I. Coherent Sequences - Department of Mathematics, University of

I. Coherent Sequences
Stevo Todorcevic
Contents
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Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Walks in the space of countable ordinals . . . . . . . . . . . . . .
The coherence of maximal weights . . . . . . . . . . . . . . . . . . . .
Oscillations of traces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Triangle inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Conditional weakly null sequences . . . . . . . . . . . . . . . . . . . . .
An ultrafilter from a coherent sequence . . . . . . . . . . . . . . .
The trace and the square-bracket operation . . . . . . . . . . .
A square-bracket operation on a tree . . . . . . . . . . . . . . . . . .
Special trees and Mahlo cardinals . . . . . . . . . . . . . . . . . . . . .
The weight function on successor cardinals . . . . . . . . . . . .
The number of steps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Square sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
The full lower trace of a square sequence . . . . . . . . . . . . . .
Special square sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Successors of regular cardinals . . . . . . . . . . . . . . . . . . . . . . . .
Successors of singular cardinals . . . . . . . . . . . . . . . . . . . . . . .
The oscillation mapping . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
The square-bracket operation . . . . . . . . . . . . . . . . . . . . . . . . .
Unbounded functions on successors of regular cardinals
Higher dimensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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0. Introduction
A transfinite sequence Cξ ⊆ ξ (ξ < θ) of sets may have a number of ’coherence properties’ and the purpose of this chapter is to study some of them as
well as some of their uses. ’Coherence’ usually means that the Cξ ’s are cho1
2
I. Coherent Sequences
sen in some canonical way, beyond the natural requirement that Cξ is closed
and unbounded in ξ for all ξ. For example, choosing a canonical ’fundamental sequence’ of sets Cξ ⊆ ξ for ξ < ε0 relying on the specific properties
of the Cantor normal form for ordinals below the first ordinal satisfying the
equation x = ω x is a basis for a number of important results in proof theory.
In set theory, one is interested in longer sequences as well and usually has
a different perspective in applications, so one is naturally led to use some
other tools beside the Cantor normal form. It turns out that the sets Cξ
can not only be used as ’ladders’ for climbing up in recursive constructions
but also as tools for ’walking’ from an ordinal to a smaller one. This notion
of a ’walk’ and the corresponding ’distance functions’ constitute the main
body of study in this chapter. We show that the resulting ’metric theory
of ordinals’ not only provides a unified approach to a number of classical
problems in set theory but also has its own intrinsic interest. For example,
from this theory one learns that the triangle inequality of an ultrametric
e(α, γ) ≤ max{e(α, β), e(β, γ)}
has three versions, depending on the natural ordering between the ordinals α, β and γ, that are of a quite different character and are occurring
in quite different places and constructions in set theory. For example, the
most frequent occurrence is the case α < β < γ when the triangle inequality becomes something that one can call ’transitivity’ of e. Considerably
more subtle is the case α < γ < β of this inequality. It is this case of
the inequality that captures most of the coherence properties found in this
article. Another thing one learns from this theory is the special role of the
first uncountable ordinal in this theory. Any natural coherency requirement
on the sets Cξ (ξ < θ) that one finds in this theory is satisfiable in the case
θ = ω1 . The first uncountable cardinal is the only cardinal on which the
theory can be carried out without relying on additional axioms of set theory.
The first uncountable cardinal is the place where the theory has its deepest
applications as well as its most important open problems. This special role
can perhaps be explained by the fact that many set-theoretical problems,
especially those coming from other fields of mathematics, are usually concerned only about the duality between the countable and the uncountable
rather than some intricate relationship between two or more uncountable
cardinalities. This is of course not to say that an intricate relationship between two or more uncountable cardinalities may not be a profitable detour
in the course of solving such a problem. In fact, this is one of the reasons
for our attempt to develop the metric theory of ordinals without restricting
ourselves only to the realm of countable ordinals.
The chapter is organized as a discussion of four basic distance functions
on ordinals, ρ, ρ0 , ρ1 and ρ2 , and the reader may choose to follow the analysis of any of these functions in various contexts. The distance functions
will naturally lead us to many other derived objects, most prominent of
1. Walks in the space of countable ordinals
3
which is the ’square-bracket operation’ that gives us a way to transfer the
quantifier ’for every unbounded set’ to the quantifier ’for every closed and
unbounded set’. This reduction of quantifiers has proven to be quite useful
in constructions of various mathematical structures, some of which have
been mentioned or reproduced here.
I wish to thank Bernhard König, Piotr Koszmider, Justin Moore and
Christine Härtl 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 space
considered in this chapter. There are many mathematical problems whose
combinatorial essence can be reformulated as problems about ω1 , which is
in some sense the smallest uncountable structure. What we mean by ’structure’ is ω1 together with system Cα (α < ω1 ) of fundamental sequences, i.e.
a system with the following two properties:
(a) Cα+1 = {α},
(b) Cα is an unbounded subset of α of order-type ω, whenever α is a
countable limit ordinal > 0.
Despite its simplicity, this structure can be used to derive virtually all
known other structures that have been defined so far on ω1 . There is a natural recursive way of picking up the fundamental sequences Cα , a recursion
that refers to the Cantor normal form which works well for, say, ordinals
< ε0 1 . For longer fundamental sequences one typically relies on some other
principles of recursive definitions and one typically works with fundamental sequences with as few extra properties as possible. We shall see that
the following assumption is what is frequently needed and will therefore be
implicitly assumed whenever necessary:
(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 ordinal
α is the minimal point of Cβ that is ≥ α. The cardinality of the set Cβ ∩ α,
or better to say the order-type of this set, is the weight of the step.
1.2 Definition. A walk (or a minimal walk ) from a countable ordinal β to
a smaller ordinal α is the sequence β = β0 > β1 > . . . > βn = α such that
for each i < n, the ordinal βi+1 is the step from βi towards α.
1 One is tempted to believe that the recursion can be stretched all the way up to ω and
1
this is probably the way P.S. Alexandroff has found his famous Pressing Down Lemma
(see [1] and [2, appendix]).
4
I. Coherent Sequences
Analysis of this notion leads to several two-place functions on ω1 that
give a rich structure with many applications. So let us expose some of these
functions.
1.3 Definition. The full code of the walk is the function ρ0 : [ω1 ]2 −→ ω <ω
defined recursively by
ρ0 (α, β) = h|Cβ ∩ α|ia ρ0 (α, min(Cβ \ α)),
with the boundary value ρ0 (α, α) = ∅2 where the symbol a refers to the
sequence obtained by concatenating the one term sequence h|Cβ ∩ α|i with
the already known finite sequence ρ0 (α, min(Cβ \ α) of integers.
Knowing ρ0 (α, β) and the ordinal β one can reconstruct two following
two traces of the walk from β to α.
1.4 Definition. The upper trace of the walk from β to α is the set
Tr(α, β) = {ξ : ρ0 (ξ, β) v ρ0 (α, β)},
where v denotes the ordering of being an initial segment among finite sequences of ordinals. One also has the recursive definition of this trace,
Tr(α, β) = {β} ∪ Tr(α, min(Cβ \ α))
with the boundary value Tr(α, α) = {α}.
1.5 Definition. The lower trace of the walk from β to α is the set
L(α, β) = {λξ (α, β) : ρ0 (ξ, β) < ρ0 (α, β)},
where < is the ordering of being a strict initial segment among the finite sequences of ordinals and where for an ordinal ξ such that ρ0 (ξ, β) < ρ0 (α, β),
λξ (α, β) = max{max(Cη ∩ α) : ρ0 (η, β) v ρ0 (ξ, β)}.
Note the following recursive definition of this trace,
L(α, β) = L(α, min(Cβ \ α)) ∪ ({max(Cβ ∩ α)} \ max(L(α, min(Cβ \ α))))
with the boundary value L(α, α) = ∅.
The following immediate facts about these notions will be frequently and
very often implicitly used.
2 Note that while ρ operates on the set [ω ]2 of unordered pairs of countable ordinals,
0
1
the formal recursive definition of ρ0 (α, β) does involves the term ρ0 (α, α) and so we have
to give it as a boundary value ρ0 (α, α) = ∅. The boundary value is always specified when
such a function is recursively defined and will typically be either 0 or the empty sequence
∅. This convention will be used even when a function e with domain such as [ω1 ]2 is
given to us beforehand and we consider its diagonal value e(α, α).
1. Walks in the space of countable ordinals
5
1.6 Lemma. For α ≤ β ≤ γ,
(a) α > L(β, γ)3 implies that ρ0 (α, γ) = ρ0 (β, γ)a ρ0 (α, β) and therefore
that Tr(α, γ) = Tr(β, γ) ∪ Tr(α, β),
(b) L(α, β) > L(β, γ) implies that L(α, γ) = L(β, γ) ∪ L(α, β).
a
1.7 Definition. The full lower trace of the minimal walk is the function
F : [ω1 ]2 −→ [ω1 ]<ω defined recursively by
[
F (α, β) = F (α, min(Cβ \ α)) ∪
F (ξ, α),
ξ∈Cβ ∩α
with the boundary value F (α, α) = {α} for all α.
Thus F (α, β) includes the traces of walks between any two ordinals ≤ α
that have ever been referred to during the walk from β to α.
1.8 Lemma. For all α ≤ β ≤ γ,
(a) F (α, γ) ⊆ F (α, β) ∪ F (β, γ),
(b) F (α, β) ⊆ F (α, γ) ∪ F (β, γ).
Proof. The proof is by induction on γ. Let F (α, γ1 ) of F (α, γ) where γ1 =
min(Cγ \ α).
To prove (a) consider first the case γ1 < β. By the inductive hypothesis,
F (α, γ1 ) ⊆ F (α, β) ∪ F (γ1 , β) ⊆ F (α, β) ∪ F (β, γ),
since γ1 ∈ Cγ ∩ β, so F (γ1 , β) ⊆ F (β, γ). If γ1 ≥ β then γ1 = min(Cγ \ β)
and again F (β, γ1 ) ⊆ F (β, γ). By the inductive hypothesis
F (α, γ1 ) ⊆ F (α, β) ∪ F (β, γ1 ) ⊆ F (α, β) ∪ F (β, γ).
Let now us consider F (ξ, α) for some ξ ∈ Cγ ∩ α. Using the inductive
hypothesis we get
F (ξ, α) ⊆ F (α, β) ∪ F (ξ, β) ⊆ F (α, β) ∪ F (β, γ),
since ξ ∈ Cγ ∩ β and F (ξ, β) is by definition included in F (β, γ).
To prove (b) consider first the case γ1 < β. By the inductive hypothesis:
F (α, β) ⊆ F (α, γ1 ) ∪ F (γ1 , β).
3 For two sets of ordinals F and G, by F < G we denote the fact that every ordinal
from F is smaller than every ordinal from G. When G = {α}, we write F < α rather
than F < {α}.
6
I. Coherent Sequences
Since γ1 ∈ Cγ ∩ β, the set F (γ1 , β) is included in F (β, γ). Similarly by the
definition of F (α, γ), the set F (α, γ1 ) is included in F (α, γ).
Suppose now that γ1 ≥ β. Note that in this case γ1 = min(Cγ ∩ β), so
F (β, γ1 ) is included in F (β, γ). Using the inductive hypothesis for α ≤ β ≤
γ1 we have
F (α, β) ⊆ F (α, γ1 ) ∪ F (β, γ1 ) ⊆ F (α, γ) ∪ F (β, γ).
a
1.9 Lemma. For all α ≤ β ≤ γ,
(a) ρ0 (α, β) = ρ0 (min(F (β, γ) \ α), β)a ρ0 (α, min(F (β, γ) \ α)),
(b) ρ0 (α, γ) = ρ0 (min(F (β, γ) \ α), γ)a ρ0 (α, min(F (β, γ) \ α)).
Proof. The proof is again by induction on γ. Let α1 = min(F (β, γ) \ α) and
γ1 = min(Cγ \ α). Let ξ1 be the minimal ξ ∈ (Cγ ∩ β) ∪ {min(Cγ \ β)} such
that α1 ∈ F (ξ, β) (or F (β, ξ)).
Applying the inductive hypothesis to α ≤ ξ1 ≤ β (or α ≤ β ≤ ξ1 ) and
noting that α1 = min(F (ξ1 , β) \ α) (or α1 = min(F (β, ξ1 ) \ α)) we get
ρ0 (α, β) = ρ0 (α1 , β)a ρ0 (α, α1 ),
(I.1)
the part (a) of Lemma 1.9. Note that depending on whether α ≤ γ1 ≤ β
or α ≤ β ≤ γ1 we have that F (γ1 , β) ⊆ F (β, γ) or F (β, γ1 ) ⊆ F (β, γ), and
therefore α2 := min(F (γ1 , β) \ α) ≥ α1 or α2 := min(F (β, γ1 ) \ α) ≥ α1 .
Applying the inductive hypothesis to α ≤ γ1 ≤ β (or α ≤ β ≤ γ1 ) we get:
ρ0 (α, β) = ρ0 (α2 , β)a ρ0 (α, α2 ),
(I.2)
ρ0 (α, γ1 ) = ρ0 (α2 , γ1 )a ρ0 (α, α2 ).
(I.3)
Comparing (I.1) and (I.2) we see that ρ0 (α, α1 ) is a tail of ρ0 (α, α2 ), so (I.3)
can be rewritten as
ρ0 (α, γ1 ) = ρ0 (α2 , γ1 )a ρ0 (α1 , α2 )a ρ0 (α, α1 ) = ρ0 (α1 , γ1 )a ρ0 (α, α1 ). (I.4)
Finally since ρ0 (α, γ) = h|Cγ ∩ α|ia ρ0 (α, γ1 ) = ρ0 (γ1 , γ)a ρ0 (α, γ1 ) holds,
the equation (I.4) gives us the desired conclusion (b) of Lemma 1.9.
a
1.10 Definition. Recall that right lexicographical ordering on ω <ω denoted
by <r refers to the total ordering defined by letting s <r t if s is an endextension of t, or if s(i) < t(i) for i = min{j : s(j) 6= t(j)}. Using the
identification between a countable ordinal α and the α-sequence (ρ0 )α of
elements of ω <ω , the ordering <r induces the following ordering <c on ω1 :
α <c β
iff
ρ0 (ξ, α) <r ρ0 (ξ, β),
where ξ = ∆(α, β) = min{η ≤ min{α, β} : ρ0 (η, α) 6= ρ0 (η, β)}.
