tγ (ξ) < tδ(ξ) = π(tδ(ξ)), contradicting the fact that π is order-reversing. a 2.8 Deﬁnition. 7 A homogeneous Eberlein compactum . The set Te(ρ1) has another natural structure, a topology generated by the family of sets of the form

Vet = {u ∈ Te(ρ1): t ⊆ u}, 0 for t a node of T (ρ1) of successor length as a clopen subbase. Let T (ρ1) denote the set of all nodes of T (ρ1) of successor length. Then Te(ρ1) can be 0 regarded as the set of all downward closed chains of the tree T (ρ1) and the topology on Te(ρ1) is simply the topology one obtains from identifying the 0 power set of T (ρ1) with the cube

0 {0, 1}T (ρ1)

8 with its Tychonoﬀ topology . Te(ρ1) being a closed subset of the cube is compact. In fact Te(ρ1) has some very strong topological properties such as the property that closed subsets of Te(ρ1) are its retracts. 7Recall that a compactum X is Eberlein if its function space C(X) can be generated by a subset which is compact in the weak topology. 8 0 This is done by identifying a subset V of T (ρ1) with its characteristic function 0 χV : T (ρ1) −→ 2. 14 I. Coherent Sequences

2.9 Lemma. Te(ρ1) is a homogeneous Eberlein compactum

Proof. The proof that Te(ρ1) is homogeneous is quite similar to the corresponding part of the proof of the Lemma 2.7. To see that Te(ρ1) is an Eberlein compactum, i.e. that the function space C(Te(ρ1)) is weak compactly generated, let {Xn} be a countable antichain decomposition of T (ρ1) and consider the set −n K = {2 χ e : n < ω, t ∈ X } ∪ {χ }. Vt n ∅

Note that K is a weakly compact subset of C(Te(ρ1)) which separates the points of Te(ρ1). a E. Borel’s notion of strong measure zero is a metric notion that has several closely related topological notions due to K.Menger, F.Rothberger and others. For example Rothberger’s property C00 is a purely topological notion which in the class of metric spaces is slightly stronger than the property C, i.e. being of strong measure zero. It says that for every sequence {Un} of open covers of a space X one can choose Un ∈ Un for each n such that [∞ X = Un. n=0 00 2.10 Lemma. T (ρ1) has Rothberger’s property C . a

2.11 Remark. The subspace T (ρ1) of Te(ρ1) has in fact a considerably stronger covering property than C00. For example, it can be shown that its function space C(T (ρ1)) with the topology of pointwise convergence is a Fr´echet space, or in other words, has the property that any family of continuous functions on T (ρ1) which pointwise accumulates to the constantly equal 0 function 0¯ contains a sequence {fn} which pointwise converges to 0.¯ In fact the function space C(T (ρ1)) with the topology of pointwise convergence has a natural decomposition as the increasing union of an ω1-sequence of closed separable metric vector subspaces. The separable metric vector subspaces are determined by the retraction maps of T (ρ1) onto its levels.

In the next application the coherent sequence ρ1α : α −→ ω (α < ω1) of ﬁnite-to-one maps needs to be turned into a coherent sequence of maps that are actually one-to-one. One way to achieve this is via the following formula:

ρ1(α,β) ρ¯1(α, β) = 2 · (2 · |{ξ ≤ α : ρ1(ξ, β) = ρ1(α, β)}| + 1).

Defineρ ¯1α fromρ ¯1 just as ρ1α was defined from ρ1; then theρ ¯1α’s are one- to-one and coherent. From ρ1 one also has a natural sequence rα (α < ω1) of elements of ωω defined as follows

rα(n) = |{ξ ≤ α : ρ1(ξ, α) ≤ n}|. 2. The coherence of maximal weights 15

Note that rβ eventually strictly dominates rα whenever α < β.

2.12 Deﬁnition. The sequences eα =ρ ¯1α (α < ω1) and rα (α < ω1) can be used in describing a functor

G 7−→ G∗,

∗ which to every graph G on ω1 associates another graph G on ω1 as follows:

∗ −1 −1 {α, β} ∈ G iff {eα (l), eβ (l)} ∈ G (I.20) for all l < ∆(rα, rβ) for which these preimages are both defined and differ- ent9. 2.13 Lemma. Suppose that every uncountable family F of pairwise disjoint 10 finite subsets of ω1 contains two sets A and B such that A ⊗ B ⊆ G . Then the same is true about G∗ provided the uncountable family F consists of finite cliques11 of G∗. Proof. Let F be a given uncountable family of pairwise disjoint finite cliques of G∗. We may assume that all members of F are of some fixed size k. Consider a countable limit ordinal δ > 0 and an A in F with all its elements above δ. Let n = n(A, δ) be the minimal integer such that for all α < β in A ∪ {δ}: 12 ∆(rα, rβ) < n, (I.21) e(ξ, α) 6= e(ξ, β) for some ξ ≤ α implies e(ξ, α), e(ξ, β) ≤ n. (I.22) Let

H(δ, A) = {ξ ≤ ω1 : e(ξ, α) ≤ n(δ, A) for some α ≥ ξ from A ∪ {δ}}. Taking the transitive collapse H¯ (δ, A) of H(δ, A), the sequence

eα ¹ H(δ, A)(α ∈ A) collapses to a k-sequences ¯(δ, A) of mappings with integer domains. Let r¯(δ, A) denote the k-sequence

rα ¹ (n(δ, A) + 1) (α ∈ A) enumerated in increasing order. Hence every A ∈ F above δ generates a quadruple (H(δ, A) ∩ δ, r¯(δ, A), s¯(δ, A), n(δ, A))

9As it will be clear from the proof of the following lemma the functor G −→ G∗ can equally be based on any other coherent sequence eα : α −→ ω (α < ω1) of one-to-one mappings and any other sequence rα (α < ω1) of pairwise distinct reals. 10Here, A ⊗ B = {{α, β} : α ∈ A, β ∈ B, α 6= β}. 11 ∗ 2 ∗ A clique of G is a subset C of ω1 with the property that [C] ⊆ G . 12Recall that for s 6= t ∈ ωω, we let ∆(s, t) = min{k : s(k) 6= t(k)}. 16 I. Coherent Sequences of parameters. Since the set of quadruples is countable, by the assumption on the graph G we can ﬁnd two distinct members Aδ and Bδ of F above δ that generate the same quadruple of parameters denoted by

(Hδ, r¯δ, s¯δ, nδ).

Moreover, we choose Aδ and Bδ to satisfy the following isomorphism condition

For every l ≤ nδ, i < k, if α is the ith member of Aδ, β the −1 −1 ith member of Bδ and if eα (l) and eβ (l) are both deﬁned (I.23) then they are G-connected to the same elements of Hδ ∩ δ.

Let mδ be the minimal integer > nδ such that

∆(rα, rβ) < mδ for all α ∈ Aδ and β ∈ Bδ. (I.24)

Let Iδ = {ξ : e(ξ, α) ≤ mδ for some α ≥ ξ in Aδ ∪ Bδ}.

Letp ¯δ be the k-sequence rα ¹ mδ(α ∈ Aδ) enumerated in increasing order and similarly letq ¯δ be the k-sequence rβ ¹ mδ(β ∈ Bδ) enumerated increasingly. Let t¯δ andu ¯δ be the transitive collapse of eα ¹ Iδ (α ∈ Aδ) and eβ ¹ Iδ(β ∈ Bδ), respectively. By the Pressing Down Lemma there is an unbounded Γ ⊆ ω1 and tuples

(H, r,¯ s,¯ n) and (I, p,¯ q,¯ t,¯ u,¯ m) such that for all δ ∈ Γ:

(Hδ, r¯δ, s¯δ, nδ) = (H, r,¯ s,¯ n), (I.25)

(Iδ ∩ δ, p¯δ, q¯δ, t¯δ, u¯δ, mδ) = (I, p,¯ q,¯ t,¯ u,¯ m). (I.26)

Moreover we assume the following analogue of (I.23) where k1 = |Iγ \ γ| for some (equivalently all) γ ∈ Γ:

For every γ, δ ∈ Γ and i < k1, if η is the ith member of Iγ and if ξ is the ith member of Iδ in their increasing enumerations, then η and ξ are G-connected to the same (I.27) elements of the root I, and if η happens to be the i∗th ∗ member of Aγ (or Bγ ) for some i < k, then ξ must be the ∗ i th member of Aδ (Bδ resp.) and vice versa. By our assumption about the graph G there exist γ < δ in Γ such that:

max(Iγ ) < δ, (I.28)

(Iγ \ γ) ⊗ (Iδ \ δ) ⊆ G. (I.29) 2. The coherence of maximal weights 17

The proof of Lemma 2.13 is ﬁnished once we show that

∗ Aγ ⊗ Bδ ⊆ G . (I.30)

Let i, j < k be given and let α be the ith member of Aγ and let β be the jth member of Bδ. Case 1: i 6= j. Pick an l < ∆(rα, rβ). By (I.21), l < n. Assume that −1 −1 eα (l) and eβ (l) are both deﬁned. −1 −1 Subcase 1.1: eα (l) < γ and eβ (l) < δ. By the ﬁrst choice of parame- −1 −1 ters, eα (l) and eβ (l) are members of the set H which is an initial part of H(γ, Aγ ) and H(δ, Bδ). Therefore, we have that the behavior of eα and eβ on H is encoded by the ith and jth term of the sequences ¯, respectively. In particular, we have −1 −1 eβ (l) = eβ00 (l), (I.31)

00 2 ∗ where β =the jth member of Aγ . By our assumption that [Aγ ] ⊆ G we infer that {α, β00} ∈ G∗. Referring to the deﬁnition (I.20) we conclude that −1 −1 eα (l) and eβ (l) must be G-connected if they are diﬀerent. −1 −1 Subcase 1.2: eα (l) < γ and eβ (l) ∈ Iδ \ δ. Let

0 β = the jth member of Bγ .

−1 By the choice of parametrization, the position of eβ0 (l) in Iγ is the same −1 as the position of eβ (l) in Iδ so by (I.27) their relationship to the point −1 eα (l) of the root I is the same. Similarly, by the choice of the first set of 00 −1 parameters, letting β be the jth member of Aγ , the position of eβ0 (l) in −1 H(γ, Bγ ) and eβ00 (l) in H(γ, Aγ ) is the same, so from (I.23) we conclude −1 that their relationship with eα (l) is the same. However, we have checked −1 −1 in the previous subcase that eα (l) and eβ00 (l) are G-connected in case they are different. −1 −1 Subcase 1.3: eα (l) ∈ Iγ \ γ and eβ (l) < δ. This is essentially symmetric to the previous subcase. −1 −1 Subcase 1.4: eα (l) ∈ Iγ \ γ and eβ (l) ∈ Iδ \ δ. The fact that −1 −1 {eα (l), eβ (l)} ∈ G in this case follows from (I.29). Case 2: i = j. Consider an l < ∆(rα, rβ). Note that now we have l < m (see (I.24)). −1 −1 −1 −1 Subcase 2.1: eα (l) < γ and eβ (l) < δ. Then eα (l) and eβ (l) are elements of I which is an initial part of both Iγ and Iδ. Therefore the jth(= ith) term ofu ¯ encodes both eβ ¹ I and eβ0 ¹ I (see the definition of β0 above). It follows that

−1 −1 eβ (l) = eβ0 (l). 18 I. Coherent Sequences

−1 If l ≤ n then eβ0 (l) belongs to H and since the jth(= ith) term ofs ¯ encodes both eα ¹ H and eβ ¹ H it follows that

−1 −1 eα (l) = eβ0 (l).

−1 −1 If l > n then eα (l) and eβ0 (l) are not members of H so from (I.22) and the deﬁnition of

H(γ, Aγ ) ∩ γ = H = H(γ, Bγ ) ∩ γ we conclude that

−1 −1 −1 −1 eγ (eα (l)) = eα(eα (l)) = l = eβ0 (eβ0 (l)) = eγ (eβ0 (l)).

Since eγ is one-to-one we conclude that

−1 −1 −1 eα (l) = eβ0 (l) = eβ (l).

−1 −1 0 Subcase 2.2: eα (l) < γ and eβ (l) ∈ Iδ \ δ. Recall that β is the ith(= jth) member of Bγ so by the ﬁrst choice of parameters the relative −1 position of eβ0 (l) in H(γ, Bγ ) must be the same as the relative position of −1 ∗ eα (l) in H(γ, Aγ ), i.e. it must belong to the root H. Note that if j is the ∗ 0 position of β in Iδ then j must also be the position of β in Iγ . It follows that the j∗th term ofu ¯ encodes both

eβ ¹ Iδ and eβ0 ¹ Iγ = eβ0 ¹ I.

−1 So it must be that eβ (l) belongs to the root I, a contradiction. So this subcase never occurs. −1 −1 Subcase 2.3: eα (l) ∈ Iγ \ γ and eβ (l) < δ. This is symmetric to the previous subcase, so it also never occurs. −1 −1 −1 −1 Subcase 2.4: eα (l) ∈ Iγ \γ and eβ (l) ∈ Iδ \δ. Then {eα (l), eβ (l)} ∈ G follows from (I.29). This completes the proof of Lemma 2.13. a

2 ∗ 2.14 Lemma. If there is uncountable Γ ⊆ ω1 such that [Γ] ⊆ G then ω1 can be decomposed into countably many sets Σ such that [Σ]2 ⊆ G.

2 ∗ Proof. Fix an uncountable Γ ⊆ ω1 such that [Γ] ⊆ G . For a ﬁnite binary sequence s of length equal to some l + 1, set

Γs = {ξ < ω1 : e(ξ, α) = l for some α in Γ with s ⊆ rα}.

2 Then the sets Γs cover ω1 and [Γs] ⊆ G for all s. a 2. The coherence of maximal weights 19

2.15 Remark. Let G be the comparability graph of some Souslin tree T . Then for every uncountable family F of pairwise disjoint cliques of G (finite chains of T ) there exist A 6= B in F such that A∪B is a clique of G (a chain of T ). However, it is not hard to see that G∗ fails to have this property (i.e. the conclusion of Lemma 2.13). This shows that some assumption on the graph G in Lemma 2.13 is necessary. There are indeed many graphs that satisfy the hypothesis of Lemma 2.13. Many examples appear when one is trying to apply Martin’s axiom to some Ramsey-theoretic problems. Note that the conclusion of Lemma 2.13 is simply saying that the poset of all finite cliques of G∗ is ccc while its hypothesis is a bit stronger than the fact that the poset of all finite cliques of G is ccc in all of its finite powers. Applying 2.14 to the case when G is the incomparability graph of some Aronszajn tree, we see that the statement saying that all Aronszajn trees are special is a purely Ramsey-theoretic statement in the same way Souslin’s hypothesis is. We finish this section by showing that the assumption (c) on a given C-sequence Cα (α < ω1) on which we base our walks and the function ρ1 is not sufficient to give us the stronger conclusion that the tree T (ρ1) can be decomposed into countably many antichains (see Corollary 2.4). To describe an example, we need the following notation given that we have fixed one C- sequence Cα (α < ω1). Let Cα(0) = 0 and for 0 < n < ω, let Cα(n)) denote the n’th element of Cα according to its increasing enumeration with the convention that Cα+1(n) = α for all n > 0. We assume that the C-sequence is chosen so that for a limit ordinal α > 0 and a positive integer n, the ordinal Cα+1(n) is at least n + 1 steps away from the closest limit ordinal bellow it. For a set D of countable ordinals, let D0 denote the set of successor ordinals in D. We also fix, for each countable ordinal α a one-to-one function eα whose domain is the set α0 of all successor ordinals < α and range included in ω which cohere in the sense that eα(ξ) = eβ(ξ) for all but finitely many successor ordinals ξ ∈ α0 ∩ β0. For each r ∈ ([ω]<ω)ω, we associate another r r r S r C-sequence Cα (α < ω1) by letting Cα = Cα ∩ [ξα, α) ∪ n∈ω Dα(n), where r 0 Dα(n) = {ξ ∈ [Cα(n),Cα(n + 1)) : eα(ξ) ∈ r(n)}, r S r and where ξα = sup( n∈ω Dα(n)). (This definition really takes place only r when α is a limit ordinal; for successor ordinals we put Cα+1 = {α}.) This r gives us a C-sequence Cα (α < ω1) and we can consider the corresponding r 2 ρ1 :[ω1] −→ ω as above. This function will have the properties stated in Lemma 2.2, so the corresponding tree

r r T (ρ1) = {(ρ1)β ¹ α : α ≤ β < ω1} is a coherent tree of ﬁnite-to-one mappings that admits a strictly increasing r function into the real line. The following fact shows that T (ρ1) will typically not be decomposable into countably many antichains. 20 I. Coherent Sequences

r 2.16 Lemma. If r is a Cohen real then T (ρ1) has no stationary antichain. Proof. The family [ω]<ω of finite subsets of ω equipped with the discrete topology and its power ([ω]<ω)ω with the corresponding product topology and the given Cohen real r is to meet all dense Gδ subset of this space r that one can explicitly define. Thus in particular, ξα = α, and therefore, r S r r Cα = n∈ω Dα(n) for every limit ordinal α. Since the tree T (ρ1) is coherent, to establish the conclusion of the Lemma, it suffices to show that that for r r every stationary Γ ⊆ ω1 there exists γ, δ ∈ Γ such that (ρ1)γ ⊂ (ρ1)δ. This in turn amounts to showing that for every stationary Γ ⊆ ω1, the set

<ω ω x x UΓ = {x ∈ ([ω] ) :(∃γ, δ ∈ Γ) (ρ1 )γ ⊂ (ρ1 )δ} is a dense-open subset of ([ω]<ω)ω. So, given a finite partial function p from ω into [ω]<ω, it is sufficient to find a finite extension q of p such that the <ω ω basic open subset of ([ω] ) determined by q is included in UΓ. Let n be the minimal integer that is bigger than all integers appearing in the domain of p or any set of the form p(j) for j ∈ dom(p). For γ ∈ Γ, set

0 Fn(γ) = {ξ ∈ γ : eγ (ξ) ≤ n}.

Applying the Pressing Down Lemma, we obtain a ﬁnite set F ⊆ ω1 and a stationary set ∆ ⊆ Γ such that Fn(γ) = F for all γ ∈ ∆ and such that if α = max(F ) + 1 then eγ ¹ α = eδ ¹ α for all γ, δ ∈ ∆. A similar application of the Pressing Down Lemma will give us an integer m > n and two ordinals γ < δ in ∆ such that Cγ (j) = Cδ(j) for all j ≤ m, Cδ(m + 1) > γ + 1, and

0 0 eγ ¹ (Cγ (m) + 1) = eδ ¹ (Cγ (m) + 1) .

Extend the partial function p to a partial function q with domain {0, 1, ..., m} such that:

(1) q(m) = {eδ(γ + 1)}, and (2) q(j) = ∅ for any j < m not belonging to dom(p). Choose any x ∈ ([ω]<ω)ω extending the partial mapping q. Then from the choices of the objects n, ∆, F ,m, γ, δ and q, we have that

x (3) γ + 1 ∈ Cδ , and x x (4) Cδ ∩ γ is an initial segment of Cγ . It follows that, given a ξ < γ, the walk from δ to ξ along the C-sequence x Cβ (β < ω1) either leads to the same finite string of the corresponding weights as the walk from γ to ξ, or else it starts with first two steps to γ + 1 x x and γ, and then follows the walk from γ to ξ. Since ρ1 (ξ, δ) and ρ1 (ξ, γ) are by definitions maximums of these two strings of weights, we conclude 3. Oscillations of traces 21

x x that ρ1 (ξ, δ) ≥ ρ1 (ξ, γ). On the other hand, note that by (4), in the second x case, the weight |Cδ ∩ ξ| of the first step from δ to ξ is less than or equal x to the weight |Cγ ∩ ξ| of the first step from γ to ξ. It follows that we have x x also the other inequality ρ1 (ξ, δ) ≤ ρ1 (ξ, γ). Hence we have shown that x x x x ρ1 (β, γ) = ρ1 (β, δ) for all β < γ, or in other words, that (ρ1 )γ ⊂ (ρ1 )δ. This finishes the proof. a 2.17 Question. What is the condition one needs to put on a given C- sequence Cα (α < ω1) that guarantees that the corresponding tree T (ρ1) can be decomposed into countably many antichains?

3. Oscillations of traces

In this section we shall consider versions of the oscillation mapping

<ω 2 osc : ([ω1] ) −→ Card defined by osc(x, y) = |x/E(x, y)|, where E(x, y) is the equivalence relation on x defined by letting αE(x, y)β if and only if the closed interval determined by α and β contains no point from y \ x. So, if x and y are disjoint, osc(x, y) is simply the number of convex pieces the set x is split by the set y. The resulting ’oscillation theory’ reproduced here in part has shown to be a quite useful and robust coding technique. We give some applications to this effect and in Section 17 below we shall extend the oscillation theory to a more general context.

3.1 Deﬁnition. For α < β < ω1, set

osc0(α, β) = osc(Tr(∆0(α, β), α), Tr(∆0(α, β), β)),

13 where ∆0(α, β) = min{ξ ≤ α : ρ0(ξ, α) 6= ρ0(ξ, β)} − 1 . The following result shows that the upper traces do realize all possible oscillations in any uncountable subset of ω1.

3.2 Lemma. For every uncountable subset Γ of ω1 and every positive integer n there is an integer l such that for all k < n there exist α, β ∈ Γ such that osc0(α, β) = l + k.

Proof. Fix a sequence Mi (i ≤ n) of continuous ∈-chains of length ω1 of countable elementary submodels of (Hω2 , ∈) containing all the relevant objects such that Mi is an element of the minimal model of Mi+1 for all

13Note that by Lemma 1.14 this is well deﬁned. 22 I. Coherent Sequences

i < n. Let Ci = {M ∩ ω1 : M ∈ Mi}. Note that each Ci is a closed and unbounded subset of ω1 and that for each i < n, Ci is an element of every model of Mi+1. Let Mn be the minimal model of Mn and pick a β ∈ Γ above δn = Mn ∩ω1. Since Cn−1 ∈ Mn we can ﬁnd Mn−1 ∈ Mn−1 ∩Mn such that δn−1 = Mn−1 ∩ω1 > L(δn, β). Similarly, we can ﬁnd Mn−2 ∈ Mn−1 ∩Mn−1 such that δn−2 = Mn−2 ∩ ω1 > L(δn−1, δn), and so on. This will give us an ∈-chain Mi (i ≤ n) of countable elementary submodels of (Hω2 , ∈) containing all relevant objects such that if we let δn+1 = β and δi = Mi ∩ω1 for i ≤ n, then δi > L(δi+1, δi+2) for all i < n. Pick a point ξ ∈ Cδ0 above L(δ0, β). Then (see Lemma 1.6 above), δi ∈ Tr(ξ, β) for all i ≤ n and in fact δ0 → ξ is the last step of the walk from β to ξ. Let t = ρ0(ξ, β), ti = ρ0(δi, β) for i ≤ n. The tn < tn−1 < ... < t0 < t. Let

Γ0 = {γ ∈ Γ: t = ρ0(ξ, γ)}.

Then Γ0 ∈ M0 and β ∈ Γ0. A simple elementarity argument using models M0 ∈ M1 ∈ ... ∈ Mn shows that there is an uncountable subset Σ of Γ0 such that Σ ∈ M0 and

≤|t| T = {x ∈ [ω1] : x v Tr(ξ, α) for some α ∈ Σ} is a tree under v in which a node x must either have only one immediate successor or there is i ≤ n and uncountably many γ < ω1 such that x∪{γ} ∈ T and Tr(γ, α) = ti for all α ∈ Σ witnessing x ∪ {γ} ∈ T . Note that {ξ} is a root of T as well as its splitting nodes. Let Ω be the uncountable set such that {ξ, γ} ∈ T for all γ ∈ Ω. Working in M0 we ﬁnd η > ξ and uncountable Ω0 ⊆ Ω such that η = ∆0(γ, β) for all γ ∈ Ω0 and therefore η = ∆0(α, β) for all α ∈ Σ witnessing {ξ, γ} ∈ T . Still working in M0 and going to an uncountable subset of Ω0 we may assume that Tr(η, γ)(γ ∈ Ω0) form a ∆-system with root x0. Let Σ0 be the set of all α ∈ Σ witnessing {ξ, γ} ∈ T for some γ ∈ Ω0. Let s = ρ0(η, β). Note that s end-extends t0 but not necessarily t. Consider now the tree

≤|s| X = {x ∈ [ω1] : x v Tr(ξ, α) for some α ∈ Σ0}.

Its root is equal to x0, the uncountably many immediate successors x0 ∪{γ} (γ ∈ Ω0) all have the property that for some fixed initial segment s0 of s we have that ρ0(γ, α) = s0 for all α ∈ Σ0 witnessing x0 ∪ {γ} ∈ X . All other splitting nodes of X are determined by the other initial segments s0 w t0 = t1 = ... = tn. Recall that all these objects are elements of M0 and therefore elements of all Mi (i ≤ n). Let y = Tr(η, β) and let l = osc(x0, y). Fix a k ≤ n. Working in M1 find γ0 ∈ Ω0 such that Tr(η, γ0) \ x0 > Tr(η, β) ∩ δ1 and find a splitting node x1 w Tr(η, γ0) corresponding to t1. Then Ω1 = {γ < ω1 : x1 ∪ {γ} ∈ X } is uncountable and ρ0(γ, α) = t1 for all α ∈ Σ0 witnessing x1 ∪ {γ} ∈ X . Working in M2 find γ1 ∈ Ω1 such 3. Oscillations of traces 23

that γ1 > Tr(η, β) ∩ δ2 and a splitting node x2 w Tr(η, γ1) corresponding to t2, and so on. Proceeding in this way we construct an increasing sequence x0 < x1 < ... < xk of splitting nodes of T such that ti ∈ Mi and

xi \ xi−1 > Tr(η, β) ∩ δ2 for all 1 ≤ i ≤ k.

Find an arbitrary α ∈ Σ0 ∩ Mk such that xk v Tr(η, α) Let x = Tr(η, α). Then

x/E(x, y) = x0/E(x0, y)) ∪ {x1 \ x0, x2 \ x1, ..., xk \ xk−1}.

This is so because the ordinal δ0 ∈ y\x separates the last class of x0/E(x0, y) from the new classes xi \ xi−1 (1 ≤ i ≤ k). It follows that osc0(α, β) = osc(x, y) = l + k, as required. a

Consider the following projection of the oscillation mapping osc0,

∗ m osc0(α, β) = max{m ∈ ω : 2 |osc0(α, β)}.

3.3 Corollary. For every uncountable Γ ⊆ ω1 and every positive integer k ∗ there exist α, β ∈ Γ such that osc0(α, β) = k. a 3.4 Corollary. Every inner model model M of set theory which correctly computes ω1 contains a partition of ω1 into inﬁnitely many pairwise disjoint subsets that are stationary in the universe V of all sets.

Proof. Our assumption about M means in particular that the C-sequence Cξ (ξ < ω1) can be chosen to be an element of M. It follows that the ∗ oscillation mapping osc0 is also an element of M. It follows that for each α < ω1, the sequence of sets

∗ Sαn = {β > α : osc0(α, β) = n} (n < ω) belongs to M. So it suﬃces to show that there must be α < ω1 such that for every n < ω the set Sαn is stationary in V. This is an immediate consequence of Corollary 3.3. a

3.5 Remark. In [46], Larson showed that Corollary 3.4 cannot be extended to partitions of ω1 into uncountably many pairwise disjoint stationary sets. He also showed that under the Proper Forcing Axiom, for every mapping 2 c :[ω1] → ω1 there is a stationary set S ⊆ ω1 such that for all α < ω1,

{c(α, β): β ∈ S \ (α + 1))} 6= ω1.

Note that by Corollary 3.3 this cannot be extended to mappings c with countable ranges. 24 I. Coherent Sequences

One can count oscillation not only between two finite subsets of ω1 but also between two finite partial functions from ω1 into ω, i.e., between two finite ’weighted sets’ x and y :

osc(x, y) = |{ξ ∈ dom(x): x(ξ) ≤ y(ξ) but x(ξ+) > y(ξ+)}|, where ξ+ denotes the minimal ordinal in dom(x) above ξ if there is one.

3.6 Deﬁnition. For α < β < ω1, set

osc1(α, β) = osc(ρ1α ¹ L(α, β), ρ1β ¹ L(α, β)). We have the following analogue of Lemma 3.2 which is now true even in its rectangular form.

3.7 Lemma. For every pair Γ and Σ of uncountable subsets of ω1 and every positive integer n there is an integer l such that for all k < n there exist α ∈ Γ and β ∈ Σ such that osc1(α, β) = l + k.

Proof. Fix a continuous ∈-chain N of length ω1 of countable elementary submodels of (Hω2 , ∈) containing all the relevant objects and fix also a countable elementary submodel M of (Hω2 , ∈) that contains N as an element. Let δ = M ∩ ω1. Fix α0 ∈ Γ \ δ and β0 ∈ Σ \ δ. Let L0 = L(δ, β0). By exchanging the steps in the following construction, we may assume that ρ1(ξ, α0) > ρ1(ξ, β0) for ξ = max(L0). Choose ξ0 ∈ Cδ such that ξ > L0 and such that the three mappings ρ1α0 ,ρ1β0 , and ρ1δ agree on the interval [ξ, δ). + + + + + By elementarity of M there will be δ > δ, α0 ∈ Γ \ δ , and β0 ∈ Σ \ δ + + realizing the same type over the parameters L0 and ξ0. Let L1 = L(δ, δ0 ). + Note that ξ0 ≤ min(L ). It follows that the mappings ρ + and ρ + agree 1 1α0 1β0 + + on L1 . Choose an N ∈ N belonging to M such that N ∩ ω1 > L1 and + choose ξ1 ∈ Cδ \N. Then we can find δ1 > δ and α1 ∈ Γ\δ1 and β1 ∈ Σ\δ1 + + realizing the same type as δ, α0 , and β0 over the objects accumulates so + ++ + far, L0, ξ0,L1 , N, and ξ1. Let L1 = L(δ, δ1) and let t1 be the restriction + + of ρ + on max(L ) + 1. Then the set of α ∈ Γ whose ρ1α end-extends t 1α1 1 1 belongs to the submodel N and is uncountable, since clearly it contains the + ordinal α1 which does not belong to N. So by the elementarity of N and + Corollary 2.3 there is α1 ∈ Γ \ δ such that ρ1α1 end-extends t1 and ++ ρ1(ξ, α1) > ρ1(ξ, β1) for all ξ ∈ L1 . (I.32) ++ Let t1 be the restriction of ρ1α on max(L1 ). Then t1 ∈ M and every α ∈ Γ whose ρ1α end-extends t1 satisfies I.32 and

+ ρ1(ξ, α) = ρ1(ξ, β1) for all ξ ∈ L1 . (I.33) + ++ Note also that L(δ, β1) = L0 ∪ L1 ∪ L1 . Using similar reasoning we can

ﬁnd t2 ∈ M ∩ T (ρ1) end-extending t1, an α2 ∈ Γ \ δ whose ρ1α2 end-extends 3. Oscillations of traces 25

++ + ++ t2, ﬁnite sets δ > max(L2 ) > max(L2 ) > max(L1 ), and β2 ∈ Σ \ δ such + ++ + ++ that L(δ, β2) = L0 ∪ L1 ∪ L1 ∪ L2 ∪ L2 and such that the analogues of I.32 and I.33 hold for all α ∈ Γ whose ρ1α end-extends t2, and so on. This + ++ + ++ procedure gives us a sequence L0 < L1 < L1 < ... < Ln < Ln of ﬁnite subsets of δ, an increasing sequence t1 < t2 < ... < tn of nodes of T (ρ0)∩M, and a sequence βk (1 ≤ k ≤ n) such that for all 1 ≤ k ≤ n, and all α ∈ Γ whose ρ1α end-extends tk,

+ ++ + ++ L(δ, βk) = L0 ∪ L1 ∪ L1 ∪ ... ∪ Lk ∪ Lk , (I.34)

++ ρ1(ξ, α) > ρ1(ξ, βk) for all ξ ∈ Lk , (I.35) + ρ1(ξ, α) = ρ1(ξ, βk) for all ξ ∈ Lk . (I.36)

Choose α ∈ Γ ∩ M whose ρ1α end-extends tn such that the set Ln = L(α, δ) ++ lies above Ln and such that all the ρ1βk ’s agree on Ln. Note that by Lemma 1.6 for each 1 ≤ k ≤ n,

+ ++ + ++ L(α, βk) = L0 ∪ L1 ∪ L1 ∪ ... ∪ Lk ∪ Lk ∪ Ln. (I.37)

Let l = osc(ρ1α ¹ (L0 ∪ Ln), ρ1β ¹ (L0 ∪ Ln)) where β is equal to one of the βk’s. Note that l does not depend on which βk we choose and that according to I.35, I.36 and I.37, for each 1 ≤ k ≤ n, we have that osc1(α, βk) = l + k, as required. a Note that in the above proof the set Γ can be replaced by any uncountable family of pairwise disjoint sets giving us the following slightly more general conclusion. 3.8 Lemma. For every uncountable family G of pairwise disjoint ﬁnite subsets of ω1 all of some ﬁxed size m, every uncountable subset Σ of ω1, and every positive integer n there exist integers li (i < m) and a ∈ G such that for all k < n there exist β ∈ Σ such that osc1(a(i), β) = li + k for all i < m.14 a

Having in mind an application we define a projection of osc1 as follows. For a real x, let [x] denote the greatest integer not bigger than x. Choose a 1 1 mapping ∗ : R → ω such that ∗(x) = n if x−[x] ∈ In, where In = ( n+3 , n+2 ]. For x in R we use the notation x∗ for ∗(x). Choose a one-to-one sequence xξ (ξ < ω1) of elements of the first half of the unit interval I = [0, 1] which together with the point 1 form a set that is linearly independent over the field of rational numbers. Finally, set

∗ ∗ osc1(α, β) = (osc1(α, β) · xα) .

∗ Then Lemma 3.8 turns into the following statement about osc1. 14Here a(i) denotes the ith member of a according to the increasing enumeration of a. 26 I. Coherent Sequences

3.9 Lemma. For every uncountable family G of pairwise disjoint finite subsets of ω1 all of some fixed size m, every uncountable subset Σ of ω1, and every mapping h : m → ω there exist a ∈ G and β ∈ Σ such that ∗ osc1(a(i), β) = h(i) for all i < m. Proof. Let ε be half of the length of the shortest interval of the form Ih(i) (i < m). Note that for a ∈ G, the points 1, xa(0), ..., xa(m−1) are ratio- nally independent, so a standard fact from number theory (see for example m [32]; XXIII) gives us that there is an integer na such that for every y ∈ I there exist k < na such that |k · xa(i) − yi| < ε (mod 1) for all i < m. Going to an uncountable subfamily of G, we may assume that there is n such that na = n for all a ∈ G. By Lemma 3.8 there exits a ∈ G and li (i < m) such that for every k < n there is βk > a such that osc1(a(i), βk) = li + k for all i < m. For i < m, let yi = li · xa(i) and let zi be the middle-point of the interval Ih(i). Find k < n such that |k · xa(i) − (zi − yi)| < ε (mod 1) for all ∗ i < m. Then osc1(a(i), βk) = h(i) for all i < m, as required. a 3.10 Corollary. There is a regular hereditarily Lindelöfspace which is not separable.

ω1 Proof. For β < ω1, let fβ ∈ {0, 1} be deﬁned by letting fβ(β) = 1, ∗ fβ(γ) = 0 for γ > β, and fβ(α) = min{1, osc1(α, β)} for α < β. Consider ω1 F = {fβ : β < ω1} as a subspace of {0, 1} . Clearly F is not separable. That F is hereditarily Lindel¨offollows easily from Lemma 3.9. a

∗ 3.11 Remark. The projection osc0 of the oscillation mapping osc0 appears in [78] as historically first such a map with more than four colors that takes all of its values on every symmetric square of an uncountable subset of ω1. ∗ The variations osc1 and osc1 are on the other hand very recent and are due to J.T. Moore [54] who made them in order to obtain the conclusion of Corollary 3.10. Concerning Corollary 3.10 we note that the dual implication behaves quite differently since assuming the Proper Forcing Axiom all hereditarily separable regular spaces are hereditarily Lindelöf(see [80]).

4. Triangle inequalities

In this section we study a characteristic of the minimal walk that satisfies certain triangle inequalities reminiscent of those found in an ultrametric space. We also present several applications of the corresponding metric- like theory of ω1. This leads us to the following variation of the oscillation mapping. 2 4.1 Definition. Define ρ :[ω1] −→ ω by recursion as follows

ρ(α, β) = max{|Cβ ∩ α|, ρ(α, min(Cβ \ α)),

ρ(ξ, α): ξ ∈ Cβ ∩ α}, 4. Triangle inequalities 27 with the boundary condition ρ(α, α) = 0 for all α.

4.2 Lemma. For all α < β < γ < ω1 and n < ω, (a) {ξ ≤ α : ρ(ξ, α) ≤ n} is ﬁnite, (b) ρ(α, γ) ≤ max{ρ(α, β), ρ(β, γ)}, (c) ρ(α, β) ≤ max{ρ(α, γ), ρ(β, γ)}.

Proof. Note that ρ(α, β) ≥ ρ1(α, β), so (a) follows from the corresponding property of ρ1. The proof of (b) and (c) is simultaneous by induction on α, β and γ: To prove (b), consider n = max{ρ(α, β), ρ(β, γ)}, and let

ξα = min(Cγ \ α) and ξβ = min(Cγ \ β).

We have to show that ρ(α, γ) ≤ n. b Case 1 : ξα = ξβ. Then by the inductive hypothesis,

ρ(α, ξα) ≤ max{ρ(α, β), ρ(β, ξβ)}.

From the deﬁnition of ρ(β, γ) we get that ρ(β, ξβ) ≤ ρ(β, γ) ≤ n, so replacing ρ(β, ξβ) by ρ(β, γ) in the above inequality we get ρ(α, ξα) ≤ n. Consider a ξ ∈ Cγ ∩ α = Cγ ∩ β. By the inductive hypothesis

ρ(ξ, α) ≤ max{ρ(ξ, β), ρ(α, β)}.

From the deﬁnition of ρ(β, γ) we see that ρ(ξ, β) ≤ ρ(β, γ), so replacing ρ(ξ, β) with ρ(β, γ) in the last inequality we get that ρ(ξ, α) ≤ n. Since |Cγ ∩ α| = |Cγ ∩ β| ≤ ρ(β, γ) ≤ n, referring to the deﬁnition of ρ(α, γ) we conclude that ρ(α, γ) ≤ n. b Case 2 : ξα < ξβ. Then ξα ∈ Cγ ∩ β, so

ρ(ξα, β) ≤ ρ(β, γ) ≤ n.

By the inductive hypothesis

ρ(α, ξα) ≤ max{ρ(α, β), ρ(ξα, β)} ≤ n.

Similarly, for every ξ ∈ Cγ ∩ α ⊆ Cγ ∩ β,

ρ(ξ, α) ≤ max{ρ(ξ, β), ρ(α, β)} ≤ n.

Finally |Cγ ∩ α| ≤ |Cγ ∩ β| ≤ ρ(β, γ) ≤ n. Combining these inequalities we get the desired conclusion ρ(α, γ) ≤ n. To prove (c), consider now n = max{ρ(α, γ), ρ(β, γ)}. We have to show that ρ(α, β) ≤ n. Let ξα and ξβ be as above and let us consider the same two cases as above. 28 I. Coherent Sequences

c ¯ Case 1 : ξα = ξβ = ξ. Then by the inductive hypothesis

ρ(α, β) ≤ max{ρ(α, ξ¯), ρ(β, ξ¯)}.

This gives the desired bound ρ(α, β) ≤ n, since ρ(α, ξα) ≤ ρ(α, γ) ≤ n and ρ(β, ξβ) ≤ ρ(β, γ) ≤ n. c Case 2 : ξα < ξβ. Applying the inductive hypothesis again we get

ρ(α, β) ≤ max{ρ(α, ξα), ρ(ξα, β)} ≤ n.

This completes the proof. a 4.3 Remark. Note the following two alternate deﬁnitions of ρ:

ρ(α, β) = max{|Cη ∩ ξ| : ξ, η ∈ F (α, β) ∪ {β} and ξ < η}.

ρ(α, β) = max{ρ1(ξ, η): ξ, η ∈ F (α, β) ∪ {β} and ξ < η}. These two deﬁnitions make it more transparent (in light of Lemma 1.9) that ρ is a subadditive function. The proof of Lemma 4.2 is given, however, because of some later generalizations of this function to higher ordinals. The following fact shows that the function ρ has a considerably ﬁner coherence property than ρ1. 4.4 Lemma. If α < β < γ and if ρ(α, β) > ρ(β, γ), then ρ(α, γ) = ρ(α, β). Proof. Applying Lemma 4.2,

ρ(α, γ) ≤ max{ρ(α, β), ρ(β, γ)} = ρ(α, β),

ρ(α, β) ≤ max{ρ(α, γ), ρ(β, γ)} = ρ(α, γ). Note that max{ρ(α, γ), ρ(β, γ)} must be equal to ρ(α, γ) rather than ρ(β, γ) since by the assumption ρ(α, β), which is bounded by the maximum, is bigger than ρ(β, γ). a 4.5 Remark. Referring to the standard definitions of metrics and ultra- metrics the properties 4.2(b) and (c) could more properly be called ultra- subadditivities though we shall keep calling them subadditivity properties of the function ρ. Recall the notion of a full lower trace F (α, β) given in 2 Definition 1.7 above. The function d :[ω1] −→ ω defined by

d(α, β) = |F (α, β)| is an example of a truly subadditive function, since by Lemma 1.8 we have the following inequalities whenever α < β < γ: (a) d(α, γ) ≤ d(α, β) + d(β, γ), 4. Triangle inequalities 29

(b) d(α, β) ≤ d(α, γ) + d(β, γ).

2 4.6 Deﬁnition. Deﬁneρ ¯ :[ω1] −→ ω as follows

ρ¯(α, β) = 2ρ(α,β) · (2 · |{ξ ≤ α : ρ(ξ, α) ≤ ρ(α, β)}| + 1).

4.7 Lemma. For all α < β < γ < ω1, (a) ρ¯(α, γ) 6=ρ ¯(β, γ),

(b) ρ¯(α, γ) ≤ max{ρ¯(α, β), ρ¯(β, γ)}, (c) ρ¯(α, β) ≤ max{ρ¯(α, γ), ρ¯(β, γ)}. Proof. Only (b) and (c) require some argument. Consider ﬁrst (b). If ρ(α, β) ≤ ρ(β, γ) then ρ(α, γ) ≤ max{ρ(α, β), ρ(β, γ)} = ρ(β, γ), so

{ξ ≤ α : ρ(ξ, α) ≤ ρ(α, γ)} ⊆ {ξ ≤ β : ρ(ξ, β) ≤ ρ(β, γ)}.

Therefore in this caseρ ¯(α, γ) ≤ ρ¯(β, γ). On the other hand, if ρ(β, γ) ≤ ρ(α, β), then ρ(α, γ) ≤ max{ρ(α, β), ρ(β, γ)} = ρ(α, β), so

{ξ ≤ α : ρ(ξ, α) ≤ ρ(α, γ)} ⊆ {ξ ≤ α : ρ(ξ, α) ≤ ρ(α, β)}.

So in this caseρ ¯(α, γ) ≤ ρ¯(α, β). This checks (b). Checking (c) is similar. a

<ω 4.8 Deﬁnition. For p ∈ ω deﬁne a binary relation

p(¯ρ(ξ, α)) = p(¯ρ(ξ, β)) for any ξ < α such thatρ ¯(ξ, α) < |p|. (I.38)

4.9 Lemma.

(a)

(b) p ⊆ q implies

4.10 Deﬁnition. For x ∈ ωω, set [

<ω 4.11 Lemma. For every p ∈ ω there is a partition of ω1 into ﬁnitely many pieces such that if α < β belong to the same piece then there is q ⊇ p <ω in ω such that α

Proof. For α < ω1, the set

F|p|(α) = {ξ ≤ α :ρ ¯(ξ, α) ≤ |p|} has size ≤ |p| + 1 so if we enrich F|p|(α) to a structure that would code the behavior ofρ ¯ on F|p|(α) one can realize only finitely many different isomorphism types. Thus it suffices to show that if α < β < ω1 with F|p|(α), F|p|(β) isomorphic, then there is an extension q of p such that α

ρ(ξ0, γ0) = n0 and ρ(ξk−1, γ2k−1) = nk, (I.39)

ρ(ξi−1, γ2i−1) = ρ(ξi, γ2i) = ni for 0 < i < k, (I.40)

γi 6= γi+1 for all i ≤ 2k − 2. (I.41)

By our assumption about the isomorphism between the structures of F|p|(γ0) and F|p|(γ1) the ordinals ξ0 ∈ F|p|(γ0) does not belong to the root

F|p|(γ0) ∩ F|p|(γ1).

It follows that n1 > |p|. Now the following general fact aboutρ ¯ (together with the properties (I.40)-(I.41)) shows that the sequence n0, n1, . . . , nk is strictly increasing, ﬁnishing thus the proof that a given n < |p| is not G- connected to any other integer < |p|.

