I. Coherent Sequences - » Department of Mathematicsstevo/cseq.pdf · 2 I. Coherent Sequences sen...

I. Coherent Sequences

Stevo Todorcevic

Contents

0 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 Walks in the space of countable ordinals . . . . . . . . . . . . . . 32 The coherence of maximal weights . . . . . . . . . . . . . . . . . . . . 93 Oscillations of traces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 214 Triangle inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 265 Conditional weakly null sequences . . . . . . . . . . . . . . . . . . . . . 396 An ultrafilter from a coherent sequence . . . . . . . . . . . . . . . 447 The trace and the square-bracket operation . . . . . . . . . . . 518 A square-bracket operation on a tree . . . . . . . . . . . . . . . . . . 629 Special trees and Mahlo cardinals . . . . . . . . . . . . . . . . . . . . . 67

10 The weight function on successor cardinals . . . . . . . . . . . . 7311 The number of steps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7512 Square sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8013 The full lower trace of a square sequence . . . . . . . . . . . . . . 9014 Special square sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9615 Successors of regular cardinals . . . . . . . . . . . . . . . . . . . . . . . . 9816 Successors of singular cardinals . . . . . . . . . . . . . . . . . . . . . . . 10917 The oscillation mapping . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11718 The square-bracket operation . . . . . . . . . . . . . . . . . . . . . . . . . 12219 Unbounded functions on successors of regular cardinals 12920 Higher dimensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 138

Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 160

0. Introduction

A transfinite sequence Cξ ⊆ ξ (ξ < θ) of sets may have a number of ’coher-ence properties’ and the purpose of this chapter is to study some of them aswell 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 closedand unbounded in ξ for all ξ. For example, choosing a canonical ’funda-mental sequence’ of sets Cξ ⊆ ξ for ξ < ε0 relying on the specific propertiesof the Cantor normal form for ordinals below the first ordinal satisfying theequation 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 hasa different perspective in applications, so one is naturally led to use someother 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 constructionsbut also as tools for ’walking’ from an ordinal to a smaller one. This notionof a ’walk’ and the corresponding ’distance functions’ constitute the mainbody of study in this chapter. We show that the resulting ’metric theoryof ordinals’ not only provides a unified approach to a number of classicalproblems 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) ≤ maxe(α, β), e(β, γ)has three versions, depending on the natural ordering between the ordi-nals α, β and γ, that are of a quite different character and are occurringin quite different places and constructions in set theory. For example, themost frequent occurrence is the case α < β < γ when the triangle inequal-ity becomes something that one can call ’transitivity’ of e. Considerablymore subtle is the case α < γ < β of this inequality. It is this case ofthe inequality that captures most of the coherence properties found in thisarticle. Another thing one learns from this theory is the special role of thefirst uncountable ordinal in this theory. Any natural coherency requirementon the sets Cξ (ξ < θ) that one finds in this theory is satisfiable in the caseθ = ω1. The first uncountable cardinal is the only cardinal on which thetheory can be carried out without relying on additional axioms of set theory.The first uncountable cardinal is the place where the theory has its deepestapplications as well as its most important open problems. This special rolecan 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 uncountablerather than some intricate relationship between two or more uncountablecardinalities. This is of course not to say that an intricate relationship be-tween two or more uncountable cardinalities may not be a profitable detourin the course of solving such a problem. In fact, this is one of the reasonsfor our attempt to develop the metric theory of ordinals without restrictingourselves only to the realm of countable ordinals.

The chapter is organized as a discussion of four basic distance functionson ordinals, ρ, ρ0, ρ1 and ρ2, and the reader may choose to follow the anal-ysis of any of these functions in various contexts. The distance functionswill 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 thequantifier ’for every unbounded set’ to the quantifier ’for every closed andunbounded set’. This reduction of quantifiers has proven to be quite usefulin constructions of various mathematical structures, some of which havebeen mentioned or reproduced here.

I wish to thank Bernhard Konig, Piotr Koszmider, Justin Moore andChristine Hartl for 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 spaceconsidered in this chapter. There are many mathematical problems whosecombinatorial essence can be reformulated as problems about ω1, which isin some sense the smallest uncountable structure. What we mean by ’struc-ture’ 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 allknown other structures that have been defined so far on ω1. There is a nat-ural recursive way of picking up the fundamental sequences Cα, a recursionthat refers to the Cantor normal form which works well for, say, ordinals< ε0

1. For longer fundamental sequences one typically relies on some otherprinciples of recursive definitions and one typically works with fundamen-tal sequences with as few extra properties as possible. We shall see thatthe following assumption is what is frequently needed and will therefore beimplicitly assumed whenever necessary:

(c) if α is a limit ordinal, then Cα does not contain limit ordinals.

1.1 Definition. A step from a countable ordinal β towards a smaller ordinalα 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 toa smaller ordinal α is the sequence β = β0 > β1 > . . . > βn = α such thatfor each i < n, the ordinal βi+1 is the step from βi towards α.

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


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

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

defined recursively by

ρ0(α, β) = 〈|Cβ ∩ α|〉aρ0(α, min(Cβ \ α)),

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

Knowing ρ0(α, β) and the ordinal β one can reconstruct two followingtwo 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 se-quences 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 se-quences of ordinals and where for an ordinal ξ such that ρ0(ξ, β) < ρ0(α, β),

λξ(α, β) = maxmax(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.2Note that while ρ0 operates on the set [ω1]2 of unordered pairs of countable ordinals,

the formal recursive definition of ρ0(α, β) does involves the term ρ0(α, α) and so we haveto give it as a boundary value ρ0(α, α) = ∅. The boundary value is always specified whensuch a function is recursively defined and will typically be either 0 or the empty sequence∅. This convention will be used even when a function e with domain such as [ω1]2 isgiven to us beforehand and we consider its diagonal value e(α, α).


1.6 Lemma. For α ≤ β ≤ γ,

(a) α > L(β, γ)3 implies that ρ0(α, γ) = ρ0(β, γ)aρ0(α, β) and thereforethat 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).

a1.7 Definition. The full lower trace of the minimal walk is the functionF : [ω1]2 −→ [ω1]<ω defined recursively by

F (α, β) = F (α, min(Cβ \ α)) ∪⋃

ξ∈Cβ∩α

F (ξ, α),

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

Thus F (α, β) includes the traces of walks between any two ordinals ≤ αthat 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 inductivehypothesis 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 ratherthan F < α.


Since γ1 ∈ Cγ ∩ β, the set F (γ1, β) is included in F (β, γ). Similarly by thedefinition 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), soF (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).

a1.9 Lemma. For all α ≤ β ≤ γ,

(a) ρ0(α, β) = ρ0(min(F (β, γ) \ α), β)aρ0(α, min(F (β, γ) \ α)),

(b) ρ0(α, γ) = ρ0(min(F (β, γ) \ α), γ)aρ0(α, min(F (β, γ) \ α)).

Proof. The proof is again by induction on γ. Let α1 = min(F (β, γ)\α) andγ1 = min(Cγ \α). Let ξ1 be the minimal ξ ∈ (Cγ ∩β)∪min(Cγ \β) suchthat α1 ∈ F (ξ, β) (or F (β, ξ)).

Applying the inductive hypothesis to α ≤ ξ1 ≤ β (or α ≤ β ≤ ξ1) andnoting that α1 = min(F (ξ1, β) \ α) (or α1 = min(F (β, ξ1) \ α)) we get

ρ0(α, β) = ρ0(α1, β)aρ0(α, α1), (I.1)

the part (a) of Lemma 1.9. Note that depending on whether α ≤ γ1 ≤ βor α ≤ β ≤ γ1 we have that F (γ1, β) ⊆ F (β, γ) or F (β, γ1) ⊆ F (β, γ), andtherefore 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:

ρ0(α, β) = ρ0(α2, β)aρ0(α, α2), (I.2)

ρ0(α, γ1) = ρ0(α2, γ1)aρ0(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

ρ0(α, γ1) = ρ0(α2, γ1)aρ0(α1, α2)aρ0(α, α1) = ρ0(α1, γ1)aρ0(α, α1). (I.4)

Finally since ρ0(α, γ) = 〈|Cγ ∩ α|〉aρ0(α, γ1) = ρ0(γ1, γ)aρ0(α, γ1) holds,the equation (I.4) gives us the desired conclusion (b) of Lemma 1.9. a1.10 Definition. Recall that right lexicographical ordering on ω<ω denotedby <r refers to the total ordering defined by letting s <r t if s is an end-extension of t, or if s(i) < t(i) for i = minj : s(j) 6= t(j). Using theidentification between a countable ordinal α and the α-sequence (ρ0)α ofelements of ω<ω, the ordering <r induces the following ordering <c on ω1:

α <c β iff ρ0(ξ, α) <r ρ0(ξ, β), (I.5)

where ξ = ∆(α, β) = minη ≤ minα, β : ρ0(η, α) 6= ρ0(η, β).


1.11 Lemma. The cartesian square of the total ordering <c of ω1 is theunion of countably many chains.

Proof. It suffices to decompose the set of all pairs (α, β) where α < β.To each such pair we associate a hereditarily finite set p(α, β) which codesthe finite structure obtained from F (α, β) ∪ β by adding relations thatdescribe the way ρ0 acts on it. To show that this parametrization works,suppose we are given two pairs (α, β) and (γ, δ) such that

p(α, β) = p(γ, δ) = p, and α <c γ.

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 toeach other in the isomorphism of the (α, β) and (γ, δ) structures. It followsthat:

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:

ρ0(∆(α, γ), α) = tαγaρ0(∆(α, γ), ξαβ), (I.8)

ρ0(∆(α, γ), γ) = tαγaρ0(∆(α, γ), ξγδ). (I.9)

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

ρ0(∆(α, γ), β) = tβδaρ0(∆(α, γ), ξαβ), (I.10)

ρ0(∆(α, γ), δ) = tβδaρ0(∆(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) <r ρ0(ξ, γ) and since these two sequenceshave tαγ as common initial part, this reduces to

ρ0(ξ, ξαβ) <r ρ0(ξ, ξγδ). (I.12)

On the other hand tβδ is a common initial part of ρ0(ξ, β) and ρ0(ξ, δ), sotheir lexicographical relationship depends on their tails which by (I.10) and(I.11) are equal to ρ0(ξ, ξαβ) and ρ0(ξ, ξγδ) respectively. Referring to (I.12)we conclude that indeed ρ0(ξ, β) <r ρ0(ξ, δ), i.e. β <c δ. a


1.12 Notation. Well-ordered sets of rationals. The set ω<ω ordered bythe right lexicographical ordering <r is a particular copy of the rationalsof the interval (0, 1] which we are going to denote by Qr or simply by Q.The next lemma shows that for a fixed α, ρ0(ξ, α) is a strictly increasingfunction from α into Qr. We let ρ0α or simply ρ0α denote this function andwe will identify it with its range, i.e. view it as a member of the tree σQr

of all well-ordered subsets of Qr, ordered by end-extension.

1.13 Lemma. ρ0(α, γ) <r ρ0(β, γ) whenever α < β < γ. aThe sequence ρ0α (α < ω1) of members of σQr naturally determines the

subtreeT (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 alllevels of T (ρ0) are countable, and therefore T (ρ0) is a particular example ofan 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 thatadmits 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 wheneverβ < 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 thewalks β → α and γ → α, respectively. Using the assumption (c) of thefundamental sequences and the fact that α > 0 is a limit ordinal we canfind 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 tailof the walks β → ξ and γ → ξ. Subtracting ρ0(ξ, α) from ρ0(ξ, β) we getρ0(α, β) and subtracting ρ0(ξ, α) from ρ0(ξ, γ) we get ρ0(α, γ). It followsthat ρ0(α, β) = ρ0(α, γ). a

It follows that T (ρ0) does not branch at limit levels. From this we canconclude that T (ρ0) is a special subtree of σQ since this is easily seen to beso for any subtree of σQ which is finitely branching at limit nodes.

1.15 Definition. Identifying the power set of Q with the particular copy2Q of the Cantor set, define for every countable ordinal α,

Gα = x ∈ 2Q : 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-subsetsof the Cantor set whose union is equal to the Cantor set. a1.17 Lemma. The set X = ρ0β : β < ω1 considered as a subset of theCantor set 2Q has universal measure zero.

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

Pt = x ∈ 2Q : x end-extends t.

Note that each Pt is a perfect subset of 2Q 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 strictlyincreasing map into the reals (as for example σQ does). It follows that Smust 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 functionρ1 : [ω1]2 −→ ω defined recursively by

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

with the boundary value ρ1(α, α) = 04. Thus ρ1(α, β) is simply the maximalinteger 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 forevery n < ω and every A ⊆ α of order-type ω there is ξ ∈ A such thatρ1(ξ, α) > n. Let η = sup(A). If η = α one chooses arbitrary ξ ∈ A withthe property that |Cα ∩ ξ| > n, so let us consider the case η < α. Letα1 = 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)

4This is another use of the convention ρ1(α, α) = 0 in order to facilitate the recursivedefinition.


Note that ρ0(ξ, α) = 〈|Cα ∩ η|〉aρ0(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-typeω there exists ξ ∈ A such that ρ1(ξ, α) = ρ1(ξ, β). Let η = sup(A) and letβ1 = 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 suchthat ρ1(ξ, α) = ρ1(ξ, β1). Then

ρ1(ξ, β) = maxn, ρ1(ξ, β1) = ρ1(ξ, β1),

so we are done. a

Note the following unboundedness property of ρ1 which immediately fol-lows from the previous lemma and which will be elaborated on in lattersections of this Chapter.

2.3 Corollary. For every pair A and B uncountable families of pairwisedisjoint finite subsets of ω1 end every positive integer n there exist uncount-able subfamilies A0 ⊆ A and B0 ⊆ B such that for every pair a ∈ A0 andb ∈ B0 such that a < b5 we have ρ1(α, β) > n for all α ∈ a and β ∈ b. a2.4 Corollary. The tree

T (ρ1) = t : α −→ ω : α < ω1, t =∗ ρ1α6

is a coherent, homogeneous, and R-embeddable Aronszajn tree. aThe 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 coher-ent 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 <l denote the lexicographicalordering of T (a). We need to decompose pairs (s, t) of T (a) into countablymany classes that are chains in the cartesian product of <l. In fact, bysymmetry it suffices to decompose pairs (s, t) with l(s) ≤ l(t) where l(s)

5For sets of ordinals a and b we write a < b whenever α < β for all α ∈ a and β ∈ b.6Here, 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.


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 heredi-tarily finite set p(s, t) which codes the behavior of the restrictions of s andt 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 <l u. We need to show that t ≤l v. By symmetry, wemay assume that l(s) ≤ l(u) and that s and u are actually different. Letα = ∆(s, u), and let us first consider the case when α < l(s), i.e., when sand u are incomparable. Note that by the choice of parametrization andthe assumption p(s, t) = p(u, v), we have

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 <l v, as required. If α ∈ F (s, t) but α /∈ F (u, v) then

t(α) ≤ n(s, t) = n(u, v) < v(α),

so t <l v also in this case. The case α /∈ F (s, t) and α ∈ F (u, v) is impossiblesince

u(α) ≤ n(u, v) = n(s, t) < s(α),

contradicts our assumption s <l u. The remaining case α ∈ F (s, t)∩F (u, v)is also impossible since the assumption p(s, t) = p(u, v) would give us s(α) =u(α) contradicting the fact that α = ∆(s, u).

Suppose now that s is a strict initial segment of u. Then F = F (s, t) ∩l(s) = F (u, v) ∩ l(s) and

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 agreebellow l(s) we conclude that t <l v, as required. If l(s) = l(t)and since tand v agree bellow l(s) we conclude that v extends t, and therefore t <l v,as required. This finishes the proof. a


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

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

If we order T (ρ1) by the right lexicographical ordering <r we get a completelinearly ordered set. It is not continuous as it contains jumps of the form

[ta〈m〉, ta〈m + 1〉a~0],

where t ∈ T (ρ1) and m < ω. Removing the right-hand points from all thejumps we get a linearly ordered continuum which we denote by A(ρ1).

2.7 Lemma. A(ρ1) is a homogeneous nonreversible ordered continuum thatcan be represented as the union of an increasing ω1-sequence of Cantor sets.

Proof. For a countable limit ordinal δ > 0, set

Aδ(ρ1) = t ∈ A(ρ1) : l(t) ≤ δ.

Then each Aδ(ρ1) is a closed subset of A(ρ1) with the level Tδ(ρ1) be-ing a countable order-dense subset. One easily concludes from this thatAδ(ρ1) (δ limit < ω1) is the required decomposition of A(ρ1).

To show that A(ρ1) is homogeneous, consider two pairs x0 < x1 andy0 < y1 of non-endpoints of A(ρ1). Choose a countable limit ordinal δstrictly above the lengths of x0, x1, y0 and y1. So we have a Cantor setAδ(ρ1), an order-dense subset Tδ(ρ1) of it and two pairs of points x0 < x1

and y0 < y1 inAδ(ρ1) \ Tδ(ρ1).

By Cantor’s theorem there is an order isomorphism σ : Aδ(ρ1) −→ Aδ(ρ1)such that:

σ′′Tδ(ρ1) = Tδ(ρ1), (I.16)

σ(xi) = yi for i < 2. (I.17)

Extend σ to the rest of A(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 : A(ρ1) −→ A(ρ1).

Since T (ρ1) with the lexicographical ordering does not have a countableorder-dense subset, the image π′′T (ρ1) must have sequences of length bigger


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

f(δ) = maxξ < δ : tδ(ξ) 6= π(tδ)(ξ)+ 1.

By the Pressing Down Lemma find an uncountable set Γ of countable limitordinals, 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 ∈ Γare antichains of the tree T (ρ1). Pick γ 6= δ in Γ and suppose for definite-ness that tγ <l tδ. Note that the ordinal ∆(tγ , tδ) where this relation isdecided is above ξ and is smaller than both γ and δ. Similarly the ordinal∆(π(tγ), π(tδ)) where the lexicographical ordering between π(tγ) and π(tδ)is decided also belongs to the interval

(ξ, minγ, δ).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. a2.8 Definition.

A homogeneous Eberlein compactum 7. The set T (ρ1) has another naturalstructure, a topology generated by the family of sets of the form

Vt = u ∈ T (ρ1) : t ⊆ u,for t a node of T (ρ1) of successor length as a clopen subbase. Let T 0(ρ1)

denote the set of all nodes of T (ρ1) of successor length. Then T (ρ1) can beregarded as the set of all downward closed chains of the tree T 0(ρ1) and thetopology on T (ρ1) is simply the topology one obtains from identifying thepower set of T 0(ρ1) with the cube

0, 1T 0(ρ1)

with its Tychonoff topology8. T (ρ1) being a closed subset of the cube iscompact. In fact T (ρ1) has some very strong topological properties such asthe property that closed subsets of T (ρ1) are its retracts.

7Recall that a compactum X is Eberlein if its function space C(X) can be generatedby a subset which is compact in the weak topology.

8This is done by identifying a subset V of T 0(ρ1) with its characteristic functionχV : T 0(ρ1) −→ 2.


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

Proof. The proof that T (ρ1) is homogeneous is quite similar to the cor-responding part of the proof of the Lemma 2.7. To see that T (ρ1) is anEberlein compactum, i.e. that the function space C(T (ρ1)) is weak com-pactly generated, let Xn be a countable antichain decomposition of T (ρ1)and consider the set

K = 2−nχeVt: n < ω, t ∈ Xn ∪ χ∅.

Note that K is a weakly compact subset of C(T (ρ1)) which separates thepoints of T (ρ1). a

E. Borel’s notion of strong measure zero is a metric notion that has sev-eral closely related topological notions due to K.Menger, F.Rothberger andothers. For example Rothberger’s property C ′′ is a purely topological notionwhich 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 ofopen covers of a space X one can choose Un ∈ Un for each n such that

X =∞⋃

n=0

Un.

2.10 Lemma. T (ρ1) has Rothberger’s property C ′′. a2.11 Remark. The subspace T (ρ1) of T (ρ1) has in fact a considerablystronger covering property than C ′′. For example, it can be shown thatits function space C(T (ρ1)) with the topology of pointwise convergence is aFrechet space, or in other words, has the property that any family of con-tinuous functions on T (ρ1) which pointwise accumulates to the constantlyequal 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 conver-gence has a natural decomposition as the increasing union of an ω1-sequenceof closed separable metric vector subspaces. The separable metric vectorsubspaces 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 mapsthat are actually one-to-one. One way to achieve this is via the followingformula:

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

Define ρ1α from ρ1 just as ρ1α was defined from ρ1; then the ρ1α’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|.


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

2.12 Definition. The sequences eα = ρ1α (α < ω1) and rα (α < ω1) canbe used in describing a functor

G 7−→ G∗,

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

α, β ∈ G∗ iff e−1α (l), e−1

β (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 disjointfinite subsets of ω1 contains two sets A and B such that A ⊗ B ⊆ G10.Then the same is true about G∗ provided the uncountable family F consistsof finite cliques11 of G∗.

Proof. Let F be a given uncountable family of pairwise disjoint finite cliquesof 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 itselements above δ. Let n = n(A, δ) be the minimal integer such that for allα < β in A ∪ δ:

∆(rα, rβ) < n, 12 (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-sequence s(δ,A) of mappings with integer domains. Letr(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 aquadruple

(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-onemappings and any other sequence rα (α < ω1) of pairwise distinct reals.

10Here, A⊗B = α, β : α ∈ A, β ∈ B, α 6= β.11A clique of G∗ is a subset C of ω1 with the property that [C]2 ⊆ G∗.12Recall that for s 6= t ∈ ωω , we let ∆(s, t) = mink : s(k) 6= t(k).


of parameters. Since the set of quadruples is countable, by the assumptionon the graph G we can find two distinct members Aδ and Bδ of F above δthat generate the same quadruple of parameters denoted by

(Hδ, rδ, sδ, nδ).

Moreover, we choose Aδ and Bδ to satisfy the following isomorphism con-dition

For every l ≤ nδ, i < k, if α is the ith member of Aδ, β theith member of Bδ and if e−1

α (l) and e−1β (l) are both defined

then they are G-connected to the same elements of Hδ ∩ δ.(I.23)

Let mδ be the minimal integer > nδ such that

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

LetIδ = ξ : e(ξ, α) ≤ mδ for some α ≥ ξ in Aδ ∪Bδ.

Let pδ be the k-sequence rα ¹ mδ(α ∈ Aδ) enumerated in increasing orderand similarly let qδ be the k-sequence rβ ¹ mδ(β ∈ Bδ) enumerated increas-ingly. Let tδ and uδ be the transitive collapse of eα ¹ Iδ (α ∈ Aδ) andeβ ¹ Iδ(β ∈ Bδ), respectively. By the Pressing Down Lemma there is anunbounded Γ ⊆ ω1 and tuples

(H, r, s, n) and (I, p, q, t, u, m)

such that for all δ ∈ Γ:

(Hδ, rδ, sδ, nδ) = (H, r, s, n), (I.25)

(Iδ ∩ δ, pδ, qδ, tδ, uδ,mδ) = (I, p, q, t, u,m). (I.26)

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

For every γ, δ ∈ Γ and i < k1, if η is the ith memberof Iγ and if ξ is the ith member of Iδ in their increasingenumerations, then η and ξ are G-connected to the sameelements of the root I, and if η happens to be the i∗thmember of Aγ (or Bγ) for some i∗ < k, then ξ must be thei∗th member of Aδ (Bδ resp.) and vice versa.

(I.27)

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)


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 thejth member of Bδ.

Case 1: i 6= j. Pick an l < ∆(rα, rβ). By (I.21), l < n. Assume thate−1α (l) and e−1

β (l) are both defined.Subcase 1.1: e−1

α (l) < γ and e−1β (l) < δ. By the first choice of parame-

ters, e−1α (l) and e−1

β (l) are members of the set H which is an initial part ofH(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 sequence s, respectively. Inparticular, we have

e−1β (l) = e−1

β′′ (l), (I.31)

where β′′ =the jth member of Aγ . By our assumption that [Aγ ]2 ⊆ G∗ weinfer that α, β′′ ∈ G∗. Referring to the definition (I.20) we conclude thate−1α (l) and e−1

β (l) must be G-connected if they are different.Subcase 1.2: e−1

α (l) < γ and e−1β (l) ∈ Iδ \ δ. Let

β′ = the jth member of Bγ .

By the choice of parametrization, the position of e−1β′ (l) in Iγ is the same

as the position of e−1β (l) in Iδ so by (I.27) their relationship to the point

e−1α (l) of the root I is the same. Similarly, by the choice of the first set of

parameters, letting β′′ be the jth member of Aγ , the position of e−1β′ (l) in

H(γ, Bγ) and e−1β′′ (l) in H(γ, Aγ) is the same, so from (I.23) we conclude

that their relationship with e−1α (l) is the same. However, we have checked

in the previous subcase that e−1α (l) and e−1

β′′ (l) are G-connected in case theyare different.

Subcase 1.3: e−1α (l) ∈ Iγ \ γ and e−1

β (l) < δ. This is essentially sym-metric to the previous subcase.

Subcase 1.4: e−1α (l) ∈ Iγ \ γ and e−1

β (l) ∈ Iδ \ δ. The fact thate−1

α (l), e−1β (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)).

Subcase 2.1: e−1α (l) < γ and e−1

β (l) < δ. Then e−1α (l) and e−1

β (l) areelements of I which is an initial part of both Iγ and Iδ. Therefore thejth(= ith) term of u encodes both eβ ¹ I and eβ′ ¹ I (see the definition ofβ′ above). It follows that

e−1β (l) = e−1

β′ (l).


If l ≤ n then e−1β′ (l) belongs to H and since the jth(= ith) term of s encodes

both eα ¹ H and eβ ¹ H it follows that

e−1α (l) = e−1

β′ (l).

If l > n then e−1α (l) and e−1

β′ (l) are not members of H so from (I.22) andthe definition of

H(γ,Aγ) ∩ γ = H = H(γ, Bγ) ∩ γ

we conclude that

eγ(e−1α (l)) = eα(e−1

α (l)) = l = eβ′(e−1β′ (l)) = eγ(e−1

β′ (l)).

Since eγ is one-to-one we conclude that

e−1α (l) = e−1

β′ (l) = e−1β (l).

Subcase 2.2: e−1α (l) < γ and e−1

β (l) ∈ Iδ \ δ. Recall that β′ is theith(= jth) member of Bγ so by the first choice of parameters the relativeposition of e−1

β′ (l) in H(γ, Bγ) must be the same as the relative position ofe−1α (l) in H(γ, Aγ), i.e. it must belong to the root H. Note that if j∗ is the

position of β in Iδ then j∗ must also be the position of β′ in Iγ . It followsthat the j∗th term of u encodes both

eβ ¹ Iδ and eβ′ ¹ Iγ = eβ′ ¹ I.

So it must be that e−1β (l) belongs to the root I, a contradiction. So this

subcase never occurs.Subcase 2.3: e−1

α (l) ∈ Iγ \ γ and e−1β (l) < δ. This is symmetric to the

previous subcase, so it also never occurs.Subcase 2.4: e−1

α (l) ∈ Iγ \γ and e−1β (l) ∈ Iδ \δ. Then e−1

α (l), e−1β (l) ∈

G follows from (I.29).This completes the proof of Lemma 2.13. a

2.14 Lemma. If there is uncountable Γ ⊆ ω1 such that [Γ]2 ⊆ G∗ then ω1

can be decomposed into countably many sets Σ such that [Σ]2 ⊆ G.

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

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

Then the sets Γs cover ω1 and [Γs]2 ⊆ G for all s. a


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 (finitechains of T ) there exist A 6= B in F such that A∪B is a clique of G (a chainof 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 onthe graph G in Lemma 2.13 is necessary. There are indeed many graphsthat satisfy the hypothesis of Lemma 2.13. Many examples appear whenone 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 posetof all finite cliques of G∗ is ccc while its hypothesis is a bit stronger thanthe fact that the poset of all finite cliques of G is ccc in all of its finitepowers. Applying 2.14 to the case when G is the incomparability graph ofsome Aronszajn tree, we see that the statement saying that all Aronszajntrees are special is a purely Ramsey-theoretic statement in the same waySouslin’s hypothesis is.

We finish this section by showing that the assumption (c) on a givenC-sequence Cα (α < ω1) on which we base our walks and the function ρ1 isnot sufficient to give us the stronger conclusion that the tree T (ρ1) can bedecomposed into countably many antichains (see Corollary 2.4). To describean 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)) denotethe n’th element of Cα according to its increasing enumeration with theconvention that Cα+1(n) = α for all n > 0. We assume that the C-sequenceis chosen so that for a limit ordinal α > 0 and a positive integer n, the ordinalCα+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 ordinalsin 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 includedin ω which cohere in the sense that eα(ξ) = eβ(ξ) for all but finitely manysuccessor ordinals ξ ∈ α0 ∩ β0. For each r ∈ ([ω]<ω)ω, we associate anotherC-sequence Cr

α (α < ω1) by letting Crα = Cα ∩ [ξr

α, α)∪⋃n∈ω Dr

α(n), where

Drα(n) = ξ ∈ [Cα(n), Cα(n + 1))0 : eα(ξ) ∈ r(n),

and where ξrα = sup(

⋃n∈ω Dr

α(n)). (This definition really takes place onlywhen α is a limit ordinal; for successor ordinals we put Cr

α+1 = α.) Thisgives us a C-sequence Cr

α (α < ω1) and we can consider the correspondingρr1 : [ω1]2 −→ ω as above. This function will have the properties stated in

Lemma 2.2, so the corresponding tree

T (ρr1) = (ρr

1)β ¹ α : α ≤ β < ω1is a coherent tree of finite-to-one mappings that admits a strictly increasingfunction into the real line. The following fact shows that T (ρr

1) will typicallynot be decomposable into countably many antichains.


2.16 Lemma. If r is a Cohen real then T (ρr1) has no stationary antichain.

Proof. The family [ω]<ω of finite subsets of ω equipped with the discretetopology and its power ([ω]<ω)ω with the corresponding product topologyand the given Cohen real r is to meet all dense Gδ subset of this spacethat one can explicitly define. Thus in particular, ξr

α = α, and therefore,Cr

α =⋃

n∈ω Drα(n) for every limit ordinal α. Since the tree T (ρr

1) is coherent,to establish the conclusion of the Lemma, it suffices to show that that forevery stationary Γ ⊆ ω1 there exists γ, δ ∈ Γ such that (ρr

1)γ ⊂ (ρr1)δ. This

in turn amounts to showing that for every stationary Γ ⊆ ω1, the set

UΓ = x ∈ ([ω]<ω)ω : (∃γ, δ ∈ Γ) (ρx1)γ ⊂ (ρx

1)δ

is a dense-open subset of ([ω]<ω)ω. So, given a finite partial function p fromω into [ω]<ω, it is sufficient to find a finite extension q of p such that thebasic open subset of ([ω]<ω)ω determined by q is included in UΓ. Let n bethe minimal integer that is bigger than all integers appearing in the domainof p or any set of the form p(j) for j ∈ dom(p). For γ ∈ Γ, set

Fn(γ) = ξ ∈ γ0 : eγ(ξ) ≤ n.

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

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

Extend the partial function p to a partial function q with domain 0, 1, ..., msuch 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 thechoices of the objects n, ∆, F ,m, γ, δ and q, we have that

(3) γ + 1 ∈ Cxδ , and

(4) Cxδ ∩ γ is an initial segment of Cx

γ .

It follows that, given a ξ < γ, the walk from δ to ξ along the C-sequenceCx

β (β < ω1) either leads to the same finite string of the correspondingweights as the walk from γ to ξ, or else it starts with first two steps to γ +1and γ, and then follows the walk from γ to ξ. Since ρx

1(ξ, δ) and ρx1(ξ, γ)

are by definitions maximums of these two strings of weights, we conclude

3. Oscillations of traces 21

that ρx1(ξ, δ) ≥ ρx

1(ξ, γ). On the other hand, note that by (4), in the secondcase, the weight |Cx

δ ∩ ξ| of the first step from δ to ξ is less than or equalto the weight |Cx

γ ∩ ξ| of the first step from γ to ξ. It follows that we havealso the other inequality ρx

1(ξ, δ) ≤ ρx1(ξ, γ). Hence we have shown that

ρx1(β, γ) = ρx

1(β, δ) for all β < γ, or in other words, that (ρx1)γ ⊂ (ρx

1)δ. Thisfinishes the proof.

a2.17 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

osc : ([ω1]<ω)2 −→ Card

defined byosc(x, y) = |x/E(x, y)|,

where E(x, y) is the equivalence relation on x defined by letting αE(x, y)βif and only if the closed interval determined by α and β contains no pointfrom y \ x. So, if x and y are disjoint, osc(x, y) is simply the number ofconvex 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 codingtechnique. We give some applications to this effect and in Section 17 belowwe 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)),

where ∆0(α, β) = minξ ≤ α : ρ0(ξ, α) 6= ρ0(ξ, β) − 113.

The following result shows that the upper traces do realize all possibleoscillations in any uncountable subset of ω1.