(I.5)
1. Walks in the space of countable ordinals
7
1.11 Lemma. The cartesian square of the total ordering <c of ω1 is the
union 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 codes
the finite structure obtained from F (α, β) ∪ {β} by adding relations that
describe 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 to
each other in the isomorphism of the (α, β) and (γ, δ) structures. It follows
that:
ρ0 (ξαβ , α) = ρ0 (ξγδ , γ)(= tα,γ ),
(I.6)
ρ0 (ξαβ , β) = ρ0 (ξγδ , δ)(= tβ,δ ).
(I.7)
Applying Lemma 1.9 we get:
ρ0 (∆(α, γ), α) = tαγ a ρ0 (∆(α, γ), ξαβ ),
(I.8)
ρ0 (∆(α, γ), γ) = tαγ a ρ0 (∆(α, γ), ξγδ ).
(I.9)
It follows that ρ0 (∆(α, γ), ξαβ ) 6= ρ0 (∆(α, γ), ξγδ ). Applying Lemma 1.9 for
β and δ and the ordinal ∆(α, γ) we get:
ρ0 (∆(α, γ), β) = tβδ a ρ0 (∆(α, γ), ξαβ ),
(I.10)
ρ0 (∆(α, γ), δ) = tβδ a ρ0 (∆(α, γ), ξγδ ).
(I.11)
It follows that ρ0 (∆(α, γ), β) 6= ρ0 (∆(α, γ), δ). This shows ∆(α, γ) ≥
∆(β, δ). A symmetrical argument shows the other inequality ∆(β, δ) ≥
∆(α, γ). It follows that
¯
∆(α, γ) = ∆(β, δ)(= ξ).
¯ α) <r ρ0 (ξ,
¯ γ) and since these two sequences
Our assumption is that ρ0 (ξ,
have tαγ as common initial part, this reduces to
¯ ξαβ ) <r ρ0 (ξ,
¯ ξγδ ).
ρ0 (ξ,
(I.12)
¯ β) and ρ0 (ξ,
¯ δ), so
On the other hand tβδ is a common initial part of ρ0 (ξ,
their lexicographical relationship depends on their tails which by (I.10) and
¯ ξαβ ) and ρ0 (ξ,
¯ ξγδ ) respectively. Referring to (I.12)
(I.11) are equal to ρ0 (ξ,
¯
¯ δ), i.e. β <c δ.
we conclude that indeed ρ0 (ξ, β) <r ρ0 (ξ,
a
8
I. Coherent Sequences
1.12 Notation. Well-ordered sets of rationals. The set ω <ω ordered by
the right lexicographical ordering <r is a particular copy of the rationals
of 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 increasing
function from α into Qr . We let ρ0 α or simply ρ0α denote this function and
we 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.
a
The sequence ρ0α (α < ω1 ) of members of σQr naturally determines the
subtree
T (ρ0 ) = {(ρ0 )β ¹ α : α ≤ β < ω1 }.
Note that for a fixed α, the restriction ρ0β ¹ α is determined by the way ρ0β
acts on the finite set F (α, β). This is the content of Lemma 1.9. Hence all
levels of T (ρ0 ) are countable, and therefore T (ρ0 ) is a particular example of
an Aronszajn tree, a tree of height ω1 with all levels and chains countable.
We shall now see that T (ρ0 ) is in fact a special Aronszajn tree, i.e. one that
admits a decomposition into countably many antichains or, equivalently,
admits a strictly increasing map into the rationals.
1.14 Lemma. {ξ < β : ρ0 (ξ, β) = ρ0 (ξ, γ)} is a closed subset of β whenever
β < γ.
Proof. Let α < β be a given accumulation point of this set and let β =
β0 > . . . > βn = α and γ = γ0 > . . . > γm = α be the traces of the
walks β → α and γ → α, respectively. Using the assumption (c) of the
fundamental sequences and the fact that α > 0 is a limit ordinal we can
find an ordinal ξ < α such that ρ0 (ξ, β) = ρ0 (ξ, γ) and
ξ > max(Cβi ∩ α) and ξ > max(Cγj ∩ α) for all i < n and j < m. (I.13)
Note that this in particular means that the walk α → ξ is a common tail
of the walks β → ξ and γ → ξ. Subtracting ρ0 (ξ, α) from ρ0 (ξ, β) we get
ρ0 (α, β) and subtracting ρ0 (ξ, α) from ρ0 (ξ, γ) we get ρ0 (α, γ). It follows
that ρ0 (α, β) = ρ0 (α, γ).
a
It follows that T (ρ0 ) does not branch at limit levels. From this we can
conclude that T (ρ0 ) is a special subtree of σQ since this is easily seen to be
so for any subtree of σQ which is finitely branching at limit nodes.
1.15 Definition. Identifying the power set of Q with the particular copy
2Q of the Cantor set, define for every countable ordinal α,
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 -subsets
of the Cantor set whose union is equal to the Cantor set.
a
1.17 Lemma. The set X = {ρ0β : β < ω1 } considered as a subset of the
Cantor set 2Q has universal measure zero.
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 strictly
increasing map into the reals (as for example σQ does). It follows that S
must be countable and so we are done.
a
2. The coherence of maximal weights
2.1 Definition. The maximal weight of the walk is the two-place function
ρ1 : [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 maximal
integer appearing in the sequence ρ0 (α, β).
2.2 Lemma. For all α < β < ω1 and n < ω,
(a) {ξ ≤ α : ρ1 (ξ, α) ≤ n} is finite,
(b) {ξ ≤ α : ρ1 (ξ, α) 6= ρ1 (ξ, β)} is finite.
Proof. The proof is by induction. To prove (a) it suffices to show that for
every n < ω and every A ⊆ α of order-type ω there is ξ ∈ A such that
ρ1 (ξ, α) > n. Let η = sup(A). If η = α one chooses arbitrary ξ ∈ A with
the property that |Cα ∩ ξ| > n, so let us consider the case η < α. Let
α1 = min(Cα \ η). By the inductive hypothesis there is ξ ∈ A such that:
ξ > max(Cα ∩ η),
(I.14)
ρ1 (ξ, α1 ) > n.
(I.15)
4 This is another use of the convention ρ (α, α) = 0 in order to facilitate the recursive
1
definition.
10
I. Coherent Sequences
Note that ρ0 (ξ, α) = h|Cα ∩ η|ia ρ0 (ξ, α1 ), and therefore
ρ1 (ξ, α) ≥ ρ1 (ξ, α1 ) > n.
To prove (b) we show by induction that for every A ⊆ α of order-type
ω there exists ξ ∈ A such that ρ1 (ξ, α) = ρ1 (ξ, β). Let η = sup(A) and let
β1 = min(Cβ \ η). Let n = |Cβ ∩ η| and let
B = {ξ ∈ A : ξ > max(Cβ ∩ η) and ρ1 (ξ, β1 ) > n}.
Then B is infinite, so by the induction hypothesis we can find ξ ∈ B such
that ρ1 (ξ, α) = ρ1 (ξ, β1 ). Then
ρ1 (ξ, β) = max{n, ρ1 (ξ, β1 )} = ρ1 (ξ, β1 ),
so we are done.
a
Note the following unboundedness property of ρ1 which immediately follows from the previous lemma and which will be elaborated on in latter
sections of this Chapter.
2.3 Corollary. For every pair A and B uncountable families of pairwise
disjoint finite subsets of ω1 end every positive integer n there exist uncountable subfamilies A0 ⊆ A and B0 ⊆ B such that for every pair a ∈ A0 and
b ∈ B0 such that a < b5 we have ρ1 (α, β) > n for all α ∈ a and β ∈ b.
a
2.4 Corollary. The tree
T (ρ1 ) = {t : α −→ ω : α < ω1 , t =∗ ρ1α }6
is a coherent, homogeneous, and R-embeddable Aronszajn tree.
a
The following is an interesting property of T (ρ1 ) or any other coherent
tree of finite-to one mappings from countable ordinals into ω.
2.5 Theorem. The cartesian square of any lexicographically ordered coherent tree of finite-to-one mappings is the union of countably many chains.
Proof. Let T (a) be a given tree given by a coherent sequence aα : α −→
ω (α < ω1 ) of finite-to-one mappings. Let <l denote the lexicographical
ordering of T (a). We need to decompose pairs (s, t) of T (a) into countably
many classes that are chains in the cartesian product of <l . In fact, by
symmetry it suffices to decompose pairs (s, t) with l(s) ≤ l(t) where l(s)
5 For
sets of ordinals a and b we write a < b whenever α < β for all α ∈ a and β ∈ b.
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.
6 Here,
2. The coherence of maximal weights
11
and l(t) are levels as well as domains of the mappings s and t, respectively.
Given such such pair (s, t), let
D(s, t) = {ξ : ξ = l(s) or ξ < l(s) and s(ξ) 6= t(ξ)},
and let n(s, t) be the maximal value s or t takes on D(s, t). Let
F (s, t) = {ξ : ξ = l(s) or ξ < l(s) and s(ξ) ≤ n(s, t) or t(ξ) ≤ n(s, t)}.
Note that D(s, t) ⊆ F (s, t). Finally, to the pair (s, t) we associate a hereditarily finite set p(s, t) which codes the behavior of the restrictions of s and
t on the set F (s, t).
Suppose we are given two pairs (s, t) and (u, v) such that p(s, t) = p(u, v)
and such that s <l u. We need to show that t ≤l v. By symmetry, we
may assume that l(s) ≤ l(u) and that s and u are actually different. Let
α = ∆(s, u), and let us first consider the case when α < l(s), i.e., when s
and u are incomparable. Note that by the choice of parametrization and
the 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 impossible
since
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 agree
bellow l(s) we conclude that t <l v, as required. If l(s) = l(t)and since t
and v agree bellow l(s) we conclude that v extends t, and therefore t <l v,
as required. This finishes the proof.
a
12
I. Coherent Sequences
2.6 Definition. Consider the following extension of T (ρ1 ):
Te(ρ1 ) = {t : α −→ ω : α < ω1 and t ¹ ξ ∈ T (ρ1 ) for all ξ < α}.
If we order Te(ρ1 ) by the right lexicographical ordering <r we get a complete
linearly ordered set. It is not continuous as it contains jumps of the form
[ta hmi, ta hm + 1ia~0],
where t ∈ T (ρ1 ) and m < ω. Removing the right-hand points from all the
e 1 ).
jumps we get a linearly ordered continuum which we denote by A(ρ
e 1 ) is a homogeneous nonreversible ordered continuum that
2.7 Lemma. A(ρ
can be represented as the union of an increasing ω1 -sequence of Cantor sets.
Proof. For a countable limit ordinal δ > 0, set
eδ (ρ1 ) = {t ∈ A(ρ
e 1 ) : l(t) ≤ δ}.
A
eδ (ρ1 ) is a closed subset of A(ρ
e 1 ) with the level Tδ (ρ1 ) beThen each A
ing a countable order-dense subset. One easily concludes from this that
eδ (ρ1 ) (δ limit < ω1 ) is the required decomposition of A(ρ
e 1 ).
A
e
To show that A(ρ1 ) is homogeneous, consider two pairs x0 < x1 and
e 1 ). Choose a countable limit ordinal δ
y0 < y1 of non-endpoints of A(ρ
strictly above the lengths of x0 , x1 , y0 and y1 . So we have a Cantor set
eδ (ρ1 ), an order-dense subset Tδ (ρ1 ) of it and two pairs of points x0 < x1
A
and y0 < y1 in
eδ (ρ1 ) \ Tδ (ρ1 ).
A
eδ (ρ1 ) −→ A
eδ (ρ1 )
By Cantor’s theorem there is an order isomorphism σ : A
such that:
σ 00 Tδ (ρ1 ) = Tδ (ρ1 ),
(I.16)
σ(xi ) = yi for i < 2.
(I.17)
e 1 ) by the formula
Extend σ to the rest of A(ρ
σ(t) = σ(t ¹ δ)a t 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
e 1 ) −→ A(ρ
e 1 ).
π : A(ρ
Since T (ρ1 ) with the lexicographical ordering does not have a countable
order-dense subset, the image π 00 T (ρ1 ) must have sequences of length bigger
2. The coherence of maximal weights
13
than any given countable ordinal. So for every limit ordinal δ we can fix tδ
in T (ρ1 ) of length ≥ δ such that π(tδ ) also has length ≥ δ. Let
f (δ) = max{ξ < δ : tδ (ξ) 6= π(tδ )(ξ)} + 1.
By the Pressing Down Lemma find an uncountable set Γ of countable limit
ordinals, a countable ordinal ξ¯ and s, t ∈ Tξ̄ (ρ1 ) such that for all δ ∈ Γ:
¯
f (δ) = ξ,
(I.18)
tδ ¹ ξ¯ = s and π(tδ ) ¹ ξ¯ = t.
(I.19)
Moreover we may assume that
{tδ ¹ δ : δ ∈ Γ} and {π(tδ ) ¹ δ : δ ∈ Γ}
are antichains of the tree T (ρ1 ). Pick γ 6= δ in Γ and suppose for definiteness that tγ <l tδ . Note that the ordinal ∆(tγ , tδ ) where this relation is
decided 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
¯ min{γ, δ}).
(ξ,
It follows that ∆(π(tγ ), π(tδ )) = ∆(tγ , tδ )(= ξ) and that
π(tγ )(ξ) = tγ (ξ) < tδ (ξ) = π(tδ (ξ)),
contradicting the fact that π is order-reversing.
a
2.8 Definition.
A homogeneous Eberlein compactum 7 . The set Te(ρ1 ) has another natural
structure, a topology generated by the family of sets of the form
Vet = {u ∈ Te(ρ1 ) : t ⊆ u},
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 Te(ρ1 ) can be
regarded as the set of all downward closed chains of the tree T 0 (ρ1 ) and the
topology on Te(ρ1 ) is simply the topology one obtains from identifying the
power set of T 0 (ρ1 ) with the cube
{0, 1}T
0
(ρ1 )
with its Tychonoff topology8 . Te(ρ1 ) being a closed subset of the cube is
compact. In fact Te(ρ1 ) has some very strong topological properties such as
the property that closed subsets of Te(ρ1 ) are its retracts.