4.12 Lemma. Suppose ηα 6= ηβ < min{α, β} and ρ¯(ηα, α) =ρ ¯(ηβ, β) = n. Then ρ¯(ηα, β), ρ¯(ηβ, α) > n.

Proof. Suppose for example that ηα < ηβ. First note that ρ(ηα, ηβ) ρ(ηβ, β) or else {ξ ≤ ηα : ρ(ξ, ηα) ≤ ρ(ηα, α)} would be a proper initial part of {ξ ≤ ηβ : ρ(ξ, ηβ) ≤ ρ(ηβ, β)} contradicting the fact that ρ¯(ηα, α) =ρ ¯(ηβ, β) yields that the two sets are of the same cardinality. Thus ρ(ηα, ηβ) > ρ(ηβ, β) = ρ(ηα, α) which together with the subadditivity properties of ρ gives us that

{ξ ≤ ηα : ρ(ξ, ηα) ≤ ρ(ηα, α)} ⊆ {ξ ≤ ηα : ρ(ξ, ηα) ≤ ρ(ηα, ηβ)}. 4. Triangle inequalities 31

This gives the inequalityρ ¯(ηα, ηβ) > ρ¯(ηα, α) = n. Combining this inequality with the subadditivities ofρ ¯ we get:

n < ρ¯(ηα, ηβ) ≤ max{ρ¯(ηα, α), ρ(ηβ, α)} =ρ ¯(ηβ, α),

n < ρ¯(ηα, ηβ) ≤ max{ρ¯(ηα, β), ρ(ηβ, β)} =ρ ¯(ηα, β).

The case ηβ ≤ ηα is considered similarly. This completes the proof. a

4.13 Theorem. For every inﬁnite subset Γ ⊆ ω1, the set

ω GΓ = {x ∈ ω : α

Proof. This is an immediate consequence of Lemma 4.11. a

4.14 Deﬁnition. For α < β < ω1, let α <ρ¯ β denote the fact thatρ ¯(ξ, α) = ρ¯(ξ, β) for all ξ < α. Then <ρ¯ is a tree ordering on ω1 obtained by identifying ω α with the memberρ ¯(·, α) of the tree T (¯ρ). Note that <ρ¯⊆

4.15 Theorem. If Γ is an inﬁnite <ρ¯-antichain, the set

ω HΓ = {x ∈ ω : α ≮x β for some α < β in Γ} is a dense open subset of the Baire space.

Proof. Let p ∈ ω<ω be given. A simple application of Ramsey’s theorem gives us a strictly increasing sequence γi (i < ω) of members of Γ such 15 that ∆(γi, γj) ≤ ∆(γj, γk) for all i < j < k. Going to a subsequence, we may assume that either ∆(γi, γi+1) < ∆(γi+1, γi+2) for all i or else for some ordinal ξ and all i < j, ∆(γi, γj) = ξ. In the ﬁrst case, by Lemma 4.7(a), we can ﬁnd i such that one of the integersρ ¯(∆(γi, γi+1), γi) orρ ¯(∆(γi, γi+1), γi+1) does not belong to the domain of p, so there will be q ⊇ p such thatρ ¯(γi, γi+1) < |q| and γi ≮q γi+1. This guarantees that γi ≮x γi+1 for every x ⊇ q. In the second case,ρ ¯(ξ, γi)(i < ω) is a one- to-one sequence of integers so there will be i < j such that neither of the integersρ ¯(ξ, γi) orρ ¯(ξ, γj) belongs to the domain of p, so there is q ⊇ p such thatρ ¯(γi, γj) < |q| and γi ≮q γj. Hence, γi ≮x γj for every x ⊇ q. a

15 For α < β < ω1, ∆(α, β) = min{ξ ≤ α :ρ ¯(ξ, α) 6=ρ ¯(ξ, β)}. 32 I. Coherent Sequences

4.16 Deﬁnition. For a family F of inﬁnite <ρ¯-antichains, we say that a ω real x ∈ ω is F-Cohen if x ∈ GΓ ∩ HΓ for all Γ ∈ F. We say that x is F-Souslin if no member of F is a

Note that since every uncountable subset of ω1 contains an uncountable <ρ¯-antichain, if a family F reﬁnes the family of all uncountable <ρ¯- antichains, then every F-Souslin real is Souslin. The following fact summa- rizes Theorems 4.13 and 4.15 and connects the two kinds of reals.

4.17 Theorem. If F is a family of inﬁnite <ρ¯-antichains, then every F- Cohen real is F-Souslin. a

4.18 Corollary. If the density of the family of all uncountable subsets of ω1 is smaller than the number of nowhere dense sets needed to cover the real line, then there is a Souslin tree. a

4.19 Remark. Recall that the density of a family F of infinite subsets of some set S is the minimal size of a family F0 of infinite subsets of S with the property that every member of F is refined by a member of F0.A special case of Corollary 4.18, when the density of the family of all uncountable subsets of ω1 is equal to ℵ1, was first observed by T. Miyamoto (unpublished).

4.20 Corollary. Every Cohen real is Souslin.

Proof. Every uncountable subset of ω1 in the Cohen extension contains an uncountable subset from the ground model. So it suﬃces to consider the family F of all inﬁnite <ρ¯-antichains from the ground model. a

If ω1 is a successor cardinal in the constructible subuniverse, thenρ ¯ can be chosen to be coanalytic and so the transformation x 7−→

4.21 Remark. We have just seen how the combination of the subadditivity properties (4.7(b),(c)) of the coherent sequenceρ ¯α : α −→ ω (α < ω1) of one-to-one mappings can be used in controlling the ﬁnite disagreement between them. It turns out that in many contexts the coherence and the subadditivities are essentially equivalent restrictions on a given sequence eα : α −→ ω (α < ω1). For example, the following construction shows that this is so for any sequence of ﬁnite-to-one mappings eα : α −→ ω (α < ω1). 4. Triangle inequalities 33

4.22 Definition. Given a coherent sequence eα : α −→ ω (α < ω1) of 2 finite-to-one mappings, define τe :[ω1] −→ ω as follows

16 τe(α, β) = max{max{e(ξ, α), e(ξ, β)} : ξ ≤ α, e(ξ, α) 6= e(ξ, β)} .

4.23 Lemma. For every α < β < γ < ω1,

(a) τe(α, β) ≥ e(α, β),

(b) τe(α, γ) ≤ max{τe(α, β), τe(β, γ)},

(c) τe(α, β) ≤ max{τe(α, γ), τe(β, γ)}. Proof. To see (a) note that it trivially holds if e(α, β) = 0 so we can concentrate on the case e(α, β) > 0. Applying the convention e(α, α) = 0 we see that e(ξ, α) 6= e(ξ, β) for ξ = α. It follows that τe(α, β) ≥ max{max{e(α, α), e(α, β)} = e(α, β). To see (b) let n = max{τe(α, β), τe(β, γ)} and let ξ ≤ α be such that e(ξ, α) 6= e(ξ, γ). We need to show that e(ξ, α) ≤ n and e(ξ, γ) ≤ n. If e(ξ, α) > n then since τe(α, β) ≤ n we must have e(ξ, α) = e(ξ, β)) and so e(ξ, β) 6= e(ξ, γ). It follows that τe(β, γ) ≥ e(ξ, β)) > n, a contradiction. Similarly, e(ξ, γ) > n yields e(ξ, β) = e(ξ, γ) 6= e(ξ, α) and therefore τe(α, β) ≥ e(ξ, β)) > n, a contradiction. The proof of (c) is similar. a

2 4.24 Definition. We say that a :[ω1] −→ ω is transitive if for α < β < γ < ω1 a(α, γ) ≤ max{a(α, β), a(β, γ)}. Transitive maps occur quite frequently in set-theoretic constructions. For example, given a sequence Aα (α < ω1) of subsets of ω that increases relative to the ordering ⊆∗ of inclusion modulo a finite set, the mapping 2 a :[ω1] −→ ω defined by

a(α, β) = min{n : Aα \ n ⊆ Aβ} is a transitive map. The transitivity condition by itself is not nearly as useful as its combination with the other subadditivity property (4.7(c)). Fortu- nately, there is a general procedure that produces a subadditive dominant to every transitive map.

2 2 4.25 Deﬁnition. For a transitive a :[ω1] −→ ω deﬁne ρa :[ω1] −→ ω recursively as follows:

ρa(α, β) = max{|Cβ ∩ α|, a(min(Cβ \ α), β), ρa(α, min(Cβ \ α)),

ρa(ξ, α): ξ ∈ Cβ ∩ α}.

16Note the occurence of the boundary value e(α, α) = 0 in this formula as well. 34 I. Coherent Sequences

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

(a) {ξ ≤ α : ρa(ξ, α) ≤ n} is ﬁnite,

(b) ρa(α, γ) ≤ max{ρa(α, β), ρa(β, γ)},

(d) ρa(α, β) ≥ a(α, β). Proof. The proof of (a),(b),(c) is quite similar to the corresponding part of the proof of Lemma 4.2. This comes of course from the fact that the definition of ρ and ρa are closely related. The occurrence of the factor a(min(Cβ \ α), β) complicates a bit the proof that ρa is subadditive, and the fact that a is transitive is quite helpful in getting rid of the additional difficulty. The details are left to the interested reader. Given α < β, for every step βn → βn+1 of the minimal walk β = β0 > β1 > . . . > βk = α, we have ρa(α, β) ≥ ρa(βn, βn+1) ≥ a(βn, βn+1) by the very definition of ρa. Applying the transitivity of a to this path of inequalities we get the conclusion (d). a

4.27 Lemma. ρa(α, β) ≥ ρa(α + 1, β) whenever 0 < α < β and α is a limit ordinal.

Proof. Recall the assumption (c) about the ﬁxed C-sequence Cξ (ξ < ω1) on which all our deﬁnitions are based: if ξ is a limit ordinal > 0, then no point of Cξ is a limit ordinal. It follows that if 0 < α < β and α is a limit ordinal, then the minimal walk β → α must pass through α + 1 and therefore ρa(α, β) ≥ ρa(α + 1, β). a

Let us now give an application of ρa to a classical phenomenon of occurrence of gaps in the quotient algebra P(ω)/fin. 4.28 Definition. A Hausdorff gap in P(ω)/fin is a pair of sequences Aα (α < ω1) and Bα (α < ω1) such that

∗ ∗ ∗ (a) Aα ⊆ Aβ ⊆ Bβ ⊆ Bα whenever α < β, but

∗ ∗ (b) there is no C such that Aα ⊆ C ⊆ Bα for all α. The following straightforward reformulation shows that a Hausdorﬀ gap is just another instance of a nontrivial coherent sequence

fα : Aα −→ 2 (α < ω1) where the domain Aα of fα is not the ordinal α itself but a subset of ω and that the corresponding sequence of domains Aα (α < ω1) is a realization ∗ of ω1 inside the quotient P(ω)/ﬁn in the sense that Aα ⊆ Aβ whenever α < β. 4. Triangle inequalities 35

4.29 Lemma. A pair of ω1-sequences Aα (α < ω1) and Bα (α < ω1) form a Hausdorﬀ gap iﬀ the pair

A¯α = Aα ∪ (ω \ Bα)(α < ω1) and B¯α = ω \ Bα (α < ω1) has the following properties:

∗ (a) A¯α ⊆ A¯β whenever α < β,

∗ (b) B¯α = B¯β ∩ A¯α whenever α < β,

∗ (c) there is no B such that B¯α = B ∩ A¯α for all α. a

∗ From now on we fix a strictly ⊆ -increasing chain Aα (α < ω1) of infinite 2 subsets of ω and let a :[ω1] −→ ω be defined by

a(α, β) = min{n : Aα \ n ⊆ Aβ}.

2 Let ρa :[ω1] −→ ω be the corresponding subadditive dominant of a deﬁned above. For α < ω1, set Dα = Aα+1 \ Aα.

4.30 Lemma. The sets Dα\ρa(α, γ) and Dβ \ρa(β, γ) are disjoint whenever 0 < α < β < γ and α and β are limit ordinals. Proof. This follows immediately from Lemmas 4.26 and 4.27. a

2 We are in a position to deﬁne a partial mapping m :[ω1] −→ ω by

m(α, β) = min(Dα \ ρa(α, β)), whenever α < β and α is a limit ordinal. 4.31 Lemma. The mapping m is coherent, i.e., m(α, β) = m(α, γ) for all but ﬁnitely many limit ordinals α < min{β, γ}.

Proof. This is by the coherence of ρa and the fact that ρa(α, β) = ρa(α, γ) already implies m(α, β) = m(α, γ). a 4.32 Lemma. m(α, γ) 6= m(β, γ) whenever α 6= β < γ and α, β limit. Proof. This follows from Lemma 4.30. a

For β < ω1, set

Bβ = {m(α, β): α < β and α limit}.

∗ 4.33 Lemma. Bβ = Bγ ∩ Aβ whenever β < γ. 36 I. Coherent Sequences

Proof. By the coherence of m. a

Note the following immediate consequence of Lemma 4.30 and the deﬁ- nition of m.

4.34 Lemma. m(α, β) = max(Bβ ∩ Dα) whenever α < β and α limit. a

∗ 4.35 Lemma. There is no B ⊆ ω such that B ∩ Aβ = Bβ for all β.

Proof. Suppose that such a B exists and for a limit ordinal α let us define g(α) = max(B ∩ Dα). Then by Lemma 4.34, g(α) = m(α, β) for all β < ω1 and all but finitely many limit ordinals α < β. By Lemma 4.32, it follows that g is a finite-to-one map, a contradiction. a

∗ 4.36 Theorem. For every strictly ⊂ -increasing chain Aα (α < ω1) of subsets of ω, there is a sequence Bα (α < ω1) of subsets of ω such that:

∗ (a) Bα = Bβ ∩ Aα whenever α < β,

(b) there is no B such that Bα = B ∩ Aα for all α.

4.37 Remark. For a given countable ordinal α let hα : Aα −→ 2 be such −1 that hα (1) = Bα. Then by Lemma 4.33, the corresponding sequence hα (α < ω1) of partial functions is coherent in the sense that for every pair α < β < ω1, the set Dh(α, β) = {n ∈ Aα ∩ Aβ : hα(n) 6= hβ(n)} is finite. By Lemma 4.35, the sequence is nontrivial in the sense that there is ∗ no g : ω −→ 2 such that hα = g ¹ Aα for all α < ω1. Hausdorff’s original sequence, when reformulated in this way, has the property that the maps dh(·, β): β −→ ω defined by

dh(α, β) = |Dh(α, β)| are ﬁnite-to-one mappings. A general principle, the P-ideal dichotomy, (see 4.45 below) asserts that every coherent and nontrivial sequence must contain a subsequence with Hausdorﬀ’s property. We shall now embark on a general construction of a very similar sequence of coherent partial functions domains included in ω1 rather than in ω.

4.38 Deﬁnition. The number of steps of the minimal walk is the two-place 2 function ρ2 :[ω1] −→ ω deﬁned recursively by

ρ2(α, β) = ρ2(α, min(Cβ \ α)) + 1, with the convention that ρ2(γ, γ) = 0 for all γ. 4. Triangle inequalities 37

This is an interesting mapping which is particularly useful on higher cardinalities and especially in situations where the more informative mappings ρ0, ρ1 and ρ lack their usual coherence properties. Later on we shall devote a whole section to ρ2 but here we need it only to succinctly express the following mapping. 4.39 Deﬁnition. The last step function of the minimal walk is the two- 2 place map ρ3 :[ω1] −→ 2 deﬁned by letting

ρ3(α, β) = 1 iﬀ ρ0(α, β)(ρ2(α, β) − 1) = ρ1(α, β).

In other words, we let ρ3(α, β) = 1 just in case the last step of the walk β → α comes with the maximal weight.

4.40 Lemma. {ξ < α : ρ3(ξ, α) 6= ρ3(ξ, β)} is finite for all α < β < ω1. Proof. It suffices to show that for every infinite Γ ⊆ α there exists ξ ∈ Γ such that ρ3(ξ, α) = ρ3(ξ, β). Shrinking Γ we may assume that for some fixedα ¯ ∈ F (α, β) and all ξ ∈ Γ:

α¯ = min(F (α, β) \ ξ), (I.42)

ρ1(ξ, α) = ρ1(ξ, β), (I.43)

ρ1(ξ, α) > ρ1(¯α, α), (I.44)

ρ1(ξ, β) > ρ1(¯α, β). (I.45) It follows (see 1.9) that for every ξ ∈ Γ:

a ρ0(ξ, α) = ρ0(¯α, α) ρ0(ξ, α¯), (I.46)

a ρ0(ξ, β) = ρ0(¯α, β) ρ0(ξ, α¯). (I.47)

So for any ξ ∈ Γ, ρ3(ξ, α) = 1 iﬀ the last term of ρ0(ξ, α¯) is its maximal term iﬀ ρ3(ξ, β) = 1. a

17 The sequence (ρ3)α : α −→ 2 (α < ω1) is therefore coherent in the ∗ sense that (ρ3)α = (ρ3)β ¹ α whenever α < β. We need to show that the sequence is not trivial, i.e. that it cannot be uniformized by a single total map from ω1 into 2. In other words, we need to show that ρ3 still contains enough information about the C-sequence Cα (α < ω1) from which it is deﬁned. For this it will be convenient to assume that Cα (α < ω1) satisﬁes the following natural condition:

(d) If α is a limit ordinal > 0 and if ξ occupies the nth place in the increasing enumeration of Cα (that starts with min(Cα) on its 0th place), then ξ = λ + n + 1 for some limit ordinal λ (possibly 0).

17Recall the way one always deﬁnes the ﬁber-functions from a two-variable function applied to the context of ρ3:(ρ3)α(ξ) = ρ3(ξ, α). 38 I. Coherent Sequences

4.41 Deﬁnition. Let Λ denote the set of all countable limit ordinals and for an integer n ∈ ω, let Λ + n = {λ + n : λ ∈ Λ}.

4.42 Lemma. ρ3(λ + n, β) = 1 for all but ﬁnitely many n with λ + n < β.

Proof. Clearly we may assume that α = λ + ω ≤ β. Then there is n0 < ω such that for every n ≥ n0 the walk β → λ + n passes through α. By (d) we know that Cα = {λ + n : 0 < n < ω} so in any such walk β → λ + n there is only one step from α to λ + n. So choosing n ≥ n0, ρ1(α, β) we will ensure that the last step of β → λ + n comes with the maximal weight, i.e. ρ3(λ + n, β) = 1. a

4.43 Lemma. For all β < ω1, n < ω, the set {λ ∈ Λ: λ + n < β and ρ3(λ + n, β) = 1} is finite. Proof. Given an infinite subset Γ of (Λ + n) ∩ β we need to find a λ + n ∈ Γ such that ρ3(λ + n, β) = 0. Shrinking Γ if necessary assume that ρ1(λ + n, β) > n + 2 for all λ + n ∈ Γ. So if ρ3(λ + n, β) = 1 for some λ + n ∈ Γ then the last step of β → λ + n would have to be of weight> n + 2 which is impossible by our assumption (d) about Cα (α < ω1). a

The meaning of these properties of ρ3 perhaps is easier to comprehend if we reformulate them in a way that resembles the original formulation of the existence of Hausdorﬀ gaps.

4.44 Lemma. Let Bα = {ξ < α : ρ3(ξ, α) = 1} for α < ω1. Then:

∗ 1. Bα = Bβ ∩ α for α < β,

2. (Λ + n) ∩ Bβ is ﬁnite for all n < ω and β < ω1,

∗ 3. {λ + n : n < ω} ⊆ Bβ whenever λ + ω ≤ β. a

∗ In particular, there is is no uncountable Γ ⊆ ω1 such that Γ∩β ⊆ Bβ for 18 all β < ω1. On the other hand, the P-ideal I generated by Bβ (β < ω1) is large as it contains all intervals of the form [λ, λ+ω). The following general dichotomy about P-ideals shows that here indeed we have quite a canonical example of a P-ideal on ω1. 4.45 Deﬁnition. The P-ideal dichotomy. For every P-ideal I of countable subsets of some set S either: 1. there is uncountable X ⊆ S such that [X]ω ⊆ I, or

18Recall that an ideal I of subsets of some set S is a P-ideal if for every sequence An (n < ω) of elements of I there is B in I such that An \ B is ﬁnite for all n < ω.A set X is orthogonal to I if X ∩ A is ﬁnite for all A in I. 5. Conditional weakly null sequences 39

2. S can be decomposed into countably many sets orthogonal to I.

4.46 Remark. It is known that the P-ideal dichotomy is a consequence of the Proper Forcing Axiom and moreover that it does not contradict the Con- tinuum Hypothesis (see [89]). This is an interesting dichotomy which will be used in this article for testing various notions of coherence as we encounter them. For example, let us consider the following notion of coherence, already encountered above at several places, and see how it is inﬂuenced by the P-ideal dichotomy.

5. Conditional weakly null sequences

In this section we mention some applications of the concept of ρ-function on ω1 satisfying the inequalities of Lemma 4.2. As we shall see any such a semi-distance function ρ can be used as a coding procedure in constructions of long weakly null sequences in Banach spaces with no unconditional basic subsequences.

5.1 Example. A weakly null sequence with no unconditional basic subsequence. We describe a weakly null normalized sequence (xξ)ξ<ω1 of members of some Banach space X which contains no infinite unconditional basic sub- sequence19. Fix an 0 < ε < 1 and an infinite set M of positive integers such that 1 ∈ M and X m {( )1/2 : m, n ∈ M, m < n} ≤ ε. (I.48) n Fix also a natural bijection p.q : HF → M, where HF denotes the family of all hereditarily finite sets, so that in particular p∅q = 1. The ρ-number od a finite set F ⊆ ω1 is the integer

ρ(F ) = max{ρ(α, β): α, β ∈ F, α ≤ β}.

For F ⊆ ω1 and k ∈ ω let

(F )k = {α ≤ max(F ): ρ(α, β) ≤ k for some β ∈ F with β ≥ α}.

19Recall that a basic sequence in a Banach space X is a seminormalized sequence (xn)n∈ω with the property that everyP x in its closed linear span span{xn : n ∈ ω} can be uniquely represented as a sum n∈ω anxn. Note that this is equivalent to saying that the norms of projection operators Pk : span{xn : n ∈ ω} → span{xn : n < k} are uniformly bounded. When the norms of projections PM : span{xn : n ∈ ω} → span{xn : n ∈ M} over all subsets M of ω are uniformly bounded, the sequence (xn) is said to be unconditional. The natural basis (en) of any of the sequence-spaces `p for p ∈ [1, ∞] is unconditional. A typical example of a basic sequence that is not unconditional is the summing basis e0, e0+e1, e0+e1+e2, ... of c0 or the summing basis e0+e1+e2+...+en+..., e1 + e2 + ... + en + ..., e2 + ... + en + ..., ... in c. 40 I. Coherent Sequences

We say that F is k-closed if (F )k = F . The ρ-closure of F is the set (F )ρ = (F )k, where k = ρ(F ). This leads us to the function

<ω <ω σ : [[ω1] ] −→ M deﬁned by letting σ(∅) = p∅q and

[n σ(F0,F1, ..., Fn) = p((πE”F0, πE”F1, ..., πE”Fn), ρ( Fi))q, i=0

Sn where E = ( i=0 Fi)ρ and where πE : E → |E| is the unique order- preserving map between the sets of ordinals E and |E| = {0, 1, ..., |E| − 1}. We say that a sequence (Ei)i

20 1. Ei < Ej for i < j < n,

2. |E0| = 1 and |Ej| = σ(E0,E1, ..., Ej−1) for 0 < j < n.

Let c00(ω1) be the normed space of all ﬁnitely supported maps from ω1 into the reals. Let (e ) be the basis of c (ω ) and let (e )∗ be the ξ ξ<ω1 00 1 ξ ξ<ω1 corresponding sequence of biorthogonal functionals. To a ﬁnite set E ⊆ ω1, we associate the following vector and functional on c00(ω1), 1 X 1 X x = e and φ = e∗ . (I.49) E |E|1/2 α E |E|1/2 α α∈E α∈E

Given a special sequence (Ei)i

k x k= sup{hf, xi : f ∈ F}. (I.50)

The Banach space X is the completion of the normed space (c00(ω1), k . k).

Note that k eα k= 1 and that (eα)α<ω1 is a weakly null sequence in X. To see this note thatP given a ﬁnite set G with |G| ∈ M and a special sequence (Ei)i

20Recall, that for two sets of ordinals E and F , by E < F we denote the fact that every ordinal from E is smaller from every ordinal from F . Sequences of sets of ordinals of this sort are called block sequences. 5. Conditional weakly null sequences 41

21 order type ω there is seminormalized block-subsequence of (eα)α∈G which represents the summing basis in c. More precisely, we shall show that if (Ei)i<ω is any infinite special sequence of finite subsets of G and if for i < ω we let vi = xEi as defined in I.49, then for every n < ω and every sequence (ai)i≤n ⊆ [−1, +1] of scalars. Xk Xn Xk max | ai |≤k aivi k≤ (3 + ε) max | ai | . (I.51) 0≤k≤n 0≤k≤n i=0 i=0 i=0 To see the first inequality, note that according to I.49 the absolute value Pn of the inner product between the spacial functional φ = i=0 φEi and the Pn Pn vector x = i=0 aivi is equal to | i=0 ai | giving us the first inequality. To Pm see the second inequality, consider a special functional ψ = i=0 φFi where (Fi)i≤m is some other special sequence of finite sets. In order to estimate the inner product hψ, xi we need to know how the two special sequences interact and this requires properties of the function ρ that controls them. To see this, letSl ≤ min(m, n) be maximalS with the property that |El| = |Fl|, and let E = ( i

2. πE”Ei = πE”Fi for i < l. It follows that E and F have the same cardinality and the same ρ-numbers and therefore by the properties 4.1 of ρ, their intersection H = E∩F is their initial segment. Hence, πE and πE agree on H. Let k be the maximal integer such that Ek ⊆ H. Then Ei = Fi for all i ≤ k, and using the properties 4.1 of ρ again, Ei ∩ Fj = ∅ for k + 1 < i, j < l. Combining all this we see that |hψ, xi| is bounded by

Xk X m | a +a hφ , x i+a hφ , x i|+ {( )1/2 : m, n ∈ M, m < n}. i k+1 Fk+1 Ek+1 l Fl El n i=0 Referring to I.48 we get that |hψ, xi| is bounded by the right hand side of the second inequality of I.51. 5.2 Remark. The first example of a weakly null sequence (of length ω) with no unconditional basic subsequence was constructed by Maurey and Rosen- thal [51]. Sixteen years latter Gowers and Maurey [27] have solved the basic unconditional sequence problem by constructing an infinite-dimensional separable reflexive Banach space with no infinite unconditional basic sequence. The following result of [3] relates to the Example 5.1 the same way the result of [27] relates to the example of [51].

21 i.e., a seminormalized sequence (xi)i<ω of ﬁnitely supported vectors of the linear span of (eα)α∈G such that supp(xi) < supp(xj )for i < j. 42 I. Coherent Sequences

5.3 Example. A reﬂexive Banach space Xω1 with a Schauder basis of length

ω1 containing no unconditional basic sequence. The space Xω1 is the completion of c00(ω1) under the norm given by the formula I.50, where now

F = Fω1 is chosen as the minimal set of ﬁnitely supported functionals on c00(ω1) such that:

∗ 1. F ⊇ {eα : α < ω1} and F is symmetric and closed under restriction on intervals of ω1. 2. For every (φ ) ⊆ F such that supp(φ ) < supp(φ ) for i < j < i i

The increasing sequences (mk) and (nk) of positive integers are deﬁned by: 4k log m3 mk = 2 and n0 = 4 and nk+1 = (4nk) 2 k+1 . The notion of a special sequence of functionals is deﬁned analogously to that in the Example 5.1 ∗ as follows. First of all we say that φ ∈ F is of type 0 if φ = ±eα Pfor some α, of type I if it obtained from a simpler ones as the sum m−1 φ k i

A sequence (φi)i

1. supp(φi) < supp(φj) for i < j < n2k+1,

2. each φi is of type I and of some ﬁxed weight w(φi) = m2ki with k0 2 even and satisfying m2k0 > n2k+1,

3. for 0 < j < n2k+1Sthere is an increasing sequence (pi)i

(The collapse ψi = πE”φi is deﬁned by the formula ψ(πE(α)) = φ(α).) The condition 1. from the deﬁnition of F provides that (eα)α<ω1 is a bimonotone

Schauder basis of Xω1 , the condition 3. is responsible for the conditional 5. Conditional weakly null sequences 43

structure of the norm of Xω1 in a similar manner as in the Example 5.1, but in order to come to the point the idea can be applied, one needs the unconditional structure given but the closure property 2. of F. This requires a rather deep tree-analysis of the functionals of F that gives the required norm estimates, an analysis that is beyond the scope of this presentation. We refer the reader to [3] for the details. Here we just list some of the most interesting properties of Xω1 . For example, the space Xω1 is reflexive, has no infinite unconditional basic sequence, and is not isomorphic to any proper subspace or a nontrivial quotient. The basis (eα)α<ω1 allows us to talk about diagonal operators of Xω1 . It turns out that all bounded diagonal operators D of Xω1 are given by a sequence (λα)α<ω1 of eigenvalues that have the property that λα = λβ whenever α ≤ β < α + ω. Moreover there is only countably many different eigenvalues among (λα)α<ω1 . Recall, than an operator between two Banach spaces is strictly singular if it is not an isomorphism when restricted to any infinite-dimensional subspace of its domain. It turns out that every bounded operator T on Xω1 has the form D+S, where D is a diagonal-step operator with countably many eigenvalues and where X is a strictly singular operator. It follows that in particular that every bounded linear operator on Xω1 is a multiple of the identity plus the operator with separable range22. It turns out that we can actually identify the space D(Xω1 ) of bounded diagonal operators as the dual of the James space over a particular reflexive Banach space T0 and ω1 as the corresponding ordered set (see [3] for the definition of this functor). One can similarly characterize the operator space of arbitrary closed subspace of

Xω1 as well obtaining thus some interesting new phenomena about operator spaces. For example one can ﬁnd a closed subspace Y of Xω1 such that every bounded operator T : Y → Y is a strictly singular perturbation of a multiple of identity while operator space L(Y,Xω1 ) is quite rich in the sense that

∗ L(Y,Xω1 ) = JT0 ⊕ S(Y,Xω1 ).

Taking the restrictions XI = span({eα : α ∈ I}) to infinite intervals I ⊆ ω1 one obtains an uncountable family of separable reflexive space which, with an appropriate adjustment on ρ, are ’asymptotic versions’ of each other. This is the new phenomenon in the separable theory not revealed by the previous methods. It essential says that any finite-dimensional subspace of one XI can be moved into any other XJ by an operator T such that k T kk T −1 k≤ 4 + ε.

5.4 Question. What is the minimal cardinal θu with property that every weakly null sequence of length θu contains an inﬁnite unconditional subsequence?

22A considerably easier example of a space with this property will be given in some detail in 7.16 below. 44 I. Coherent Sequences

We have seen above that θu ≥ ω2. From a result of Ketonen [39] we infer that θu is not bigger than the ﬁrst Ramsey cardinal. Not much more about θu seems to be known.

6. An ultraﬁlter from a coherent sequence

2 6.1 Deﬁnition. A mapping a :[ω1] −→ ω is coherent if for every α < β < ω1 there exist only ﬁnitely many ξ < α such that a(ξ, α) 6= a(ξ, β), or ∗ 23 in other words, aα = aβ ¹ α . We say that a is nontrivial if there is no ∗ h : ω1 −→ ω such that h ¹ α = aα for all α < ω1.

2 Note that the existence of a coherent and nontrivial a :[ω1] −→ 2 (such as, for example, the function ρ3 deﬁned above) is something that corresponds to the notion of a Hausdorﬀ gap in this context. Notice moreover, that this notion is also closely related to the notion of an Aronszajn tree since ∗ T (a) = {t : α −→ ω : α < ω1 and t = aα} 2 is such an Aronszajn tree whenever a :[ω1] −→ ω is coherent and non- trivial24. In fact, we shall call an arbitrary Aronszajn tree T coherent if T 2 is isomorphic to T (a) for some coherent and nontrivial a :[ω1] −→ ω. In case the range of the map a is actually smaller than ω, e.g. equal to some integer k, then it is natural to let T (a) be the collection of all t : α −→ k ∗ such that α < ω1 and t = aα. This way, we have coherent binary, ternary, etc. Aronszajn trees rather than only ω-ary coherent Aronszajn trees.

2 6.2 Deﬁnition. The support of a map a :[ω1] −→ ω is the sequence supp(aα) = {ξ < α : a(ξ, α) 6= 0} (α < ω1) of subsets of ω1. A set Γ is orthogonal to a if supp(aα) ∩ Γ is ﬁnite for all α < ω1. We say that 2 a :[ω1] −→ ω is nowhere dense if there is no uncountable Γ ⊆ ω1 such that ∗ Γ ∩ α ⊆ supp(aα) for all α < ω1.

Note that ρ3 is an example of a nowhere dense coherent map for the simple reason that ω1 can be covered by countably many sets Λ+n (n < ω) that are orthogonal to ρ3. The following immediate fact shows that ρ3 is 2 indeed a prototype of a nowhere dense and coherent map a :[ω1] −→ ω. 6.3 Proposition. Under the P-ideal dichotomy, for every nowhere dense 2 and coherent map a :[ω1] −→ ω the domain ω1 can be decomposed into countably many sets orthogonal to a. a

23 2 A mapping a :[ω1] −→ ω is naturally identified with a sequence aα (α < ω1), where aα : α −→ ω is defined by aα(ξ) = a(ξ, α). 24Similarities between the notion of a Hausdorff gap and the notion of an Aronszajn tree have been further explained recently in the two papers of Talayco ([72], [73]), where it is shown that they naturally correspond to first cohomology groups over a pair of very similar spaces. 6. An ultrafilter from a coherent sequence 45

2 6.4 Notation. To every a :[ω1] −→ ω associate the corresponding ∆- 2 function ∆a :[ω1] −→ ω as follows:

∆a(α, β) = min{ξ < α : a(ξ, α) 6= a(ξ, β)} with the convention that ∆a(α, β) = α whenever a(ξ, α) = a(ξ, β) for all ξ < α. Given this notation, it is natural to let

∆a(Γ) = {∆a(α, β): α, β ∈ Γ, α < β} for an arbitrary set Γ ⊆ ω1. 6.5 Lemma. Suppose that a is nontrivial and coherent and that every uncountable subset of T (a) contains an uncountable antichain. Then for every pair Σ,Ω of uncountable subsets of ω1 there exists an uncountable subset Γ of ω1 such that ∆a(Γ) ⊆ ∆a(Σ) ∩ ∆a(Ω).

Proof. For each limit ordinal ξ < ω1 we ﬁx α(ξ) ∈ Σ and β(ξ) ∈ Ω above ξ and let

Fξ = {η < ξ : a(η, α(ξ)) 6= a(η, ξ) or a(η, β(ξ)) 6= a(η, ξ)}.

By the Pressing Down Lemma there is a stationary set Γ ⊆ ω1 and a ﬁnite set F such that Fξ = F for all ξ ∈ Γ. Let η0 = max(F ) + 1. By the assumption about T (a), ﬁnd an uncountable Γ0 ⊆ Γ such that aξ (ξ ∈ Γ0) is an antichain of T (a). Moreover, we may assume that for some s, t and u in T (a):

aα(ξ) ¹ η0 = s, aβ(ξ) ¹ η0 = t and aξ ¹ η0 = u for all ξ ∈ Γ0. (I.52)

It follows that for all ξ < η in Γ0:

∆a(α(ξ), α(η)) = ∆a(ξ, η) = ∆a(β(ξ), β(η)). (I.53)

So in particular, ∆a(Γ0) ⊆ ∆a(Σ) ∩ ∆a(Ω). a

2 6.6 Notation. For a :[ω1] −→ ω, set

U(a) = {A ⊆ ω1 : A ⊇ ∆a(Γ) for some uncountable Γ ⊆ ω1}.

By Lemma 6.5, U(a) is a uniform ﬁlter on ω1 for every nontrivial coherent 2 a :[ω1] −→ ω for which T (a) contains no Souslin subtrees. It turns out that under some very mild assumption, U(a) is in fact a uniform ultraﬁlter on ω1.

6.7 Theorem. Under MAω1 , the ﬁlter U(a) is an ultraﬁlter for every non- 2 trivial and coherent a :[ω1] −→ ω. 46 I. Coherent Sequences

Proof. Let A be a given subset of ω1 and let P be the collection of all ﬁnite subsets p of ω1 such that ∆a(p) ⊆ A. If P is a ccc poset then an application of MAω1 would give us an uncountable Γ ⊆ ω1 such that ∆a(Γ) ⊆ A, showing thus that A belongs to U(a). So let us consider the alternative when P is not a ccc poset. Let X be an uncountable subset of P consisting of pairwise incompatible conditions of P. Going to an uncountable ∆-subsystem of X and noticing that the root does not contribute to the incompatibility, one sees that without loss of generality, we may assume that the members of X are in fact pairwise disjoint. Thus for each limit ordinal ξ < ω1 we can ﬁx pξ in X lying entirely above ξ. For ξ < ω1, let

Fξ = {η < ξ : a(η, ξ) 6= a(η, α) for some α ∈ pξ}.

By the Pressing Down Lemma ﬁnd a stationary set Γ ⊆ ω1, a ﬁnite set F ⊆ ω1, an integer n and t0, . . . , tn−1 in T (a) of height η0 = max(F ) + 1 such that for all ξ ∈ Γ: Fξ = F, (I.54)

aα ¹ η0 = ti with α the ith member of pξ for some i < n. (I.55)

By MAω1 the tree T (a) is special, so going to an uncountable subset of Γ we may assume that aξ (ξ ∈ Γ) is an antichain of T (a), and moreover that for every ξ < η in Γ:

aα ¹ ξ * aβ ¹ η for all α ∈ pξ and β ∈ pη. (I.56)

It follows that for every ξ < η in Γ:

∆a(α, β) = ∆a(ξ, η) for all α ∈ pξ and β ∈ pη. (I.57)

By our assumption that conditions from X are incompatible in P, we conclude that for every ξ < η in Γ, there exist α ∈ pξ and β ∈ pη such that ∆a(α, β) ∈/ A. So from (I.57) we conclude that ∆a(Γ) ∩ A = ∅, showing thus that the complement of A belongs to U(a). a

6.8 Remark. One may find Theorem 6.7 a bit surprising in view of the fact that it gives us an ultrafilter U(a) on ω1 that is Σ1-definable over the structure (Hω2 , ∈). It is well-known that there is no ultrafilter on ω that is

Σ1-deﬁnable over the structure (Hω1 , ∈). It turns out that the transformation a 7−→ U(a) captures some of the essential properties of the corresponding and more obvious transformation a 7−→ T (a). To state this we need some standard deﬁnitions.

6.9 Definition. For two trees S and T , by S ≤ T we denote the fact that there is a strictly increasing map f : S −→ T . Let S < T whenever S ≤ T and T S and let S ≡ T whenever S ≤ T and T ≤ S. In general, 6. An ultrafilter from a coherent sequence 47 the equivalence relation ≡ on trees is very far from the finer relation =∼, the isomorphism relation. However, the following fact shows that in the realm of trees T (a), these two relations may coincide and moreover, that the mapping T (a) 7−→ U(a) reduces ≡ and =∼ to the equality relation among ultrafilters on ω1 (see [90]).

6.10 Theorem. Assuming MAω1 , for every pair of coherent and nontrivial 2 2 mappings a :[ω1] −→ ω and b :[ω1] −→ ω, the trees T (a) and T (b) are isomorphic iﬀ T (a) ≡ T (b) iﬀ U(a) = U(b). Proof. We only prove that T (a) ≡ T (b) is equivalent to U(a) = U(b), referring the reader to [90] for the remaining part of the argument. Choose a pair of strictly increasing mappings f : T (a) −→ T (b) and g : T (b) −→ T (a). Replacing f by the mapping t 7−→ f(t) ¹ dom(t) and g by the mapping t 7−→ g(t) ¹ dom(t), we may assume that f and g are also level-preserving. Recall our notation aδ(ξ) = a(ξ, δ) and bδ(ξ) = b(ξ, δ) that gives us particular representatives aδ and bδ of the δth level of T (a) and T (b) respectively. For δ ∈ Λ, set

Fδ = {ξ < δ : aδ(ξ) 6= g(f(aδ))(ξ) or bδ(ξ) 6= f(aδ)(ξ)}.

Then Fδ is a ﬁnite subset of δ. Find a stationary subset Σ of Λ and F ⊆ ω1 such that Fδ = F for all δ ∈ Σ. Find now an uncountable Γ ⊆ Σ such that aδ (δ ∈ Γ) is an antichain of T (a), bδ (δ ∈ Γ) is an antichain of T (b), and the mapping ¯ ¯ ¯ ¯ δ 7−→ (aδ ¹ ξ, f(aδ) ¹ ξ, g(f(aδ)) ¹ ξ, bδ ¹ ξ) is constant on Γ, where ξ¯ = max(F ) + 1. It follows that for γ 6= δ in Γ,

∆(aγ , aδ) = ∆(g(f(aγ )), g(f(aδ))), (I.58)

∆(bγ , bδ) = ∆(f(aγ ), f(aδ)). (I.59)

Since f is strictly increasing, we must have ∆(aγ , aδ) ≤ ∆(f(aγ ), f(aδ)) = ∆(bγ , bδ). Since g is strictly increasing, we have that

∆(bγ , bδ) = ∆(f(aγ ), f(aδ)) ≤ ∆(g(f(aγ )), g(f(aδ))) = ∆(aγ , aδ).

It follows that ∆(aγ , aδ) = ∆(bγ , bδ) for all γ 6= δ in Γ. From the proof of Lemma 6.5 we conclude that ∆a(Ω) (Ω ⊆ Γ, Ω uncountable) generates the filter U(a) and similarly that ∆b(Ω) (Ω ⊆ Γ, Ω uncountable) generates the ultrafilter U(b). It follows that U(a) = U(b). Suppose now that U(a) = U(b). By symmetry, it suffices to show that T (a) ≤ T (b). Note again that the argument in the proof of Lemma 6.5 shows 2 that there is a stationary set Γ ⊆ Λ such that ∆a and ∆b coincide on [Γ] . Let P be the poset of all finite partial strictly increasing level-preserving 48 I. Coherent Sequences mappings p from T (a) into T (b) which (uniquely) extend to strictly increasing level-preserving mappings on the downward-closures of their domains.

By MAω1 , it suﬃces to show that P satisﬁes the countable chain condition. So let pδ (δ ∈ Γ) be a given sequence of conditions of P. For δ ∈ Γ, set

Eδ = {ξ < δ : aδ(ξ) 6= t(ξ) or bδ(ξ) 6= p(t)(ξ)

for some t ∈ dom(pδ) of height ≥ δ}.

Then Eδ is a finite subset of δ for each δ ∈ Γ. Find a stationary set Σ ⊆ Γ and E ⊆ ω1 such that Eδ = E for all δ ∈ Σ. Shrinking Σ, we may assume that the mapping δ 7−→ pδ ¹ δ is constant on Σ. Letη ¯ be the minimal ordinal strictly above max(E) and the height of every node of dom(pδ) ¹ δ for some (all) δ ∈ Σ. For δ ∈ Σ, letp ˆδ be the extension of pδ on the downward-closure of its domain in T (a). Find a stationary Ω ⊆ Σ such that each of the projections δ 7−→ pˆδ ¹ η¯, δ 7−→ aδ ¹ η¯ and δ 7−→ bδ ¹ η¯ is constant on Ω. Find γ < δ in Ω such that aγ 6= aδ ¹ γ, bγ 6= bδ ¹ γ andp ˆγ andp ˆδ are isomorphic via the isomorphism that fixes their common restriction to the η¯th levels of T (a) and T (b). It suffices to check that q = pγ ∪ pδ satisfies

∆(q(x), q(y)) ≥ ∆(x, y) (I.60) for every x ∈ dom(pγ ) and y ∈ dom(pδ). If x and y correspond to each other in the isomorphism betweenp ˆγ andp ˆδ, then either they belong to the common part of pγ and pδ in which case (I.60) follows by the assumption that pγ and pδ are strictly increasing level-preserving, or we have

∆(x, y) = ∆(aγ , aδ) = ∆(bγ , bδ) = ∆(pγ (x), pδ(y)). (I.61)

If x is diﬀerent from the copyy ¯ ∈ dom(pγ ) of y ∈ dom(pδ), then ∆(x, y) = ∆(x, y¯) and ∆(pγ (x), pδ(y)) = ∆(pγ (x), pγ (y)), so we conclude that (I.60) follows from the fact that pγ is strictly increasing level-preserving. This ﬁnishes the proof. a

6.11 Remark. It turns out that the ultraﬁlters of the form U(a) for a : 2 [ω1] −→ ω nontrivial and coherent are all Rudin-Keisler equivalent. The proof of this fact can be found in [90].