3.2 Lemma. For every uncountable subset Γ of ω1 and every positive inte-ger n there is an integer l such that for all k < n there exist α, β ∈ Γ suchthat 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 relevantobjects 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.


i < n. Let Ci = M ∩ ω1 : M ∈ Mi. Note that each Ci is a closed andunbounded subset of ω1 and that for each i < n, Ci is an element of everymodel ofMi+1. Let Mn be the minimal model ofMn and pick a β ∈ Γ aboveδn = Mn∩ω1. Since Cn−1 ∈ Mn we can find Mn−1 ∈Mn−1∩Mn such thatδn−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 giveus 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 aboveL(δ0, β). Then (see Lemma 1.6 above), δi ∈ Tr(ξ, β) for all i ≤ n and in factδ0 → 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 modelsM0 ∈ M1 ∈ ... ∈ Mn shows that there is an uncountable subset Σ of Γ0

such that Σ ∈ M0 and

T = x ∈ [ω1]≤|t| : x v Tr(ξ, α) for some α ∈ Σ

is a tree under v in which a node x must either have only one immediatesuccessor 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 ξis a root of T as well as its splitting nodes. Let Ω be the uncountable setsuch that ξ, γ ∈ T for all γ ∈ Ω. Working in M0 we find η > ξ anduncountable Ω0 ⊆ Ω such that η = ∆0(γ, β) for all γ ∈ Ω0 and thereforeη = ∆0(α, β) for all α ∈ Σ witnessing ξ, γ ∈ T . Still working in M0 andgoing 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 witnessingξ, γ ∈ T for some γ ∈ Ω0. Let s = ρ0(η, β). Note that s end-extends t0but not necessarily t. Consider now the tree

X = x ∈ [ω1]≤|s| : 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 swe have that ρ0(γ, α) = s0 for all α ∈ Σ0 witnessing x0 ∪ γ ∈ X . Allother splitting nodes of X are determined by the other initial segmentss0 w t0 = t1 = ... = tn. Recall that all these objects are elements of M0 andtherefore 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 forall α ∈ Σ0 witnessing x1 ∪ γ ∈ X . Working in M2 find γ1 ∈ Ω1 such


that γ1 > Tr(η, β)∩ δ2 and a splitting node x2 w Tr(η, γ1) corresponding tot2, and so on. Proceeding in this way we construct an increasing sequencex0 < 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.

aConsider the following projection of the oscillation mapping osc0,

osc∗0(α, β) = maxm ∈ ω : 2m|osc0(α, β).

3.3 Corollary. For every uncountable Γ ⊆ ω1 and every positive integer kthere exist α, β ∈ Γ such that osc∗0(α, β) = k. a3.4 Corollary. Every inner model model M of set theory which correctlycomputes ω1 contains a partition of ω1 into infinitely many pairwise disjointsubsets that are stationary in the universe V of all sets.

Proof. Our assumption about M means in particular that the C-sequenceCξ (ξ < ω1) can be chosen to be an element of M. It follows that theoscillation mapping osc∗0 is also an element of M. It follows that for eachα < ω1, the sequence of sets

Sαn = β > α : osc∗0(α, β) = n (n < ω)

belongs to M. So it suffices to show that there must be α < ω1 such that forevery n < ω the set Sαn is stationary in V. This is an immediate consequenceof Corollary 3.3. a3.5 Remark. In [46], Larson showed that Corollary 3.4 cannot be extendedto partitions of ω1 into uncountably many pairwise disjoint stationary sets.He also showed that under the Proper Forcing Axiom, for every mappingc : [ω1]2 → 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 withcountable ranges.


One can count oscillation not only between two finite subsets of ω1 butalso between two finite partial functions from ω1 into ω, i.e., between twofinite ’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 inits rectangular form.

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

Proof. Fix a continuous ∈-chain N of length ω1 of countable elementarysubmodels of (Hω2 ,∈) containing all the relevant objects and fix also acountable elementary submodel M of (Hω2 ,∈) that contains N as an ele-ment. 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 thatρ1(ξ, α0) > ρ1(ξ, β0) for ξ = max(L0). Choose ξ0 ∈ Cδ such that ξ > L0 andsuch that the three mappings ρ1α0 ,ρ1β0 , and ρ1δ agree on the interval [ξ, δ).By elementarity of M there will be δ+ > δ, α+

0 ∈ Γ \ δ+, and β+0 ∈ Σ \ δ+

realizing the same type over the parameters L0 and ξ0. Let L+1 = L(δ, δ+

0 ).Note that ξ0 ≤ min(L+

1 ). It follows that the mappings ρ1α+0

and ρ1β+0

agreeon L+

1 . Choose an N ∈ N belonging to M such that N ∩ ω1 > L+1 and

choose ξ1 ∈ Cδ \N. Then we can find δ1 > δ and α+1 ∈ Γ\δ1 and β1 ∈ Σ\δ1

realizing the same type as δ, α+0 , and β+

0 over the objects accumulates sofar, L0, ξ0, L+

1 , N , and ξ1. Let L++1 = L(δ, δ1) and let t+1 be the restriction

of ρ1α+1

on max(L+1 ) + 1. Then the set of α ∈ Γ whose ρ1α end-extends t+1

belongs to the submodel N and is uncountable, since clearly it contains theordinal α+

1 which does not belong to N. So by the elementarity of N andCorollary 2.3 there is α1 ∈ Γ \ δ such that ρ1α1 end-extends t+1 and

ρ1(ξ, α1) > ρ1(ξ, β1) for all ξ ∈ L++1 . (I.32)

Let t1 be the restriction of ρ1α on max(L++1 ). 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 ∈ L+1 . (I.33)

Note also that L(δ, β1) = L0 ∪ L+1 ∪ L++

1 . Using similar reasoning we canfind t2 ∈ M ∩T (ρ1) end-extending t1, an α2 ∈ Γ\ δ whose ρ1α2 end-extends


t2, finite sets δ > max(L++2 ) > max(L+

2 ) > max(L++1 ), and β2 ∈ Σ \ δ such

that L(δ, β2) = L0 ∪ L+1 ∪ L++

1 ∪ L+2 ∪ L++

2 and such that the analogues ofI.32 and I.33 hold for all α ∈ Γ whose ρ1α end-extends t2, and so on. Thisprocedure gives us a sequence L0 < L+

1 < L++1 < ... < L+

n < L++n 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 ∈ Γwhose ρ1α end-extends tk,

L(δ, βk) = L0 ∪ L+1 ∪ L++

1 ∪ ... ∪ L+k ∪ L++

k , (I.34)

ρ1(ξ, α) > ρ1(ξ, βk) for all ξ ∈ L++k , (I.35)

ρ1(ξ, α) = ρ1(ξ, βk) for all ξ ∈ L+k . (I.36)

Choose α ∈ Γ∩M whose ρ1α end-extends tn such that the set Ln = L(α, δ)lies above L++

n 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 ∪ L+1 ∪ L++

1 ∪ ... ∪ L+k ∪ L++

k ∪ Ln. (I.37)

Let l = osc(ρ1α ¹ (L0∪Ln), ρ1β ¹ (L0∪Ln)) where β is equal to one of theβk’s. Note that l does not depend on which βk we choose and that accordingto 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 uncountablefamily of pairwise disjoint sets giving us the following slightly more generalconclusion.

3.8 Lemma. For every uncountable family G of pairwise disjoint finite sub-sets of ω1 all of some fixed size m, every uncountable subset Σ of ω1, andevery positive integer n there exist integers li (i < m) and a ∈ G such thatfor all k < n there exist β ∈ Σ such that osc1(a(i), β) = li + k for alli < 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 amapping ∗ : R→ ω such that ∗(x) = n if x−[x] ∈ In, where In = ( 1

n+3 , 1n+2 ].

For x in R we use the notation x∗ for ∗(x). Choose a one-to-one sequencexξ (ξ < ω1) of elements of the first half of the unit interval I = [0, 1] whichtogether with the point 1 form a set that is linearly independent over thefield of rational numbers. Finally, set

osc∗1(α, β) = (osc1(α, β) · xα)∗.

Then Lemma 3.8 turns into the following statement about osc∗1.14Here a(i) denotes the ith member of a according to the increasing enumeration of a.


3.9 Lemma. For every uncountable family G of pairwise disjoint finite sub-sets 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 thatosc∗1(a(i), β) = h(i) for all i < m.

Proof. Let ε be half of the length of the shortest interval of the formIh(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[32]; XXIII) gives us that there is an integer na such that for every y ∈ Im

there exist k < na such that |k ·xa(i)− yi| < ε (mod 1) for all i < m. Goingto an uncountable subfamily of G, we may assume that there is n such thatna = n for all a ∈ G. By Lemma 3.8 there exits a ∈ G and li (i < m) suchthat for every k < n there is βk > a such that osc1(a(i), βk) = li + k for alli < m. For i < m, let yi = li · xa(i) and let zi be the middle-point of theinterval Ih(i). Find k < n such that |k · xa(i) − (zi − yi)| < ε (mod 1) for alli < m. Then osc∗1(a(i), βk) = h(i) for all i < m, as required. a3.10 Corollary. There is a regular hereditarily Lindelof space which is notseparable.

Proof. For β < ω1, let fβ ∈ 0, 1ω1 be defined by letting fβ(β) = 1,fβ(γ) = 0 for γ > β, and fβ(α) = min1, osc∗1(α, β) for α < β. ConsiderF = fβ : β < ω1 as a subspace of 0, 1ω1 . Clearly F is not separable.That F is hereditarily Lindelof follows easily from Lemma 3.9. a3.11 Remark. The projection osc∗0 of the oscillation mapping osc0 appearsin [78] as historically first such a map with more than four colors that takesall of its values on every symmetric square of an uncountable subset of ω1.The variations osc1 and osc∗1 are on the other hand very recent and aredue to J.T. Moore [54] who made them in order to obtain the conclusion ofCorollary 3.10. Concerning Corollary 3.10 we note that the dual implica-tion behaves quite differently since assuming the Proper Forcing Axiom allhereditarily separable regular spaces are hereditarily Lindelof (see [80]).

4. Triangle inequalities

In this section we study a characteristic of the minimal walk that satisfiescertain triangle inequalities reminiscent of those found in an ultrametricspace. We also present several applications of the corresponding metric-like theory of ω1. This leads us to the following variation of the oscillationmapping.

4.1 Definition. Define ρ : [ω1]2 −→ 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) ≤ maxρ(α, β), ρ(β, γ),(c) ρ(α, β) ≤ maxρ(α, γ), ρ(β, γ).

Proof. Note that ρ(α, β) ≥ ρ1(α, β), so (a) follows from the correspondingproperty of ρ1. The proof of (b) and (c) is simultaneous by induction on α,β and γ:

To prove (b), consider n = maxρ(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.Case 1b: ξα = ξβ . Then by the inductive hypothesis,

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

ρ(ξ, α) ≤ maxρ(ξ, β), ρ(α, β).From the definition of ρ(β, γ) we see that ρ(ξ, β) ≤ ρ(β, γ), so replacingρ(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) weconclude that ρ(α, γ) ≤ n.

Case 2b: ξα < ξβ . Then ξα ∈ Cγ ∩ β, so

ρ(ξα, β) ≤ ρ(β, γ) ≤ n.

By the inductive hypothesis

ρ(α, ξα) ≤ maxρ(α, β), ρ(ξα, β) ≤ n.

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

ρ(ξ, α) ≤ maxρ(ξ, β), ρ(α, β) ≤ n.

Finally |Cγ ∩ α| ≤ |Cγ ∩ β| ≤ ρ(β, γ) ≤ n. Combining these inequalities weget the desired conclusion ρ(α, γ) ≤ n.

To prove (c), consider now n = maxρ(α, γ), ρ(β, γ). We have to showthat ρ(α, β) ≤ n. Let ξα and ξβ be as above and let us consider the sametwo cases as above.


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

ρ(α, β) ≤ maxρ(α, ξ), ρ(β, ξ).

This gives the desired bound ρ(α, β) ≤ n, since ρ(α, ξα) ≤ ρ(α, γ) ≤ n andρ(β, ξβ) ≤ ρ(β, γ) ≤ n.

Case 2c: ξα < ξβ . Applying the inductive hypothesis again we get

ρ(α, β) ≤ maxρ(α, ξα), ρ(ξα, β) ≤ n.

This completes the proof. a4.3 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) = maxρ1(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 finercoherence 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) ≤ maxρ(α, β), ρ(β, γ) = ρ(α, β),

ρ(α, β) ≤ maxρ(α, γ), ρ(β, γ) = ρ(α, γ).

Note that maxρ(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, isbigger than ρ(β, γ). a4.5 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 propertiesof the function ρ. Recall the notion of a full lower trace F (α, β) given inDefinition 1.7 above. The function d : [ω1]2 −→ ω defined by

d(α, β) = |F (α, β)|

is an example of a truly subadditive function, since by Lemma 1.8 we havethe following inequalities whenever α < β < γ:

(a) d(α, γ) ≤ d(α, β) + d(β, γ),


(b) d(α, β) ≤ d(α, γ) + d(β, γ).

4.6 Definition. Define ρ : [ω1]2 −→ 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) ≤ maxρ(α, β), ρ(β, γ),(c) ρ(α, β) ≤ maxρ(α, γ), ρ(β, γ).

Proof. Only (b) and (c) require some argument. Consider first (b). Ifρ(α, β) ≤ ρ(β, γ) then ρ(α, γ) ≤ maxρ(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 case ρ(α, γ) ≤ ρ(β, γ). On the other hand, if ρ(β, γ) ≤ρ(α, β), then ρ(α, γ) ≤ maxρ(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 case ρ(α, γ) ≤ ρ(α, β). This checks (b). Checking (c) is similar.a

4.8 Definition. For p ∈ ω<ω define a binary relation <p on ω1 by lettingα <p β iff α < β, ρ(α, β) ∈ |p| and

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

4.9 Lemma.

(a) <p is a tree ordering on ω1 of height ≤ |p|+ 1,

(b) p ⊆ q implies <p⊆<q.

Proof. This follows immediately from Lemma 4.7. a4.10 Definition. For x ∈ ωω, set

<x=⋃<x¹n: n < ω.

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


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 codethe behavior of ρ on F|p|(α) one can realize only finitely many differentisomorphism 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 <q β.

Consider the graph G of all unordered pairs of integers< ρ(α, β) of theform

ρ(ξ, α), ρ(ξ, β)for some ξ < α. It suffices to show that every connected component ofG has at most one point < |p|. For if this is true, extend p to a mappingq : ρ(α, β) −→ ω which is constant on each component of G. Clearly, α <q βfor any such q.

Consider an integer n < |p| and a G-path n = n0, n1, . . . , nk which startswith n. Pick ordinals ξ0, ξ1, . . . , ξk−1 < α and γi ∈ α, β for i ≤ 2k − 1which witness the connections, i.e.

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 about ρ (togetherwith the properties (I.40)-(I.41)) shows that the sequence n0, n1, . . . , nk isstrictly 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 < minα, 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 ini-tial part of ξ ≤ ηβ : ρ(ξ, ηβ) ≤ ρ(ηβ , β) contradicting the fact thatρ(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 subadditivityproperties of ρ gives us that

ξ ≤ ηα : ρ(ξ, ηα) ≤ ρ(ηα, α) ⊆ ξ ≤ ηα : ρ(ξ, ηα) ≤ ρ(ηα, ηβ).


This gives the inequality ρ(ηα, ηβ) > ρ(ηα, α) = n. Combining this inequal-ity with the subadditivities of ρ we get:

n < ρ(ηα, ηβ) ≤ maxρ(ηα, α), ρ(ηβ , α) = ρ(ηβ , α),

n < ρ(ηα, ηβ) ≤ maxρ(ηα, β), ρ(ηβ , β) = ρ(ηα, β).

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

a

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

GΓ = x ∈ ωω : α <x β for some α, β ∈ Γ

is a dense open subset of the Baire space.

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 that ρ(ξ, α) =ρ(ξ, β) for all ξ < α. Then <ρ is a tree ordering on ω1 obtained by identifyingα with the member ρ(·, α) of the tree T (ρ). Note that <ρ⊆<x for all x ∈ ωω

and that there exists x ∈ ωω such that <x=<ρ (e.g., one such x is theidentity map id : ω −→ ω).

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

HΓ = x ∈ ωω : 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 theoremgives us a strictly increasing sequence γi (i < ω) of members of Γ suchthat ∆(γi, γj) ≤ ∆(γj , γk)15 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 elsefor some ordinal ξ and all i < j, ∆(γi, γj) = ξ. In the first case, byLemma 4.7(a), we can find i such that one of the integers ρ(∆(γi, γi+1), γi)or ρ(∆(γi, γi+1), γi+1) does not belong to the domain of p, so there will beq ⊇ p such that ρ(γi, γi+1) < |q| and γi ≮q γi+1. This guarantees thatγi ≮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 theintegers ρ(ξ, γi) or ρ(ξ, γj) belongs to the domain of p, so there is q ⊇ psuch that ρ(γi, γj) < |q| and γi ≮q γj . Hence, γi ≮x γj for every x ⊇ q. a

15For α < β < ω1, ∆(α, β) = minξ ≤ α : ρ(ξ, α) 6= ρ(ξ, β).


4.16 Definition. For a family F of infinite <ρ-antichains, we say that areal x ∈ ωω is F-Cohen if x ∈ GΓ ∩ HΓ for all Γ ∈ F . We say that x isF-Souslin if no member of F is a <x-chain or a <x-antichain. We say thata real x ∈ ωω is Souslin if the tree ordering <x on ω1 has no uncountablechains nor antichains, i.e. when x is F-Souslin for F equal to the family ofall uncountable subsets of ω1.

Note that since every uncountable subset of ω1 contains an uncount-able <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. a4.18 Corollary. If the density of the family of all uncountable subsets ofω1 is smaller than the number of nowhere dense sets needed to cover thereal line, then there is a Souslin tree. a4.19 Remark. Recall that the density of a family F of infinite subsets ofsome set S is the minimal size of a family F0 of infinite subsets of S withthe property that every member of F is refined by a member of F0. Aspecial case of Corollary 4.18, when the density of the family of all un-countable 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 anuncountable subset from the ground model. So it suffices to consider thefamily F of all infinite <ρ-antichains from the ground model. a

If ω1 is a successor cardinal in the constructible subuniverse, then ρ canbe chosen to be coanalytic and so the transformation x 7−→<x will transfercombinatorial notions of Souslin, Aronszajn or special Aronszajn trees intothe corresponding classes of reals that lie in the third level of the projectivehierarchy. This transformation has been explored in several places in theliterature (see, e.g. [4],[33]).

4.21 Remark. We have just seen how the combination of the subadditivityproperties (4.7(b),(c)) of the coherent sequence ρα : α −→ ω (α < ω1)of one-to-one mappings can be used in controlling the finite disagreementbetween them. It turns out that in many contexts the coherence and thesubadditivities are essentially equivalent restrictions on a given sequenceeα : α −→ ω (α < ω1). For example, the following construction shows thatthis is so for any sequence of finite-to-one mappings eα : α −→ ω (α < ω1).


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

τe(α, β) = maxmaxe(ξ, α), e(ξ, β) : ξ ≤ α, e(ξ, α) 6= e(ξ, β)16.4.23 Lemma. For every α < β < γ < ω1,

(a) τe(α, β) ≥ e(α, β),

(b) τe(α, γ) ≤ maxτe(α, β), τe(β, γ),(c) τe(α, β) ≤ maxτe(α, γ), τe(β, γ).

Proof. To see (a) note that it trivially holds if e(α, β) = 0 so we canconcentrate 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) ≥maxmaxe(α, α), e(α, β) = e(α, β).

To see (b) let n = maxτe(α, β), τe(β, γ) and let ξ ≤ α be such thate(ξ, α) 6= e(ξ, γ). We need to show that e(ξ, α) ≤ n and e(ξ, γ) ≤ n. Ife(ξ, α) > n then since τe(α, β) ≤ n we must have e(ξ, α) = e(ξ, β)) andso e(ξ, β) 6= e(ξ, γ). It follows that τe(β, γ) ≥ e(ξ, β)) > n, a contradic-tion. Similarly, e(ξ, γ) > n yields e(ξ, β) = e(ξ, γ) 6= e(ξ, α) and thereforeτe(α, β) ≥ e(ξ, β)) > n, a contradiction.

The proof of (c) is similar. a4.24 Definition. We say that a : [ω1]2 −→ ω is transitive if for α < β <γ < ω1

a(α, γ) ≤ maxa(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 increasesrelative to the ordering ⊆∗ of inclusion modulo a finite set, the mappinga : [ω1]2 −→ ω defined by

a(α, β) = minn : Aα \ n ⊆ Aβis a transitive map. The transitivity condition by itself is not nearly as usefulas its combination with the other subadditivity property (4.7(c)). Fortu-nately, there is a general procedure that produces a subadditive dominantto every transitive map.

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


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) ≤ maxρa(α, β), ρa(β, γ),(c) ρa(α, β) ≤ maxρa(α, γ), ρa(β, γ),(d) ρa(α, β) ≥ a(α, β).

Proof. The proof of (a),(b),(c) is quite similar to the corresponding partof the proof of Lemma 4.2. This comes of course from the fact that thedefinition of ρ and ρa are closely related. The occurrence of the factora(min(Cβ \ α), β) complicates a bit the proof that ρa is subadditive, andthe fact that a is transitive is quite helpful in getting rid of the additionaldifficulty. The details are left to the interested reader. Given α < β, forevery 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 ofρa. Applying the transitivity of a to this path of inequalities we get theconclusion (d). a4.27 Lemma. ρa(α, β) ≥ ρa(α+1, β) whenever 0 < α < β and α is a limitordinal.

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 nopoint of Cξ is a limit ordinal. It follows that if 0 < α < β and α is alimit ordinal, then the minimal walk β → α must pass through α + 1 andtherefore ρa(α, β) ≥ ρa(α + 1, β). a

Let us now give an application of ρa to a classical phenomenon of occur-rence of gaps in the quotient algebra P(ω)/fin.

4.28 Definition. A Hausdorff gap in P(ω)/fin is a pair of sequencesAα (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 gapis 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 andthat the corresponding sequence of domains Aα (α < ω1) is a realizationof ω1 inside the quotient P(ω)/fin in the sense that Aα ⊆∗ Aβ wheneverα < β.


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

Aα = Aα ∪ (ω \Bα) (α < ω1) and Bα = ω \Bα (α < ω1)

has the following properties:

(a) Aα ⊆∗ Aβ whenever α < β,

(b) Bα =∗ Bβ ∩ Aα whenever α < β,

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

aFrom now on we fix a strictly ⊆∗-increasing chain Aα (α < ω1) of infinite

subsets of ω and let a : [ω1]2 −→ ω be defined by

a(α, β) = minn : Aα \ n ⊆ Aβ.

Let ρa : [ω1]2 −→ ω be the corresponding subadditive dominant of a definedabove. For α < ω1, set

Dα = Aα+1 \Aα.

4.30 Lemma. The sets Dα\ρa(α, γ) and Dβ\ρa(β, γ) are disjoint whenever0 < α < β < γ and α and β are limit ordinals.

Proof. This follows immediately from Lemmas 4.26 and 4.27. aWe are in a position to define a partial mapping m : [ω1]2 −→ 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 allbut finitely many limit ordinals α < minβ, 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). a4.32 Lemma. m(α, γ) 6= m(β, γ) whenever α 6= β < γ and α, β limit.

Proof. This follows from Lemma 4.30. aFor β < ω1, set

Bβ = m(α, β) : α < β and α limit.

4.33 Lemma. Bβ =∗ Bγ ∩Aβ whenever β < γ.


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 defineg(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 followsthat g is a finite-to-one map, a contradiction. a

4.36 Theorem. For every strictly ⊂∗-increasing chain Aα (α < ω1) ofsubsets 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.

a

4.37 Remark. For a given countable ordinal α let hα : Aα −→ 2 be suchthat h−1

α (1) = Bα. Then by Lemma 4.33, the corresponding sequencehα (α < ω1) of partial functions is coherent in the sense that for everypair α < β < ω1, the set Dh(α, β) = n ∈ Aα ∩ Aβ : hα(n) 6= hβ(n) isfinite. By Lemma 4.35, the sequence is nontrivial in the sense that there isno g : ω −→ 2 such that hα =∗ g ¹ Aα for all α < ω1. Hausdorff’s originalsequence, when reformulated in this way, has the property that the mapsdh(·, β) : β −→ ω defined by

dh(α, β) = |Dh(α, β)|

are finite-to-one mappings. A general principle, the P-ideal dichotomy, (see4.45 below) asserts that every coherent and nontrivial sequence must containa subsequence with Hausdorff’s property. We shall now embark on a generalconstruction of a very similar sequence of coherent partial functions domainsincluded in ω1 rather than in ω.

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

ρ2(α, β) = ρ2(α, min(Cβ \ α)) + 1,

with the convention that ρ2(γ, γ) = 0 for all γ.


This is an interesting mapping which is particularly useful on higher car-dinalities and especially in situations where the more informative mappingsρ0, ρ1 and ρ lack their usual coherence properties. Later on we shall devotea whole section to ρ2 but here we need it only to succinctly express thefollowing mapping.

4.39 Definition. The last step function of the minimal walk is the two-place map ρ3 : [ω1]2 −→ 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 walkβ → 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 ∈ Γsuch that ρ3(ξ, α) = ρ3(ξ, β). Shrinking Γ we may assume that for somefixed α ∈ 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:

ρ0(ξ, α) = ρ0(α, α)aρ0(ξ, α), (I.46)

ρ0(ξ, β) = ρ0(α, β)aρ0(ξ, α). (I.47)

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

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

(d) If α is a limit ordinal > 0 and if ξ occupies the nth place in theincreasing enumeration of Cα (that starts with min(Cα) on its 0thplace), 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 functionapplied to the context of ρ3: (ρ3)α(ξ) = ρ3(ξ, α).


4.41 Definition. Let Λ denote the set of all countable limit ordinals andfor 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 < ωsuch 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 + nthere is only one step from α to λ+n. So choosing n ≥ n0, ρ1(α, β) we willensure that the last step of β → λ + n comes with the maximal weight, i.e.ρ3(λ + n, β) = 1. a4.43 Lemma. For all β < ω1, n < ω, the set λ ∈ Λ : λ + n < β andρ3(λ + n, β) = 1 is finite.

Proof. Given an infinite subset Γ of (Λ+n)∩β we need to find a λ+n ∈ Γsuch 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 ∈ Γthen the last step of β → λ + n would have to be of weight> n + 2 which isimpossible by our assumption (d) about Cα (α < ω1). a

The meaning of these properties of ρ3 perhaps is easier to comprehendif we reformulate them in a way that resembles the original formulation ofthe 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.

aIn particular, there is is no uncountable Γ ⊆ ω1 such that Γ∩β ⊆∗ Bβ for

all β < ω1. On the other hand, the P-ideal18 I generated by Bβ (β < ω1) islarge as it contains all intervals of the form [λ, λ+ω). The following generaldichotomy about P-ideals shows that here indeed we have quite a canonicalexample 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, or18Recall 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. Aset 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 ofthe Proper Forcing Axiom and moreover that it does not contradict the Con-tinuum Hypothesis (see [89]). This is an interesting dichotomy which will beused in this article for testing various notions of coherence as we encounterthem. For example, let us consider the following notion of coherence, al-ready encountered above at several places, and see how it is influenced bythe P-ideal dichotomy.

5. Conditional weakly null sequences

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

5.1 Example. A weakly null sequence with no unconditional basic subse-quence. We describe a weakly null normalized sequence (xξ)ξ<ω1 of membersof 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 suchthat 1 ∈ M and

∑(m

n)1/2 : m,n ∈ M, m < n ≤ ε. (I.48)

Fix also a natural bijection p.q : HF → M, where HF denotes the family ofall hereditarily finite sets, so that in particular p∅q = 1. The ρ-number oda finite set F ⊆ ω1 is the integer

ρ(F ) = maxρ(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 every x in its closed linear span spanxn : n ∈ ω canbe uniquely represented as a sum

Pn∈ω anxn. Note that this is equivalent to saying

that the norms of projection operators Pk : spanxn : n ∈ ω → spanxn : n < k areuniformly bounded. When the norms of projections PM : spanxn : n ∈ ω → spanxn :n ∈ M over all subsets M of ω are uniformly bounded, the sequence (xn) is said tobe 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 thesumming basis e0, e0+e1, e0+e1+e2, ... of c0 or the summing basis e0+e1+e2+...+en+...,e1 + e2 + ... + en + ..., e2 + ... + en + ..., ... in c.


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

σ : [[ω1]<ω]<ω −→ M

defined by letting σ(∅) = p∅q and

σ(F0, F1, ..., Fn) = p((πE”F0, πE”F1, ..., πE”Fn), ρ(n⋃

i=0

Fi))q,

where E = (⋃n

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<n of finite subsets of ω1 is special if

1. Ei < Ej20 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 (eξ)ξ<ω1 be the basis of c00(ω1) and let (eξ)∗ξ<ω1be the

corresponding sequence of biorthogonal functionals. To a finite set E ⊆ ω1,we associate the following vector and functional on c00(ω1),

xE =1

|E|1/2

∑

α∈E

eα and φE =1

|E|1/2

∑

α∈E

e∗α. (I.49)

Given a special sequence (Ei)i<n of finite subsets of ω1, the correspondingspecial functional is defined by

∑i<n φEi . Let F be the family of all special

functionals of c00(ω1). This family induces the following norm on c00(ω1),

‖ x ‖= sup〈f, x〉 : f ∈ F. (I.50)

The Banach space X is the completion of the normed space (c00(ω1), ‖ . ‖).Note that ‖ eα ‖= 1 and that (eα)α<ω1 is a weakly null sequence in X. Tosee this note that given a finite set G with |G| ∈ M and a special sequence(Ei)i<n, if φ =

∑i<n φEi is the corresponding special functional, then the

inner product between φ and the average |G|−1∑

α∈G eα = |G|−1/2xG isnot bigger than |G|−1/2(1 +

∑|G|6=m∈M min((m/|G|)1/2, (|G|/m)1/2) and

therefore not bigger than |G|−1/2(1 + ε) by our assumption I.48. It followsthat the averages ‖ |G|−1

∑α∈G eα ‖ tend to 0 as |G| ∈ M goes to infinity,

so (eα)α<ω1 must be weakly null in X. To show that no infinite subsequenceof (eα)α<ω1 is unconditional it suffices to show that for every set G ⊆ ω1 of

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


order type ω there is seminormalized block-subsequence21 of (eα)α∈G whichrepresents 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 < ωwe let vi = xEi

as defined in I.49, then for every n < ω and every sequence(ai)i≤n ⊆ [−1, +1] of scalars.

max0≤k≤n

|k∑

i=0

ai |≤‖n∑

i=0

aivi ‖≤ (3 + ε) max0≤k≤n

|k∑

i=0

ai | . (I.51)

To see the first inequality, note that according to I.49 the absolute valueof the inner product between the spacial functional φ =

∑ni=0 φEi and the

vector x =∑n

i=0 aivi is equal to | ∑ni=0 ai | giving us the first inequality. To

see the second inequality, consider a special functional ψ =∑m

i=0 φFiwhere

(Fi)i≤m is some other special sequence of finite sets. In order to estimatethe inner product 〈ψ, x〉 we need to know how the two special sequencesinteract and this requires properties of the function ρ that controls them.To see this, let l ≤ min(m,n) be maximal with the property that |El| = |Fl|,and let E = (

⋃i<l Ei)ρ and F = (

⋃i<l Fi)ρ. By the definition of a special

sequence and |El| = |Fl|, we know that:

1. ρ(⋃

i<l Ei) = ρ(⋃

i<l Fi),

2. πE”Ei = πE”Fi for i < l.

It follows that E and F have the same cardinality and the same ρ-numbersand therefore by the properties 4.1 of ρ, their intersection H = E∩F is theirinitial segment. Hence, πE and πE agree on H. Let k be the maximal integersuch that Ek ⊆ H. Then Ei = Fi for all i ≤ k, and using the properties 4.1of ρ again, Ei ∩ Fj = ∅ for k + 1 < i, j < l. Combining all this we see that|〈ψ, x〉| is bounded by

|k∑

i=0

ai+ak+1〈φFk+1 , xEk+1〉+al〈φFl, xEl

〉|+∑

(m

n)1/2 : m,n ∈ M, m < n.

Referring to I.48 we get that |〈ψ, x〉| is bounded by the right hand side ofthe second inequality of I.51.

5.2 Remark. The first example of a weakly null sequence (of length ω) withno unconditional basic subsequence was constructed by Maurey and Rosen-thal [51]. Sixteen years latter Gowers and Maurey [27] have solved the basicunconditional sequence problem by constructing an infinite-dimensional sep-arable reflexive Banach space with no infinite unconditional basic sequence.The following result of [3] relates to the Example 5.1 the same way theresult of [27] relates to the example of [51].