7 Recall that a compactum X is Eberlein if its function space C(X) can be generated
by a subset which is compact in the weak topology.
8 This is done by identifying a subset V of T 0 (ρ ) with its characteristic function
1
χV : T 0 (ρ1 ) −→ 2.
14
I. Coherent Sequences
2.9 Lemma. Te(ρ1 ) is a homogeneous Eberlein compactum
Proof. The proof that Te(ρ1 ) is homogeneous is quite similar to the corresponding part of the proof of the Lemma 2.7. To see that Te(ρ1 ) is an
Eberlein compactum, i.e. that the function space C(Te(ρ1 )) is weak compactly generated, let {Xn } be a countable antichain decomposition of T (ρ1 )
and consider the set
K = {2−n χVet : n < ω, t ∈ Xn } ∪ {χ∅ }.
Note that K is a weakly compact subset of C(Te(ρ1 )) which separates the
points of Te(ρ1 ).
a
E. Borel’s notion of strong measure zero is a metric notion that has several closely related topological notions due to K.Menger, F.Rothberger and
others. For example Rothberger’s property C 00 is a purely topological notion
which in the class of metric spaces is slightly stronger than the property C,
i.e. being of strong measure zero. It says that for every sequence {Un } of
open covers of a space X one can choose Un ∈ Un for each n such that
X=
∞
[
Un .
n=0
2.10 Lemma. T (ρ1 ) has Rothberger’s property C 00 .
a
2.11 Remark. The subspace T (ρ1 ) of Te(ρ1 ) has in fact a considerably
stronger covering property than C 00 . For example, it can be shown that
its function space C(T (ρ1 )) with the topology of pointwise convergence is a
Fréchet space, or in other words, has the property that any family of continuous functions on T (ρ1 ) which pointwise accumulates to the constantly
equal 0 function 0̄ contains a sequence {fn } which pointwise converges to 0̄.
In fact the function space C(T (ρ1 )) with the topology of pointwise convergence has a natural decomposition as the increasing union of an ω1 -sequence
of closed separable metric vector subspaces. The separable metric vector
subspaces are determined by the retraction maps of T (ρ1 ) onto its levels.
In the next application the coherent sequence ρ1α : α −→ ω (α < ω1 )
of finite-to-one maps needs to be turned into a coherent sequence of maps
that are actually one-to-one. One way to achieve this is via the following
formula:
ρ̄1 (α, β) = 2ρ1 (α,β) · (2 · |{ξ ≤ α : ρ1 (ξ, β) = ρ1 (α, β)}| + 1).
Define ρ̄1α from ρ̄1 just as ρ1α was defined from ρ1 ; then the ρ̄1α ’s are oneto-one and coherent. From ρ1 one also has a natural sequence rα (α < ω1 )
of elements of ω ω defined as follows
rα (n) = |{ξ ≤ α : ρ1 (ξ, α) ≤ n}|.
2. The coherence of maximal weights
15
Note that rβ eventually strictly dominates rα whenever α < β.
2.12 Definition. The sequences eα = ρ̄1α (α < ω1 ) and rα (α < ω1 ) can
be used in describing a functor
G 7−→ G∗ ,
which to every graph G on ω1 associates another graph G∗ on ω1 as follows:
−1
{α, β} ∈ G∗ iff {e−1
α (l), eβ (l)} ∈ G
(I.20)
for all l < ∆(rα , rβ ) for which these preimages are both defined and different9 .
2.13 Lemma. Suppose that every uncountable family F of pairwise disjoint
finite 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 consists
of finite cliques11 of G∗ .
Proof. Let F be a given uncountable family of pairwise disjoint finite cliques
of G∗ . We may assume that all members of F are of some fixed size k.
Consider a countable limit ordinal δ > 0 and an A in F with all its
elements above δ. Let n = n(A, δ) be the minimal integer such that for all
α < β in A ∪ {δ}:
∆(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 H̄(δ, A) of H(δ, A), the sequence
eα ¹ H(δ, A) (α ∈ A)
collapses to a k-sequence s̄(δ, A) of mappings with integer domains. Let
r̄(δ, A) denote the k-sequence
rα ¹ (n(δ, A) + 1) (α ∈ A)
enumerated in increasing order. Hence every A ∈ F above δ generates a
quadruple
(H(δ, A) ∩ δ, r̄(δ, A), s̄(δ, A), n(δ, A))
9 As it will be clear from the proof of the following lemma the functor G −→ G∗ can
equally be based on any other coherent sequence eα : α −→ ω (α < ω1 ) of one-to-one
mappings and any other sequence rα (α < ω1 ) of pairwise distinct reals.
10 Here, A ⊗ B = {{α, β} : α ∈ A, β ∈ B, α 6= β}.
11 A clique of G∗ is a subset C of ω with the property that [C]2 ⊆ G∗ .
1
12 Recall that for s 6= t ∈ ω ω , we let ∆(s, t) = min{k : s(k) 6= t(k)}.
16
I. Coherent Sequences
of parameters. Since the set of quadruples is countable, by the assumption
on the graph G we can find two distinct members Aδ and Bδ of F above δ
that generate the same quadruple of parameters denoted by
(Hδ , r̄δ , s̄δ , nδ ).
Moreover, we choose Aδ and Bδ to satisfy the following isomorphism condition
For every l ≤ nδ , i < k, if α is the ith member of Aδ , β the
−1
ith member of Bδ and if e−1
α (l) and eβ (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)
Let
Iδ = {ξ : e(ξ, α) ≤ mδ for some α ≥ ξ in Aδ ∪ Bδ }.
Let p̄δ be the k-sequence rα ¹ mδ (α ∈ Aδ ) enumerated in increasing order
and similarly let q̄δ be the k-sequence rβ ¹ mδ (β ∈ Bδ ) enumerated increasingly. Let t̄δ and ūδ be the transitive collapse of eα ¹ Iδ (α ∈ Aδ ) and
eβ ¹ Iδ (β ∈ Bδ ), respectively. By the Pressing Down Lemma there is an
unbounded Γ ⊆ ω1 and tuples
(H, r̄, s̄, n) and (I, p̄, q̄, t̄, ū, m)
such that for all δ ∈ Γ:
(Hδ , r̄δ , s̄δ , nδ ) = (H, r̄, s̄, n),
(I.25)
(Iδ ∩ δ, p̄δ , q̄δ , t̄δ , ūδ , mδ ) = (I, p̄, q̄, t̄, ū, m).
(I.26)
Moreover we assume the following analogue of (I.23) where k1 = |Iγ \ γ| for
some (equivalently all) γ ∈ Γ:
For every γ, δ ∈ Γ and i < k1 , if η is the ith member
of Iγ and if ξ is the ith member of Iδ in their increasing
enumerations, then η and ξ are G-connected to the same
elements of the root I, and if η happens to be the i∗ th
member of Aγ (or Bγ ) for some i∗ < k, then ξ must be the
i∗ th member of Aδ (Bδ resp.) and vice versa.
(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)
2. The coherence of maximal weights
17
The proof of Lemma 2.13 is finished once we show that
Aγ ⊗ Bδ ⊆ G∗ .
(I.30)
Let i, j < k be given and let α be the ith member of Aγ and let β be the
jth member of Bδ .
Case 1: i 6= j. Pick an l < ∆(rα , rβ ). By (I.21), l < n. Assume that
−1
e−1
α (l) and eβ (l) are both defined.
−1
Subcase 1.1: e−1
α (l) < γ and eβ (l) < δ. By the first choice of parame−1
−1
ters, eα (l) and eβ (l) are members of the set H which is an initial part of
H(γ, Aγ ) and H(δ, Bδ ). Therefore, we have that the behavior of eα and eβ
on H is encoded by the ith and jth term of the sequence s̄, respectively. In
particular, we have
−1
e−1
(I.31)
β (l) = eβ 00 (l),
where β 00 =the jth member of Aγ . By our assumption that [Aγ ]2 ⊆ G∗ we
infer that {α, β 00 } ∈ G∗ . Referring to the definition (I.20) we conclude that
−1
e−1
α (l) and eβ (l) must be G-connected if they are different.
−1
Subcase 1.2: e−1
α (l) < γ and eβ (l) ∈ Iδ \ δ. Let
β 0 = the jth member of Bγ .
By the choice of parametrization, the position of e−1
β 0 (l) in Iγ is the same
as the position of e−1
(l)
in
I
so
by
(I.27)
their
relationship
to the point
δ
β
e−1
(l)
of
the
root
I
is
the
same.
Similarly,
by
the
choice
of
the
first set of
α
parameters, letting β 00 be the jth member of Aγ , the position of e−1
β 0 (l) in
−1
H(γ, Bγ ) and eβ 00 (l) in H(γ, Aγ ) is the same, so from (I.23) we conclude
that their relationship with e−1
α (l) is the same. However, we have checked
−1
in the previous subcase that e−1
α (l) and eβ 00 (l) are G-connected in case they
are different.
−1
Subcase 1.3: e−1
α (l) ∈ Iγ \ γ and eβ (l) < δ. This is essentially symmetric to the previous subcase.
−1
Subcase 1.4: e−1
α (l) ∈ Iγ \ γ and eβ (l) ∈ Iδ \ δ. The fact that
−1
−1
{eα (l), eβ (l)} ∈ G in this case follows from (I.29).
Case 2: i = j. Consider an l < ∆(rα , rβ ). Note that now we have l < m
(see (I.24)).
−1
−1
−1
Subcase 2.1: e−1
α (l) < γ and eβ (l) < δ. Then eα (l) and eβ (l) are
elements of I which is an initial part of both Iγ and Iδ . Therefore the
jth(= ith) term of ū encodes both eβ ¹ I and eβ 0 ¹ I (see the definition of
β 0 above). It follows that
−1
e−1
β (l) = eβ 0 (l).
18
I. Coherent Sequences
If l ≤ n then e−1
β 0 (l) belongs to H and since the jth(= ith) term of s̄ encodes
both eα ¹ H and eβ ¹ H it follows that
−1
e−1
α (l) = eβ 0 (l).
−1
If l > n then e−1
α (l) and eβ 0 (l) are not members of H so from (I.22) and
the definition of
H(γ, Aγ ) ∩ γ = H = H(γ, Bγ ) ∩ γ
we conclude that
−1
−1
−1
eγ (e−1
α (l)) = eα (eα (l)) = l = eβ 0 (eβ 0 (l)) = eγ (eβ 0 (l)).
Since eγ is one-to-one we conclude that
−1
−1
e−1
α (l) = eβ 0 (l) = eβ (l).
−1
0
Subcase 2.2: e−1
α (l) < γ and eβ (l) ∈ Iδ \ δ. Recall that β is the
ith(= jth) member of Bγ so by the first choice of parameters the relative
position of e−1
β 0 (l) in H(γ, Bγ ) must be the same as the relative position of
−1
eα (l) in H(γ, Aγ ), i.e. it must belong to the root H. Note that if j ∗ is the
position of β in Iδ then j ∗ must also be the position of β 0 in Iγ . It follows
that the j ∗ th term of ū encodes both
eβ ¹ Iδ and eβ 0 ¹ Iγ = eβ 0 ¹ I.
So it must be that e−1
β (l) belongs to the root I, a contradiction. So this
subcase never occurs.
−1
Subcase 2.3: e−1
α (l) ∈ Iγ \ γ and eβ (l) < δ. This is symmetric to the
previous subcase, so it also never occurs.
−1
−1
−1
Subcase 2.4: e−1
α (l) ∈ Iγ \ γ and eβ (l) ∈ Iδ \δ. Then {eα (l), eβ (l)} ∈
G follows from (I.29).
This completes the proof of Lemma 2.13.
a
2.14 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 binary
sequence s of length equal to some l + 1, set
Γs = {ξ < ω1 : e(ξ, α) = l for some α in Γ with s ⊆ rα }.
Then the sets Γs cover ω1 and [Γs ]2 ⊆ G for all s.
a
2. The coherence of maximal weights
19
2.15 Remark. Let G be the comparability graph of some Souslin tree T .
Then for every uncountable family F of pairwise disjoint cliques of G (finite
chains of T ) there exist A 6= B in F such that A ∪ B is a clique of G (a chain
of T ). However, it is not hard to see that G∗ fails to have this property
(i.e. the conclusion of Lemma 2.13). This shows that some assumption on
the graph G in Lemma 2.13 is necessary. There are indeed many graphs
that satisfy the hypothesis of Lemma 2.13. Many examples appear when
one is trying to apply Martin’s axiom to some Ramsey-theoretic problems.
Note that the conclusion of Lemma 2.13 is simply saying that the poset
of all finite cliques of G∗ is ccc while its hypothesis is a bit stronger than
the fact that the poset of all finite cliques of G is ccc in all of its finite
powers. Applying 2.14 to the case when G is the incomparability graph of
some Aronszajn tree, we see that the statement saying that all Aronszajn
trees are special is a purely Ramsey-theoretic statement in the same way
Souslin’s hypothesis is.
We finish this section by showing that the assumption (c) on a given
C-sequence Cα (α < ω1 ) on which we base our walks and the function ρ1 is
not sufficient to give us the stronger conclusion that the tree T (ρ1 ) can be
decomposed into countably many antichains (see Corollary 2.4). To describe
an example, we need the following notation given that we have fixed one Csequence Cα (α < ω1 ). Let Cα (0) = 0 and for 0 < n < ω, let Cα (n)) denote
the n’th element of Cα according to its increasing enumeration with the
convention that Cα+1 (n) = α for all n > 0. We assume that the C-sequence
is chosen so that for a limit ordinal α > 0 and a positive integer n, the ordinal
Cα+1 (n) is at least n + 1 steps away from the closest limit ordinal bellow it.