2 6.12 Deﬁnition. The shift of a :[ω1] −→ ω is deﬁned to be the mapping (1) 2 a :[ω1] −→ ω determined by

a(1)(α, β) = a(α + 1, βˆ), where βˆ = min{λ ∈ Λ: λ ≥ β}. The n-fold iteration of the shift operation is deﬁned recursively by the formula a(n+1) = (a(n))(1). 6. An ultraﬁlter from a coherent sequence 49

The following fact about this shift operation, together with Lemma 4.44, can be used to show that Aronszajn trees are not well-quasi-ordered under the quasi-ordering ≤ (see [90]).

6.13 Theorem. If a is nontrivial, coherent and orthogonal to Λ, then it holds that T (a) > T (a(1)).

Proof. By the assumption a ⊥ Λ, the subtree S of T (a) consisting of all t’s that take the value 0 at every limit ordinal in their domains is an uncountable initial part of T (a). Note that every node of T (a) is equal to the restriction of some ﬁnite change of a node from S. For a node t at some limit level λ of S, let t(1) : λ −→ ω be determined by the formula t(1)(α) = t(α + 1). Let S(1) = {t(1) : t ∈ S ¹ λ}. Every node of T (a(1)) is equal to the restriction of some ﬁnite change of a node of S(1). Note also that t 7−→ t(1) is a one-to-one map on S ¹ Λ and that its inverse naturally extends to a strictly increasing map from T (a(1)) into T . This shows that T (a(1)) ≤ T (a). It remains to establish T (a) T (a(1)). Note that if there is a strictly increasing g : T (a) −→ T (a(1)), then there is one that is moreover level- preserving. For example f(t) = g(t) ¹ |t| is one such map. So let f : T (a) −→ T (a(1)) be a given level-preserving map. For a countable limit ordinal ξ, set

(1) Fξ = (supp(aξ) ∩ Λ) ∪ {α < ξ : aξ (α) 6= f(aξ)(α)}. (I.62)

Then Fξ is a ﬁnite set of ordinals < ξ, so by the Pressing Down Lemma there is a stationary set Γ ⊆ Λ and F ⊆ ω1 such that Fξ = F for all ξ ∈ Γ. Let α0 = max(F ) + 1. Shrinking Γ if necessary, assume that for some s, t : α0 −→ 2,

aξ ¹ α0 = s and f(aξ) ¹ α0 = t for all ξ ∈ Γ. (I.63)

Choose ξ < η in Γ such that aξ 6= aη ¹ ξ and let δ = ∆(aξ, aη)(= min{α < ξ : aξ(α) 6= aη(α)}). Then by (I.62) and (I.63), δ is not a limit ordinal and

(1) (1) δ − 1 = ∆(aξ , aη ) = ∆(f(aξ), f(aη)). (I.64)

It follows that aξ and aη split in T (a) after their images f(aξ) and f(aη) split in T (a(1)), which cannot happen if f is strictly increasing. This ﬁnishes the proof. a

6.14 Corollary. If a is nontrivial, coherent and orthogonal to Λ + n for all n < ω, then T (a(n)) > T (a(m)) whenever n < m < ω.

Proof. Note that if a is orthogonal to Λ + n for all n < ω, then so is every of its ﬁnite shifts a(m). a 50 I. Coherent Sequences

(n) (m) 6.15 Corollary. T (ρ3 ) > T (ρ3 ) whenever n < m < ω. a Somewhat unexpectedly, with very little extra assumptions we can say much more about ≤ in the domain of coherent Aronszajn trees (see [90]).

6.16 Theorem. Under MAω1 , the family of coherent Aronszajn trees is totally ordered under ≤. Proof. Given two nontrivial coherent sequences a and b, we need to show that either T (a) ≤ T (b) or T (b) ≤ T (a). It turns out that these two alternatives correspond to the following two cases. Case 1: There is uncountable Ω ⊆ ω1 such that ∆a is dominated by ∆b on [Ω]2. Let P be the poset of ﬁnite partial level-preserving functions p from T (a) into T (b) that can be extended to strictly increasing level-preserving maps from Dˆ p into T (b), where Dˆ p is the downward closure of the domain Dp of p. Let Rp denote the range of p and Rˆp its downward closure in T (b).

If P is a ccc poset, then applying MAω1 to P and the natural sequence of dense open subsets of P would give us the desired strictly increasing level-preserving map from T (a) into T (b). So let pξ (ξ < ω1) be a given one-to-one sequence of members of P. We may assume that the pξ’s form a ∆-system with root r. Let Dξ = Dpξ \ Dr and Rξ = Rpξ \ Rr for ξ < ω1. For limit ξ ﬁx g(ξ) < ξ and h(ξ) ∈ Ω \ ξ such that for every t ∈ Dξ with height ≥ ξ:

t ≡ aξ ≡ ah(ξ) on [g(ξ), ξ), (I.65)

pξ(t) ≡ bξ ≡ bh(ξ) on [g(ξ), ξ). (I.66)

By the Pressing Down Lemma, there is a stationary Σ ⊆ ω1 such that g is constantly equal to η0 on Σ and moreover that the projection maps of Dξ ∪ {aξ, ah(ξ)} onto the η0th level of T (a) and Rξ ∪ {bξ, bh(ξ)} onto the η0th level of T (b) are constant as long as ξ belongs to Σ. Moreover we may assume for ξ 6= η, every node of Dξ ∪ {aξ, ah(ξ)} is incomparable to every node of Dη ∪ {aη, ah(η)}, and similarly, every node of Rξ ∪ {bξ, bh(ξ)} is incomparable to every node of Rη ∪ {bη, bh(η)}. Note that we may also assume that every node of Dξ, for ξ ∈ Σ, has height ≥ ξ. So pick ξ 6= η in Σ. Then for all s ∈ Dξ and t ∈ Dη:

∆(s, t) = ∆(ah(ξ), ah(η)) = ∆a(h(ξ), h(η)), (I.67)

∆(pξ(s), pη(t)) = ∆(bh(ξ), bh(η)) = ∆b(h(ξ), h(η)). (I.68) 2 Since {h(ξ), h(η)} belongs to [Ω] , we conclude that for all s ∈ Dξ and t ∈ Dη, s and t split before pξ(s) and pη(t). So pξ ∪ pη extends to a strictly increasing level-preserving function from the downward closure of Dpξ ∪Dpη into the downward closure of Rpξ ∪ Rpη . This ﬁnishes the proof that P is a ccc forcing relation. 7. The trace and the square-bracket operation 51

2 Case 2: For every uncountable Ω ⊆ ω1 there exists {α, β} in [Ω] such that ∆a(α, β) > ∆b(α, β). Let Q be the poset of all ﬁnite subsets q of ω1 2 such that ∆a dominates ∆b on [q] . If Q is a ccc poset, then an application of

MAω1 would give us an uncountable subset Ω of ω1 such that ∆a dominates 2 ∆b on [Ω] , so by Case 1 we would conclude that T (b) ≤ T (a). To show that Q is a ccc poset, let qξ (ξ < ω1) be a given one-to-one sequence of members of Q. We may assume the qξ’s form a ∆-system. Note that the root does not contribute to the incompatibility of two conditions of the ∆-system and so removing it, we assume that all qξ’s are in fact pairwise disjoint. For a limit ξ < ω1, ﬁx g(ξ) < ξ such that for all α ∈ qξ \ ξ:

aα ≡ aξ and bα ≡ bξ on [g(ξ), ξ). (I.69)

Applying the Pressing Down Lemma and working as above, we ﬁnd a stationary set Ω ⊆ ω1 such that for all ξ 6= η in Ω, and α ∈ qξ, β ∈ qη:

∆a(α, β) = ∆a(ξ, η) and ∆b(α, β) = ∆b(ξ, η). (I.70)

By the assumption of Case 2, we can ﬁnd ξ 6= η in Ω such that ∆a(ξ, η) > ∆b(ξ, η). It follows that ∆a dominates ∆b on pξ ∪ pη. This completes the proof that Q is a ccc poset and therefore the proof of Theorem 6.16. a

6.17 Remark. While under MAω1 , the class of coherent Aronszajn trees is totally ordered by ≤, Corollary 6.15 gives us that this chain of trees is not well-ordered. This should be compared with an old result of Ohkuma [56] that the class of all scattered trees is well-ordered by ≤ (see also [47]). It turns out that the class of all Aronszajn trees is not totally ordered under ≤, i.e. there exist Aronszajn trees S and T such that S T and T S. The reader is referred to [90] for more information on this and other related results that we chose not to reproduce here.

7. The trace and the square-bracket operation

Recall the notion of a minimal walk from a countable ordinal β to a smaller ordinal α along the ﬁxed C-sequence Cξ (ξ < ω1): β = β0 > β1 > ··· >

βn = α where βi+1 = min(Cβi \ α). Recall also the notion of a trace

Tr(α, β) = {β0, β1, ··· , βn}, the ﬁnite set of places visited in the minimal walk from β to α. The following simple fact about the trace lies at the heart of all known deﬁnitions of square-bracket operations not only on ω1 but also at higher cardinalities.

7.1 Lemma. For every uncountable subset Γ of ω1 the union of Tr(α, β) for α < β in Γ contains a closed and unbounded subset of ω1. 52 I. Coherent Sequences

Proof. It suﬃces to show that the union of traces contains every countable limit ordinal δ such that sup(Γ ∩ δ) = δ. Pick an arbitrary β ∈ Γ \ δ and let

β = β0 > β1 > ··· > βk = δ be the minimal walk from β to δ. Let γ < δ be an upper bound of all sets of the form Cβi ∩ δ for i < k. By the choice of δ there is α ∈ Γ ∩ δ above γ. Then the minimal walk from β to α starts as β0 > β1 > ··· > βk, so in particular δ belongs to Tr(α, β). a We shall now see that it is possible to pick a single place [αβ] in Tr(α, β) so that Lemma 7.1 remains valid with [αβ] in place of Tr(α, β). Recall that by Lemma 1.14,

∆(α, β) = min{ξ ≤ α : ρ0(ξ, α) 6= ρ0(ξ, β)} is a successor ordinal. We shall be interested in its predecessor,

7.2 Deﬁnition. ∆0(α, β) = ∆(α, β) − 1.

Thus, if ξ = ∆0(α, β), then ρ0(ξ, α) = ρ0(ξ, β) and so there is a natural isomorphism between Tr(ξ, α) and Tr(ξ, β). We shall deﬁne [αβ] by comparing the three sets Tr(α, β), Tr(ξ, α) and Tr(ξ, β).

7.3 Deﬁnition. The square-bracket operation on ω1 is deﬁned as follows:

[αβ] = min(Tr(α, β) ∩ Tr(∆0(α, β), β))

= min(Tr(∆0(α, β), β) \ α).

2 <ω Next, recall the function ρ0 :[ω1] → ω defined from the C-sequence Cξ (ξ < ω1) and the corresponding tree T (ρ0). For γ < ω1 let (ρ0)γ be the <ω fiber-mapping : γ → ω defined by (ρ0)γ (α) = ρ0(α, γ).

7.4 Lemma. For every uncountable subset Γ of ω1 the set of all ordinals of the form [αβ] for some α < β in Γ contains a closed and unbounded subset of ω1.

Proof. For t ∈ T (ρ0) let

Γt = {γ ∈ Γ:(ρ0)γ end-extends t}.

Let S be the collection of all t ∈ T (ρ0) for which Γt is uncountable. Clearly, S is a downward closed uncountable subtree of T . The Lemma is proved once we prove that every countable limit ordinal δ > 0 with the following two properties can be represented as [αβ] for some α < β in Γ:

sup(Γt ∩ δ) = δ for every t ∈ S of length < δ, (I.71) 7. The trace and the square-bracket operation 53

every t ∈ S of length < δ has two incomparable successors (I.72) in S both of length < δ.

Fix such a δ and choose β ∈ Γ \ δ such that (ρ0)β ¹ δ ∈ S and consider the minimal walk from β to δ:

β = β0 > β1 > ··· > βk = δ.

Let γ < δ be an upper bound of all sets of the form Cβi ∩ δ for i < k. Since the restriction t = (ρ0)β ¹ γ belongs to S, by (I.72) we can ﬁnd one of its end-extensions s in S which is incomparable with (ρ0)β. It follows that for α ∈ Γs, the ordinal ∆0(α, β) has the ﬁxed value

ξ = min{ξ < |s| : s(ξ) 6= ρ0(ξ, β)} − 1.

Note that ξ ≥ γ, so the walk β → δ is a common initial part of walks β → ξ and β → α for every α ∈ Γs ∩ δ. Hence if we choose α ∈ Γs ∩ δ above min(Cδ \ ξ) (which we can by (I.71)), we get that the walks β → ξ and β → α never meet after δ. In other words for any such α, the ordinal δ is the minimum of Tr(α, β) ∩ Tr(ξ, β), or equivalently δ is the minimum of Tr(ξ, β) \ α. a

It should be clear that the above argument can easily be adjusted to give us the following slightly more general fact about the square-bracket operation.

7.5 Lemma. For every uncountable family A of pairwise disjoint finite subsets of ω1, all of the same size n, the set of all ordinals of the form [a(1)b(1)] = [a(2)b(2)] = ··· = [a(n)b(n)] 25 for some a 6= b in A contains a closed and unbounded subset of ω1. a It turns out that the square-bracket operation can be used in constructions of various mathematical objects of complex behaviour where all known previous constructions needed the Continuum Hypothesis or stronger enumeration principles. The usefulness of [··] in these constructions is based on the fact that [··] reduces the quantification over uncountable subsets of ω1 to the quantification over closed unbounded subsets of ω1. For example composing [··] with a unary operation ∗ : ω1 −→ ω1 which takes each of the values stationary many times one gets the following fact about the mapping c(α, β) = [αβ]∗.

2 7.6 Theorem. There is a mapping c :[ω1] −→ ω1 which takes all the 2 values from ω1 on any square [Γ] of some uncountable subset Γ of ω1. a

25For a ﬁnite set x of ordinals of size n we use the notation x(1), x(2), . . . , x(n) or x(0), x(1), . . . , x(n − 1), depending on the context, for the enumeration of x according to the natural ordering on the ordinals. 54 I. Coherent Sequences

This result gives a deﬁnitive limitation on any form of a Ramsey Theorem for the uncountable which tries to say something about the behaviors of restrictions of colorings on an uncountable square. Recall that in Section 3.1 we have seen that there are mappings of this sort with complex behavior on sets of the form

Γ0 ⊗ Γ1 = {{α0, α1} : α0 ∈ Γ0, α1 ∈ Γ1, α0 < α1}, where Γ0 and Γ1 are two uncountable subsets of ω1. It turns out that the square bracket operation has a simple behavior on Γ0 × Γ1 for every pair Γ0 and Γ1 of suﬃciently thin uncountable subsets of ω1. This can be made precise as follows. 7.7 Lemma. For every uncountable family A of pairwise disjoint subsets of ω1, all of the same size n, there exist a partition I1,...,Ik of {1, . . . , n}, an uncountable B ⊆ A, and an hb : n × n −→ ω1 for each b ∈ B such that 26 for all a < b in B, [a(p)b(q)] = hb(p, q) whenever p ∈ Ii, q ∈ Ij and (i, j) ∈ k × k \ ∆27. Moreover the set of ordinals δ for which there exist a < b in B such that [a(p)b(q)] = δ for all p, q ∈ Ii and all i = 1, . . . , k contains a closed and unbounded subset of ω1.

Proof. Identifying an ordinal α with (ρ0)α, we can think of A as a family of n-tuples of nodes of the tree T (ρ0). For a limit ordinal δ < ω1, choose aδ in A such that every node from aδ has height ≥ δ. Let cδ be the projection of aδ on the δth level of T (ρ0). Let h(δ) < δ be such that the projection of cδ on the h(δ)th level has size equal to the size of cδ, i.e.

∆(s, t) < h(δ) for all s 6= t in cδ. (I.73)

Find a stationary set Γ ⊆ ω1 such that h takes a constant value γ0 on Γ and that the γ0th projection takes the constant value on cδ (δ ∈ Γ). Fix δ < ε in Γ and consider α ∈ aδ and β ∈ aε that project to different elements of the γ0th level of T (ρ0). Then ∆(α, β) is independent of α and β or δ and ε as it is equal to ∆((ρ0)α ¹ γ0, (ρ0)β ¹ γ0). So the trace Tr(∆0(α, β), β) depends only on β and the projection (ρ0)α ¹ γ0. Thus, going to a subset of Γ, we may assume that its size depends only on the position of β in some aε and that the minimal point of Tr(∆0(α, β), β) \ α is the same for all α in some aδ, for δ < ε, having a fixed projection on the γ0th level of T (ρ0). This is the content of the first part of the conclusion of Lemma 7.7. To see the second part of the conclusion, we shrink Γ even further to assure that the union of cδ (δ ∈ Γ) is an antichain of T (ρ0). Now note that for two 0 0 0 0 typical δ < ε in Γ, [αβ] = [α β ] for all α, α ∈ aδ and β, β ∈ aε with the property that (ρ0)α ¹ γ0 = (ρ0)α0 ¹ γ0 = (ρ0)β ¹ γ0 = (ρ0)β0 ¹ γ0. So the rest of the proof reduces to the proof of Lemma 7.5 above. a 26For a pair a and b of sets of ordinals, a < b denotes the fact that α < β for all α ∈ a and β ∈ b. 27By ∆, we denote the diagonal of k × k. 7. The trace and the square-bracket operation 55

7.8 Remark. To see the point of Lemma 7.7, consider the projection [αβ]∗ of the square-bracket operation generated by a mapping ∗ from ω1 onto ω ∗ rather than ω1. Then the sequence hb : n×n −→ ω (b ∈ B) of corresponding compositions will have the same constant value h : n × n −→ ω modulo, of course, shrinking B to some uncountable subset. Noting that the proof of Lemma 7.7 applied to a typical A will actually result in the partition Ii = {i} (i = 1, . . . , n), we conclude that if α and β are coming from two ∗ diﬀerent positions of some aδ and aε in B, then [αβ] will depend on the positions themselves rather than α and β. 7.9 Remark. It turns out that there is a natural variation of the square- bracket operation (see the mapping b of [78, p.287]), where, in Lemma 7.7, we always have the ﬁnest possible partition Ii = {i} (i = 1, . . . , n) for every uncountable family A of pairwise disjoint n-tuples of countable ordinals. This property of the projection of the square-bracket operation has an interesting application in the Quadratic Form Theory, given by Baumgartner- Spinas [6] and Shelah-Spinas [67].

Note that the basic C-sequence Cα (α < ω1) which we have fixed at the beginning of this chapter can be used to actually define a unary operation ∗ : ω1 −→ ω1 which takes each of the ordinals from ω1 stationarily many times. So the projection [αβ]∗ can actually be defined in our basic structure

(ω1, ω, C~ ). We are now at the point to see that our basic structure is actually rigid.

7.10 Lemma. The algebraic structure (ω1, [··], ∗) has no nontrivial automorphisms.

Proof. Let h be a given automorphism of (ω1, [··], ∗). If the set Γ of ﬁxed points of h is uncountable, h must be the identity map. To see this, consider ∗ a ξ < ω1. By the property of the map c(α, β) = [αβ] stated in Theorem 7.6 there exist γ < δ in Γ such that [γδ]∗ = ξ. Applying h to this equation we get

h(ξ) = h([γδ]∗) = (h([γδ]))∗ = [h(γ)h(δ)]∗ = [γδ]∗ = ξ.

It follows that ∆ = {δ < ω1 : h(δ) 6= δ} is in particular uncountable. Shrinking ∆ and replacing h by h−1, if necessary, we may safely assume that h(δ) > δ for all δ ∈ ∆. Consider a ξ < ω1 and let Sξ be the set of all ∗ α < ω1 such that α = ξ. By our choice of ∗ the set Sξ is stationary. By Lemma 7.5 applied to the family A = {{δ, h(δ)} : δ ∈ ∆} we can ﬁnd γ < δ in ∆ such that [γδ] = [h(γ)h(δ)] belongs to Sξ, or in other words, [γδ]∗ = [h(γ)h(δ)]∗ = ξ.

Since [h(γ)h(δ)]∗ = h([γδ]∗) we conclude that h(ξ) = ξ. Since ξ was an arbitrary countable ordinal, this shows that h is the identity map. a 56 I. Coherent Sequences

We give now an application of this rigidity result to a problem in Model Theory about the quantiﬁer Qx =’there exist uncountably many x’ and its n higher dimensional analogues Q x1 ··· xn =’there exist an uncountable n- cube many x1, ··· , xn’. By a result of Ebbinghaus and Flum [19] (see also [57]) every model of every sentence of L(Q) has nontrivial automorphisms. However we shall now see that this is no longer true about the quantiﬁer Q2.

7.11 Example. The sentence φ will talk about one unary relation N, one binary relation < and two binary functional symbols C and E. It is the conjunction of the following seven sentences

(φ1) Qx x = x,

(φ2) ¬Qx N(x),

(φ3) < is a total ordering,

(φ4) E is a symmetric binary operation,

(φ5) ∀x < y N(E(x, y)),

(φ6) ∀x < y < z E(x, z) 6= E(y, z),

2 0 0 0 0 0 0 (φ7) ∀x∀n{N(n) → ¬Q uv[∃u < u∃v < v(u 6= v ∧ E(u , u) = E(v , v) = n) ∧ ∀u0 < u ∀v0 < v(E(u0, u) = E(v0, v) = n → (C(u0, v0) 6= x ∨ C(u, v) 6= x))]}.

The model of φ that we have in mind is the model (ω1, ω, <, c, e) where ∗ 2 c(α, β) = [αβ] and e :[ω1] −→ ω is any mapping such that e(α, γ) 6= e(β, γ) whenever α < β < γ (e.g. we can take e =ρ ¯1 or e =ρ ¯). The sentence φ7 is simply saying that for every ξ < ω1 and every uncountable family A of pairwise disjoint unordered pairs of countable ordinals there exist a 6= b in A such that

c(min a, min b) = c(max a, max b) = ξ.

∗ This is a consequence of Lemma 7.5 and the fact that Sξ = {α : α = ξ} is a stationary subset of ω1. These are the properties of [··] and ∗ which we have used in the proof of Lemma 7.10 in order to prove that (ω1, [··], ∗) is a rigid structure. So a quite analogous proof will show that any model (M, N, <, C, E) of φ must be rigid. a

The crucial property of [··] stated in Lemma 7.5 can also be used to provide a negative answer to the basis problem for uncountable graphs by constructing a large family of pairwise orthogonal uncountable graphs. 7. The trace and the square-bracket operation 57

7.12 Deﬁnition. For a subset Γ of ω1, let GΓ be the graph whose vertex-set is ω1 and whose edge-set is equal to

{{α, β} :[αβ] ∈ Γ}.

7.13 Lemma. If the symmetric diﬀerence between Γ and ∆ is a stationary subset of ω1, then the corresponding graphs GΓ and G∆ are orthogonal to each other, i.e. they do not contain uncountable isomorphic subgraphs. a

We have seen above that comparing [··] with a map π : ω1 −→ I where I is some set of mathematical objects/requirements in such a way that each object/requirement is given a stationary preimage, gives us a way to meet each of these objects/requirements in the square of any uncountable subset of ω1. This observation is the basis of all known applications of the square-bracket operation. A careful choice of I and π : ω1 −→ I gives us a projection of the square-bracket operation that can be quite useful. So let us illustrate this on yet another example.

n 7.14 Deﬁnition. Let H be the collection of all maps h : 2 −→ ω1 where n is a positive integer denoted by n(h). Choose a mapping π : ω1 −→ H which takes each value from H stationarily many times. Choose also a one- ω to-one sequence rα (α < ω1) of elements of the Cantor set 2 . Note that both these objects can actually be deﬁned in our basic structure (ω1, ω, C~ ). Consider the following projection of the square-bracket operation:

[[αβ]] = π([αβ])(rα ¹ n(π([αβ]))).

It is easily checked that the property of [··] stated in Lemma 7.5 corresponds to the following property of the projection [[αβ]]: 7.15 Lemma. For every uncountable family A of pairwise disjoint ﬁnite subsets of ω1, all of the same size n, and for every n-sequence ξ1, . . . , ξn of countable ordinals there exist a and b in A such that [[a(i)b(i)]] = ξi for i = 1, . . . , n. a This projection of [··] leads to an interesting example of a Banach space with ’few’ operators, which we will now describe. 7.16 Example. A nonseparable reﬂexive Banach space E with the property that every bounded linear operator T : E −→ E can be expressed as T = λI + S where λ is a scalar, I the identity operator of E, and S an operator with separable range. <ω Let I = 3×[ω1] and let us identify the index-set I with ω1, i.e. pretend that [[··]] takes its values in I rather than ω1. Let [[··]]0 and [[··]]1 be the two projections of [[··]]. 58 I. Coherent Sequences

<ω 2 G = {G ∈ [ω1] : [[αβ]]0 = 0 for all {α, β} ∈ [G] }, <ω 2 H = {H ∈ [ω1] : [[αβ]]0 = 1 for all {α, β} ∈ [H] }.

Let K be the collection of all ﬁnite sets {{αi, βi} : i < k} of pairs of countable ordinals such that for all i < j < k:

max{αi, βi} < min{αj, βj}, (I.74)

[[αiαj]]0 = [[βiβj]]0 = 2, (I.75)

[[αiαj]]1 = [[βiβj]]1 = {αl : l < i} ∪ {βl : l < i}. (I.76) The following properties of G, H and K should be clear:

G and H contain all the singletons, are closed under subsets and they are 1-orthogonal to each other in the sense that (I.77) G ∩ H contains no doubleton. G and H are both 2-orthogonal to the family of the unions (I.78) of members of K. If K and L are two distinct members of K, then there are no more than 5 ordinals α such that {α, β} ∈ K and {α, γ} ∈ L (I.79) for some β 6= γ.

For every sequence {αξ, βξ} (ξ < ω1) of pairwise disjoint pairs of countable ordinals there exist arbitrarily large ﬁnite (I.80) sets Γ, ∆ ⊆ ω1 such that {αξ : ξ ∈ Γ} ∈ G, {βξ : ξ ∈ Γ} ∈ H and {{αξ, βξ} : ξ ∈ ∆} ∈ K.

For a function x from ω1 into R, set X 2 1 kxkH,2 = sup{( x(α) ) 2 : H ∈ H}, α∈H X 2 1 kxkK,2 = sup{( (x(α) − x(β)) ) 2 : K ∈ K}. {α,β}∈K

Let k · k = max{k · k∞, k · kH,2, k · kK,2} and deﬁne E¯2 = {x : kxk < ∞}. Let 1α be the characteristic function of {α}. Finally, let E2 be the closure of the linear span of {1α : α ∈ ω1} inside (E¯2, k · k). The following facts about the norm k · k are easy to establish using the properties of the families G, H and K listed above

If x is supported by some G ∈ G, then kxk ≤ 2 · kxk∞. (I.81) [ If x is supported by K for some K in K, then kxk ≤ 10 · kxk∞. (I.82) 7. The trace and the square-bracket operation 59

The role of the seminorm k·kH,2 is to ensure that every bounded operator T : E2 −→ E2 can be expressed as D+S, where D is a diagonal operator relative 28 to the basis 1α (α < ω1) and where S has separable range. Namely, if this were not so, we could ﬁnd a sequence {αξ, βξ} (ξ < ω1) of pairwise disjoint pairs from ω1 such that T (1αξ )(βξ) 6= 0 for all ξ < ω1. Thinning the sequence, we may assume that for some ﬁxed r > 0 and all ξ < ω1,

|T (1αξ )(βξ)| ≥ r. By (I.80) for every n < ω we can ﬁnd a subset Γ of ω1 of size n such that G = {αξ : ξ ∈ Γ} ∈ G and H = {βξ : ξ ∈ Γ} ∈ H. P 1 √ 2 2 Then, by (I.81), 2 · kT k ≥ kT (χG)k ≥ ( β∈H T (χG)(β) ) ≥ r · n. Since n is arbitrary, this contradicts the fact that T is bounded. The role of k · kK,2 is to represent every bounded diagonal operator D(x)(ξ) = λξx(ξ) as λI + S for some scalar λ and operator S with separable range. This reduces to the fact that the sequence λξ (ξ < ω1) is eventually constant. If this were false, we would be able to extract a sequence {αξ, βξ} (ξ < ω1) of pairwise disjoint pairs of countable ordinals and an r > 0 such that

|λαξ − λβξ | ≥ r for all ξ < ω1. By (I.80), for every n < ω, we can ﬁnd a subset ∆ of ω1 of size n such that K = {{αξ, βξ} : ξ ∈ ∆} belongs to K. Then, by (I.82),

S 10 · kDk ≥ kD(χ K )k X S S 2 1 ≥ ( (D(χ K )(αξ) − D(χ K )(βξ)) ) 2 ξ∈∆ X 1 2 2 = ( (λαξ − λβξ ) ) ξ∈∆ √ ≥ r · n.

Since n was arbitrary this contradicts the fact that D is a bounded operator. Note that kxk ≤ 2kxk2 for all x ∈ `2(ω1). It follows that `2(ω1) ⊆ E2 and the inclusion is a bounded linear operator. Note also that `2(ω1) is a dense subset of E2. Therefore E2 is a weak compactly generated space. For example, W = {x ∈ `2(ω1): kxk2 ≤ 1} is a weakly compact subset of E2 and its linear span is dense in E2. To get a reﬂexive example out of E2 one uses an interpolation method of Davis, Figiel, Johnson and Pelczynski [11] as n −n 29 follows. Let pn be the Minkowski functional of the set 2 W +2 Ball(E2). Let X∞ 2 1 E = {x ∈ E2 : kxkE = ( pn(x) ) 2 < ∞}. n=0

28 Indeed it can be shown that 1α (α < ω1) is a ’transﬁnite basis’ of E2 in the sense of [70]. So every vector x of E2 has a unique representation as Σα<ω1 x(α)1α and the projection operators Pβ : E2 → E2 ¹ β (β < ω1) are uniformly bounded. 29 I.e. pn(x) = inf{λ > 0 : x ∈ λB}, where B denotes this set. 60 I. Coherent Sequences

By [11, Lemma 1], E is a reﬂexive Banach space and `2(ω1) ⊆ E ⊆ E2 are continuous inclusions. Note that pn(x) < r iﬀ x = y + z for some y ∈ E2 −n n and z ∈ `2(ω1) such that kyk < 2 r and kzk2 < 2 r. The following two facts about E correspond to the facts (I.81) and (I.82) and they are used in the proof that every bounded operator T : E −→ E has the form λI + S in a very similar way.

If x is the vector supported by some G ∈ G which can be split into n blocks such that the k-th one has size 24k and (I.83) x takes the value 2−2k on each term of the k-th block, then kxkE ≤ 2.

If {{αξ, βξ} : ξ ∈ ∆} is a member of K so that ∆ can be split into n blocks such that the k-th one ∆ has size 24k S k (I.84) and if x is the vector supported by K for ξ ∈ ∆k, x(αξ) = −2k −2k 2 and x(βξ) = 2 , then kxkE ≤ 12. We leave the checking of (I.83) and (I.84) as well as the corresponding proof that E has ’few’ operators to the interested reader. a

7.17 Remark. The above example is reproduced from Wark [94] who based his example on a previous construction due to Shelah and Steprans [68]. The reader is referred to these sources for more information.

We only mention yet another interesting application of the square-bracket operation, given recently by Erd˝os,Jackson and Mauldin [21]:

7.18 Example. For every positive integer n there exist collections H and X of hyperplanes and points of Rn, respectively, and a coloring P : H −→ ω such that: any n hyperplanes of distinct colors meet in at most one (I.85) point, and

there is no coloring Q : X −→ ω such that for every H ∈ H there exist at most n − 1 points x in X ∩ H such that (I.86) Q(x) = P (H). Finally we mention yet another projection of the square-bracket operation.

n n 7.19 Deﬁnition. Let H now be the collection of all maps h : 2 ×2 −→ ω1 where n = n(h) < ω and let π be a mapping from ω1 onto H that takes each of the values stationarily many times. Deﬁne a new operation on ω1 by |αβ| = π([αβ])(rα ¹ n(π([αβ])), rβ ¹ n(π([αβ]))). (I.87) 7. The trace and the square-bracket operation 61

7.20 Lemma. For every positive integer n, every uncountable subset Γ of ω1 and every symmetric n × n-matrix M of countable ordinals there is a one-to-one φ : n −→ Γ such that |φ(i)φ(j)| = M(i, j) for i, j < n.

Proof. Let Γ be an uncountable subset of ω1 and let M be a given n × n matrix on countable ordinals. Then for each α we can find tα in the αth level of the tree T (ρ0) and a finite set Fα ⊆ Γ \ α of size n such that tα ⊆ (ρ0)γ for all γ ∈ Fα. Let nα be the minimal integer such that rγ ¹ nα 6= rδ ¹ nα whenever γ 6= δ are members of Fα. For a limit ordinal α let ξ(α) < α be an upper bound of all sets of the form Cγi ∩ α where γ ∈ Fα, i < k < ω and γ = γ0 > . . . > γk = α is the minimal walk from γ to α along the C- sequence fixed at the beginning of this chapter. Then there is a stationary m set ∆ ⊆ ω1, m < ω, s0, . . . , sn−1 ∈ 2 and ξ < ω1 such that:

nα = m and ξα = ξ for all α ∈ ∆, (I.88)

tα (α ∈ ∆) form an antichain of T (ρ0), (I.89) r ¹ m = s whenever i < n and γ occupies the ith place in γ i (I.90) some Fα for α ∈ ∆. m m Let h : 2 × 2 −→ ω1 be an arbitrary map subject to the requirement that h :(si, sj) = M(i, j) for all i, j < n. Let Σh = {δ < ω1 : π(δ) = h}. By the choice of π the set Σh is stationary in ω1. By the proof of the basic property 7.4 of the square bracket operation we can ﬁnd ν ∈ Σh, α ∈ ∆ ∩ ν and β ∈ ∆ \ ν such that tα ¹ ξ = tβ ¹ ξ, Fα ⊆ ν and [αβ] = ν. We claim that [γδ] = ν for all γ ∈ Fα and δ ∈ Fβ. (I.91)

To see this, ﬁrst note that ∆0(γ, δ) = ∆0(α, β) = σ for all γ ∈ Fα and δ ∈ Fβ. By the choice of ξ and the fact that σ ≥ ξ the walk from some δ ∈ Fβ to σ goes through β, so for any γ ∈ Fα and δ ∈ Fβ we have [γδ] = min(Tr(σ, δ) \ γ) = min(Tr(σ, β) \ α) = [αβ].

In fact the proof of Lemma 7.4 shows that we can ﬁnd ν0 ∈ Σh, α0 ∈ ∆ ∩ ν and an uncountable set ∆1 ⊆ ∆ \ ν0 such that Fα0 ⊆ ν0, tα0 ¹ ξ = tβ ¹ ξ and [α0β] = γ0 for all β ∈ ∆1. It follows that

[γδ] = ν0 for all γ ∈ Fα0 , δ ∈ Fβ and β ∈ ∆1. (I.92)

So we can repeat the procedure for ∆1 in place of ∆ and get ν1 ∈ Σh, α1 ∈ ∆1 ∩ ν0 and uncountable ∆2 ⊆ ∆1 \ γ1 such that [α1β] = ν1 for all β ∈ ∆2, and so on. Repeating this n times we get a sequence α0, . . . , αn−1 such that for all i < j < n,

π([γδ]) = h for all γ ∈ Fαi and δ ∈ Fβ with β ∈ ∆j. (I.93)

Deﬁne φ : n −→ ω1 by letting φ(i) be the ith member of Fαi . Going back to the deﬁnition of the projection | · ·| of [··] one sees that indeed |φ(i)φ(j)| = M(i, j) holds for all i < j < n. a 62 I. Coherent Sequences

8. A square-bracket operation on a tree

In this section we try to show that the basic idea of the square-bracket operation on ω1 can perhaps be more easily grasped by working on an arbitrary special Aronszajn tree rather than T (ρ0). So let T = hT,

Fn(t) = {s ≤T t : s = t or a(s) ≤ n}.

Finally, for s, t ∈ T with ht(s) ≤ ht(t), let

[st]T = min{v ∈ Fa(s∧t)(t): ht(v) ≥ ht(s)}.

(If ht(s) ≥ ht(t) we let [st]T = [ts]T .) The following fact corresponds to Lemma 7.4 when T = T (ρ0).

8.1 Lemma. If X is an uncountable subset of T , the set of nodes of T of the form [st]T for some s, t ∈ X intersects a closed and unbounded set of levels of T . a

We do not give a proof of this fact as it is almost identical to the proof of Lemma 7.4 which deals with the special case T = T (ρ0). But one can go further and show that [··]T shares all the other properties of the square-bracket operation [··] described in the previous section. Some of these properties, however, are easier to visualize and prove in the general context. For example, consider the following fact which in the case T = T (ρ0) is the essence of Lemma 7.20.

8.2 Lemma. Suppose A ⊆ T is an uncountable antichain and that for each t ∈ A be given a ﬁnite set Ft of its successors. Then for every stationary set Γ ⊆ ω1 there exists an arbitrarily large ﬁnite set B ⊆ A such that the height of [xy]T belongs to Γ whenever x ∈ Fs and y ∈ Ft for some s =6 t in B. a

Let us now examine in more detail the collection of graphs GΓ(Γ ⊆ ω1) of 7.12 but in the present more general context.

8.3 Deﬁnition. For Γ ⊆ ω1, let

2 KΓ = {{s, t} ∈ [T ] : ht([st]T ) ∈ Γ}. 8. A square-bracket operation on a tree 63

Working as in 7.13 one shows that (T,KΓ) and (T,K∆) have no isomorphic uncountable subgraph whenever the symmetric diﬀerence between Γ and ∆ is a stationary subset of ω1, i.e. whenever they represent diﬀerent members of the quotient algebra P(ω1)/NS. In particular, KΓ contains no square [X]2 of an uncountable set X ⊆ T whenever Γ contains no closed and unbounded subset of ω1. The following fact is a sort of converse to this.

8.4 Lemma. If Γ contains a closed and unbounded subset of ω1 then there is a proper forcing notion introducing an uncountable set X ⊆ T such that 2 [X] ⊆ KΓ.

Proof. The forcing notion P is deﬁned to be the set of all pairs p = hXp, Npi where N is a ﬁnite ∈-chain of countable elementary submodels p (I.94) of Hω2 containing all the relevant objects.

X is a ﬁnite subset of T such that ht([st] ) ∈ Γ for every p T (I.95) pair {s, t} of elements of Xp. for all s 6= t in X there exist N ∈ N such that s ∈ N iﬀ p p (I.96) t∈ / N. for every s 6= t in X , N ∈ N , if t∈ / N then the height of p p (I.97) min(Fa(s∧t)(t) \ N) belongs to Γ. The ordering of P is the coordinatewise inclusion. To show that P is a proper forcing notion, let M be a given countable elementary submodel of some large enough Hθ containing all the relevant objects and let p ∈ P ∩ M. We shall show that

q = hXp, Np ∪ {M ∩ Hω2 }i is an M-generic condition extending p. So let D ∈ M be a dense-open subset of P and let r be a given extension of q. Extending r we may assume that r ∈ D. Let p0 = r ∩ M. Note that p0 ∈ P ∩ M. Let m0 be the maximum of the union of the following two ﬁnite sets of integers

{a(s ∧ t): s, t ∈ Xr},

{a(t ¹ δ): t ∈ Xr, δ = N ∩ ω1 for some N ∈ Np such that t∈ / N}.

Let k be the size of the projection of the set Xr \ M on the level TM∩ω1 of the tree T . Let F be the family of all k-element subsets F of T for which one can ﬁndr ¯ ∈ D realizing the same type as r over p0 such that, if N is the minimal model of Nr¯ \Np0 then

F = {x ¹ (N ∩ ω1): x ∈ Xr¯ \ Xp0 }. 64 I. Coherent Sequences

Since the projection of Xr¯ \ M onto TM∩ω1 belongs to F, the family is a rather large subset of [T ]k. It follows that the family E of all k-element subsets E of T for which we can ﬁnd δ < ω1 such that

FE = {F ∈ F : {x ¹ δ : x ∈ F } = E} is uncountable, is also quite large. In particular, we can ﬁnd an ordinal α0 ∈ M ∩ ω1 above the heights of all nodes from the set [

{M ∩ Fm0 (t): t ∈ Xr} and E1 ∈ E ∩ M such that

{x ¹ α0 : x ∈ E1} = {x ¹ α0 : x ∈ Xr \ M} and such that every node of Xr \ M is incomparable with every node of E1. Let + m = max{a(x ∧ y): x ∈ Xr \ M, y ∈ E1}, − m = min{a(x ∧ y): x ∈ Xr \ M, y ∈ E1 and a(x ∧ y) > m0}.

Let β0 be an ordinal of M ∩ ω1 above α0 which bounds the heights of all nodes from the set [ {M ∩ Fm+ (t): t ∈ Xr \ M}.

By the deﬁnitions of E and F there isr ¯ ∈ D∩M realizing the same type over p0 as r such that if N is the minimal model of Nr¯ \Np0 then N ∩ ω1 ≥ β0 and the projection

F = {x ¹ (N ∩ ω1): x ∈ Xr¯ \ Xp0 } dominates the set E1. We claim that

q¯ = hXr¯ ∪ Xr, Nr¯ ∪ Nri is a condition of P. This will finish the proof of properness of P since clearly q¯ extends bothr ¯ and r. Clearly, only (I.95) and (I.97) forq ¯ require some argumentation. We check first (I.97), so let s and t be a pair of distinct members of Xr¯ ∪ Xr and let N ∈ Nr¯ ∪ Nr be a model such that t∈ / N. 0 0 Let m = a(s ∧ t). If t ∈ Xr¯ and if there is s ∈ Xr¯ such that s 6= t and 0 s ∧ t = s ∧ t the conclusion that the height of min(Fm(t) \ N) belongs to Γ follows from the fact thatr ¯ satisfies (I.97). Similarly one considers the case 0 0 0 when t ∈ Xr \ M and when s ∧ t = s ∧ t for some s ∈ Xr, s 6= t. For if N belongs to Nr the conclusion follows from the fact that r satisfies (I.97) and if N ∈ Nr¯ \Nr then by the choice ofr ¯,

Fm(t) \ N = Fm(t) \ M, 8. A square-bracket operation on a tree 65

and so the conclusion that the height of min(Fm(t)\N) belongs to Γ follows 0 from the fact that r satisﬁes (I.97) for s and t from Xr and M ∩ Hω2 from 0 Nr. Let us now consider the case t ∈ Xr \ M, s ∈ Xr¯ \ Xr and s ∧ t 6= s ∧ t 0 for all s ∈ Xr \{t}. Note that in this case we have the following inequalities:

− + m0 < m ≤ m = a(s ∧ t) ≤ m .

0 0 So by the choice of m0 the set Fm(t) contains t ¹ (N ∩ω1) for every N ∈ Nr. 0 0 Since every ordinal of the form N ∩ ω1 (N ∈ Nr) belongs to Γ we are done in case N belongs to Nr. So suppose N ∈ Nr¯ \Nr. Then by the choice of the bound β0 and the choice ofr ¯,

Fm(t) \ N = Fm(t) \ M

so the conclusion of (I.97) follows again from the fact that M ∩ Hω2 belongs to Nr. Consider now the remaining case t ∈ Xr¯ \ Xr and s ∈ Xr \ Xr¯. Note 0 that in this case the model N must belong to Nr so if s ∧ t = s ∧ t for 0 some s ∈ Xr¯, s 6= t the conclusion follows from the fact thatr ¯ satisﬁes the condition (I.97). So let us consider the subcase when s0 ∧ t 6= s ∧ t for all 0 0 s ∈ Xr¯, s 6= t. Thus, s ∧ t is the new splitting so by the choice of α0 and E0 we have the inequalities:

− m0 < m ≤ m = a(s ∧ t).

Recall thatr ¯ was chosen to realize the same type over p0 as r so we in particular have that Fm0 (t) and therefore its superset Fm(t) contains the projection t ¹ (N ∩ ω1). It follows that min(Fm(t) \ N) = t ¹ (N ∩ ω1), so its height is indeed a member of Γ (as Γ ∈ N and Γ contains a closed and unbounded subset of ω1). It remains to check (I.95) forq ¯. So let s and t be a given pair of distinct members of Xr¯ ∪Xr. We need to show that the height of [st]T belongs to Γ. Clearly the only nontrivial case is when, say, s ∈ Xr¯ \ Xr and t ∈ Xr \ Xr¯. 0 0 0 Suppose ﬁrst that s ∧ t = s ∧ t for some s ∈ Xr, s 6= t. Since M ∩ Hω2 belongs to Nr and since r satisﬁes (I.97), the height of min(Fa(s0∧t)(t) \ M) belongs to Γ. Since the height of s is above the bound β0 for Fm+ (t) ∩ M,

[st]T = min(Fa(s∧t)(t) \ M) = min(Fa(s0∧t)(t) \ M), so we are done in this subcase. Suppose now that s ∧ t is a new splitting, i.e. that + m0 < m = a(s ∧ t) ≤ m .

So again we know that in this subcase Fm(t) contains the projection t ¹ (M ∩ ω1). By our choice of the bound β0 and the conditionr ¯ we know that this projection is the minimal point of Fm(t) whose height is bigger or 66 I. Coherent Sequences equal to the height of s. By the deﬁnition of the square-bracket operation, we have [st]T = t ¹ (M ∩ ω1).