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


5.3 Example. A reflexive Banach space Xω1 with a Schauder basis of lengthω1 containing no unconditional basic sequence. The space Xω1 is the com-pletion of c00(ω1) under the norm given by the formula I.50, where nowF = Fω1 is chosen as the minimal set of finitely supported functionals onc00(ω1) such that:

1. F ⊇ e∗α : α < ω1 and F is symmetric and closed under restrictionon intervals of ω1.

2. For every (φi)i<n2k⊆ F such that supp(φi) < supp(φj) for i < j <

n2k, the functional m−12k

∑i<n2k

φi belongs to F .

3. For every special sequence (φi)i<n2k+1 ⊆ F the corresponding specialfunctional m−1

2k+1

∑i<n2k+1

φi belongs to F .

4. F is closed under rational convex combinations.

The increasing sequences (mk) and (nk) of positive integers are defined by:mk = 24k

and n0 = 4 and nk+1 = (4nk)log2m3k+1 . The notion of a special

sequence of functionals is defined analogously to that in the Example 5.1as follows. First of all we say that φ ∈ F is of type 0 if φ = ±e∗α for someα, of type I if it obtained from a simpler ones as the sum m−1

k

∑i<nk

φi

restricted to some interval and we call the integer mk, the weight of φ, anddenote it w(φ). Note that a given functional of type I can have more thanone representation and therefore more than one weight. We say that φ isof type II if it is a rational convex combination of functionals of type 0and type I. The definition of a special sequence of functionals will be againbased on ρ and an injection p.q : HF → 2k : k odd , with the propertythat for every finite sequence (φi, wi, pi)i<j where wi, pi ∈ ω and where φi’sare finite partial maps from ω into Q such that p(φi, wi, pi)i<jq is biggerthan the maximum of support of any of the φi’s, the square of any integerof the form wi or pi, and the ratio of the square of any nonzero value of a φi.A sequence (φi)i<n2k+1 of members of F is a special sequence of functionals,if

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

even and satisfying m2k0 > n22k+1,

3. for 0 < j < n2k+1 there is an increasing sequence (pi)i<j of integerssuch that pi ≥ ρ(

⋃i′<i supp(φi′)) and w(φj) = mp(πE”φi,w(φi),pi)i<jq,

where E = (⋃

i<j supp(φi))ρ.

(The collapse ψi = πE”φi is defined by the formula ψ(πE(α)) = φ(α).) Thecondition 1. from the definition of F provides that (eα)α<ω1 is a bimonotoneSchauder basis of Xω1 , the condition 3. is responsible for the conditional


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 theunconditional structure given but the closure property 2. of F . This requiresa rather deep tree-analysis of the functionals of F that gives the requirednorm 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 themost interesting properties of Xω1 . For example, the space Xω1 is reflexive,has no infinite unconditional basic sequence, and is not isomorphic to anyproper subspace or a nontrivial quotient. The basis (eα)α<ω1 allows us totalk about diagonal operators of Xω1 . It turns out that all bounded diagonaloperators D of Xω1 are given by a sequence (λα)α<ω1 of eigenvalues thathave the property that λα = λβ whenever α ≤ β < α + ω. Moreoverthere is only countably many different eigenvalues among (λα)α<ω1 . Recall,than an operator between two Banach spaces is strictly singular if it is notan isomorphism when restricted to any infinite-dimensional subspace of itsdomain. It turns out that every bounded operator T on Xω1 has the formD+S, where D is a diagonal-step operator with countably many eigenvaluesand where X is a strictly singular operator. It follows that in particularthat every bounded linear operator on Xω1 is a multiple of the identityplus the operator with separable range22. It turns out that we can actuallyidentify the space D(Xω1) of bounded diagonal operators as the dual ofthe James space over a particular reflexive Banach space T0 and ω1 as thecorresponding ordered set (see [3] for the definition of this functor). Onecan similarly characterize the operator space of arbitrary closed subspace ofXω1 as well obtaining thus some interesting new phenomena about operatorspaces. For example one can find a closed subspace Y of Xω1 such that everybounded operator T : Y → Y is a strictly singular perturbation of a multipleof identity while operator space L(Y,Xω1) is quite rich in the sense that

L(Y,Xω1) = J∗T0⊕ 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, withan appropriate adjustment on ρ, are ’asymptotic versions’ of each other.This is the new phenomenon in the separable theory not revealed by theprevious methods. It essential says that any finite-dimensional subspaceof one XI can be moved into any other XJ by an operator T such that‖ T ‖‖ T−1 ‖≤ 4 + ε.

5.4 Question. What is the minimal cardinal θu with property that everyweakly null sequence of length θu contains an infinite unconditional subse-quence?

22A considerably easier example of a space with this property will be given in somedetail in 7.16 below.


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

6. An ultrafilter from a coherent sequence

6.1 Definition. A mapping a : [ω1]2 −→ 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), orin other words, aα =∗ aβ ¹ α23. We say that a is nontrivial if there is noh : ω1 −→ ω such that h ¹ α =∗ aα for all α < ω1.

Note that the existence of a coherent and nontrivial a : [ω1]2 −→ 2 (suchas, for example, the function ρ3 defined above) is something that corre-sponds 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 treesince

T (a) = t : α −→ ω : α < ω1 and t =∗ aαis such an Aronszajn tree whenever a : [ω1]2 −→ ω is coherent and non-trivial24. In fact, we shall call an arbitrary Aronszajn tree T coherent if Tis isomorphic to T (a) for some coherent and nontrivial a : [ω1]2 −→ ω. Incase the range of the map a is actually smaller than ω, e.g. equal to someinteger k, then it is natural to let T (a) be the collection of all t : α −→ ksuch that α < ω1 and t =∗ aα. This way, we have coherent binary, ternary,etc. Aronszajn trees rather than only ω-ary coherent Aronszajn trees.

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

Note that ρ3 is an example of a nowhere dense coherent map for thesimple 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 isindeed a prototype of a nowhere dense and coherent map a : [ω1]2 −→ ω.

6.3 Proposition. Under the P-ideal dichotomy, for every nowhere denseand coherent map a : [ω1]2 −→ ω the domain ω1 can be decomposed intocountably many sets orthogonal to a. a

23A mapping a : [ω1]2 −→ ω is naturally identified with a sequence aα (α < ω1), whereaα : α −→ ω is defined by aα(ξ) = a(ξ, α).

24Similarities between the notion of a Hausdorff gap and the notion of an Aronszajntree have been further explained recently in the two papers of Talayco ([72], [73]), whereit is shown that they naturally correspond to first cohomology groups over a pair of verysimilar spaces.

6. An ultrafilter from a coherent sequence 45

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

∆a(α, β) = minξ < α : a(ξ, α) 6= a(ξ, β)

with the convention that ∆a(α, β) = α whenever a(ξ, α) = a(ξ, β) for allξ < 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 un-countable subset of T (a) contains an uncountable antichain. Then for everypair Σ,Ω of uncountable subsets of ω1 there exists an uncountable subset Γof 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 ξand 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 finiteset F such that Fξ = F for all ξ ∈ Γ. Let η0 = max(F ) + 1. By theassumption 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 uin 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). a6.6 Notation. For a : [ω1]2 −→ 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 coherenta : [ω1]2 −→ ω for which T (a) contains no Souslin subtrees. It turns outthat under some very mild assumption, U(a) is in fact a uniform ultrafilteron ω1.

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


Proof. Let A be a given subset of ω1 and let P be the collection of all finitesubsets p of ω1 such that ∆a(p) ⊆ A. If P is a ccc poset then an applicationof MAω1 would give us an uncountable Γ ⊆ ω1 such that ∆a(Γ) ⊆ A, show-ing thus that A belongs to U(a). So let us consider the alternative when P isnot a ccc poset. Let X be an uncountable subset of P consisting of pairwiseincompatible conditions of P. Going to an uncountable ∆-subsystem of Xand noticing that the root does not contribute to the incompatibility, onesees that without loss of generality, we may assume that the members of Xare in fact pairwise disjoint. Thus for each limit ordinal ξ < ω1 we can fixpξ 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 setF ⊆ ω1, an integer n and t0, . . . , tn−1 in T (a) of height η0 = max(F ) + 1such 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 Γwe may assume that aξ (ξ ∈ Γ) is an antichain of T (a), and moreover thatfor 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 con-clude 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 = ∅, showingthus that the complement of A belongs to U(a). a6.8 Remark. One may find Theorem 6.7 a bit surprising in view of thefact that it gives us an ultrafilter U(a) on ω1 that is Σ1-definable over thestructure (Hω2 ,∈). It is well-known that there is no ultrafilter on ω that isΣ1-definable over the structure (Hω1 ,∈).

It turns out that the transformation a 7−→ U(a) captures some of theessential properties of the corresponding and more obvious transformationa 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 thatthere is a strictly increasing map f : S −→ T . Let S < T whenever S ≤ Tand T S and let S ≡ T whenever S ≤ T and T ≤ S. In general,


the equivalence relation ≡ on trees is very far from the finer relation ∼=,the isomorphism relation. However, the following fact shows that in therealm of trees T (a), these two relations may coincide and moreover, thatthe mapping T (a) 7−→ U(a) reduces ≡ and ∼= to the equality relation amongultrafilters on ω1 (see [90]).

6.10 Theorem. Assuming MAω1 , for every pair of coherent and nontrivialmappings a : [ω1]2 −→ ω and b : [ω1]2 −→ ω, the trees T (a) and T (b) areisomorphic 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), refer-ring the reader to [90] for the remaining part of the argument.

Choose a pair of strictly increasing mappings f : T (a) −→ T (b) andg : T (b) −→ T (a). Replacing f by the mapping t 7−→ f(t) ¹ dom(t) andg by the mapping t 7−→ g(t) ¹ dom(t), we may assume that f and g arealso 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 thataδ (δ ∈ Γ) is an antichain of T (a), bδ (δ ∈ Γ) is an antichain of T (b), andthe 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 ofLemma 6.5 we conclude that ∆a(Ω) (Ω ⊆ Γ, Ω uncountable) generates thefilter U(a) and similarly that ∆b(Ω) (Ω ⊆ Γ, Ω uncountable) generates theultrafilter U(b). It follows that U(a) = U(b).

Suppose now that U(a) = U(b). By symmetry, it suffices to show thatT (a) ≤ T (b). Note again that the argument in the proof of Lemma 6.5 showsthat there is a stationary set Γ ⊆ Λ such that ∆a and ∆b coincide on [Γ]2.Let P be the poset of all finite partial strictly increasing level-preserving


mappings p from T (a) into T (b) which (uniquely) extend to strictly increas-ing 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 ⊆ Γand E ⊆ ω1 such that Eδ = E for all δ ∈ Σ. Shrinking Σ, we may assumethat the mapping δ 7−→ pδ ¹ δ is constant on Σ. Let η be the minimalordinal strictly above max(E) and the height of every node of dom(pδ) ¹ δfor some (all) δ ∈ Σ. For δ ∈ Σ, let pδ be the extension of pδ on thedownward-closure of its domain in T (a). Find a stationary Ω ⊆ Σ such thateach of the projections δ 7−→ pδ ¹ η, δ 7−→ aδ ¹ η and δ 7−→ bδ ¹ η is constanton Ω. Find γ < δ in Ω such that aγ 6= aδ ¹ γ, bγ 6= bδ ¹ γ and pγ and pδ areisomorphic via the isomorphism that fixes their common restriction to theηth 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 eachother in the isomorphism between pγ and pδ, then either they belong to thecommon part of pγ and pδ in which case (I.60) follows by the assumptionthat 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 copy y ∈ 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. Thisfinishes the proof. a

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

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

a(1)(α, β) = a(α + 1, β),

where β = minλ ∈ Λ : λ ≥ β. The n-fold iteration of the shift operationis defined recursively by the formula a(n+1) = (a(n))(1).


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 underthe quasi-ordering ≤ (see [90]).

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

Proof. By the assumption a ⊥ Λ, the subtree S of T (a) consisting of allt’s that take the value 0 at every limit ordinal in their domains is an un-countable initial part of T (a). Note that every node of T (a) is equal tothe restriction of some finite change of a node from S. For a node t atsome limit level λ of S, let t(1) : λ −→ ω be determined by the formulat(1)(α) = t(α + 1). Let S(1) = t(1) : t ∈ S ¹ λ. Every node of T (a(1)) isequal to the restriction of some finite change of a node of S(1). Note alsothat t 7−→ t(1) is a one-to-one map on S ¹ Λ and that its inverse naturallyextends to a strictly increasing map from T (a(1)) into T . This shows thatT (a(1)) ≤ T (a).

It remains to establish T (a) T (a(1)). Note that if there is a strictlyincreasing 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 limitordinal ξ, set

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

Then Fξ is a finite set of ordinals < ξ, so by the Pressing Down Lemmathere is a stationary set Γ ⊆ Λ and F ⊆ ω1 such that Fξ = F for allξ ∈ Γ. Let α0 = max(F ) + 1. Shrinking Γ if necessary, assume that forsome 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)(= minα <ξ : aξ(α) 6= aη(α)). Then by (I.62) and (I.63), δ is not a limit ordinal and

δ − 1 = ∆(a(1)ξ , a(1)

η ) = ∆(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 finishesthe proof. a6.14 Corollary. If a is nontrivial, coherent and orthogonal to Λ+n for alln < 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 everyof its finite shifts a(m). a


6.15 Corollary. T (ρ(n)3 ) > T (ρ(m)

3 ) whenever n < m < ω. aSomewhat 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 istotally ordered under ≤.

Proof. Given two nontrivial coherent sequences a and b, we need to showthat either T (a) ≤ T (b) or T (b) ≤ T (a). It turns out that these twoalternatives 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 fromT (a) into T (b) that can be extended to strictly increasing level-preservingmaps from Dp into T (b), where Dp is the downward closure of the domainDp of p. Let Rp denote the range of p and Rp its downward closure in T (b).If P is a ccc poset, then applying MAω1 to P and the natural sequenceof dense open subsets of P would give us the desired strictly increasinglevel-preserving map from T (a) into T (b). So let pξ (ξ < ω1) be a givenone-to-one sequence of members of P. We may assume that the pξ’s forma ∆-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 withheight ≥ 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 gis constantly equal to η0 on Σ and moreover that the projection maps ofDξ ∪ aξ, ah(ξ) onto the η0th level of T (a) and Rξ ∪ bξ, bh(ξ) onto theη0th level of T (b) are constant as long as ξ belongs to Σ. Moreover wemay assume for ξ 6= η, every node of Dξ ∪ aξ, ah(ξ) is incomparable toevery 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 alsoassume that every node of Dξ, for ξ ∈ Σ, has height ≥ ξ. So pick ξ 6= η inΣ. 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)

Since h(ξ), h(η) belongs to [Ω]2, we conclude that for all s ∈ Dξ andt ∈ Dη, s and t split before pξ(s) and pη(t). So pξ ∪ pη extends to a strictlyincreasing 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

Case 2: For every uncountable Ω ⊆ ω1 there exists α, β in [Ω]2 suchthat ∆a(α, β) > ∆b(α, β). Let Q be the poset of all finite subsets q of ω1

such that ∆a dominates ∆b on [q]2. IfQ is a ccc poset, then an application ofMAω1 would give us an uncountable subset Ω of ω1 such that ∆a dominates∆b on [Ω]2, so by Case 1 we would conclude that T (b) ≤ T (a). To show thatQ is a ccc poset, let qξ (ξ < ω1) be a given one-to-one sequence of membersof Q. We may assume the qξ’s form a ∆-system. Note that the root doesnot contribute to the incompatibility of two conditions of the ∆-system andso removing it, we assume that all qξ’s are in fact pairwise disjoint. For alimit 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 sta-tionary 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 theproof that Q is a ccc poset and therefore the proof of Theorem 6.16. a6.17 Remark. While under MAω1 , the class of coherent Aronszajn trees istotally ordered by ≤, Corollary 6.15 gives us that this chain of trees is notwell-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]). Itturns 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 relatedresults 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 smallerordinal 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 followingsimple fact about the trace lies at the heart of all known definitions ofsquare-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.


Proof. It suffices to show that the union of traces contains every countablelimit 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 setsof the form Cβi ∩ δ for i < k. By the choice of δ there is α ∈ Γ ∩ δ aboveγ. Then the minimal walk from β to α starts as β0 > β1 > · · · > βk, so inparticular 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 thatby Lemma 1.14,

∆(α, β) = minξ ≤ 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 nat-ural isomorphism between Tr(ξ, α) and Tr(ξ, β). We shall define [αβ] bycomparing 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).

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

7.4 Lemma. For every uncountable subset Γ of ω1 the set of all ordinals ofthe form [αβ] for some α < β in Γ contains a closed and unbounded subsetof 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 provedonce we prove that every countable limit ordinal δ > 0 with the followingtwo 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)


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

Fix such a δ and choose β ∈ Γ \ δ such that (ρ0)β ¹ δ ∈ S and considerthe 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. Sincethe restriction t = (ρ0)β ¹ γ belongs to S, by (I.72) we can find one of itsend-extensions s in S which is incomparable with (ρ0)β . It follows that forα ∈ Γs, the ordinal ∆0(α, β) has the fixed value

ξ = minξ < |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 → ξand β → α for every α ∈ Γs ∩ δ. Hence if we choose α ∈ Γs ∩ δ abovemin(Cδ \ ξ) (which we can by (I.71)), we get that the walks β → ξ andβ → α never meet after δ. In other words for any such α, the ordinal δis the minimum of Tr(α, β) ∩ Tr(ξ, β), or equivalently δ is the minimum ofTr(ξ, β) \ α. a

It should be clear that the above argument can easily be adjusted togive us the following slightly more general fact about the square-bracketoperation.

7.5 Lemma. For every uncountable family A of pairwise disjoint finitesubsets 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 aclosed and unbounded subset of ω1. a

It turns out that the square-bracket operation can be used in construc-tions of various mathematical objects of complex behaviour where all knownprevious constructions needed the Continuum Hypothesis or stronger enu-meration principles. The usefulness of [··] in these constructions is basedon the fact that [··] reduces the quantification over uncountable subsets ofω1 to the quantification over closed unbounded subsets of ω1. For examplecomposing [··] with a unary operation ∗ : ω1 −→ ω1 which takes each of thevalues stationary many times one gets the following fact about the mappingc(α, β) = [αβ]∗.

7.6 Theorem. There is a mapping c : [ω1]2 −→ ω1 which takes all thevalues from ω1 on any square [Γ]2 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) orx(0), x(1), . . . , x(n− 1), depending on the context, for the enumeration of x according tothe natural ordering on the ordinals.


This result gives a definitive limitation on any form of a Ramsey Theoremfor the uncountable which tries to say something about the behaviors ofrestrictions of colorings on an uncountable square. Recall that in Section3.1 we have seen that there are mappings of this sort with complex behavioron 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 thesquare bracket operation has a simple behavior on Γ0 × Γ1 for every pairΓ0 and Γ1 of sufficiently thin uncountable subsets of ω1. This can be madeprecise as follows.

7.7 Lemma. For every uncountable family A of pairwise disjoint subsetsof 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 thatfor all a < b 26 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 exista < b in B such that [a(p)b(q)] = δ for all p, q ∈ Ii and all i = 1, . . . , kcontains a closed and unbounded subset of ω1.

Proof. Identifying an ordinal α with (ρ0)α, we can think of A as a familyof 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 projectionof aδ on the δth level of T (ρ0). Let h(δ) < δ be such that the projection ofcδ 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 andthat the γ0th projection takes the constant value on cδ (δ ∈ Γ). Fix δ < εin Γ and consider α ∈ aδ and β ∈ aε that project to different elements ofthe γ0th level of T (ρ0). Then ∆(α, β) is independent of α and β or δ andε 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 ofΓ, we may assume that its size depends only on the position of β in someaε and that the minimal point of Tr(∆0(α, β), β) \ α is the same for all αin 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 seethe second part of the conclusion, we shrink Γ even further to assure thatthe union of cδ (δ ∈ Γ) is an antichain of T (ρ0). Now note that for twotypical δ < ε in Γ, [αβ] = [α′β′] for all α, α′ ∈ aδ and β, β′ ∈ aε with theproperty that (ρ0)α ¹ γ0 = (ρ0)α′ ¹ γ0 = (ρ0)β ¹ γ0 = (ρ0)β′ ¹ γ0. So therest 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 ∈ aand β ∈ b.

27By ∆, we denote the diagonal of k × k.


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 ωrather than ω1. Then the sequence h∗b : n×n −→ ω (b ∈ B) of correspondingcompositions will have the same constant value h : n × n −→ ω modulo,of course, shrinking B to some uncountable subset. Noting that the proofof Lemma 7.7 applied to a typical A will actually result in the partitionIi = i (i = 1, . . . , n), we conclude that if α and β are coming from twodifferent positions of some aδ and aε in B, then [αβ]∗ will depend on thepositions 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 everyuncountable family A of pairwise disjoint n-tuples of countable ordinals.This property of the projection of the square-bracket operation has an in-teresting 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 thebeginning of this chapter can be used to actually define a unary operation∗ : ω1 −→ ω1 which takes each of the ordinals from ω1 stationarily manytimes. 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 actuallyrigid.

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

Proof. Let h be a given automorphism of (ω1, [··], ∗). If the set Γ of fixedpoints of h is uncountable, h must be the identity map. To see this, considera ξ < ω1. By the property of the map c(α, β) = [αβ]∗ stated in Theorem7.6 there exist γ < δ in Γ such that [γδ]∗ = ξ. Applying h to this equationwe 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 assumethat h(δ) > δ for all δ ∈ ∆. Consider a ξ < ω1 and let Sξ be the set of allα < ω1 such that α∗ = ξ. By our choice of ∗ the set Sξ is stationary. ByLemma 7.5 applied to the family A = δ, h(δ) : δ ∈ ∆ we can find γ < δin ∆ 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 anarbitrary countable ordinal, this shows that h is the identity map. a


We give now an application of this rigidity result to a problem in ModelTheory about the quantifier Qx =’there exist uncountably many x’ and itshigher dimensional analogues Qnx1 · · ·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 quantifierQ2.

7.11 Example. The sentence φ will talk about one unary relation N , onebinary relation < and two binary functional symbols C and E. It is theconjunction 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),

(φ7) ∀x∀nN(n) → ¬Q2uv[∃u′ < u∃v′ < v(u′ 6= v′ ∧ E(u′, u) = E(v′, v) =n) ∧ ∀u′ < u ∀v′ < v(E(u′, u) = E(v′, v) = n → (C(u′, v′) 6= x ∨C(u, v) 6= x))].

The model of φ that we have in mind is the model (ω1, ω, <, c, e) wherec(α, β) = [αβ]∗ and e : [ω1]2 −→ ω is any mapping such that e(α, γ) 6=e(β, γ) whenever α < β < γ (e.g. we can take e = ρ1 or e = ρ). Thesentence φ7 is simply saying that for every ξ < ω1 and every uncountablefamily A of pairwise disjoint unordered pairs of countable ordinals thereexist 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∗ = ξis a stationary subset of ω1. These are the properties of [··] and ∗ whichwe 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 toprovide a negative answer to the basis problem for uncountable graphs byconstructing a large family of pairwise orthogonal uncountable graphs.


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

α, β : [αβ] ∈ Γ.

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

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

7.14 Definition. Let H be the collection of all maps h : 2n −→ ω1 wheren is a positive integer denoted by n(h). Choose a mapping π : ω1 −→ Hwhich takes each value from H stationarily many times. Choose also a one-to-one sequence rα (α < ω1) of elements of the Cantor set 2ω. Note thatboth 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 corre-sponds to the following property of the projection [[αβ]]:

7.15 Lemma. For every uncountable family A of pairwise disjoint finitesubsets 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 fori = 1, . . . , n. a

This projection of [··] leads to an interesting example of a Banach spacewith ’few’ operators, which we will now describe.

7.16 Example. A nonseparable reflexive Banach space E with the propertythat 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 operatorwith separable range.

Let I = 3×[ω1]<ω and let us identify the index-set I with ω1, i.e. pretendthat [[··]] takes its values in I rather than ω1. Let [[··]]0 and [[··]]1 be the twoprojections of [[··]].


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

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

maxαi, βi < minαj , 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 subsetsand they are 1-orthogonal to each other in the sense thatG ∩ H contains no doubleton.

(I.77)

G and H are both 2-orthogonal to the family of the unionsof members of K.

(I.78)

If K and L are two distinct members of K, then there are nomore than 5 ordinals α such that α, β ∈ K and α, γ ∈ Lfor some β 6= γ.

(I.79)

For every sequence αξ, βξ (ξ < ω1) of pairwise disjointpairs of countable ordinals there exist arbitrarily large finitesets Γ,∆ ⊆ ω1 such that αξ : ξ ∈ Γ ∈ G, βξ : ξ ∈ Γ ∈ Hand αξ, βξ : ξ ∈ ∆ ∈ K.

(I.80)

For a function x from ω1 into R, set

‖x‖H,2 = sup(∑

α∈H

x(α)2)12 : H ∈ H,

‖x‖K,2 = sup(∑

α,β∈K

(x(α)− x(β))2)12 : K ∈ K.

Let ‖ · ‖ = max‖ · ‖∞, ‖ · ‖H,2, ‖ · ‖K,2 and define E2 = x : ‖x‖ < ∞. Let1α be the characteristic function of α. Finally, let E2 be the closure ofthe linear span of 1α : α ∈ ω1 inside (E2, ‖ · ‖). The following facts aboutthe norm ‖ · ‖ are easy to establish using the properties of the families G,Hand K listed above

If x is supported by some G ∈ G, then ‖x‖ ≤ 2 · ‖x‖∞. (I.81)

If x is supported by⋃

K for some K in K, then ‖x‖ ≤ 10 · ‖x‖∞. (I.82)


The role of the seminorm ‖·‖H,2 is to ensure that every bounded operator T :E2 −→ E2 can be expressed as D+S, where D is a diagonal operator relativeto the basis28 1α (α < ω1) and where S has separable range. Namely, ifthis were not so, we could find a sequence αξ, βξ (ξ < ω1) of pairwisedisjoint pairs from ω1 such that T (1αξ

)(βξ) 6= 0 for all ξ < ω1. Thinningthe 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.Then, by (I.81), 2 · ‖T‖ ≥ ‖T (χG)‖ ≥ (

∑β∈H T (χG)(β)2)

12 ≥ r · √n. Since

n is arbitrary, this contradicts the fact that T is bounded. The role of‖ · ‖K,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. Thisreduces to the fact that the sequence λξ (ξ < ω1) is eventually constant. Ifthis 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),

10 · ‖D‖ ≥ ‖D(χSK)‖≥ (

∑

ξ∈∆

(D(χSK)(αξ)−D(χSK)(βξ))2)12

= (∑

ξ∈∆

(λαξ− λβξ

)2)12

≥ r · √n.

Since n was arbitrary this contradicts the fact that D is a bounded op-erator.

Note that ‖x‖ ≤ 2‖x‖2 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 adense subset of E2. Therefore E2 is a weak compactly generated space. Forexample, W = x ∈ `2(ω1) : ‖x‖2 ≤ 1 is a weakly compact subset of E2

and its linear span is dense in E2. To get a reflexive example out of E2 oneuses an interpolation method of Davis, Figiel, Johnson and Pelczynski [11] asfollows. Let pn be the Minkowski functional of the set 2nW +2−nBall(E2).29

Let

E = x ∈ E2 : ‖x‖E = (∞∑

n=0

pn(x)2)12 < ∞.

28Indeed it can be shown that 1α (α < ω1) is a ’transfinite basis’ of E2 in the senseof [70]. So every vector x of E2 has a unique representation as Σα<ω1x(α)1α and theprojection operators Pβ : E2 → E2 ¹ β (β < ω1) are uniformly bounded.

29I.e. pn(x) = infλ > 0 : x ∈ λB, where B denotes this set.


By [11, Lemma 1], E is a reflexive Banach space and `2(ω1) ⊆ E ⊆ E2 arecontinuous inclusions. Note that pn(x) < r iff x = y + z for some y ∈ E2

and z ∈ `2(ω1) such that ‖y‖ < 2−nr and ‖z‖2 < 2nr. The following twofacts about E correspond to the facts (I.81) and (I.82) and they are used inthe proof that every bounded operator T : E −→ E has the form λI + S ina very similar way.

If x is the vector supported by some G ∈ G which can besplit into n blocks such that the k-th one has size 24k andx takes the value 2−2k on each term of the k-th block, then‖x‖E ≤ 2.

(I.83)

If αξ, βξ : ξ ∈ ∆ is a member of K so that ∆ can besplit into n blocks such that the k-th one ∆k has size 24k

and if x is the vector supported by⋃

K for ξ ∈ ∆k, x(αξ) =2−2k and x(βξ) = 2−2k, then ‖x‖E ≤ 12.

(I.84)

We leave the checking of (I.83) and (I.84) as well as the correspondingproof that E has ’few’ operators to the interested reader. a7.17 Remark. The above example is reproduced from Wark [94] who basedhis example on a previous construction due to Shelah and Steprans [68]. Thereader is referred to these sources for more information.

We only mention yet another interesting application of the square-bracketoperation, given recently by Erdos, Jackson and Mauldin [21]:

7.18 Example. For every positive integer n there exist collections H andX of hyperplanes and points of Rn, respectively, and a coloring P : H −→ ωsuch that:

any n hyperplanes of distinct colors meet in at most onepoint, and (I.85)

there is no coloring Q : X −→ ω such that for every H ∈ Hthere exist at most n − 1 points x in X ∩ H such thatQ(x) = P (H).

(I.86)

Finally we mention yet another projection of the square-bracket opera-tion.

7.19 Definition. Let H now be the collection of all maps h : 2n×2n −→ ω1

where n = n(h) < ω and let π be a mapping from ω1 onto H that takeseach 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)


7.20 Lemma. For every positive integer n, every uncountable subset Γ ofω1 and every symmetric n × n-matrix M of countable ordinals there is aone-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 × nmatrix on countable ordinals. Then for each α we can find tα in the αth levelof 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 bean upper bound of all sets of the form Cγi ∩ α where γ ∈ Fα, i < k < ωand 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 stationaryset ∆ ⊆ ω1, m < ω, s0, . . . , sn−1 ∈ 2m 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)rγ ¹ m = si whenever i < n and γ occupies the ith place insome Fα for α ∈ ∆.

(I.90)

Let h : 2m × 2m −→ ω1 be an arbitrary map subject to the requirementthat 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 basicproperty 7.4 of the square bracket operation we can find ν ∈ Σh, α ∈ ∆∩ νand β ∈ ∆ \ ν such that tα ¹ ξ = tβ ¹ ξ, Fα ⊆ ν and [αβ] = ν. We claimthat

[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 andδ ∈ Fβ . By the choice of ξ and the fact that σ ≥ ξ the walk from someδ ∈ 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 ∈ ∆∩ νand an uncountable set ∆1 ⊆ ∆ \ ν0 such that Fα0 ⊆ ν0, tα0 ¹ ξ = tβ ¹ ξand [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 allβ ∈ ∆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 . Goingback 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


8. A square-bracket operation on a tree

In this section we try to show that the basic idea of the square-bracketoperation on ω1 can perhaps be more easily grasped by working on anarbitrary special Aronszajn tree rather than T (ρ0). So let T = 〈T,<T 〉 bea fixed special Aronszajn tree and let a : T −→ ω be a fixed map witnessingthis, i.e. a mapping with the property that a−1(n) is an antichain of T forall n < ω. We shall assume that for every s, t ∈ T the greatest lower bounds ∧ t exists in T . For t ∈ T and n < ω, set

Fn(t) = s ≤T t : s = t or a(s) ≤ n.

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

[st]T = minv ∈ 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 ofthe form [st]T for some s, t ∈ X intersects a closed and unbounded set oflevels of T . a

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

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

KΓ = s, t ∈ [T ]2 : 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 isomor-phic uncountable subgraph whenever the symmetric difference between Γand ∆ is a stationary subset of ω1, i.e. whenever they represent differentmembers of the quotient algebra P(ω1)/NS. In particular, KΓ contains nosquare [X]2 of an uncountable set X ⊆ T whenever Γ contains no closedand 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 thereis a proper forcing notion introducing an uncountable set X ⊆ T such that[X]2 ⊆ KΓ.

Proof. The forcing notion P is defined to be the set of all pairs p = 〈Xp,Np〉where

Np is a finite ∈-chain of countable elementary submodelsof Hω2 containing all the relevant objects. (I.94)

Xp is a finite subset of T such that ht([st]T ) ∈ Γ for everypair s, t of elements of Xp.

(I.95)

for all s 6= t in Xp there exist N ∈ Np such that s ∈ N ifft /∈ N .

(I.96)

for every s 6= t in Xp, N ∈ Np, if t /∈ N then the height ofmin(Fa(s∧t)(t) \N) belongs to Γ. (I.97)

The ordering of P is the coordinatewise inclusion. To show that P is a properforcing notion, let M be a given countable elementary submodel of somelarge enough Hθ containing all the relevant objects and let p ∈ P ∩M . Weshall show that

q = 〈Xp,Np ∪ M ∩Hω2〉is an M -generic condition extending p. So let D ∈ M be a dense-open subsetof P and let r be a given extension of q. Extending r we may assume thatr ∈ D. Let p0 = r ∩M . Note that p0 ∈ P∩M . Let m0 be the maximum ofthe 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 whichone can find r ∈ D realizing the same type as r over p0 such that, if N isthe minimal model of Nr \ Np0 then

F = x ¹ (N ∩ ω1) : x ∈ Xr \Xp0.