For a set D of countable ordinals, let D0 denote the set of successor ordinals
in D. We also fix, for each countable ordinal α a one-to-one function eα
whose domain is the set α0 of all successor ordinals < α and range included
in ω which cohere in the sense that eα (ξ) = eβ (ξ) for all but finitely many
successor ordinals ξ ∈ α0 ∩ β 0 . For each r ∈ ([ω]<ω )ω , we
S associate another
C-sequence Cαr (α < ω1 ) by letting Cαr = Cα ∩ [ξαr , α) ∪ n∈ω Dαr (n), where
Dαr (n) = {ξ ∈ [Cα (n), Cα (n + 1))0 : eα (ξ) ∈ r(n)},
S
and where ξαr = sup( n∈ω Dαr (n)). (This definition really takes place only
r
= {α}.) This
when α is a limit ordinal; for successor ordinals we put Cα+1
r
gives us a C-sequence Cα (α < ω1 ) and we can consider the corresponding
ρr1 : [ω1 ]2 −→ ω as above. This function will have the properties stated in
Lemma 2.2, so the corresponding tree
T (ρr1 ) = {(ρr1 )β ¹ α : α ≤ β < ω1 }
is a coherent tree of finite-to-one mappings that admits a strictly increasing
function into the real line. The following fact shows that T (ρr1 ) will typically
not be decomposable into countably many antichains.
20
I. Coherent Sequences
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 discrete
topology and its power ([ω]<ω )ω with the corresponding product topology
and the given Cohen real r is to meet all dense Gδ subset of this space
that one
define. Thus in particular, ξαr = α, and therefore,
S can explicitly
r
r
Cα = n∈ω Dα (n) for every limit ordinal α. Since the tree T (ρr1 ) is coherent,
to establish the conclusion of the Lemma, it suffices to show that that for
every stationary Γ ⊆ ω1 there exists γ, δ ∈ Γ such that (ρr1 )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 ⊂ (ρx1 )δ }
is a dense-open subset of ([ω]<ω )ω . So, given a finite partial function p from
ω into [ω]<ω , it is sufficient to find a finite extension q of p such that the
basic open subset of ([ω]<ω )ω determined by q is included in UΓ . Let n be
the minimal integer that is bigger than all integers appearing in the domain
of p or any set of the form p(j) for j ∈ dom(p). For γ ∈ Γ, set
Fn (γ) = {ξ ∈ γ 0 : eγ (ξ) ≤ n}.
Applying the Pressing Down Lemma, we obtain a finite set F ⊆ ω1 and a
stationary set ∆ ⊆ Γ such that Fn (γ) = F for all γ ∈ ∆ and such that if
α = max(F ) + 1 then eγ ¹ α = eδ ¹ α for all γ, δ ∈ ∆. A similar application
of the Pressing Down Lemma will give us an integer m > n and two ordinals
γ < δ in ∆ such that Cγ (j) = Cδ (j) for all j ≤ m, Cδ (m + 1) > γ + 1, and
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, ..., m}
such that:
(1) q(m) = {eδ (γ + 1)}, and
(2) q(j) = ∅ for any j < m not belonging to dom(p).
Choose any x ∈ ([ω]<ω )ω extending the partial mapping q. Then from the
choices of the objects n, ∆, F ,m, γ, δ and q, we have that
(3) γ + 1 ∈ Cδx , and
(4) Cδx ∩ γ is an initial segment of Cγx .
It follows that, given a ξ < γ, the walk from δ to ξ along the C-sequence
Cβx (β < ω1 ) either leads to the same finite string of the corresponding
weights as the walk from γ to ξ, or else it starts with first two steps to γ + 1
and γ, and then follows the walk from γ to ξ. Since ρx1 (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) ≥ ρx1 (ξ, γ). On the other hand, note that by (4), in the second
case, the weight |Cδx ∩ ξ| of the first step from δ to ξ is less than or equal
to the weight |Cγx ∩ ξ| of the first step from γ to ξ. It follows that we have
also the other inequality ρx1 (ξ, δ) ≤ ρx1 (ξ, γ). Hence we have shown that
ρx1 (β, γ) = ρx1 (β, δ) for all β < γ, or in other words, that (ρx1 )γ ⊂ (ρx1 )δ . This
finishes the proof.
a
2.17 Question. What is the condition one needs to put on a given Csequence 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 by
osc(x, y) = |x/E(x, y)|,
where E(x, y) is the equivalence relation on x defined by letting αE(x, y)β
if and only if the closed interval determined by α and β contains no point
from y \ x. So, if x and y are disjoint, osc(x, y) is simply the number of
convex pieces the set x is split by the set y. The resulting ’oscillation theory’
reproduced here in part has shown to be a quite useful and robust coding
technique. We give some applications to this effect and in Section 17 below
we shall extend the oscillation theory to a more general context.
3.1 Definition. For α < β < ω1 , set
osc0 (α, β) = osc(Tr(∆0 (α, β), α), Tr(∆0 (α, β), β)),
where ∆0 (α, β) = min{ξ ≤ α : ρ0 (ξ, α) 6= ρ0 (ξ, β)} − 113 .
The following result shows that the upper traces do realize all possible
oscillations in any uncountable subset of ω1 .
3.2 Lemma. For every uncountable subset Γ of ω1 and every positive integer n there is an integer l such that for all k < n there exist α, β ∈ Γ such
that osc0 (α, β) = l + k.
Proof. Fix a sequence Mi (i ≤ n) of continuous ∈-chains of length ω1
of countable elementary submodels of (Hω2 , ∈) containing all the relevant
objects such that Mi is an element of the minimal model of Mi+1 for all
13 Note
that by Lemma 1.14 this is well defined.
22
I. Coherent Sequences
i < n. Let Ci = {M ∩ ω1 : M ∈ Mi }. Note that each Ci is a closed and
unbounded subset of ω1 and that for each i < n, Ci is an element of every
model of Mi+1 . Let Mn be the minimal model of Mn and pick a β ∈ Γ above
δn = Mn ∩ ω1 . Since Cn−1 ∈ Mn we can find Mn−1 ∈ Mn−1 ∩ Mn such that
δn−1 = Mn−1 ∩ ω1 > L(δn , β). Similarly, we can find Mn−2 ∈ Mn−1 ∩ Mn−1
such that δn−2 = Mn−2 ∩ ω1 > L(δn−1 , δn ), and so on. This will give
us an ∈-chain Mi (i ≤ n) of countable elementary submodels of (Hω2 , ∈)
containing all relevant objects such that if we let δn+1 = β and δi = Mi ∩ ω1
for i ≤ n, then δi > L(δi+1 , δi+2 ) for all i < n. Pick a point ξ ∈ Cδ0 above
L(δ0 , β). Then (see Lemma 1.6 above), δi ∈ Tr(ξ, β) for all i ≤ n and in fact
δ0 → ξ is the last step of the walk from β to ξ. Let t = ρ0 (ξ, β), ti = ρ0 (δi , β)
for i ≤ n. The tn < tn−1 < ... < t0 < t. Let
Γ0 = {γ ∈ Γ : t = ρ0 (ξ, γ)}.
Then Γ0 ∈ M0 and β ∈ Γ0 . A simple elementarity argument using models
M0 ∈ M1 ∈ ... ∈ Mn shows that there is an uncountable subset Σ of Γ0
such that Σ ∈ M0 and
T = {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 immediate
successor or there is i ≤ n and uncountably many γ < ω1 such that x∪{γ} ∈
T and Tr(γ, α) = ti for all α ∈ Σ witnessing x ∪ {γ} ∈ T . Note that {ξ}
is a root of T as well as its splitting nodes. Let Ω be the uncountable set
such that {ξ, γ} ∈ T for all γ ∈ Ω. Working in M0 we find η > ξ and
uncountable Ω0 ⊆ Ω such that η = ∆0 (γ, β) for all γ ∈ Ω0 and therefore
η = ∆0 (α, β) for all α ∈ Σ witnessing {ξ, γ} ∈ T . Still working in M0 and
going to an uncountable subset of Ω0 we may assume that Tr(η, γ) (γ ∈ Ω0 )
form a ∆-system with root x0 . Let Σ0 be the set of all α ∈ Σ witnessing
{ξ, γ} ∈ T for some γ ∈ Ω0 . Let s = ρ0 (η, β). Note that s end-extends t0
but not necessarily t. Consider now the tree
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 s
we have that ρ0 (γ, α) = s0 for all α ∈ Σ0 witnessing x0 ∪ {γ} ∈ X . All
other splitting nodes of X are determined by the other initial segments
s0 w t0 = t1 = ... = tn . Recall that all these objects are elements of M0 and
therefore elements of all Mi (i ≤ n). Let y = Tr(η, β) and let l = osc(x0 , y).
Fix a k ≤ n. Working in M1 find γ0 ∈ Ω0 such that Tr(η, γ0 ) \ x0 >
Tr(η, β) ∩ δ1 and find a splitting node x1 w Tr(η, γ0 ) corresponding to t1 .
Then Ω1 = {γ < ω1 : x1 ∪ {γ} ∈ X } is uncountable and ρ0 (γ, α) = t1 for
all α ∈ Σ0 witnessing x1 ∪ {γ} ∈ X . Working in M2 find γ1 ∈ Ω1 such
3. Oscillations of traces
23
that γ1 > Tr(η, β) ∩ δ2 and a splitting node x2 w Tr(η, γ1 ) corresponding to
t2 , and so on. Proceeding in this way we construct an increasing sequence
x0 < x1 < ... < xk of splitting nodes of T such that ti ∈ Mi and
xi \ xi−1 > Tr(η, β) ∩ δ2 for all 1 ≤ i ≤ k.
Find an arbitrary α ∈ Σ0 ∩ Mk such that xk v Tr(η, α) Let x = Tr(η, α).
Then
x/E(x, y) = x0 /E(x0 , y)) ∪ {x1 \ x0 , x2 \ x1 , ..., xk \ xk−1 }.
This is so because the ordinal δ0 ∈ y\x separates the last class of x0 /E(x0 , y)
from the new classes xi \ xi−1 (1 ≤ i ≤ k). It follows that osc0 (α, β) =
osc(x, y) = l + k, as required.
a
Consider the following projection of the oscillation mapping osc0 ,
osc∗0 (α, β) = max{m ∈ ω : 2m |osc0 (α, β)}.
3.3 Corollary. For every uncountable Γ ⊆ ω1 and every positive integer k
a
there exist α, β ∈ Γ such that osc∗0 (α, β) = k.
3.4 Corollary. Every inner model model M of set theory which correctly
computes ω1 contains a partition of ω1 into infinitely many pairwise disjoint
subsets that are stationary in the universe V of all sets.
Proof. Our assumption about M means in particular that the C-sequence
Cξ (ξ < ω1 ) can be chosen to be an element of M. It follows that the
oscillation mapping osc∗0 is also an element of M. It follows that for each
α < 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 for
every n < ω the set Sαn is stationary in V. This is an immediate consequence
of Corollary 3.3.
a
3.5 Remark. In [46], Larson showed that Corollary 3.4 cannot be extended
to partitions of ω1 into uncountably many pairwise disjoint stationary sets.
He also showed that under the Proper Forcing Axiom, for every mapping
c : [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 with
countable ranges.
24
I. Coherent Sequences
One can count oscillation not only between two finite subsets of ω1 but
also between two finite partial functions from ω1 into ω, i.e., between two
finite ’weighted sets’ x and y :
osc(x, y) = |{ξ ∈ dom(x) : x(ξ) ≤ y(ξ) but x(ξ + ) > y(ξ + )}|,
where ξ + denotes the minimal ordinal in dom(x) above ξ if there is one.
3.6 Definition. For α < β < ω1 , set
osc1 (α, β) = osc(ρ1α ¹ L(α, β), ρ1β ¹ L(α, β)).
We have the following analogue of Lemma 3.2 which is now true even in
its rectangular form.
3.7 Lemma. For every pair Γ and Σ of uncountable subsets of ω1 and every
positive integer n there is an integer l such that for all k < n there exist
α ∈ Γ and β ∈ Σ such that osc1 (α, β) = l + k.
Proof. Fix a continuous ∈-chain N of length ω1 of countable elementary
submodels of (Hω2 , ∈) containing all the relevant objects and fix also a
countable elementary submodel M of (Hω2 , ∈) that contains N as an element. Let δ = M ∩ ω1 . Fix α0 ∈ Γ \ δ and β0 ∈ Σ \ δ. Let L0 = L(δ, β0 ).
By exchanging the steps in the following construction, we may assume that
ρ1 (ξ, α0 ) > ρ1 (ξ, β0 ) for ξ = max(L0 ). Choose ξ0 ∈ Cδ such that ξ > L0 and
such that the three mappings ρ1α0 ,ρ1β0 , and ρ1δ agree on the interval [ξ, δ).
By elementarity of M there will be δ + > δ, α0+ ∈ Γ \ δ + , and β0+ ∈ Σ \ δ +
+
realizing the same type over the parameters L0 and ξ0 . Let L+
1 = L(δ, δ0 ).
+
Note that ξ0 ≤ min(L1 ). It follows that the mappings ρ1α+ and ρ1β + agree
0
0
+
on L+
1 . Choose an N ∈ N belonging to M such that N ∩ ω1 > L1 and
+
choose ξ1 ∈ Cδ \ N. Then we can find δ1 > δ and α1 ∈ Γ \ δ1 and β1 ∈ Σ \ δ1
realizing the same type as δ, α0+ , and β0+ over the objects accumulates so
++
far, L0 , ξ0 , L+
= L(δ, δ1 ) and let t+
1 , N , and ξ1 . Let L1
1 be the restriction
+
of ρ1α+ on max(L1 ) + 1. Then the set of α ∈ Γ whose ρ1α end-extends t+
1
1
belongs to the submodel N and is uncountable, since clearly it contains the
ordinal α1+ which does not belong to N. So by the elementarity of N and
Corollary 2.3 there is α1 ∈ Γ \ δ such that ρ1α1 end-extends 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 ∪ L1 . Using similar reasoning we can
find t2 ∈ M ∩ T (ρ1 ) end-extending t1 , an α2 ∈ Γ \ δ whose ρ1α2 end-extends
3. Oscillations of traces
25
+
++
t2 , finite sets δ > max(L++
2 ) > max(L2 ) > max(L1 ), and β2 ∈ Σ \ δ such
+
++
+
++
that L(δ, β2 ) = L0 ∪ L1 ∪ L1 ∪ L2 ∪ L2 and such that the analogues of
I.32 and I.33 hold for all α ∈ Γ whose ρ1α end-extends t2 , and so on. This
++
++
procedure gives us a sequence L0 < L+
< ... < L+
of finite
n < Ln
1 < L1
subsets of δ, an increasing sequence t1 < t2 < ... < tn of nodes of T (ρ0 ) ∩ M,
and a sequence βk (1 ≤ k ≤ n) such that for all 1 ≤ k ≤ n, and all α ∈ Γ
whose ρ1α end-extends tk ,
++
++
L(δ, βk ) = L0 ∪ L+
∪ ... ∪ L+
1 ∪ L1
k ∪ Lk ,
(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++
and such that all the ρ1βk ’s agree on Ln . Note that by
n
Lemma 1.6 for each 1 ≤ k ≤ n,
++
++
L(α, βk ) = L0 ∪ L+
∪ ... ∪ L+
∪ Ln .