Since M ∩ ω1 belongs to Γ this finishes our checking thatq ¯ satisfies (I.95). Moreover, this also finishes the proof that the forcing notion P is proper. It is clear that P forces that the square of the union X˙ of the Xp’s for p belonging to the generic filter is included in KΓ. What is not so clear is that P forces that the set X˙ is uncountable as promised. To ensure this we replace P with its restriction P(≤ r) to the condition

r = h{t}, {M ∩ Hω2 }i, where M is an arbitrary countable elementary submodel of some large enough Hθ containing all the relevant objects and where t is any node of T of height at least M ∩ ω1. Note that the above proof of properness of P starting from p = h∅, ∅i shows that

q = h∅, {M ∩ Hω2 }i is an M-generic condition of P so in particular its extension r is also M- generic. It follows that if we let M[G˙ ] be the name for the set formed by interpreting all P-names belonging to M itself. Then r forces that X˙ is an element of the countable elementary submodel M[G˙ ] of Hθ[G˙ ] which contains a point t of height above M[G˙ ] ∩ ω1. It follows that r forces X˙ to be uncountable. This ﬁnishes the proof of Lemma 8.4. a

8.5 Corollary. The graph KΓ contains the square of some uncountable subset of T in some ω1-preserving forcing extension if and only if Γ is a stationary subset of ω1.

Proof. If Γ is disjoint from a closed and unbounded subset then in any ω1- preserving forcing extension its complement ∆ = ω1 \ Γ will be a stationary subset of ω1. So by the basic property 8.1 of the square-bracket operation no such a forcing extension will contain an uncountable set X ⊆ T such 2 that [X] ⊆ KΓ. On the other hand, if Γ is a stationary subset of ω1, going ﬁrst to some standard ω1-preserving forcing extension in which Γ contains a closed and unbounded subset of ω1 and then applying Lemma 8.4, we get an ω1-preserving forcing extension having an uncountable set X ⊆ T such 2 that [X] ⊆ KΓ. a 8.6 Remark. Corollary 8.5 gives us a further indication of the extreme complexity of the class of graphs on the vertex-set ω1. It also bears some relevance to the recent work of Woodin [98] who, working in his Pmax-forcing extension, was able to associate a stationary subset of ω1 to any partition 2 of [ω1] into two pieces. So one may view Corollary 8.5 as some sort of 9. Special trees and Mahlo cardinals 67

converse to this since in the Pmax-extension one is able to get a suﬃciently generic ﬁlter to the forcing notion P = PΓ of Lemma 8.4 that would give us 2 an uncountable X ⊆ T such that [X] ⊆ KΓ. In other words, under a bit of PFA or Woodin’s axiom (*), a set Γ ⊆ ω1 contains a closed and unbounded 2 subset of ω1 if and only if KΓ contains [X] for some uncountable X ⊆ T .

9. Special trees and Mahlo cardinals

One of the most basic questions frequently asked about set-theoretical trees is the question whether they contain any cofinal branch, a branch that intersects each level of the tree. The fundamental importance of this question has already been realized in the work of Kurepa [45] and then later in the works of Erd˝osand Tarski in their respective attempts to develop the theory of partition calculus and large cardinals (see [23]). A tree T of height equal to some regular cardinal θ may not have a cofinal branch for a very special reason as the following definition indicates.

9.1 Deﬁnition. For a tree T = hT,

This definition in case θ = ω1 reduces indeed to the old definition of special tree, a tree that can be decomposed into countably many antichains. More generally we have the following: 9.2 Lemma. If θ is a successor cardinal then a tree T of height θ is special if and only if T is the union of < θ antichains. a The new definition, however, seems to be the right notion of speciality as it makes sense even if θ is a limit cardinal. 9.3 Definition. A tree T of height θ is Aronszajn if T has no cofinal branches and if every level of T has size < θ. Recall the well-known characterization of weakly compact cardinals due to Tarski and his collaborators: a strongly inaccessible cardinal θ is weakly compact if and only if there are no Aronszajn trees of height θ. We supple- ment this with the following: 9.4 Theorem. The following are equivalent for a strongly inaccessible cardinal θ: (1) θ is Mahlo. (2) there are no special Aronszajn trees of height θ. 68 I. Coherent Sequences

Proof. Suppose θ is a Mahlo cardinal and let T be a given tree of height θ all of whose levels have size < θ. To show that T is not special let f : T −→ T be a given regressive mapping. By our assumption of θ there is an elementary submodel M of some large enough structure Hκ such that T, f ∈ M and λ = M ∩ θ is a regular cardinal < θ. Note that T ¹ λ is a subset of M and since this tree of height λ is clearly not special, there is t ∈ T ¹ λ such that the preimage f −1(t) is not the union of < λ antichains. Using the elementarity of M we conclude that f −1(t) is actually not the union of < θ antichains. The proof that (2) implies (1) uses the method of minimal walks in a rather crucial way. So suppose to the contrary that our cardinal contains a closed and unbounded subset C consisting of singular strong limit cardinals. Using C, we choose a C-sequence Cα (α < θ) such that: Cα+1 = {α}, Cα = (¯α, α) for α limit such thatα ¯ = sup(C ∩ α) < α but if α = sup(C ∩ α) then take Cα such that:

tp(Cα) = cf(α) < min(Cα), (I.98)

ξ = sup(Cα ∩ ξ) implies ξ ∈ C, (I.99)

ξ ∈ Cα and ξ > sup(Cα ∩ ξ) imply ξ = η + 1 for some η ∈ C. (I.100)

Given the C-sequence Cα (α < θ) we have the notion of minimal walk along the sequence and various distance functions deﬁned above. In this proof we 2 are particularly interested in the function ρ0 from [θ] into the set Qθ of all ﬁnite sequences of ordinals from θ:

a ρ0(α, β) = htp(Cβ ∩ α)i ρ0(α, min(Cβ \ α)) where we stipulate that ρ0(γ, γ) = 0 for all γ. We would like to show that the tree T (ρ0) = {(ρ0)β ¹ α : α ≤ β < θ} is a special Aronszajn tree of height θ. Note that the size of the αth level (T (ρ0))α of T (ρ0) is controlled in the following way:

|(T (ρ0))α| ≤ |{Cβ ∩ α : α ≤ β < θ}| + |α| + ℵ0. (I.101)

So under the present assumption that θ is a strongly inaccessible cardinal, all levels of T (ρ0) do indeed have size < θ. It remains to define the regressive map f : T (ρ0) −→ T (ρ0) that will witness speciality of T (ρ0). Note that it really suffices defining f on all levels whose index belong to our club C of singular cardinals. So let t = (ρ0)β ¹ α be a given node of T such that α ∈ C and α ≤ β < θ. Note that by our choice of the C-sequence every term of the finite sequence of 9. Special trees and Mahlo cardinals 69

ordinals ρ0(α, β) is strictly smaller than α. So, if we let f(t) = t ¹ pρ0(α, β)q, where p·q is a standard coding of ﬁnite sequences of ordinals by ordinals, we get a regressive map. To show that f is one-to-one on chains of T (ρ0), which would be more than suﬃcient, suppose ti = (ρ0)βi ¹ αi (i < 2) are two nodes such that t0 $ t1. Our choice of the C-sequence allows us to deduce the following general fact about the corresponding ρ0-function as in the case θ = ω1 dealt with above in Lemma 1.14. If α ≤ β ≤ γ, α is a limit ordinal, and if ρ (ξ, β) = ρ (ξ, γ) 0 0 (I.102) for all ξ < α, then ρ0(α, β) = ρ0(α, γ).

Applying this to the triple of ordinals α0, β0 and β1 we conclude that ρ0(α0, β0) = ρ0(α0, β1). Now observe another fact about the ρ0-function whose proof is identical to that of θ = ω1 dealt with above in Lemma 1.13.

If α ≤ β ≤ γ then ρ0(α, γ)

Applying this to the triple α0 < α1 ≤ β1 we in particular have that ρ0(α0, β1) 6= ρ0(α1, β1). Combining this with the above equality gives us that ρ0(α0, β0) 6= ρ0(α1, β1) and therefore that f(t0) 6= f(t1). a This leads us to the following Ramsey-theoretic characterization of Mahlo cardinals. 9.5 Theorem. A cardinal θ is a Mahlo cardinal if and only if every regressive map f defined on a cube [C]3 of a closed and unbounded subset of θ has an infinite min-homogeneous set X ⊆ C.30 Proof. Again, the direct implication is simple so we leave it to the interested reader. For the converse, note that the statement about regressive maps easily implies that θ must be a strongly inaccessible cardinal. So assume θ is a strongly inaccessible non-Mahlo cardinal. Let C be a closed and unbounded subset of θ consisting of strong limit singular cardinals. We shall find a regressive map on [C]3 without an infinite min-homogeneous set. This will give us an opportunity to further expose the tree T (ρ0) and the regressive map f : T (ρ0) ¹ C −→ T (ρ0) defined in the proof of Theorem 9.4. We have already observed that the length of every branching node t must be of the form α + 1 for some α and in fact the following more precise description is true: If t is a branching node of T (ρ ) then its length is equal to 0 (I.104) α + 1 for some α ∈ C. Let α− = max(C ∩ (α + 1)) and α+ = min(C \ (α + 1)).

30Recall, that X is min-homogeneous for f if f(α, β, γ) = f(α0, β0, γ0) for every pair α < β < γ and < α0 < β0 < γ0 of triples of elements of X such that α = α0. 70 I. Coherent Sequences

Let β, γ > α + 1 be two ordinals such that (ρ0)β and (ρ0)γ split at t. Let β = β0 > . . . > βk = α and γ = γ0 > . . . > γl = α be the walks from β to α and γ to α respectively. From the fact that (ρ0)β and (ρ0)γ agree below α we conclude that k = l and Cβi ∩ α = Cγi ∩ α for all i ≤ k. By the choice of the C-sequence and the inequality α > α− we can deduce a sequence + + of stronger agreements Cβi ∩ α = Cγi ∩ α for all i ≤ k. However, that would mean that ρ0(α + 1, β) = ρ0(α + 1, γ), contrary to our assumption − that (ρ0)β and (ρ0)γ split at t. It follows that α = α which was to be shown. Going back to the proof of Theorem 9.5, recall the regressive map f : T (ρ0) ¹ C −→ T (ρ0) deﬁned in 9.4:

f((ρ0)β ¹ α) = (ρ0)β ¹ pρ0(α, β)q.

We extend f to the whole tree by letting f(t) = t ¹ α where α is the maximal element of C that is less than or equal to the length of t. Then we know that f is one-to-one on any chain whose nodes are separated by C. From f we shall derive three other regressive maps p :[C]3 −→ θ, q :[C]3 −→ ω 2 and r :[C] −→ θ. First note that for α 6= β in C the functions (ρ0)α and (ρ0)β are incomparable, so it is natural to denote by α∧β the node of T (ρ0) where they split. For α < β < γ < θ, let

p(α, β, γ) = length(f n(β ∧ γ)), where n is the minimal in- n (I.105) teger such that f (β ∧ γ) ⊆ (ρ0)α.

q(α, β, γ) = min{n : f n(β ∧ γ) = ∅}. (I.106)

r(α, β) = phρ0(ξ, α), ρ0(ξ, β)iq, where ξ = length(α ∧ β). (I.107) Now we claim that no inﬁnite subset of C can be simultaneously min- homogeneous with respect to all three regressive maps. For, given an inﬁnite increasing sequence {αi} of elements of C using Ramsey’s theorem we may assume that either

αi ∧ αj = αk ∧ αl for all i 6= j and k 6= l in ω, or (I.108)

αi ∧ αi+1 $ αi+1 ∧ αi+2 for all i < ω. (I.109) Note that (I.108) would violate min-homogeneity with respect to the mapping r. If (I.109) holds and if the sequence is min-homogeneous with respect to p and q another application of Ramsey’s theorem would give us an integer m and an ordinal ξ such that for all i < j < k < ω:

n m = min{n : f (αj ∧ αk) ⊆ (ρ0)αi }, (I.110)

m f (αj ∧ αk) = (ρ0)α0 ¹ ξ(= t). (I.111) 9. Special trees and Mahlo cardinals 71

By (I.104) the chain {αi ∧ αi+1 : i ≥ 1} is separated by C, so using the basic property of f we conclude that m > 1. Then f −1(t) includes the set m−1 {f (αi ∧ αi+1): i ≥ 1}. By (I.110) we have,

m−1 αi ∧ αi+1 $ f (αi+1 ∧ αi+2) ⊆ αi+1 ∧ αi+2, (I.112) so {fm−1(αi ∧ αi+1): i ≥ 1} is an inﬁnite chain of T (ρ0) separated by C, a contradiction. This ﬁnishes the proof of Theorem 9.5. a

Starting from the case n = 1 one can now easily deduce the following characterization of n-Mahlo cardinals.

9.6 Theorem. The following are equivalent for an uncountable cardinal θ and a positive integer n:

(1) θ is n-Mahlo.

(2) Every regressive map deﬁned on [C]n+2 for some closed and unbounded subset C of θ has an inﬁnite min-homogeneous subset.

Proof. For a set X of ordinals and n ∈ ω, let W (X, n) denote the statement that there exist f :[X]n+2 −→ ω and regressive g[X]n+3 −→ sup(X) such that if H ⊆ X is min-homogeneous for g and f”[H]n+2 = {k} for some k ∈ ω, then |H| ≤ n + k + 5. Note that in the previous proof we have established W (C, 0) for every closed unbounded subset C of some limit ordinal θ such that C contains no inaccessible cardinals (the mapping q is really a map from [C]2 into ω). The proof of the Theorem is ﬁnished once we show that in general W (C, n) holds for every set of ordinals C of limit order type that is closed in its supremum and which contains no n-Mahlo cardinals. So let C be a given closed and unbounded subset of some limit ordinal θ and suppose that C contains no n + 1-Mahlo cardinals. We need to show that W (C, n + 1) holds. Note that for this it suﬃces to show by induction that W (C ∩ δ, n) holds for all δ ∈ C since there is a natural n+2 n+3 way to combine pairs of maps fδ :[C] −→ ω and gδ :[C] −→ θ for δ ∈ C into maps f :[C]n+3 −→ ω and g :[C]n+4 −→ θ witnessing W (C, n + 1). The successor steps of the induction follow from the easily checked fact that W (X, n) implies W (Y, n) for every end-extension Y of X of cardinality at most 2|sup(X)|. Assume now that δ is a limit point of C and that W (C∩γ, n) holds for all δ ∈ C∩δ. Since δ is not an n+1-Mahlo cardinal there is a closed and unbounded set D ⊆ C ∩ δ containing no n-Mahlo cardinals. By the induction hypothesis, W (D, n) holds. Now, there is a n+2 n+3 very natural way to combine maps fγ :[C] −→ ω and gγ :[C] −→ θ n+2 for γ ∈ C ∩ δ witnessing W (C ∩ γ, n) and maps fδ :[D] −→ ω and n+3 n+2 gδ :[D] −→ θ witnessing W (D, n) into maps Fδ :[C ∩ δ] −→ ω and n+3 Gδ :[C ∩ δ] −→ θ witnessing W (C ∩ δ). a 72 I. Coherent Sequences

9.7 Remark. Theorems 9.5 and 9.6 are due to Hajnal, Kanamori and She- lah [30] who were improving an earlier characterization of this sort due to Schmerl [64]. Using Theorem 9.6 one can also prove the following characterization theorem essentially given in [64](see also [30]) which has some interesting metamathematical applications (see [26] and [38]).

9.8 Theorem. The following are equivalent for an uncountable cardinal θ and a positive integer n:

(1) θ is n-Mahlo.

(2) Every regressive map deﬁned on [C]n+3 for some closed and unbounded subset C of θ has a min-homogeneous set of size n + 5.

We ﬁnish this section by showing how the idea of proof of Theorem 9.4 leads us naturally to the following well-known fact, ﬁrst established by Silver (see [52]) when θ is a successor of a regular cardinal.

9.9 Theorem. If θ is a regular uncountable cardinal which is not Mahlo in the constructible universe, then there is a constructible special Aronszajn tree of height θ.

Proof. Working in L we choose a closed and unbounded subset C of θ consisting of singular ordinals and a C-sequence Cα (α < θ) such that Cα+1 = {α}, Cα = (¯α, α) whenα ¯ = sup(C ∩ α) < α, while if α is a limit point of C we take Cα to have the following properties:

ξ = sup(Cα ∩ ξ) implies ξ ∈ C (I.113)

ξ > sup(Cα ∩ ξ) implies ξ = η + 1 for some η ∈ C. (I.114) We choose the C-sequence to also have the following crucial property:

|{Cα ∩ ξ : ξ ≤ α < θ}| ≤ |ξ| + ℵ0 for all ξ < θ. (I.115)

It is clear then that the tree T (ρ0), where ρ0 is the ρ0-function of Cα (α < θ), is a constructible special Aronszajn tree of height θ. a

We are also in a position to deduce the following well-known fact.

9.10 Theorem. The following are equivalent for a successor cardinal θ:

(a) there is a special Aronszajn tree of height θ.

− (b) there is a C-sequence Cα (α < θ) such that tp(Cα) ≤ θ for all α and − such that {Cα ∩ ξ : α < θ} has size ≤ θ for all ξ < θ. 10. The weight function on successor cardinals 73

Proof. If Cα (α < θ) is a C-sequence satisfying (b) and if ρ0 is the associated ρ0-function then T (ρ0) is a special Aronszajn tree of height θ. Suppose

αλ = sup{αξ : ξ < λ} for λ limit < η, (I.116)

α0 =the

αξ+1 =the f(αζ+1) for all (I.118) ζ < ξ, η is the limit ordinal ≤ θ− where the process stops, i.e. − (I.119) sup{f(αξ+1): ξ < η} = θ .

Note that if α and β are two limit points of C and if γ

9.11 Corollary. If θ<θ = θ then there exists a special Aronszajn tree of height θ+. a

9.12 Corollary. In the constructible universe, special Aronszajn trees of any regular uncountable non-Mahlo height exist. a

9.13 Remark. In a large portion of the literature on this subject the notion of a special Aronszajn tree of height equal to some successor cardinal θ+ is somewhat weaker, equivalent to the fact that the tree can be embedded inside the tree {f : α −→ θ : α < θ & f is 1 − 1}. One would get our notion of speciality by restricting the tree on successor ordinals losing thus the frequently useful property of a tree that diﬀerent nodes of the same limit height have diﬀerent sets of predecessors. The result 9.11 in this weaker form is due to Specker [71], while the result 9.12 is essentially due to Jensen [35].

10. The weight function on successor cardinals

+ + In this section we assume that θ = κ and we ﬁx a C-sequence Cα (α < κ ) such that + tp(Cα) ≤ κ for all α < κ . (I.120) 74 I. Coherent Sequences

+ 2 Let ρ1 :[κ ] −→ κ be deﬁned recursively by

ρ1(α, β) = max{tp(Cβ ∩ α), ρ1(α, min(Cβ \ α))} where we stipulate that ρ1(γ, γ) = 0 for all γ. + 10.1 Lemma. |{ξ ≤ α : ρ1(ξ, α) ≤ ν}| ≤ |ν|+ℵ0 for all α < κ and ν < κ. Proof. Let ν+ be the first infinite cardinal above the ordinal ν. The proof of the conclusion is by induction on α. So let Γ ⊆ α be a given set of order- + type ν . We need to find ξ ∈ Γ such that ρ1(ξ, α) > ν. This will clearly be true if there is ξ ∈ Γ such that tp(Cα ∩ ξ) > ν. So, we may assume that

tp(Cα ∩ ξ) ≤ ν for all ξ ∈ Γ. (I.121)

Then there must be an ordinal α1 ∈ Cα such that

Γ1 = {ξ ∈ Γ: α1 = min(Cα \ ξ)}

+ has size ν . By the inductive hypothesis there is ξ ∈ Γ1 such that (see (I.121)) ρ1(ξ, α1) > ν ≥ tp(Cα ∩ ξ). (I.122) It follows that

ρ1(ξ, α) = max{tp(Cα ∩ ξ), ρ1(ξ, α1)} = ρ1(ξ, α1) > ν. This ﬁnishes the proof. a

10.2 Lemma. If κ is regular, then {ξ ≤ α : ρ1(ξ, α) 6= ρ1(ξ, β)} has size < κ for all α < β < κ+. Proof. The proof is by induction on α and β. Let Γ ⊆ α be a given set of order-type κ. We need to ﬁnd ξ ∈ Γ such that ρ1(ξ, α) = ρ1(ξ, β). Let γ = sup(Γ) and

γ0 = max(Cβ ∩ γ), β0 = min(Cβ \ γ). (I.123) Note that by our assumption on κ and the C-sequence, these two ordinals are well-deﬁned and γ0 < γ ≤ β0 < β. (I.124)

By Lemma 10.1 and the inductive hypothesis there is ξ in Γ ∩ (γ0, γ) such that ρ1(ξ, α) = ρ1(ξ, β0) > tp(Cβ ∩ γ). (I.125)

It follows that Cβ ∩ γ = Cβ ∩ ξ and β0 = min(Cβ \ ξ), and so

ρ1(ξ, β) = max{tp(Cβ ∩ ξ), ρ1(ξ, β0)} = ρ1(ξ, β0) = ρ1(ξ, α). This completes the proof. a 11. The number of steps 75

10.3 Remark. The assumption about the regularity of κ in Lemma 10.2 is essential. For example, it can be seen (see [7, p.72]) that the conclusion of this lemma fails if κ is a singular limit of supercompact cardinals.

+ 2 10.4 Deﬁnition. Deﬁneρ ¯1 :[κ ] −→ κ by

ρ1(α,β) ρ¯1(α, β) = 2 · (2 · tp{ξ ≤ α : ρ1(ξ, β) = ρ1(α, β)} + 1).

The following fact shows that this stretching of ρ1 keeps the basic coherence property stated in Lemma 10.2 but it gives us also the injectivity property that could be quite usefull. 10.5 Lemma. If κ is a regular cardinal then

+ (a) ρ¯1(α, γ) 6=ρ ¯1(β, γ) whenever α < β < γ < κ ,

+ (b) |{ξ ≤ α :ρ ¯1(ξ, α) 6=ρ ¯1(ξ, β)}| < κ whenever α < β < κ . a 10.6 Remark. Note that Lemma 10.5 gives an alternative proof of Corol- <κ lary 9.11 since under the assumption κ = κ the tree T (¯ρ1) will have levels of size at most κ. It should be noted that the coherent sequence + (¯ρ)α (α < κ ) of one-to-one mappings is an object of independent interest which can be particularly useful in stepping-up combinatorial properties of κ to κ+. It is also an object that has interpretations in such areas as the theory of Cech-Stoneˇ compactiﬁcations of discrete spaces (see, e.g. [95], [10], [60], [44], [16]). We have already noted that if κ is singular then we may no longer have the coherence property of Lemma 10.2. To get this property, one needs to make some additional assumption on the C-sequence + Cα (α < κ ), an assumption about the coherence of the C-sequence. This will be subject of some of the following chapters where we will concentrate on the ﬁner function ρ rather than ρ1.

11. The number of steps

The purpose of this section is to isolate a condition on C-sequences Cα (α < θ) on regular uncountable cardinals θ as weak as possible subject to a requirement that the corresponding function

ρ2(α, β) = ρ2(α, min(Cβ \ α)) + 1 is in some sense nontrivial, i.e. far from being constant. Without doubt the C-sequence Cα = α (α < θ) is the most trivial choice and the corresponding ρ0-function gives no information whatsoever about the cardinal θ. The following notion of the triviality of a C-sequence on θ seems to be only marginally diﬀerent. 76 I. Coherent Sequences

11.1 Deﬁnition. A C-sequence Cα (α < θ) on a regular uncountable cardinal θ is trivial if there is a closed and unbounded set C ⊆ θ such that for every α < θ there is β ≥ α with C ∩ α ⊆ Cβ.

11.2 Theorem. The following are equivalent for any C-sequence Cα (α < θ) on a regular uncountable cardinal θ and the corresponding function ρ2:

(i) Cα (α < θ) is nontrivial. (ii) For every family A of θ pairwise disjoint finite subsets of θ and every integer n there is a subfamily B of A of size θ such that ρ2(α, β) > n for all α ∈ a, β ∈ b and a 6= b in B. Proof. Note that if a closed and unbounded set C ⊆ θ witnesses the triviality of the C-sequence and if for α < θ we let β(α) ≥ α be the minimal β such that C ∩ α ⊆ Cβ(α), then any disjoint subfamily of A = {{α, β(α)} : α ∈ C} of size θ violates (ii) for n = 1. Conversely, assume we are given a nontrivial C-sequence Cα (α < θ). Consider the following statement: (iii) For every pair Γ and ∆ of unbounded subsets of θ and every integer n there exist a tail subset Γ0 of Γ, an unbounded subset ∆0 of ∆ and a regressive strictly increasing map f : ∆0 −→ θ such that ρ2(α, β) > n for all α ∈ Γ0 and β ∈ ∆0 with α < f(β). To get (ii) from (iii) one assumes that the family A consists of finite sets of some fixed size m and performs m2 successive applications of (iii): this will give us an unbounded A¯ ⊆ A and regressive strictly increasing maps fij : πj[A¯] −→ θ (i, j < m), where πj is the projection mapping that takes b to the jth member of b. We have then that for all a, b ∈ A¯ and i, j < m:

if max(a) < fij (πj(b)), then ρ2(πi(a), πj(b)) > n. Going from there, we get a B as in (ii) by thinning out A¯ once again. Let us prove (iii) from (i) by induction on n. So suppose Γ and ∆ are given two unbounded subsets of θ and that (iii) be true for n − 1. Let Mξ (ξ < θ) be a continuous ∈-chain of submodels of Hθ+ containing all relevant objects such that δξ = Mξ ∩ θ ∈ θ. Let

C = {ξ < θ : δξ = ξ}. Then C is a closed and unbounded subset of θ, so by our assumption (i) on Cα (α < θ) there is δ ∈ C such that C ∩ δ * Cβ for all β ≥ δ. Pick an arbitrary β ∈ ∆ above δ. Then there is ξ ∈ C ∩ δ such that ξ∈ / Cβ. Thenα ¯ = sup(Cβ ∩ ξ) < ξ. Using the elementarity of the submodel Mξ we conclude that for every α ∈ Γ \ α¯ there is β(α) ∈ ∆ \ (α + 1) such that sup(Cβ(α) ∩ α) =α ¯. Applying the inductive hypothesis to the two sets

Γ \ α¯ and {min(Cβ(α) \ α): α ∈ Γ \ α¯} 11. The number of steps 77

we get a tail subset Γ0 of Γ \ α¯, an unbounded subset Γ1 of Γ \ α¯ and a strictly increasing map g :Γ1 −→ θ such that g(γ) < min(Cβ(γ) \ γ) and

ρ2(α, min(Cβ(γ) \ γ)) > n − 1 for all α ∈ Γ0, γ ∈ Γ1 with α < g(γ). (I.126)

Let ∆0 = {β(γ): γ ∈ Γ1} and f : ∆0 −→ θ be deﬁned by

f(β(γ)) = min{γ, g(γ)}.

Then for α ∈ Γ0 and γ ∈ Γ1 with α < f(β(γ)) ≤ γ < β(γ) we have

ρ2(α, β(γ)) = ρ2(α, min(Cβ(γ) \ α)) + 1

= ρ2(α, min(Cβ(γ) \ γ)) + 1 > (n − 1) + 1 = n.

This ﬁnishes the proof. a

11.3 Corollary. Suppose that Cα (α < θ) is a nontrivial C-sequence and let T (ρ0) be the corresponding tree (see Section 9 above). Then every subset of T (ρ0) of size θ contains an antichain of size θ.

2 Proof. Consider a subset K of [θ] of size θ which gives us a subset of T (ρ0) of size θ as follows: ρ0(·, β) ¹ (α + 1) ({α, β} ∈ K). Here, we are assuming without loss of generality that the set consists of successor nodes of T (ρ0). Clearly, we may also assume that the set takes at most one point from a given level of T (ρ0). Shrinking K further, we obtain that ρ2 is constant on K. Let n be the constant value of ρ2 ¹ K. Applying 11.2(ii) to K and n, we get K0 ⊆ K of size θ such that ρ2(α, δ) > n for all {α, β} and {γ, δ} from K0 with properties α < β, γ < δ and α < γ. Then ρ0(·, β) ¹ (α + 1) ({α, β} ∈ K0) is an antichain in T (ρ0). a 11.4 Remark. It should be clear that nontrivial C-sequences exist on any successor cardinal. Indeed, with very little extra work one can show that nontrivial C-sequences exist for some inaccessible cardinals quite high in the Mahlo-hierarchy. To show how close this is to the notion of weak compactness, we will give the following characterization of it which is of independent interest. 31 11.5 Theorem. The following are equivalent for an inaccessible cardinal θ: (i) θ is weakly compact.

(ii) For every C-sequence Cα (α < θ) there is a closed and unbounded set C ⊆ θ such that for all α < θ there is β ≥ α such that Cβ ∩α = C ∩α.

31It turns out that every C-sequence on θ being trivial is not quite as strong as the weak compactness of θ. As pointed out to us by Donder and K¨onig,one can show this using a model of Kunen [43, §3]. 78 I. Coherent Sequences

Proof. To see the implication from (i) to (ii), let us assume that Cα (α < θ) 2 is a C-sequence and let ρ0 :[θ] −→ Qθ be the corresponding ρ0-function. Set T (ρ0) = {(ρ0)β ¹ α : α ≤ β < θ}.

Since θ is inaccessible, the levels of T (ρ0) have size < θ, so by the tree- property of θ (see [37]) the tree T (ρ0) must contain a coﬁnal branch b. For each α < θ ﬁx β(α) ≥ α such that the restriction of (ρ0)β(α) to α belongs to b. For limit α < θ let γ(α) be the largest member γ appearing in the walk

β(α) = β0(α) > β1(α) > . . . > βk(α)(α) = α along the C-sequence with the property that Cγ ∩ α is unbounded in α. (Thus γ = α or γ = βk(α)−1(α).) By the Pressing Down Lemma there is a stationary set Γ ⊆ θ and an integer k, a sequence t in Qθ and an ordinal α¯ < θ such that for all α ∈ Γ: k(α) = k and sup(C ∩ α) ≤ α¯ for all i < k for which βi(α) (I.127) this set is bounded in α. tp(C ∩ α) = t(i) for all i < k for which this set is βi(α) (I.128) bounded in α. Shrinking Γ we may assume that either for all α ∈ Γ, γ(α) = α in which case t is chosen to be a k-sequence, or for all α ∈ Γ, γ(α) = βk−1(α) in which case t is chosen to be a (k − 1)-sequence. It follows that

a ρ0(ξ, β(α)) = t ρ0(ξ, γ(α)) for all α ∈ Γ and ξ ∈ [¯α, α). (I.129) 0 It follows that for α < α in Γ the mappings (ρ0)γ(α) and (ρ0)γ(α0) have the same restrictions on the interval [¯α, α). So in particular we have that 0 Cγ(α) ∩ [¯α, α) = Cγ(α0) ∩ [¯α, α) whenever α < α in Γ. (I.130)

Going to a stationary subset of Γ we may assume that all Cγ(α) (α ∈ Γ) have the same intersection withα ¯. It is now clear that the union of Cγ(α) ∩α (α ∈ Γ) is a closed and unbounded subset of θ satisfying the conclusion of (ii). To prove the implication from (ii) to (i), let T = (θ,

Cα = {ξ < α : ξ

We have already remarked that every successor cardinal θ = κ+ admits a nontrivial C-sequence Cα (α < θ). It suﬃces to take the Cα’s to be all of order-type ≤ κ. It turns out that for such a C-sequence the corresponding ρ2-function has a property that is considerably stronger than 11.2(ii).

11.6 Theorem. For every inﬁnite cardinal κ there is a C-sequence on κ+ such that the corresponding ρ2-function has the following unboundedness property: for every family A of κ+ pairwise disjoint subsets of κ+, all of size < κ, and for every n < ω there exists B ⊆ A of size κ+ such that ρ2(α, β) > n whenever α ∈ a and β ∈ b for some a 6= b in B.

+ Proof. We shall show that this is true for every C-sequence Cα (α < κ ) + with the property that tp(Cα) ≤ κ for all α < κ . Let us ﬁrst consider the case when κ is a regular cardinal. The proof is by induction on n. Suppose the conclusion is true for some integer n and let A be a given family of κ+ + + + pairwise disjoint subsets of κ , all of size < κ . Let Mξ (ξ < κ ) be a continuous ∈-chain of elementary submodels of Hκ++ such that for every ξ the Mξ contains all the relevant objects and has the property that δξ = + + + Mξ∩κ ∈ κ . Let C = {δξ : ξ < κ }. Choose ξ = δξ ∈ C of coﬁnality κ and b ∈ A such that β > ξ for all β ∈ b. Then γ = sup{max(Cβ ∩ξ): β ∈ b} < ξ. + By the elementarity of Mξ we conclude that for every η ∈ (γ, κ ) there is bη in A above η with γ = sup{max(Cβ ∩ η): β ∈ bη}. Consider the following + + family of subsets of κ :a ˆη = bη ∪ {min(Cβ \ η): β ∈ bη} (η ∈ (γ, κ )). By the inductive hypothesis there is an unbounded Γ ⊆ κ+ such that for all η < ζ in Γ, bη ⊆ ζ and ρ2(α, β) > n for all α ∈ aˆη and β ∈ aˆζ . Let B = {bη : η ∈ Γ} and let bη and bζ for η < ζ be two given members of B. Then ρ2(α, β) = ρ2(α, min(Cβ \ α)) + 1 ≥ n + 1 for all α ∈ bη and β ∈ bζ . This completes the inductive step and therefore the proof of the Theorem when κ is a regular cardinal. If κ is a singular cardinal, going to a subfamily of A, we may assume that there is a regular cardinal λ < κ and an unbounded set Γ ⊆ κ+ with the property that A can be enumerated as aη (η ∈ Γ) such that for all η ∈ Γ:

|aη| < λ, (I.131)

β > η for all β ∈ aη, (I.132)

|Cβ ∩ η| < λ for all β ∈ aη. (I.133)

Choose ξ = δξ ∈ C of coﬁnality λ andη ˆ ∈ Γ above ξ and proceed as in the previous case. This completes the inductive step and the proof of the theorem 11.6. a

Note that the combination of the ideas from the proofs of Theorem 11.2 and 11.6 gives us the following variation. 80 I. Coherent Sequences

11.7 Theorem. Suppose that a regular uncountable cardinal θ supports a nontrivial C-sequence and let ρ2 be the associated function. Then for every integer n and every pair of θ-sized families A0 and A1, where the members of A0 are pairwise disjoint bounded subsets of θ and the members of A1 are pairwise disjoint ﬁnite subsets of θ, there exist B0 ⊆ A0 and B1 ⊆ A1 of size θ such that ρ2(α, β) > n whenever α ∈ a and β ∈ b for some a ∈ B0 and b ∈ B1 such that sup(a) < min(b). a An interesting application of this result has recently been given by Gru- enhage [28]. To state his result, given a map from a space X onto a space Y , let ](f) be the order-type of the shortestT chain C of closed subsets of X such that f 00H = Y for all H ∈ C, while f 00( C) 6= Y . If such a chain does not exist, set ](f) = ∞.

11.8 Theorem. Let f : X −→ Y be a given closed surjection. Assume that the space X has the property that every open cover of X has a point countable open reﬁnement and that the space Y has the property that every open subspace of Y either contains an isolated point or a sequence of open sets whose intersection is not open. Then ](f) does not support a nontrivial C-sequence. a

This shows that under some very mild assumption on a pair of spaces X and Y , if their sizes are smaller than the ﬁrst regular uncountable cardinal θ not supporting a nontrivial C-sequence, then for every closed map f from X onto Y , there is a closed subspace H of X on which f is irreducible, i.e. f 00K 6= Y for every proper closed subset K of H.

12. Square sequences

12.1 Deﬁnition. A C-sequence Cα (α < θ) is a square-sequence if and only if it is coherent, i.e. it has the property that Cα = Cβ ∩ α whenever α is a limit point of Cβ. Note that the nontriviality conditions appearing in Deﬁnition 11.1 and Theorem 11.5 coincide in the realm of square-sequences:

12.2 Lemma. A square-sequence Cα (α < θ) is trivial if and only if there is a closed and unbounded subset C of θ such that Cα = C ∩ α whenever α is a limit point of C. a

To a given square-sequence Cα (α < θ) one naturally associates a tree ordering <2 on θ as follows:

2 α < β if and only if α is a limit point of Cβ. (I.134) 12. Square sequences 81

The triviality of Cα (α < θ) is then equivalent to the statement that the tree (θ, <2) has a chain of size θ. In fact, one can characterize the tree orderings

12.3 Lemma. A tree ordering

(i) α

(ii) Pβ = {α : α ω and

(iii) minimal as well as successor nodes of the tree

Proof. For each ordinal α < θ of countable cofinality we fix a subset Sα ⊆ α of order-type ω cofinal with α. Given a tree ordering

It is easily checked that this deﬁnes a square-sequence Cβ (β < θ) with the property that α

12.4 Remark. It should be clear that the proof of Lemma 12.3 shows that the exact analogue of this result is true for any coﬁnality κ < θ rather than ω.

An important result about square sequences is the following result that can be deduced on the basis of the well-known construction of square sequences in the constructible universe due to Jensen [35] (see 1.10 of [78]).

12.5 Theorem. If a regular uncountable cardinal θ is not weakly compact in the constructible subuniverse then there is a nontrivial square sequence on θ which is moreover constructible. a

12.6 Corollary. If a regular uncountable cardinal θ is not weakly compact in the constructible subuniverse then there is a constructible Aronszajn tree on θ. 82 I. Coherent Sequences

Proof. Let Cα (α < θ) be a ﬁxed nontrivial square sequence which is constructible. Changing the Cα’s a bit, we may assume that if β is a limit ordinal with α = min Cβ or if α ∈ Cβ but sup(Cβ ∩α) < α then α must be a 2 successor ordinal in θ. Consider the corresponding function ρ0 :[θ] −→ Qθ

a ρ0(α, β) = tp(Cβ ∩ α) ρ0(α, min(Cβ \ α)), where ρ0(γ, γ) = ∅ for all γ < θ. Consider the tree

T (ρ0) = {(ρ0)β ¹ α : α ≤ β < θ}.

Clearly T (ρ0) is constructible. By (I.101) the αth level of T (ρ0) is bounded by the size of the set {Cβ ∩ α : β ≥ α}. Since the intersection of the form Cβ ∩ α is determined by its maximal limit point modulo a ﬁnite subset of α, we conclude that the αth level of T (ρ0) has size ≤ |α| + ℵ0. Since the sequence Cα (α < θ) is nontrivial, the proof of Theorem 11.5 shows that T (ρ0) has no coﬁnal branches. a

2 12.7 Lemma. Suppose Cα (α < θ) is a square sequence on θ, < the associated tree ordering on θ and T (ρ0) = {(ρ0)β ¹ α : α ≤ β < θ} where 2 ρ0 :[θ] −→ Qθ is the associated ρ0-function. Then α 7−→ (ρ0)α is a strictly 2 increasing map from the tree (θ, < ) into the tree T (ρ0).

Proof. If α is a limit point of Cβ then Cα = Cβ ∩ α so the walks α → ξ and β → ξ for ξ < α get the same code ρ0(ξ, α) = ρ0(ξ, β). a The purpose of this section, however, is to analyze a family of ρ-functions associated with a square sequence Cα (α < θ) on some regular uncountable cardinal θ, both fixed from now on. Recall that an ordinal α divides an ordinal γ if there is β such that γ = α · β, i.e. γ can be written as the union of an increasing β-sequence of intervals of type α. Let κ ≤ θ be a fixed 2 infinite regular cardinal. Let Λκ :[θ] −→ θ be defined by

Λκ(α, β) = max{ξ ∈ Cβ ∩ (α + 1) : κ divides tp(Cβ ∩ ξ)}. (I.136)

Finally we are ready to deﬁne the main object of study in this section:

2 ρκ :[θ] −→ κ (I.137) deﬁned recursively by

ρκ(α, β) = sup{tp(Cβ ∩ [Λκ(α, β), α)), ρκ(α, min(Cβ \ α)),

ρκ(ξ, α): ξ ∈ Cβ ∩ [Λκ(α, β), α)}, where we stipulate that ρκ(γ, γ) = 0 for all γ. The following consequence of the coherence property of Cα (α < θ) will be quite useful. 12. Square sequences 83

12.8 Lemma. If α is a limit point of Cβ then ρκ(ξ, α) = ρκ(ξ, β) for all ξ < α. a

2 Note that ρκ is something that corresponds to the function ρ :[ω1] −→ ω considered in Deﬁnition 4.1 (see also Section 14) and that the ρκ’s are simply various local versions of the key deﬁnition. We shall show that they all have the crucial subadditive properties.

12.9 Lemma. If α < β < γ < θ then

(a) ρκ(α, γ) ≤ max{ρκ(α, β), ρκ(β, γ)},

(b) ρκ(α, β) ≤ max{ρκ(α, γ), ρκ(β, γ)}. Proof. The proof is by induction an α, β and γ. We ﬁrst prove the inductive step for (a). Let ν = max{ρκ(α, β), ρκ(β, γ)}. We need to show that ρκ(α, γ) ≤ ν. To this end, let

γα = min(Cγ \ α) and γβ = min(Cγ \ β). (I.138)

a Case 1 : α < Λκ(β, γ). Then by Lemma 12.8, ρκ(α, γ) = ρκ(α, Λκ(β, γ)). By the deﬁnition of ρκ(β, γ) we have

ρκ(Λκ(β, γ), β) ≤ ρκ(β, γ) ≤ ν. (I.139)

Applying the inductive hypothesis (b) for the triple α < Λκ(β, γ) ≤ β we conclude that ρκ(α, Λκ(β, γ)) ≤ ν. a Case 2 : α ≥ Λκ(β, γ). Then Λκ(β, γ) = Λκ(α, γ) = λ. a Subcase 2 .1: γα = γβ =γ ¯. By the inductive hypothesis (a) for the triple α < β ≤ γ¯ we get

ρκ(α, γ¯) ≤ max{ρκ(α, β), ρκ(β, γ¯)} ≤ ν, (I.140) since ρκ(β, γ¯) ≤ ρκ(β, γ). Note that [λ, α)∩Cγ is an initial part of [λ, β)∩Cγ so its order-type is bounded by ρκ(β, γ) and therefore by ν. Consider a ξ ∈ [λ, α) ∩ Cγ . Applying the inductive hypothesis (b) for the triple ξ < α < β we conclude that

ρκ(ξ, α) ≤ max{ρκ(ξ, β), ρκ(α, β)} ≤ ν, (I.141) since ρκ(ξ, β) ≤ ρκ(β, γ) by the deﬁnition of ρκ(β, γ). This shows that all ordinals appearing in the formula for ρκ(α, γ) are ≤ ν and so ρκ(α, γ) ≤ ν. a Subcase 2 .2: γα < γβ. Then γα belongs to Cγ ∩ [λ, β) and therefore ρκ(γα, β) ≤ ρκ(β, γ) ≤ ν. Applying the inductive hypothesis (b) for α ≤ γα < β gives us

ρκ(α, γα) ≤ max{ρκ(α, β), ρκ(γα, β)} ≤ ν. (I.142) 84 I. Coherent Sequences

Given ξ ∈ Cγ ∩ [λ, α) using the inductive hypothesis (b) for ξ < α < β we get that ρκ(ξ, α) ≤ max{ρκ(ξ, β), ρκ(α, β)} ≤ ν, (I.143) since ρκ(ξ, β) ≤ ρκ(β, γ). Since tp(Cγ ∩[λ, α)) ≤ tp(Cγ ∩[λ, β)) ≤ ρκ(β, γ) ≤ ν we see again that all the ordinals appearing in the formula for ρκ(α, γ) are ≤ ν. Let us now concentrate on the inductive step for (b). This time let ν = max{ρκ(α, γ), ρκ(β, γ)} and let γα and γβ be as in (I.138). We need to show that ρκ(α, β) ≤ ν. b Case 1 : α < Λκ(β, γ). By Lemma 12.8,

ρκ(α, Λκ(β, γ)) = ρκ(α, γ) ≤ ν. (I.144)

From the formula for ρκ(β, γ) we see that ρκ(Λκ(β, γ), β) ≤ ρκ(β, γ) ≤ ν. Applying the inductive hypothesis (a) for the triple α < Λκ(β, γ) ≤ β we get

ρκ(α, β) ≤ max{ρκ(α, Λκ(β, γ)), ρκ(Λκ(β, γ), β)} ≤ ν. (I.145)

b Case 2 : α ≥ Λκ(β, γ). b Subcase 2 .1: γα = γβ =γ ¯. Then ρκ(α, γ¯) ≤ ρκ(α, γ) ≤ ν and ρκ(β, γ¯) ≤ ρκ(β, γ) ≤ ν. Applying the inductive hypothesis (b) for α < β < γ¯ we get ρκ(α, β) ≤ max{ρκ(α, γ¯), ρκ(β, γ¯)} ≤ ν. (I.146) b Subcase 2 .2: γα < γβ. Then γα ∈ Cγ ∩ [λ, β) and so ρκ(γα, β) appears in the formula for ρκ(β, γ) and so we conclude that ρκ(γα, β) ≤ ρκ(β, γ) ≤ ν. Similarly, ρκ(α, γα) ≤ ρκ(α, γ) ≤ ν. Applying the inductive hypothesis (a) for the triple α ≤ γα < β we get

ρκ(α, β) ≤ max{ρκ(α, γα), ρκ(γα, β)} ≤ ν. (I.147) This finishes the proof. a The following is an immediate consequence of the fact that the definition of ρκ is closely tied to the notion of a walk along the fixed square sequence.