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

FE = F ∈ F : x ¹ δ : x ∈ F = Eis uncountable, is also quite large. In particular, we can find an ordinalα0 ∈ 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 \Mand such that every node of Xr \M is incomparable with every node of E1.Let

m+ = maxa(x ∧ y) : x ∈ Xr \M,y ∈ E1,m− = mina(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 allnodes from the set

⋃M ∩ Fm+(t) : t ∈ Xr \M.

By the definitions of E and F there is r ∈ D∩M realizing the same type overp0 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 \Xp0dominates the set E1. We claim that

q = 〈Xr ∪Xr,Nr ∪Nr〉is a condition of P. This will finish the proof of properness of P since clearlyq extends both r and r. Clearly, only (I.95) and (I.97) for q require someargumentation. We check first (I.97), so let s and t be a pair of distinctmembers of Xr ∪ Xr and let N ∈ Nr ∪ Nr be a model such that t /∈ N .Let m = a(s ∧ t). If t ∈ Xr and if there is s′ ∈ Xr such that s′ 6= t ands′ ∧ t = s∧ t the conclusion that the height of min(Fm(t) \N) belongs to Γfollows from the fact that r satisfies (I.97). Similarly one considers the casewhen t ∈ Xr \M and when s′ ∧ t = s ∧ t for some s′ ∈ Xr, s′ 6= t. For ifN belongs to Nr the conclusion follows from the fact that r satisfies (I.97)and if N ∈ Nr \ Nr then by the choice of r,

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 followsfrom the fact that r satisfies (I.97) for s′ and t from Xr and M ∩Hω2 fromNr. Let us now consider the case t ∈ Xr \M , s ∈ Xr \Xr and s′ ∧ t 6= s∧ tfor all s′ ∈ Xr\t. Note that in this case we have the following inequalities:

m0 < m− ≤ m = a(s ∧ t) ≤ m+.

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

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

so the conclusion of (I.97) follows again from the fact that M ∩Hω2 belongsto Nr. Consider now the remaining case t ∈ Xr \Xr and s ∈ Xr \Xr. Notethat in this case the model N must belong to Nr so if s′ ∧ t = s ∧ t forsome s′ ∈ Xr, s 6= t the conclusion follows from the fact that r satisfies thecondition (I.97). So let us consider the subcase when s′ ∧ t 6= s ∧ t for alls′ ∈ Xr, s′ 6= t. Thus, s ∧ t is the new splitting so by the choice of α0 andE0 we have the inequalities:

m0 < m− ≤ m = a(s ∧ t).

Recall that r was chosen to realize the same type over p0 as r so we inparticular have that Fm0(t) and therefore its superset Fm(t) contains theprojection t ¹ (N ∩ ω1). It follows that min(Fm(t) \N) = t ¹ (N ∩ ω1), soits height is indeed a member of Γ (as Γ ∈ N and Γ contains a closed andunbounded subset of ω1).

It remains to check (I.95) for q. So let s and t be a given pair of distinctmembers 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.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(s′∧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(s′∧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 condition r we knowthat this projection is the minimal point of Fm(t) whose height is bigger or


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 that q 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 forp belonging to the generic filter is included in KΓ. What is not so clear isthat P forces that the set X is uncountable as promised. To ensure this wereplace P with its restriction P(≤ r) to the condition

r = 〈t, M ∩Hω2〉,

where M is an arbitrary countable elementary submodel of some largeenough Hθ containing all the relevant objects and where t is any node ofT of height at least M ∩ ω1. Note that the above proof of properness of Pstarting from p = 〈∅, ∅〉 shows that

q = 〈∅, M ∩Hω2〉

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 byinterpreting all P-names belonging to M itself. Then r forces that X isan element of the countable elementary submodel M [G] of Hθ[G] whichcontains a point t of height above M [G] ∩ ω1. It follows that r forces X tobe uncountable. This finishes the proof of Lemma 8.4. a8.5 Corollary. The graph KΓ contains the square of some uncountable sub-set of T in some ω1-preserving forcing extension if and only if Γ is a sta-tionary 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 stationarysubset of ω1. So by the basic property 8.1 of the square-bracket operationno such a forcing extension will contain an uncountable set X ⊆ T suchthat [X]2 ⊆ KΓ. On the other hand, if Γ is a stationary subset of ω1, goingfirst to some standard ω1-preserving forcing extension in which Γ containsa closed and unbounded subset of ω1 and then applying Lemma 8.4, we getan ω1-preserving forcing extension having an uncountable set X ⊆ T suchthat [X]2 ⊆ KΓ. a8.6 Remark. Corollary 8.5 gives us a further indication of the extremecomplexity of the class of graphs on the vertex-set ω1. It also bears somerelevance to the recent work of Woodin [98] who, working in his Pmax-forcingextension, was able to associate a stationary subset of ω1 to any partitionof [ω1]2 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 sufficientlygeneric filter to the forcing notion P = PΓ of Lemma 8.4 that would give usan uncountable X ⊆ T such that [X]2 ⊆ KΓ. In other words, under a bit ofPFA or Woodin’s axiom (*), a set Γ ⊆ ω1 contains a closed and unboundedsubset of ω1 if and only if KΓ contains [X]2 for some uncountable X ⊆ T .

9. Special trees and Mahlo cardinals

One of the most basic questions frequently asked about set-theoretical treesis the question whether they contain any cofinal branch, a branch that in-tersects each level of the tree. The fundamental importance of this questionhas already been realized in the work of Kurepa [45] and then later in theworks of Erdos and Tarski in their respective attempts to develop the theoryof partition calculus and large cardinals (see [23]). A tree T of height equalto some regular cardinal θ may not have a cofinal branch for a very specialreason as the following definition indicates.

9.1 Definition. For a tree T = 〈T, <T 〉, a function f : T → T is regressiveif f(t) <T t for every t ∈ T that is not a minimal node of T. A tree T ofheight θ is special if there is a regressive map f : T −→ T with the propertythat the f -preimage of every point of T can be written as the union of < θantichains of T .

This definition in case θ = ω1 reduces indeed to the old definition ofspecial 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 specialif and only if T is the union of < θ antichains. a

The new definition, however, seems to be the right notion of specialityas 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 cofinalbranches and if every level of T has size < θ.

Recall the well-known characterization of weakly compact cardinals dueto Tarski and his collaborators: a strongly inaccessible cardinal θ is weaklycompact 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 car-dinal θ:

(1) θ is Mahlo.

(2) there are no special Aronszajn trees of height θ.


Proof. Suppose θ is a Mahlo cardinal and let T be a given tree of heightθ all of whose levels have size < θ. To show that T is not special letf : T −→ T be a given regressive mapping. By our assumption of θ thereis an elementary submodel M of some large enough structure Hκ such thatT, f ∈ M and λ = M ∩ θ is a regular cardinal < θ. Note that T ¹ λ is asubset of M and since this tree of height λ is clearly not special, there ist ∈ 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 theunion of < θ antichains.

The proof that (2) implies (1) uses the method of minimal walks in arather crucial way. So suppose to the contrary that our cardinal contains aclosed 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 that α = 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 alongthe sequence and various distance functions defined above. In this proof weare particularly interested in the function ρ0 from [θ]2 into the set Qθ of allfinite sequences of ordinals from θ:

ρ0(α, β) = 〈tp(Cβ ∩ α)〉aρ0(α, min(Cβ \ α))

where we stipulate that ρ0(γ, γ) = 0 for all γ. We would like to show thatthe tree

T (ρ0) = (ρ0)β ¹ α : α ≤ β < θis 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 regressivemap

f : T (ρ0) −→ T (ρ0)

that will witness speciality of T (ρ0). Note that it really suffices defining fon all levels whose index belong to our club C of singular cardinals. So lett = (ρ0)β ¹ α be a given node of T such that α ∈ C and α ≤ β < θ. Notethat by our choice of the C-sequence every term of the finite sequence of


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) aretwo nodes such that t0 $ t1. Our choice of the C-sequence allows us todeduce the following general fact about the corresponding ρ0-function as inthe case θ = ω1 dealt with above in Lemma 1.14.

If α ≤ β ≤ γ, α is a limit ordinal, and if ρ0(ξ, β) = ρ0(ξ, γ)for all ξ < α, then ρ0(α, β) = ρ0(α, γ). (I.102)

Applying this to the triple of ordinals α0, β0 and β1 we conclude thatρ0(α0, β0) = ρ0(α0, β1). Now observe another fact about the ρ0-functionwhose proof is identical to that of θ = ω1 dealt with above in Lemma 1.13.

If α ≤ β ≤ γ then ρ0(α, γ) <r ρ0(β, γ). (I.103)

Applying this to the triple α0 < α1 ≤ β1 we in particular have thatρ0(α0, β1) 6= ρ0(α1, β1). Combining this with the above equality gives usthat 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 Mahlocardinals.

9.5 Theorem. A cardinal θ is a Mahlo cardinal if and only if every regres-sive map f defined on a cube [C]3 of a closed and unbounded subset of θ hasan infinite min-homogeneous set X ⊆ C.30

Proof. Again, the direct implication is simple so we leave it to the interestedreader. For the converse, note that the statement about regressive mapseasily implies that θ must be a strongly inaccessible cardinal. So assumeθ is a strongly inaccessible non-Mahlo cardinal. Let C be a closed andunbounded subset of θ consisting of strong limit singular cardinals. Weshall find a regressive map on [C]3 without an infinite min-homogeneousset. This will give us an opportunity to further expose the tree T (ρ0) andthe regressive map f : T (ρ0) ¹ C −→ T (ρ0) defined in the proof of Theorem9.4. We have already observed that the length of every branching node tmust be of the form α +1 for some α and in fact the following more precisedescription is true:

If t is a branching node of T (ρ0) then its length is equal toα + 1 for some α ∈ C.

(I.104)

Letα− = max(C ∩ (α + 1)) and α+ = min(C \ (α + 1)).

30Recall, that X is min-homogeneous for f if f(α, β, γ) = f(α′, β′, γ′) for every pairα < β < γ and < α′ < β′ < γ′ of triples of elements of X such that α = α′.


Let β, γ > α + 1 be two ordinals such that (ρ0)β and (ρ0)γ split at t. Letβ = β0 > . . . > βk = α and γ = γ0 > . . . > γl = α be the walks from β to αand γ to α respectively. From the fact that (ρ0)β and (ρ0)γ agree below αwe 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 sequenceof 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 assumptionthat (ρ0)β and (ρ0)γ split at t. It follows that α = α− which was to beshown.

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 maximalelement of C that is less than or equal to the length of t. Then we knowthat f is one-to-one on any chain whose nodes are separated by C. Fromf we shall derive three other regressive maps p : [C]3 −→ θ, q : [C]3 −→ ωand r : [C]2 −→ 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(fn(β ∧ γ)), where n is the minimal in-teger such that fn(β ∧ γ) ⊆ (ρ0)α. (I.105)

q(α, β, γ) = minn : fn(β ∧ γ) = ∅. (I.106)

r(α, β) = p〈ρ0(ξ, α), ρ0(ξ, β)〉q, 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 infiniteincreasing sequence αi of elements of C using Ramsey’s theorem we mayassume 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 map-ping r. If (I.109) holds and if the sequence is min-homogeneous with respectto p and q another application of Ramsey’s theorem would give us an integerm and an ordinal ξ such that for all i < j < k < ω:

m = minn : fn(αj ∧ αk) ⊆ (ρ0)αi, (I.110)

fm(αj ∧ αk) = (ρ0)α0 ¹ ξ(= t). (I.111)


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

αi ∧ αi+1 $ fm−1(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, acontradiction. This finishes the proof of Theorem 9.5. a

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

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

(1) θ is n-Mahlo.

(2) Every regressive map defined on [C]n+2 for some closed and unboundedsubset 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 statementthat there exist f : [X]n+2 −→ ω and regressive g[X]n+3 −→ sup(X) suchthat if H ⊆ X is min-homogeneous for g and f”[H]n+2 = k for somek ∈ ω, then |H| ≤ n + k + 5. Note that in the previous proof we haveestablished W (C, 0) for every closed unbounded subset C of some limitordinal θ such that C contains no inaccessible cardinals (the mapping q isreally a map from [C]2 into ω). The proof of the Theorem is finished oncewe show that in general W (C, n) holds for every set of ordinals C of limitorder type that is closed in its supremum and which contains no n-Mahlocardinals. So let C be a given closed and unbounded subset of some limitordinal θ and suppose that C contains no n + 1-Mahlo cardinals. We needto show that W (C, n + 1) holds. Note that for this it suffices to show byinduction that W (C ∩ δ, n) holds for all δ ∈ C since there is a naturalway to combine pairs of maps fδ : [C]n+2 −→ ω and gδ : [C]n+3 −→ θfor δ ∈ C into maps f : [C]n+3 −→ ω and g : [C]n+4 −→ θ witnessingW (C, n + 1). The successor steps of the induction follow from the easilychecked fact that W (X, n) implies W (Y, n) for every end-extension Y of Xof cardinality at most 2|sup(X)|. Assume now that δ is a limit point of C andthat W (C∩γ, n) holds for all δ ∈ C∩δ. Since δ is not an n+1-Mahlo cardinalthere is a closed and unbounded set D ⊆ C ∩ δ containing no n-Mahlocardinals. By the induction hypothesis, W (D, n) holds. Now, there is avery natural way to combine maps fγ : [C]n+2 −→ ω and gγ : [C]n+3 −→ θfor γ ∈ C ∩ δ witnessing W (C ∩ γ, n) and maps fδ : [D]n+2 −→ ω andgδ : [D]n+3 −→ θ witnessing W (D,n) into maps Fδ : [C ∩ δ]n+2 −→ ω andGδ : [C ∩ δ]n+3 −→ θ witnessing W (C ∩ δ). a


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 toSchmerl [64]. Using Theorem 9.6 one can also prove the following charac-terization theorem essentially given in [64](see also [30]) which has someinteresting metamathematical applications (see [26] and [38]).

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

(1) θ is n-Mahlo.

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

aWe 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 Mahloin the constructible universe, then there is a constructible special Aronszajntree of height θ.

Proof. Working in L we choose a closed and unbounded subset C of θconsisting of singular ordinals and a C-sequence Cα (α < θ) such thatCα+1 = α, Cα = (α, α) when α = sup(C ∩ α) < α, while if α is a limitpoint 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 andsuch 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 associatedρ0-function then T (ρ0) is a special Aronszajn tree of height θ. Suppose <T

is a special Aronszajn tree ordering on θ such that [θ− ·α, θ− · (α+1)) is itsαth level. Let C be the club of ordinals < θ divisible by θ−. Let f : θ −→ θ−

be such that the f -preimage of every ordinal < θ− is an antichain of thetree (θ, <T ). We choose a C-sequence Cα (α < θ) such that Cα+1 = α,Cα = (α, α) for α limit with the property that α = sup(C ∩α) < α, but if αis a limit point of C we take Cα more carefully as follows: Cα = αξ : ξ < ηwhere

αλ = supαξ : ξ < λ for λ limit < η, (I.116)

α0 =the <T -predecessor of α with minimal f -image, (I.117)

αξ+1 =the <T -predecessor of α with minimal f -image sub-ject to the requirement that f(αξ+1) > f(αζ+1) for allζ < ξ,

(I.118)

η is the limit ordinal ≤ θ− where the process stops, i.e.supf(αξ+1) : ξ < η = θ−. (I.119)

Note that if α and β are two limit points of C and if γ <T α, β thenCα ∩ γ = Cβ ∩ γ. From this one concludes that the C-sequence is locallysmall, i.e. that Cα ∩ γ : γ ≤ α < θ has size ≤ θ− for all γ < θ. a

9.11 Corollary. If θ<θ = θ then there exists a special Aronszajn tree ofheight θ+. a9.12 Corollary. In the constructible universe, special Aronszajn trees ofany regular uncountable non-Mahlo height exist. a9.13 Remark. In a large portion of the literature on this subject the notionof a special Aronszajn tree of height equal to some successor cardinal θ+

is somewhat weaker, equivalent to the fact that the tree can be embeddedinside the tree f : α −→ θ : α < θ & f is 1− 1. One would get our notionof speciality by restricting the tree on successor ordinals losing thus thefrequently useful property of a tree that different nodes of the same limitheight have different sets of predecessors. The result 9.11 in this weakerform 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)


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

ρ1(α, β) = maxtp(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 proofof 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 betrue 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) = maxtp(Cα ∩ ξ), ρ1(ξ, α1) = ρ1(ξ, α1) > ν.

This finishes the proof. a10.2 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 setof order-type κ. We need to find ξ ∈ Γ such that ρ1(ξ, α) = ρ1(ξ, β). Letγ = 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 ordinalsare well-defined and

γ0 < γ ≤ β0 < β. (I.124)

By Lemma 10.1 and the inductive hypothesis there is ξ in Γ ∩ (γ0, γ) suchthat

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) = maxtp(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 isessential. For example, it can be seen (see [7, p.72]) that the conclusion ofthis lemma fails if κ is a singular limit of supercompact cardinals.

10.4 Definition. Define ρ1 : [κ+]2 −→ κ by

ρ1(α, β) = 2ρ1(α,β) · (2 · tpξ ≤ α : ρ1(ξ, β) = ρ1(α, β)+ 1).

The following fact shows that this stretching of ρ1 keeps the basic co-herence property stated in Lemma 10.2 but it gives us also the injectivityproperty that could be quite usefull.

10.5 Lemma. If κ is a regular cardinal then

(a) ρ1(α, γ) 6= ρ1(β, γ) whenever α < β < γ < κ+,

(b) |ξ ≤ α : ρ1(ξ, α) 6= ρ1(ξ, β)| < κ whenever α < β < κ+.

a10.6 Remark. Note that Lemma 10.5 gives an alternative proof of Corol-lary 9.11 since under the assumption κ<κ = κ the tree T (ρ1) will havelevels 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 interestwhich can be particularly useful in stepping-up combinatorial properties ofκ to κ+. It is also an object that has interpretations in such areas as thetheory of Cech-Stone compactifications of discrete spaces (see, e.g. [95],[10], [60], [44], [16]). We have already noted that if κ is singular then wemay no longer have the coherence property of Lemma 10.2. To get thisproperty, one needs to make some additional assumption on the C-sequenceCα (α < κ+), an assumption about the coherence of the C-sequence. Thiswill be subject of some of the following chapters where we will concentrateon 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 re-quirement 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 theC-sequence Cα = α (α < θ) is the most trivial choice and the correspondingρ0-function gives no information whatsoever about the cardinal θ. Thefollowing notion of the triviality of a C-sequence on θ seems to be onlymarginally different.


11.1 Definition. A C-sequence Cα (α < θ) on a regular uncountable car-dinal θ is trivial if there is a closed and unbounded set C ⊆ θ such that forevery 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 everyinteger n there is a subfamily B of A of size θ such that ρ2(α, β) > nfor all α ∈ a, β ∈ b and a 6= b in B.

Proof. Note that if a closed and unbounded set C ⊆ θ witnesses the trivialityof the C-sequence and if for α < θ we let β(α) ≥ α be the minimal β suchthat C ∩α ⊆ Cβ(α), then any disjoint subfamily of A = α, β(α) : α ∈ Cof 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 nthere exist a tail subset Γ0 of Γ, an unbounded subset ∆0 of ∆ and aregressive strictly increasing map f : ∆0 −→ θ such that ρ2(α, β) > nfor all α ∈ Γ0 and β ∈ ∆0 with α < f(β).

To get (ii) from (iii) one assumes that the family A consists of finite setsof some fixed size m and performs m2 successive applications of (iii): thiswill give us an unbounded A ⊆ A and regressive strictly increasing mapsfij : πj [A] −→ θ (i, j < m), where πj is the projection mapping that takesb 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. LetMξ (ξ < θ) be a continuous ∈-chain of submodels of Hθ+ containing allrelevant 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. Pickan arbitrary β ∈ ∆ above δ. Then there is ξ ∈ C ∩ δ such that ξ /∈ Cβ .Then α = 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 thatsup(Cβ(α) ∩ α) = α. Applying the inductive hypothesis to the two sets

Γ \ α and min(Cβ(α) \ α) : α ∈ Γ \ α


we get a tail subset Γ0 of Γ \ α, an unbounded subset Γ1 of Γ \ α and astrictly 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)) = minγ, 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. a11.3 Corollary. Suppose that Cα (α < θ) is a nontrivial C-sequence andlet T (ρ0) be the corresponding tree (see Section 9 above). Then every subsetof T (ρ0) of size θ contains an antichain of size θ.

Proof. Consider a subset K of [θ]2 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 assumingwithout 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 agiven level of T (ρ0). Shrinking K further, we obtain that ρ2 is constanton K. Let n be the constant value of ρ2 ¹ K. Applying 11.2(ii) to Kand n, we get K0 ⊆ K of size θ such that ρ2(α, δ) > n for all α, βand γ, δ from K0 with properties α < β, γ < δ and α < γ. Thenρ0(·, β) ¹ (α + 1) (α, β ∈ K0) is an antichain in T (ρ0). a11.4 Remark. It should be clear that nontrivial C-sequences exist on anysuccessor cardinal. Indeed, with very little extra work one can show thatnontrivial C-sequences exist for some inaccessible cardinals quite high in theMahlo-hierarchy. To show how close this is to the notion of weak compact-ness, we will give the following characterization of it which is of independentinterest. 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 setC ⊆ 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 theweak compactness of θ. As pointed out to us by Donder and Konig, one can show thisusing a model of Kunen [43, §3].


Proof. To see the implication from (i) to (ii), let us assume that Cα (α < θ)is a C-sequence and let ρ0 : [θ]2 −→ 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. Foreach α < θ fix β(α) ≥ α such that the restriction of (ρ0)β(α) to α belongs tob. 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 astationary set Γ ⊆ θ and an integer k, a sequence t in Qθ and an ordinalα < θ such that for all α ∈ Γ:

k(α) = k and sup(Cβi(α) ∩ α) ≤ α for all i < k for whichthis set is bounded in α.

(I.127)

tp(Cβi(α) ∩ α) = t(i) for all i < k for which this set isbounded in α.

(I.128)

Shrinking Γ we may assume that either for all α ∈ Γ, γ(α) = α in whichcase t is chosen to be a k-sequence, or for all α ∈ Γ, γ(α) = βk−1(α) inwhich case t is chosen to be a (k − 1)-sequence. It follows that

ρ0(ξ, β(α)) = taρ0(ξ, γ(α)) for all α ∈ Γ and ξ ∈ [α, α). (I.129)

It follows that for α < α′ in Γ the mappings (ρ0)γ(α) and (ρ0)γ(α′) have thesame restrictions on the interval [α, α). So in particular we have that

Cγ(α) ∩ [α, α) = Cγ(α′) ∩ [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) havethe same intersection with α. 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, <T ) be a tree onθ whose levels are intervals of θ or more precisely, there is a closed andunbounded set ∆ of limit ordinals in θ such that if δξ (ξ < θ) is its increasingenumeration then [δξ, δξ+1) is equal to the ξth level of T . Choose a C-sequence Cα (α < θ) as follows. First of all let Cα+1 = α and Cα = (α, α)if α is limit and α = sup(∆ ∩ α) < α. If α is a limit point of ∆, let

Cα = ξ < α : ξ <T α.By (ii) there is a closed and unbounded set C ⊆ θ such that for all α < θthere is β(α) ≥ α with Cβ(α) ∩ α = C ∩ α. Let Γ be the set of all successorordinals ξ < θ such that C ∩ [δξ, δξ+1) 6= ∅. For ξ ∈ Γ let tξ = min(C ∩[δξ, δξ+1)). Then it is seen that tξ : ξ ∈ Γ is a chain of T of size θ. Thisfinishes the proof. a


We have already remarked that every successor cardinal θ = κ+ admitsa nontrivial C-sequence Cα (α < θ). It suffices to take the Cα’s to be all oforder-type ≤ κ. It turns out that for such a C-sequence the correspondingρ2-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 unboundednessproperty: for every family A of κ+ pairwise disjoint subsets of κ+, all ofsize < κ, and for every n < ω there exists B ⊆ A of size κ+ such thatρ2(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 thecase when κ is a regular cardinal. The proof is by induction on n. Supposethe 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 acontinuous ∈-chain of elementary submodels of Hκ++ such that for everyξ 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 andb ∈ A such that β > ξ for all β ∈ b. Then γ = supmax(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 = supmax(Cβ ∩ η) : β ∈ bη. Consider the followingfamily of subsets of κ+: aη = bη ∪ min(Cβ \ η) : β ∈ bη (η ∈ (γ, κ+)).By the inductive hypothesis there is an unbounded Γ ⊆ κ+ such that forall η < ζ in Γ, bη ⊆ ζ and ρ2(α, β) > n for all α ∈ aη and β ∈ aζ . LetB = 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 Theoremwhen κ is a regular cardinal.

If κ is a singular cardinal, going to a subfamily of A, we may assume thatthere is a regular cardinal λ < κ and an unbounded set Γ ⊆ κ+ with theproperty 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 and η ∈ Γ above ξ and proceed as inthe previous case. This completes the inductive step and the proof of thetheorem 11.6. a

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


11.7 Theorem. Suppose that a regular uncountable cardinal θ supports anontrivial C-sequence and let ρ2 be the associated function. Then for everyinteger n and every pair of θ-sized families A0 and A1, where the membersof A0 are pairwise disjoint bounded subsets of θ and the members of A1 arepairwise disjoint finite subsets of θ, there exist B0 ⊆ A0 and B1 ⊆ A1 ofsize 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). aAn 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 spaceY , let ](f) be the order-type of the shortest chain C of closed subsets of Xsuch that f ′′H = Y for all H ∈ C, while f ′′(

⋂ C) 6= Y . If such a chain doesnot exist, set ](f) = ∞.

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

This shows that under some very mild assumption on a pair of spaces Xand Y , if their sizes are smaller than the first regular uncountable cardinalθ not supporting a nontrivial C-sequence, then for every closed map f fromX onto Y , there is a closed subspace H of X on which f is irreducible, i.e.f ′′K 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 onlyif it is coherent, i.e. it has the property that Cα = Cβ ∩ α whenever α is alimit point of Cβ .

Note that the nontriviality conditions appearing in Definition 11.1 andTheorem 11.5 coincide in the realm of square-sequences:

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

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

α <2 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 thatthe tree (θ, <2) has a chain of size θ. In fact, one can characterize the treeorderings <T on θ for which there exists a square sequence Cα (α < θ) suchthat for all α < β < θ:

α <T β if and only if α is a limit point of Cβ . (I.135)

12.3 Lemma. A tree ordering <T on θ admits a square sequence Cα (α <θ) satisfying (I.135) if and only if

(i) α <T β can hold only for limit ordinals α and β such that α < β,

(ii) Pβ = α : α <T β is a closed subset of β, which is unbounded in βwhenever cf(β) > ω and

(iii) minimal as well as successor nodes of the tree <T on θ are ordinalsof cofinality ω.

Proof. For each ordinal α < θ of countable cofinality we fix a subset Sα ⊆ αof order-type ω cofinal with α. Given a tree ordering <T on θ with properties(i)-(iii), for a limit ordinal β < θ let P+

β be the set of all successor nodes fromPβ ∪ β including the minimal one. For α ∈ P+

β let α− be its immediatepredecessor in Pβ . Finally, set

Cβ = Pβ ∪⋃Sα ∩ [α−, α) : α ∈ P+

β .

It is easily checked that this defines a square-sequence Cβ (β < θ) with theproperty that α <T β holds if and only if α is a limit point of Cβ . a

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

An important result about square sequences is the following result thatcan be deduced on the basis of the well-known construction of square se-quences 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 compactin the constructible subuniverse then there is a nontrivial square sequenceon θ which is moreover constructible. a

12.6 Corollary. If a regular uncountable cardinal θ is not weakly compactin the constructible subuniverse then there is a constructible Aronszajn treeon θ.


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

ρ0(α, β) = tp(Cβ ∩ α)aρ0(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 boundedby the size of the set Cβ ∩ α : β ≥ α. Since the intersection of the formCβ ∩ α is determined by its maximal limit point modulo a finite subset ofα, we conclude that the αth level of T (ρ0) has size ≤ |α| + ℵ0. Since thesequence Cα (α < θ) is nontrivial, the proof of Theorem 11.5 shows thatT (ρ0) has no cofinal branches. a12.7 Lemma. Suppose Cα (α < θ) is a square sequence on θ, <2 theassociated tree ordering on θ and T (ρ0) = (ρ0)β ¹ α : α ≤ β < θ whereρ0 : [θ]2 −→ Qθ is the associated ρ0-function. Then α 7−→ (ρ0)α is a strictlyincreasing map from the tree (θ, <2) into the tree T (ρ0).

Proof. If α is a limit point of Cβ then Cα = Cβ ∩α so the walks α → ξ andβ → 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-functionsassociated with a square sequence Cα (α < θ) on some regular uncountablecardinal θ, both fixed from now on. Recall that an ordinal α divides anordinal γ if there is β such that γ = α ·β, i.e. γ can be written as the unionof an increasing β-sequence of intervals of type α. Let κ ≤ θ be a fixedinfinite regular cardinal. Let Λκ : [θ]2 −→ θ be defined by

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

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

ρκ : [θ]2 −→ κ (I.137)

defined recursively by

ρκ(α, β) = suptp(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.


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

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

12.9 Lemma. If α < β < γ < θ then

(a) ρκ(α, γ) ≤ maxρκ(α, β), ρκ(β, γ),(b) ρκ(α, β) ≤ maxρκ(α, γ), ρκ(β, γ).

Proof. The proof is by induction an α, β and γ. We first prove the induc-tive step for (a). Let ν = maxρκ(α, β), ρκ(β, γ). We need to show thatρκ(α, γ) ≤ ν. To this end, let

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

Case 1a: 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 weconclude that ρκ(α, Λκ(β, γ)) ≤ ν.

Case 2a: α ≥ Λκ(β, γ). Then Λκ(β, γ) = Λκ(α, γ) = λ.Subcase 2a.1: γα = γβ = γ. By the inductive hypothesis (a) for the

triple α < β ≤ γ we get

ρκ(α, γ) ≤ maxρκ(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 < βwe conclude that

ρκ(ξ, α) ≤ maxρκ(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 allordinals appearing in the formula for ρκ(α, γ) are ≤ ν and so ρκ(α, γ) ≤ ν.

Subcase 2a.2: γα < γβ . Then γα belongs to Cγ ∩ [λ, β) and thereforeρκ(γα, β) ≤ ρκ(β, γ) ≤ ν. Applying the inductive hypothesis (b) for α ≤γα < β gives us

ρκ(α, γα) ≤ maxρκ(α, β), ρκ(γα, β) ≤ ν. (I.142)


Given ξ ∈ Cγ ∩ [λ, α) using the inductive hypothesis (b) for ξ < α < β weget that

ρκ(ξ, α) ≤ maxρκ(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 letν = maxρκ(α, γ), ρκ(β, γ) and let γα and γβ be as in (I.138). We need toshow that ρκ(α, β) ≤ ν.

Case 1b: 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 weget

ρκ(α, β) ≤ maxρκ(α, Λκ(β, γ)), ρκ(Λκ(β, γ), β) ≤ ν. (I.145)

Case 2b: α ≥ Λκ(β, γ).Subcase 2b.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) ≤ maxρκ(α, γ), ρκ(β, γ) ≤ ν. (I.146)

Subcase 2b.2: γα < γβ . Then γα ∈ Cγ ∩ [λ, β) and so ρκ(γα, β) appearsin 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) ≤ maxρκ(α, γα), ρκ(γα, β) ≤ ν. (I.147)

This finishes the proof. aThe 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 tothe trace of the walk from γ to α. a12.11 Lemma. Suppose β ≤ γ < θ and that β is a limit ordinal > 0. Thenρκ(α, γ) ≥ ρκ(α, β) for coboundedly many α < β.

Proof. Let γ = γ0 > γ1 > . . . > γn−1 > γn = β be the trace of the walkfrom γ to β. Let γ = γn−1 if β is a limit point of Cγn−1 , otherwise let γ = 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)


Let β < β be an upper bound of all Cγi∩ β (i < n) which are bounded in

β. Then γ is a member of the trace of any walk from γ to some ordinal αin 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 theconclusion of the Lemma. a12.12 Lemma. The set Pκ

ν (β) = ξ < β : ρκ(ξ, β) ≤ ν is a closed subsetof β for every β < θ and ν < κ.