1 ∪ L1
k ∪ Lk
(I.37)
Let l = osc(ρ1α ¹ (L0 ∪ Ln ), ρ1β ¹ (L0 ∪ Ln )) where β is equal to one of the
βk ’s. Note that l does not depend on which βk we choose and that according
to I.35, I.36 and I.37, for each 1 ≤ k ≤ n, we have that osc1 (α, βk ) = l + k,
as required.
a
Note that in the above proof the set Γ can be replaced by any uncountable
family of pairwise disjoint sets giving us the following slightly more general
conclusion.
3.8 Lemma. For every uncountable family G of pairwise disjoint finite subsets of ω1 all of some fixed size m, every uncountable subset Σ of ω1 , and
every positive integer n there exist integers li (i < m) and a ∈ G such that
for all k < n there exist β ∈ Σ such that osc1 (a(i), β) = li + k for all
i < m.14
a
Having in mind an application we define a projection of osc1 as follows.
For a real x, let [x] denote the greatest integer not bigger than x. Choose a
1
1
mapping ∗ : R → ω such that ∗(x) = n if x−[x] ∈ In , where In = ( n+3
, n+2
].
∗
For x in R we use the notation x for ∗(x). Choose a one-to-one sequence
xξ (ξ < ω1 ) of elements of the first half of the unit interval I = [0, 1] which
together with the point 1 form a set that is linearly independent over the
field of rational numbers. Finally, set
osc∗1 (α, β) = (osc1 (α, β) · xα )∗ .
Then Lemma 3.8 turns into the following statement about osc∗1 .
14 Here
a(i) denotes the ith member of a according to the increasing enumeration of a.
26
I. Coherent Sequences
3.9 Lemma. For every uncountable family G of pairwise disjoint finite subsets of ω1 all of some fixed size m, every uncountable subset Σ of ω1 ,
and every mapping h : m → ω there exist a ∈ G and β ∈ Σ such that
osc∗1 (a(i), β) = h(i) for all i < m.
Proof. Let ε be half of the length of the shortest interval of the form
Ih(i) (i < m). Note that for a ∈ G, the points 1, xa(0) , ..., xa(m−1) are rationally 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. Going
to an uncountable subfamily of G, we may assume that there is n such that
na = n for all a ∈ G. By Lemma 3.8 there exits a ∈ G and li (i < m) such
that for every k < n there is βk > a such that osc1 (a(i), βk ) = li + k for all
i < m. For i < m, let yi = li · xa(i) and let zi be the middle-point of the
interval Ih(i) . Find k < n such that |k · xa(i) − (zi − yi )| < ε (mod 1) for all
i < m. Then osc∗1 (a(i), βk ) = h(i) for all i < m, as required.
a
3.10 Corollary. There is a regular hereditarily Lindelöf space which is not
separable.
Proof. For β < ω1 , let fβ ∈ {0, 1}ω1 be defined by letting fβ (β) = 1,
fβ (γ) = 0 for γ > β, and fβ (α) = min{1, osc∗1 (α, β)} for α < β. Consider
F = {fβ : β < ω1 } as a subspace of {0, 1}ω1 . Clearly F is not separable.
That F is hereditarily Lindelöf follows easily from Lemma 3.9.
a
3.11 Remark. The projection osc∗0 of the oscillation mapping osc0 appears
in [78] as historically first such a map with more than four colors that takes
all of its values on every symmetric square of an uncountable subset of ω1 .
The variations osc1 and osc∗1 are on the other hand very recent and are
due to J.T. Moore [54] who made them in order to obtain the conclusion of
Corollary 3.10. Concerning Corollary 3.10 we note that the dual implication behaves quite differently since assuming the Proper Forcing Axiom all
hereditarily separable regular spaces are hereditarily Lindelöf (see [80]).
4. Triangle inequalities
In this section we study a characteristic of the minimal walk that satisfies
certain triangle inequalities reminiscent of those found in an ultrametric
space. We also present several applications of the corresponding metriclike theory of ω1 . This leads us to the following variation of the oscillation
mapping.
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) ≤ max{ρ(α, β), ρ(β, γ)},
(c) ρ(α, β) ≤ max{ρ(α, γ), ρ(β, γ)}.
Proof. Note that ρ(α, β) ≥ ρ1 (α, β), so (a) follows from the corresponding
property of ρ1 . The proof of (b) and (c) is simultaneous by induction on α,
β and γ:
To prove (b), consider n = max{ρ(α, β), ρ(β, γ)}, and let
ξα = min(Cγ \ α) and ξβ = min(Cγ \ β).
We have to show that ρ(α, γ) ≤ n.
Case 1b : ξα = ξβ . Then by the inductive hypothesis,
ρ(α, ξα ) ≤ max{ρ(α, β), ρ(β, ξβ )}.
From the definition of ρ(β, γ) we get that ρ(β, ξβ ) ≤ ρ(β, γ) ≤ n, so replacing ρ(β, ξβ ) by ρ(β, γ) in the above inequality we get ρ(α, ξα ) ≤ n. Consider
a ξ ∈ Cγ ∩ α = Cγ ∩ β. By the inductive hypothesis
ρ(ξ, α) ≤ max{ρ(ξ, β), ρ(α, β)}.
From the definition of ρ(β, γ) we see that ρ(ξ, β) ≤ ρ(β, γ), so replacing
ρ(ξ, β) with ρ(β, γ) in the last inequality we get that ρ(ξ, α) ≤ n. Since
|Cγ ∩ α| = |Cγ ∩ β| ≤ ρ(β, γ) ≤ n, referring to the definition of ρ(α, γ) we
conclude that ρ(α, γ) ≤ n.
Case 2b : ξα < ξβ . Then ξα ∈ Cγ ∩ β, so
ρ(ξα , β) ≤ ρ(β, γ) ≤ n.
By the inductive hypothesis
ρ(α, ξα ) ≤ max{ρ(α, β), ρ(ξα , β)} ≤ n.
Similarly, for every ξ ∈ Cγ ∩ α ⊆ Cγ ∩ β,
ρ(ξ, α) ≤ max{ρ(ξ, β), ρ(α, β)} ≤ n.
Finally |Cγ ∩ α| ≤ |Cγ ∩ β| ≤ ρ(β, γ) ≤ n. Combining these inequalities we
get the desired conclusion ρ(α, γ) ≤ n.
To prove (c), consider now n = max{ρ(α, γ), ρ(β, γ)}. We have to show
that ρ(α, β) ≤ n. Let ξα and ξβ be as above and let us consider the same
two cases as above.
28
I. Coherent Sequences
¯ Then by the inductive hypothesis
Case 1c : ξα = ξβ = ξ.
¯ ρ(β, ξ)}.
¯
ρ(α, β) ≤ max{ρ(α, ξ),
This gives the desired bound ρ(α, β) ≤ n, since ρ(α, ξα ) ≤ ρ(α, γ) ≤ n and
ρ(β, ξβ ) ≤ ρ(β, γ) ≤ n.
Case 2c : ξα < ξβ . Applying the inductive hypothesis again we get
ρ(α, β) ≤ max{ρ(α, ξα ), ρ(ξα , β)} ≤ n.
This completes the proof.
a
4.3 Remark. Note the following two alternate definitions of ρ:
ρ(α, β) = max{|Cη ∩ ξ| : ξ, η ∈ F (α, β) ∪ {β} and ξ < η}.
ρ(α, β) = max{ρ1 (ξ, η) : ξ, η ∈ F (α, β) ∪ {β} and ξ < η}.
These two definitions make it more transparent (in light of Lemma 1.9)
that ρ is a subadditive function. The proof of Lemma 4.2 is given, however,
because of some later generalizations of this function to higher ordinals.
The following fact shows that the function ρ has a considerably finer
coherence property than ρ1 .
4.4 Lemma. If α < β < γ and if ρ(α, β) > ρ(β, γ), then ρ(α, γ) = ρ(α, β).
Proof. Applying Lemma 4.2,
ρ(α, γ) ≤ max{ρ(α, β), ρ(β, γ)} = ρ(α, β),
ρ(α, β) ≤ max{ρ(α, γ), ρ(β, γ)} = ρ(α, γ).
Note that max{ρ(α, γ), ρ(β, γ)} must be equal to ρ(α, γ) rather than ρ(β, γ)
since by the assumption ρ(α, β), which is bounded by the maximum, is
bigger than ρ(β, γ).
a
4.5 Remark. Referring to the standard definitions of metrics and ultrametrics the properties 4.2(b) and (c) could more properly be called ultrasubadditivities though we shall keep calling them subadditivity properties
of the function ρ. Recall the notion of a full lower trace F (α, β) given in
Definition 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 have
the following inequalities whenever α < β < γ:
(a) d(α, γ) ≤ d(α, β) + d(β, γ),
4. Triangle inequalities
29
(b) d(α, β) ≤ d(α, γ) + d(β, γ).
4.6 Definition. Define ρ̄ : [ω1 ]2 −→ ω as follows
ρ̄(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) ρ̄(α, γ) 6= ρ̄(β, γ),
(b) ρ̄(α, γ) ≤ max{ρ̄(α, β), ρ̄(β, γ)},
(c) ρ̄(α, β) ≤ max{ρ̄(α, γ), ρ̄(β, γ)}.
Proof. Only (b) and (c) require some argument. Consider first (b). If
ρ(α, β) ≤ ρ(β, γ) then ρ(α, γ) ≤ max{ρ(α, β), ρ(β, γ)} = ρ(β, γ), so
{ξ ≤ α : ρ(ξ, α) ≤ ρ(α, γ)} ⊆ {ξ ≤ β : ρ(ξ, β) ≤ ρ(β, γ)}.
Therefore in this case ρ̄(α, γ) ≤ ρ̄(β, γ). On the other hand, if ρ(β, γ) ≤
ρ(α, β), then ρ(α, γ) ≤ max{ρ(α, β), ρ(β, γ)} = ρ(α, β), so
{ξ ≤ α : ρ(ξ, α) ≤ ρ(α, γ)} ⊆ {ξ ≤ α : ρ(ξ, α) ≤ ρ(α, β)}.
So in this case ρ̄(α, γ) ≤ ρ̄(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 letting
α <p β iff α < β, ρ̄(α, β) ∈ |p| and
p(ρ̄(ξ, α)) = p(ρ̄(ξ, β)) for any ξ < α such that ρ̄(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.
a
4.10 Definition. For x ∈ ω ω , set
[
<x = {<x¹n : n < ω}.
4.11 Lemma. For every p ∈ ω <ω there is a partition of ω1 into finitely
many pieces such that if α < β belong to the same piece then there is q ⊇ p
in ω <ω such that α <q β.
30
I. Coherent Sequences
Proof. For α < ω1 , the set
F|p| (α) = {ξ ≤ α : ρ̄(ξ, α) ≤ |p|}
has size ≤ |p| + 1 so if we enrich F|p| (α) to a structure that would code
the behavior of ρ̄ on F|p| (α) one can realize only finitely many different
isomorphism types. Thus it suffices to show that if α < β < ω1 with F|p| (α),
F|p| (β) isomorphic, then there is an extension q of p such that α <q β.
Consider the graph G of all unordered pairs of integers< ρ̄(α, β) of the
form
{ρ̄(ξ, α), ρ̄(ξ, β)}
for some ξ < α. It suffices to show that every connected component of
G has at most one point < |p|. For if this is true, extend p to a mapping
q : ρ̄(α, β) −→ ω which is constant on each component of G. Clearly, α <q β
for any such q.
Consider an integer n < |p| and a G-path n = n0 , n1 , . . . , nk which starts
with n. Pick ordinals ξ0 , ξ1 , . . . , ξk−1 < α and γi ∈ {α, β} for i ≤ 2k − 1
which 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 ρ̄ (together
with the properties (I.40)-(I.41)) shows that the sequence n0 , n1 , . . . , nk is
strictly increasing, finishing thus the proof that a given n < |p| is not Gconnected to any other integer < |p|.
4.12 Lemma. Suppose ηα 6= ηβ < min{α, β} and ρ̄(ηα , α) = ρ̄(ηβ , β) = n.
Then ρ̄(ηα , β), ρ̄(ηβ , α) > n.
Proof. Suppose for example that ηα < ηβ . First note that ρ(ηα , ηβ ) ρ(ηβ , β) or else {ξ ≤ ηα : ρ(ξ, ηα ) ≤ ρ(ηα , α)} would be a proper initial part of {ξ ≤ ηβ : ρ(ξ, ηβ ) ≤ ρ(ηβ , β)} contradicting the fact that
ρ̄(ηα , α) = ρ̄(ηβ , β) yields that the two sets are of the same cardinality.
Thus ρ(ηα , ηβ ) > ρ(ηβ , β) = ρ(ηα , α) which together with the subadditivity
properties of ρ gives us that
{ξ ≤ ηα : ρ(ξ, ηα ) ≤ ρ(ηα , α)} ⊆ {ξ ≤ ηα : ρ(ξ, ηα ) ≤ ρ(ηα , ηβ )}.