12.10 Lemma. ρκ(α, γ) ≥ ρκ(α, β) whenever α ≤ β ≤ γ and β belongs to the trace of the walk from γ to α. a 12.11 Lemma. Suppose β ≤ γ < θ and that β is a limit ordinal > 0. Then ρκ(α, γ) ≥ ρκ(α, β) for coboundedly many α < β.

Proof. Let γ = γ0 > γ1 > . . . > γn−1 > γn = β be the trace of the walk from γ to β. Letγ ¯ = γn−1 if β is a limit point of Cγn−1 , otherwise letγ ¯ = β. Note that by Lemma 12.8, in any case we have that

ρκ(α, β) = ρκ(α, γ¯) for all α < β. (I.148) 12. Square sequences 85

¯ Let β < β be an upper bound of all Cγi ∩ β (i < n) which are bounded in β. Thenγ ¯ is a member of the trace of any walk from γ to some ordinal α in the interval [β,¯ β). Applying Lemma 12.10 to this fact gives us ¯ ρκ(α, γ) ≥ ρκ(α, γ¯) for all α ∈ [β, β). (I.149)

Since ρκ(α, γ¯) = ρκ(α, β) for all α < β (see (I.148)), this gives us the conclusion of the Lemma. a

κ 12.12 Lemma. The set Pν (β) = {ξ < β : ρκ(ξ, β) ≤ ν} is a closed subset of β for every β < θ and ν < κ. Proof. The proof is by induction on β. So let α < β be a limit point κ of Pν (β). Let β1 = min(Cβ \ α) and let λ = Λκ(α, β). If λ = α then κ ρκ(α, β) = 0 and therefore α ∈ Pν (β). So we may assume that λ < α. Then Λκ(ξ, β) = λ for every ξ ∈ [λ, α). (I.150)

Case 1: α¯ = sup(Cβ ∩ α) < α. Then β1 = min(Cβ \ ξ) for every κ ξ ∈ Pν (β) ∩ (¯α, α). It follows that κ κ Pν (β) ∩ (¯α, α) ⊆ Pν (β1), (I.151) κ κ so α is also a limit point of Pν (β1). By the inductive hypothesis α ∈ Pν (β1), i.e. ρκ(α, β1) ≤ ν. Since Λκ(ξ, β) = λ for all ξ ∈ [λ, α), we get κ tp(Cβ ∩ [λ, α)) = sup{tp(Cβ ∩ [λ, ξ)) : ξ ∈ Pν (β) ∩ [λ, α)} ≤ ν. (I.152) κ Consider η ∈ Cβ ∩ [λ, α). Then η ∈ Cβ ∩ [λ, ξ) for some ξ ∈ Pν (β) ∩ [¯α, α) so by the deﬁnition of ρκ(ξ, β) we conclude that ρκ(η, ξ) ≤ ρκ(ξ, β) ≤ ν. By (I.151) we also have ρκ(ξ, β1) ≤ ν so by 12.9(a), ρκ(η, β1) ≤ ν. Applying 12.9(b) to η < α < β1, we conclude that

ρκ(η, α) ≤ max{ρκ(η, β1), ρκ(α, β1)} ≤ ν. (I.153)

This shows that all the ordinals appearing in the formula for ρκ(α, β) are ≤ ν and therefore that ρκ(α, β) ≤ ν. Case 2: α is a limit point of Cβ. As in the previous case, to show that ρκ(α, β) ≤ ν we need to have tp(Cβ ∩ [λ, α)), ρκ(α, β1) and ρκ(η, α) with η ∈ Cβ ∩ [λ, α) all ≤ ν. Note that β1 = α so ρκ(α, β1) = 0. Similarly as in the ﬁrst case we conclude tp(Cβ ∩ [λ, α)) ≤ ν. By Lemma 12.8, κ ρκ(ξ, α) = ρκ(ξ, β) ≤ ν for all ξ ∈ Pν (β) ∩ α. (I.154) κ Consider an η ∈ Cβ ∩ [λ, α). Since α is a limit point of Pν (β) there is ξ from this set above η. By (I.150) we have that η ∈ Cβ ∩ [Λκ(ξ, β), ξ) so ρκ(η, ξ) ≤ ρκ(ξ, β) ≤ ν. Applying 12.9(a) to η < ξ < α we get

ρκ(η, α) ≤ max{ρκ(η, ξ), ρκ(ξ, α)} ≤ ν. (I.155) This ﬁnishes the proof. a 86 I. Coherent Sequences

For α < β < θ and ν < κ set

κ α <ν β if and only if ρκ(α, β) ≤ ν. (I.156)

12.13 Lemma.

κ (1) <ν is a tree ordering on θ, κ κ (2) <ν ⊆<µ whenever ν < µ < κ, S κ (3) ∈¹ θ = ν<κ <ν . Proof. This follows immediately from Lemma 12.9. a

Recall the notion of a special tree of height θ from Section 9, a tree T for which one can ﬁnd a T -regressive map f : T −→ T with the property that the preimage of any point is the union of < θ antichains. By a tree on θ we mean a tree of the form (θ,

12.14 Lemma. If a tree T on θ is special, then there is an ordinal-regressive map f : θ −→ θ and a closed and unbounded set C ⊆ θ such that f is one- to-one on all chains separated by C.

Proof. Let g : θ −→ θ be a T -regressive map such that for each ξ < θ the −1 preimage g (ξ) can be written as a union of a sequence Aδ(ξ)(δ < λξ) of antichains, where λξ < θ. Let C be the collection of all limit α < θ with the property that λξ < α for all ξ < α. Let p., .q denote the G¨odel pairing function which in particular has the property that pα, βq < δ for δ ∈ C and α, β, δ. Deﬁne ordinal-regressive map f : θ −→ θ by letting f(α) = max(C ∪ α) if α∈ / C and for α ∈ C, let f(α) = pg(α), γq where γ < λg(α)(< α) is minimal ordinal with property that α ∈ Aγ (g(α)). Then f is as required. a

κ By Lemma 12.13 we have a sequence <ν (ν < κ) of tree orderings on θ. The following Lemma tells us that they are frequently quite large orderings.

12.15 Lemma. If θ > κ is not a successor of a cardinal of cofinality κ then κ there must be ν < κ such that (θ, <ν ) is a nonspecial tree on θ. Proof. Suppose to the contrary that all trees are special. By Lemma 12.14 we may choose ordinal-regressive maps fν : θ −→ θ for all ν < κ and a single closed and unbounded set C ⊆ θ such that each of the maps fν is κ one-to-one on <ν -chains separated by C. Using the Pressing Down Lemma we find a stationary set Γ of cofinality κ+ ordinals < θ and λ < θ such that + fν (γ) < λ for all γ ∈ Γ and ν < κ. If |λ| < θ, let ∆ = λ, Γ = Γ0 and if + |λ| = θ, represent λ as the increasing union of a sequence ∆ξ (ξ < cf(|λ|)) 12. Square sequences 87 of sets of size < |λ|. Since κ 6= cf(|λ|) there is a ξ¯ < cf(|λ|) and a stationary Γ0 ⊆ Γ such that for all γ ∈ Γ0, fν (γ) ∈ ∆ξ¯ for κ many ν < κ. Let ∆ = ∆ξ¯. This gives us subsets ∆ and Γ0 of θ such that

+ |∆| < θ and Γ0 is stationary in θ, (I.157)

Σγ = {ν < κ : fν (γ) ∈ ∆} is unbounded in κ for all γ ∈ Γ0. (I.158) ¯ + + ¯ Let θ = κ ·|∆| . Then θ < θ and so we can find β ∈ Γ0 such that Γ0 ∩C ∩β ¯ ¯ has size θ. Then there will be ν0 < κ and Γ1 ⊆ Γ0 ∩ C ∩ β of size θ such that ρκ(α, β) ≤ ν0 for all α ∈ Γ1. By (I.158) we can find Γ2 ⊆ Γ1 of size ¯ θ and ν1 ≥ ν0 such that fν1 (α) ∈ ∆ for all α ∈ Γ2. Note that Γ2 is a κ <ν1 -chain separated by C, so fν1 is one-to-one on Γ2. However, this gives us the desired contradiction since the set ∆, in which fν1 embeds Γ2 has size smaller than the size of Γ2. This finishes the proof. a

It is now natural to ask the following question: under which assumption on the square sequence Cα (α < θ) can we conclude that neither of the trees κ (θ, <ν ) will have a branch of size θ?

12.16 Lemma. If the set Γκ = {α < θ : tp Cα = κ} is stationary in θ, κ then none of the trees (θ, <ν ) has a branch of size θ.

κ Proof. Assume that B is a <ν -branch of size θ. By Lemma 12.12, B is a closed and unbounded subset of θ. Pick a limit point β of B which belongs to Γκ. Pick α ∈ B ∩ β such that tp(Cβ ∩ α) > ν. By deﬁnition of ρκ(α, β) we have that ρκ(α, β) ≥ tp(Cβ ∩ α) > ν since clearly Λκ(α, β) = 0. This κ contradicts the fact that α <ν β and ﬁnishes the proof. a

12.17 Deﬁnition. A square sequence on θ is special if the corresponding tree (θ, <2) is special, i.e. there is a <2-regressive map f : θ −→ θ with the property that the f-preimage of every ξ < θ is the union of < θ antichains of (θ, <2).

12.18 Theorem. Suppose κ < θ are regular cardinals such that θ is not a successor of a cardinal of coﬁnality κ. Then to every square sequence Cα (α < θ) for which there exist stationarily many α such that tp Cα = κ, one can associate a sequence Cαν (α < θ, ν < κ) such that:

(i) Cαν ⊆ Cαµ for all α and ν ≤ µ, S (ii) α = ν<κ Cαν for all limit α,

(iii) Cαν (α < θ) is a nonspecial (and nontrivial) square sequence on θ for all ν < κ. 88 I. Coherent Sequences

Proof. Fix ν < κ and deﬁne Cαν by induction on α < θ. So suppose β is a κ limit ordinal < θ and that Cαν is deﬁned for all α < β. If Pν (β) is bounded ¯ κ ¯ in β, let β be the maximal limit point of Pν (β)(β = 0 if the set has no limit points) and let

κ κ Cβν = Cβν¯ ∪ Pν (β) ∪ (Cβ ∩ [max(Pν (β)), β)). (I.159)

κ If Pν (β) is unbounded in β, let [ κ κ κ Cβν = Pν (β) ∪ {Cαν : α ∈ Pν (β)& α = sup(Pν (β) ∩ α)}. (I.160)

By Lemmas 12.9 and 12.12, Cβν (β < θ) is well deﬁned and it forms a square sequence on θ. The properties (i) and (ii) are also immediate. To see that for each ν < κ the sequence Cβν (β < θ) is nontrivial, one uses Lemma 12.16 and the fact that if α is a limit point of Cβν occupying a place in Cβν κ that is divisible by κ, then α <ν β. By Lemma 12.15, or rather its proof, we conclude that there isν ¯ < κ such that Cβν (β < θ) is nonspecial for all ν ≥ ν¯. This ﬁnishes the proof. a

12.19 Lemma. For every pair of regular cardinals κ < θ, every special square sequence Cα (α < θ) can be reﬁned to a special square sequence C¯α (α < θ) with the property that tp C¯α = κ for stationarily many α < θ.

2 Proof. Let < be the tree ordering associated with Cα (α < θ) and let f : θ −→ θ be a <2-regressive map such that f −1(ξ) is the union of < θ antichains of (θ, <2) for all ξ < θ. Case 1: κ = ω. By the Pressing Down Lemma there is a stationary set Γ of cofinality ω ordinals < θ on which f takes some constant value. Thus Γ can be decomposed into < θ antichains of (θ, <2) and so Γ contains a 2 stationary subset Γ0 of pairwise < -incomparable ordinals. For α ∈ Γ0 let C¯α be any ω-sequence of successor ordinals converging to α. If α is a limit 2 ordinal not in Γ0 but there is (a unique!)α ¯ < α in Γ0, let C¯α = Cα \ α¯. In all other cases, let C¯α = Cα. It is easily seen that C¯α (α < θ) is as required. Case 2: κ > ω. By Lemma 12.14 we may find a closed and unbounded subset C of θ and an ordinal-regressive map g : C −→ θ which is one-to-one on <2-chains included in C (and which has the property that for α ∈ C g(α) is the Gödelpairing of f(α) and the index of the antichain which contains α in the fixed antichain-decomposition of the preimage f −1(f(α)).By the argument from Case 1 there is a stationary set Γ of limit points of C consisting of cofinality κ ordinals < θ such that Γ is an antichain of (θ, <2). Given α ∈ Γ, let h(α) be the minimal ordinal λ < θ such that g(ξ) < λ for unboundedly many limit points ξ of Cα that also belong to C. Since g is regressive and since κ, the cofinality of α, is uncountable, the Pressing Down Lemma gives us that h(α) < α. Choose a stationary set Γ0 ⊆ Γ such that h takes some constant value λ on Γ0. Note that since g is one-to one 12. Square sequences 89 on <2-chains included in C the constant value λ cannot be a successor ordinal. It follows that λ must be a limit ordinal of cofinality equal to κ. Let λν (ν < κ) be a strictly increasing sequence which converges to λ. We first define C¯α for α ∈ Γ0 as a strictly increasing continuous sequence cν (α)(ν < κ) of limit points of Cα that also belong to C and satisfy the following requirements at each ν < κ:

cµ(α) = supν<µcν (α) for µ limit (I.161)

cν+1(α) > cν (α) is minimal subject to being a limit point of Cα, belonging to C, and being > γ for every limit point (I.162) γ of Cα such that γ ∈ C and g(γ) < λν+1.

If α is a limit point of C¯β for some β ∈ Γ0, set C¯α = C¯β ∩ α. Note that by the canonicity of the deﬁnition of cν (β)(ν < κ) for β ∈ Γ0, we would get ¯ ¯ ¯ the same result, as Cβ ∩ α = Cβ¯ ∩ α for any other β ∈ Γ0 for which α is a ¯ limit point of Cβ¯. If α is a limit ordinal which is not a limit point of any C¯β for β ∈ Γ0 and 2 there is (a unique!)α ¯ < α in Γ0, let C¯α = Cα \ α¯. 2 If α is a limit ordinal such that α < α¯ for someα ¯ ∈ Γ0 but α is not a limit point on any C¯β for β ∈ Γ0, let C¯α = Cα \ max(C¯α¯ ∩ α). Note again that there is no ambiguity in this deﬁnition, as

C¯β ∩ α = C¯γ ∩ α

2 2 holds for every β and γ in Γ0 such that α < β and α < γ. In all other cases set C¯α = Cα. It should be clear that C¯α (α < θ) is a nontrivial square sequence satisfying the conclusion of Lemma 12.19. a

Finally we can state the main result of this Section which follows from Theorem 12.18 and Lemma 12.19.

12.20 Theorem. If regular uncountable cardinal θ 6= ω1 carries a nontrivial square sequence then it also carries a nontrivial square-sequence which is moreover nonspecial. a

12.21 Corollary. If a regular uncountable cardinal θ 6= ω1 is not weakly compact in the constructible subuniverse then there is a nonspecial Aron- szajn tree of height θ.

Proof. By Theorem 12.5, θ carries a nontrivial square sequence Cα (α < θ). By Theorem 12.20 we may assume that the sequence is moreover nonspecial. Let ρ0 be the associated ρ0-function and consider the tree T (ρ0). As in Corollary 12.6 we conclude that T (ρ0) is an Aronszajn tree of height θ. By 2 Lemma 12.7 there is a strictly increasing map from (θ, < ) into T (ρ0), so T (ρ0) must be nonspecial. a 90 I. Coherent Sequences

12.22 Remark. The assumption θ 6= ω1 in Theorem 12.20 is essential as there is always a nontrivial square sequence on ω1 but it is possible to have a situation where all Aronszajn trees on ω1 are special. For example MAω1 implies this. In [48], Laver and Shelah have shown that any model with a weakly compact cardinal admits a forcing extension satisfying CH and the statement that all Aronszajn trees on ω2 are special. A well-known open problem in this area asks whether one can have GCH rather than CH in a model where all Aronszajn trees on ω2 are special.

13. The full lower trace of a square sequence

In this section θ is a regular uncountable cardinal and Cα (α < θ) is a 2 nontrivial square sequence on θ. Recall the function Λ = Λω :[θ] −→ θ:

Λ(α, β) = maximal limit point of Cβ ∩ (α + 1).

(Λ(α, β) = 0 if Cβ ∩ (α + 1) has no limit points.) The purpose of this section is to study the following recursive trace formula, describing a mapping F :[θ]2 −→ [θ]<ω: [ F (α, β) = F (α, min(Cβ \ α)) ∪ {F (ξ, α): ξ ∈ Cβ ∩ [Λ(α, β), α)},

where F (γ, γ) = {γ} for all γ. 13.1 Lemma. For all α ≤ β ≤ γ, (a) F (α, γ) ⊆ F (α, β) ∪ F (β, γ), (b) F (α, β) ⊆ F (α, γ) ∪ F (β, γ). Proof. The proof of the Lemma is by simultaneous induction. We ﬁrst check (a). Let γα = min(Cγ \ α), γβ = min(Cγ \ β) and λ = Λ(β, γ). If α < λ then Cλ = Cγ ∩ λ and therefore F (α, λ) = F (α, γ). Applying the inductive hypothesis (b) for α < λ ≤ β we get the required conclusion

F (α, γ) = F (α, λ) ⊆ F (α, β) ∪ F (λ, β) ⊆ F (α, β) ∪ F (β, γ). (I.163)

So we may assume that λ ≤ α. Then Λ(α, γ) = λ. If γα < β, applying the inductive hypothesis (b) for α ≤ γα < β, we have

F (α, γα) ⊆ F (α, β) ∪ F (γα, β) ⊆ F (α, β) ∪ F (β, γ), (I.164) since γα ∈ Cγ ∩ [Λ(β, γ), β). If γα ≥ β then γα = γβ, and applying the inductive hypothesis (a) for α < β ≤ γα = γβ, we have

F (α, γα) ⊆ F (α, β) ∪ F (β, γβ) ⊆ F (α, β) ∪ F (β, γ). (I.165) 13. The full lower trace of a square sequence 91

Consider now ξ ∈ Cγ ∩[λ, α). By the inductive hypothesis (b) for ξ < α ≤ β we have

F (ξ, α) ⊆ F (α, β) ∪ F (ξ, β) ⊆ F (α, β) ∪ F (β, γ), (I.166) since ξ ∈ Cγ ∩ [Λ(β, γ), β). It follows that all the factors of F (α, γ) are included in the union F (α, β) ∪ F (β, γ), so this completes the checking of (a). To prove (b) deﬁne γα, γβ and λ as above and consider ﬁrst the case λ ≥ α. Then Cλ = Cγ ∩ λ and therefore F (α, λ) = F (α, γ). Applying the inductive hypothesis (a) for α < λ ≤ β we get

F (α, β) ⊆ F (α, λ) ∪ F (λ, β) ⊆ F (α, γ) ∪ F (β, γ). (I.167)

So we may assume that λ < α in which case we also know that Λ(α, γ) = λ. Consider ﬁrst the case γα < β. Applying the inductive hypothesis (a) for α ≤ γα < β we have

F (α, β) ⊆ F (α, γα) ∪ F (γα, β) ⊆ F (α, γ) ∪ F (β, γ), (I.168) since γα ∈ Cγ ∩ [Λ(β, γ), β). Suppose now that γα ≥ β. Then γα = γβ and applying the inductive hypothesis (b) for the triple α ≤ β ≤ γα = γβ we have

F (α, β) ⊆ F (α, γα) ∪ F (β, γβ) ⊆ F (α, γ) ∪ F (β, γ). (I.169)

This completes the proof. a 13.2 Lemma. For all α ≤ β ≤ γ,

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

a (b) ρ0(α, γ) = ρ0(min(F (β, γ) \ α), γ) ρ0(α, min(F (β, γ) \ α)). Proof. The proof is by induction on α, β and γ. Let

λ = Λ(β, γ), γ1 = min(Cγ \ α) and α1 = min(F (β, γ) \ α).

Pick ξ ∈ {min(Cγ \ β)} ∪ (Cγ ∩ [λ, β)) such that α1 ∈ F (ξ, β). Then α1 = min(F (ξ, β) \ α), so applying the inductive hypothesis (b) for α ≤ ξ < β we get a ρ0(α, β) = ρ0(α1, β) ρ0(α, α1) (I.170) which checks (a) for α ≤ β ≤ γ. Suppose ﬁrst that λ > α. Note that F (λ, β) is a subset of F (β, γ) so α2 = min(F (λ, β) \ α) ≥ α1. Applying the inductive hypothesis for α ≤ λ ≤ β we get

a ρ0(α, β) = ρ0(α2, β) ρ0(α, α2), (I.171) 92 I. Coherent Sequences

a ρ0(α, λ) = ρ0(α2, λ) ρ0(α, α2). (I.172)

Combining (I.170) and (I.171) we infer that ρ0(α, α1) is a tail of ρ0(α, α2), so (I.172) can be written as

a a a ρ0(α, λ) = ρ0(α2, λ) ρ0(α1, α2) ρ0(α, α1) = ρ0(α1, λ) ρ0(α, α1). (I.173)

Since λ is a limit point of Cγ , we have ρ0(α, γ) = ρ0(α, λ), so (I.173) gives us the conclusion (b) of Lemma 13.2. Consider now the case λ ≤ α. Then γ1 is either equal to min(Cγ \ β) or it belongs to [λ, β) ∩ Cγ . In any case we have that F {γ1, β} ⊆ F (β, γ), so

γ2 = min(F {γ1, β}\ α) ≥ α1. (I.174)

Applying the inductive hypothesis for α, γ1 and β we get

a ρ0(α, β) = ρ0(γ2, β) ρ0(α, γ2), (I.175)

a ρ0(α, γ1) = ρ0(γ2, γ1) ρ0(α, γ2). (I.176)

Combining (I.170) and (I.175) we see that ρ0(α, α1) is a tail of ρ0(α, γ2), so (I.176) can be rewritten as

a a a ρ0(α, γ1) = ρ0(γ2, γ1) ρ0(α1, γ2) ρ0(α, α1) = ρ0(α1, γ1) ρ0(α, α1). (I.177) a Since ρ0(α, γ) = tp(Cγ ∩ α) ρ0(α, γ1), (I.177) gives us the conclusion (b) of Lemma 13.2. This ﬁnishes the proof. a

2 Recall the function ρ2 :[θ] −→ ω which counts the number of steps in the walk along the ﬁxed C-sequence Cα (α < θ) which in this Section is assumed to be moreover a square sequence:

ρ2(α, β) = ρ2(α, min(Cβ \ α)) + 1, where we let ρ2(γ, γ) = 0 for all γ. Thus ρ2(α, β) + 1 is simply equal to the cardinality of the trace Tr(α, β) of the minimal walk from β to α.

13.3 Lemma. supξ<α|ρ2(ξ, α) − ρ2(ξ, β)| < ∞ for all α < β < θ. Proof. By Lemma 13.2,

supξ<α|ρ2(ξ, α) − ρ2(ξ, β)| ≤ supξ∈F (α,β)|ρ2(ξ, α) − ρ2(ξ, β)|. a 13.4 Deﬁnition. Set I to be the set of all countable Γ ⊆ θ such that supξ∈∆ ρ2(ξ, α) = ∞ for all α < θ and inﬁnite ∆ ⊆ Γ ∩ α. 13.5 Lemma. I is a P-ideal of countable subsets of θ. 13. The full lower trace of a square sequence 93

Proof. Let Γn (n < ω) be a given sequence of members of I and fix β < θ ∗ such that Γn ⊆ β for all n. For n < ω set Γn = {ξ ∈ Γn : ρ2(ξ, β) ≥ n}. ∗ S ∗ Since Γn belongs to I,Γn is a cofinite subset of Γn. Let Γ∞ = n<ω Γn. Then Γ∞ is a member of I such that Γn \ Γ∞ is finite for all n. a 13.6 Theorem. The P-ideal dichotomy implies that a nontrivial square sequence can exist only on θ = ω1. Proof. Applying the P-ideal dichotomy on I from 13.4 we get the two alternatives (see 4.45):

there is an uncountable ∆ ⊆ θ such that [∆]ω ⊆ I, or (I.178) S there is a decomposition θ = Σ such that Σ ⊥I for n<ω n n (I.179) all n. By Lemma 13.3, if (I.178) holds, then ∆∩α must be countable for all α < θ and so the coﬁnality of θ must be equal to ω1. Since we are working only with regular uncountable cardinals, we see that (I.178) gives us that θ = ω1 must hold. Suppose now (I.179) holds and pick k < ω such that Σk is unbounded in θ. Since Σk⊥I we have that (ρ2)α is bounded on Σk ∩ α for all α < θ. So there is an unbounded set Γ ⊆ θ and an integer n such that for each α ∈ Γ the restriction of (ρ2)α on Σk ∩ α is bounded by n. By Theorem 11.2 we conclude that the square sequence Cα (α < θ) we started with must be trivial. a

13.7 Deﬁnition. By Sθ we denote the sequential fan with θ edges, i.e. the space on (θ × ω) ∪ {∗} with ∗ as the only nonisolated point, while a typical neighborhood of ∗ has the form Uf = {(α, n): n ≥ f(α)} ∪ {∗} where f : θ −→ ω.

13.8 Theorem. If there is a nontrivial square sequence on θ then the square 32 of the sequential fan Sθ has tightness equal to θ. The proof will be given after a sequence of deﬁnitions and lemmas.

13.9 Deﬁnition. Given a square sequence Cα (α < θ) and its number of 2 2 steps function ρ2 :[θ] −→ ω we deﬁne d :[θ] −→ ω by letting

d(α, β) = sup |ρ2(ξ, α) − ρ2(ξ, β)|. ξ≤α

13.10 Lemma. For all α ≤ β ≤ γ,

(a) d(α, β) ≥ ρ2(α, β),

32The tightness of a point x in a space X is equal to θ if θ is the minimal cardinal such that, if a set W ⊆ X \{x} accumulates to x, then there is a subset of W of size ≤ θ that accumulates to x. 94 I. Coherent Sequences

(b) d(α, γ) ≤ d(α, β) + d(β, γ),

Proof. The conclusion (a) follows from the fact that we allow ξ = α in the deﬁnition of d(α, β). The conclusions (b) and (c) are consequences of the triangle inequalities of the `∞-norm and the fact that in both inequalities we have that the domain of functions on the left-hand side is included in the domain of functions on the right-hand side. a

13.11 Deﬁnition. For γ ≤ θ, let

Wγ = {((α, d(α, β)), (β, d(α, β))) : α < β < γ}.

The following Lemma establishes that the tightness of the point (∗, ∗) of 2 Sθ is equal to θ, giving us the proof of Theorem 13.8.

13.12 Lemma. (∗, ∗) ∈ W¯ θ but (∗, ∗) ∈/ W¯ γ for all γ < θ.

2 Proof. To see that Wθ accumulates to (∗, ∗), let Uf be a given neighborhood of (∗, ∗). Fix an unbounded set Γ ⊆ θ on which f is constant. By Theorem 11.2 and Lemma 13.10(a) there exist α < β in Γ such that d(α, β) ≥ f(α) = 2 f(β). Then ((α, d(α, β)), (β, d(α, β))) belongs to the intersection Wθ ∩ Uf . To see that for a given γ < θ the set Wγ does not accumulate to (∗, ∗), choose g : θ −→ ω such that

g(α) = 2d(α, γ) + 1 for α < γ.

2 Suppose Wγ ∩ Ug is nonempty and choose ((α, d(α, β)), (β, d(α, β))) from this set. Then

d(α, β) ≥ 2d(α, γ) + 1 and d(α, β) ≥ 2d(β, γ) + 1. (I.180)

It follows that d(α, β) ≥ d(α, γ)+d(β, γ)+1 contradicting Lemma 13.10(c). a

Since θ = ω1 admits a nontrivial square sequence, Theorem 13.8 leads to the following result of Gruenhage and Tanaka [29].

13.13 Corollary. The square of the sequential fan with ω1 edges is not countably tight. a

13.14 Question. What is the tightness of the square of the sequential fan with ω2 edges? 13.15 Corollary. If a regular uncountable cardinal θ is not weakly compact in the constructible subuniverse then the square of the sequential fan with θ edges has tightness equal to θ. a 13. The full lower trace of a square sequence 95

2 The way we have proved that Sθ has tightness equal to θ is of indepen- 2 dent interest. We have found a point (∗, ∗) of Sθ and a set Wθ of isolated points which accumulates to (∗, ∗) but has no other accumulation points and moreover, no subset of Wθ of size < θ has accumulation points at all. This is interesting in view of the following result of Dow and Watson [17], where C denotes the category of spaces that contain the converging sequence ω+1 and is closed under ﬁnite products, quotients and arbitrary topological sums.

13.16 Theorem. If for every regular cardinal θ there is a space X in C which has a set of isolated points of size θ witnessing tightness θ of its unique accumulation point in X, then every space belongs to C. a

13.17 Corollary. If every regular uncountable cardinal supports a nontrivial square sequence then every topological space X can be obtained from the converging sequence after ﬁnitely many steps of taking (ﬁnite) products, quotients, and arbitrary topological sums.

Proof. This follows from combining (the proof of) Theorem 13.8 and The- orem 13.16. a

If κ is a strongly compact cardinal then the tightness of the product of every pair of spaces of tightness < κ is < κ and so C contains no space of tightness κ or larger. It has been shown in [17] that if the Lebesque measure extends to a countably additive measure on all sets of reals then there even is a countable space which does not belong to C. On the other hand, by Theorem 12.5 and Corollary 13.17, if no regular cardinal above ω1 is weakly compact in the constructible subuniverse, then every space can be generated by the converging sequence by taking ﬁnite products, quotients and arbitrary topological sums.

13.18 Remark. We have seen that the case θ = ω1 is quite special when one considers the problem of existence of various nontrivial square sequences on θ. It should be noted that a similar result about the problem of the tightness 2 of Sθ is not available. In particular, it is not known whether the P-ideal dichotomy or a similar consistent hypothesis of set theory implies that the tightness of the square of, say, Sω2 is smaller than ω2. It is interesting that considerably more is known about the dual question, the question of initial compactness of the Tychonoff cube Nθ. For example, if one defines θ Bαβ = {f ∈ N : f(α), f(β) ≤ d(α, β)} (α < β < θ) one gets an open cover of Nθ without a subcover of size < θ. However, for small θ such θ as θ = ω2 one is able to find such a cover of N without any additional set-theoretic assumption and in particular without the assumption that θ carries a nontrivial square sequence. 96 I. Coherent Sequences

14. Special square sequences

The following well-known result of Jensen [35] supplements the corresponding result for weakly compact cardinals listed above as Theorem 12.5.

14.1 Theorem. If a regular uncountable cardinal θ is not Mahlo in the constructible subuniverse then there is a special square sequence on θ which is moreover constructible. a

Today we know many more inner models with sufficient amount of fine structure necessary for building special square sequences. So the existence of special square sequences, especially at successors of strong-limit singular cardinals, is tied to the existence of some other large cardinal axioms. The reader is referred to the relevant chapters of this Handbook for the specific information. In this section we give the combinatorial analysis of walks along special square sequences and the corresponding distance functions. Let us start by restating some results of Section 12.

14.2 Theorem. Suppose κ < θ are regular cardinals and that θ carries a special square sequence. Then there exist Cαν (α < θ, ν < κ) such that:

(1) Cαν ⊆ Cαµ for all α and ν < µ, S (2) α = ν<κ Cα for all limit α,

(3) Cαν (α < θ) is a nontrivial square sequence on θ for all ν < κ. Moreover, if θ is not a successor of a cardinal of coﬁnality κ then each of the square sequences can be chosen to be nonspecial. a

14.3 Theorem. Suppose κ < θ are regular cardinals and that θ carries a special square sequence. Then there exist <ν (ν < κ) such that:

(i) <ν is a closed tree ordering of θ for each ν < κ, S (ii) ∈¹ θ = ν<κ <ν ,

(iii) no tree (θ, <ν ) has a chain of size θ. a

14.4 Lemma. The following are equivalent when θ is a successor of some cardinal κ:

(1) there is a special square sequence on θ,

(2) there is a square sequence Cα (α < θ) such that tp(Cα) ≤ κ for all α < θ. 14. Special square sequences 97

+ Proof. Let Dα (α < κ ) be a given special square sequence. By Lemma 9.2 the corresponding tree (κ+, <2) can be decomposed into κ antichains so let f : κ+ −→ κ be a fixed map such that f −1(ξ) is a <2-antichain for all + ξ < κ. Let α < κ be a given limit ordinal. If Dα has a maximal limit point 2 α¯ < α, let Cα = Dα \ α¯. Suppose now that {ξ : ξ < α} is unbounded in α and define a strictly increasing continuous sequence cα(ξ)(ξ < ν(α)) of 2 its elements as follows. Let cα(0) = min{ξ : ξ < α}, cα(η) = supξ<ηcα(ξ) 2 for η limit, and cα(ξ + 1) is the minimal < -predecessor γ of α such that 2 γ > cα(ξ) and has the minimal f-image among all < -predecessors that are > cα(ξ). The ordinal ν(α) is defined as the place where the process stops, i.e. when α = supξ<ν(α)cα(ξ). Let Cα = {cα(ξ): ξ < ν(α)}. It is easily + checked that this gives a square sequence Cα (α < κ ) with the property + that tp(Cα) ≤ κ for all α < κ . a

+ Square sequences Cα (α < κ ) that have the property tp(Cα) ≤ κ for + + all α < κ are usually called ¤κ-sequences. So let Cα (α < κ ) be a ¤κ-sequence ﬁxed from now on. Let

Λ(α, β) = maximal limit point of Cβ ∩ (α + 1) (I.181) when such a limit point exists; otherwise Λ(α, β) = 0. The purpose of this section is to analyze the following distance function:

ρ :[κ+]2 −→ κ deﬁned recursively by (I.182)

ρ(α, β) = max{tp(Cβ ∩ α), ρ(α, min(Cβ \ α)),

ρ(ξ, α): ξ ∈ Cβ ∩ [Λ(α, β), α)}, where we stipulate that ρ(γ, γ) = 0 for all γ. Clearly ρ(α, β) ≥ ρ1(α, β), so by Lemma 10.1 we have

+ 14.5 Lemma. |{ξ ≤ α : ρ(ξ, α) ≤ ν}| ≤ |ν| + ℵ0 for all α < κ and ν < κ. a The following two crucial subadditive properties of ρ have proofs that are almost identical to the proof of the corresponding properties of, say, the function ρω discussed above in Section 12. 14.6 Lemma. For all α ≤ β ≤ γ, (a) ρ(α, γ) ≤ max{ρ(α, β), ρ(β, γ)}, (b) ρ(α, β) ≤ max{ρ(α, γ), ρ(β, γ)}. a The following immediate fact will also be quite useful. 98 I. Coherent Sequences

14.7 Lemma. If α is a limit point of Cβ, then ρ(ξ, α) = ρ(ξ, β) for all ξ < α. a The following as well is an immediate consequence of the fact that the deﬁnition of ρ is closely tied to the notion of a minimal walk along the square sequence. 14.8 Lemma. ρ(α, γ) ≥ max{ρ(α, β), ρ(β, γ)} whenever α ≤ β ≤ γ and β belongs to the trace of the walk from γ to α. a Using Lemmas 14.7 and 14.8 one proves the following fact exactly as in the case of ρκ of Section 12 (the proof of Lemma 12.11). 14.9 Lemma. If 0 < β ≤ γ and if β is a limit ordinal, then there is β¯ < β such that ρ(α, γ) ≥ ρ(α, β) for all α in the interval [β,¯ β). a The proof of the following fact is also completely analogous to the proof of the corresponding fact for the local version ρκ considered above in Section 12 (the proof of Lemma 12.12).

14.10 Lemma. Pν (γ) = {β < γ : ρ(β, γ) ≤ ν} is a closed subset of γ for all γ < κ+ and ν < κ. a The discussion of ρ :[κ+]2 −→ κ now splits naturally into two cases depending on whether κ is a regular or a singular cardinal (with the case cf(κ) = ω of special importance). This is done in the following two sections.

15. Successors of regular cardinals

+ In this section κ is a ﬁxed regular cardinal, Cα (α < κ ) a ﬁxed ¤κ-sequence and ρ :[κ+]2 −→ κ the corresponding ρ-function.

For ν < κ and α < β < κ+ set α < β if and only if ν (I.183) ρ(α, β) ≤ ν.

The following is an immediate consequence of the analysis of ρ given in the previous section. 15.1 Lemma.

+ (a) <ν is a closed tree ordering of κ of height ≤ κ for all ν < κ,

(b) <ν ⊆<µ whenever ν ≤ µ < κ,

+ S (c) ∈¹ κ = ν<κ <ν . a 15. Successors of regular cardinals 99

15.2 Remark. Recall that by Lemma 14.3 for every regular cardinal λ ≤ κ the ∈-relation of κ+ can also be written as an increasing union of a λ- sequence <ν (ν < λ) of closed tree orderings. When λ < κ, however, we + can no longer insist that the trees (κ , <ν ) have heights ≤ κ. Using an additional set-theoretic assumption (compatible with the existence of a ¤κ- sequence) one can strengthen (a) of Lemma 15.1 and have that the height of + (κ , <ν ) is < κ for all ν < κ. Without additional set-theoretic assumptions, these trees, however, do have some properties of smallness not covered by Lemma 15.1.

+ 15.3 Lemma. If κ > ω, then no tree (κ , <ν ) has a branch of size κ.

+ Proof. Suppose towards a contradiction that some tree (κ , <ν ) does have a branch of size κ and let B be one such fixed branch (maximal chain). By Lemmas 14.5 and 14.10, if γ = sup(B) then B is a closed and unbounded subset of γ of order-type κ. Since κ is regular and uncountable, Cγ ∩ B is unbounded in Cγ , so in particular we can find α ∈ Cγ ∩ B such that tp(Cγ ∩ α) > ν. Reading off the definition of ρ(α, γ) we conclude that ρ(α, β) = tp(Cγ ∩ α) > ν. Similarly we can find a β > α belonging to the intersection of lim(Cγ ) and B. Then Cβ = Cγ ∩ β so α ∈ Cβ and therefore ρ(α, β) = tp(Cβ ∩ α) > ν contradicting the fact that α <ν β. a

+ 15.4 Lemma. If κ > ω, then no tree (κ , <ν ) has a Souslin subtree of height κ.

+ Proof. Forcing with a Souslin subtree of (κ , <ν ) of height κ would produce an ordinal γ of coﬁnality κ and a closed and unbounded subset B of Cγ + forming a chain of the tree (κ , <ν ). It is well-known that in this case B would contain a ground model subset of size κ contradicting Lemma 15.3. a

15.5 Lemma. If κ > ω then for every ν < κ and every family A of κ + pairwise disjoint ﬁnite subsets of κ there exists A0 ⊆ A of size κ such that for all a 6= b in A0 and all α ∈ a, β ∈ b we have ρ(α, β) > ν. Proof. We may assume that for some n and all a ∈ A we have |a| = n. Let a(0), . . . , a(n − 1) enumerate a given element a of A increasingly. By Lemma 15.4, shrinking A, we may assume that a(i)(a ∈ A) is an antichain + of (κ , <ν ) for all i < n. Going to a subfamily of A of equal size we may assume to have a well-ordering

2 f is constantly equal to (i, j) on [A1] . The ﬁrst alternative is what we want, so let us see that the second one is impossible. Otherwise, choose a

+ 15.6 Lemma. Suppose κ > 0, let γ < κ and let {αξ, βξ} (ξ < κ) be a sequence of pairwise disjoint elements of [κ+]≤2. Then there is an unbounded set Γ ⊆ κ such that ρ{αξ, βη} ≥ min{ρ{αξ, γ}, ρ{βη, γ}} for all ξ 6= η in Γ.33

Proof. Clearly we may assume that the sequences αξ (ξ < κ) and βξ (ξ < κ) are strictly increasing. For deﬁniteness we assume that αξ ≤ βξ for all ξ. The case αξ > βξ for all ξ is considered similarly. Let

α = sup αξ and β = sup βξ. (I.184)

Then α and β are ordinals of coﬁnality κ, so the order-types of Cα and Cβ are both equal to κ. It will be convenient to assume that the two sequences {αξ} and {βξ} are actually indexed by Cα rather than κ. It is then clear we may assume that

βξ ≥ αξ ≥ ξ for all ξ ∈ Cα. (I.185)

Case 1: α = β. By Lemma 14.9, for each limit point ν of Cα there is f(ν) < ν in Cα such that

ρ(ξ, αν ), ρ(ξ, βν ) ≥ ρ(ξ, ν) = ρ(ξ, α) for all ξ ∈ [f(ν), ν). (I.186)

Fix a stationary Γ ⊆ lim(Cα) such that f is constant on Γ. Then

ρ(αξ, βη) ≥ ρ(αξ, α) for all ξ < η in Γ, and (I.187)

ρ(βη, αξ) ≥ ρ(βη, α) for all η < ξ in Γ. (I.188) Subcase 1a: γ ≥ α. Using the two subadditive properties of ρ in Lemma 14.6 one easily concludes that ρ(ξ, α) = ρ(ξ, γ) for any ξ < α such that ρ(ξ, α) > ρ(α, γ). So, going to a tail of Γ we may assume that

ρ(αξ, α) = ρ(αξ, γ) and ρ(βξ, α) = ρ(βξ, γ) for all ξ ∈ Γ. (I.189)

33Here, and everywhere else later in this Chapter, the convention is that, ρ{α, β} is meant to be equal to ρ(α, β) if α < β, equal to ρ(β, α) if β < α, and equal to 0 if α = β. 15. Successors of regular cardinals 101

Consider ξ < η in Γ. Combining (I.187) and the first equality of (I.189) gives us ρ(αξ, βη) ≥ ρ(αξ, α) = ρ(αξ, γ), which is sufficient for the conclusion of Lemma 15.6. Suppose ξ > η are chosen from Γ in this order. Combining (I.188) and the second equality of (I.189) gives us ρ(βη, αξ) ≥ ρ(βη, α) = ρ(βη, γ), which is also sufficient for the conclusion of Lemma 15.6. Subcase 1b: γ < α. Using Lemma 14.5 and going to a tail of the set Γ we may assume that Γ lies above γ and that

ρ(αξ, α), ρ(βξ, α) > ρ(γ, α) for all ξ ∈ Γ. (I.190)

Applying the subadditive property 14.6(b) of ρ we get for all ξ ∈ Γ:

ρ(γ, αξ) ≤ max{ρ(γ, α), ρ(αξ, α)} = ρ(αξ, α), and (I.191)

ρ(γ, βξ) ≤ max{ρ(γ, α), ρ(βξ, α)} = ρ(βξ, α). (I.192) If ξ < η are chosen from Γ in this order then (I.187) and (I.191) give us

ρ(αξ, βη) ≥ ρ(αξ, α) ≥ ρ(γ, αξ), which is suﬃcient for the conclusion of Lemma 15.6. If ξ > η are chosen from Γ in this order, then combining (I.188) and (I.192) give us

ρ(βη, αξ) ≥ ρ(βη, α) ≥ ρ(γ, βη), which is again suﬃcient for the conclusion of Lemma 15.6. Case 2: α < β. We may assume all βξ > α for all ξ and working as in the Case 1 we ﬁnd a stationary set Γ of limit points of Cα such that

ρ(αξ, βη) ≥ ρ(αξ, α) for all ξ < η in Γ. (I.193)

By Lemma 14.5 for each η ∈ Γ there is g(η) ∈ Cα such that ρ(ξ, α) > ρ(α, βη) for all ξ in the interval [g(η), α). Using the two subadditive properties of ρ from Lemma 14.6, one concludes that ρ(ξ, α) = ρ(α, βη) for all ξ ∈ [g(η), α). Intersecting Γ with the closed and unbounded subset of Cα of all ordinals that are closed under the mapping g we may assume

ρ(αξ, βη) = ρ(αξ, α) for all ξ > η in Γ. (I.194)

Subcase 2a: γ < α. Applying Lemma 14.5 again and going to a tail of Γ we may assume that Γ lies above γ and

ρ(αξ, α) > ρ(γ, α) for all ξ ∈ Γ. (I.195) 102 I. Coherent Sequences

Applying the subadditive properties of ρ we get

ρ(γ, αξ) ≤ max{ρ(γ, α), ρ(αξ, α)} = ρ(αξ, α) for all ξ ∈ Γ. (I.196)

If ξ < η are chosen from Γ in this order, then combining the inequalities (I.193) and (I.196) we get

ρ(αξ, βη) ≥ ρ(αξ, α) ≥ ρ(γ, αξ), which is suﬃcient to give us the conclusion of Lemma 15.6. If ξ > η are chosen from Γ in this order, then combining (I.194) and (I.196) we get

ρ(αξ, βη) = ρ(αξ, α) ≥ ρ(γ, αξ), which is again suﬃcient for the conclusion of Lemma 15.6. b Subcase 2 : γ ≥ α. Going to a tail of Γ we may assume that ρ(αξ, α) > ρ(α, γ), so as before the subadditive properties give us that

ρ(αξ, α) = ρ(αξ, γ) for all ξ ∈ Γ. (I.197)

If ξ < η are chosen from Γ in this order then combining the inequality (I.193) and equality (I.197) we have

ρ(αξ, βη) ≥ ρ(αξ, α) = ρ(αξ, γ), which is sufficient to give us the conclusion of Lemma 15.6. If ξ > η are chosen from Γ in this order, then combining the equalities (I.194) and (I.197) we have ρ(αξ, βη) = ρ(αξ, α) = ρ(αξ, γ), which is again sufficient to give us the conclusion of Lemma 15.6. This finishes the proof. a

15.7 Lemma. Suppose κ is λ-inaccessible 34 for some λ < κ and that A is a family of size κ of subsets of κ+, all of size < λ. Then for every ordinal ν < κ there is a subfamily B of A of size κ such that for all a and b in B:

(a) ρ{α, β} > ν for all α ∈ a \ b and β ∈ b \ a.