Proof. The proof is by induction on β. So let α < β be a limit pointof Pκ

ν (β). Let β1 = min(Cβ \ α) and let λ = Λκ(α, β). If λ = α thenρκ(α, β) = 0 and therefore α ∈ Pκ

ν (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 everyξ ∈ Pκ

ν (β) ∩ (α, α). It follows that

Pκν (β) ∩ (α, α) ⊆ Pκ

ν (β1), (I.151)

so α is also a limit point of Pκν (β1). By the inductive hypothesis α ∈ Pκ

ν (β1),i.e. ρκ(α, β1) ≤ ν. Since Λκ(ξ, β) = λ for all ξ ∈ [λ, α), we get

tp(Cβ ∩ [λ, α)) = suptp(Cβ ∩ [λ, ξ)) : ξ ∈ Pκν (β) ∩ [λ, α) ≤ ν. (I.152)

Consider η ∈ Cβ ∩ [λ, α). Then η ∈ Cβ ∩ [λ, ξ) for some ξ ∈ Pκν (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. Applying12.9(b) to η < α < β1, we conclude that

ρκ(η, α) ≤ maxρκ(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 thatρκ(α, β) ≤ ν we need to have tp(Cβ ∩ [λ, α)), ρκ(α, β1) and ρκ(η, α) withη ∈ Cβ ∩ [λ, α) all ≤ ν. Note that β1 = α so ρκ(α, β1) = 0. Similarly as inthe first case we conclude tp(Cβ ∩ [λ, α)) ≤ ν. By Lemma 12.8,

ρκ(ξ, α) = ρκ(ξ, β) ≤ ν for all ξ ∈ Pκν (β) ∩ α. (I.154)

Consider an η ∈ Cβ ∩ [λ, α). Since α is a limit point of Pκν (β) there is ξ

from this set above η. By (I.150) we have that η ∈ Cβ ∩ [Λκ(ξ, β), ξ) soρκ(η, ξ) ≤ ρκ(ξ, β) ≤ ν. Applying 12.9(a) to η < ξ < α we get

ρκ(η, α) ≤ maxρκ(η, ξ), ρκ(ξ, α) ≤ ν. (I.155)

This finishes the proof. a


For α < β < θ and ν < κ set

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

12.13 Lemma.

(1) <κν is a tree ordering on θ,

(2) <κν⊆<κ

µ whenever ν < µ < κ,

(3) ∈¹ θ =⋃

ν<κ <κν .

Proof. This follows immediately from Lemma 12.9. aRecall 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 thatthe preimage of any point is the union of < θ antichains. By a tree on θwe mean a tree of the form (θ, <T ) with the property that α <T β impliesα < β.

12.14 Lemma. If a tree T on θ is special, then there is an ordinal-regressivemap 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 thepreimage g−1(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 < θwith the property that λξ < α for all ξ < α. Let p., .q denote the Godelpairing function which in particular has the property that pα, βq < δ forδ ∈ C and α, β, δ. Define ordinal-regressive map f : θ −→ θ by lettingf(α) = max(C ∪ α) if α /∈ C and for α ∈ C, let f(α) = pg(α), γq whereγ < λg(α)(< α) is minimal ordinal with property that α ∈ Aγ(g(α)). Thenf is as required. a

By Lemma 12.13 we have a sequence <κν (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 thenthere must be ν < κ such that (θ, <κ

ν ) is a nonspecial tree on θ.

Proof. Suppose to the contrary that all trees are special. By Lemma 12.14we may choose ordinal-regressive maps fν : θ −→ θ for all ν < κ and asingle closed and unbounded set C ⊆ θ such that each of the maps fν isone-to-one on <κ

ν -chains separated by C. Using the Pressing Down Lemmawe find a stationary set Γ of cofinality κ+ ordinals < θ and λ < θ such thatfν(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|))


of sets of size < |λ|. Since κ 6= cf(|λ|) there is a ξ < cf(|λ|) and a stationaryΓ0 ⊆ 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∩βhas size θ. Then there will be ν0 < κ and Γ1 ⊆ Γ0 ∩ C ∩ β of size θ suchthat ρκ(α, β) ≤ ν0 for all α ∈ Γ1. By (I.158) we can find Γ2 ⊆ Γ1 of sizeθ 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 hassize smaller than the size of Γ2. This finishes the proof. a

It is now natural to ask the following question: under which assumptionon the square sequence Cα (α < θ) can we conclude that neither of the trees(θ,<κ

ν ) will have a branch of size θ?

12.16 Lemma. If the set Γκ = α < θ : tp Cα = κ is stationary in θ,then none of the trees (θ,<κ

ν ) has a branch of size θ.

Proof. Assume that B is a <κν -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 belongsto 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. Thiscontradicts the fact that α <κ

ν β and finishes the proof. a

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

12.18 Theorem. Suppose κ < θ are regular cardinals such that θ is nota successor of a cardinal of cofinality κ. Then to every square sequenceCα (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,

(ii) α =⋃

ν<κ Cαν for all limit α,

(iii) Cαν (α < θ) is a nonspecial (and nontrivial) square sequence on θ forall ν < κ.


Proof. Fix ν < κ and define Cαν by induction on α < θ. So suppose β is alimit ordinal < θ and that Cαν is defined for all α < β. If Pκ

ν (β) is boundedin β, let β be the maximal limit point of Pκ

ν (β) (β = 0 if the set has nolimit points) and let

Cβν = Cβν ∪ Pκν (β) ∪ (Cβ ∩ [max(Pκ

ν (β)), β)). (I.159)

If Pκν (β) is unbounded in β, let

Cβν = Pκν (β) ∪

⋃Cαν : α ∈ Pκ

ν (β) & α = sup(Pκν (β) ∩ α). (I.160)

By Lemmas 12.9 and 12.12, Cβν (β < θ) is well defined and it forms asquare sequence on θ. The properties (i) and (ii) are also immediate. To seethat for each ν < κ the sequence Cβν (β < θ) is nontrivial, one uses Lemma12.16 and the fact that if α is a limit point of Cβν occupying a place in Cβν

that is divisible by κ, then α <κν β. By Lemma 12.15, or rather its proof,

we conclude that there is ν < κ such that Cβν (β < θ) is nonspecial for allν ≥ ν. This finishes the proof. a12.19 Lemma. For every pair of regular cardinals κ < θ, every specialsquare sequence Cα (α < θ) can be refined to a special square sequenceCα (α < θ) with the property that tp Cα = κ for stationarily many α < θ.

Proof. Let <2 be the tree ordering associated with Cα (α < θ) and letf : θ −→ θ be a <2-regressive map such that f−1(ξ) is the union of < θantichains of (θ, <2) for all ξ < θ.

Case 1: κ = ω. By the Pressing Down Lemma there is a stationary setΓ of cofinality ω ordinals < θ on which f takes some constant value. ThusΓ can be decomposed into < θ antichains of (θ, <2) and so Γ contains astationary subset Γ0 of pairwise <2-incomparable ordinals. For α ∈ Γ0 letCα be any ω-sequence of successor ordinals converging to α. If α is a limitordinal not in Γ0 but there is (a unique!) α <2 α in Γ0, let Cα = Cα \ α. Inall other cases, let Cα = Cα. It is easily seen that Cα (α < θ) is as required.

Case 2: κ > ω. By Lemma 12.14 we may find a closed and unboundedsubset C of θ and an ordinal-regressive map g : C −→ θ which is one-to-oneon <2-chains included in C (and which has the property that for α ∈ Cg(α) is the Godel pairing of f(α) and the index of the antichain which con-tains α in the fixed antichain-decomposition of the preimage f−1(f(α)).Bythe argument from Case 1 there is a stationary set Γ of limit points of Cconsisting 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) < λfor unboundedly many limit points ξ of Cα that also belong to C. Since gis regressive and since κ, the cofinality of α, is uncountable, the PressingDown Lemma gives us that h(α) < α. Choose a stationary set Γ0 ⊆ Γ suchthat h takes some constant value λ on Γ0. Note that since g is one-to one


on <2-chains included in C the constant value λ cannot be a successor or-dinal. It follows that λ must be a limit ordinal of cofinality equal to κ. Letλν (ν < κ) be a strictly increasing sequence which converges to λ.

We first define Cα for α ∈ Γ0 as a strictly increasing continuous sequencecν(α) (ν < κ) of limit points of Cα that also belong to C and satisfy thefollowing 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 pointof Cα, belonging to C, and being > γ for every limit pointγ of Cα such that γ ∈ C and g(γ) < λν+1.

(I.162)

If α is a limit point of Cβ for some β ∈ Γ0, set Cα = Cβ ∩ α. Note that bythe canonicity of the definition of cν(β) (ν < κ) for β ∈ Γ0, we would getthe same result, as Cβ ∩ α = Cβ ∩ α for any other β ∈ Γ0 for which α is alimit point of Cβ .

If α is a limit ordinal which is not a limit point of any Cβ for β ∈ Γ0 andthere is (a unique!) α <2 α in Γ0, let Cα = Cα \ α.

If α is a limit ordinal such that α <2 α for some α ∈ Γ0 but α is not alimit point on any Cβ for β ∈ Γ0, let Cα = Cα \max(Cα ∩ α). Note againthat there is no ambiguity in this definition, as

Cβ ∩ α = Cγ ∩ α

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

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

12.20 Theorem. If regular uncountable cardinal θ 6= ω1 carries a nontrivialsquare sequence then it also carries a nontrivial square-sequence which ismoreover nonspecial. a12.21 Corollary. If a regular uncountable cardinal θ 6= ω1 is not weaklycompact 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 inCorollary 12.6 we conclude that T (ρ0) is an Aronszajn tree of height θ. ByLemma 12.7 there is a strictly increasing map from (θ, <2) into T (ρ0), soT (ρ0) must be nonspecial. a


12.22 Remark. The assumption θ 6= ω1 in Theorem 12.20 is essential asthere is always a nontrivial square sequence on ω1 but it is possible to havea 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 aweakly compact cardinal admits a forcing extension satisfying CH and thestatement that all Aronszajn trees on ω2 are special. A well-known openproblem in this area asks whether one can have GCH rather than CH in amodel 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 anontrivial square sequence on θ. Recall the function Λ = Λω : [θ]2 −→ 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 for-

mula, 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.


(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 < λthen Cλ = Cγ ∩ λ and therefore F (α, λ) = F (α, γ). Applying the inductivehypothesis (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 theinductive 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 theinductive 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 ≤ βwe 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) areincluded 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 caseλ ≥ α. Then Cλ = Cγ ∩ λ and therefore F (α, λ) = F (α, γ). Applying theinductive 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) forα ≤ 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 andapplying the inductive hypothesis (b) for the triple α ≤ β ≤ γα = γβ wehave

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

This completes the proof. a13.2 Lemma. For all α ≤ β ≤ γ,

(a) ρ0(α, β) = ρ0(min(F (β, γ) \ α), β)aρ0(α, min(F (β, γ) \ α)),

(b) ρ0(α, γ) = ρ0(min(F (β, γ) \ α), γ)aρ0(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 weget

ρ0(α, β) = ρ0(α1, β)aρ0(α, α1) (I.170)

which checks (a) for α ≤ β ≤ γ. Suppose first that λ > α. Note thatF (λ, β) is a subset of F (β, γ) so α2 = min(F (λ, β) \ α) ≥ α1. Applying theinductive hypothesis for α ≤ λ ≤ β we get

ρ0(α, β) = ρ0(α2, β)aρ0(α, α2), (I.171)


ρ0(α, λ) = ρ0(α2, λ)aρ0(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

ρ0(α, λ) = ρ0(α2, λ)aρ0(α1, α2)aρ0(α, α1) = ρ0(α1, λ)aρ0(α, α1). (I.173)

Since λ is a limit point of Cγ , we have ρ0(α, γ) = ρ0(α, λ), so (I.173) givesus the conclusion (b) of Lemma 13.2.

Consider now the case λ ≤ α. Then γ1 is either equal to min(Cγ \ β) orit belongs to [λ, β) ∩ Cγ . In any case we have that Fγ1, β ⊆ F (β, γ), so

γ2 = min(Fγ1, β \ α) ≥ α1. (I.174)

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

ρ0(α, β) = ρ0(γ2, β)aρ0(α, γ2), (I.175)

ρ0(α, γ1) = ρ0(γ2, γ1)aρ0(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

ρ0(α, γ1) = ρ0(γ2, γ1)aρ0(α1, γ2)aρ0(α, α1) = ρ0(α1, γ1)aρ0(α, α1).(I.177)

Since ρ0(α, γ) = tp(Cγ ∩α)aρ0(α, γ1), (I.177) gives us the conclusion (b) ofLemma 13.2. This finishes the proof. a

Recall the function ρ2 : [θ]2 −→ ω which counts the number of steps inthe walk along the fixed C-sequence Cα (α < θ) which in this Section isassumed 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 thecardinality 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)|.

a13.4 Definition. Set I to be the set of all countable Γ ⊆ θ such thatsupξ∈∆ ρ2(ξ, α) = ∞ for all α < θ and infinite ∆ ⊆ Γ ∩ α.

13.5 Lemma. I is a P-ideal of countable subsets of θ.


Proof. Let Γn (n < ω) be a given sequence of members of I and fix β < θsuch that Γn ⊆ β for all n. For n < ω set Γ∗n = ξ ∈ Γn : ρ2(ξ, β) ≥ n.Since Γn belongs to I, Γ∗n is a cofinite subset of Γn. Let Γ∞ =

⋃n<ω Γ∗n.

Then Γ∞ is a member of I such that Γn \ Γ∞ is finite for all n. a13.6 Theorem. The P-ideal dichotomy implies that a nontrivial squaresequence can exist only on θ = ω1.

Proof. Applying the P-ideal dichotomy on I from 13.4 we get the two al-ternatives (see 4.45):

there is an uncountable ∆ ⊆ θ such that [∆]ω ⊆ I, or (I.178)

there is a decomposition θ =⋃

n<ω Σn such that Σn⊥I forall n.

(I.179)

By Lemma 13.3, if (I.178) holds, then ∆∩α must be countable for all α < θand so the cofinality of θ must be equal to ω1. Since we are working onlywith 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 isunbounded in θ. Since Σk⊥I we have that (ρ2)α is bounded on Σk ∩ αfor all α < θ. So there is an unbounded set Γ ⊆ θ and an integer n suchthat for each α ∈ Γ the restriction of (ρ2)α on Σk ∩ α is bounded by n. ByTheorem 11.2 we conclude that the square sequence Cα (α < θ) we startedwith must be trivial. a13.7 Definition. By Sθ we denote the sequential fan with θ edges, i.e. thespace on (θ× ω)∪ ∗ with ∗ as the only nonisolated point, while a typicalneighborhood of ∗ has the form Uf = (α, n) : n ≥ f(α) ∪ ∗ wheref : θ −→ ω.

13.8 Theorem. If there is a nontrivial square sequence on θ then the squareof the sequential fan Sθ has tightness 32 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 ofsteps function ρ2 : [θ]2 −→ ω we define d : [θ]2 −→ ω by letting

d(α, β) = supξ≤α

|ρ2(ξ, α)− ρ2(ξ, β)|.


(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 suchthat, if a set W ⊆ X \ x accumulates to x, then there is a subset of W of size ≤ θ thataccumulates to x.


(b) d(α, γ) ≤ d(α, β) + d(β, γ),

(c) d(α, β) ≤ d(α, γ) + d(β, γ).

Proof. The conclusion (a) follows from the fact that we allow ξ = α in thedefinition of d(α, β). The conclusions (b) and (c) are consequences of thetriangle inequalities of the `∞-norm and the fact that in both inequalitieswe have that the domain of functions on the left-hand side is included inthe domain of functions on the right-hand side. a13.11 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 (∗, ∗) ofS2

θ is equal to θ, giving us the proof of Theorem 13.8.

13.12 Lemma. (∗, ∗) ∈ Wθ but (∗, ∗) /∈ Wγ for all γ < θ.

Proof. To see that Wθ accumulates to (∗, ∗), let U2f be a given neighborhood

of (∗, ∗). Fix an unbounded set Γ ⊆ θ on which f is constant. By Theorem11.2 and Lemma 13.10(a) there exist α < β in Γ such that d(α, β) ≥ f(α) =f(β). Then ((α, d(α, β)), (β, d(α, β))) belongs to the intersection Wθ ∩ U2

f .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.

Suppose Wγ ∩ U2g 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 tothe following result of Gruenhage and Tanaka [29].

13.13 Corollary. The square of the sequential fan with ω1 edges is notcountably tight. a13.14 Question. What is the tightness of the square of the sequential fanwith ω2 edges?

13.15 Corollary. If a regular uncountable cardinal θ is not weakly compactin the constructible subuniverse then the square of the sequential fan with θedges has tightness equal to θ. a


The way we have proved that S2θ has tightness equal to θ is of indepen-

dent interest. We have found a point (∗, ∗) of S2θ and a set Wθ of isolated

points which accumulates to (∗, ∗) but has no other accumulation pointsand 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 sequenceω+1 and is closed under finite products, quotients and arbitrary topologicalsums.

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

13.17 Corollary. If every regular uncountable cardinal supports a nontriv-ial square sequence then every topological space X can be obtained fromthe 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 ofevery pair of spaces of tightness < κ is < κ and so C contains no spaceof tightness κ or larger. It has been shown in [17] that if the Lebesquemeasure extends to a countably additive measure on all sets of reals thenthere even is a countable space which does not belong to C. On the otherhand, 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 begenerated by the converging sequence by taking finite products, quotientsand arbitrary topological sums.

13.18 Remark. We have seen that the case θ = ω1 is quite special when oneconsiders the problem of existence of various nontrivial square sequences onθ. It should be noted that a similar result about the problem of the tightnessof S2

θ is not available. In particular, it is not known whether the P-idealdichotomy or a similar consistent hypothesis of set theory implies that thetightness of the square of, say, Sω2 is smaller than ω2. It is interestingthat considerably more is known about the dual question, the question ofinitial compactness of the Tychonoff cube Nθ. For example, if one definesBαβ = f ∈ Nθ : f(α), f(β) ≤ d(α, β) (α < β < θ) one gets an opencover of Nθ without a subcover of size < θ. However, for small θ suchas θ = ω2 one is able to find such a cover of Nθ without any additionalset-theoretic assumption and in particular without the assumption that θcarries a nontrivial square sequence.


14. Special square sequences

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

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

Today we know many more inner models with sufficient amount of finestructure necessary for building special square sequences. So the existenceof special square sequences, especially at successors of strong-limit singularcardinals, is tied to the existence of some other large cardinal axioms. Thereader is referred to the relevant chapters of this Handbook for the specificinformation. In this section we give the combinatorial analysis of walksalong 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 aspecial square sequence. Then there exist Cαν (α < θ, ν < κ) such that:

(1) Cαν ⊆ Cαµ for all α and ν < µ,

(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 ofthe square sequences can be chosen to be nonspecial. a14.3 Theorem. Suppose κ < θ are regular cardinals and that θ carries aspecial square sequence. Then there exist <ν (ν < κ) such that:

(i) <ν is a closed tree ordering of θ for each ν < κ,

(ii) ∈¹ θ =⋃

ν<κ <ν ,

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

a14.4 Lemma. The following are equivalent when θ is a successor of somecardinal 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 allα < θ.

14. Special square sequences 97

Proof. Let Dα (α < κ+) be a given special square sequence. By Lemma9.2 the corresponding tree (κ+, <2) can be decomposed into κ antichains solet f : κ+ −→ κ be a fixed map such that f−1(ξ) is a <2-antichain for allξ < κ. Let α < κ+ be a given limit ordinal. If Dα has a maximal limit pointα < α, let Cα = Dα \ α. Suppose now that ξ : ξ <2 α is unbounded inα and define a strictly increasing continuous sequence cα(ξ) (ξ < ν(α)) ofits elements as follows. Let cα(0) = minξ : ξ <2 α, cα(η) = supξ<ηcα(ξ)for η limit, and cα(ξ + 1) is the minimal <2-predecessor γ of α such thatγ > cα(ξ) and has the minimal f -image among all <2-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 easilychecked that this gives a square sequence Cα (α < κ+) with the propertythat tp(Cα) ≤ κ for all α < κ+. a

Square sequences Cα (α < κ+) that have the property tp(Cα) ≤ κ forall 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 thissection is to analyze the following distance function:

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

ρ(α, β) = maxtp(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), soby 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 thatare almost identical to the proof of the corresponding properties of, say, thefunction ρω discussed above in Section 12.


(a) ρ(α, γ) ≤ maxρ(α, β), ρ(β, γ),(b) ρ(α, β) ≤ maxρ(α, γ), ρ(β, γ).

aThe following immediate fact will also be quite useful.


14.7 Lemma. If α is a limit point of Cβ, then ρ(ξ, α) = ρ(ξ, β) for allξ < α.

aThe 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 thesquare sequence.

14.8 Lemma. ρ(α, γ) ≥ maxρ(α, β), ρ(β, γ) whenever α ≤ β ≤ γ and βbelongs 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 inthe 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 < βsuch 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 proofof the corresponding fact for the local version ρκ considered above in Section12 (the proof of Lemma 12.12).

14.10 Lemma. Pν(γ) = β < γ : ρ(β, γ) ≤ ν is a closed subset of γ forall γ < κ+ and ν < κ. a

The discussion of ρ : [κ+]2 −→ κ now splits naturally into two casesdepending on whether κ is a regular or a singular cardinal (with the casecf(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-sequenceand ρ : [κ+]2 −→ κ the corresponding ρ-function.

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

(I.183)

The following is an immediate consequence of the analysis of ρ given inthe 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,

(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 ≤ κthe ∈-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, wecan no longer insist that the trees (κ+, <ν) have heights ≤ κ. Using anadditional 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 byLemma 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 havea branch of size κ and let B be one such fixed branch (maximal chain). ByLemmas 14.5 and 14.10, if γ = sup(B) then B is a closed and unboundedsubset of γ of order-type κ. Since κ is regular and uncountable, Cγ ∩ Bis unbounded in Cγ , so in particular we can find α ∈ Cγ ∩ B such thattp(Cγ ∩ α) > ν. Reading off the definition of ρ(α, γ) we conclude thatρ(α, β) = tp(Cγ ∩ α) > ν. Similarly we can find a β > α belonging to theintersection of lim(Cγ) and B. Then Cβ = Cγ ∩ β so α ∈ Cβ and thereforeρ(α, β) = tp(Cβ ∩ α) > ν contradicting the fact that α <ν β. a15.4 Lemma. If κ > ω, then no tree (κ+, <ν) has a Souslin subtree ofheight κ.

Proof. Forcing with a Souslin subtree of (κ+, <ν) of height κ would producean 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 caseB would contain a ground model subset of size κ contradicting Lemma15.3. a15.5 Lemma. If κ > ω then for every ν < κ and every family A of κpairwise disjoint finite subsets of κ+ there exists A0 ⊆ A of size κ such thatfor 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. ByLemma 15.4, shrinking A, we may assume that a(i) (a ∈ A) is an antichainof (κ+, <ν) for all i < n. Going to a subfamily of A of equal size we mayassume to have a well-ordering <w of A with the property that if a <w bthen no node from a is above a node from b in the tree ordering <ν . Definef : [A]2 −→ 0 ∪ (n× n) by letting f(a, b) = 0 if a ∪ b is an <ν-antichain;otherwise, assuming a <w b, let f(a, b) = (i, j) where (i, j) is the minimalpair such that a(i) <ν b(j). By the Dushnik-Miller partition theorem ([18]),either there exists A0 ⊆ A of size κ such that f is constantly equal to 0on [A0]2 or there exist (i, j) ∈ n × n and an infinite A1 ⊆ A such that


f is constantly equal to (i, j) on [A1]2. The first alternative is what wewant, so let us see that the second one is impossible. Otherwise, choosea <w b <w c in A1. Then a(i) and b(i) are both <ν-dominated by c(j),so they must be <ν-comparable, contradicting our initial assumption aboutA. This completes the proof. a

The unboundedness property of Lemma 15.5 can be quite useful in de-signing forcing notions satisfying good chain conditions. Having such ap-plications in mind, we shall now work on refining further this kind of un-boundedness property of the ρ-function.

15.6 Lemma. Suppose κ > 0, let γ < κ+ and let αξ, βξ (ξ < κ) be a se-quence of pairwise disjoint elements of [κ+]≤2. Then there is an unboundedset Γ ⊆ κ such that ραξ, βη ≥ minραξ, γ, ρβη, γ for all ξ 6= η inΓ.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 sequencesαξ and βξ are actually indexed by Cα rather than κ. It is then clearwe 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 isf(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 Lemma14.6 one easily concludes that ρ(ξ, α) = ρ(ξ, γ) for any ξ < α such thatρ(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, ρα, β ismeant 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.


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 arechosen 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) ≤ maxρ(γ, α), ρ(αξ, α) = ρ(αξ, α), and (I.191)

ρ(γ, βξ) ≤ maxρ(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 chosenfrom 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 prop-erties of ρ from Lemma 14.6, one concludes that ρ(ξ, α) = ρ(α, βη) for allξ ∈ [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 ofΓ we may assume that Γ lies above γ and

ρ(αξ, α) > ρ(γ, α) for all ξ ∈ Γ. (I.195)


Applying the subadditive properties of ρ we get

ρ(γ, αξ) ≤ maxρ(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 arechosen 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.Subcase 2b: 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 arechosen 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. Thisfinishes the proof. a15.7 Lemma. Suppose κ is λ-inaccessible 34 for some λ < κ and that A isa family of size κ of subsets of κ+, all of size < λ. Then for every ordinalν < κ there is a subfamily B of A of size κ such that for all a and b in B:

(a) ρα, β > ν for all α ∈ a \ b and β ∈ b \ a.

(b) ρα, β ≥ minρα, γ, ρβ, γ for all α ∈ a \ b, β ∈ b \ a andγ ∈ a ∩ b.

Proof. Consider first the case λ = ω. Going to a subfamily, assume A formsa ∆-system with root r and that for some fixed integer n ≥ 1 and all a ∈ Awe have |a \ r| = n. By Lemma 15.5, going to a subfamily we may assumethat 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.


enumeration of a \ r. Going to a subfamily, we may assume that A canbe enumerated as aξ (ξ < κ) in such a way that aξ(i) (ξ < κ) is strictlyincreasing for all i < n. By n2 · |r| successive applications of Lemma 15.6we find a single unbounded set Γ ⊆ κ such that for all γ ∈ r, (i, j) ∈ n× nand ξ 6= η in Γ:

ρaξ(i), aη(j) ≥ minρaξ(i), γ, ρaη(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 membersof A have some fixed order-type, i.e. that for some fixed ordinal µ < λwe 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 anenumeration aξ (ξ < κ) of A such that aξ(i) (ξ < κ) is a strictly increasingsequence 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 asmaller cardinal we may assume that λ is a regular cardinal < κ. ApplyingLemma 14.9 simultaneously µ times gives us a regressive map f : κ −→ κsuch that for every η < κ of cofinality λ, all ξ ∈ [cδ(i)(f(η)), cδ(i)(η)) andi < 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) foreach i < µ. Pick a stationary set Γ of cofinality λ ordinals < κ such that fis constant on Γ. It follows that for all i, j < µ, if δ(i) = δ(j) then

ρaξ(i), aη(j) ≥ ρ(aξ(i), δ(i)) for ξ < η in Γ, (I.200)

ρaη(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

ρaξ(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

ρaξ(i), aη(j) ≥ minξ, η for all i, j < µ and ξ 6= η in Γ. (I.205)


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. Letαξ = 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 ofLemma 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 thisproof of Lemma 15.6, once the set Γ was obtained by a single application ofthe 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 ofthe closed and unbounded filter to Γ. In other words, we can now proceedas in the case λ = ω and successively apply this refining procedure for eachγ ∈ r and (i, j) ∈ µ×µ obtaining a set Γ0 ⊆ Γ with the property that Γ\Γ0

is nonstationary and

ρaξ(i), aη(j) ≥ minρaξ(i), γ, ρaη(j), γ for all γ ∈ r;i, j < µ and ξ 6= η in Γ0.

(I.206)

Since this is clearly equivalent to the conclusion (b) of Lemma 15.7, theproof is completed. a15.8 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 theformula

Dα, β = ξ ≤ minα, β : ρ(ξ, α) ≤ ρα, βas our definition of Dα, β when there is no any implicit assumption aboutthe 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 familyA of size κ of subsets of κ+, all of size < λ, there exists B ⊆ A of size κsuch 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 Dα, γ ∪Dβ, γ ⊆ Dα, β,(b) β > γ implies Dα, γ ⊆ Dα, β,(c) α > γ implies Dβ, γ ⊆ Dα, β,(d) γ > α, β implies Dα, γ ⊆ Dα, β or Dβ, γ ⊆ Dα, 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 theconclusion of 15.7(b), we have

ρα, β ≥ minρα, γ, ρβ, γ. (I.207)


a. Suppose α, β > γ. Note that in this case a single inequality ρ(γ, α) ≤ρα, β or ρ(γ, β) ≤ ρα, β given to us by (I.207) implies that we actuallyhave both inequalities simultaneously holding. The subadditivity of ρ givesus ρ(ξ, α) ≤ ρα, β, or equivalently ρ(ξ, β) ≤ ρα, β for any ξ ≤ γ withρ(ξ, γ) ≤ ρ(γ, α) or ρ(ξ, γ) ≤ ρ(γ, β). This is exactly the conclusion of15.9(a).

b. Suppose that β > γ > α. Using the subadditivity of ρ we see thatin both cases given to us by (I.207) we have that ρ(α, γ) ≤ ρα, β. So theinclusion Dα, γ ⊆ Dα, β follows immediately.

c. Suppose that α > γ > β. The conclusion Dβ, γ ⊆ Dα, β followsfrom the previous case by symmetry.

d. Suppose that γ > α, β. Then ρ(α, γ) ≤ ρα, β gives Dα, γ ⊆Dα, β while ρ(β, γ) ≤ ρα, β gives us Dβ, γ ⊆ Dα, β.

This completes the proof. a15.10 Remark. Note that minx, y ∈ Dx, y for every x, y ∈ [κ+]2, sothe conclusion (a) of Lemma 15.9 in particular means that γ < minα, βimplies γ ∈ Dα, β. In applications, one usually needs this consequence of15.9(a) rather than 15.9(a) itself.

15.11 Definition. The ∆-function of some family F of subsets of someordinal 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) = minξ : 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 objectsthat live on κ to objects on F . This is especially interesting when F is ofsize 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 shallnow 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-closedκ-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 pin P, if p is a one-to-one function from a subset of κ+ of size < λ into thefamily 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)


∆(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-chaincondition, let A be a given subset of P of size κ. By the assumption that κis λ-inaccessible, going to a subfamily of A of size κ, we may assume thatA forms a ∆-system, or 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 naturalisomorphism that fixes the roots. By Lemma 15.7 there exists B ⊆ A of sizeκ 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):

ρα, β > max(c), (I.210)

ρα, β ≥ minρα, γ, ρβ, γ. (I.211)

Suppose 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 suchthat ∆(q(β), q(γ)) ∈ b, the γ must be unique. Put r(β) = p(γ) ∪ q(β) inthis case. If such γ does not exist, put r(β) = q(β) ∪ max(c) + 1. Forα ∈ dom(p) \ dom(q), put r(α) = p(α). Note that this choice of r does notchange the behaviour of ∆ on pairs coming from [dom(p)]2 or [dom(q)]2, soin 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 ∈ dwe have that ∆(r(α), r(β)) = ∆(p(α), p(γ)) and that ∆(q(β), q(γ)) ∈ b. Itfollows that

∆(r(α), r(β)) = min∆(p(α), p(γ)), ∆(q(β), q(γ))≤ minρα, γ, ρβ, γ.

(I.212)

Applying (I.211) we get the desired inequality ∆(r(α), r(β)) ≤ ρα, β. Ifr(β) was chosen to be q(β)∪max(c)+1. Then ∆(r(α), r(β)) ≤ max(c)+1and this is ≤ ρα, β by (I.210).

Let G be the generic filter of P and let F = fα : α < κ+, where fα isthe 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 Mahlocardinal in the constructible universe, then there is a property K poset, forc-ing the Kurepa hypothesis. a


15.16 Remark. This is a variation on a result of Jensen who proved thatunder ¤ω1 there is a ccc poset forcing the Kurepa hypothesis. It should benoted that Jensen also proved (see [36]) that in the Levy collapse of a Mahlocardinal to ω2 there is no ccc poset forcing the Kurepa hypothesis. It shouldalso be noted that Velickovic [91] was the first to use ρ to reprove Jensen’sresult. He has also shown (see [91]) that ρ easily yields an example of afunction 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 functionthat satisfies some of the properties of the function D of Lemma 15.9 anda function that forms a ground for the well-known forcing construction ofa locally compact scattered topology on ω2 all of whose Cantor-Bendixsonranks are countable. We now use ρ to improve the chain-condition of theBaumgartner-Shelah forcing.

15.17 Theorem. If ¤ω1 holds then there is a property K forcing notion thatintroduces a locally compact scattered topology on ω2 all of whose Cantor-Bendixson ranks are countable.