4. Triangle inequalities
31
This gives the inequality ρ̄(ηα , ηβ ) > ρ̄(ηα , α) = n. Combining this inequality with the subadditivities of ρ̄ we get:
n < ρ̄(ηα , ηβ ) ≤ max{ρ̄(ηα , α), ρ(ηβ , α)} = ρ̄(ηβ , α),
n < ρ̄(ηα , ηβ ) ≤ max{ρ̄(ηα , β), ρ(ηβ , β)} = ρ̄(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 ∈ ω ω : 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 <ρ̄ β denote the fact that ρ̄(ξ, α) =
ρ̄(ξ, β) for all ξ < α. Then <ρ̄ is a tree ordering on ω1 obtained by identifying
α with the member ρ̄(·, α) of the tree T (ρ̄). Note that <ρ̄ ⊆<x for all x ∈ ω ω
and that there exists x ∈ ω ω such that <x =<ρ̄ (e.g., one such x is the
identity map id : ω −→ ω).
4.15 Theorem. If Γ is an infinite <ρ̄ -antichain, the set
HΓ = {x ∈ ω ω : α ≮x β for some α < β in Γ}
is a dense open subset of the Baire space.
Proof. Let p ∈ ω <ω be given. A simple application of Ramsey’s theorem
gives us a strictly increasing sequence γi (i < ω) of members of Γ such
that ∆(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 else
for some ordinal ξ and all i < j, ∆(γi , γj ) = ξ. In the first case, by
Lemma 4.7(a), we can find i such that one of the integers ρ̄(∆(γi , γi+1 ), γi )
or ρ̄(∆(γi , γi+1 ), γi+1 ) does not belong to the domain of p, so there will be
q ⊇ p such that ρ̄(γi , γi+1 ) < |q| and γi ≮q γi+1 . This guarantees that
γi ≮x γi+1 for every x ⊇ q. In the second case, ρ̄(ξ, γi ) (i < ω) is a oneto-one sequence of integers so there will be i < j such that neither of the
integers ρ̄(ξ, γi ) or ρ̄(ξ, γj ) belongs to the domain of p, so there is q ⊇ p
such that ρ̄(γi , γj ) < |q| and γi ≮q γj . Hence, γi ≮x γj for every x ⊇ q. a
15 For
α < β < ω1 , ∆(α, β) = min{ξ ≤ α : ρ̄(ξ, α) 6= ρ̄(ξ, β)}.
32
I. Coherent Sequences
4.16 Definition. For a family F of infinite <ρ̄ -antichains, we say that a
real x ∈ ω ω is F-Cohen if x ∈ GΓ ∩ HΓ for all Γ ∈ F . We say that x is
F-Souslin if no member of F is a <x -chain or a <x -antichain. We say that
a real x ∈ ω ω is Souslin if the tree ordering <x on ω1 has no uncountable
chains nor antichains, i.e. when x is F-Souslin for F equal to the family of
all uncountable subsets of ω1 .
Note that since every uncountable subset of ω1 contains an uncountable <ρ̄ -antichain, if a family F refines the family of all uncountable <ρ̄ antichains, then every F-Souslin real is Souslin. The following fact summarizes Theorems 4.13 and 4.15 and connects the two kinds of reals.
4.17 Theorem. If F is a family of infinite <ρ̄ -antichains, then every FCohen real is F-Souslin.
a
4.18 Corollary. If the density of the family of all uncountable subsets of
ω1 is smaller than the number of nowhere dense sets needed to cover the
real line, then there is a Souslin tree.
a
4.19 Remark. Recall that the density of a family F of infinite subsets of
some set S is the minimal size of a family F0 of infinite subsets of S with
the property that every member of F is refined by a member of F0 . A
special case of Corollary 4.18, when the density of the family of all uncountable subsets of ω1 is equal to ℵ1 , was first observed by T. Miyamoto
(unpublished).
4.20 Corollary. Every Cohen real is Souslin.
Proof. Every uncountable subset of ω1 in the Cohen extension contains an
uncountable subset from the ground model. So it suffices to consider the
family F of all infinite <ρ̄ -antichains from the ground model.
a
If ω1 is a successor cardinal in the constructible subuniverse, then ρ̄ can
be chosen to be coanalytic and so the transformation x 7−→<x will transfer
combinatorial notions of Souslin, Aronszajn or special Aronszajn trees into
the corresponding classes of reals that lie in the third level of the projective
hierarchy. This transformation has been explored in several places in the
literature (see, e.g. [4],[33]).
4.21 Remark. We have just seen how the combination of the subadditivity
properties (4.7(b),(c)) of the coherent sequence ρ̄α : α −→ ω (α < ω1 )
of one-to-one mappings can be used in controlling the finite disagreement
between them. It turns out that in many contexts the coherence and the
subadditivities are essentially equivalent restrictions on a given sequence
eα : α −→ ω (α < ω1 ). For example, the following construction shows that
this is so for any sequence of finite-to-one mappings eα : α −→ ω (α < ω1 ).
4. Triangle inequalities
33
4.22 Definition. Given a coherent sequence eα : α −→ ω (α < ω1 ) of
finite-to-one mappings, define τe : [ω1 ]2 −→ ω as follows
τe (α, β) = max{max{e(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) ≤ max{τe (α, β), τe (β, γ)},
(c) τe (α, β) ≤ max{τe (α, γ), τe (β, γ)}.
Proof. To see (a) note that it trivially holds if e(α, β) = 0 so we can
concentrate on the case e(α, β) > 0. Applying the convention e(α, α) =
0 we see that e(ξ, α) 6= e(ξ, β) for ξ = α. It follows that τe (α, β) ≥
max{max{e(α, α), e(α, β)} = e(α, β).
To see (b) let n = max{τe (α, β), τe (β, γ)} and let ξ ≤ α be such that
e(ξ, α) 6= e(ξ, γ). We need to show that e(ξ, α) ≤ n and e(ξ, γ) ≤ n. If
e(ξ, α) > n then since τe (α, β) ≤ n we must have e(ξ, α) = e(ξ, β)) and
so e(ξ, β) 6= e(ξ, γ). It follows that τe (β, γ) ≥ e(ξ, β)) > n, a contradiction. Similarly, e(ξ, γ) > n yields e(ξ, β) = e(ξ, γ) 6= e(ξ, α) and therefore
τe (α, β) ≥ e(ξ, β)) > n, a contradiction.
The proof of (c) is similar.
a
4.24 Definition. We say that a : [ω1 ]2 −→ ω is transitive if for α < β <
γ < ω1
a(α, γ) ≤ max{a(α, β), a(β, γ)}.
Transitive maps occur quite frequently in set-theoretic constructions. For
example, given a sequence Aα (α < ω1 ) of subsets of ω that increases
relative to the ordering ⊆∗ of inclusion modulo a finite set, the mapping
a : [ω1 ]2 −→ ω defined by
a(α, β) = min{n : Aα \ n ⊆ Aβ }
is a transitive map. The transitivity condition by itself is not nearly as useful
as its combination with the other subadditivity property (4.7(c)). Fortunately, there is a general procedure that produces a subadditive dominant
to every transitive map.
4.25 Definition. For a transitive a : [ω1 ]2 −→ ω define ρa : [ω1 ]2 −→ ω
recursively as follows:
ρa (α, β)
16 Note
= max{|Cβ ∩ α|, a(min(Cβ \ α), β), ρa (α, min(Cβ \ α)),
ρa (ξ, α) : ξ ∈ Cβ ∩ α}.
the occurence of the boundary value e(α, α) = 0 in this formula as well.
34
I. Coherent Sequences
4.26 Lemma. For all α < β < γ < ω1 and n < ω,
(a) {ξ ≤ α : ρa (ξ, α) ≤ n} is finite,
(b) ρa (α, γ) ≤ max{ρa (α, β), ρa (β, γ)},
(c) ρa (α, β) ≤ max{ρa (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 part
of the proof of Lemma 4.2. This comes of course from the fact that the
definition of ρ and ρa are closely related. The occurrence of the factor
a(min(Cβ \ α), β) complicates a bit the proof that ρa is subadditive, and
the fact that a is transitive is quite helpful in getting rid of the additional
difficulty. The details are left to the interested reader. Given α < β, for
every step βn → βn+1 of the minimal walk β = β0 > β1 > . . . > βk = α,
we have ρa (α, β) ≥ ρa (βn , βn+1 ) ≥ a(βn , βn+1 ) by the very definition of
ρa . Applying the transitivity of a to this path of inequalities we get the
conclusion (d).
a
4.27 Lemma. ρa (α, β) ≥ ρa (α + 1, β) whenever 0 < α < β and α is a limit
ordinal.
Proof. Recall the assumption (c) about the fixed C-sequence Cξ (ξ < ω1 )
on which all our definitions are based: if ξ is a limit ordinal > 0, then no
point of Cξ is a limit ordinal. It follows that if 0 < α < β and α is a
limit ordinal, then the minimal walk β → α must pass through α + 1 and
therefore ρa (α, β) ≥ ρa (α + 1, β).
a
Let us now give an application of ρa to a classical phenomenon of occurrence of gaps in the quotient algebra P(ω)/fin.
4.28 Definition. A Hausdorff gap in P(ω)/fin is a pair of sequences
Aα (α < ω1 ) and Bα (α < ω1 ) such that
(a) Aα ⊆∗ Aβ ⊆∗ Bβ ⊆∗ Bα whenever α < β, but
(b) there is no C such that Aα ⊆∗ C ⊆∗ Bα for all α.
The following straightforward reformulation shows that a Hausdorff gap
is just another instance of a nontrivial coherent sequence
fα : Aα −→ 2 (α < ω1 )
where the domain Aα of fα is not the ordinal α itself but a subset of ω and
that the corresponding sequence of domains Aα (α < ω1 ) is a realization
of ω1 inside the quotient P(ω)/fin in the sense that Aα ⊆∗ Aβ whenever
α < β.
4. Triangle inequalities
35
4.29 Lemma. A pair of ω1 -sequences Aα (α < ω1 ) and Bα (α < ω1 ) form
a Hausdorff gap iff the pair
Āα = Aα ∪ (ω \ Bα ) (α < ω1 ) and B̄α = ω \ Bα (α < ω1 )
has the following properties:
(a) Āα ⊆∗ Āβ whenever α < β,
(b) B̄α =∗ B̄β ∩ Āα whenever α < β,
(c) there is no B such that B̄α =∗ B ∩ Āα for all α.
a
From 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) = min{n : Aα \ n ⊆ Aβ }.
Let ρa : [ω1 ]2 −→ ω be the corresponding subadditive dominant of a defined
above. For α < ω1 , set
Dα = Aα+1 \ Aα .
4.30 Lemma. The sets Dα \ρa (α, γ) and Dβ \ρa (β, γ) are disjoint whenever
0 < α < β < γ and α and β are limit ordinals.
Proof. This follows immediately from Lemmas 4.26 and 4.27.
a
We 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 all
but finitely many limit ordinals α < min{β, γ}.
Proof. This is by the coherence of ρa and the fact that ρa (α, β) = ρa (α, γ)
already implies m(α, β) = m(α, γ).
a
4.32 Lemma. m(α, γ) 6= m(β, γ) whenever α 6= β < γ and α, β limit.
Proof. This follows from Lemma 4.30.
For β < ω1 , set
Bβ = {m(α, β) : α < β and α limit}.
4.33 Lemma. Bβ =∗ Bγ ∩ Aβ whenever β < γ.
a
36
I. Coherent Sequences
Proof. By the coherence of m.
a
Note the following immediate consequence of Lemma 4.30 and the definition of m.
4.34 Lemma. m(α, β) = max(Bβ ∩ Dα ) whenever α < β and α limit.
a
4.35 Lemma. There is no B ⊆ ω such that B ∩ Aβ =∗ Bβ for all β.
Proof. Suppose that such a B exists and for a limit ordinal α let us define
g(α) = max(B ∩ Dα ). Then by Lemma 4.34, g(α) = m(α, β) for all β < ω1
and all but finitely many limit ordinals α < β. By Lemma 4.32, it follows
that g is a finite-to-one map, a contradiction.
a
4.36 Theorem. For every strictly ⊂∗ -increasing chain Aα (α < ω1 ) of
subsets of ω, there is a sequence Bα (α < ω1 ) of subsets of ω such that:
(a) Bα =∗ Bβ ∩ Aα whenever α < β,
(b) there is no B such that Bα = B ∩ Aα for all α.
a
4.37 Remark. For a given countable ordinal α let hα : Aα −→ 2 be such
that h−1
α (1) = Bα . Then by Lemma 4.33, the corresponding sequence
hα (α < ω1 ) of partial functions is coherent in the sense that for every
pair α < β < ω1 , the set Dh (α, β) = {n ∈ Aα ∩ Aβ : hα (n) 6= hβ (n)} is
finite. By Lemma 4.35, the sequence is nontrivial in the sense that there is
no g : ω −→ 2 such that hα =∗ g ¹ Aα for all α < ω1 . Hausdorff’s original
sequence, when reformulated in this way, has the property that the maps
dh (·, β) : β −→ ω defined by
dh (α, β) = |Dh (α, β)|
are finite-to-one mappings. A general principle, the P-ideal dichotomy, (see
4.45 below) asserts that every coherent and nontrivial sequence must contain
a subsequence with Hausdorff’s property. We shall now embark on a general
construction of a very similar sequence of coherent partial functions domains
included in ω1 rather than in ω.
4.38 Definition. The number of steps of the minimal walk is the two-place
function 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.
4. Triangle inequalities
37
This is an interesting mapping which is particularly useful on higher cardinalities and especially in situations where the more informative mappings
ρ0 , ρ1 and ρ lack their usual coherence properties. Later on we shall devote
a whole section to ρ2 but here we need it only to succinctly express the
following mapping.
4.39 Definition. The last step function of the minimal walk is the twoplace 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 walk
β → α comes with the maximal weight.
4.40 Lemma. {ξ < α : ρ3 (ξ, α) 6= ρ3 (ξ, β)} is finite for all α < β < ω1 .
Proof. It suffices to show that for every infinite Γ ⊆ α there exists ξ ∈ Γ
such that ρ3 (ξ, α) = ρ3 (ξ, β). Shrinking Γ we may assume that for some
fixed ᾱ ∈ F (α, β) and all ξ ∈ Γ:
ᾱ = min(F (α, β) \ ξ),
(I.42)
ρ1 (ξ, α) = ρ1 (ξ, β),
(I.43)
ρ1 (ξ, α) > ρ1 (ᾱ, α),
(I.44)
ρ1 (ξ, β) > ρ1 (ᾱ, β).