(b) ρ{α, β} ≥ min{ρ{α, γ}, ρ{β, γ}} for all α ∈ a \ b, β ∈ b \ a and γ ∈ a ∩ b.

Proof. Consider first the case λ = ω. Going to a subfamily, assume A forms a ∆-system with root r and that for some fixed integer n ≥ 1 and all a ∈ A we have |a \ r| = n. By Lemma 15.5, going to a subfamily we may assume that A satisfies (a). For a ∈ A let a(0), . . . , a(n − 1) be the increasing

34I.e. ντ < κ for all ν < κ and τ < λ. 15. Successors of regular cardinals 103 enumeration of a \ r. Going to a subfamily, we may assume that A can be enumerated as aξ (ξ < κ) in such a way that aξ(i)(ξ < κ) is strictly increasing for all i < n. By n2 · |r| successive applications of Lemma 15.6 we ﬁnd a single unbounded set Γ ⊆ κ such that for all γ ∈ r,(i, j) ∈ n × n and ξ 6= η in Γ:

ρ{aξ(i), aη(j)} ≥ min{ρ{aξ(i), γ}, ρ{aη(j), γ}}. (I.198)

This gives us the conclusion (b) of the Lemma. Let us now consider the general case. Using the λ-inaccessibility of κ we may assume that A forms a ∆-system with root r and that members of A have some fixed order-type, i.e. that for some fixed ordinal µ < λ we can enumerate each a \ r for a ∈ A in increasing order as a(i)(i < µ). Using λ-inaccessibility again and refining A we may assume that there is an enumeration aξ (ξ < κ) of A such that aξ(i)(ξ < κ) is a strictly increasing sequence for all i < µ. For i < µ, set

δ(i) = supξ aξ(i).

Then cf(δ(i)) = κ and therefore tp Cδ(i) = κ for all i < µ. Going to a smaller cardinal we may assume that λ is a regular cardinal < κ. Applying Lemma 14.9 simultaneously µ times gives us a regressive map f : κ −→ κ such that for every η < κ of coﬁnality λ, all ξ ∈ [cδ(i)(f(η)), cδ(i)(η)) and i < µ: ρ(ξ, aη(i)) ≥ ρ(ξ, cδ(i)(η)) = ρ(ξ, δ(i)). (I.199)

Here cδ(i)(ξ)(ξ < κ) refers to the increasing enumeration of the set Cδ(i) for each i < µ. Pick a stationary set Γ of coﬁnality λ ordinals < κ such that f is constant on Γ. It follows that for all i, j < µ, if δ(i) = δ(j) then

ρ{aξ(i), aη(j)} ≥ ρ(aξ(i), δ(i)) for ξ < η in Γ, (I.200)

ρ{aη(j), aξ(i)} ≥ ρ(aη(j), δ(i)) for ξ > η in Γ. (I.201) On the other hand, if i, j < µ such that δ(i) < δ(j) we have

ρ{aξ(i), aη(j)} ≥ ρ(aξ(i), δ(i)) for all ξ < η in Γ. (I.202)

Intersecting Γ with a closed and unbounded subset of κ we also assume that

aξ(i) ≥ cδ(i)(ξ) for all i < µ and ξ ∈ Γ. (I.203)

aξ(i) ≤ aξ(j) for all ξ ∈ Γ and i, j < µ such that δ(i) ≤ δ(j). (I.204)

Since ρ(α, δ(i)) ≥ tp(Cδ(i) ∩ α) for every i < µ and α < δ(i), combining (I.200)-(I.204) we see that

ρ{aξ(i), aη(j)} ≥ min{ξ, η} for all i, j < µ and ξ 6= η in Γ. (I.205) 104 I. Coherent Sequences

So replacing Γ with Γ \ (ν + 1) gives us the conclusion (a) of Lemma 15.7. Suppose now we are given two coordinates i, j < µ and an ordinal γ in the root r. Let αξ = aξ(i) and βξ = aξ(j) for ξ ∈ Γ. Note that (I.200) and (I.201) give us the conditions (I.187) and (I.188) of the Case 1 (i.e. δ(i) = α = β = δ(j)) in the proof of Lemma 15.6 while (I.202) gives us the condition (I.193) of the Case 2 (i.e. δ(i) = α < β = δ(j)) in the proof of Lemma 15.6. Observe that in this proof of Lemma 15.6, once the set Γ was obtained by a single application of the Pressing Down Lemma to finally give us (I.187) and (I.188) of Case 1 or (I.193) of Case 2, the rest of the refinements of Γ all lie in the restriction of the closed and unbounded filter to Γ. In other words, we can now proceed as in the case λ = ω and successively apply this refining procedure for each γ ∈ r and (i, j) ∈ µ×µ obtaining a set Γ0 ⊆ Γ with the property that Γ\Γ0 is nonstationary and ρ{a (i), a (j)} ≥ min{ρ{a (i), γ}, ρ{a (j), γ}} for all γ ∈ r; ξ η ξ η (I.206) i, j < µ and ξ 6= η in Γ0. Since this is clearly equivalent to the conclusion (b) of Lemma 15.7, the proof is completed. a 15.8 Definition. D :[κ+]2 −→ [κ+]<κ is defined by D(α, β) = {ξ ≤ α : ρ(ξ, α) ≤ ρ(α, β)}. (Note that D(α, β) = {ξ ≤ α : ρ(ξ, β) ≤ ρ(α, β)}, so we could take the formula D{α, β} = {ξ ≤ min{α, β} : ρ(ξ, α) ≤ ρ{α, β}} as our definition of D{α, β} when there is no any implicit assumption about the ordering between α and β as there is whenever we write D(α, β). ) 15.9 Lemma. If κ is λ-inaccessible for some λ < κ, then for every family A of size κ of subsets of κ+, all of size < λ, there exists B ⊆ A of size κ such that for all a and b in B and all α ∈ a \ b, β ∈ b \ a and γ ∈ a ∩ b: (a) α, β > γ implies D{α, γ} ∪ D{β, γ} ⊆ D{α, β}, (b) β > γ implies D{α, γ} ⊆ D{α, β}, (c) α > γ implies D{β, γ} ⊆ D{α, β}, (d) γ > α, β implies D{α, γ} ⊆ D{α, β} or D{β, γ} ⊆ D{α, β}. Proof. Choose B ⊆ A of size κ satisfying the conclusion (b) of Lemma 15.7. Pick a 6= b in B and consider α ∈ a \ b, β ∈ b \ a and γ ∈ a ∩ b. By the conclusion of 15.7(b), we have ρ{α, β} ≥ min{ρ{α, γ}, ρ{β, γ}}. (I.207) 15. Successors of regular cardinals 105

a. Suppose α, β > γ. Note that in this case a single inequality ρ(γ, α) ≤ ρ{α, β} or ρ(γ, β) ≤ ρ{α, β} given to us by (I.207) implies that we actually have both inequalities simultaneously holding. The subadditivity of ρ gives us ρ(ξ, α) ≤ ρ{α, β}, or equivalently ρ(ξ, β) ≤ ρ{α, β} for any ξ ≤ γ with ρ(ξ, γ) ≤ ρ(γ, α) or ρ(ξ, γ) ≤ ρ(γ, β). This is exactly the conclusion of 15.9(a). b. Suppose that β > γ > α. Using the subadditivity of ρ we see that in both cases given to us by (I.207) we have that ρ(α, γ) ≤ ρ{α, β}. So the inclusion D{α, γ} ⊆ D{α, β} follows immediately. c. Suppose that α > γ > β. The conclusion D{β, γ} ⊆ D{α, β} follows from the previous case by symmetry. d. Suppose that γ > α, β. Then ρ(α, γ) ≤ ρ{α, β} gives D{α, γ} ⊆ D{α, β} while ρ(β, γ) ≤ ρ{α, β} gives us D{β, γ} ⊆ D{α, β}. This completes the proof. a 15.10 Remark. Note that min{x, y} ∈ D{x, y} for every {x, y} ∈ [κ+]2, so the conclusion (a) of Lemma 15.9 in particular means that γ < min{α, β} implies γ ∈ D{α, β}. In applications, one usually needs this consequence of 15.9(a) rather than 15.9(a) itself. 15.11 Deﬁnition. The ∆-function of some family F of subsets of some ordinal κ (respectively, a family of functions with domain κ) is the function ∆ : [F]2 −→ κ deﬁned by

∆(f, g) = min(f 4 g),

(respectively, ∆(f, g) = min{ξ : f(ξ) 6= g(ξ)}). Note the following property of ∆: 15.12 Lemma. ∆(f, g) ≥ min{∆(f, h), ∆(g, h)} for all {f, g, h} ∈ [F]3. a 15.13 Remark. This property can be very useful when transferring objects that live on κ to objects on F. This is especially interesting when F is of size larger than κ while all of its restrictions F ¹ ν = {f ∩ν : f ∈ F} (ν < κ) have size < κ, i.e. when F is a Kurepa family (see for example [12]). We shall + now see that it is possible to have a Kurepa family F = {fα : α < κ } whose + ∆-function is dominated by ρ, i.e. ∆(fα, fβ) ≤ ρ(α, β) for all α < β < κ .

15.14 Theorem. If ¤κ holds and κ is λ-inaccessible then there is a λ-closed κ-cc forcing notion P that introduces a Kurepa family on κ. Proof. Going to a larger cardinal, we may assume that λ is regular. Put p in P, if p is a one-to-one function from a subset of κ+ of size < λ into the family of all subsets of κ of size < λ such that for all α and β in dom(p):

p(α) ∩ p(β) is an initial part of p(α) and of p(β), (I.208) 106 I. Coherent Sequences

∆(p(α), p(β)) ≤ ρ(α, β) provided that α 6= β. (I.209) Let p ≤ q whenever dom(p) ⊇ dom(q) and p(α) ⊇ q(α) for all α ∈ dom(q). Clearly P is a λ-closed forcing notion. To see that it satisﬁes the κ-chain condition, let A be a given subset of P of size κ. By the assumption that κ is λ-inaccessible, going to a subfamily of A of size κ, we may assume that A forms a ∆-system, orS more precisely, that dom(p)(p ∈ A) forms a ∆- system with root d, that rng(p)(p ∈ A) forms a ∆-system with root c and that any two members of A generate isomorphic structures via the natural isomorphism that ﬁxes the roots. By Lemma 15.7 there exists B ⊆ A of size κ such that for all p, q ∈ B, all α ∈ dom(p)\dom(q), all β ∈ dom(q)\dom(p) and all γ ∈ dom(p) ∩ dom(q)(= d):

ρ{α, β} > max(c), (I.210)

ρ{α, β} ≥ min{ρ{α, γ}, ρ{β, γ}}. (I.211) S SupposeS p, q ∈ B and that a = ( rng(p)) \ c lies entirely below b = ( rng(q)) \ c. Define r : dom(p) ∪ dom(q) −→ [κ]<λ as follows. For γ ∈ d, let r(γ) = p(γ) ∪ q(γ). If β ∈ dom(q) \ dom(p) and if there is γ ∈ d such that ∆(q(β), q(γ)) ∈ b, the γ must be unique. Put r(β) = p(γ) ∪ q(β) in this case. If such γ does not exist, put r(β) = q(β) ∪ {max(c) + 1}. For α ∈ dom(p) \ dom(q), put r(α) = p(α). Note that this choice of r does not change the behaviour of ∆ on pairs coming from [dom(p)]2 or [dom(q)]2, so in checking that r satisfies (I.208) it suffices to assume α ∈ dom(p) \ dom(q) and β ∈ dom(q) \ dom(p). If r(β) was defined as p(γ) ∪ q(β) for some γ ∈ d we have that ∆(r(α), r(β)) = ∆(p(α), p(γ)) and that ∆(q(β), q(γ)) ∈ b. It follows that

∆(r(α), r(β)) = min{∆(p(α), p(γ)), ∆(q(β), q(γ))} (I.212) ≤ min{ρ{α, γ}, ρ{β, γ}}.

Applying (I.211) we get the desired inequality ∆(r(α), r(β)) ≤ ρ{α, β}. If r(β) was chosen to be q(β)∪{max(c)+1}. Then ∆(r(α), r(β)) ≤ max(c)+1 and this is ≤ ρ{α, β} by (I.210). ˙ + ˙ Let G˙ be the generic ﬁlter of P and let F˙ = {fα : α < κ }, where fα is ˙ the union of p(α)(p ∈ G˙ ). A simple genericity argument shows that each fα intersects every interval if the form [ν, ν +λ)(ν < κ). It follows (see (I.208)) that F˙ ¹ ν has size at most λ for all ν < κ. This ﬁnishes the proof. a

15.15 Corollary. If ¤ω1 holds, and so in particular if ω2 is not a Mahlo cardinal in the constructible universe, then there is a property K poset, forcing the Kurepa hypothesis. a 15. Successors of regular cardinals 107

15.16 Remark. This is a variation on a result of Jensen who proved that under ¤ω1 there is a ccc poset forcing the Kurepa hypothesis. It should be noted that Jensen also proved (see [36]) that in the Levy collapse of a Mahlo cardinal to ω2 there is no ccc poset forcing the Kurepa hypothesis. It should also be noted that Veliˇcković[91] was the first to use ρ to reprove Jensen’s result. He has also shown (see [91]) that ρ easily yields an example of a function with the ’∆-property’ in the sense of Baumgartner and Shelah [5]. Recall that a function with the ∆-property in the sense of [5] is a function that satisfies some of the properties of the function D of Lemma 15.9 and a function that forms a ground for the well-known forcing construction of a locally compact scattered topology on ω2 all of whose Cantor-Bendixson ranks are countable. We now use ρ to improve the chain-condition of the Baumgartner-Shelah forcing.

15.17 Theorem. If ¤ω1 holds then there is a property K forcing notion that introduces a locally compact scattered topology on ω2 all of whose Cantor- Bendixson ranks are countable.

Proof. Let P be the set of all p = hDp, ≤p,Mpi where Dp is a ﬁnite subset of ω2, where ≤p is a partial ordering of Dp compatible with the well-ordering 2 <ω and Mp :[Dp] −→ [ω2] has the following properties:

Mp{α, β} ⊆ D{α, β} ∩ Dp, (I.213)

Mp{α, β} = {α} if α ≤p β and Mp{α, β} = {β} if β ≤p α, (I.214)

γ ≤p α, β for all γ ∈ Mp{α, β}, (I.215)

for every δ ≤p α, β there is γ ∈ Mp{α, β} such that δ ≤p γ. (I.216) 2 We let p ≤ q if and only if Dp ⊇ Dq, ≤p¹ Dq =≤q and Mp ¹ [Dq] = Mq. To verify that P satisﬁes Knaster’s chain condition, let A be a given uncountable subset of P. Shrinking A we may assume that A forms a ∆-system with root R and that the conditions of A generate isomorphic structures over R. By Lemma 15.9 there is an uncountable subset B of A such that for all p and q in B and all α ∈ Dp \ Dq, β ∈ Dq \ Dp and γ ∈ R = Dp ∩ Dq:

α, β > γ implies D{α, γ} ∪ D{β, γ} ⊆ D{α, β}, (I.217) β > γ implies D{α, γ} ⊆ D{α, β}, (I.218) α > γ implies D{β, γ} ⊆ D{α, β}. (I.219)

Consider two conditions p and q from B. Let r = hDr, ≤r,Mri be deﬁned as follows: Dr = Dp ∪ Dq and α ≤r β if and only if α ≤p β, or α ≤q β, or there is γ ∈ R such that α ≤p γ ≤q β or α ≤q γ ≤p β. It is easily checked that ≤r is a partial ordering on Dr compatible with the well-ordering such 108 I. Coherent Sequences

2 <ω that ≤r¹ Dp =≤p and ≤r¹ Dq =≤q. Deﬁne Mr :[Dr] −→ [ω2] by 2 2 letting Mr ¹ [Dp] = Mp, Mr ¹ [Dq] = Mq, Mr{α, β} = {α} if α ≤r β, Mr{α, β} = {β} if β ≤r α and if α ∈ Dp \ R and β ∈ Dq \ R are ≤r- incomparable, set

Mr{α, β} = {γ ∈ D{α, β} ∩ Dr : γ ≤r α, β}. (I.220) Only nontrivial to check is that a so defined r satisfies (I.216). So let δ ≤r α, β be given. If α, β ∈ Dp \ Dq and δ ∈ Dq \ Dp then there exist γα, γβ ∈ R such that δ ≤q γα ≤p α and δ ≤q γβ ≤p β. By (I.216) for q there exists γ ∈ Mq{γα, γβ} such that δ ≤q γ. Since Mq{γα, γβ} ⊆ R we conclude that γ ∈ R. Thus γ ≤p α, β, so by (I.216) there isγ ¯ ∈ Mp{α, β} such that γ ≤p γ¯ and therefore δ ≤r γ¯. The case α ∈ R, β ∈ Dp \ D and δ ∈ Dq \ Dp is quite similar. Consider now the case when α ∈ Dp \ Dq, β ∈ Dq \ Dp and they are ≤r-incomparable, i.e. Mr{α, β} is given by (I.220). For symmetry we may consider only the subcase δ ≤r α, β with δ ∈ Dp. Then δ ≤p α and δ ≤p γ ≤q β for some γ ∈ R. By (I.216) for p there isγ ¯ ∈ Mp{α, γ} such that δ ≤p γ¯. It remains to show thatγ ¯ was put in Mr{α, β} in (I.220), i.e. thatγ ¯ belongs to D{α, β}. Note that since ≤r agrees with the well-ordering, we have that β > γ, so by (I.218) we have that D{α, γ} ⊆ D{α, β}. By (I.213) for p we have thatγ ¯ ∈ Mp{α, γ} ⊆ D{α, γ} so we can conclude that γ¯ ∈ D{α, β}. The rest of the cases are symmetric to the ones above. ˙ ˙ Let G be a generic filter of P and let ≤G˙ be the union of ≤p (p ∈ G). For ˙ α < ω2 set Bα = {ξ : ξ ≤G˙ α}. Let τ be the topology on ω2 generated by B˙ α (α < ω2) as a clopen subbasis. It is easily checked that τ is a Hausdorff locally compact scattered topology on ω2. A simple density argument and induction on α shows that the interval [ωα, ωα + ω) is the αth Cantor- Bendixson rank of the space (ω2, τ). a

2 ≤ω 15.18 Deﬁnition. A function f :[ω2] −→ [ω2] has property ∆ if for every uncountable set A of ﬁnite subsets of ω2 there exist a and b in A such that for all α ∈ a \ b, β ∈ b \ a and γ ∈ a ∩ b:

(a) α, β > γ implies γ ∈ f{α, β}, (b) β > γ implies f{α, γ} ⊆ f{α, β}, (c) α > γ implies f{β, γ} ⊆ f{α, β}.

15.19 Remark. This deﬁnition is due to Baumgartner and Shelah [5] who have used it in their well-known forcing construction. It should be noted that they were also able to force a function with the property ∆ using a σ-closed ω2-cc poset. As shown above (a fact ﬁrst checked in [5, p.129]), the function

D{α, β} = {ξ ≤ min{α, β} : ρ(ξ, α) ≤ ρ{α, β}} 16. Successors of singular cardinals 109 has property ∆. However, Lemma 15.9 shows that the function D has many other properties that are of independent interest and that are likely to be needed in other forcing constructions of this sort. The reader is referred to papers of Koszmider [41] and Rabus [58] for further work in this area.

16. Successors of singular cardinals

In the previous section we have witnessed the fact that the function ρ : + 2 + [κ ] −→ κ defined from a ¤κ-sequence Cα (α < κ ) can be a quite useful tool in stepping-up objects from κ to κ+. In this section we analyse the stepping-up power of ρ under the assumption that κ is a singular cardinal of cofinality ω. So let κn (n < ω) be a strictly increasing sequence of regular cardinals converging to κ fixed from now on. This immediately gives rise to + a rather striking tree decomposition

+ S (1) ∈¹ κ = n<ω

(2)

+ + (3) (κ ,

16.2 Deﬁnition. For α < κ+ and n < ω set

Fn(α) = {ξ ≤ α : ρ(ξ, α) ≤ κn},

fα(n) = tp(Fn(α)).

+ Let L = {fα : α < κ }, considered as a linearly ordered set with the lexicographical ordering. Since L is a subset of ωκ, it has an order-dense subset of size κ, so in particular it contains no well-ordered subset of size κ+. The following result shows that, however, every subset of L of smaller size is the union of countably many well-ordered subsets.

+ 16.3 Lemma. For each β < κ , Lβ = {fα : α < β} can be decomposed into countably many well-ordered subsets.

Proof. Let Lβn = {fα : α ∈ Fn(β)} for n < ω. Note that the projection f 7−→ f ¹ (n + 1) is one-to-one on Lβn so each Lβn is lexicographically well-ordered. a 110 I. Coherent Sequences

+ 16.4 Remark. Note that K = {{(n, fα(n)) : n < ω} : α < κ } is a family of countable subsets of ω ×κ which has the property that K ¹ X = {K ∩X : K ∈ K} has size ≤ |X| + ℵ0 for every X ⊆ ω × κ of size < κ. We now give an application of the existence of such a family of countable sets, due to Fremlin [25]. This requires a few standard notions from measure theory and real analysis.

16.5 Definition. A Radon measure space is a Hausdorff topological space X with a σ-finite measure µ defined on a field of subsets of X containing open sets with the property that the measure of every set is equal to the supremum of measures of its compact subsets. When µ(X) = 1 then (X, µ) is called Radon probability space. Finally recall that the Maharam type of (X, µ) is the least cardinality of a subset of the measure algebra of µ which completely generates the algebra.

16.6 Deﬁnition. A directed set D is Tukey reducible to a directed set E, written as D ≤ E, if there is f : D −→ E which maps unbounded sets to unbounded sets. In case D and E are ideals of subsets, this is equivalent to saying that there is q : E −→ D which is monotone and has coﬁnal range. We say that D and E are Tukey equivalent, D ≡ E, whenever D ≤ E and E ≤ D.

16.7 Theorem. If there is a locally countable subset of [κ]ω of size cf[κ]ω then the null ideal of every atomless Radon probability space of Maharam type κ is Tukey equivalent to the product N × [κ]ω where N is the ideal of measure zero subsets of the unit interval.

Proof. We shall give an argument for the case X = {0, 1}κ and µ the Haar measure of X, referring the reader to the general result from [25] that the null ideal of any atomless probability space of Maharam type κ is Tukey κ equivalent to the null ideal Nκ of {0, 1} . The only nontrivial direction is ω ω to produce a Tukey map from Nκ into Nω × [κ] . Let K ⊆ [κ] of size cf[κ]ω with the property that K ¹ Γ = {a ∩ Γ: a ∈ K} is countable for all countable Γ ⊆ κ. Let h : K −→ [κ]ω be a mapping whose range is cofinal in [κ]ω when considered as a directed set ordered by the inclusion. Let C = {a ∪ h(a): a ∈ K}. Then C is a cofinal subset of [κ]ω with the property that C(b) = {a ∈ C : a ⊆ b} is countable for every b ∈ [κ]ω. For c ∈ C fix a bijection ec : c −→ ω. This gives us a way to define a function κ ω πc mapping cylinders of {0, 1} with supports included in c into 2 in the cl κ natural way. Let Nκ be the family of all null subsets of {0, 1} that are κ cl countably supported cylinders of {0, 1} . Clearly, Nκ is cofinal in Nκ, so cl ω cl it suffices to produce a Tukey map f : Nκ −→ Nω × [κ] . Given N ∈ Nκ , find c(N) ∈ C such that supp(N) ⊆ c(N) and put

00 f(N) = (πc(N) N, c(N)). 16. Successors of singular cardinals 111

cl It is straightforward to check that f maps unbounded subsets of Nκ to ω unbounded subsets of the product Nω × [κ] . a 16.8 Remark. Note that the hypothesis of Fremlin’s theorem is true when ω [κ] has cofinality κ and so it is always satisfied for κ < ℵω. By remark 16.4 the hypothesis of Theorem 16.7 is satisfied for all κ if one assumes ¤κ and cf[κ]ω = κ+ for every κ of uncountable cofinality. Note that the transfer from a locally countable K to a cofinal C = {a ∪ h(a): a ∈ K} in the above proof does not preserve local countability. To obtain a locally countable cofinal C ⊆ [κ]ω one needs to work a bit harder along the same lines. 16.9 Definition. If a family K ⊆ [S]ω is at the same time locally countable and cofinal in [S]ω then we call it a cofinal Kurepa family (cofinal K-family in short). Two cofinal K-families H and K are compatible if H ∩ K ∈ H ∩ K for all H ∈ H and K ∈ K. We say that K extends H if they are compatible and if H ⊆ K. 16.10 Remark. Note that the size of any cofinal K-family K on a set S is equal to the cofinality of [S]ω. Note also that for every X ⊆ S there is Y ⊇ X of size cf[X]ω such that K ∩ Y ∈ K for all K ∈ K. 16.11 Definition. Define CK(θ) to be the statement that every sequence Kn (n < ω) of comparable cofinal K-families with domains included in θ which are closed under ∪, ∩ and \ can be extended to a single cofinal K-family on θ, which is also closed under these three operations.

16.12 Lemma. CK(ω1) is true and if CK(θ) is true for some θ such that cf[θ]ω = θ then CK(θ+) is also true.

Proof. The easy proof of CK(ω1) is left to the reader. Suppose CK(θ) and let Kn (n < ω) be a given sequence of compatible cofinal K-families as in the hypothesis of CK(θ+). By Remark 16.10 there + is a strictly increasing sequence δξ (ξ < θ ) of ordinals < θ such that + + Kn ¹ δξ ⊆ Kn for all ξ < θ and n < ω. Recursively on ξ < θ we + construct a chain Hξ (ξ < θ ) of cofinal K-families as follows. If ξ = 0 or ξ = η + 1 for some η, using CK(δξ) we can find a cofinal K-family Hξ on δξ extending Hξ−1 and Kn ¹ δξ (n < ω). If ξ has uncountable ¯ S cofinality then the union of Hξ = η<ξ Hη is a cofinal K-family with domain included in δξ, so using CK(δξ) we can find a cofinal K-family Hξ on δξ extending H¯ξ and Kn ¹ δξ (n < ω). If ξ has countable cofinality, pick a sequence {ξn} converging to ξ and use CK(δξ) to find a cofinal K-family

Hξ on δξ extending Hξn (n < ω) and Kn (n < ω). When the recursion is S + done, set H = ξ<θ+ Hξ. Then H is a coﬁnal K-family on θ extending Kn (n < ω). a

16.13 Corollary. For each n < ω there is a coﬁnal Kurepa family on ωn. a 112 I. Coherent Sequences

16.14 Definition. Let κ be a cardinal of cofinality ω.A Jensen matrix on + + + κ is a matrix Jαn (α < κ , n < ω) of subsets of κ with the following properties, where κn (n < ω) is some fixed increasing sequence of cardinals converging to κ:

+ (1) |Jαn| ≤ κn for all α < κ and n < ω,

(2) for all α < β and n < ω there is m < ω such that Jαn ⊆ Jβm,

S ω S S ω (3) n<ω[Jβn] = α<β n<ω[Jαn] whenever cf(β) > ω,

+ ω S S ω (4) [κ ] = α<κ+ n<ω[Jαn] . 16.15 Remark. The notion of a Jensen matrix is the combinatorial essence behind Silver’s proof of Jensen’s model-theoretic two-cardinal transfer theorem in the constructible universe (see [35, appendix]), so the matrix could equally well be called ’Silver matrix’. It has been implicitly or explicitly used on several places in the literature. The reader is referred to the paper of Foreman and Magidor [24] which gives a quite complete discussion of this notion and its occurrence in the literature.

16.16 Lemma. Suppose some cardinal κ of countable coﬁnality carries a + Jensen matrix Jαn (α < κ , n < ω) relative to some sequence of cardinals + κn (n < ω) that converge to κ. If CK(κn) holds for all n < ω then CK(κ ) is also true.

Proof. Let Kn (n < ω) be a given sequence of compatible cofinal K-families + with domains included in κ . Given Jαn, there is a natural continuous ξ + 0 chain Jαn (ξ < ω1) of subsets of κ ofS size ≤ κn such that Jαn = Jαn and Jξ+1 equal to the union of all K ∈ K which intersect Jξ . Let αnS n<ω n αn J∗ = Jξ . It is easily seen that J∗ (α < κ+, n < ω) is also a Jensen αn ξ<ω1 αn αn + matrix. By recursion on α and n we define a sequence Hαn (α < κ , n < ω) of compatible cofinal K-families as follows. If α = β + 1 or α = 0 and ∗ n < ω using CK(κn) we can find a cofinal K-family Hαn with domain Jαn ∗ compatible with Hαm (m < n), H(α−1)m (m < ω) and Km ¹ Jαn (m < ω). If cf(α) = ω let αn (n < ω) be an increasing sequence of ordinals converging to α. Using CK(κn) we can find a cofinal K-family Hαn which extends ∗ Hαm (m < n), Km ¹ Jαn (m < ω) and each of the families Hαik (i < ω, k < ω and J∗ ⊆ J∗ ). Finally suppose that cf(α) > ω. For n < ω, set αik αn [ [ ∗ ω Hαn = [Jαn] ∩ ( Hξm). ξ<α m<ω

Using the properties of the Jensen matrix (especially (3)) as well as the compatibility of Hξm (ξ < α, m < ω) one easily checks that Hαn is a coﬁnal ∗ K-family with domain Jαn which extends each member of Hαm (m < n) 16. Successors of singular cardinals 113

∗ and Km ¹ Jαn (m < ω) and which is compatible with all of the previously constructed families Hξm (ξ < α, m < ω). When the recursion is done we set [ [ H = Hαn. α<κ+ n<ω ∗ + Using the property (4) of Jαn (α < κ , n < ω), it follows easily that H is a + cofinal K-family on κ extending Kn (n < ω). a 16.17 Theorem. If a Jensen matrix exists on any successor of a cardinal of cofinality ω, then a cofinal Kurepa family exists on any domain. a

+ 2 The ρ-function ρ :[κ ] −→ κ associated with a ¤κ-sequence Cα (α < κ+) for some singular cardinal κ of coﬁnality ω leads to the matrix

+ Fn(α) = {ξ < α : ρ(ξ, α) ≤ n}(α < κ , n < ω) (I.222) which has the properties (1)-(3) of 16.14 as well as some other properties not captured by the deﬁnition of a Jensen matrix. If one additionally has + + + ω a sequence aα (α < κ ) of countable subsets of κ that is coﬁnal in [κ ] one can extend the matrix (I.222) as follows: [ + Mβn = (aα ∪ {α})(β < κ , n < ω). (I.223)

+ (Recall that

(1) α

(2) Mαm ⊆ Mαn whenever m < n, S (3) if β = sup{α : α < β} then M = M , n βn α

+ (5) M = {Mβn : β < κ , n < ω} is a locally countable family if we have + started with a locally countable K = {aα : α < κ }. a

+ 16.19 Remark. One can think of the matrix M = {Mβn : β < κ , n < ω} as a version of a ’morass’ for the singular cardinal κ (see [92]). It would be interesting to see how far one can go in this analogy. We give a few applications just to illustrate the usefulness of the families we have constructed so far. 114 I. Coherent Sequences

16.20 Deﬁnition. A Bernstein decomposition of a topological space X is a function f : X −→ 2N with the property that f takes all the values from 2N on any subset of X homeomorphic to the Cantor set.

16.21 Remark. The classical construction of Bernstein [8] can be interpreted by saying that every space of size at most continuum admits a Bern- stein decomposition. For larger spaces one must assume Hausdorff’s sep- aration axiom, a result of Neˇsetriland R˝odl(see [55]). In this context Malykhin was able to extend Bernstein’s result to all spaces of size < c+ω (see [50]). To extend this to all Hausdorff spaces, some use of square sequences seems natural. In fact, the first Bernstein decompositions of an ω + arbitrary Hausdorff space have been constructed using ¤κ and κ = κ for every κ > c of cofinality ω by Weiss [96] and Wolfsdorf [97]. We shall now see that cofinal K-families are quite natural tools in constructions of Bernstein decompositions.

16.22 Theorem. Suppose every regular θ > c supports a coﬁnal Kurepa family of size θ. Then every Hausdorﬀ space admits a Bernstein decomposition.

Proof. Let us say that a subset S of a topological space X is sequentially closed in X if it contains all limit points of sequences {xn} ⊆ S that converge in X. We say S is σ-sequentially closed in X if it can be decomposed into countably many sequentially closed sets. For a given cardinal θ, let B(θ) denote the statement that for every Hausdorff space X of size ≤ θ, every σ-sequentially closed subset S of X and every Bernstein decomposition f : S −→ 2N there is a Bernstein decomposition g : X −→ 2N extending f. Using the fact that a Hausdorff space of size ≤ c has at most c copies of the Cantor set and following the idea of Bernstein’s original proof, one sees that B(c) holds. Note also that by our assumption θℵ0 = θ for every cardinal θ ≥ c of uncountable cofinality. So if we have B(θ) for such θ + + one proves B(θ ) by constructing a chain fξ (ξ < θ ) of partial Bernstein decompositions on any Hausdorff topology τ on θ+ using the fact that the set C of all δ < θ+ which are σ-sequentially closed in (θ+, τ) is closed and + + unbounded in θ . Thus if δξ (ξ < θ ) enumerates C increasingly we will N have gξ : δξ −→ 2 and gξ compatible with the given partial Bernstein decomposition f of a given σ-sequentially closed set S ⊆ θ+. Note that S ∩ δξ will be σ-sequentially closed, so the use of B(δξ) to get fξ will be possible. This inductive procedure encounters a crucial difficulty only when θ = κ+ for some κ > c of cofinality ω. This is where a cofinal K-family K on κ+ is useful. Let τ be a given Hausdorff topology on θ+ and for K ∈ K let K0 be the collection of all limits of τ-convergent sequences of elements of K. Note that if a set Γ ⊆ κ+ is K-closed in the sense that K ¹ Γ ⊆ K (i.e. K ∩ Γ ∈ K for all K ∈ K) then Γ0 will be equal to the union of all K0 (K ∈ K ¹ Γ). So if |Γ| > c is a cardinal of uncountable cofinality then 16. Successors of singular cardinals 115

0 + |Γ | = |Γ|. It follows that every δ < θ with the property that δ = lim δn for some sequence δn (n < ω) of ordinals ≤ δ with the property that K ¹ δn ⊆ K for all n can be decomposed into countably many sets Γδn (n < ω) such that for each n, the set Γδn is σ-sequentially closed in τ and |Γδn| is a cardinal > c of uncountable cofinality. However, note that the set C of all such δ < θ+ is closed and unbounded in θ+ so as above we use our inductive N + hypothesis B(θ)(θ < κ) to construct a chain gξ : δξ −→ 2 (ξ < θ ) of Bernstein decompositions. This completes the proof. a It is interesting that various less pathological classes of spaces admit a local version of Theorem 16.22. 16.23 Theorem. A sigma-product 35 of the unit interval admits a Bern- stein decomposition if its index set carries a cofinal Kurepa family. So in particular, any metric space that carries a cofinal Kurepa family admits a Bernstein decomposition.

Proof. So let K be a ﬁxed well-founded coﬁnal K-family on Γ and let

1. There is an uncountable ∆ ⊆ Γ such that fa ¹ ∆ is ﬁnite-to-one for all a ∈ I, or

∗ 2. There is g :Γ −→ ω such that g ¹ a = fa for all a ∈ I. Proof. Let L be the family of all countable subsets b of Γ for which one can find an a in I such that b \ a is finite and fa is finite-to-one on b. To see that L is a P-ideal, let {bn} be a given sequence of members of L and for

35Recall that a sigma-product is the subspace of some Tychonoﬀ cube [0, 1]Γ consisting of all mappings x such that supp(x) = {ξ ∈ Γ: x(ξ) 6= 0} is countable. The set Γ is its index-set . 116 I. Coherent Sequences

each n fix a member an of I such that fan is finite-to-one on bn. Since I is a P-ideal, we can find a ∈ I such that an \ a is finite for all n. Note that for all n, bn \ a is finite and that fa is finite-to-one on bn. For n < ω, let

∗ bn = {ξ ∈ bn ∩ a : fa(ξ) > n}.

∗ Then bn is a cofinite subset of bn for each n, so if we set b to be equal to the ∗ union of the bn’s, we get a subset of a which almost includes each bn and on which fa is finite-to-one. It follows that b belongs to L. This finishes the proof that L is a P-ideal. Applying the P-ideal dichotomy to L, we get the two alternatives that translate into the alternatives 1 and 2 of Theorem 16.25. a

This leads to the natural question whether for any set Γ one can construct ω a family {fa : a −→ ω} of ﬁnite-to-one mappings indexed by [Γ] . This question was answered by Koszmider [40] using the notion of a Jensen matrix discussed above. We shall present Koszmider’s result using the notion of a coﬁnal Kurepa family instead.

16.26 Theorem. If Γ carries a coﬁnal Kurepa family then there is a co- ω herent family fa : a −→ ω (a ∈ [Γ] ) of ﬁnite-to-one mappings.

Proof. Let K be a ﬁxed well-founded coﬁnal K-family on Γ and let

16.27 Corollary. For every nonnegative integer n there is a coherent family ω fa : a −→ ω (a ∈ [ωn] ) of ﬁnite-to-one mappings. a

16.28 Remark. It is interesting that ’finite-to-one’ cannot be replaced by ’one-to-one’ in these results. For example, there is no coherent family of + ω one-to-one mappings fa : a −→ ω (a ∈ [c ] ). We finish this section with a typical application of coherent families of finite-to-one mappings discovered by Scheepers [61].

ω 16.29 Theorem. If there is a coherent family fa : a −→ ω (a ∈ [Γ] ) of ﬁnite-to-one mappings, then there is F : [[Γ]ω]2 −→ [Γ]<ω with the property that for every strictly ⊆-increasing sequence an (n < ω) of countable subsets of Γ, the union of F (an, an+1)(n < ω) covers the union of an (n < ω). 17. The oscillation mapping 117

ω Proof. For a ∈ [Γ] let xa : ω −→ ω be defined by letting xa(n) = |{ξ ∈ a : fa(ξ) ≤ n}|. Note that xa is eventually dominated by xb whenever a is a proper subset of b. Choose Φ : ωω −→ ωω with the property that x <∗ y implies Φ(y) <∗ Φ(x), where <∗ is the ordering of eventual dominance on ωω (i.e. x <∗ y if x(n) < y(n) for all but finitely many n’s). Define another ω family of functions ga : a −→ ω (a ∈ [Γ] ) by letting

ga(ξ) = Φ(xa)(fa(ξ)). (I.224)

ω Note the following interesting property of ga (a ∈ [Γ] ):

ω F (a, b) = {ξ ∈ a : gb(ξ) ≥ ga(ξ)} is ﬁnite for all a $ b in [Γ] . (I.225)

So if an (n < ω) is a strictly ⊆-increasing sequence of countable subsets of Γ ¯ ¯ and ξ belongs to some an¯ then the sequence of integers gan (ξ) (¯n ≤ n < ω) ¯ ¯ must have some place n ≥ n¯ with the property that gan (ξ) < gan+1 (ξ), i.e. a place n ≥ n¯ such that ξ ∈ F (an, an+1). a 16.30 Remark. Note that if κ is a singular cardinal of cofinality ω with the property that cf[θ]ω < κ for all θ < κ, then the existence of a cofinal Kurepa family on κ+ implies the existence of a Jensen matrix on κ+. So these two notions appear to be quite close to each other. The three basic properties of the function ρ :[κ+] −→ κ (14.5 and 14.6(a),(b)) seem much stronger in view of the fact that the linear ordering as in 16.3 cannot exist for κ above a supercompact cardinal and the fact that Foreman and Magidor [24] have produced a model with a supercompact cardinal that carries a Jensen matrix on any successor of a singular cardinal of cofinality ω. Note that + Chang’s conjecture of the form (κ , κ) ³ (ω1, ω) for some singular κ of cofinality ω implies that every locally countable family K ⊆ [κ]ω must have size ≤ κ. So, one of the models of set theory that has no cofinal K-family on, say ℵω+1, is the model of Levinski, Magidor and Shelah [49]. It seems still unknown whether the conclusion of Theorem 16.29 can be proved without additional set-theoretic assumptions.

17. The oscillation mapping

In what follows, θ will be a ﬁxed regular inﬁnite cardinal.

osc : P(θ)2 −→ Card (I.226) is defined by osc(x, y) = |x \ (sup(x ∩ y) + 1)/ ∼ |, where ∼ is the equivalence relation on x \ (sup(x ∩ y) + 1) defined by letting α ∼ β iff the closed interval determined by α and β contains no point from 118 I. Coherent Sequences y. So, if x and y are disjoint, osc(x, y) is simply the number of convex pieces the set x is split by the set y. The oscillation mapping has proven to be a useful device in various schemes for coding information. It usefulness in a given context depend very much of the corresponding ’oscillation theory’, a set of definitions and lemmas that disclose when it is possible to achieve a given number as oscillation between two sets x and y in a given family X . The following definition reveals the notion of largeness relevant to the oscillation theory that we develop in this section.

17.1 Deﬁnition. A family X ⊆ P(θ) is unbounded if for every closed and unbounded subset C of θ there exist x ∈ X and an increasing sequence {δn : n < ω} ⊆ C such that sup(x ∩ δn) < δn and [δn, δn+1) ∩ x 6= ∅ for all n < ω.

This notion of unboundedness has proven to be the key behind a number of results asserting the complex behaviour of the oscillation mapping on X 2. The case θ = ω seems to contain the deeper part of the oscillation theory known so far (see [79],[80, §1] and [88]), though in this section we shall only consider the case θ > ω. We shall also restrict ourselves to the family K(θ) of all closed bounded subsets of θ rather than the whole power-set of θ. The following is the basic result about the behaviour of the oscillation mapping in this context.

17.2 Lemma. If X is an unbounded subfamily of K(θ) then for every positive integer n there exist x and y in X such that osc(x, y) = n.

Proof. Choose an ∈-chain M of length θ of elementary submodels M of some large enough structure of the form Hλ such that X ∈ M and M ∩ θ ∈ θ. Applying the fact that X is unbounded with respect to the closed and unbounded set {M ∩ θ : M ∈ M} we can ﬁnd y ∈ X such that:

δi = Mi ∩ θ∈ / y for all i < n, (I.227)

δ0 ∩ y 6= ∅ and y \ δn−1 6= ∅, (I.228)

[δi−1, δi) ∩ y 6= ∅ for all 0 < i < n. (I.229) Let J0 = [0, sup(y ∩ δ0)], (I.230)

Ji = [δi−1, max(y ∩ δi)] for 0 < i < n, (I.231)

Jn = [δn−1, max(y)]. (I.232)

Let Fn be the collection of all increasing sequences I0 < I1 < . . . < In of closed intervals of θ such that I0 = J0 and such that there is x = xI in X such that x ⊆ I0 ∪ I1 ∪ ... ∪ In, (I.233) 17. The oscillation mapping 119

max(Ii) ∈ x for all i ≤ n. (I.234)

Clearly, Fn ∈ M0 and hJ0,J1,...,Jni ∈ Fn. Let Fn−1 be the collection of all hI0,...,In−1i such that for every α < θ there exists an interval I of θ above α such that hI0,...,In−1,Ii ∈ Fn. Note that Fn−1 ∈ M0 and that using the elementarity of Mn−1 one proves that hJ0,...,Jn−1i ∈ Fn−1. Let Fn−2 be the collection of all hI0,...,In−2i such that for every α < θ there exists interval I above α such that hI0,...,In−2,Ii ∈ Fn−1. Then again Fn−2 ∈ M0 and using the elementarity of Mn−2 one shows that the restriction hJ0,...,Jn−2i belongs to Fn−2. Continuing in this way we construct families Fn, Fn−1,..., F0 such that:

Fi ∈ M0 for all i ≤ n, (I.235)

F0 = {J0}. (I.236) hI ,...,I i ∈ F implies that for all α < θ there exists 0 n−i n−i (I.237) an interval I above α such that hI0,...,In−i,Ii ∈ Fn−i+1.