Proof. Let P be the set of all p = 〈Dp,≤p,Mp〉 where Dp is a finite subset ofω2, where ≤p is a partial ordering of Dp compatible with the well-orderingand Mp : [Dp]2 −→ [ω2]<ω has the following properties:

Mpα, β ⊆ Dα, β ∩Dp, (I.213)

Mpα, β = α if α ≤p β and Mpα, β = β if β ≤p α, (I.214)

γ ≤p α, β for all γ ∈ Mpα, β, (I.215)

for every δ ≤p α, β there is γ ∈ Mpα, β such that δ ≤p γ. (I.216)

We let p ≤ q if and only if Dp ⊇ Dq, ≤p¹ Dq =≤q and Mp ¹ [Dq]2 =Mq. To verify that P satisfies Knaster’s chain condition, let A be a givenuncountable subset of P. Shrinking A we may assume that A forms a∆-system with root R and that the conditions of A generate isomorphicstructures over R. By Lemma 15.9 there is an uncountable subset B ofA such that for all p and q in B and all α ∈ Dp \ Dq, β ∈ Dq \ Dp andγ ∈ R = Dp ∩Dq:

α, β > γ implies Dα, γ ∪Dβ, γ ⊆ Dα, β, (I.217)β > γ implies Dα, γ ⊆ Dα, β, (I.218)α > γ implies Dβ, γ ⊆ Dα, β. (I.219)

Consider two conditions p and q from B. Let r = 〈Dr,≤r,Mr〉 be definedas follows: Dr = Dp ∪Dq and α ≤r β if and only if α ≤p β, or α ≤q β, orthere is γ ∈ R such that α ≤p γ ≤q β or α ≤q γ ≤p β. It is easily checkedthat ≤r is a partial ordering on Dr compatible with the well-ordering such


that ≤r¹ Dp =≤p and ≤r¹ Dq =≤q. Define Mr : [Dr]2 −→ [ω2]<ω byletting Mr ¹ [Dp]2 = Mp, Mr ¹ [Dq]2 = Mq, Mrα, β = α if α ≤r β,Mrα, β = β if β ≤r α and if α ∈ Dp \ R and β ∈ Dq \ R are ≤r-incomparable, set

Mrα, β = γ ∈ Dα, β ∩Dr : γ ≤r α, β. (I.220)

Only nontrivial to check is that a so defined r satisfies (I.216). So letδ ≤r α, β be given. If α, β ∈ Dp \ Dq and δ ∈ Dq \ Dp then there existγα, γβ ∈ R such that δ ≤q γα ≤p α and δ ≤q γβ ≤p β. By (I.216) for qthere exists γ ∈ Mqγα, γβ such that δ ≤q γ. Since Mqγα, γβ ⊆ R weconclude that γ ∈ R. Thus γ ≤p α, β, so by (I.216) there is γ ∈ Mpα, βsuch that γ ≤p γ and therefore δ ≤r γ. The case α ∈ R, β ∈ Dp \D andδ ∈ Dq \Dp is quite similar.

Consider now the case when α ∈ Dp \ Dq, β ∈ Dq \ Dp and they are≤r-incomparable, i.e. Mrα, β is given by (I.220). For symmetry we mayconsider only the subcase δ ≤r α, β with δ ∈ Dp. Then δ ≤p α andδ ≤p γ ≤q β for some γ ∈ R. By (I.216) for p there is γ ∈ Mpα, γ suchthat δ ≤p γ. It remains to show that γ was put in Mrα, β in (I.220), i.e.that γ belongs to Dα, β. Note that since ≤r agrees with the well-ordering,we have that β > γ, so by (I.218) we have that Dα, γ ⊆ Dα, β. By(I.213) for p we have that γ ∈ Mpα, γ ⊆ Dα, γ so we can conclude thatγ ∈ Dα, 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). Forα < ω2 set Bα = ξ : ξ ≤G α. Let τ be the topology on ω2 generated byBα (α < ω2) as a clopen subbasis. It is easily checked that τ is a Hausdorfflocally compact scattered topology on ω2. A simple density argument andinduction on α shows that the interval [ωα, ωα + ω) is the αth Cantor-Bendixson rank of the space (ω2, τ). a15.18 Definition. A function f : [ω2]2 −→ [ω2]≤ω has property ∆ if forevery uncountable set A of finite subsets of ω2 there exist a and b in A suchthat for all α ∈ a \ b, β ∈ b \ a and γ ∈ a ∩ b:

(a) α, β > γ implies γ ∈ fα, β,(b) β > γ implies fα, γ ⊆ fα, β,(c) α > γ implies fβ, γ ⊆ fα, β.

15.19 Remark. This definition is due to Baumgartner and Shelah [5] whohave used it in their well-known forcing construction. It should be notedthat they were also able to force a function with the property ∆ using aσ-closed ω2-cc poset. As shown above (a fact first checked in [5, p.129]),the function

Dα, β = ξ ≤ minα, β : ρ(ξ, α) ≤ ρα, β

16. Successors of singular cardinals 109

has property ∆. However, Lemma 15.9 shows that the function D has manyother properties that are of independent interest and that are likely to beneeded in other forcing constructions of this sort. The reader is referred topapers 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 :[κ+]2 −→ κ defined from a ¤κ-sequence Cα (α < κ+) can be a quite usefultool in stepping-up objects from κ to κ+. In this section we analyse thestepping-up power of ρ under the assumption that κ is a singular cardinalof cofinality ω. So let κn (n < ω) be a strictly increasing sequence of regularcardinals converging to κ fixed from now on. This immediately gives rise toa rather striking tree decomposition <n (n < ω) of the ∈-relation on κ+:

α <n β if and only if ρ(α, β) ≤ κn. (I.221)

16.1 Lemma.

(1) ∈¹ κ+ =⋃

n<ω <n,

(2) <n⊆<n+1,

(3) (κ+, <n) is a tree of height ≤ κ+n .

a16.2 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 thelexicographical ordering. Since L is a subset of ωκ, it has an order-densesubset of size κ, so in particular it contains no well-ordered subset of sizeκ+. The following result shows that, however, every subset of L of smallersize is the union of countably many well-ordered subsets.

16.3 Lemma. For each β < κ+, Lβ = fα : α < β can be decomposedinto countably many well-ordered subsets.

Proof. Let Lβn = fα : α ∈ Fn(β) for n < ω. Note that the projectionf 7−→ f ¹ (n + 1) is one-to-one on Lβn so each Lβn is lexicographicallywell-ordered. a


16.4 Remark. Note that K = (n, fα(n)) : n < ω : α < κ+ is a familyof 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 nowgive an application of the existence of such a family of countable sets, dueto Fremlin [25]. This requires a few standard notions from measure theoryand real analysis.

16.5 Definition. A Radon measure space is a Hausdorff topological spaceX with a σ-finite measure µ defined on a field of subsets of X containingopen sets with the property that the measure of every set is equal to thesupremum 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 whichcompletely 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 tounbounded sets. In case D and E are ideals of subsets, this is equivalent tosaying 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 andE ≤ 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 Maharamtype κ is Tukey equivalent to the product N × [κ]ω where N is the ideal ofmeasure zero subsets of the unit interval.

Proof. We shall give an argument for the case X = 0, 1κ and µ the Haarmeasure of X, referring the reader to the general result from [25] that thenull ideal of any atomless probability space of Maharam type κ is Tukeyequivalent to the null ideal Nκ of 0, 1κ. The only nontrivial direction isto produce a Tukey map from Nκ into Nω × [κ]ω. Let K ⊆ [κ]ω of sizecf[κ]ω with the property that K ¹ Γ = a ∩ Γ : a ∈ K is countable forall countable Γ ⊆ κ. Let h : K −→ [κ]ω be a mapping whose range iscofinal 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 theproperty that C(b) = a ∈ C : a ⊆ b is countable for every b ∈ [κ]ω. Forc ∈ C fix a bijection ec : c −→ ω. This gives us a way to define a functionπc mapping cylinders of 0, 1κ with supports included in c into 2ω in thenatural way. Let N cl

κ be the family of all null subsets of 0, 1κ that arecountably supported cylinders of 0, 1κ. Clearly, N cl

κ is cofinal in Nκ, soit suffices to produce a Tukey map f : N cl

κ −→ Nω × [κ]ω. Given N ∈ N clκ ,

find c(N) ∈ C such that supp(N) ⊆ c(N) and put

f(N) = (πc(N)′′N, c(N)).


It is straightforward to check that f maps unbounded subsets of N clκ to

unbounded subsets of the product Nω × [κ]ω. a16.8 Remark. Note that the hypothesis of Fremlin’s theorem is true when[κ]ω has cofinality κ and so it is always satisfied for κ < ℵω. By remark 16.4the hypothesis of Theorem 16.7 is satisfied for all κ if one assumes ¤κ andcf[κ]ω = κ+ for every κ of uncountable cofinality. Note that the transferfrom a locally countable K to a cofinal C = a ∪ h(a) : a ∈ K in the aboveproof does not preserve local countability. To obtain a locally countablecofinal 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 countableand cofinal in [S]ω then we call it a cofinal Kurepa family (cofinal K-familyin short). Two cofinal K-families H and K are compatible if H ∩K ∈ H∩Kfor all H ∈ H and K ∈ K. We say that K extends H if they are compatibleand if H ⊆ K.

16.10 Remark. Note that the size of any cofinal K-family K on a set Sis equal to the cofinality of [S]ω. Note also that for every X ⊆ S there isY ⊇ 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 sequenceKn (n < ω) of comparable cofinal K-families with domains included inθ which are closed under ∪, ∩ and \ can be extended to a single cofinalK-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 thatcf[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 thereis a strictly increasing sequence δξ (ξ < θ+) of ordinals < θ such thatKn ¹ δξ ⊆ Kn for all ξ < θ+ and n < ω. Recursively on ξ < θ+ weconstruct a chain Hξ (ξ < θ+) of cofinal K-families as follows. If ξ = 0or ξ = η + 1 for some η, using CK(δξ) we can find a cofinal K-familyHξ on δξ extending Hξ−1 and Kn ¹ δξ (n < ω). If ξ has uncountablecofinality 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 Hξ and Kn ¹ δξ (n < ω). If ξ has countable cofinality, pick asequence ξn converging to ξ and use CK(δξ) to find a cofinal K-familyHξ on δξ extending Hξn (n < ω) and Kn (n < ω). When the recursion isdone, set H =

⋃ξ<θ+ Hξ. Then H is a cofinal K-family on θ+ extending

Kn (n < ω). a16.13 Corollary. For each n < ω there is a cofinal Kurepa family on ωn.

a


16.14 Definition. Let κ be a cardinal of cofinality ω. A Jensen matrix onκ+ is a matrix Jαn (α < κ+, n < ω) of subsets of κ+ with the followingproperties, where κn (n < ω) is some fixed increasing sequence of cardinalsconverging 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,

(3)⋃

n<ω[Jβn]ω =⋃

α<β

⋃n<ω[Jαn]ω whenever cf(β) > ω,

(4) [κ+]ω =⋃

α<κ+

⋃n<ω[Jαn]ω.

16.15 Remark. The notion of a Jensen matrix is the combinatorial essencebehind Silver’s proof of Jensen’s model-theoretic two-cardinal transfer the-orem in the constructible universe (see [35, appendix]), so the matrix couldequally well be called ’Silver matrix’. It has been implicitly or explicitlyused on several places in the literature. The reader is referred to the paperof Foreman and Magidor [24] which gives a quite complete discussion of thisnotion and its occurrence in the literature.

16.16 Lemma. Suppose some cardinal κ of countable cofinality carries aJensen matrix Jαn (α < κ+, n < ω) relative to some sequence of cardinalsκn (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-familieswith domains included in κ+. Given Jαn, there is a natural continuouschain Jξ

αn (ξ < ω1) of subsets of κ+ of size ≤ κn such that J0αn = Jαn

and Jξ+1αn equal to the union of all K ∈ ⋃

n<ω Kn which intersect Jξαn. Let

J∗αn =⋃

ξ<ω1Jξ

αn. It is easily seen that J∗αn (α < κ+, n < ω) is also a Jensenmatrix. 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 andn < ω using CK(κn) we can find a cofinal K-family Hαn with domain J∗αn

compatible with Hαm (m < n), H(α−1)m (m < ω) and Km ¹ J∗αn (m < ω).If cf(α) = ω let αn (n < ω) be an increasing sequence of ordinals convergingto α. Using CK(κn) we can find a cofinal K-family Hαn which extendsHαm (m < n), Km ¹ J∗αn (m < ω) and each of the families Hαik (i < ω, k <ω and J∗αik

⊆ J∗αn). Finally suppose that cf(α) > ω. For n < ω, set

Hαn = [J∗αn]ω ∩ (⋃

ξ<α

⋃m<ω

Hξm).

Using the properties of the Jensen matrix (especially (3)) as well as thecompatibility of Hξm (ξ < α,m < ω) one easily checks that Hαn is a cofinalK-family with domain J∗αn which extends each member of Hαm (m < n)


and Km ¹ J∗αn (m < ω) and which is compatible with all of the previouslyconstructed families Hξm (ξ < α,m < ω). When the recursion is done weset

H =⋃

α<κ+

⋃n<ω

Hαn.

Using the property (4) of J∗αn (α < κ+, n < ω), it follows easily that H is acofinal K-family on κ+ extending Kn (n < ω). a16.17 Theorem. If a Jensen matrix exists on any successor of a cardinalof cofinality ω, then a cofinal Kurepa family exists on any domain. a

The ρ-function ρ : [κ+]2 −→ 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 propertiesnot captured by the definition of a Jensen matrix. If one additionally hasa sequence aα (α < κ+) of countable subsets of κ+ that is cofinal in [κ+]ω

one can extend the matrix (I.222) as follows:

Mβn =⋃

α<nβ

(aα ∪ α) (β < κ+, n < ω). (I.223)

(Recall that <n is the tree ordering on κ+ defined by the formula α <n β iffρ(α, β) ≤ κn where κn is a fixed increasing sequence of cardinals convergingto κ.) The matrix Mβn (β < κ+, n < ω) has properties not captured byDefinition 16.14 that are of independent interest.

16.18 Lemma.

(1) α <n β implies Mαn ⊆ Mβn,

(2) Mαm ⊆ Mαn whenever m < n,

(3) if β = supα : α <n β then Mβn =⋃

α<nβ Mαn,

(4) every countable subset of κ+ is covered by some Mβn.

(5) M = Mβn : β < κ+, n < ω is a locally countable family if we havestarted with a locally countable K = aα : α < κ+.

a16.19 Remark. One can think of the matrix M = Mβn : β < κ+, n < ωas a version of a ’morass’ for the singular cardinal κ (see [92]). It would beinteresting to see how far one can go in this analogy. We give a few appli-cations just to illustrate the usefulness of the families we have constructedso far.


16.20 Definition. A Bernstein decomposition of a topological space X isa function f : X −→ 2N with the property that f takes all the values from2N on any subset of X homeomorphic to the Cantor set.

16.21 Remark. The classical construction of Bernstein [8] can be inter-preted 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 Nesetril and Rodl (see [55]). In this contextMalykhin 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 se-quences seems natural. In fact, the first Bernstein decompositions of anarbitrary Hausdorff space have been constructed using ¤κ and κω = κ+

for every κ > c of cofinality ω by Weiss [96] and Wolfsdorf [97]. We shallnow see that cofinal K-families are quite natural tools in constructions ofBernstein decompositions.

16.22 Theorem. Suppose every regular θ > c supports a cofinal Kurepafamily of size θ. Then every Hausdorff space admits a Bernstein decompo-sition.

Proof. Let us say that a subset S of a topological space X is sequentiallyclosed in X if it contains all limit points of sequences xn ⊆ S that convergein X. We say S is σ-sequentially closed in X if it can be decomposedinto countably many sequentially closed sets. For a given cardinal θ, letB(θ) denote the statement that for every Hausdorff space X of size ≤ θ,every σ-sequentially closed subset S of X and every Bernstein decompositionf : S −→ 2N there is a Bernstein decomposition g : X −→ 2N extendingf . Using the fact that a Hausdorff space of size ≤ c has at most c copiesof the Cantor set and following the idea of Bernstein’s original proof, onesees that B(c) holds. Note also that by our assumption θℵ0 = θ for everycardinal θ ≥ c of uncountable cofinality. So if we have B(θ) for such θone proves B(θ+) by constructing a chain fξ (ξ < θ+) of partial Bernsteindecompositions on any Hausdorff topology τ on θ+ using the fact that theset C of all δ < θ+ which are σ-sequentially closed in (θ+, τ) is closed andunbounded in θ+. Thus if δξ (ξ < θ+) enumerates C increasingly we willhave gξ : δξ −→ 2N and gξ compatible with the given partial Bernsteindecomposition f of a given σ-sequentially closed set S ⊆ θ+. Note thatS ∩ δξ will be σ-sequentially closed, so the use of B(δξ) to get fξ will bepossible. This inductive procedure encounters a crucial difficulty only whenθ = κ+ for some κ > c of cofinality ω. This is where a cofinal K-family Kon κ+ is useful. Let τ be a given Hausdorff topology on θ+ and for K ∈ Klet K ′ be the collection of all limits of τ -convergent sequences of elementsof 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 Γ′ will be equal to the union of allK ′ (K ∈ K ¹ Γ). So if |Γ| > c is a cardinal of uncountable cofinality then


|Γ′| = |Γ|. It follows that every δ < θ+ with the property that δ = lim δn forsome sequence δn (n < ω) of ordinals ≤ δ with the property that K ¹ δn ⊆ Kfor all n can be decomposed into countably many sets Γδn (n < ω) such thatfor 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 suchδ < θ+ is closed and unbounded in θ+ so as above we use our inductivehypothesis B(θ) (θ < κ) to construct a chain gξ : δξ −→ 2N (ξ < θ+) ofBernstein decompositions. This completes the proof. a

It is interesting that various less pathological classes of spaces admit alocal 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 inparticular, any metric space that carries a cofinal Kurepa family admits aBernstein decomposition.

Proof. So let K be a fixed well-founded cofinal K-family on Γ and let <w

be a well-ordering of K compatible with ⊆. For K ∈ K fix a Bernsteindecomposition fK : XK −→ 2N where XK = x ∈ X : supp(x) ⊆ K.Define f : X −→ 2N by letting f(x) = fK(x) where K is the <w-minimalK ∈ K such that supp(x) ⊆ K. It is easily checked that f is a Bernsteindecomposition of the sigma-product. a16.24 Definition. Recall the notion of a coherent family of partial functionsindexed by some ideal I, a family of the form fa : a −→ ω (a ∈ I) with theproperty that x ∈ a ∩ b : fa(x) 6= fb(x) is finite for all a, b ∈ I.

It can be seen (see [89]) that the P-ideal dichotomy (see 4.45) has a stronginfluence on such families provided I is a P-ideal of countable subsets ofsome set Γ.

16.25 Theorem. Assuming the P-ideal dichotomy, for every coherent fam-ily of functions fa : a −→ ω (a ∈ I) indexed by some P-ideal I of countablesubsets of some set Γ, either

1. There is an uncountable ∆ ⊆ Γ such that fa ¹ ∆ is finite-to-one forall 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 canfind an a in I such that b \ a is finite and fa is finite-to-one on b. To seethat 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 consistingof all mappings x such that supp(x) = ξ ∈ Γ : x(ξ) 6= 0 is countable. The set Γ is itsindex-set .


each n fix a member an of I such that fanis 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 thatfor all n, bn \ a is finite and that fa is finite-to-one on bn. For n < ω, let

b∗n = ξ ∈ bn ∩ a : fa(ξ) > n.

Then b∗n is a cofinite subset of bn for each n, so if we set b to be equal to theunion of the b∗n’s, we get a subset of a which almost includes each bn andon which fa is finite-to-one. It follows that b belongs to L. This finishesthe proof that L is a P-ideal. Applying the P-ideal dichotomy to L, we getthe two alternatives that translate into the alternatives 1 and 2 of Theorem16.25. a

This leads to the natural question whether for any set Γ one can constructa family fa : a −→ ω of finite-to-one mappings indexed by [Γ]ω. Thisquestion was answered by Koszmider [40] using the notion of a Jensen matrixdiscussed above. We shall present Koszmider’s result using the notion of acofinal Kurepa family instead.

16.26 Theorem. If Γ carries a cofinal Kurepa family then there is a co-herent family fa : a −→ ω (a ∈ [Γ]ω) of finite-to-one mappings.

Proof. Let K be a fixed well-founded cofinal K-family on Γ and let <w bea well-ordering of K compatible with ⊆. It suffices to produce a coherentfamily of finite-to-one mappings indexed by K. This is done by inductionon <w. Suppose K ∈ K and fH : H −→ ω is determined for all H ∈ Kwith H <w K. Let Hn (n < ω) be a sequence of elements of K that are<w K and have the property that for every H ∈ K with H <w K there isn < ω such that H ∩K =∗ Hn ∩K. So it suffices to construct a finite-to-one fK : K −→ ω which coheres with each fHn (n < ω), a straightforwardtask. a

16.27 Corollary. For every nonnegative integer n there is a coherent familyfa : a −→ ω (a ∈ [ωn]ω) 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 ofone-to-one mappings fa : a −→ ω (a ∈ [c+]ω). We finish this section with atypical application of coherent families of finite-to-one mappings discoveredby Scheepers [61].

16.29 Theorem. If there is a coherent family fa : a −→ ω (a ∈ [Γ]ω) offinite-to-one mappings, then there is F : [[Γ]ω]2 −→ [Γ]<ω with the propertythat for every strictly ⊆-increasing sequence an (n < ω) of countable subsetsof Γ, the union of F (an, an+1) (n < ω) covers the union of an (n < ω).

17. The oscillation mapping 117

Proof. For a ∈ [Γ]ω 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 aproper subset of b. Choose Φ : ωω −→ ωω with the property that x <∗ yimplies Φ(y) <∗ Φ(x), where <∗ is the ordering of eventual dominance onωω (i.e. x <∗ y if x(n) < y(n) for all but finitely many n’s). Define anotherfamily of functions ga : a −→ ω (a ∈ [Γ]ω) by letting

ga(ξ) = Φ(xa)(fa(ξ)). (I.224)

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]ω. (I.225)

So if an (n < ω) is a strictly ⊆-increasing sequence of countable subsets of Γand 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). a16.30 Remark. Note that if κ is a singular cardinal of cofinality ω with theproperty that cf[θ]ω < κ for all θ < κ, then the existence of a cofinal Kurepafamily on κ+ implies the existence of a Jensen matrix on κ+. So these twonotions appear to be quite close to each other. The three basic propertiesof the function ρ : [κ+] −→ κ (14.5 and 14.6(a),(b)) seem much strongerin view of the fact that the linear ordering as in 16.3 cannot exist for κabove a supercompact cardinal and the fact that Foreman and Magidor [24]have produced a model with a supercompact cardinal that carries a Jensenmatrix on any successor of a singular cardinal of cofinality ω. Note thatChang’s conjecture of the form (κ+, κ) ³ (ω1, ω) for some singular κ ofcofinality ω implies that every locally countable family K ⊆ [κ]ω must havesize ≤ 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 stillunknown whether the conclusion of Theorem 16.29 can be proved withoutadditional 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 byosc(x, y) = |x \ (sup(x ∩ y) + 1)/ ∼ |,

where ∼ is the equivalence relation on x\ (sup(x∩y)+1) defined by lettingα ∼ β iff the closed interval determined by α and β contains no point from


y. So, if x and y are disjoint, osc(x, y) is simply the number of convex piecesthe set x is split by the set y. The oscillation mapping has proven to be auseful device in various schemes for coding information. It usefulness in agiven context depend very much of the corresponding ’oscillation theory’,a set of definitions and lemmas that disclose when it is possible to achievea given number as oscillation between two sets x and y in a given familyX . The following definition reveals the notion of largeness relevant to theoscillation theory that we develop in this section.

17.1 Definition. A family X ⊆ P(θ) is unbounded if for every closed andunbounded subset C of θ there exist x ∈ X and an increasing sequenceδn : n < ω ⊆ C such that sup(x ∩ δn) < δn and [δn, δn+1) ∩ x 6= ∅ for alln < ω.

This notion of unboundedness has proven to be the key behind a numberof 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 theoryknown so far (see [79],[80, §1] and [88]), though in this section we shall onlyconsider 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. Thefollowing is the basic result about the behaviour of the oscillation mappingin this context.

17.2 Lemma. If X is an unbounded subfamily of K(θ) then for every pos-itive integer n there exist x and y in X such that osc(x, y) = n.

Proof. Choose an ∈-chainM of length θ of elementary submodels M of somelarge 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 andunbounded 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)

LetJ0 = [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 ofclosed intervals of θ such that I0 = J0 and such that there is x = xI in Xsuch that

x ⊆ I0 ∪ I1 ∪ . . . ∪ In, (I.233)


max(Ii) ∈ x for all i ≤ n. (I.234)

Clearly, Fn ∈ M0 and 〈J0, J1, . . . , Jn〉 ∈ Fn. Let Fn−1 be the collectionof all 〈I0, . . . , In−1〉 such that for every α < θ there exists an interval Iof θ above α such that 〈I0, . . . , In−1, I〉 ∈ Fn. Note that Fn−1 ∈ M0 andthat using the elementarity of Mn−1 one proves that 〈J0, . . . , Jn−1〉 ∈ Fn−1.Let Fn−2 be the collection of all 〈I0, . . . , In−2〉 such that for every α < θthere exists interval I above α such that 〈I0, . . . , In−2, I〉 ∈ Fn−1. Thenagain Fn−2 ∈ M0 and using the elementarity of Mn−2 one shows thatthe restriction 〈J0, . . . , Jn−2〉 belongs to Fn−2. Continuing in this way weconstruct families Fn,Fn−1, . . . ,F0 such that:

Fi ∈ M0 for all i ≤ n, (I.235)

F0 = J0. (I.236)

〈I0, . . . , In−i〉 ∈ Fn−i implies that for all α < θ there existsan interval I above α such that 〈I0, . . . , In−i, I〉 ∈ Fn−i+1.

(I.237)

Recursively in i ≤ n we choose intervals I0 < I1 < . . . < In such that:

〈I0, . . . , Ii〉 ∈ 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 elemen-tarity of the submodels Mi−1 and the condition (I.237). By the definition〈I0, . . . In〉 ∈ 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 forall but finitely many positive integers n there exist x ∈ X and y ∈ Y suchthat osc(x, y) = n. a

Recall the notion of a nontrivial C-sequence Cα (α < θ) on θ from Section11, a C-sequence with the property that for every closed and unboundedsubset C of θ there is a limit point δ of C such that C ∩ δ * Cα for allα < θ.

17.4 Definition. For a subset D of θ let lim(D) denote the set of all α < θwith the property that α = sup(D∩α). A subsequence Cα (α ∈ Γ) of someC-sequence Cα (α < θ) is stationary if the union of all lim(Cα) (α ∈ Γ) is astationary subset of θ.


17.5 Lemma. A stationary subsequence of a nontrivial C-sequence on θ isan unbounded family of subsets of θ.

Proof. Let Cα (α ∈ Γ) be a given stationary subsequence of a nontrivialC-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 assumptionthat Cα (α ∈ Γ) is a nontrivial C-subsequence, we can find ξ ∈ ∆ ∩ lim(C)such that

C ∩ [η, ξ) * Cαξfor all η < ξ. (I.241)

If such ξ cannot be found using the stationarity of the set ∆ ∩ lim(C) wewould be able to use the Pressing Down Lemma on the regressive mappingthat would give us an η < ξ violating (I.241) and get that a tail of Ctrivializes Cα (α ∈ Γ). Using (I.241) and the fact that Cαξ

∩ξ is unboundedin ξ we can find a strictly increasing sequence δn (n < ω) of elements of(C∩ξ)\Cαξ

such that [δn, δn+1)∩Cαξ6= ∅ for all n. So the set Cαξ

satisfiesthe conclusion of Definition 17.1 for the given closed and unbounded setC. a

Recall that Qθ denotes the set of all finite sequences of ordinals < θ andthat we consider it ordered by the right lexicographical ordering. We needthe 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)

α〈ξ〉 = the ξth member of Cα if ξ < tp(Cα); otherwise α〈ξ〉 = α, (I.245)

αta〈ξ〉 = (αt)〈ξ〉. (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 theC-sequence, each member of the trace Tr(α, β) = β0, β1, . . . , βn has theform 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 forevery unbounded set Γ ⊆ θ and positive integer n there exist α < β in Γ andt v ρ0(α, β) such that


(a) osc(αt, βt) = n,

(b) osc(αs, βs) = 1 for all s < t.

Proof. Choose an elementary submodel M of some large enough structureof the form Hλ such that δ = M ∩ θ ∈ θ and M contains all the relevantobjects. 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 ifCβk−1

∩ δ is unbounded; otherwise k = k. Recall our implicit assumptionabout essentially all C-sequences that we consider here: if γ is a limit ordinalthen 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 mustbe that βk−1 is a limit ordinal and δ = sup(Cβk−1

∩ δ) or βk−1 = δ + 1.Itfollows 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 = βt¹ifor i ≤ k. Let Γ0 be the set of all β ∈ Γ such that:

β = β∅ > βt¹1 > . . . > βt, (I.247)

sup(Cβt¹i ∩ βt) = sup(Cβi∩ βk) for all i < k. (I.248)

Note that Γ0 ∈ M and β ∈ Γ0. Let ∆ = βt : β ∈ Γ0. Then ∆ ∈ Mand βk ∈ ∆. By the very choice of k, δ belongs to lim(Cβk

), so using theelementarity of M one concludes that Cδ (δ ∈ ∆) is a stationary subsequenceof our original sequence. Shrinking ∆ a bit if necessary, we may assume thatfor every γ ∈ ∆ an α ∈ Γ0 with αt = γ can be found below the next memberof ∆ above γ. By Lemma 17.5, Cδ (δ ∈ ∆) is an unbounded family of subsetsof θ and so we can apply the basic oscillation Lemma 17.2 and find γ < δsuch that osc(αt, βt) = n. Using the fact α < δ and the fact that α and βsatisfy (I.248) we conclude that osc(αt¹i, βt¹i) = 1 for all i < k. This finishesthe proof. a17.10 Corollary. Suppose a regular uncountable cardinal θ carries a non-trivial C-sequence. Then there is f : [θ]2 −→ ω which takes all the valuesfrom 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


17.11 Remark. The class of all regular cardinals θ that carry a nontrivialC-sequence is quite extensive. It includes not only all successor cardinalsbut also some inaccessible as well as hyperinaccessible cardinals such as forexample, first inaccessible or first Mahlo cardinals. In view of the well-known Ramsey-theoretic characterization of weak compactness, Corollary17.10 leads us to the following natural question.

17.12 Question. Can the weak compactness of a strong limit regularuncountable cardinal be characterized by the fact that for every f : [θ]2 −→ω there exists an unbounded set Γ ⊆ θ such that f ′′[Γ]2 6= ω? This is truewhen ω is replaced by 2, but can any other number beside 2 be used in thischaracterization?

18. The square-bracket operation

In this Section we show that the basic idea of the square-bracket operationon ω1 introduced in Definition 7.3 extends to a general setting on an ar-bitrary uncountable regular cardinal θ that carries a nontrivial C-sequenceCα (α < θ). The basic idea is based on the oscillation map defined in theprevious section and, in particular, on the property of this map exposed inTheorem 17.9: for α < β < θ we set

[αβ] = βt, where t v ρ0(α, β) is such that osc(αt, βt) ≥ 2but osc(αs, βs) = 1 for all s < t; if such t does not exist, welet [αβ] = α.

(I.250)

Thus, [αβ] is the first place visited by β on its walk to α where a nontrivialoscillation with the corresponding step of α occurs. What Theorem 17.9is telling us is that the nontrivial oscillation indeed happens most of thetime. Results that would say that the set of values [αβ] : α, β ∈ [Γ]2is 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-bracketoperation on ω1. It turns out that this is indeed possible and to describe itwe 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 avoidsa 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 inΓ 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 anylimit 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, aninteger 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 assumefor 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 astationary subset of the cardinal, in particular, must be nontrivial. ThusCδ (δ ∈ Ω0) is a stationary subsequence of a nontrivial C-sequence, andso by Lemma 17.5, unbounded as a family of subsets of θ. By the basicoscillation 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 familyof size θ consisting of pairwise disjoint finite sets, all of some fixed sizen. 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 notsurprising that [··] strongly depends on the behaviour of Tr. For examplethe following should be clear from the proof of Lemma 18.2.

18.4 Lemma. The following are equivalent for a given C-sequence and aset Ω:

(a) Ω \ [αβ] : α, β ∈ [Γ]2 is nonstationary in θ,

(b) Ω \⋃Tr(α, β) : α, β ∈ [Γ]2 is nonstationary in θ.

aThis fact suggests the following definition.

18.5 Definition. The trace filter of a given C-sequence Cα (α < θ) is thenormal filter on θ generated by sets of the form

⋃Tr(α, β) : α, β ∈ [Γ]2where Γ is an unbounded subset of θ.