(I.45)
It follows (see 1.9) that for every ξ ∈ Γ:
ρ0 (ξ, α) = ρ0 (ᾱ, α)a ρ0 (ξ, ᾱ),
(I.46)
ρ0 (ξ, β) = ρ0 (ᾱ, β)a ρ0 (ξ, ᾱ).
(I.47)
So for any ξ ∈ Γ, ρ3 (ξ, α) = 1 iff the last term of ρ0 (ξ, ᾱ) is its maximal
term iff ρ3 (ξ, β) = 1.
a
The sequence (ρ3 )α : α −→ 2 (α < ω1 )17 is therefore coherent in the
sense that (ρ3 )α =∗ (ρ3 )β ¹ α whenever α < β. We need to show that the
sequence is not trivial, i.e. that it cannot be uniformized by a single total
map from ω1 into 2. In other words, we need to show that ρ3 still contains
enough information about the C-sequence Cα (α < ω1 ) from which it is
defined. For this it will be convenient to assume that Cα (α < ω1 ) satisfies
the following natural condition:
(d) If α is a limit ordinal > 0 and if ξ occupies the nth place in the
increasing enumeration of Cα (that starts with min(Cα ) on its 0th
place), then ξ = λ + n + 1 for some limit ordinal λ (possibly 0).
17 Recall the way one always defines the fiber-functions from a two-variable function
applied to the context of ρ3 : (ρ3 )α (ξ) = ρ3 (ξ, α).
38
I. Coherent Sequences
4.41 Definition. Let Λ denote the set of all countable limit ordinals and
for an integer n ∈ ω, let Λ + n = {λ + n : λ ∈ Λ}.
4.42 Lemma. ρ3 (λ + n, β) = 1 for all but finitely many n with λ + n < β.
Proof. Clearly we may assume that α = λ + ω ≤ β. Then there is n0 < ω
such that for every n ≥ n0 the walk β → λ + n passes through α. By (d)
we know that Cα = {λ + n : 0 < n < ω} so in any such walk β → λ + n
there is only one step from α to λ + n. So choosing n ≥ n0 , ρ1 (α, β) we will
ensure that the last step of β → λ + n comes with the maximal weight, i.e.
ρ3 (λ + n, β) = 1.
a
4.43 Lemma. For all β < ω1 , n < ω, the set {λ ∈ Λ : λ + n < β and
ρ3 (λ + n, β) = 1} is finite.
Proof. Given an infinite subset Γ of (Λ + n) ∩ β we need to find a λ + n ∈ Γ
such that ρ3 (λ + n, β) = 0. Shrinking Γ if necessary assume that ρ1 (λ +
n, β) > n + 2 for all λ + n ∈ Γ. So if ρ3 (λ + n, β) = 1 for some λ + n ∈ Γ
then the last step of β → λ + n would have to be of weight> n + 2 which is
impossible by our assumption (d) about Cα (α < ω1 ).
a
The meaning of these properties of ρ3 perhaps is easier to comprehend
if we reformulate them in a way that resembles the original formulation of
the existence of Hausdorff gaps.
4.44 Lemma. Let Bα = {ξ < α : ρ3 (ξ, α) = 1} for α < ω1 . Then:
1. Bα =∗ Bβ ∩ α for α < β,
2. (Λ + n) ∩ Bβ is finite for all n < ω and β < ω1 ,
3. {λ + n : n < ω} ⊆∗ Bβ whenever λ + ω ≤ β.
a
In particular, there is is no uncountable Γ ⊆ ω1 such that Γ∩β ⊆∗ Bβ for
all β < ω1 . On the other hand, the P-ideal18 I generated by Bβ (β < ω1 ) is
large as it contains all intervals of the form [λ, λ + ω). The following general
dichotomy about P-ideals shows that here indeed we have quite a canonical
example of a P-ideal on ω1 .
4.45 Definition.
The P-ideal dichotomy. For every P-ideal I of countable subsets of
some set S either:
1. there is uncountable X ⊆ S such that [X]ω ⊆ I, or
18 Recall that an ideal I of subsets of some set S is a P-ideal if for every sequence
An (n < ω) of elements of I there is B in I such that An \ B is finite for all n < ω. A
set X is orthogonal to I if X ∩ A is finite for all A in I.
5. Conditional weakly null sequences
39
2. S can be decomposed into countably many sets orthogonal to I.
4.46 Remark. It is known that the P-ideal dichotomy is a consequence of
the Proper Forcing Axiom and moreover that it does not contradict the Continuum Hypothesis (see [89]). This is an interesting dichotomy which will be
used in this article for testing various notions of coherence as we encounter
them. For example, let us consider the following notion of coherence, already encountered above at several places, and see how it is influenced by
the P-ideal dichotomy.
5. Conditional weakly null sequences
In this section we mention some applications of the concept of ρ-function
on ω1 satisfying the inequalities of Lemma 4.2. As we shall see any such a
semi-distance function ρ can be used as a coding procedure in constructions
of long weakly null sequences in Banach spaces with no unconditional basic
subsequences.
5.1 Example. A weakly null sequence with no unconditional basic subsequence. We describe a weakly null normalized sequence (xξ )ξ<ω1 of members
of some Banach space X which contains no infinite unconditional basic subsequence19 . Fix an 0 < ε < 1 and an infinite set M of positive integers such
that 1 ∈ M and
X m
{( )1/2 : m, n ∈ M, m < n} ≤ ε.
(I.48)
n
Fix also a natural bijection p.q : HF → M, where HF denotes the family of
all hereditarily finite sets, so that in particular p∅q = 1. The ρ-number od
a finite set F ⊆ ω1 is the integer
ρ(F ) = max{ρ(α, β) : α, β ∈ F, α ≤ β}.
For F ⊆ ω1 and k ∈ ω let
(F )k = {α ≤ max(F ) : ρ(α, β) ≤ k for some β ∈ F with β ≥ α}.
19 Recall that a basic sequence in a Banach space X is a seminormalized sequence
(xn )n∈ω with the property that every
P x in its closed linear span span{xn : n ∈ ω} can
be uniquely represented as a sum
n∈ω an xn . Note that this is equivalent to saying
that the norms of projection operators Pk : span{xn : n ∈ ω} → span{xn : n < k} are
uniformly bounded. When the norms of projections PM : span{xn : n ∈ ω} → span{xn :
n ∈ M } over all subsets M of ω are uniformly bounded, the sequence (xn ) is said to
be unconditional. The natural basis (en ) of any of the sequence-spaces `p for p ∈ [1, ∞]
is unconditional. A typical example of a basic sequence that is not unconditional is the
summing basis e0 , e0 +e1 , e0 +e1 +e2 , ... of c0 or the summing basis e0 +e1 +e2 +...+en +...,
e1 + e2 + ... + en + ..., e2 + ... + en + ..., ... in c.
40
I. Coherent Sequences
We say that F is k-closed if (F )k = F . The ρ-closure of F is the set
(F )ρ = (F )k , where k = ρ(F ). This leads us to the function
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
[
Fi ))q,
i=0
Sn
where E = ( i=0 Fi )ρ and where πE : E → |E| is the unique orderpreserving 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 < Ej 20 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<ω1 be the
corresponding sequence of biorthogonal functionals. To a finite set E ⊆ ω1 ,
we associate the following vector and functional on c00 (ω1 ),
xE =
1 X
1 X ∗
eα and φE =
e .
1/2
|E|
|E|1/2 α∈E α
α∈E
(I.49)
Given a special sequence (Ei )i<n
P of finite subsets of ω1 , the corresponding
special 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 ),
k x k= sup{hf, xi : f ∈ F}.
(I.50)
The Banach space X is the completion of the normed space (c00 (ω1 ), k . k).
Note that k eα k= 1 and that (eα )α<ω1 is a weakly null sequence in X. To
see this note that
Pgiven 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
P
inner product between φ and the average |G|−1 α∈G eα = |G|−1/2 xG is
P
not bigger than |G|−1/2 (1 + |G|6=m∈M min((m/|G|)1/2 , (|G|/m)1/2 ) and
therefore not bigger than P
|G|−1/2 (1 + ε) by our assumption I.48. It follows
−1
that the averages k |G|
α∈G eα k tend to 0 as |G| ∈ M goes to infinity,
so (eα )α<ω1 must be weakly null in X. To show that no infinite subsequence
of (eα )α<ω1 is unconditional it suffices to show that for every set G ⊆ ω1 of
20 Recall, that for two sets of ordinals E and F , by E < F we denote the fact that
every ordinal from E is smaller from every ordinal from F . Sequences of sets of ordinals
of this sort are called block sequences.
5. Conditional weakly null sequences
41
order type ω there is seminormalized block-subsequence21 of (eα )α∈G which
represents the summing basis in c. More precisely, we shall show that if
(Ei )i<ω is any infinite special sequence of finite subsets of G and if for i < ω
we let vi = xEi as defined in I.49, then for every n < ω and every sequence
(ai )i≤n ⊆ [−1, +1] of scalars.
max |
0≤k≤n
k
X
i=0
ai |≤k
n
X
ai vi k≤ (3 + ε) max |
i=0
0≤k≤n
k
X
ai | .
(I.51)
i=0
To see the first inequality, note that according to I.49 the
Pnabsolute value
of the innerPproduct between the P
spacial functional φ = i=0 φEi and the
n
n
vector x = i=0 ai vi is equal to | i=0 ai | giving us the first
To
Pinequality.
m
see the second inequality, consider a special functional ψ = i=0 φFi where
(Fi )i≤m is some other special sequence of finite sets. In order to estimate
the inner product hψ, xi we need to know how the two special sequences
interact and this requires properties of the function ρ that controls them.
To see this, letSl ≤ min(m, n) be maximal
with the property that |El | = |Fl |,
S
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:
S
S
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-numbers
and therefore by the properties 4.1 of ρ, their intersection H = E ∩F is their
initial segment. Hence, πE and πE agree on H. Let k be the maximal integer
such that Ek ⊆ H. Then Ei = Fi for all i ≤ k, and using the properties 4.1
of ρ again, Ei ∩ Fj = ∅ for k + 1 < i, j < l. Combining all this we see that
|hψ, xi| is bounded by
|
k
X
X m
ai +ak+1 hφFk+1 , xEk+1 i+al hφFl , xEl i|+ {( )1/2 : m, n ∈ M, m < n}.
n
i=0
Referring to I.48 we get that |hψ, xi| is bounded by the right hand side of
the second inequality of I.51.
5.2 Remark. The first example of a weakly null sequence (of length ω) with
no unconditional basic subsequence was constructed by Maurey and Rosenthal [51]. Sixteen years latter Gowers and Maurey [27] have solved the basic
unconditional sequence problem by constructing an infinite-dimensional separable reflexive Banach space with no infinite unconditional basic sequence.
The following result of [3] relates to the Example 5.1 the same way the
result of [27] relates to the example of [51].
21 i.e., a seminormalized sequence (x )
i i<ω of finitely supported vectors of the linear
span of (eα )α∈G such that supp(xi ) < supp(xj )for i < j.
42
I. Coherent Sequences
5.3 Example. A reflexive Banach space Xω1 with a Schauder basis of length
ω1 containing no unconditional basic sequence. The space Xω1 is the completion of c00 (ω1 ) under the norm given by the formula I.50, where now
F = Fω1 is chosen as the minimal set of finitely supported functionals on
c00 (ω1 ) such that:
1. F ⊇ {e∗α : α < ω1 } and F is symmetric and closed under restriction
on intervals of ω1 .
2. For every (φi )i<n2k ⊆ FP
such that supp(φi ) < supp(φj ) for i < j <
n2k , the functional m−1
i<n2k φi belongs to F.
2k
3. For every special sequence (φi )i<n2k+1 ⊆ F the corresponding special
P
functional m−1
i<n2k+1 φi belongs to F.
2k+1
4. F is closed under rational convex combinations.
The increasing sequences (mk ) and (nk ) of positive integers are defined by:
3
k
mk = 24 and n0 = 4 and nk+1 = (4nk )log2 mk+1 . The notion of a special
sequence of functionals is defined analogously to that in the Example 5.1
as follows. First of all we say that φ ∈ F is of type 0 if φ = ±e∗α P
for some
α, of type I if it obtained from a simpler ones as the sum m−1
i<nk φi
k
restricted to some interval and we call the integer mk , the weight of φ, and
denote it w(φ). Note that a given functional of type I can have more than
one representation and therefore more than one weight. We say that φ is
of type II if it is a rational convex combination of functionals of type 0
and type I. The definition of a special sequence of functionals will be again
based on ρ and an injection p.q : HF → {2k : k odd }, with the property
that for every finite sequence (φi , wi , pi )i<j where wi , pi ∈ ω and where φi ’s
are finite partial maps from ω into Q such that p(φi , wi , pi )i<j q is bigger
than the maximum of support of any of the φi ’s, the square of any integer
of 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+1Sthere is an increasing sequence (pi )i<j of integers
such that pi S
≥ ρ( i0 <i supp(φi0 )) and w(φj ) = mp(πE ”φi ,w(φi ),pi )i<j q ,
where E = ( i<j supp(φi ))ρ .
(The collapse ψi = πE ”φi is defined by the formula ψ(πE (α)) = φ(α).) The
condition 1. from the definition of F provides that (eα )α<ω1 is a bimonotone
Schauder basis of Xω1 , the condition 3. is responsible for the conditional
5. Conditional weakly null sequences
43
structure of the norm of Xω1 in a similar manner as in the Example 5.1,
but in order to come to the point the idea can be applied, one needs the
unconditional structure given but the closure property 2. of F. This requires
a rather deep tree-analysis of the functionals of F that gives the required
norm estimates, an analysis that is beyond the scope of this presentation.