Recursively in i ≤ n we choose intervals I0 < I1 < . . . < In such that:

hI0,...,Iii ∈ Fi for all i ≤ n, (I.238)

Ji−1 < Ii ∈ Mi−1 for 0 < i ≤ n. (I.239) Clearly, there is no problem in choosing the sequence using the elementarity of the submodels Mi−1 and the condition (I.237). By the deﬁnition hI0,...Ini ∈ Fn means that there is x ∈ X satisfying (I.233) and (I.234). It follows that max(x ∩ y) = max(I0) and that

x ∩ Ii 6= ∅ (0 < i ≤ n) (I.240) are the convex pieces of x \ (max(x ∩ y) + 1) to which this set is split by y. It follows that osc(x, y) = n. This ﬁnishes the proof. a Lemma 17.2 also has a rectangular form. 17.3 Lemma. If X and Y are two unbounded subfamilies of K(θ) then for all but ﬁnitely many positive integers n there exist x ∈ X and y ∈ Y such that osc(x, y) = n. a

Recall the notion of a nontrivial C-sequence Cα (α < θ) on θ from Section 11, a C-sequence with the property that for every closed and unbounded subset C of θ there is a limit point δ of C such that C ∩ δ * Cα for all α < θ. 17.4 Deﬁnition. For a subset D of θ let lim(D) denote the set of all α < θ with the property that α = sup(D ∩ α). A subsequence Cα (α ∈ Γ) of some C-sequence Cα (α < θ) is stationary if the union of all lim(Cα)(α ∈ Γ) is a stationary subset of θ. 120 I. Coherent Sequences

17.5 Lemma. A stationary subsequence of a nontrivial C-sequence on θ is an unbounded family of subsets of θ.

Proof. Let Cα (α ∈ Γ) be a given stationary subsequence of a nontrivial C-sequence on θ. Let C be a given closed and unbounded subset of θ. Let ∆ be the union of all lim(Cα)(α ∈ Γ). Then ∆ is a stationary subset of θ.

For ξ ∈ ∆ choose αξ ∈ Γ such that ξ ∈ lim(Cαξ ). Applying the assumption that Cα (α ∈ Γ) is a nontrivial C-subsequence, we can ﬁnd ξ ∈ ∆ ∩ lim(C) such that

C ∩ [η, ξ) * Cαξ for all η < ξ. (I.241) If such ξ cannot be found using the stationarity of the set ∆ ∩ lim(C) we would be able to use the Pressing Down Lemma on the regressive mapping that would give us an η < ξ violating (I.241) and get that a tail of C trivializes Cα (α ∈ Γ). Using (I.241) and the fact that Cαξ ∩ξ is unbounded in ξ we can ﬁnd a strictly increasing sequence δn (n < ω) of elements of

(C ∩ξ)\Cαξ such that [δn, δn+1)∩Cαξ 6= ∅ for all n. So the set Cαξ satisﬁes the conclusion of Deﬁnition 17.1 for the given closed and unbounded set C. a

Recall that Qθ denotes the set of all ﬁnite sequences of ordinals < θ and that we consider it ordered by the right lexicographical ordering. We need the following two further orderings on Qθ: s v t if and only if s is an initial segment of t. (I.242)

s < t if and only if s is a proper initial part of t. (I.243)

17.6 Deﬁnition. Given a C-sequence Cα (α < θ) we can deﬁne an action (α, t) 7−→ αt of Qθ on θ recursively on the ordering v of Qθ as follows:

α∅ = α, (I.244)

αhξi = the ξth member of Cα if ξ < tp(Cα); otherwise αhξi = α, (I.245)

αtahξi = (αt)hξi. (I.246)

17.7 Remark. Note that if ρ0(α, β) = t for some α < β < θ then βt = α. In fact, if β = β0 > . . . > βn = α is the walk from β to α along the C-sequence, each member of the trace Tr(α, β) = {β0, β1, . . . , βn} has the form βs where s is the uniquely determined initial part of t. Note, however, that in general βt = α does not imply that ρ0(α, β) = t.

17.8 Notation. Given a C-sequence Cα (α < θ) on θ we shall use osc(α, β) to denote osc(Cα,Cβ).

17.9 Theorem. If Cα (α < θ) is a nontrivial C-sequence on θ, then for every unbounded set Γ ⊆ θ and positive integer n there exist α < β in Γ and t v ρ0(α, β) such that 17. The oscillation mapping 121

(a) osc(αt, βt) = n,

(b) osc(αs, βs) = 1 for all s < t. Proof. Choose an elementary submodel M of some large enough structure ¯ of the form Hλ such that δ = M ∩ θ ∈ θ and M contains all the relevant objects. Choose β¯ ∈ Γ above δ¯ and let ¯ ¯ ¯ ¯ ¯ β = β0 > β1 > . . . > βk = δ be the minimal walk from β¯ to δ¯ along the C-sequence. Let k¯ = k − 1 if ¯ ¯ ¯ Cβk−1 ∩ δ is unbounded; otherwise k = k. Recall our implicit assumption about essentially all C-sequences that we consider here: if γ is a limit ordinal then a ξ ∈ Cγ occupying a successor place in Cγ must be a successor ordinal. ¯ ¯ In this case, since δ is a limit ordinal and since δ ∈ Cβk−1 , either it must ¯ ¯ ¯ ¯ ¯ ¯ be that βk−1 is a limit ordinal and δ = sup(Cβk−1 ∩ δ) or βk−1 = δ + 1.It ¯ ¯ ¯ follows that βk¯ is a limit ordinal. Set t = ρ0(βk¯, β). Let ∅ = t ¹ 0, t ¹ 1, . . . , t ¹ k¯ = t ¯ ¯ be our notation for all the restrictions of the sequence t. Note that βi = βt¹i for i ≤ k. Let Γ0 be the set of all β ∈ Γ such that:

β = β∅ > βt¹1 > . . . > βt, (I.247) ¯ ¯ sup(Cβt¹i ∩ βt) = sup(Cβ¯i ∩ βk¯) for all i < k. (I.248) ¯ Note that Γ0 ∈ M and β ∈ Γ0. Let ∆ = {βt : β ∈ Γ0}. Then ∆ ∈ M and β¯¯ ∈ ∆. By the very choice of k¯, δ¯ belongs to lim(C ¯ ), so using the k βk¯ elementarity of M one concludes that Cδ (δ ∈ ∆) is a stationary subsequence of our original sequence. Shrinking ∆ a bit if necessary, we may assume that for every γ ∈ ∆ an α ∈ Γ0 with αt = γ can be found below the next member of ∆ above γ. By Lemma 17.5, Cδ (δ ∈ ∆) is an unbounded family of subsets of θ and so we can apply the basic oscillation Lemma 17.2 and ﬁnd γ < δ such that osc(αt, βt) = n. Using the fact α < δ and the fact that α and β ¯ satisfy (I.248) we conclude that osc(αt¹i, βt¹i) = 1 for all i < k. This ﬁnishes the proof. a 17.10 Corollary. Suppose a regular uncountable cardinal θ carries a nontrivial C-sequence. Then there is f :[θ]2 −→ ω which takes all the values from ω on any set of the form [Γ]2 for an unbounded subset of θ.

Proof. Given α < β < θ, if there is t v ρ0(α, β) satisfying 17.9(a),(b) put

f(α, β) = osc(αt, βt) − 2 (I.249) otherwise put f(α, β) = 0. a 122 I. Coherent Sequences

17.11 Remark. The class of all regular cardinals θ that carry a nontrivial C-sequence is quite extensive. It includes not only all successor cardinals but also some inaccessible as well as hyperinaccessible cardinals such as for example, ﬁrst inaccessible or ﬁrst Mahlo cardinals. In view of the well- known Ramsey-theoretic characterization of weak compactness, Corollary 17.10 leads us to the following natural question. 17.12 Question. Can the weak compactness of a strong limit regular uncountable cardinal be characterized by the fact that for every f :[θ]2 −→ ω there exists an unbounded set Γ ⊆ θ such that f 00[Γ]2 6= ω? This is true when ω is replaced by 2, but can any other number beside 2 be used in this characterization?

18. The square-bracket operation

In this Section we show that the basic idea of the square-bracket operation on ω1 introduced in Deﬁnition 7.3 extends to a general setting on an arbitrary uncountable regular cardinal θ that carries a nontrivial C-sequence Cα (α < θ). The basic idea is based on the oscillation map deﬁned in the previous section and, in particular, on the property of this map exposed in Theorem 17.9: for α < β < θ we set

[αβ] = βt, where t v ρ0(α, β) is such that osc(αt, βt) ≥ 2 but osc(αs, βs) = 1 for all s < t; if such t does not exist, we (I.250) let [αβ] = α.

Thus, [αβ] is the ﬁrst place visited by β on its walk to α where a nontrivial oscillation with the corresponding step of α occurs. What Theorem 17.9 is telling us is that the nontrivial oscillation indeed happens most of the time. Results that would say that the set of values {[αβ]: {α, β} ∈ [Γ]2} is in some sense large no matter how small the unbounded set Γ ⊆ θ is, would correspond to the results of Lemmas 7.4-7.5 about the square-bracket operation on ω1. It turns out that this is indeed possible and to describe it we need the following deﬁnition.

18.1 Deﬁnition. A C-sequence Cα (α < θ) on θ avoids a given subset ∆ of θ if Cα ∩ ∆ = ∅ for all limit ordinals α < θ.

18.2 Lemma. Suppose Cα (α < θ) is a given C-sequence on θ that avoids a set ∆ ⊆ θ. Then for every unbounded set Γ ⊆ θ, the set of elements of ∆ not of the form [αβ] for some α < β in Γ is nonstationary in θ. Proof. Let Ω be a given stationary subset of ∆. We need to ﬁnd α < β in Γ such that [αβ] ∈ Ω. For a limit δ ∈ Ω ﬁx a β = β(δ) ∈ Γ above δ and let

β = β0 > β1 > . . . > βk(δ) = δ 18. The square-bracket operation 123

be the walk from β to δ along the sequence Cα (α < θ). Since δ∈ / Cα for any limit α and since Cα+1 = {α} for all α, we must have that βk(δ)−1 = δ + 1. In particular,

Cβi ∩ δ is bounded in δ for all i < k(δ). (I.251)

Applying the Pressing Down Lemma gives us a stationary set Ω0 ⊆ Ω, an integer k, ordinals ξi (i < k), and a sequence t ∈ Qθ such that for all δ ∈ Ω0:

k(δ) = k and ρ0(δ, β(δ)) = t, (I.252)

max(Cβ(δ)i ∩ δ) = ξi for all i < k. (I.253)

Intersecting Ω0 with a closed and unbounded subset of θ, we may assume for every δ ∈ Ω0 that β(δ) is smaller than the next member of Ω0 above δ. Note that any C-sequence on a regular uncountable cardinal that avoids a stationary subset of the cardinal, in particular, must be nontrivial. Thus Cδ (δ ∈ Ω0) is a stationary subsequence of a nontrivial C-sequence, and so by Lemma 17.5, unbounded as a family of subsets of θ. By the basic oscillation Lemma 17.2, we can ﬁnd γ < δ in Ω0 such that osc(Cγ ,Cδ) = 2. Then by the choice of Ω, if α = β(γ) and β = β(δ) then γ = αt, δ = βt and α < δ. Comparing this with (I.252) and (I.253) we conclude that [αβ] = βt = δ ∈ Ω. This ﬁnishes the proof. a It is clear that the argument gives the following more general result.

18.3 Lemma. Suppose Cα (α < θ) avoids ∆ ⊆ θ and let A be a family of size θ consisting of pairwise disjoint finite sets, all of some fixed size n. Then the set of all elements of ∆ that are not of the form [a(1)b(1)] = [a(2)b(2)] = ... = [a(n)b(n)] for some a 6= b in A is nonstationary in θ. a Since [αβ] belongs to the trace Tr(α, β) of the walk from β to α it is not surprising that [··] strongly depends on the behaviour of Tr. For example the following should be clear from the proof of Lemma 18.2. 18.4 Lemma. The following are equivalent for a given C-sequence and a set Ω: (a) Ω \{[αβ]: {α, β} ∈ [Γ]2} is nonstationary in θ, S (b) Ω \ {Tr(α, β): {α, β} ∈ [Γ]2} is nonstationary in θ. a This fact suggests the following definition.

18.5 Definition. The trace filter of a given C-sequenceS Cα (α < θ) is the normal filter on θ generated by sets of the form {Tr(α, β): {α, β} ∈ [Γ]2} where Γ is an unbounded subset of θ. 124 I. Coherent Sequences

18.6 Remark. Having a proper (i.e. 6= P(θ)) trace filter is a strengthening of the nontriviality requirement on a given C-sequence Cα (α < θ). For example, if a C-sequence avoids a stationary set Ω ⊆ θ, then its trace filter is nontrivial and in fact no stationary subset of Γ is a member of it. Note the following analogue of Lemma 18.4: the trace filter of a given C-sequence is the normal filter generated by sets of the form {[αβ]: {α, β} ∈ [Γ]2} where Γ is an unbounded subset of θ. So to obtain the analogues of the results of Section 7 about the square-bracket operation on ω1 one needs a C-sequence Cα (α < θ) on θ whose trace filter is not only nontrivial but also not θ-saturated, i.e. it allows a family of θ pairwise disjoint positive sets. It turns out that the hypothesis of Lemma 18.2 is sufficient for both of these conclusions.

18.7 Lemma. If a C-sequence on θ avoids a stationary subset of θ, then there exist θ pairwise disjoint subsets of θ that are positive with respect to its trace ﬁlter. 36

Proof. This follows from the well-known fact (see [37]) that if there is a normal, nontrivial and θ-saturated ﬁlter on θ, then for every stationary Ω ⊆ θ there exists λ < θ such that Ω ∩ λ is stationary in λ (and the fact that the stationary set which is avoided by the C-sequence does not reﬂect in this way). a

18.8 Corollary. If a regular cardinal θ admits a nonreﬂecting stationary subset then there is c :[θ]2 −→ θ which takes all the values from θ on any set of the form [Γ]2 for some unbounded set Γ ⊆ θ. a

To get such a c, one composes the square-bracket operation of some C- sequence, that avoids a stationary subset of θ, with a mapping ∗ : θ −→ θ with the property that the ∗-preimage of each point from θ is positive with respect to the trace ﬁlter of the square sequence. In other words c is equal to the composition of [··] and ∗, i.e. c(α, β) = [αβ]∗. Note that, as in Section 7, the property of the square-bracket operation from Lemma 18.3 leads to the following rigidity result which corresponds to Lemma 7.10.

18.9 Lemma. The algebraic structure (θ, [··], ∗) has no nontrivial automorphisms. a

18.10 Remark. Note that every θ which is a successor of a regular cardinal κ admits a nonreﬂecting stationary set. For example Ω = {δ < θ : cf δ = κ} is such a set. Thus any C-sequence on θ that avoids Ω leads to a square bracket operation which allows analogues of all the results from Section 7 about the square-bracket operation on ω1. The reader is urged to examine

36A subset A of the domain of some ﬁlter F is positive with respect to F if A ∩ F 6= ∅ for every F ∈ F. 18. The square-bracket operation 125 these analogues. We ﬁnish this section by giving a projection of the square- bracket operation (the analogue of 7.14) which has been used by Shelah and Steprans [69] in order to construct a family of nonisomorphic extraspecial p-groups that contain no large abelian subgroups.

Suppose now that θ = κ+ for some regular cardinal κ and fix a C-sequence + + Cα (α < κ ) on κ such that tp(Cα) ≤ κ for all α, or equivalently, such + + that Cα (α < κ ) avoids the set Ωκ = {δ < κ : cf δ = κ}. Let [··] be the corresponding square-bracket operation. Let λ be the minimal cardinal λ + + such that 2 ≥ κ . Choose a sequence rξ (ξ < κ ) of distinct subsets of λ. Let H be the collection of all maps h : P(D(h)) −→ κ+ where D(h) is a finite subset of λ. Let π : κ+ −→ H be a map with the property that −1 π (h) ∩ Ωκ is stationary for all h ∈ H. Finally, define an operation [[··]] on κ+ as follows: [[αβ]] = π([αβ])(rα ∩ D(π([αβ]))). (I.254) The following is a simple consequence of the property 18.3 of the square- bracket operation.

18.11 Lemma. For every family A of size κ+ consisting of pairwise disjoint + ﬁnite subsets of κ all of some ﬁxed size n and every sequence ξ0, . . . , ξn−1 + of ordinals < κ there exist a 6= b in A such that [[a(i)b(i)]] = ξi for all i < n. a

18.12 Remark. Shelah and Steprans consider the function Φ : [κ+]2 −→ 2 deﬁned by Φ(α, β) = 0 iﬀ [[αβ]] = 0 in [69] which they then incorporate into a general construction, due to Ehrenfeucht and Faber, to obtain an extraspecial p-group GΓ associated to essentially every unbounded subset + + Γ of κ . They show that GΓ contains no abelian subgroup of size κ using not only Lemma 18.11 but also the behaviour of [[a(i)b(j)]] for i 6= j < n, a 6= b ∈ A that follows from the behaviour of the square-bracket operation, exposed above in Lemma 7.7.

For suﬃciently large cardinals θ we have the following variation on the theme ﬁrst encountered above in Theorem 11.2.

18.13 Theorem. Suppose θ is bigger than the continuum and carries a C- sequence avoiding a stationary set Γ of cofinality > ω ordinals in θ. Let A be a family of θ pairwise disjoint finite subsets of θ, all of some fixed size n. n Then for every stationary Γ0 ⊆ Γ there exist s, t ∈ ω and a positive integer 37 k such that for every l < ω there exist a < b in A and δ0 > δ1 > . . . > δl in Γ0 ∩ (max(a), min(b)) such that:

(1) ρ2(δi+1, δi) = k for all i < l,

37Recall that if a and b are two sets of ordinals, then the notation a < b means that max(a) < min(b). 126 I. Coherent Sequences

a a a a (2) ρ0(a(i), b(j)) = ρ0(δ0, b(j)) ρ0(δ1, δ0) ... ρ0(δl, δl−1) ρ0(a(i), δl) for all i, j < n,

(3) ρ2(δ0, b(j)) = tj and ρ2(a(i), δl) = si for all i, j < n. Proof. For ∆ ⊆ Γ and δ < θ set \ Sδ(A) = {hρ2(a(i), δ): i < ni : a ∈ A ¹ [γ, δ)}, (I.255) γ<δ \ Tδ(∆) = {ρ2(α, δ): α ∈ ∆ ∩ [γ, δ)}. (I.256) γ<δ n Note that Sδ(A) and Tδ(∆) range over subsets of ω and ω, respectively, so if δ has uncountable cofinality and if A and ∆ are unbounded in δ then both of these sets, Sδ(A) and Tδ(∆), are nonempty. Let Γ0 be a given stationary subset of Γ. Since θ has size bigger than the continuum, starting from Γ0 one canT find an infinite sequence Γ0 ⊇ Γ1 ⊇ ... of subsets of Γ such that Γω = i<ω Γi is stationary in θ and

Sδ(A) = Sε(A) 6= ∅ for all δ, ε ∈ Γ1, (I.257)

Tδ(Γi) = Tε(Γi) 6= ∅ for all δ, ε ∈ Γi+1 and i < ω. (I.258)

Pick δ ∈ Γω such that Γω ∩ δ is unbounded in δ and choose b ∈ A above δ. Let tj = ρ2(δ, bj) for j < n. Now let s be an arbitrary member of Sδ(A) and k an arbitrary member of Tδ(Γω). To check that these objects satisfy the conclusion of the Theorem, let l < ω be given. Let δ0 = δ and let γ0 < δ0 be an upper bound for all sets of the form Cξ ∩ δ0 (ξ ∈ Tr(δ0, b(j)), ξ 6= δ0, j < n). Then the walk from any b(j) to any α ∈ [γ0, δ0) must pass through δ0. Since k ∈ Tδ(Γω) ⊆ Tδ(Γl) we can ﬁnd δ1 ∈ Γl ∩ (γ0, δ0) such that ρ2(δ1, δ0) = k. Choose γ1 ∈ (γ0, δ1) such that the walk from δ0 to any

α ∈ [γ1, δ1) must pass through δ1. Since k ∈ Tδ(Γω) ⊆ Tδ(Γl−1) = Tδ1 (Γl−1) there exists δ2 ∈ Γl−1∩(γ1, δ1) such that ρ2(δ2, δ1) = k. Choose γ2 ∈ (γ1, δ2) such that the walk from δ1 to any α ∈ [γ2, δ2) must pass through δ2, and so on. Proceeding in this way, we ﬁnd δ = δ0 > δ1 > . . . > δl > γl > γl−1 > . . . > γ0 such that: δi ∈ Γl−i+1 for all 0 < i ≤ l, (I.259)

ρ2(δi+1, δi) = k for all i < l, (I.260) for every i < l the walk from δ to some α ∈ [γ , δ ) i i+1 i+1 (I.261) passes through δi+1.

Since s ∈ Sδ(A) = Sδl (A), we can ﬁnd a ∈ A ¹ [γl, δl) such that ρ2(a(i), δl) = s(i) for all i < n. Now, given i, j < n, the walk from b(j) to a(i) ﬁrst passes through all the δi (i ≤ l), so ρ0(a(i), b(j)) is simply equal to the concatenation of ρ0(δ0, b(j)), ρ0(δ1, δ0), ..., ρ0(δl, δl−1), ρ0(a(i), δl). This together with (I.260) gives us the three conclusions of Theorem 18.13. a 18. The square-bracket operation 127

From now on θ is assumed to be a fixed cardinal satisfying the hypotheses of Theorem 18.13. It turns out that Theorem 18.13 gives us a way to define another square-bracket operation which has complex behavior not only on squares of unbounded subsets of θ but also on rectangles formed by two unbounded subsets of θ. To define this new operation we choose a mapping h : ω −→ ω such that: for every k, m, n, p < ω and s ∈ ωn there exist l < ω such (I.262) that h(m + l · k + s(i)) = m + p for all i < n.

2 18.14 Deﬁnition. [··]h :[θ] −→ θ is deﬁned by letting [αβ]h = βt where t = ρ0(α, β) ¹ h(ρ2(α, β)).

Thus, [αβ]h is the h(ρ2(α, β))th place that β visits on its walk to α. It is clear that Theorem 18.13 and the choice of h in (I.262) give us the following conclusion.

18.15 Lemma. Let A be a family of θ pairwise disjoint ﬁnite subsets of θ, all of some ﬁxed size n, and let Ω be an unbounded subset of θ. Then almost 38 every δ ∈ Γ has the form [a(0)β]h = [a(1)β]h = ... = [a(n − 1)β]h for some a ∈ A, β ∈ Ω, a < β. a

In fact, one can get a projection of this square-bracket operation with seemingly even more complex behavior. Keeping the notation of Theorem 18.13, pick a function ξ 7−→ ξ∗ from θ to ω such that {ξ ∈ Γ: ξ∗ = n} is stationary for all n. This gives us a way to consider the following projection of the trace function Tr∗ :[θ]2 −→ ω<ω:

∗ ∗ a ∗ Tr (α, β) = hmin(Cβ \ α) i Tr (α, min(Cβ \ α)), (I.263) where we stipulate that Tr∗(γ, γ) = hγ∗i for all γ. It is clear that the proof of Theorem 18.13 allows us to add the following conclusions: 18.13∗ Theorem. Under the hypothesis of Theorem 18.13, its conclusion can be extended by adding the following two new statements:

∗ ∗ (4) Tr (δ1, δ0) = ... = Tr (δl, δl−1),

∗ ∗ (5) The maximal term of the sequence Tr (δ1, δ0) = ... = Tr (δl, δl−1) is ∗ bigger than the maximal term of any of the sequences Tr (δ0, b(j)) or ∗ Tr (a(i), δl) for i, j < n. a

∗ 18.16 Deﬁnition. For α < β < θ, let [αβ] = βt for t the minimal initial ∗ ∗ part of ρ0(α, β) such that βt = max(Tr (α, β)).

38Here ’almost every’ is to be interpreted by ’all except a nonstationary set’. 128 I. Coherent Sequences

Thus [αβ]∗ is the first place in the walk from β to α where the function ∗ reaches its maximum among all other places visited during the walk. Note that combining the conclusions (1)-(5) of Theorem 18.13(∗) we get: 18.13∗∗ Theorem. Under the hypothesis of Theorem 18.13, its conclusion can be extended by adding the following: ∗ ∗ (6) [a(i)b(j)] = [δ1δ0] for all i, j < n. a Having in mind the property of [··]h stated in Lemma 18.15, the following variation is now quite natural. ∗ ∗ 18.17 Definition. [αβ]h = [α[αβ] ]h for α < β < θ. Using Theorem 18.13(∗∗)(1)-(6) one easily gets the following conclusion. 18.18 Lemma. Let A be a family of θ pairwise disjoint finite subsets of θ, all of some fixed size n. Then for all but nonstationarily many δ ∈ Γ one ∗ can find a < b in A such that [a(i)b(j)]h = δ for all i, j < n. a ∗ 18.19 Remark. Composing [··]h with a mapping π : θ −→ θ with the property that π−1(ξ) ∩ Γ is stationary for all ξ < θ, one gets a projection ∗ of [··]h for which the conclusion of Lemma 18.18 is true for all δ < κ. Assuming that θ is moreover a successor of a regular cardinal κ (of size at least continuum), in which case Γ can be taken to be {δ < κ+ : cf δ = κ}, ∗ and proceeding as in 18.11 above we get a projection [[··]]h with the following property: 18.20 Lemma. For every family A of pairwise disjoint finite subsets of κ+ all of some fixed size n and for every q : n × n −→ κ+ there exist a < b in ∗ A such that [[a(i)b(j)]]h = q(i, j) for all i, j < n. a 18.21 Remark. The first example of a cardinal with such a complex binary operation was given by the author [79] using the oscillation mapping described above in Section 17. It was the cardinal b, the minimal cardinality of an unbounded subset of ωω under the ordering of eventual dominance. The oscillation mapping restricted to some well-ordered unbounded subset W of ωω is perhaps still the most interesting example of this kind due to the fact that its properties are preserved in forcing extensions that do not change the unboundedness of W (although they can collapse cardinals and therefore destroy the properties of the square-bracket operations on them). This absoluteness of osc is the key feature behind its applications in various coding procedures (see, e.g. [85]). 18.22 Theorem. For every regular cardinal κ of size at least the continuum, the κ+-chain condition is not productive, i.e., there exist two partially + ordered sets P0 and P1 satisfying the κ -chain condition but their product P0 × P1 fails to have this property. 19. Unbounded functions on successors of regular cardinals 129

+ Proof. Fix two disjoint stationary subsets Γ0 and Γ1 of {δ < κ : cf δ = κ}. + Let Pi be the collection of all finite subsets p of κ with the property that ∗ + [αβ]h ∈ Γi for all α < β in p. By Lemma 18.18, P0 and P1 are κ -cc posets. + Their product P0 × P1, however, contains a family h{α}, {α}i (α < κ ) of pairwise incomparable conditions. a 18.23 Remark. Theorem 18.22 is due to Shelah [65] who proved it using similar methods. The first ZFC-examples of non-productiveness of the κ+- chain condition were given by the author in [77] using what is today known under the name ’pcf theory’. After the full development of pcf theory it became apparent that the basic construction from [77] applies to every successor of a singular cardinal [66]. A quite different class of cardinals θ with θ-cc non-productive was given by the author in [76]. For example, θ = cf c is one of these cardinals. For an overview of recent advances in this area, the reader is referred to [53]. The following problem seems still open: 18.24 Question. Suppose that θ is a regular strong limit cardinal and the θ-chain condition is productive. Is θ necessarily a weakly compact cardinal?

19. Unbounded functions on successors of regular cardinals

+ In this section κ is a regular cardinal and Cα (α < κ ) is a ﬁxed sequence + ∗ + 2 with tp(Cα) ≤ κ for all α < κ . Deﬁne ρ :[κ ] −→ κ by

ρ∗(α, β) = sup{tp(C ∩ α), ρ∗(α, min(C \ α)), β β (I.264) ∗ ρ (ξ, α): ξ ∈ Cβ ∩ α}, where we stipulate that ρ∗(γ, γ) = 0 for all γ < κ+. Since ρ∗(α, β) ≥ + ρ1(α, β) for all α < β < κ by Lemma 10.1 we have the following:

+ ∗ 19.1 Lemma. For ν < κ, α < κ the set Pν (α) = {ξ ≤ α : ρ (ξ, α) ≤ ν} has size no more than |ν| + ℵ0. a The proof of the following subadditivity properties of ρ∗ is very similar to the proof of the corresponding fact for the function ρ from Section 14. 19.2 Lemma. For all α ≤ β ≤ γ, (a) ρ∗(α, γ) ≤ max{ρ∗(α, β), ρ∗(β, γ)}, (b) ρ∗(α, β) ≤ max{ρ∗(α, γ), ρ∗(β, γ)}. a 130 I. Coherent Sequences

We mention a typical application of this function to the problem of exis- tance of partial square sequences which, for example, have some applications in the pcf theory (see [9]). 19.3 Theorem. For every regular uncountable cardinal λ < κ and stationary Γ ⊆ {δ < κ+ : cf(δ) = λ}, there is a stationary set Σ ⊆ Γ and a sequence Cα (α ∈ Σ) such that:

(1) Cα is a closed and unbounded subset of α,

(2) Cα ∩ ξ = Cβ ∩ ξ for every ξ ∈ Cα ∩ Cβ.

Proof. For each δ ∈ Γ, choose ν = ν(δ) < κ such that the set P<ν (δ) = {ξ < δ : ρ∗(ξ, δ) < ν} is unbounded in δ and closed under taking suprema of sequences of size < λ. Then there isν, ¯ µ¯ < κ and stationary Σ ⊆ Γ such that ν(δ) =ν ¯ and tp(P<ν¯(δ)) =µ ¯ for all δ ∈ Σ. Let C be a ﬁxed closed and unbounded subset ofµ ¯ of order-type λ. Finally, for δ ∈ Γ set

Cδ = {α ∈ P<ν¯(δ) : tp(P<ν¯(α)) ∈ C}.

Using Lemma 19.2, one easily checks that Cα (α ∈ Σ) satisfies the conditions (1) and (2). a Another application concerns the fact exposed above in Section 16, that the inequalities 19.2(a),(b) are particularly useful, when κ has cofinality ω. Also consider the well-known phenomenon first discovered by K.Prikry (see [37]), that in some cases, the cofinality of a regular cardinal κ can be changed to ω, while preserving all cardinals. 19.4 Theorem. In any cardinal-preserving extension of the universe, which has no new bounded subsets of κ, but in which κ has a cofinal ω-sequence diagonalizing the filter of closed and unbounded subsets of κ restricted to + the ordinals of cofinality > ω, there is a sequence Cαn(α ∈ lim (κ ), n < ω) such that for all α < β in lim(κ+):

(1) Cαn is a closed subset of α for all n,

(2) Cαn ⊆ Cαm, whenever n ≤ m, S (3) α = n<ω Cαn,

(4) α ∈ lim(Cβn) implies Cαn = Cβn ∩ α.

+ Proof. For α < κ , let Dα be the collection of all ν < κ for which P<ν (α) is σ-closed, i.e., closed under supremums of all of its countable bounded subsets. Clearly, Dα contains a closed unbounded subset of κ, restricted to coﬁnality > ω ordinals. Note that

∗ ν ∈ Dβ and ρ (α, β) < ν imply that ν ∈ Dα. (I.265) 19. Unbounded functions on successors of regular cardinals 131

In the extended universe, pick a strictly increasing sequence νn (n < ω), which converges to κ and has the property that for each α < κ+, there is n < ω such that νm ∈ Dα for all m ≥ n. Let n(α) be the minimal such integer n. Fix α ∈ lim(κ+) and n < ω. If there is γ ≥ α such that n ≥ n(γ) and

α ∈ lim(P<νn (γ)) (I.266) let γ(α) be the minimal such γ and let

Cαn = P<νn (γ(α)) ∩ α. (I.267)

If such a γ ≥ α does not exist, let Cαn = ∅ for n < n(α) and let

Cαn = P<νn (α) ∩ α. (I.268) for n ≥ n(α). Clearly, this choice of Cαn’s satisfies (1),(2) and (3). To verify (4), let α be a limit point of some Cβn. Suppose first that cf(α) = ω. If Cβn was defined according to the case I.267 so is Cαn. Since P<νn (γ(β)) is σ-closed we conclude that α ∈

P<νn (γ(β)) which by 19.2(a),(b) means that P<ν (α) = P<ν (γ(β)) ∩ α for all ν ≥ νn and therefore Dγ(β) \ νn ⊆ Dα. It follows that n(α) ≥ n and therefore that γ(α) = α. Hence, Cαn = Cβn ∩ α. If Cβn was defined according to the case I.268 then our assumption α ∈ lim(Cβn) and the fact that n ≥ n(β) means that Cαn was defined according to I.267. Similarly as above we conclude that actually α ∈ P<νn (β) which gives us n ≥ n(α) and γ(α) = α. It follows that Cαn = Cβn ∩ α in this case as well. Suppose now that cf(α) > ω. If Cβn was defined according to I.267 then so is Cαn and so Pνn (γ(α)) ∩ α and Pνn (γ(β)) ∩ α are two σ-closed and unbounded subsets of α. So their intersection is unbounded in α which by

19.2(a),(b) gives us that Pνn (γ(α)) ∩ α = Pνn (γ(β)) ∩ α. It follows that Cαn = Cβn ∩ α in this case. Suppose now that Cβn was defined according to I.268. Then γ = β satisfies n ≥ n(γ) and I.266. So Cαn was defined according to the case I.267. It follows that Pνn (γ(α))∩α and Pνn (β)∩α are two σ-closed and unbounded subsets of α so their intersection is unbounded in α. Applying 19.2(a),(b) we get that Pνn (γ(α))∩α = Pνn (β)∩α. It follows that Cαn = Cβn ∩ α in this case as well. a 19.5 Remark. The combinatorial principle appearing in the statement of Theorem 19.4 is a member of a family of square principles that has been studied systematically by Schimmerling and others (see e.g. [62]). It is definitely a principle sufficient for all of the applications of ¤κ appearing in Section 16 above. 19.6 Definition. A function f :[κ+]2 −→ κ is unbounded if f 00[Γ]2 is unbounded in κ for every Γ ⊆ κ+ of size κ. We shall say that such an f is 132 I. Coherent Sequences strongly unbounded if for every family A of size κ+, consisting of pairwise + disjoint finite subsets of κ , and every ν < κ there exists A0 ⊆ A of size κ such that f(α, β) > ν for all α ∈ a, β ∈ b and a 6= b in A0. 19.7 Lemma. If f :[κ+]2 −→ κ is unbounded and subadditive (i.e. it satisfies the two inequalities 19.2(a),(b)), then f is strongly unbounded.

+ Proof. For α < β < κ , set α <ν β if and only if f(α, β) ≤ ν. Then our assumption about f satisfying 19.2(a) and (b) reduces to the fact that + each <ν is a tree ordering on κ compatible with the usual ordering on κ+. Note that the unboundedness property of f is preserved by any forcing + notion satisfying the κ-chain condition, so in particular no tree (κ , <ν ) can contain a Souslin subtree of height κ. In the proof of Lemma 15.5 above + we have seen that this property of (κ , <ν ) alone is suﬃcient to conclude that every family A of κ many pairwise disjoint subsets of κ+ contains a subfamily A0 of size κ such that for every a 6= b in A0 every α ∈ a is <ν -incomparable to every β ∈ b, which is exactly the conclusion of f being strongly unbounded. a

19.8 Lemma. The following are equivalent:

+ (1) There is a structure (κ , κ, <, Rn)n<ω with no elementary substructure B of size κ such that B ∩ κ is bounded in κ.

(2) There is an unbounded function f :[κ+]2 −→ κ,

(3) There is a strongly unbounded, subadditive function f :[κ+]2 −→ κ.

Proof. Only the implication from (2) to (3) needs some argument. So let f :[κ+]2 −→ κ be a given unbounded function. Increasing f we may assume + that each of its sections fα : α −→ κ is one-to-one. For a set Γ ⊆ κ , let the f-closure of Γ, CLf (Γ), be the minimal ∆ ⊇ Γ with the property that 00 2 −1 f [∆] ⊆ ∆ and fα (ξ) ∈ ∆, whenever α ∈ ∆ and ξ < sup(∆ ∩ κ). Deﬁne e :[κ+]2 −→ κ by letting

∗ e(α, β) = sup(κ ∩ CLf (ρ (α, β) ∪ Pρ∗(α,β)(α))). (I.269)

We show the two subadditive properties for e:

e(α, γ) ≤ max{e(α, β), e(β, γ)} whenever α ≤ β ≤ γ. (I.270)

∗ ∗ By 19.2(a) either ρ (α, γ) ≤ ρ (α, β) in which case Pρ∗(α,γ)(α) ⊆ Pρ∗(α,β)(α) ∗ ∗ holds, or ρ (α, γ) ≤ ρ (β, γ) when we have Pρ∗(α,γ)(α) ⊆ Pρ∗(β,γ)(β). By the deﬁnition of e, the ﬁrst case gives us the inequality e(α, γ) ≤ e(α, β) and the second gives us e(α, γ) ≤ e(β, γ).

e(α, β) ≤ max{e(α, γ), e(β, γ)} whenever α ≤ β ≤ γ. (I.271) 19. Unbounded functions on successors of regular cardinals 133

∗ ∗ By 19.2(b) either ρ (α, β) ≤ ρ (α, γ) in which case Pρ∗(α,β)(α) ⊆ Pρ∗(α,γ)(α) ∗ ∗ holds, or ρ (α, β) ≤ ρ (β, γ) when we have Pρ∗(α,β)(α) ⊆ Pρ∗(β,γ)(β). In the first case we get e(α, β) ≤ e(α, γ) and in the second e(α, β) ≤ e(β, γ). By Lemma 19.7, to get the full conclusion of (3) it suffices to show that e is unbounded. So let Γ be a given subset of κ+ of order-type κ and let ν < κ be a given ordinal. Since e(α, β) ≥ ρ∗(α, β) for all α < β < κ+ we may assume that ρ∗ is bounded on Γ, or more precisely, that for someν ¯ ≤ ν and all α < β in Γ, ρ∗(α, β) ≤ ν¯. For α ∈ Γ, let β(α) be the minimal point of Γ above α. Then there is µ ≤ ν¯ and ∆ ⊆ Γ of size κ such that ρ∗(α, β(α)) = µ for all α ∈ ∆. Note that the sequence of sets Pµ(α)(α ∈ ∆) forms a chain under end-extension. Let Γ∗ be the union of this sequence of sets. Since f is unbounded, there exist ξ < η in Γ∗ such that f(ξ, η) > ν. Pick α ∈ ∆ above η. Then ξ, η ∈ Pµ(α) and therefore e(α, β(α)) ≥ f(ξ, η) > ν. This finishes the proof. a

19.9 Remark. Recall that Chang’s conjecture is the model-theoretic transfer principle asserting that every structure of the form (ω2, ω1, <, . . .) with a countable signature has an uncountable elementary submodel B with the property that B ∩ ω1 is countable. This principle shows up in many con- siderations including the ﬁrst two uncountable cardinals ω1 and ω2. For example, it is known that it is preserved by ccc forcing extensions, that it holds in the Silver collapse of an ω1-Erd˝oscardinal, and that it in turn implies that ω2 is an ω1-Erd˝oscardinal in the core model of Dodd and Jensen (see, e.g. [13],[37]).

19.10 Corollary. The negation of Chang’s conjecture is equivalent to the 2 statement that there exists e :[ω2] −→ ω1 such that: (a) e(α, γ) ≤ max{e(α, β), e(β, γ)} whenever α ≤ β ≤ γ,

(b) e(α, β) ≤ max{e(α, γ), e(β, γ)} whenever α ≤ β ≤ γ,

(c) For every uncountable family A of pairwise disjoint ﬁnite subsets of ω2 and every ν < ω1 there exists an uncountable A0 ⊆ A such that e(α, β) > ν whenever α ∈ a and β ∈ b for every a 6= b ∈ A0. a

2 19.11 Remark. Note that if a mapping e :[ω2] −→ ω1 has properties 2 ℵ0 (a),(b) and (c) of Corollary 19.10, then De :[ω2] −→ [ω2] defined by De{α, β} = {ξ ≤ min{α, β} : e(ξ, α) ≤ e{α, β}} satisfies the weak form of Definition 15.18, where only the conclusion (a) is kept. The full form of Definition 15.18, however, cannot be achieved assuming only the negation of Chang’s conjecture. This shows that the function ρ, based on a ¤ω1 - 2 sequence, is a considerably deeper object than an e :[ω2] −→ ω1 satisfying 19.10(a),(b),(c). 134 I. Coherent Sequences

Recall that the successor of the continuum is characterized as the minimal cardinal θ with the property that every f :[θ]2 −→ ω is constant on the square of some infinite set. We shall now see that in slightly weak- ening the partition property by replacing squares by rectangles one gets a characterization of a quite different sort. To see this, let us use the arrow notation µ ¶ µ ¶1,1 θ ω −→ θ ω ω to succinctly express the statement that for every map f : θ × θ −→ ω, there exist infinite sets A, B ⊆ θ such that f is constant on their product. Let θ2 be the minimal θ which fails to satisfy this property. Note that + ω1 < θ2 ≤ c . The following result shows that θ2 can have the minimal possible value ω2, as well as that θ2 can be considerably smaller than the continuum.

19.12 Theorem. Chang’s conjecture is equivalent to the statement that µ ¶ µ ¶1,1 ω ω 2 −→ holds in every ccc forcing extension. ω2 ω ω

Proof. We have already noted that Chang’s conjecture is preserved by ccc forcing extensions, so in order to prove the direct implication, it suffices to show that Chang’s conjecture implies that every f : ω2 × ω2 −→ ω is constant on the product of two infinite subsets of ω2. Given such an f, using Chang’s conjecture it is possible to find an elementary submodel M of Hω3 such that f ∈ M, δ = M ∩ ω1 ∈ ω1 and Γ = M ∩ ω2 is uncountable. Choose i ∈ ω and uncountable ∆ ⊆ Γ such that f(δ, β) = i for all β ∈ ∆. Choose α0 < δ such that ∆0 = {β ∈ ∆ : f(α0, β) = i} is uncountable. Such α0 exists by elementarity of M. Pick β0 ∈ ∆0 above ω1. Note that by the elementarity of M, for every β ∈ ∆0 there exist unboundedly many α < δ such that f(α, β0) = f(α, β) = i. So we can choose α1 < δ above α0 such that the set ∆1 = {β ∈ ∆0 : f(α1, β) = i} is uncountable. Then again for every β ∈ ∆1 there exist unboundedly many α < δ such that f(α, β0) = f(α, β1) = f(α, β) = i, and so on. Continuing in this way, we build two increasing sequences {αn} ⊆ δ and {βn} ⊆ Γ \ ω1 such that f(αn, βm) = i for all n, m < ω. 2 Assume now that Chang’s conjecture fails and fix e :[ω2] −→ ω as in Corollary 19.10. We shall describe a ccc forcing notion P which forces an f : ω2 × ω2 −→ ω which is not constant on the product of any two infinite subsets of ω2. The conditions of P are simply maps of the form 2 p :[Dp] −→ ω where Dp is some finite subset of ω2 such that

p(ξ, α) 6= p(ξ, β) whenever ξ < α < β belong to D and p (I.272) have the property that e(ξ, α) > e(α, β). 19. Unbounded functions on successors of regular cardinals 135

We order P by letting p extend q if and only if p extends q as a function and p(ξ, α) 6= p(ξ, β) whenever α < β belong to D and ξ be- q (I.273) longs to Dp \ Dq.