18.6 Remark. Having a proper (i.e. 6= P(θ)) trace filter is a strengtheningof the nontriviality requirement on a given C-sequence Cα (α < θ). Forexample, if a C-sequence avoids a stationary set Ω ⊆ θ, then its trace filteris nontrivial and in fact no stationary subset of Γ is a member of it. Note thefollowing analogue of Lemma 18.4: the trace filter of a given C-sequenceis the normal filter generated by sets of the form [αβ] : α, β ∈ [Γ]2where Γ is an unbounded subset of θ. So to obtain the analogues of theresults of Section 7 about the square-bracket operation on ω1 one needs aC-sequence Cα (α < θ) on θ whose trace filter is not only nontrivial butalso not θ-saturated, i.e. it allows a family of θ pairwise disjoint positivesets. It turns out that the hypothesis of Lemma 18.2 is sufficient for bothof these conclusions.

18.7 Lemma. If a C-sequence on θ avoids a stationary subset of θ, thenthere exist θ pairwise disjoint subsets of θ that are positive with respect toits trace filter. 36

Proof. This follows from the well-known fact (see [37]) that if there is anormal, nontrivial and θ-saturated filter on θ, then for every stationaryΩ ⊆ θ there exists λ < θ such that Ω ∩ λ is stationary in λ (and the factthat the stationary set which is avoided by the C-sequence does not reflectin this way). a

18.8 Corollary. If a regular cardinal θ admits a nonreflecting stationarysubset then there is c : [θ]2 −→ θ which takes all the values from θ on anyset 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 −→ θwith the property that the ∗-preimage of each point from θ is positive withrespect to the trace filter of the square sequence. In other words c is equalto the composition of [··] and ∗, i.e. c(α, β) = [αβ]∗. Note that, as in Section7, the property of the square-bracket operation from Lemma 18.3 leads tothe following rigidity result which corresponds to Lemma 7.10.

18.9 Lemma. The algebraic structure (θ, [··], ∗) has no nontrivial automor-phisms. a18.10 Remark. Note that every θ which is a successor of a regular cardinalκ admits a nonreflecting stationary set. For example Ω = δ < θ : cf δ = κis such a set. Thus any C-sequence on θ that avoids Ω leads to a squarebracket operation which allows analogues of all the results from Section 7about 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 .


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 andSteprans [69] in order to construct a family of nonisomorphic extraspecialp-groups that contain no large abelian subgroups.

Suppose now that θ = κ+ for some regular cardinal κ and fix a C-sequenceCα (α < κ+) on κ+ such that tp(Cα) ≤ κ for all α, or equivalently, suchthat Cα (α < κ+) avoids the set Ωκ = δ < κ+ : cf δ = κ. Let [··] bethe corresponding square-bracket operation. Let λ be the minimal cardinalsuch that 2λ ≥ κ+. Choose a sequence rξ (ξ < κ+) of distinct subsets ofλ. Let H be the collection of all maps h : P(D(h)) −→ κ+ where D(h) isa finite subset of λ. Let π : κ+ −→ H be a map with the property thatπ−1(h) ∩Ωκ is stationary for all h ∈ H. Finally, define an operation [[··]] onκ+ 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 disjointfinite 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 alli < n. a18.12 Remark. Shelah and Steprans consider the function Φ : [κ+]2 −→ 2defined by Φ(α, β) = 0 iff [[αβ]] = 0 in [69] which they then incorporateinto a general construction, due to Ehrenfeucht and Faber, to obtain anextraspecial p-group GΓ associated to essentially every unbounded subsetΓ of κ+. They show that GΓ contains no abelian subgroup of size κ+ usingnot 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 thetheme 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 Abe a family of θ pairwise disjoint finite subsets of θ, all of some fixed size n.Then for every stationary Γ0 ⊆ Γ there exist s, t ∈ ωn and a positive integerk such that for every l < ω there exist a < b 37 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 thatmax(a) < min(b).


(2) ρ0(a(i), b(j)) = ρ0(δ0, b(j))aρ0(δ1, δ0)a . . . aρ0(δl, δl−1)aρ0(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) =⋂

γ<δ

〈ρ2(a(i), δ) : i < n〉 : a ∈ A ¹ [γ, δ), (I.255)

Tδ(∆) =⋂

γ<δ

ρ2(α, δ) : α ∈ ∆ ∩ [γ, δ). (I.256)

Note that Sδ(A) and Tδ(∆) range over subsets of ωn and ω, respectively, soif δ has uncountable cofinality and if A and ∆ are unbounded in δ then bothof these sets, Sδ(A) and Tδ(∆), are nonempty. Let Γ0 be a given stationarysubset of Γ. Since θ has size bigger than the continuum, starting from Γ0

one can find an infinite sequence Γ0 ⊇ Γ1 ⊇ . . . of subsets of Γ such thatΓω =

⋂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) andk an arbitrary member of Tδ(Γω). To check that these objects satisfy theconclusion 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 passthrough δ0. Since k ∈ Tδ(Γω) ⊆ Tδ(Γl) we can find δ1 ∈ Γl ∩ (γ0, δ0) suchthat ρ2(δ1, δ0) = k. Choose γ1 ∈ (γ0, δ1) such that the walk from δ0 to anyα ∈ [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 soon. 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 δi to some α ∈ [γi+1, δi+1)passes through δi+1.

(I.261)

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) firstpasses through all the δi (i ≤ l), so ρ0(a(i), b(j)) is simply equal to theconcatenation of ρ0(δ0, b(j)), ρ0(δ1, δ0), . . ., ρ0(δl, δl−1), ρ0(a(i), δl). Thistogether with (I.260) gives us the three conclusions of Theorem 18.13. a


From now on θ is assumed to be a fixed cardinal satisfying the hypothesesof Theorem 18.13. It turns out that Theorem 18.13 gives us a way to defineanother square-bracket operation which has complex behavior not only onsquares of unbounded subsets of θ but also on rectangles formed by twounbounded subsets of θ. To define this new operation we choose a mappingh : ω −→ ω such that:

for every k,m, n, p < ω and s ∈ ωn there exist l < ω suchthat h(m + l · k + s(i)) = m + p for all i < n. (I.262)

18.14 Definition. [··]h : [θ]2 −→ θ is defined by letting [αβ]h = βt wheret = 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 isclear that Theorem 18.13 and the choice of h in (I.262) give us the followingconclusion.

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 almostevery 38 δ ∈ Γ has the form [a(0)β]h = [a(1)β]h = . . . = [a(n − 1)β]h forsome a ∈ A, β ∈ Ω, a < β. a

In fact, one can get a projection of this square-bracket operation withseemingly even more complex behavior. Keeping the notation of Theorem18.13, pick a function ξ 7−→ ξ∗ from θ to ω such that ξ ∈ Γ : ξ∗ = n isstationary for all n. This gives us a way to consider the following projectionof the trace function Tr∗ : [θ]2 −→ ω<ω:

Tr∗(α, β) = 〈min(Cβ \ α)∗〉aTr∗(α, min(Cβ \ α)), (I.263)

where we stipulate that Tr∗(γ, γ) = 〈γ∗〉 for all γ. It is clear that the proofof Theorem 18.13 allows us to add the following conclusions:

18.13∗ Theorem. Under the hypothesis of Theorem 18.13, its conclusioncan 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) isbigger than the maximal term of any of the sequences Tr∗(δ0, b(j)) orTr∗(a(i), δl) for i, j < n.

a18.16 Definition. For α < β < θ, let [αβ]∗ = βt for t the minimal initialpart of ρ0(α, β) such that β∗t = max(Tr∗(α, β)).

38Here ’almost every’ is to be interpreted by ’all except a nonstationary set’.


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. Notethat combining the conclusions (1)-(5) of Theorem 18.13(∗) we get:

18.13∗∗ Theorem. Under the hypothesis of Theorem 18.13, its conclusioncan be extended by adding the following:

(6) [a(i)b(j)]∗ = [δ1δ0]∗ for all i, j < n.

aHaving 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 onecan find a < b in A such that [a(i)b(j)]∗h = δ for all i, j < n. a18.19 Remark. Composing [··]∗h with a mapping π : θ −→ θ with theproperty that π−1(ξ) ∩ Γ is stationary for all ξ < θ, one gets a projectionof [··]∗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 atleast 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 followingproperty:

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 inA such that [[a(i)b(j)]]∗h = q(i, j) for all i, j < n. a18.21 Remark. The first example of a cardinal with such a complex bi-nary operation was given by the author [79] using the oscillation mappingdescribed above in Section 17. It was the cardinal b, the minimal cardinalityof an unbounded subset of ωω under the ordering of eventual dominance.The oscillation mapping restricted to some well-ordered unbounded subsetW of ωω is perhaps still the most interesting example of this kind due tothe fact that its properties are preserved in forcing extensions that do notchange the unboundedness of W (although they can collapse cardinals andtherefore destroy the properties of the square-bracket operations on them).This absoluteness of osc is the key feature behind its applications in variouscoding procedures (see, e.g. [85]).

18.22 Theorem. For every regular cardinal κ of size at least the contin-uum, the κ+-chain condition is not productive, i.e., there exist two partiallyordered sets P0 and P1 satisfying the κ+-chain condition but their productP0 × 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 〈α, α〉 (α < κ+) ofpairwise incomparable conditions. a18.23 Remark. Theorem 18.22 is due to Shelah [65] who proved it usingsimilar methods. The first ZFC-examples of non-productiveness of the κ+-chain condition were given by the author in [77] using what is today knownunder the name ’pcf theory’. After the full development of pcf theory itbecame apparent that the basic construction from [77] applies to every suc-cessor of a singular cardinal [66]. A quite different class of cardinals θ withθ-cc non-productive was given by the author in [76]. For example, θ = cf cis 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 theθ-chain condition is productive. Is θ necessarily a weakly compact cardinal?

19. Unbounded functions on successors of reg-ular cardinals

In this section κ is a regular cardinal and Cα (α < κ+) is a fixed sequencewith tp(Cα) ≤ κ for all α < κ+. Define ρ∗ : [κ+]2 −→ κ by

ρ∗(α, β) = suptp(Cβ ∩ α), ρ∗(α, min(Cβ \ α)),ρ∗(ξ, α) : ξ ∈ Cβ ∩ α,

(I.264)

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) ≤ νhas size no more than |ν|+ ℵ0. a

The proof of the following subadditivity properties of ρ∗ is very similarto the proof of the corresponding fact for the function ρ from Section 14.


(a) ρ∗(α, γ) ≤ maxρ∗(α, β), ρ∗(β, γ),(b) ρ∗(α, β) ≤ maxρ∗(α, γ), ρ∗(β, γ).

a


We mention a typical application of this function to the problem of exis-tance of partial square sequences which, for example, have some applicationsin the pcf theory (see [9]).

19.3 Theorem. For every regular uncountable cardinal λ < κ and station-ary Γ ⊆ δ < κ+ : cf(δ) = λ, there is a stationary set Σ ⊆ Γ and asequence 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 supremaof sequences of size < λ. Then there is ν, µ < κ and stationary Σ ⊆ Γ suchthat ν(δ) = ν and tp(P<ν(δ)) = µ for all δ ∈ Σ. Let C be a fixed closedand unbounded subset of µ 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, thatthe inequalities 19.2(a),(b) are particularly useful, when κ has cofinalityω. 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 bechanged to ω, while preserving all cardinals.

19.4 Theorem. In any cardinal-preserving extension of the universe, whichhas no new bounded subsets of κ, but in which κ has a cofinal ω-sequencediagonalizing the filter of closed and unbounded subsets of κ restricted tothe 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,

(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 boundedsubsets. Clearly, Dα contains a closed unbounded subset of κ, restricted tocofinality > ω ordinals. Note that

ν ∈ Dβ and ρ∗(α, β) < ν imply that ν ∈ Dα. (I.265)


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 isn < ω such that νm ∈ Dα for all m ≥ n. Let n(α) be the minimal suchinteger 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 thecase 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 forall ν ≥ νn and therefore Dγ(β) \ νn ⊆ Dα. It follows that n(α) ≥ n andtherefore that γ(α) = α. Hence, Cαn = Cβn ∩ α. If Cβn was defined ac-cording to the case I.268 then our assumption α ∈ lim(Cβn) and the factthat n ≥ n(β) means that Cαn was defined according to I.267. Similarly asabove we conclude that actually α ∈ P<νn(β) which gives us n ≥ n(α) andγ(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 thenso is Cαn and so Pνn(γ(α)) ∩ α and Pνn(γ(β)) ∩ α are two σ-closed andunbounded subsets of α. So their intersection is unbounded in α which by19.2(a),(b) gives us that Pνn(γ(α)) ∩ α = Pνn(γ(β)) ∩ α. It follows thatCαn = Cβn ∩ α in this case. Suppose now that Cβn was defined accordingto I.268. Then γ = β satisfies n ≥ n(γ) and I.266. So Cαn was definedaccording to the case I.267. It follows that Pνn(γ(α))∩α and Pνn(β)∩α aretwo σ-closed and unbounded subsets of α so their intersection is unboundedin α. Applying 19.2(a),(b) we get that Pνn(γ(α))∩α = Pνn(β)∩α. It followsthat Cαn = Cβn ∩ α in this case as well. a19.5 Remark. The combinatorial principle appearing in the statement ofTheorem 19.4 is a member of a family of square principles that has beenstudied systematically by Schimmerling and others (see e.g. [62]). It isdefinitely a principle sufficient for all of the applications of ¤κ appearing inSection 16 above.

19.6 Definition. A function f : [κ+]2 −→ κ is unbounded if f ′′[Γ]2 isunbounded in κ for every Γ ⊆ κ+ of size κ. We shall say that such an f is


strongly unbounded if for every family A of size κ+, consisting of pairwisedisjoint finite subsets of κ+, and every ν < κ there exists A0 ⊆ A of size κsuch 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. itsatisfies 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. Thenour assumption about f satisfying 19.2(a) and (b) reduces to the fact thateach <ν is a tree ordering on κ+ compatible with the usual ordering onκ+. Note that the unboundedness property of f is preserved by any forcingnotion satisfying the κ-chain condition, so in particular no tree (κ+, <ν) cancontain a Souslin subtree of height κ. In the proof of Lemma 15.5 abovewe have seen that this property of (κ+, <ν) alone is sufficient to concludethat every family A of κ many pairwise disjoint subsets of κ+ contains asubfamily 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 beingstrongly unbounded. a19.8 Lemma. The following are equivalent:

(1) There is a structure (κ+, κ,<, Rn)n<ω with no elementary substruc-ture 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 letf : [κ+]2 −→ κ be a given unbounded function. Increasing f we may assumethat each of its sections fα : α −→ κ is one-to-one. For a set Γ ⊆ κ+, letthe f -closure of Γ, CLf (Γ), be the minimal ∆ ⊇ Γ with the property thatf ′′[∆]2 ⊆ ∆ and f−1

α (ξ) ∈ ∆, whenever α ∈ ∆ and ξ < sup(∆ ∩ κ). Definee : [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) ≤ maxe(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). Bythe 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) ≤ maxe(α, γ), e(β, γ) whenever α ≤ β ≤ γ. (I.271)


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). Inthe 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 eis unbounded. So let Γ be a given subset of κ+ of order-type κ and let ν < κbe a given ordinal. Since e(α, β) ≥ ρ∗(α, β) for all α < β < κ+ we mayassume that ρ∗ is bounded on Γ, or more precisely, that for some ν ≤ ν andall α < β in Γ, ρ∗(α, β) ≤ ν. For α ∈ Γ, let β(α) be the minimal point of Γabove α. Then there is µ ≤ ν and ∆ ⊆ Γ of size κ such that ρ∗(α, β(α)) = µfor all α ∈ ∆. Note that the sequence of sets Pµ(α) (α ∈ ∆) forms a chainunder end-extension. Let Γ∗ be the union of this sequence of sets. Since fis 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. Thisfinishes the proof. a19.9 Remark. Recall that Chang’s conjecture is the model-theoretic trans-fer principle asserting that every structure of the form (ω2, ω1, <, . . .) witha countable signature has an uncountable elementary submodel B with theproperty that B ∩ ω1 is countable. This principle shows up in many con-siderations including the first two uncountable cardinals ω1 and ω2. Forexample, it is known that it is preserved by ccc forcing extensions, that itholds in the Silver collapse of an ω1-Erdos cardinal, and that it in turn im-plies that ω2 is an ω1-Erdos cardinal 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 thestatement that there exists e : [ω2]2 −→ ω1 such that:

(a) e(α, γ) ≤ maxe(α, β), e(β, γ) whenever α ≤ β ≤ γ,

(b) e(α, β) ≤ maxe(α, γ), e(β, γ) whenever α ≤ β ≤ γ,

(c) For every uncountable family A of pairwise disjoint finite subsets ofω2 and every ν < ω1 there exists an uncountable A0 ⊆ A such thate(α, β) > ν whenever α ∈ a and β ∈ b for every a 6= b ∈ A0.

a19.11 Remark. Note that if a mapping e : [ω2]2 −→ ω1 has properties(a),(b) and (c) of Corollary 19.10, then De : [ω2]2 −→ [ω2]ℵ0 defined byDeα, β = ξ ≤ minα, β : e(ξ, α) ≤ eα, β satisfies the weak form ofDefinition 15.18, where only the conclusion (a) is kept. The full form ofDefinition 15.18, however, cannot be achieved assuming only the negationof Chang’s conjecture. This shows that the function ρ, based on a ¤ω1-sequence, is a considerably deeper object than an e : [ω2]2 −→ ω1 satisfying19.10(a),(b),(c).


Recall that the successor of the continuum is characterized as the min-imal cardinal θ with the property that every f : [θ]2 −→ ω is constant onthe square of some infinite set. We shall now see that in slightly weak-ening the partition property by replacing squares by rectangles one gets acharacterization of a quite different sort. To see this, let us use the arrownotation (

θθ

)−→

(ωω

)1,1

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 thatω1 < θ2 ≤ c+. The following result shows that θ2 can have the minimalpossible value ω2, as well as that θ2 can be considerably smaller than thecontinuum.

19.12 Theorem. Chang’s conjecture is equivalent to the statement that(ω2

ω2

)−→

(ωω

)1,1

ω

holds in every ccc forcing extension.

Proof. We have already noted that Chang’s conjecture is preserved by cccforcing extensions, so in order to prove the direct implication, it sufficesto show that Chang’s conjecture implies that every f : ω2 × ω2 −→ ω isconstant 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 Mof 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 thatby the elementarity of M , for every β ∈ ∆0 there exist unboundedly manyα < δ such that f(α, β0) = f(α, β) = i. So we can choose α1 < δ aboveα0 such that the set ∆1 = β ∈ ∆0 : f(α1, β) = i is uncountable. Thenagain for every β ∈ ∆1 there exist unboundedly many α < δ such thatf(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 thatf(αn, βm) = i for all n, m < ω.

Assume now that Chang’s conjecture fails and fix e : [ω2]2 −→ ω asin Corollary 19.10. We shall describe a ccc forcing notion P which forcesan f : ω2 × ω2 −→ ω which is not constant on the product of any twoinfinite subsets of ω2. The conditions of P are simply maps of the formp : [Dp]2 −→ ω where Dp is some finite subset of ω2 such that

p(ξ, α) 6= p(ξ, β) whenever ξ < α < β belong to Dp andhave the property that e(ξ, α) > e(α, β). (I.272)


We order P by letting p extend q if and only if p extends q as a functionand

p(ξ, α) 6= p(ξ, β) whenever α < β belong to Dq and ξ be-longs to Dp \Dq.

(I.273)

Let A be a given uncountable subset of P. Refining A, we may assume thatDp (p ∈ A) forms a ∆-system with root D and that Fp = e′′[Dp]2 (p ∈ A)form an increasing ∆-system on ω1 with root F . Moreover, we may assumethat for every p and q in A, the finite structures they generate together withe are isomorphic via the isomorphism that fixes D and F . By 19.10(c) thereis 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 seethis, extend p∪ q to a mapping r : [Dp ∪Dq]2 −→ ω which is one-to-one onnew pairs, avoiding the old values. To see that r is indeed a member of P weneed to check (I.272) for r. So let ξ < α < β be three given ordinals fromDr 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.that they belong to [Dp]2 ∪ [Dq]2. Since p and q both satisfy (I.272), wemay consider only the case when (say) α ∈ Dp \ Dq, β ∈ Dq \ Dp andξ ∈ 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) belongsto F contradicting (I.274). To see that r extends p and q, by symmetry itsuffices to check that r extends, say, p. So suppose ξ < α < β are givensuch that α < β belong to Dp and ξ belongs to Dq \Dp. We need to showthat r(ξ, α) 6= r(ξ, β). By the choice on new pairs, this is automatic if oneof the two pairs is new. So we may consider the case when both α andβ are in D. Since r(ξ, α) = q(ξ, α) and r(ξ, β) = q(ξ, β) in this case theinequality r(ξ, α) 6= r(ξ, β) would follow from (I.272) for q if we show thate(ξ, α) > e(α, β). Otherwise e(ξ, α) would belong to the root F and by thefact 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 theproperty 19.10(c) of e since ξ(s) 6= ξ(t) when s 6= t.

Forcing with P gives us a mapping g : [ω2]2 −→ ω with the property thatξ < α : g(ξ, α) = g(ξ, β) is finite for all α < β < ω2. Define f : ω2×ω2 −→ω by letting f(α, β) = 2gα, β + 2 when α < β, f(α, β) = 2gα, β + 1when α > β, and f(α, β) = 0 when α = β. Then f is not constant on anyproduct of two infinite sets. a

19.13 Remark. The relative size of θ2 (or its higher-dimensional analoguesθ3, θ4, . . .) in comparison to the sequence of cardinals ω2, ω3, ω4, . . . is ofconsiderable interest, both in Set Theory and Model Theory (see e.g. [63],


[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 thestatement that within every decomposition of the usual ordering on ω2 asan increasing chain of tree orderings, one of the trees has an uncountablechain. 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 tothis question is equivalent to a different well-known combinatorial statementabout ω2 rather than Chang’s conjecture itself. Recall that f : [κ+]2 −→ κis transitive if

f(α, γ) ≤ maxf(α, β), f(β, γ) whenever α ≤ β ≤ γ. (I.275)

Given a transitive map f : [κ+]2 −→ κ, one defines ρf : [κ+]2 −→ κrecursively 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),3. ρf (α, min(Cβ \ α)),4. ρf (ξ, α) (ξ ∈ Cβ ∩ α).

(I.276)

19.15 Lemma. For every transitive map f : [κ+]2 −→ κ the correspondingρf : [κ+]2 −→ κ has the following properties:

(a) ρf (α, γ) ≤ maxρf (α, β), ρf (β, γ) whenever α ≤ β ≤ γ,

(b) ρf (α, β) ≤ maxρf (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 proofsfor the function ρ itself and so we avoid repetition. The inequality (d) isdeduced from (I.276) by using the transitivity of f as follows:

ρf (α, β) ≥ maxi<n

f(βi+1, βi) ≥ f(α, β),

where β = β0 > β1 > . . . > βn−1 > βn = β is the walk from β to α alongthe given C-sequence. a19.16 Remark. Transitive maps are frequently used combinatorial objects,especially when one works with quotient structures. Adding the extra sub-additivity condition 19.15(b), one obtains a considerably more subtle objectwhich is much less understood.


19.17 Definition. Let fα : κ −→ κ (α < κ+) be a given sequence offunctions 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 mapf : [κ+]2 −→ κ is defined by

f(α, β) = minµ < κ : fα(ν) < fβ(ν) for all ν ≥ µ. (I.277)

Let ρf be the corresponding ρ-function that dominates this particular fand for ν < κ let <f

ν be the corresponding tree ordering of κ+, i.e.

α <fν 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 ofeverywhere dominance. Then for every ν < κ the tree (κ+, <f

ν ) has height≤ g(ν).

Proof. Let P be a maximal chain of (κ+, <fν ). 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 sequencefα (α < κ+) in the ordering <∗ of eventual dominance, then the newsequence fα = minfα, g (α < κ+) is still strictly <∗-increasing but nowbounded by g even in the ordering of everywhere dominance. So this provesthe following result of Galvin (see [34],[59]).

19.19 Corollary. The following two conditions are equivalent for everyregular cardinal κ.

(1) There is a sequence fα : κ −→ κ (α < κ+) which is strictly increasingand bounded in the ordering of eventual dominance.

(2) The usual order-relation of κ+ can be decomposed into an increasingκ-sequence of tree orderings of heights < κ.

a19.20 Remark. The assertion that every strictly <∗-increasing κ+-sequen-ce of functions from κ to κ is <∗-unbounded is strictly weaker than Chang’sconjecture and in the literature it is usually referred to as weak Chang’sconjecture. This statement still has considerable large cardinal strength(see [14]). Also note the following consequence of Corollary 19.19 which canbe 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 theusual order-relation of κ+ can be decomposed into an increasing κ-sequenceof tree orderings of heights < κ. a


20. Higher dimensions

The reader must have noticed already that in this Chapter so far, we haveonly considered functions of the form f : [θ]2 −→ I or equivalently se-quences fα : α −→ I (α < θ) of one-place functions. To obtain analo-gous results about functions defined on higher-dimensional cubes [θ]n oneusually develops some form of stepping-up procedure that lifts a functionof 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 sequenceeα : α −→ θ (α < θ+) of one-to-one mappings and wishes to defineg : [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 andthen 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 re-strictions eδ ¹ (Γ∩δ) (δ ∈ Γ) may not cohere, so we cannot produce a subsetof θ that would correspond to Γ and on which we would like to apply someproperty of f . It turns out that the definition (I.279) is basically correct ex-cept that we need to replace eαn by eτ(αn−2,αn−1,αn), where τ : [θ+]3 −→ 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-sequenceCα (α < θ+) on θ+. The following result shows that if the C-sequence iscarefully 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 alongthe ¤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 DownLemma there is a stationary set Σ ⊆ lim(Cε) and δ ∈ Cε such that f(δ) ≤ δfor all δ ∈ Σ. Shrinking Σ we may assume that ξγ < δ whenever γ < δ aremembers 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 definitionof εδ for δ ∈ lim(Cε), δ belongs to Cεδ

and so by the implicit assumption

20. Higher dimensions 139

that nonlimit points of Cεδare 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. aRecall 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 ofordinals < θ. It will be convenient to identify Qθ with θ via, for example,Godel’s coding of sequences of ordinals by ordinals. This identification givesus a well-ordering <w of Qθ of length θ . We use this identification to lift-up of an arbitrary map f : [θ]n −→ I (really, f : [Qθ]n −→ I) to a mapg : [θ+]n+1 −→ I by the following formula:

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 particularexample.

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 amap 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 inplace of n.

Proof. We assume that the domain of f is actually equal to the set [Qθ]n

rather than [θ]n and define g by

g(α0, . . . , αn−1, αn) = (ρ0)−1ε (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 tosatisfy this condition. To check (2), let Γ ⊆ θ be a given set of size κ. ByLemma 20.1, shrinking Γ we may assume that Γ has order-type κ and thatif 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

g(α0, . . . , αn−1, αn) = (ρ0)−1ε (f(ρ0(α0, ε), . . . , ρ0(αn−1, ε))). (I.285)


Shrinking Γ further we assume that the mapping (ρ0)ε, as a map fromthe ordered set (θ+,∈) into the ordered set (Qθ, <w) is strictly increasingwhen restricted to Γ. Given an ordinal ν < κ, we apply (2) for f to theset ∆ = ρ0(α, ε) : α ∈ Γ and find a ∈ [∆]n and A ∈ f(a) such thattp(A) ≥ ν and A ⊆ ∆. Let α0, . . . , αn−1 be the increasing enumerationof the preimage (ρ0)ε

−1(a) and pick αn ∈ Γ above αn−1. Let B be thepreimage (ρ0)ε

−1(A). Then B ∈ g(α0, . . . , αn−1, αn), tp(B) ≥ ν and B ⊆ Γ.This finishes the proof. a20.3 Remark. It should be noted that Velleman [93] was the first to at-tempt to step-up the combinatorial property appearing in Theorem 20.2using his version of the gap-2 morass. Unfortunately the stepping up pro-cedure of [93] worked only up to the value n = 3 so it remains unclear if thehigher-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,Hand K of Example 7.16 for ω2 instead of ω1. To make this precise, wefirst fix a ¤ω1 -sequence Cα (α < ω2) and consider the corresponding mapsρ0 : [ω2]2 −→ ω<ω

1 and ρ : [ω2]2 −→ ω1. Let ρ : [ω2]2 −→ ω1 be defined by

ρ(α, β) = 2ρ(α,β) · (2 · tpξ ≤ α : ρ(ξ, α) ≤ ρ(α, β)+ 1).

Then for α < ω2, the mapping ρα : α −→ ω1 that sends ξ to ρ(ξ, α) is1-1 and we have the following coherence property:

ρα = ρβ ¹ α whenever α ∈ lim(Cβ). (I.286)

20.4 Definition. The ternary operation [[[··]]] : [ω2]3 −→ ω2 is defined asfollows:

[[[αβγ]]] = (ρτ(α,β,γ))−1[[ρτ(α,β,γ)(α) ρτ(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 disjointsubsets 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 havethat 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. aAs in Example 7.16, let us identify ω2 and 3 × [ω2]<ω and view [[[··]]] as

having the range 3× [ω2]<ω rather than ω2. Let [[[··]]]0 and [[[··]]]1 be the twoprojections of [[[··]]]. Set

G = G ∈ [ω2]<ω : [[[αβγ]]]0 = 0 for all α, β, γ ∈ [G]3,


H = H ∈ [ω2]<ω : [[[αβγ]]]0 = 1 for all α, β, γ ∈ [H]3.Note the following immediate consequence of Lemma 20.5.

20.6 Lemma. For every uncountable family A of disjoint 2-element subsetsof ω2 there is an arbitrarily large finite subset B of A with b(0) : b ∈ B ∈ Gand 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. Weuse it to define an analogue of the family K of Example 7.16 as follows. Inthis case, let K be the collection of all finite sets ai : i < n of elements of[ω2]2 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 =⋃

l<i

al. (I.289)

The following property of K is a consequence of Lemma 20.5 and thesubadditivity of ρ.

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 thereexist ξ0 < ξ1 < . . . < ξn−1 < ω1 such that aξi : i < n ∈ K. a

The particular definition of K is made in order to have the followingproperty of K which is analogous to (I.79) of Section 7:

If K and L are two distinct members of K, then there are nomore than 9 ordinals α such that α, β ∈ K and α, γ ∈ Lfor some β 6= γ.

(I.290)

The definitions of G,H and K immediately lead to the following analoguesof (I.77) and (I.78):

G and H contain all singletons, are closed under subsetsand are 3-orthogonal to each other. (I.291)

G and H are both 5-orthogonal to the family of all unionsof members of K.

(I.292)

So we can proceed as in Example 7.16 and for a function x from ω2 intoR, define


‖x‖H,2 = sup(∑

α∈Hx(α)2)

12 : H ∈ H,

‖x‖K,2 = sup(∑

α,β∈K(x(α)− x(β))2)

12 : K ∈ K.

Let ‖·‖ = max‖·‖∞, ‖·‖H,2, ‖·‖K,2, let E2 = x : ‖x‖ < ∞, and let E2

be the closure of the linear span of 1α : α ∈ ω2 inside (E2, ‖ · ‖). Then asin Example 7.16 one shows, using the properties of G, H and K listed abovein Lemmas 20.6 and 20.7 and in (I.287-I.292), that E2 is a Banach spacewith 1α : α ∈ ω2 as its transitive basis and with the property that everybounded linear operator T : E2 −→ E2 can be written as S +D, where S isan operator with separable range and D is a diagonal operator relative tothe 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 projectionson 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, usingLemma 20.6, that T can be written as a sum of one such projection and adiagonal operator relative to 1α : α ∈ ω2.

Using the interpolation method of [11], one can turn E2 into a reflexiveexample. Thus, we have the following:

20.8 Example. Assuming ¤ω1 , there is a reflexive Banach space E with atransitive basis of type ω2 with the property that every bounded operatorT : E −→ E can be written as a sum of an operator with a separable rangeand a diagonal operator (relative to the basis) with only countably manychanges of constants. a20.9 Remark. In [42], P.Koszmider has shown that such a space cannotbe constructed on the basis of the usual axioms of set theory. We referthe reader to that paper for more details about these kinds of examples ofBanach spaces.

For the rest of this section we shall examine the stepping-up method withless 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 suchthat log θ+ = θ. 39

(1) There is a substructure of the form (θ++, θ+, <, . . .) with no substruc-ture B of size θ+ with B ∩ θ+ of size θ.

(2) There is f : [θ++]3 −→ θ+ which takes all the possible values on thecube of any subset Γ of θ++ of size θ+.

39log κ = minλ : 2λ ≥ κ.


Proof. To prove the nontrivial direction from (1) to (2), we use Lemma 19.8and choose a strongly unbounded and subadditive e : [θ++]2 −→ θ+. Wealso 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 inI.264. Finally we choose a one-to-one sequence rα (α < θ++) of elements of0, 1θ+

and consider the corresponding function ∆ : [θ++]2 −→ θ+:

∆(α, β) = ∆(rα, rβ) = minν : rα(ν) 6= rβ(ν). (I.293)

The definition of f : [θ++]3 −→ θ is given according to the following tworules 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 or rα >lex

rβ >lex rγ , let

f(α, β, γ) = min(Pν(∆(β, γ)) \∆(α, β)),

where ν = ρ∗(minξ ≤ ∆(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), whereγ 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++ ofsize θ+ there exist α < β < γ in Γ such that f(α, β, γ) ∈ Σ. Clearly we mayassume 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 assumethat

∆(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 tripleα < β < γ of elements of Γ. Let ∆ = ∆(rα, rβ) : α, β ∈ [Γ]2. Then ∆is a subset of θ+ of size θ+ and the range of f on [Γ]3 includes the rangeof [··] on [∆]2, where [··] is the square-bracket operation on θ+ defined asfollows:

[αβ] = min(Pν(β) \ α), (I.295)

where ν = ρ∗(minξ ≤ α : ρ∗(ξ, α) 6= ρ∗(ξ, β), β). Note that this is theexact analogue of the square-bracket operation that has been analysed abovein Section 7. So exactly as in Section 7, one establishes the following fact.

For ∆ ⊆ θ+ unbounded, the set [αβ] : α, β ∈ [∆]2contains almost all ordinals < θ+ of cofinality θ.

(I.296)


Case 2: RΓ contains no subsets of size θ+ which are lexicographicallywell-ordered or conversely well-ordered. For t ∈ 0, 1<θ+

, let Γt = α ∈Γ : t ⊆ rα. Let T be the set of all t ∈ 0, 1<θ+

for which Γt has size θ+.Shrinking Γ we may assume that T is either a downwards closed subtree of0, 1<θ+

of size θ and height λ < θ+, or a downwards closed Aronszajnsubtree of 0, 1<θ+

of height θ+. For β ∈ Γ, let

eβ′′Γ+ = eβ, γ : γ ∈ Γ, rβ <lex rγ. (I.297)

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 enoughstructure of the form Hκ such that δ = M ∩ θ+ ∈ Σ and M contains allthe relevant objects. By the elementarity of M , strong unboundedness ofe and the fact that T contains no θ+-chain, there exist β and γ in Γ \Msuch that rγ <lex rβ , ε = eβ, γ ≥ δ and ∆(rβ , rγ) < δ. Let ν = ρ∗(δ, ε).Note that there exist θ many ξ > ∆(rβ , rγ) such that (rβ ¹ ξ)a0 belongs toM 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 alsotrue in case T has height λ < θ+ and has size θ, since from our assumptionlog θ+ = θ the ordinal λ must have cofinality exactly θ. So by the property19.1 of ρ∗ there is one such ξ such that ν = ρ∗(ξ, ε) ≥ ν. Let t = (rβ ¹ ξ)a0.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 mustbe an α in Γt ∩M such that e(α, β) > sup(Pν(ε)∩ δ). Since e(α, β) belongsto eα

′′Γ+, a set which, by the assumption of Case 2.1, is bounded in θ+, weconclude that e(α, β) < δ. It follows that f(α, β, γ) is defined according toRule 2, and therefore

f(α, β, γ) = min(Pν(ε) \ e(α, β)) = δ. (I.298)

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 andδ = M ∩ θ+ ∈ Σ be chosen as before. If T is Aronszajn, using elementarityof M , there must be β ∈ Γ ∩M and δ < δ of cofinality θ such that thereexist unboundedly many ξ below δ such that the set

Ωξ = eβ, χ : χ ∈ Γ, rβ <lex rχ,∆(rβ , rχ) = ξ (I.299)

is unbounded in θ+. If T is a tree of height λ < θ+, δ = λ will satisfy thisfor all but θ many β ∈ Γ, so in particular, we can pick one such β ∈ Γ∩M .Choose γ ∈ Γ such that rβ <lex rγ , ε = e(β, γ) ≥ δ and ∆(rβ , rγ) < δ. Letν = ρ∗(δ, ε). By the property 19.1 of ρ∗, there is µ < δ such that ρ∗(ξ, ε) ≥ νfor all ξ ∈ [µ, δ). Choose ξ ∈ [µ, δ) above ∆(rβ , rγ) such that the set Ωξ isunbounded in θ+. Let ν = ρ∗(ξ, ε). Then ν ≥ ν. The set Pν(ε) ∩ δ beingof size < θ is bounded in δ. Since Ωξ ∈ M , there is α ∈ Γ ∩ M above β


such that ∆(rβ , rα) = ξ, rβ <lex rα and e(β, α) > sup(Pν(ε)∩δ). Note thatrα <lex rγ , so the definition of fα, β, γ obeys Rule 2. Applying this rule,we get

fα, β, γ = min(Pν(ε) \ e(β, α)) = δ. (I.300)

This finishes the proof of Theorem 20.10. a20.11 Theorem. If θ is a regular strong limit cardinal carrying a nonre-flecting stationary set, then there is f : [θ+]3 −→ θ which takes all the valuesfrom θ on the cube of any subset of θ+ of size θ.

Proof. This is really a corollary of the proof of Theorem 20.10, so let usonly indicate the adjustments. By Corollary 19.21 and Lemma 19.7, wecan choose a strongly unbounded subadditive map e : [θ+]2 −→ θ. Bythe assumption about θ we can choose a C-sequence Cα (α < θ) avoidinga stationary set Σ ⊆ θ and consider the corresponding notion of a walk,trace, ρ0-function and the square-bracket operation [··] as defined in (I.250)above. As in the proof of Theorem 20.10, we choose a one-to-one sequencerα (α < θ+) of elements of 0, 1θ and consider the corresponding function∆ : [θ+]2 −→ θ:

∆(α, β) = ∆(rα, rβ) = minν < 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 , or rα >lex rβ >lex rγ , let

fα, β, γ = [∆(α, β)∆(β, γ)].

Rule 2: If α ∈ x is such that rα is lexicographically between the othertwo rξ’s for ξ ∈ x, if β ∈ x \ α is such that ∆(rα, rβ) > ∆(rα, rγ), whereγ is the remaining element of x, and they do not satisfy the conditions ofRule 1, set

fα, β, γ = min(Tr(∆(α, β), eβ, γ) \ eα, β),i.e. fα, β, γ is the minimal point on the trace of the walk from eβ, γto ∆(α, β) above the ordinal eα, β; if such a point does not exist, setfα, β, γ = 0.

The proof of the Theorem is finished once we show that for every sta-tionary Ω ⊆ Σ and every Γ ⊆ θ+ of size θ, there exist x ∈ [Γ]3 such thatf(x) ∈ Ω. This is done, as in the proof of Theorem 20.10, by consideringthe two cases. Case 1 is reduced to the property 18.2 of the square-bracketoperation. The treatment of Case 2 is similar and a bit simpler than in theproof of Theorem 20.10, since the case that T is of height < θ is impossibledue to our assumption that θ is a strong limit cardinal. This finishes theproof. a


Since log ω1 = ω, the following is an immediate consequence of Theorem20.10.

20.12 Theorem. Chang’s conjecture is equivalent to the statement that forevery f : [ω2]3 −→ ω1 there is uncountable Γ ⊆ ω2 such that f ′′[Γ]3 6= ω1.

a20.13 Remark. Since this same statement is stronger for functions fromhigher dimensional cubes [ω2]n into ω1 the Theorem 20.12 shows that theyare all equivalent to Chang’s conjecture. Note also that n = 3 is the minimaldimension for which this equivalence holds, since the case n = 2 follows fromthe Continuum Hypothesis, which has no relationship to Chang’s conjecture.

For the rest of this section we examine the stepping-up procedure withoutthe assumption that some form of Chang’s conjecture is false. So let θ bea given regular uncountable cardinal and let Cα (α < θ+) be a fixed C-sequence such that tp(Cα) ≤ κ for all α < θ+. Let ρ∗ : [θ+]2 −→ θ bethe 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 functionτ : [θ+]3 −→ θ+ defined by the formula (I.280). Without the assumption ofCα (α < θ+) being a ¤θ-sequence, the following related function turns outto 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 walksγ → 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. Thenρ∗(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. Weneed to show that χ is unbounded on [Ω]3. Therefore we may assume thatΩ 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:

χ(α, β, γ) ≥ n for all α, β, γ ∈ [Ωn]3. (I.303)

Given Ωn, choose an ∈-chain M of length θ of elementary submodels M ofsome large enough structure of the form Hκ such that M ∩ θ ∈ θ and Mcontains all the relevant objects. Now choose another elementary submodelN of Hκ which contains M as well as all the other relevant objects suchthat N ∩ θ ∈ θ. Let λN = sup(N ∩ λ). Choose γ ∈ Ωn above λN . Letγ = γ0 > . . . > γk = λN be the walk from γ to λN . Let γ be the γi with


the smallest index such that Cγi∩λN is unbounded in λN . Then γ is either

equal to λN or γk−1. Choose λ0 ∈ N ∩ λ such that:

λ0 > max(Cγi∩ λN ) for all i < k for which Cγi

∩ λN isbounded in λN .

(I.304)

Then γi : i < k \ γ ⊆ Tr(α, γ) for all α ∈ N ∩ Ωn above λ0. So if wechoose 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 hassize 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 ourassumption that ρ∗ is bounded by ν on [Ω]2, we get:

tp(Cγ ∩ λN ) ≤ suptp(Cγ ∩ α) : α ∈ Ω ∩ [λ0, λN )≤ supρ∗(α, γ) : α ∈ Ω ∩ [λ0, λN )≤ ν.

(I.306)

Since C ∩ [λ0, λN ) has order-type > ν there must be M ∈ M ∩ N suchthat λM /∈ Cγ and λM > λ0. Thus, we have an elementary submodel Mof Hκ which contains all the relevant objects and a γ ∈ Ωn above λM suchthat 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 if γis the γi (i ≤ l) with the smallest index, subject to the requirement thatCγ ∩ λM is unbounded in λM , then γi : i < l \ γ has size at least n + 1.Let λ0 ∈ λ ∩M be such that:

λ0 > max(Cγi ∩ λM ) for all i < l for which Cγi ∩ λM isbounded in λM .

(I.307)

Then γi : i < l \ γ ⊆ Tr(α, γ) for all α ∈ M ∩ Ωn above λ0. Using theelementarity of M we conclude that for every δ < λ there exist ε ∈ Ωn \ δand a set aε ⊆ [δ, ε) of size at least n + 1 such that aε ⊆ Tr(α, ε) for allα ∈ Ωn ∩ [λ0, δ). Hence we can choose a closed and unbounded set D of[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:

aε(δ) ⊆ Tr(α, ε(δ)) for α ∈ [λ0, δ), (I.308)

ε(δ) < min(D \ (δ + 1)). (I.309)


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. a20.16 Definition. The 3-dimensional version of the oscillation mapping,osc : [θ+]3 −→ ω, is defined on the basis of the 2-dimensional version ofSection 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 twowalks γ → α and γ → β. Then we consider the walks γ = γ0 > . . . > γk = αand β = β0 > . . . > βl = α from γ to α and β to α respectively; if both kand l are bigger than n (and note that k ≥ n), i.e. if γn and βn are bothdefined, we let osc(α, β, γ) be equal to the oscillation of the two sets Cβn \αand 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 un-countable cardinal, and that every subset of Γ of size κ is unstable. Then forevery 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++ suchthat M ∩ κ ∈ κ and M contains all the relevant objects. For M ∈ M, letλM = 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. LetλN = sup(λ ∩N) and choose γ ∈ Γ above λN and let γ = γ0 > . . . > γk =λN be the walk from γ to λN . Let γ and λ0 be chosen as above in (I.304).Then γ = γk or γ = γk−1 and

Tr(α, γ) ∩ Tr(β, γ) = γi : i ≤ k \ γ for all α < β inΓ ∩ [λ0, λN ) such that Cγ ∩ [α, β) 6= ∅. (I.311)

So if a tail of the set λM : M ∈M∩N is included in Cγ , then using theelementarity of N , we would be able to construct a cofinal subset Γ0 ⊆ Γsuch that χ is constantly equal to k on [Γ0]3, where k is the place of γ inthe sequence γ0, . . . , γk. This, of course, would contradict our assumptionthat Γ contains no large stable subsets. It follows that the set of all λN forM ∈ M ∩ N which do not belong to Cγ is cofinal in λN . So we can pickM1 ∈ 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)


[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 Ii = [λi, λi], andlet In+1 = [λn, γ]. Now set F to be the collection of all increasing sequencesI1 < I2 < . . . < In+1 of closed intervals of ordinals < λ with the propertythat there is β = β( ~I ) in Γ ∩ In+1 such that, if β = β0 > . . . > βl = α isthe walk from β to α, then:

l > k and βk = min(Tr(α, β) ∩ In+1), (I.314)

Cβk\ α ⊆

n+1⋃

i=1

Ii, (I.315)

Cβk∩ Ii 6= ∅ for all 1 ≤ n ≤ n + 1. (I.316)

Clearly, F ∈ M1 and 〈I1, . . . , In+1〉 ∈ F , as γ is a witness of this. Soworking as in the proof of Lemma 17.2, we can find F0 ⊆ F in M1 suchthat, 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 intervalK included in [ξ, λ) such that the concatenation ~JaK also end-extends toa sequence in F0. Working inductively on 1 ≤ i ≤ n + 1 we can select asequence 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)

〈I1, . . . , Ii〉 end-extends to a member of F0. (I.319)

Clearly, there is no problem in choosing these objects, using the elementarityof the models Mi (1 ≤ i ≤ n+1) and the splitting property of F0. Workingin Mn+1, we find an interval In+1 such that 〈I0, . . . , In, In+1〉 belongs toF0. Let β = β(〈I0, . . . , In+1〉). 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 bydefinition

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 concludethat I1∩Cβk

, . . . , In−1∩Cβkand (In∪In+1)∩Cβk

are the n convex pieces inwhich 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 aninteresting dichotomy:

20.18 Lemma. Every Γ ⊆ θ+ of size θ can be refined to a subset Ω of sizeθ such that either:


(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.

aWe 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 cubeof any subset of θ++ of size θ+.

Proof. We choose two C-sequences Cα (α < θ+) and C+α (α < θ++) on θ+

and θ++ respectively, such that tp(Cα) ≤ θ for all α < θ+ and tp(C+α ) ≤

θ+ for all α < θ++. Let ρ∗ : [θ+]2 −→ θ and ρ∗+ : [θ++]2 −→ θ+ bethe corresponding ρ∗-functions defined above in I.264. Also choose a one-to-one sequence rα (α < θ++) of elements of 0, 1θ+

and consider thecorresponding function ∆ : [θ++]2 −→ θ+ defined in (I.293). We definef : [θ++]3 −→ θ+ according to the following two cases for a given tripleα < 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 leastχ(α, β, γ).

Rule 1: If ∆(rα, rβ) < ∆(rβ , rγ) and rα <lex rβ <lex rγ or rα >lex

rβ >lex rγ , set

f(α, β, γ) = min(Pν(∆(β, γ)) \∆(α, β)),

where ν = ρ∗(minξ ≤ ∆(α, β) : ρ∗(ξ, ∆(α, β)) 6= ρ∗(ξ, ∆(β, γ)), ∆(β, γ)).Rule 2: If α ∈ 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), where γ is the remaining member of α, β, γ, andif 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 leastχ(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, letf(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 ′′[Γ]3 either contains all positive integers or almostall ordinals < θ+ of cofinality θ.


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 tripleα < β < γ from Ω. Note that in this case, the definition follows exactly thedefinition 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 proofof Theorem 20.10 we conclude that in this case the image f ′′[Ω]3 containsalmost all ordinals < θ+ of cofinality θ.

Suppose now that no subset of Γ of size θ+ is stable. By Lemma 20.17 (infact, its proof) for every integer n ≥ 1 there exist α < β < γ in Γ such thatfor 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. a20.20 Corollary. There is f : [ω2]3 −→ ω which takes all the values on thecube of any uncountable subset of ω2. a20.21 Remark. Note that the dimension 3 in this Corollary cannot belowered to 2 as long as one does not use some additional axioms to constructsuch f . Note also that the range ω cannot be replaced by a set of biggersize, as this would contradict Chang’s conjecture. We have seen above thatChang’s conjecture is equivalent to the statement that for every f : [ω2]3 −→ω1 there is an uncountable set Γ ⊆ ω2 such that f ′′[Γ]3 6= ω1. Is there asimilar reformulation of the Continuum Hypothesis? More precisely, onecan ask the following question.

20.22 Question. Is CH equivalent to the statement that for everyf : [ω2]2 −→ ω there exists an uncountable Γ ⊆ ω1 with f ′′[Γ]2 6= ω?

Bibliography

[1] P.S. Alexandroff. Pages from an autobiography. Russian Math. Surveys,34:267–302, 1979.

[2] P.S. Alexandroff and P. Urysohn. Memoire sur les espaces topologiquecompacts. Verh. Akad. Wetensch., 14:VIII+96, 1929.

[3] S. A. Argyros, J. Lopez-Abad, and S. Todorcevic. A class of Ba-nach spaces with few non-strictly singular operators. J. Funct. Anal.,222(2):306–384, 2005.

[4] J. Bagaria. Fragments of Martin’s axiom and ∆13 sets of reals. Ann.

Pure Appl. Logic, 69:1–25, 1994.

[5] James Baumgartner and Saharon Shelah. Remarks on superatomicBoolean algebras. Ann. Pure Appl. Logic, 33:109–129, 1987.

[6] James Baumgartner and Otmar Spinas. Independence and consistencyproofs in quadratic form theory. J. Symbolic Logic, 56:1195–1211, 1991.

[7] M. Bekkali. Topics in set theory, volume 1476 of Lect. notes in Math.Springer-Verlag, 1992.

[8] Felix Bernstein. Zur Theorie der trigonometrischen Reihen. BerichteVerh. Konigl. Sachs. Gesellschaft, 60:325–338, 1908.

[9] M. Burke and Menachem Magidor. Shelah’s pcf theory and its appli-cations. Ann. Pure Appl. Logic, 50:207–254, 1990.

[10] W.W. Comfort and S. Negrepontis. Ultrafilters. Springer-Verlag, 1974.

[11] W.J. Davis, T. Figiel, W.B. Johnson, and A. Pelczynski. Factoringweakly compact operators. J. Functional Analysis, 17:311–327, 1974.

[12] K.J. Devlin. Constructibility. Springer-Verlag, 1984.

[13] H.-D. Donder, Ronald Jensen, and Koppelberg B. Some applicationsof the core model. In Set theory & Model theory, volume 872 of Lect.notes in Math., pages 55–97. Springer-Verlag, 1979.

153


[14] H.-D. Donder and P. Koepke. On the consistency strength of ’accessi-ble’ Jonsson cardinals and of the weak Chang conjecture. Ann. PureAppl. Logic, 25:233–261, 1983.

[15] H.-D. Donder and J.P. Levinski. Some principles related to Chang’sconjecture. Ann. Pure Appl. Logic, 45:39–101, 1989.

[16] A. Dow. PFA and ω∗1 . Topology and its Appl., 28:127–140, 1988.

[17] A. Dow and S. Watson. A subcategory of TOP. Trans. Amer. Math.Soc., 337:825–837, 1993.

[18] B. Dushnik and E.W. Miller. Partially ordered sets. Amer. J. Math.,63:600–610, 1941.

[19] H.-D. Ebbinghaus and J. Flum. Eine Maximalitatseigenschaft derPradikatenlogik erster Stufe. Math.-Phys. Semesterber., 21:182–202,1974.

[20] P. Erdos, S. Jackson, and R.D. Mauldin. On partitions of lines andspaces. Fund. Math., 145:101–119, 1994.

[21] P. Erdos, S. Jackson, and R.D. Mauldin. On infinite partitions of linesand spaces. Fund. Math., 152:75–95, 1997.

[22] P. Erdos and A. Tarski. On families of mutually exclusive sets. Ann.of Math., 44:315–329, 1943.

[23] P. Erdos and A. Tarski. On some problems involving inaccessible car-dinals. In Y. Bar-Hillel, E. Poznanski, M. Rabin, and A. Robinson, ed-itors, Essays on the Foundations of Mathematics, pages 50–82. MagnesPress, 1961.

[24] Matthew Foreman and Menachem Magidor. A very weak square prin-ciple. J. Symbolic Logic, 62:175–196, 1997.

[25] D.H. Fremlin. The partially ordered sets of measure theory and Tukey’sordering. Note di Matematica, XI:177–214, 1991.

[26] Harvey Friedman. On the necessary use of abstract set theory. Adv. inMath., 41(3):209–280, 1981.

[27] W. T. Gowers and B. Maurey. The unconditional basic sequence prob-lem. J. Amer. Math. Soc., 6(4):851–874, 1993.

[28] G. Gruenhage. Irreducible restrictions of closed mappings. Topologyand its Appl., 85:127–135, 1998.

Bibliography 155

[29] G. Gruenhage and Y. Tanaka. Products of k-spaces. Trans. Amer.Math. Soc., 273:299–308, 1982.

[30] Andras Hajnal, Akihiro Kanamori, and Saharon Shelah. Regressivepartition relations for infinite cardinals. Trans. Amer. Math. Soc.,299:145–154, 1987.

[31] William Hanf. On a problem of Erdos and Tarski. Fund. Math., 53:325–334, 1964.

[32] G.H. Hardy and E.M.Wright. An introduction to the theory of numbers.Clarendon Press, Oxford, 1979.

[33] Leo Harrington and Saharon Shelah. Some exact equiconsistency re-sults in set theory. Notre Dame J. Formal Logic, 26:178–187, 1985.

[34] W. Hodges and Saharon Shelah. Infinite games and reduced products.Annals of Mathematical Logic, 20:77–108, 1981.

[35] Ronald Jensen. The fine structure of the constructible hierarchy. An-nals of Mathematical Logic, 4:229–308, 1972.

[36] Ronald Jensen and Karl Schlechta. Results on the generic Kurepahypothesis. Archive for Math. Logic, 30:13–27, 1990.

[37] Akihiro Kanamori. The Higher Infinite. Perspectives in Math. Logic.Springer-Verlag, 1997.

[38] Akihiro Kanamori and Kenneth McAloon. On Godel incompletenessand finite combinatorics. Ann. Pure Appl. Logic, 33(1):23–41, 1987.

[39] Jussi Ketonen. Banach spaces and large cardinals. Fund. Math.,81:291–303, 1974. Collection of articles dedicated to Andrzej Mostowskion the occasion of his sixtieth birthday, IV.

[40] Piotr Koszmider. On coherent families of finite-to-one functions. J.Symbolic Logic, 58:128–138, 1993.

[41] Piotr Koszmider. Forcing minimal extensions of Boolean algebras.Trans. Amer. Math. Soc., 351:3073–3117, 1999.

[42] Piotr Koszmider. On Banach spaces of large density but few operators.preprint, 2000.

[43] K. Kunen. Saturated ideals. J. Symbolic Logic, 43:65–76, 1978.

[44] K. Kunen. An introduction to independence proofs. North-Holland,1980.


[45] D. Kurepa. Ensembles ordonnes et ramifies. Publ. Math. Univ. Bel-grade, 4:1–138, 1935.

[46] Paul Larson. The nonstationary ideal in the Pmax-extension. preprint,2005.

[47] Richard Laver. Better-quasi-orderings and a class of trees. In G.C.Rota, editor, Studies in Foundations and Combinatorics, volume 1 ofAdvances in Math. Suppl. Stud., pages 31–48. Academic Press, 1978.

[48] Richard Laver and Saharon Shelah. The ℵ2-Souslin hypothesis. Trans.Amer. Math. Soc., 264:411–417, 1981.

[49] J.-P. Levinski, Menachem Magidor, and Saharon Shelah. Chang’s con-jecture for ℵω. Israel J. Math., 69:161–172, 1990.

[50] V.I. Malykhin. On Ramsey spaces. Sov. Math. Dokl., 20:894–898, 1979.

[51] B. Maurey and H. P. Rosenthal. Normalized weakly null sequence withno unconditional subsequence. Studia Math., 61(1):77–98, 1977.

[52] William Mitchell. Aronszajn trees and the independence of the transferpropery. Annals of Mathematical Logic, 5:21–46, 1972.

[53] J. Donald Monk. Cardinal invariants on Boolean algebras. BirkhauserVerlag, Basel, 1996.

[54] Justin T. Moore. A solution to the l-space problem. Journal of theAmer Math. Soc., 2005.

[55] J. Nesetril and Rodl. Ramsey topological spaces. In J. Novak, editor,General toplogy and its relation to modern algebra, volume IV Part B,pages 333–337. Society of Czechoslovak Mathematicians and Physicists,Prague, 1977.

[56] T. Ohkuma. Comparability between ramified sets. Proc. Japan Acad.Sci., 36:383–388, 1960.

[57] Martin Otto. EM constructions for a class of generalized quantifiers.Archive for Math. Logic, 31:355–371, 1992.

[58] Mariusz Rabus. An ω2-minimal Boolean algebra. Trans. Amer. Math.Soc., 348:3235–3244, 1996.

[59] J. Roitman. A thin atomic Boolean algebra. Algebra Universalis,21:137–142, 1985.

[60] M.E. Rudin. Lectures in set-theoretic topology. Amer. Math. Soc. Prov-idence R.I., 1975.

Bibliography 157

[61] Marion Scheepers. Concerning n-tactics in the countable-finite game.J. Symbolic Logic, 56:786–794, 1991.

[62] E. Schimmerling. Combinatorial set theory and inner models. In C.A.DiPrisco et al., editors, Set Theory: Techniques and Applications, pages207–212. Kluwer Academic Publishers, 1998.

[63] J. Schmerl. Transfer theorems and their applications to logics.In J. Barwise et al., editors, Model-theoretic logics, pages 177–209.Springer-Verlag, 1985.

[64] James Schmerl. A partition property characterizing cardinals hyperi-naccessible of finite type. Trans. Amer. Math. Soc., 188:281–291, 1974.

[65] Saharon Shelah. Strong negative partition above the continuum. J.Symbolic Logic, 55:21–31, 1990.

[66] Saharon Shelah. Cardinal Arithmetic, volume 29 of Oxford LogicGuides. Oxford University Press, 1994.

[67] Saharon Shelah and Otmar Spinas. Gross spaces. Trans. Amer. Math.Soc., 348:4257–4277, 1996.

[68] Saharon Shelah and Juris Steprans. A Banach space on which thereare few operators. Proc. Amer. Math. Soc., 104:101–105, 1986.

[69] Saharon Shelah and Juris Steprans. Extraspecial p-groups. Ann. PureAppl. Logic, 34:87–97, 1987.

[70] I. Singer. Bases in Banach Spaces, II. Springer-Verlag, 1981.

[71] E. Specker. Sur un probleme de Sikorski. Colloquium Mathematicum,2:9–12, 1949.

[72] D.A. Talayco. Application of Cohomology to Set Theory I. Ann. PureAppl. Logic, 71:69–107, 1995.

[73] D.A. Talayco. Application of Cohomology to Set Theory II. Ann. PureAppl. Logic, 77:279–299, 1996.

[74] Stevo Todorcevic. Trees and linearly ordered sets. In K. Kunen and J.E.Vaughan, editors, Handbook of Set-theoretic topology, pages 235–293.North-Holland, 1984.

[75] Stevo Todorcevic. Aronszajn trees and partitions. Israel J. Math.,52:53–58, 1985.

[76] Stevo Todorcevic. Remarks on chain conditions in products. Compo-sitio Math., 55:295–302, 1985.


[77] Stevo Todorcevic. Remarks on cellularity in products. CompositioMath., 57:357–372, 1986.

[78] Stevo Todorcevic. Partitioning pairs of countable ordinals. Acta Math.,159:261–294, 1987.

[79] Stevo Todorcevic. Oscillations of real numbers. In Logic colloquium’86, pages 325–331. North-Holland, Amsterdam, 1988.

[80] Stevo Todorcevic. Partition problems in topology. Amer. Math. Soc.Providence R.I., 1989.

[81] Stevo Todorcevic. Special square sequences. Proc. Amer. Math. Soc.,105:199–205, 1989.

[82] Stevo Todorcevic. Remarks on Martin’s axiom and the ContinuumHypothesis. Canadian J. Math., 43:832–851, 1991.

[83] Stevo Todorcevic. Some partitions of three-dimensional combinatorialcubes. J. Combin. Theory Ser. A, 68:410–437, 1994.

[84] Stevo Todorcevic. Aronszajn orderings. Publ. Inst. Math. Belgrade,57(71):29–46, 1995.

[85] Stevo Todorcevic. Comparing the continuum with the first two un-countable cardinals. In M.L. Dalla Chiara et al., editors, Logic andScientific Methods, pages 145–155. Kluwer Academic Publishers, 1997.

[86] Stevo Todorcevic. Basis problems in combinatorial set theory. InProceedings of the International Congress of Mathematicians, Vol. II(Berlin), pages 43–52, 1998.

[87] Stevo Todorcevic. Countable chain condition in partition calculus.Disr. Math., 188:205–223, 1998.

[88] Stevo Todorcevic. Oscillations of sets of integers. Adv. in Appl. Math.,20:220–252, 1998.

[89] Stevo Todorcevic. A dichotomy for P-ideals of countable sets. Fund.Math., 166:251–267, 2000.

[90] Stevo Todorcevic. Lipschitz maps on trees. preprint, 2000.

[91] Boban Velickovic. PhD thesis, University of Wisconsin, 1986.

[92] Dan Velleman. Simplified morasses. J. Symbolic Logic, 49:257–271,1984.

[93] Dan Velleman. On a combinatorial principle of Hajnal and Komjath.J. Symbolic Logic, 51:1056–1060, 1986.

Bibliography 159

[94] H.M. Wark. A non-separable reflexive Banach space on which thereare few operators. J. London Math. Soc., 64:675–689, 2001.

[95] N.M. Warren. Extending continuous functions in zero-dimensionalspaces. Proc. Amer. Math. Soc., 52:414–416, 1975.

[96] William Weiss. Partitioning topological spaces. In J. Nesetril andV. Rodl, editors, Mathematics of Ramsey theory, pages 155–171.Springer-Verlag, 1990.

[97] K. Wolfsdorf. Farbungen großer Wurfel mit bunten Wegen. Arch.Math., 40:569–576, 1983.

[98] W. Hugh Woodin. The Axiom of Determinacy, Forcing Axioms, andthe Nonstationary Ideal. Walter de Gruyter & Co., Berlin, 1999.

Index

<c, 6<p, 29<r, 6<x, 29<ρ, 31D, 104F , 5Fn(α), 109G 7−→ G∗, 15Gα, 8KΓ, 62Sθ, 93T (ρ0), 8Wγ , 94[αβ], 52[αβ]∗, 127[αβ]∗h, 128[··]h, 127∆, 105∆-function, 105Λ, 38Λ + n, 38αt, 120ρ, 29ρ1, 75χV , 13≤, 46F-Cohen, 32F-Souslin, 32GΓ, 57I, 92Uf , 93| αβ |, 60ρ, 26ρ0, 4

ρ1, 9ρ2, 36ρ3, 37ρa, 33σ-sequentially closed, 114σ(α, β), 52τe, 33[[αβ]], 57A(ρ1), 12T (ρ1), 12Vt, 13a(n), 48d, 93fα(n), 109CK(θ), 111lim(D), 119osc, 148

action, 120Aronszajn tree, 8, 44, 67avoid ∆, 122

Bernstein decomposition, 114

Chang’s conjecture, 133clique, 15cofinal branch, 67coherent C-sequence, 80coherent family, 115coherent function sequence, 36coherent mapping, 44coherent tree, 44corresponding transitive map, 137

division on ordinals, 82

160

INDEX 161

Eberlein compactum, 13

full code, 4full lower trace, 5

Hausdorff gap, 34, 44

irreducible, 80

Jensen matrix, 112

Kurepa family, 105Kurepa family, cofinal, 111Kurepa family, compatibility, 111Kurepa family, extendibility, 111

last step, 37lower trace, 4

Maharam type, 110maximal weight, 9metric, 28minimal walk, 3, 51

nontrivial, 36, 44nowhere dense, 44number of steps, 36

orthogonal, 38, 44

P-ideal, 38P-ideal dichotomy, 38positive with respect to a filter, 124property ∆, 108property C, 14property C ′′, 14

Radon measure space, 110Radon probability space, 110right lexicographical ordering, 6

sequential fan, 93sequentially closed, 114shift, 48sigma-product, 115sigma-product, index-set, 115

Souslin, 32special Aronszajn tree, 8special square sequence, 87special tree, 67square-bracket operation, 52square-sequence, 80stable, 146stationary subsequence, 119step, 3stepping-up, 138strong measure zero, 14strongly unbounded function, 132subadditivity, 28support, 44

tightness, 93trace, 4, 51trace filter, 123transitive map, 33, 136tree on θ, 86trivial C-sequence, 76Tukey equivalent, 110Tukey reducible, 110

ultra-metric, 28ultra-subadditivity, 28unbounded family, 118unbounded function, 131upper trace, 4

walk, 3weak Chang’s conjecture, 137weight, 3

I. Coherent Sequences - » Department of Mathematicsstevo/cseq.pdf · 2 I. Coherent Sequences sen...

Documents

Transcript of I. Coherent Sequences - » Department of Mathematicsstevo/cseq.pdf · 2 I. Coherent Sequences sen...