Let A be a given uncountable subset of P. Refining A, we may assume that 00 2 Dp (p ∈ A) forms a ∆-system with root D and that Fp = e [Dp] (p ∈ A) form an increasing ∆-system on ω1 with root F . Moreover, we may assume that for every p and q in A, the finite structures they generate together with e are isomorphic via the isomorphism that fixes D and F . By 19.10(c) there is uncountable A0 ⊆ A such that for all p, q ∈ A0,

e(α, β) > max(F ) whenever α ∈ Dp \ D and β ∈ Dq \ D. (I.274)

We claim that every two conditions p and q from A0 are compatible. To see 2 this, extend p ∪ q to a mapping r :[Dp ∪ Dq] −→ ω which is one-to-one on new pairs, avoiding the old values. To see that r is indeed a member of P we need to check (I.272) for r. So let ξ < α < β be three given ordinals from Dr such that e(ξ, α) > e(α, β). We have to prove that r(ξ, α) 6= r(ξ, β). By the choice of r we may assume that {ξ, α} and {ξ, β} are old pairs, i.e. 2 2 that they belong to [Dp] ∪ [Dq] . Since p and q both satisfy (I.272), we may consider only the case when (say) α ∈ Dp \ Dq, β ∈ Dq \ Dp and ξ ∈ D. Recall that e(ξ, α) > e(α, β) and the subadditive properties of e (see 19.10(a),(b)) give us the equality e(ξ, α) = e(ξ, β), i.e. e(ξ, α) belongs to F contradicting (I.274). To see that r extends p and q, by symmetry it suffices to check that r extends, say, p. So suppose ξ < α < β are given such that α < β belong to Dp and ξ belongs to Dq \ Dp. We need to show that r(ξ, α) 6= r(ξ, β). By the choice on new pairs, this is automatic if one of the two pairs is new. So we may consider the case when both α and β are in D. Since r(ξ, α) = q(ξ, α) and r(ξ, β) = q(ξ, β) in this case the inequality r(ξ, α) 6= r(ξ, β) would follow from (I.272) for q if we show that e(ξ, α) > e(α, β). Otherwise e(ξ, α) would belong to the root F and by the fact that all conditions are isomorphic, e(ξ(s), α) ∈ F holds for all s ∈ A0, where ξ(s) ∈ Ds (s ∈ A0) are the copies of ξ. This of course contradicts the property 19.10(c) of e since ξ(s) 6= ξ(t) when s 6= t. 2 Forcing with P gives us a mapping g :[ω2] −→ ω with the property that {ξ < α : g(ξ, α) = g(ξ, β)} is finite for all α < β < ω2. Define f : ω2 ×ω2 −→ ω by letting f(α, β) = 2g{α, β} + 2 when α < β, f(α, β) = 2g{α, β} + 1 when α > β, and f(α, β) = 0 when α = β. Then f is not constant on any product of two infinite sets. a

19.13 Remark. The relative size of θ2 (or its higher-dimensional analogues θ3, θ4,...) in comparison to the sequence of cardinals ω2, ω3, ω4,... is of considerable interest, both in Set Theory and Model Theory (see e.g. [63], 136 I. Coherent Sequences

[82], [87]). On the other hand, even the following most simple questions, left open by Theorem 19.12, are still unanswered.

19.14 Question. Can one prove any of the bounds like θ2 ≤ ω3, θ3 ≤ ω4, θ4 ≤ ω5, etc. without appealing to additional axioms? Note that by Corollary 19.10, Chang’s conjecture is equivalent to the statement that within every decomposition of the usual ordering on ω2 as an increasing chain of tree orderings, one of the trees has an uncountable chain. Is it possible to have decompositions of ∈¹ ω2 into an increasing ω1- chain of tree orderings of countable heights? It turns out that the answer to this question is equivalent to a diﬀerent well-known combinatorial statement + 2 about ω2 rather than Chang’s conjecture itself. Recall that f :[κ ] −→ κ is transitive if f(α, γ) ≤ max{f(α, β), f(β, γ)} whenever α ≤ β ≤ γ. (I.275)

+ 2 + 2 Given a transitive map f :[κ ] −→ κ, one deﬁnes ρf :[κ ] −→ κ recursively on α ≤ β < κ as follows:

ρf (α, β) is equal to the supremum of:

1. f(min(Cβ \ α), β), 2. tp(Cβ ∩ α), (I.276) 3. ρf (α, min(Cβ \ α)), 4. ρf (ξ, α)(ξ ∈ Cβ ∩ α). 19.15 Lemma. For every transitive map f :[κ+]2 −→ κ the corresponding + 2 ρf :[κ ] −→ κ has the following properties:

(a) ρf (α, γ) ≤ max{ρf (α, β), ρf (β, γ)} whenever α ≤ β ≤ γ,

(b) ρf (α, β) ≤ max{ρf (α, γ), ρf (β, γ)} whenever α ≤ β ≤ γ, + (c) |{ξ ≤ α : ρf (ξ, α) ≤ ν}| ≤ |ν| + ℵ0 for ν < κ and α < κ , + (d) ρf (α, β) ≥ f(α, β) for all α < β < κ . Proof. Proofs of (a)-(c) are almost identical to the corresponding proofs for the function ρ itself and so we avoid repetition. The inequality (d) is deduced from (I.276) by using the transitivity of f as follows:

ρf (α, β) ≥ max f(βi+1, βi) ≥ f(α, β), i β1 > . . . > βn−1 > βn = β is the walk from β to α along the given C-sequence. a 19.16 Remark. Transitive maps are frequently used combinatorial objects, especially when one works with quotient structures. Adding the extra subadditivity condition 19.15(b), one obtains a considerably more subtle object which is much less understood. 19. Unbounded functions on successors of regular cardinals 137

+ 19.17 Deﬁnition. Let fα : κ −→ κ (α < κ ) be a given sequence of ∗ functions such that fα < fβ whenever α < β (i.e. {ν < κ : fα(ν) ≥ fβ(ν)} is bounded in κ whenever α < β). Then the corresponding transitive map f :[κ+]2 −→ κ is deﬁned by

f(α, β) = min{µ < κ : fα(ν) < fβ(ν) for all ν ≥ µ}. (I.277)

Let ρf be the corresponding ρ-function that dominates this particular f f + and for ν < κ let <ν be the corresponding tree ordering of κ , i.e.

f α <ν β if and only if ρf (α, β) ≤ ν. (I.278)

+ 19.18 Lemma. Suppose fα ≤ g for all α < κ where ≤ is the ordering of + f everywhere dominance. Then for every ν < κ the tree (κ , <ν ) has height ≤ g(ν).

+ f Proof. Let P be a maximal chain of (κ , <ν ). f(α, β) ≤ ρf (α, β) ≤ ν for every α < β in P . It follows that fα(ν) < fβ(ν) ≤ g(ν) for all α < β in P . So P has order-type ≤ g(ν). a

Note that if we have a function g : κ −→ κ which bounds the sequence + ∗ fα (α < κ ) in the ordering < of eventual dominance, then the new ¯ + ∗ sequence fα = min{fα, g} (α < κ ) is still strictly < -increasing but now bounded by g even in the ordering of everywhere dominance. So this proves the following result of Galvin (see [34],[59]).

19.19 Corollary. The following two conditions are equivalent for every regular cardinal κ.

+ (1) There is a sequence fα : κ −→ κ (α < κ ) which is strictly increasing and bounded in the ordering of eventual dominance.

(2) The usual order-relation of κ+ can be decomposed into an increasing κ-sequence of tree orderings of heights < κ.

19.20 Remark. The assertion that every strictly <∗-increasing κ+-sequence of functions from κ to κ is <∗-unbounded is strictly weaker than Chang’s conjecture and in the literature it is usually referred to as weak Chang’s conjecture. This statement still has considerable large cardinal strength (see [14]). Also note the following consequence of Corollary 19.19 which can be deduced from Lemmas 19.1 and 19.2 above as well.

19.21 Corollary. If κ is a regular limit cardinal (e.g. κ = ω), then the usual order-relation of κ+ can be decomposed into an increasing κ-sequence of tree orderings of heights < κ. a 138 I. Coherent Sequences

20. Higher dimensions

The reader must have noticed already that in this Chapter so far, we have only considered functions of the form f :[θ]2 −→ I or equivalently sequences fα : α −→ I (α < θ) of one-place functions. To obtain analogous results about functions deﬁned on higher-dimensional cubes [θ]n one usually develops some form of stepping-up procedure that lifts a function of the form f :[θ]n −→ I to a function of the form g :[θ+]n+1 −→ I. The basic idea seems quite simple. One starts with a coherent sequence + eα : α −→ θ (α < θ ) of one-to-one mappings and wishes to deﬁne g :[θ+]n+1 −→ I as follows:

g(α0, α1, . . . , αn) = f(e(α0, αn), . . . , e(αn−1, αn)). (I.279)

In other words, we use eαn to send {α0, . . . , αn−1} to the domain of f and then apply f to the resulting n-tuple. The problem with such a simple- minded definition is that for a typical subset Γ of θ+, the sequence of restrictions eδ ¹ (Γ∩δ)(δ ∈ Γ) may not cohere, so we cannot produce a subset of θ that would correspond to Γ and on which we would like to apply some property of f. It turns out that the definition (I.279) is basically correct ex- + 3 + cept that we need to replace eαn by eτ(αn−2,αn−1,αn), where τ :[θ ] −→ θ is defined as follows (see Definition 17.6):

τ(α, β, γ) = γt, where t = ρ0(α, γ) ∩ ρ0(β, γ). (I.280)

The function ρ0 to which (I.280) refers is of course based on some C-sequence + + Cα (α < θ ) on θ . The following result shows that if the C-sequence is carefully chosen, the function τ will serve as a stepping-up tool.

20.1 Lemma. Suppose ρ0 and τ are based on some ¤θ-sequence Cα (α < θ+) and let κ be a regular uncountable cardinal ≤ θ. Then every set Γ ⊆ θ+ of order-type κ contains a coﬁnal subset ∆ such that, if ε = sup(Γ) = sup(∆), then ρ0(ξ, ε) = ρ0(ξ, τ(α, β, γ)) for all ξ < α < β < γ in ∆.

Proof. For δ ∈ lim(Cε), let ξδ be the minimal element of Γ above δ. Let εδ be the maximal point of the trace Tr(δ, ξδ) of the walk from ξδ to δ along + the ¤θ-sequence Cα (α < θ ). Thus εδ = δ or εδ = min(Tr(δ, ξδ) \ (δ + 1)) and Cξ ∩ δ is bounded in δ for all ξ ∈ Tr(δ, ξδ) strictly above εδ. Let f(δ) be a member of Cε ∩ δ which bounds all these sets. By the Pressing Down ¯ ¯ Lemma there is a stationary set Σ ⊆ lim(Cε) and δ ∈ Cε such that f(δ) ≤ δ for all δ ∈ Σ. Shrinking Σ we may assume that ξγ < δ whenever γ < δ are members of Σ. Note that

εγ = min(Tr(ξα, ξγ ) ∩ Tr(ξβ, ξγ )) for every α < β < γ in Σ. (I.281)

In other words, τ(ξα, ξβ, ξγ ) = εγ for all α < β < γ in Σ. By the deﬁnition of εδ for δ ∈ lim(Cε), δ belongs to Cεδ and so by the implicit assumption 20. Higher dimensions 139

that nonlimit points of Cεδ are successor ordinals, we conclude that δ is a limit point of Cεδ . Thus the section (ρ0)δ of the function ρ0 is an initial part of both the section (ρ0)ε and (ρ0)εδ , which is exactly what is needed for the conclusion of Lemma 20.1. a

+ Recall that for a given C-sequence Cα (α < θ ) such that tp(Cα) ≤ θ + for all α < θ , the range of ρ0 is the collection of all finite sequences of ordinals < θ. It will be convenient to identify Qθ with θ via, for example, Gödel’scoding of sequences of ordinals by ordinals. This identification gives us a well-ordering

g(α0, . . . , αn−1, αn) = f(ρ0(α0, ε), . . . , ρ0(αn−1, ε)), (I.282)

where ε = τ(αn−2, αn−1, αn). Let us examine how this stepping-up procedure works on a particular example.

20.2 Theorem. Suppose θ is an arbitrary cardinal for which ¤θ holds. Suppose further that for some regular κ > ω and integer n ≥ 2 there is a map f :[θ]n −→ [[θ]<κ]<κ such that: (1) A ⊆ min(a) for all a ∈ [θ]n and A ∈ f(a). (2) For all ν < κ and Γ ⊆ θ of size κ there exist a ∈ [Γ]n and A ∈ f(a) such that tp(A) ≥ ν and A ⊆ Γ. Then θ+ and κ satisfy the same combinatorial property, but with n + 1 in place of n.

n Proof. We assume that the domain of f is actually equal to the set [Qθ] rather than [θ]n and deﬁne g by

−1 g(α0, . . . , αn−1, αn) = (ρ0)ε (f(ρ0(α0, ε), . . . , ρ0(αn−1, ε))), (I.283) where ε = τ(αn−2, αn−1, αn), and where ρ0 and τ are based on some ﬁxed ¤θ-sequence. Note that the transformation does not necessarily preserve (1), so we intersect each member of a given g(a) with min(a) in order to satisfy this condition. To check (2), let Γ ⊆ θ be a given set of size κ. By Lemma 20.1, shrinking Γ we may assume that Γ has order-type κ and that if ε = sup(Γ), then

ρ0(ξ, ε) = ρ0(ξ, τ(α, β, γ)) for all α < β < γ in Γ. (I.284) It follows that g restricted to [Γ]n+1 satisﬁes the formula

−1 g(α0, . . . , αn−1, αn) = (ρ0)ε (f(ρ0(α0, ε), . . . , ρ0(αn−1, ε))). (I.285) 140 I. Coherent Sequences

Shrinking Γ further we assume that the mapping (ρ0)ε, as a map from + the ordered set (θ , ∈) into the ordered set (Qθ,

20.3 Remark. It should be noted that Velleman [93] was the ﬁrst to attempt to step-up the combinatorial property appearing in Theorem 20.2 using his version of the gap-2 morass. Unfortunately the stepping up procedure of [93] worked only up to the value n = 3 so it remains unclear if the higher-gap morass could be used to reach all the other values of n.

If we apply this stepping-up procedure to the projection [[··]] of the square- bracket operation deﬁned in 7.14, one obtains analogues of families G, H and K of Example 7.16 for ω2 instead of ω1. To make this precise, we

first fix a ¤ω1 -sequence Cα (α < ω2) and consider the corresponding maps 2 <ω 2 2 ρ0 :[ω2] −→ ω1 and ρ :[ω2] −→ ω1. Letρ ¯ :[ω2] −→ ω1 be defined by

ρ¯(α, β) = 2ρ(α,β) · (2 · tp{ξ ≤ α : ρ(ξ, α) ≤ ρ(α, β)} + 1).

Then for α < ω2, the mappingρ ¯α : α −→ ω1 that sends ξ toρ ¯(ξ, α) is 1-1 and we have the following coherence property:

ρ¯α =ρ ¯β ¹ α whenever α ∈ lim(Cβ). (I.286)

3 20.4 Deﬁnition. The ternary operation [[[··]]] : [ω2] −→ ω2 is deﬁned as follows: −1 [[[αβγ]]] = (¯ρτ(α,β,γ)) [[¯ρτ(α,β,γ)(α)ρ ¯τ(α,β,γ)(β)]]. It should be clear that the stepping-up method lifts the property 7.15 of [[··]] to the following property of the ternary operation [[[··]]]:

20.5 Lemma. Let A be a given uncountable family of pairwise disjoint subsets of ω2, all of some ﬁxed size n. Let ξ1, . . . , ξn be a sequence of ordinals < ω2 such that for all i = 1, . . . , n and all but countably many a ∈ A we have that ξi < a(i). Then there exist a,b and c in A such that [[[a(i)b(i)c(i)]]] = ξi for i = 1, . . . , n. a

<ω As in Example 7.16, let us identify ω2 and 3 × [ω2] and view [[[··]]] as <ω having the range 3 × [ω2] rather than ω2. Let [[[··]]]0 and [[[··]]]1 be the two projections of [[[··]]]. Set

<ω 3 G = {G ∈ [ω2] : [[[αβγ]]]0 = 0 for all {α, β, γ} ∈ [G] }, 20. Higher dimensions 141

<ω 3 H = {H ∈ [ω2] : [[[αβγ]]]0 = 1 for all {α, β, γ} ∈ [H] }. Note the following immediate consequence of Lemma 20.5. 20.6 Lemma. For every uncountable family A of disjoint 2-element subsets of ω2 there is an arbitrarily large ﬁnite subset B of A with {b(0) : b ∈ B} ∈ G and {b(1) : b ∈ B} ∈ H. a

Recall that the parameter in all our definitions so far is a fixed ¤ω1 - sequence, so we also have the corresponding ρ-function at our disposal. We use it to define an analogue of the family K of Example 7.16 as follows. In this case, let K be the collection of all finite sets {ai : i < n} of elements of 2 [ω2] such that for all i < j < k < n:

ai < aj < ak, ρ(α, β) ≤ ρ(β, γ) holds if α ∈ ai, β ∈ aj and γ ∈ ak, (I.287)

[[[ai(0)aj(0)ak(0)]]]0 = [[[ai(1)aj(1)ak(1)]]]0 = 2, (I.288) [ [[[ai(0)aj(0)ak(0)]]]1 = [[[ai(1)aj(1)ak(1)]]]1 = al. (I.289) l

20.7 Lemma. For every sequence aξ (ξ < ω1) of 2-element subsets of ω2 such that aξ < aη whenever ξ < η and for every positive integer n there exist ξ0 < ξ1 < . . . < ξn−1 < ω1 such that {aξi : i < n} ∈ K. a The particular deﬁnition of K is made in order to have the following property of K which is analogous to (I.79) of Section 7:

If K and L are two distinct members of K, then there are no more than 9 ordinals α such that {α, β} ∈ K and {α, γ} ∈ L (I.290) for some β 6= γ. The deﬁnitions of G, H and K immediately lead to the following analogues of (I.77) and (I.78):

G and H contain all singletons, are closed under subsets (I.291) and are 3-orthogonal to each other.

G and H are both 5-orthogonal to the family of all unions (I.292) of members of K.

So we can proceed as in Example 7.16 and for a function x from ω2 into R, deﬁne 142 I. Coherent Sequences

X 2 1 kxkH,2 = sup{( x(α) ) 2 : H ∈ H}, α∈H X 2 1 kxkK,2 = sup{( (x(α) − x(β)) ) 2 : K ∈ K}. {α,β}∈K

Let k·k = max{k·k∞, k·kH,2, k·kK,2}, let E¯2 = {x : kxk < ∞}, and let E2 be the closure of the linear span of {1α : α ∈ ω2} inside (E¯2, k · k). Then as in Example 7.16 one shows, using the properties of G, H and K listed above in Lemmas 20.6 and 20.7 and in (I.287-I.292), that E2 is a Banach space with {1α : α ∈ ω2} as its transitive basis and with the property that every bounded linear operator T : E2 −→ E2 can be written as S + D, where S is an operator with separable range and D is a diagonal operator relative to the basis {1α : α ∈ ω2} with only countably many changes of the constants. The new feature here is the use of (I.287) in checking that the projections on countable sets of the form {α < β : ρ(α, β) < ν} (α < ω2, ν < ω1) are (uniformly) bounded. This is needed among other things in showing, using Lemma 20.6, that T can be written as a sum of one such projection and a diagonal operator relative to {1α : α ∈ ω2}. Using the interpolation method of [11], one can turn E2 into a reﬂexive example. Thus, we have the following:

20.8 Example. Assuming ¤ω1 , there is a reﬂexive Banach space E with a transitive basis of type ω2 with the property that every bounded operator T : E −→ E can be written as a sum of an operator with a separable range and a diagonal operator (relative to the basis) with only countably many changes of constants. a

20.9 Remark. In [42], P.Koszmider has shown that such a space cannot be constructed on the basis of the usual axioms of set theory. We refer the reader to that paper for more details about these kinds of examples of Banach spaces.

For the rest of this section we shall examine the stepping-up method with + less restrictions on the given C-sequence Cα (α < θ ) on which it is based. 20.10 Theorem. The following are equivalent for a regular cardinal θ such that log θ+ = θ. 39

(1) There is a substructure of the form (θ++, θ+, <, . . .) with no substructure B of size θ+ with B ∩ θ+ of size θ.

(2) There is f :[θ++]3 −→ θ+ which takes all the possible values on the cube of any subset Γ of θ++ of size θ+.

39log κ = min{λ : 2λ ≥ κ}. 20. Higher dimensions 143

Proof. To prove the nontrivial direction from (1) to (2), we use Lemma 19.8 and choose a strongly unbounded and subadditive e :[θ++]2 −→ θ+. We + + also choose a C-sequence Cα (α < θ ) such that tp(Cα) ≤ θ for all α < θ and consider the corresponding function ρ∗ :[θ+]2 −→ θ deﬁned above in ++ I.264. Finally we choose a one-to-one sequence rα (α < θ ) of elements of + {0, 1}θ and consider the corresponding function ∆ : [θ++]2 −→ θ+:

∆(α, β) = ∆(rα, rβ) = min{ν : rα(ν) 6= rβ(ν)}. (I.293)

The deﬁnition of f :[θ++]3 −→ θ is given according to the following two rules applied to a given triple x = {α, β, γ} ∈ [θ++]3 (α < β < γ): Rule 1: If ∆(rα, rβ) < ∆(rβ, rγ ) and rα lex rβ >lex rγ , let

f(α, β, γ) = min(Pν (∆(β, γ)) \ ∆(α, β)), where ν = ρ∗(min{ξ ≤ ∆(α, β): ρ∗(ξ, ∆(α, β)) 6= ρ∗(ξ, ∆(β, γ))}, ∆(β, γ)). Rule 2: If α ∈ x is such that rα is lexicographically between the other two rξ’s for ξ ∈ x, if β ∈ x \{α} is such that ∆(rα, rβ) > ∆(rα, rγ ), where γ is the remaining element of x and if x does not fall under Rule 1, let

f(α, β, γ) = min(Pν (e(β, γ)) \ e(α, β)), where ν = ρ∗{∆(α, β), e(β, γ)}. The proof of the Theorem is finished once we show the following: for every stationary set Σ of cofinality θ ordinals < θ+ and every Γ ⊆ θ++ of size θ+ there exist α < β < γ in Γ such that f(α, β, γ) ∈ Σ. Clearly we may + assume that Γ has order-type θ . Let RΓ = {rα : α ∈ Γ}. + Case 1: RΓ contains a subset of size θ which is lexicographically well- ordered or conversely well-ordered. Going to a subset of Γ, we may assume that ∆(rα, rβ) = ∆(rα, rγ ) for all α < β < γ in Γ. (I.294) It follows that Rule 1 applies in the definition of f(α, β, γ) for any triple 2 α < β < γ of elements of Γ. Let ∆ = {∆(rα, rβ): {α, β} ∈ [Γ] }. Then ∆ is a subset of θ+ of size θ+ and the range of f on [Γ]3 includes the range of [··] on [∆]2, where [··] is the square-bracket operation on θ+ defined as follows: [αβ] = min(Pν (β) \ α), (I.295) where ν = ρ∗(min{ξ ≤ α : ρ∗(ξ, α) 6= ρ∗(ξ, β)}, β). Note that this is the exact analogue of the square-bracket operation that has been analysed above in Section 7. So exactly as in Section 7, one establishes the following fact.

For ∆ ⊆ θ+ unbounded, the set {[αβ]: {α, β} ∈ [∆]2} (I.296) contains almost all ordinals < θ+ of coﬁnality θ. 144 I. Coherent Sequences

+ Case 2: RΓ contains no subsets of size θ which are lexicographically <θ+ well-ordered or conversely well-ordered. For t ∈ {0, 1} , let Γt = {α ∈ <θ+ + Γ: t ⊆ rα}. Let T be the set of all t ∈ {0, 1} for which Γt has size θ . Shrinking Γ we may assume that T is either a downwards closed subtree of + {0, 1}<θ of size θ and height λ < θ+, or a downwards closed Aronszajn + subtree of {0, 1}<θ of height θ+. For β ∈ Γ, let

00 + eβ Γ = {e{β, γ} : γ ∈ Γ, rβ

+ 00 + Case 2.1: There exist θ many β ∈ Γ such that the set eβ Γ is bounded for all β ∈ Γ. Choose an elementary submodel M of some large enough + structure of the form Hκ such that δ = M ∩ θ ∈ Σ and M contains all the relevant objects. By the elementarity of M, strong unboundedness of e and the fact that T contains no θ+-chain, there exist β and γ in Γ \ M ∗ such that rγ ∆(rβ, rγ ) such that (rβ ¹ ξ) 0 belongs to M and rβ(ξ) = 1. This is clear in the case when T is an Aronszajn tree, by the elementarity of M and the fact that cf(δ) = θ. However, this is also true in case T has height λ < θ+ and has size θ, since from our assumption log θ+ = θ the ordinal λ must have coﬁnality exactly θ. So by the property ∗ ∗ a 19.1 of ρ there is one such ξ such that ν = ρ (ξ, ε) ≥ ν¯. Let t = (rβ ¹ ξ) 0. + Then t ∈ T ∩ M and so Γt is a subset of Γ of size θ which belongs to M. Using the strong unboundedness of e and the elementarity of M, there must be an α in Γt ∩ M such that e(α, β) > sup(Pν (ε) ∩ δ). Since e(α, β) belongs 00 + + to eα Γ , a set which, by the assumption of Case 2.1, is bounded in θ , we conclude that e(α, β) < δ. It follows that f(α, β, γ) is deﬁned according to Rule 2, and therefore

f(α, β, γ) = min(Pν (ε) \ e(α, β)) = δ. (I.298)

00 + Case 2.2: For every β in some tail of Γ the set eβ Γ is unbounded in θ+. Going to a tail, we assume that this is true for all β ∈ Γ. Let M and δ = M ∩ θ+ ∈ Σ be chosen as before. If T is Aronszajn, using elementarity of M, there must be β ∈ Γ ∩ M and δ¯ < δ of coﬁnality θ such that there exist unboundedly many ξ below δ¯ such that the set

Ωξ = {e{β, χ} : χ ∈ Γ, rβ

such that ∆(rβ, rα) = ξ, rβ sup(Pν (ε)∩δ). Note that rα

∆(α, β) = ∆(rα, rβ) = min{ν < θ : rα(ν) 6= rβ(ν)}. (I.301) The deﬁnition of f :[θ+]3 −→ θ is given according to the following rules, applied to a given x ∈ [θ+]3. Rule 1: If x = {α < β < γ}, ∆(rα, rβ) < ∆(rβ, rγ ) and rα lex rβ >lex rγ , let f{α, β, γ} = [∆(α, β)∆(β, γ)].

Rule 2: If α ∈ x is such that rα is lexicographically between the other two rξ’s for ξ ∈ x, if β ∈ x \{α} is such that ∆(rα, rβ) > ∆(rα, rγ ), where γ is the remaining element of x, and they do not satisfy the conditions of Rule 1, set f{α, β, γ} = min(Tr(∆(α, β), e{β, γ}) \ e{α, β}), i.e. f{α, β, γ} is the minimal point on the trace of the walk from e{β, γ} to ∆(α, β) above the ordinal e{α, β}; if such a point does not exist, set f{α, β, γ} = 0. The proof of the Theorem is ﬁnished once we show that for every stationary Ω ⊆ Σ and every Γ ⊆ θ+ of size θ, there exist x ∈ [Γ]3 such that f(x) ∈ Ω. This is done, as in the proof of Theorem 20.10, by considering the two cases. Case 1 is reduced to the property 18.2 of the square-bracket operation. The treatment of Case 2 is similar and a bit simpler than in the proof of Theorem 20.10, since the case that T is of height < θ is impossible due to our assumption that θ is a strong limit cardinal. This ﬁnishes the proof. a 146 I. Coherent Sequences

Since log ω1 = ω, the following is an immediate consequence of Theorem 20.10.

20.12 Theorem. Chang’s conjecture is equivalent to the statement that for 3 00 3 every f :[ω2] −→ ω1 there is uncountable Γ ⊆ ω2 such that f [Γ] 6= ω1. a

20.13 Remark. Since this same statement is stronger for functions from n higher dimensional cubes [ω2] into ω1 the Theorem 20.12 shows that they are all equivalent to Chang’s conjecture. Note also that n = 3 is the minimal dimension for which this equivalence holds, since the case n = 2 follows from the Continuum Hypothesis, which has no relationship to Chang’s conjecture.

For the rest of this section we examine the stepping-up procedure without the assumption that some form of Chang’s conjecture is false. So let θ be + a given regular uncountable cardinal and let Cα (α < θ ) be a fixed C- + ∗ + 2 sequence such that tp(Cα) ≤ κ for all α < θ . Let ρ :[θ ] −→ θ be ∗ + the ρ -function defined above in (I.264). Recall that, in case Cα (α < θ ) is a ¤θ-sequence, the key to our stepping-up procedure was the function τ :[θ+]3 −→ θ+ defined by the formula (I.280). Without the assumption of + Cα (α < θ ) being a ¤θ-sequence, the following related function turns out to be a good substitute: χ :[θ+]3 −→ ω defined by

χ(α, β, γ) = |ρ0(α, γ) ∩ ρ0(β, γ)|. (I.302)

Thus χ(α, β, γ) is equal to the length of the common part of the walks γ → α and γ → β.

20.14 Deﬁnition. A subset Γ of θ+ is stable if χ is bounded on [Γ]3.

20.15 Lemma. Suppose that Γ is a stable subset of θ+ of size θ. Then {ρ∗(α, β): {α, β} ∈ [Ω]2} is unbounded in θ for every Ω ⊆ Γ of size θ.

Proof. Suppose ρ∗ is bounded by ν on [Ω]2 for some Ω ⊆ θ+ of size θ. We need to show that χ is unbounded on [Ω]3. Therefore we may assume that Ω is of order-type θ and let λ = sup(Ω). Construct a sequence Ω = Ω0 ⊇ Ω1 ⊇ ... of subsets of Ω of order-type θ such that:

3 χ(α, β, γ) ≥ n for all {α, β, γ} ∈ [Ωn] . (I.303)

Given Ωn, choose an ∈-chain M of length θ of elementary submodels M of some large enough structure of the form Hκ such that M ∩ θ ∈ θ and M contains all the relevant objects. Now choose another elementary submodel N of Hκ which contains M as well as all the other relevant objects such that N ∩ θ ∈ θ. Let λN = sup(N ∩ λ). Choose γ ∈ Ωn above λN . Let γ = γ0 > . . . > γk = λN be the walk from γ to λN . Letγ ¯ be the γi with 20. Higher dimensions 147

the smallest index such that Cγi ∩ λN is unbounded in λN . Thenγ ¯ is either equal to λN or γk−1. Choose λ0 ∈ N ∩ λ such that:

λ > max(C ∩ λ ) for all i < k for which C ∩ λ is 0 γi N γi N (I.304) bounded in λN .

Then {γi : i < k}\ γ¯ ⊆ Tr(α, γ) for all α ∈ N ∩ Ωn above λ0. So if we choose α < β in N ∩ Ω0 such that Cγ¯ ∩ [α, β) 6= ∅, then

Tr(α, γ) ∩ Tr(β, γ) = {γi : i ≤ k}\ γ.¯ (I.305)

From this and our assumption χ(α, β, γ) ≥ n we conclude that this set has size at least n. For M ∈ M, let λM = sup(M ∩λ) and C = {λM : M ∈ M}. Then C is a closed and unbounded subset of λ of order-type θ. By our assumption that ρ∗ is bounded by ν on [Ω]2, we get:

tp(Cγ¯ ∩ λN ) ≤ sup{tp(Cγ¯ ∩ α): α ∈ Ω ∩ [λ0, λN )} ∗ (I.306) ≤ sup{ρ (α, γ): α ∈ Ω ∩ [λ0, λN )} ≤ ν.

Since C ∩ [λ0, λN ) has order-type > ν there must be M ∈ M ∩ N such that λM ∈/ Cγ¯ and λM > λ0. Thus, we have an elementary submodel M of Hκ which contains all the relevant objects and a γ ∈ Ωn above λM such that the walk γ → λM contains all the points from the set {γi : i < k}\ γ¯, but includes at least the point min(Cγ¯ \ λM ) which is strictly above λM . Hence, if γ = γ0 > γ1 > . . . > γl = λM is the trace of γ → λM and ifγ ˆ is the γi (i ≤ l) with the smallest index, subject to the requirement that Cγˆ ∩ λM is unbounded in λM , then {γi : i < l}\ γˆ has size at least n + 1. ˆ Let λ0 ∈ λ ∩ M be such that:

λˆ > max(C ∩ λ ) for all i < l for which C ∩ λ is 0 γi M γi M (I.307) bounded in λM . ˆ Then {γi : i < l}\ γ¯ ⊆ Tr(α, γ) for all α ∈ M ∩ Ωn above λ0. Using the elementarity of M we conclude that for every δ < λ there exist ε ∈ Ωn \ δ and a set aε ⊆ [δ, ε) of size at least n + 1 such that aε ⊆ Tr(α, ε) for all ˆ α ∈ Ωn ∩ [λ0, δ). Hence we can choose a closed and unbounded set D of ˆ [λ0, λ) of order-type θ, for each δ ∈ D an ε(δ) ∈ Ωn\δ and an aε(δ) ⊆ [δ, ε(δ)) such that for every δ ∈ D: ˆ aε(δ) ⊆ Tr(α, ε(δ)) for α ∈ [λ0, δ), (I.308)

ε(δ) < min(D \ (δ + 1)). (I.309) 148 I. Coherent Sequences

Finally let Ωn+1 = {ε(δ): δ ∈ D}. Then

aε(γ) ⊆ Tr(ε(α), ε(γ)) ∩ Tr(ε(β), ε(γ)) for all α < β < γ in ∆. (I.310)

This gives us that χ(α, β, γ) ≥ n + 1 for every triple α < β < γ in Ωn+1, finishing the inductive step and therefore the proof of Lemma 20.15. a 20.16 Definition. The 3-dimensional version of the oscillation mapping, osc : [θ+]3 −→ ω, is defined on the basis of the 2-dimensional version of Section 17 as follows

osc(α, β, γ) = osc(Cβs \ α, Cγt \ α), where s = ρ0(α, β) ¹ χ(α, β, γ) and t = ρ0(α, γ) ¹ χ(α, β, γ). In other words, we let n be the length of the common part of the two walks γ → α and γ → β. Then we consider the walks γ = γ0 > . . . > γk = α and β = β0 > . . . > βl = α from γ to α and β to α respectively; if both k and l are bigger than n (and note that k ≥ n), i.e. if γn and βn are both deﬁned, we let osc(α, β, γ) be equal to the oscillation of the two sets Cβn \ α and Cγn \ α. If not, we let osc(α, β, γ) = 0. 20.17 Lemma. Suppose that Γ is a subset of θ+ of size κ, a regular uncountable cardinal, and that every subset of Γ of size κ is unstable. Then for every positive integer n, there exist α < β < γ in Γ such that osc(α, β, γ) = n. Proof. We may assume that Γ is actually of order-type κ and let λ = sup(Γ). Let M be an ∈-chain of length κ of elementary submodels M of Hθ++ such that M ∩ κ ∈ κ and M contains all the relevant objects. For M ∈ M, let λM = sup(λ ∩ M). Now choose another elementary submodel N of Hθ++ containing M as well as all the other relevant objects with N ∩ κ ∈ κ. Let λN = sup(λ ∩ N) and choose γ ∈ Γ above λN and let γ = γ0 > . . . > γk = λN be the walk from γ to λN . Letγ ¯ and λ0 be chosen as above in (I.304). Thenγ ¯ = γk orγ ¯ = γk−1 and Tr(α, γ) ∩ Tr(β, γ) = {γ : i ≤ k}\ γ¯ for all α < β in i (I.311) Γ ∩ [λ0, λN ) such that Cγ¯ ∩ [α, β) 6= ∅.

So if a tail of the set {λM : M ∈ M ∩ N} is included in Cγ¯, then using the elementarity of N, we would be able to construct a coﬁnal subset Γ0 ⊆ Γ ¯ 3 ¯ such that χ is constantly equal to k on [Γ0] , where k is the place ofγ ¯ in the sequence γ0, . . . , γk. This, of course, would contradict our assumption that Γ contains no large stable subsets. It follows that the set of all λN for M ∈ M ∩ N which do not belong to Cγ¯ is coﬁnal in λN . So we can pick

M1 ∈ M2 ∈ ... ∈ Mn+1 in M ∩ N such that, if λi = λMi (1 ≤ i ≤ n + 1), then: λi ∈ [λ0, λ) \ Cγ¯ for all 1 ≤ i ≤ n + 1, (I.312) 20. Higher dimensions 149

[λi, λi+1) ∩ Cγ¯ 6= ∅ for all 1 ≤ i ≤ n. (I.313) ¯ Fix α ∈ Γ ∩ M1 above max(Cγ¯ ∩ λ1). For 1 ≤ i ≤ n, pick λi ∈ [λi, λi+1) ∩ ¯ ¯ Mi+1 such that λi ≥ max(Cγ¯ ∩ λi+1). For 1 ≤ i ≤ n, let I¯i = [λi, λi], and let I¯n+1 = [λn, γ]. Now set F to be the collection of all increasing sequences I1 < I2 < . . . < In+1 of closed intervals of ordinals < λ with the property that there is β = β( I~ ) in Γ ∩ In+1 such that, if β = β0 > . . . > βl = α is the walk from β to α, then: ¯ l > k and βk¯ = min(Tr(α, β) ∩ In+1), (I.314) n[+1

Cβk¯ \ α ⊆ Ii, (I.315) i=1

Cβk¯ ∩ Ii 6= ∅ for all 1 ≤ n ≤ n + 1. (I.316)

Clearly, F ∈ M1 and hI¯1,..., I¯n+1i ∈ F, as γ is a witness of this. So working as in the proof of Lemma 17.2, we can ﬁnd F0 ⊆ F in M1 such that, if some sequence J~ of length ≤ n (including the case J~ = ∅) end- extends to a member of F0, then for every ξ < λ there is a closed interval K included in [ξ, λ) such that the concatenation J~aK also end-extends to a sequence in F0. Working inductively on 1 ≤ i ≤ n + 1 we can select a sequence I1,I2,...,In such that:

I1,...,Ii ∈ Mi+1 for 1 ≤ i ≤ n, (I.317) ¯ Ii ⊆ [λi, λi+1) for 1 ≤ i ≤ n. (I.318)

hI1,...,Iii end-extends to a member of F0. (I.319) Clearly, there is no problem in choosing these objects, using the elementarity of the models Mi (1 ≤ i ≤ n + 1) and the splitting property of F0. Working in Mn+1, we ﬁnd an interval In+1 such that hI0,...,In,In+1i belongs to ¯ F0. Let β = β(hI0,...,In+1i). Since Cγ¯ intersects the interval [λ1, λ1) it separates α from β, so by (I.311) we get that τ(α, β, γ) = k¯, hence by deﬁnition

osc(α, β, γ) = osc(Cβk¯ \ α, Cγk¯ \ α). (I.320)

Recall that γk¯ =γ ¯, so referring to (I.313),(I.315) and (I.316) we conclude that I1 ∩Cβk¯ ,...,In−1 ∩Cβk¯ and (In ∪In+1)∩Cβk¯ are the n convex pieces in which the set Cγ¯ \ α splits the set Cβk¯ \ α. Hence osc(Cβk¯ \ α, Cγk¯ \ α) = n. This completes the proof of Lemma 20.17. a Applying the last two Lemmas to the subsets of θ+ of size θ, we get an interesting dichotomy: 20.18 Lemma. Every Γ ⊆ θ+ of size θ can be reﬁned to a subset Ω of size θ such that either: 150 I. Coherent Sequences

(1) ρ∗ is unbounded and therefore strongly unbounded on Ω, or (2) the oscillation mapping takes all its possible values on the cube of Ω. a We ﬁnish the section with a typical application of this dichotomy. 20.19 Theorem. Suppose θ is a regular cardinal such that log θ+ = θ. Then there is f :[θ++]3 −→ ω which takes all the values from ω on the cube of any subset of θ++ of size θ+.

+ + ++ + Proof. We choose two C-sequences Cα (α < θ ) and Cα (α < θ ) on θ ++ + + and θ respectively, such that tp(Cα) ≤ θ for all α < θ and tp(Cα ) ≤ θ+ for all α < θ++. Let ρ∗ :[θ+]2 −→ θ and ρ∗+ :[θ++]2 −→ θ+ be the corresponding ρ∗-functions defined above in I.264. Also choose a one- ++ θ+ to-one sequence rα (α < θ ) of elements of {0, 1} and consider the corresponding function ∆ : [θ++]2 −→ θ+ defined in (I.293). We define f :[θ++]3 −→ θ+ according to the following two cases for a given triple α < β < γ of elements of θ++.

Case 1: (Cβs ∩ Cγt ) \ α 6= ∅, where s = ρ0(α, β) ¹ χ(α, β, γ) and t = ρ0(α, γ) ¹ χ(α, β, γ) of course assuming that ρ0(α, β) has length at least χ(α, β, γ). Rule 1: If ∆(rα, rβ) < ∆(rβ, rγ ) and rα lex rβ >lex rγ , set

f(α, β, γ) = min(Pν (∆(β, γ)) \ ∆(α, β)), where ν = ρ∗(min{ξ ≤ ∆(α, β): ρ∗(ξ, ∆(α, β)) 6= ρ∗(ξ, ∆(β, γ))}, ∆(β, γ)). Rule 2: Ifα ¯ ∈ {α, β, γ} is such that rα¯ is lexicographically between ¯ the other two rξ’s for ξ ∈ {α, β, γ}, if β ∈ {α, β, γ}\{α¯} is such that ∆(rα¯, rβ¯) > ∆(rα¯, rγ¯), whereγ ¯ is the remaining member of {α, β, γ}, and if {α, β, γ} does not fall under Rule 1, let

∗+ ¯ ∗+ f(α, β, γ) = min(Pν (ρ {β, γ¯}) \ ρ (α, β)), where ν = ρ∗{∆(α, β), ρ∗+(β, γ)}.

Case 2: (Cβs ∩ Cγt ) \ α = ∅, where s = ρ0(α, β) ¹ χ(α, β, γ) and t = ρ0(α, γ) ¹ χ(α, β, γ) of course assuming that ρ0(α, β) has length at least χ(α, β, γ). Let f(α, β, γ) = osc(α, β, γ). If a given triple α < β < γ does not fall into one of these two cases, let f(α, β, γ) = 0. The proof of Theorem 20.19 is ﬁnished if we show that for every Γ ⊆ θ++ of size θ+, the image f 00[Γ]3 either contains all positive integers or almost all ordinals < θ+ of coﬁnality θ. 20. Higher dimensions 151

So let Γ be a given subset of θ++ of order-type θ+ and let λ = sup(Γ). By the proof of Lemma 20.15, if Γ is stable, then it must contain a (stable) subset Ω such that for some n < ω and all α < β < γ in Ω:

χ(α, β, γ) = n, (I.321)

(Cβs ∩ Cγt ) \ α 6= ∅ where s = ρ0(α, β) ¹ n and t = ρ0(α, γ) ¹ n. (I.322) In other words, Case 1 of the definition of f(α, β, γ) applies to any triple α < β < γ from Ω. Note that in this case, the definition follows exactly the definition of the analogous function in the proof of Theorem 20.10 with ρ∗+ in the role of the strongly unbounded subadditive function e :[θ++]2 −→ θ+. By Lemma 20.15, ρ∗+ is strongly unbounded on subsets of Ω, so by the proof of Theorem 20.10 we conclude that in this case the image f 00[Ω]3 contains almost all ordinals < θ+ of cofinality θ. Suppose now that no subset of Γ of size θ+ is stable. By Lemma 20.17 (in fact, its proof) for every integer n ≥ 1 there exist α < β < γ in Γ such that for s = ρ0(α, β) ¹ χ(α, β, γ) and t = ρ0(α, γ) ¹ χ(α, β, γ), where ρ0(α, β) has length > χ(α, β, γ): osc(α, β, γ) = n, (I.323)

(Cβs ∩ Cγt ) \ α = ∅. (I.324) This ﬁnishes the proof. a

3 20.20 Corollary. There is f :[ω2] −→ ω which takes all the values on the cube of any uncountable subset of ω2. a 20.21 Remark. Note that the dimension 3 in this Corollary cannot be lowered to 2 as long as one does not use some additional axioms to construct such f. Note also that the range ω cannot be replaced by a set of bigger size, as this would contradict Chang’s conjecture. We have seen above that 3 Chang’s conjecture is equivalent to the statement that for every f :[ω2] −→ 00 3 ω1 there is an uncountable set Γ ⊆ ω2 such that f [Γ] 6= ω1. Is there a similar reformulation of the Continuum Hypothesis? More precisely, one can ask the following question.

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160 INDEX 161

Eberlein compactum, 13 Souslin, 32 special Aronszajn tree, 8 full code, 4 special square sequence, 87 full lower trace, 5 special tree, 67 square-bracket operation, 52 Hausdorff gap, 34, 44 square-sequence, 80 stable, 146 irreducible, 80 stationary subsequence, 119 Jensen matrix, 112 step, 3 stepping-up, 138 Kurepa family, 105 strong measure zero, 14 Kurepa family, cofinal, 111 strongly unbounded function, 132 Kurepa family, compatibility, 111 subadditivity, 28 Kurepa family, extendibility, 111 support, 44 last step, 37 tightness, 93 lower trace, 4 trace, 4, 51 trace filter, 123 Maharam type, 110 transitive map, 33, 136 maximal weight, 9 tree on θ, 86 metric, 28 trivial C-sequence, 76 minimal walk, 3, 51 Tukey equivalent, 110 Tukey reducible, 110 nontrivial, 36, 44 nowhere dense, 44 ultra-metric, 28 number of steps, 36 ultra-subadditivity, 28 unbounded family, 118 orthogonal, 38, 44 unbounded function, 131 upper trace, 4 P-ideal, 38 P-ideal dichotomy, 38 walk, 3 positive with respect to a filter, 124 weak Chang’s conjecture, 137 property ∆, 108 weight, 3 property C, 14 property C00, 14

Radon measure space, 110 Radon probability space, 110 right lexicographical ordering, 6 sequential fan, 93 sequentially closed, 114 shift, 48 sigma-product, 115 sigma-product, index-set, 115