We refer the reader to [3] for the details. Here we just list some of the
most interesting properties of Xω1 . For example, the space Xω1 is reflexive,
has no infinite unconditional basic sequence, and is not isomorphic to any
proper subspace or a nontrivial quotient. The basis (eα )α<ω1 allows us to
talk about diagonal operators of Xω1 . It turns out that all bounded diagonal
operators D of Xω1 are given by a sequence (λα )α<ω1 of eigenvalues that
have the property that λα = λβ whenever α ≤ β < α + ω. Moreover
there is only countably many different eigenvalues among (λα )α<ω1 . Recall,
than an operator between two Banach spaces is strictly singular if it is not
an isomorphism when restricted to any infinite-dimensional subspace of its
domain. It turns out that every bounded operator T on Xω1 has the form
D+S, where D is a diagonal-step operator with countably many eigenvalues
and where X is a strictly singular operator. It follows that in particular
that every bounded linear operator on Xω1 is a multiple of the identity
plus the operator with separable range22 . It turns out that we can actually
identify the space D(Xω1 ) of bounded diagonal operators as the dual of
the James space over a particular reflexive Banach space T0 and ω1 as the
corresponding ordered set (see [3] for the definition of this functor). One
can similarly characterize the operator space of arbitrary closed subspace of
Xω1 as well obtaining thus some interesting new phenomena about operator
spaces. For example one can find a closed subspace Y of Xω1 such that every
bounded operator T : Y → Y is a strictly singular perturbation of a multiple
of identity while operator space L(Y, Xω1 ) is quite rich in the sense that
L(Y, Xω1 ) = JT∗0 ⊕ S(Y, Xω1 ).
Taking the restrictions XI = span({eα : α ∈ I}) to infinite intervals I ⊆ ω1
one obtains an uncountable family of separable reflexive space which, with
an appropriate adjustment on ρ, are ’asymptotic versions’ of each other.
This is the new phenomenon in the separable theory not revealed by the
previous methods. It essential says that any finite-dimensional subspace
of one XI can be moved into any other XJ by an operator T such that
k T kk T −1 k≤ 4 + ε.
5.4 Question. What is the minimal cardinal θu with property that every
weakly null sequence of length θu contains an infinite unconditional subsequence?
22 A considerably easier example of a space with this property will be given in some
detail in 7.16 below.
44
I. Coherent Sequences
We have seen above that θu ≥ ω2 . From a result of Ketonen [39] we infer
that θu is not bigger than the first Ramsey cardinal. Not much more about
θu seems to be known.
6. An ultrafilter from a coherent sequence
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), or
in other words, aα =∗ aβ ¹ α23 . We say that a is nontrivial if there is no
h : 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 (such
as, for example, the function ρ3 defined above) is something that corresponds to the notion of a Hausdorff gap in this context. Notice moreover,
that this notion is also closely related to the notion of an Aronszajn tree
since
T (a) = {t : α −→ ω : α < ω1 and t =∗ aα }
is such an Aronszajn tree whenever a : [ω1 ]2 −→ ω is coherent and nontrivial24 . In fact, we shall call an arbitrary Aronszajn tree T coherent if T
is isomorphic to T (a) for some coherent and nontrivial a : [ω1 ]2 −→ ω. In
case the range of the map a is actually smaller than ω, e.g. equal to some
integer k, then it is natural to let T (a) be the collection of all t : α −→ k
such that α < ω1 and t =∗ aα . This way, we have coherent binary, ternary,
etc. Aronszajn trees rather than only ω-ary coherent Aronszajn trees.
6.2 Definition. The support of a map a : [ω1 ]2 −→ ω is the sequence
supp(aα ) = {ξ < α : a(ξ, α) 6= 0} (α < ω1 ) of subsets of ω1 . A set Γ
is orthogonal to a if supp(aα ) ∩ Γ is finite for all α < ω1 . We say that
a : [ω1 ]2 −→ ω is nowhere dense if there is no uncountable Γ ⊆ ω1 such that
Γ ∩ α ⊆∗ supp(aα ) for all α < ω1 .
Note that ρ3 is an example of a nowhere dense coherent map for the
simple reason that ω1 can be covered by countably many sets Λ + n (n < ω)
that are orthogonal to ρ3 . The following immediate fact shows that ρ3 is
indeed a prototype of a nowhere dense and coherent map a : [ω1 ]2 −→ ω.
6.3 Proposition. Under the P-ideal dichotomy, for every nowhere dense
and coherent map a : [ω1 ]2 −→ ω the domain ω1 can be decomposed into
countably many sets orthogonal to a.
a
23 A mapping a : [ω ]2 −→ ω is naturally identified with a sequence a (α < ω ), where
α
1
1
aα : α −→ ω is defined by aα (ξ) = a(ξ, α).
24 Similarities between the notion of a Hausdorff gap and the notion of an Aronszajn
tree have been further explained recently in the two papers of Talayco ([72], [73]), where
it is shown that they naturally correspond to first cohomology groups over a pair of very
similar spaces.
6. An ultrafilter from a coherent sequence
45
6.4 Notation. To every a : [ω1 ]2 −→ ω associate the corresponding ∆function ∆a : [ω1 ]2 −→ ω as follows:
∆a (α, β) = min{ξ < α : a(ξ, α) 6= a(ξ, β)}
with the convention that ∆a (α, β) = α whenever a(ξ, α) = a(ξ, β) for all
ξ < α. Given this notation, it is natural to let
∆a (Γ) = {∆a (α, β) : α, β ∈ Γ, α < β}
for an arbitrary set Γ ⊆ ω1 .
6.5 Lemma. Suppose that a is nontrivial and coherent and that every uncountable subset of T (a) contains an uncountable antichain. Then for every
pair Σ,Ω of uncountable subsets of ω1 there exists an uncountable subset Γ
of ω1 such that ∆a (Γ) ⊆ ∆a (Σ) ∩ ∆a (Ω).
Proof. For each limit ordinal ξ < ω1 we fix α(ξ) ∈ Σ and β(ξ) ∈ Ω above ξ
and let
Fξ = {η < ξ : a(η, α(ξ)) 6= a(η, ξ) or a(η, β(ξ)) 6= a(η, ξ)}.
By the Pressing Down Lemma there is a stationary set Γ ⊆ ω1 and a finite
set F such that Fξ = F for all ξ ∈ Γ. Let η0 = max(F ) + 1. By the
assumption about T (a), find an uncountable Γ0 ⊆ Γ such that aξ (ξ ∈ Γ0 )
is an antichain of T (a). Moreover, we may assume that for some s, t and u
in T (a):
aα(ξ) ¹ η0 = s, aβ(ξ) ¹ η0 = t and aξ ¹ η0 = u for all ξ ∈ Γ0 .
(I.52)
It follows that for all ξ < η in Γ0 :
∆a (α(ξ), α(η)) = ∆a (ξ, η) = ∆a (β(ξ), β(η)).
So in particular, ∆a (Γ0 ) ⊆ ∆a (Σ) ∩ ∆a (Ω).
(I.53)
a
6.6 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 coherent
a : [ω1 ]2 −→ ω for which T (a) contains no Souslin subtrees. It turns out
that under some very mild assumption, U(a) is in fact a uniform ultrafilter
on ω1 .
6.7 Theorem. Under MAω1 , the filter U(a) is an ultrafilter for every nontrivial and coherent a : [ω1 ]2 −→ ω.
46
I. Coherent Sequences
Proof. Let A be a given subset of ω1 and let P be the collection of all finite
subsets p of ω1 such that ∆a (p) ⊆ A. If P is a ccc poset then an application
of MAω1 would give us an uncountable Γ ⊆ ω1 such that ∆a (Γ) ⊆ A, showing thus that A belongs to U(a). So let us consider the alternative when P is
not a ccc poset. Let X be an uncountable subset of P consisting of pairwise
incompatible conditions of P. Going to an uncountable ∆-subsystem of X
and noticing that the root does not contribute to the incompatibility, one
sees that without loss of generality, we may assume that the members of X
are in fact pairwise disjoint. Thus for each limit ordinal ξ < ω1 we can fix
pξ in X lying entirely above ξ. For ξ < ω1 , let
Fξ = {η < ξ : a(η, ξ) 6= a(η, α) for some α ∈ pξ }.
By the Pressing Down Lemma find a stationary set Γ ⊆ ω1 , a finite set
F ⊆ ω1 , an integer n and t0 , . . . , tn−1 in T (a) of height η0 = max(F ) + 1
such that for all ξ ∈ Γ:
Fξ = F,
(I.54)
aα ¹ η0 = ti with α the ith member of pξ for some i < n.
(I.55)
By MAω1 the tree T (a) is special, so going to an uncountable subset of Γ
we may assume that aξ (ξ ∈ Γ) is an antichain of T (a), and moreover that
for every ξ < η in Γ:
aα ¹ ξ * aβ ¹ η for all α ∈ pξ and β ∈ pη .
(I.56)
It follows that for every ξ < η in Γ:
∆a (α, β) = ∆a (ξ, η) for all α ∈ pξ and β ∈ pη .
(I.57)
By our assumption that conditions from X are incompatible in P, we conclude that for every ξ < η in Γ, there exist α ∈ pξ and β ∈ pη such that
∆a (α, β) ∈
/ A. So from (I.57) we conclude that ∆a (Γ) ∩ A = ∅, showing
thus that the complement of A belongs to U(a).
a
6.8 Remark. One may find Theorem 6.7 a bit surprising in view of the
fact that it gives us an ultrafilter U (a) on ω1 that is Σ1 -definable over the
structure (Hω2 , ∈). It is well-known that there is no ultrafilter on ω that is
Σ1 -definable over the structure (Hω1 , ∈).
It turns out that the transformation a 7−→ U(a) captures some of the
essential properties of the corresponding and more obvious transformation
a 7−→ T (a). To state this we need some standard definitions.
6.9 Definition. For two trees S and T , by S ≤ T we denote the fact that
there is a strictly increasing map f : S −→ T . Let S < T whenever S ≤ T
and T S and let S ≡ T whenever S ≤ T and T ≤ S. In general,
6. An ultrafilter from a coherent sequence
47
the equivalence relation ≡ on trees is very far from the finer relation ∼
=,
the isomorphism relation. However, the following fact shows that in the
realm of trees T (a), these two relations may coincide and moreover, that
the mapping T (a) 7−→ U(a) reduces ≡ and ∼
= to the equality relation among
ultrafilters on ω1 (see [90]).
6.10 Theorem. Assuming MAω1 , for every pair of coherent and nontrivial
mappings a : [ω1 ]2 −→ ω and b : [ω1 ]2 −→ ω, the trees T (a) and T (b) are
isomorphic iff T (a) ≡ T (b) iff U(a) = U(b).
Proof. We only prove that T (a) ≡ T (b) is equivalent to U(a) = U(b), referring the reader to [90] for the remaining part of the argument.
Choose a pair of strictly increasing mappings f : T (a) −→ T (b) and
g : T (b) −→ T (a). Replacing f by the mapping t 7−→ f (t) ¹ dom(t) and
g by the mapping t 7−→ g(t) ¹ dom(t), we may assume that f and g are
also level-preserving. Recall our notation aδ (ξ) = a(ξ, δ) and bδ (ξ) = b(ξ, δ)
that gives us particular representatives aδ and bδ of the δth level of T (a)
and T (b) respectively. For δ ∈ Λ, set
Fδ = {ξ < δ : aδ (ξ) 6= g(f (aδ ))(ξ) or bδ (ξ) 6= f (aδ )(ξ)}.
Then Fδ is a finite subset of δ. Find a stationary subset Σ of Λ and F ⊆ ω1
such that Fδ = F for all δ ∈ Σ. Find now an uncountable Γ ⊆ Σ such that
aδ (δ ∈ Γ) is an antichain of T (a), bδ (δ ∈ Γ) is an antichain of T (b), and
the mapping
¯ f (aδ ) ¹ ξ,
¯ g(f (aδ )) ¹ ξ,
¯ bδ ¹ ξ)
¯
δ 7−→ (aδ ¹ ξ,
is constant on Γ, where ξ¯ = max(F ) + 1. It follows that for γ 6= δ in Γ,
∆(aγ , aδ ) = ∆(g(f (aγ )), g(f (aδ ))),
(I.58)
∆(bγ , bδ ) = ∆(f (aγ ), f (aδ )).
(I.59)
Since f is strictly increasing, we must have ∆(aγ , aδ ) ≤ ∆(f (aγ ), f (aδ )) =
∆(bγ , bδ ). Since g is strictly increasing, we have that
∆(bγ , bδ ) = ∆(f (aγ ), f (aδ )) ≤ ∆(g(f (aγ )), g(f (aδ ))) = ∆(aγ , aδ ).
It follows that ∆(aγ , aδ ) = ∆(bγ , bδ ) for all γ 6= δ in Γ. From the proof of
Lemma 6.5 we conclude that ∆a (Ω) (Ω ⊆ Γ, Ω uncountable) generates the
filter U (a) and similarly that ∆b (Ω) (Ω ⊆ Γ, Ω uncountable) generates the
ultrafilter U (b). It follows that U(a) = U(b).
Suppose now that U (a) = U(b). By symmetry, it suffices to show that
T (a) ≤ T (b). Note again that the argument in the proof of Lemma 6.5 shows
that 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
48
I. Coherent Sequences
mappings p from T (a) into T (b) which (uniquely) extend to strictly increasing level-preserving mappings on the downward-closures of their domains.
By MAω1 , it suffices to show that P satisfies the countable chain condition.
So let pδ (δ ∈ Γ) be a given sequence of conditions of P. For δ ∈ Γ, set
Eδ = {ξ < δ : aδ (ξ) 6= t(ξ) or bδ (ξ) 6= p(t)(ξ)
for some t ∈ dom(pδ ) of height ≥ δ}.
Then Eδ is a finite subset of δ for each δ ∈ Γ. Find a stationary set Σ ⊆ Γ
and E ⊆ ω1 such that Eδ = E for all δ ∈ Σ. Shrinking Σ, we may assume
that the mapping δ 7−→ pδ ¹ δ is constant on Σ. Let η̄ be the minimal
ordinal strictly above max(E) and the height of every node of dom(pδ ) ¹ δ
for some (all) δ ∈ Σ. For δ ∈ Σ, let p̂δ be the extension of pδ on the
downward-closure of its domain in T (a). Find a stationary Ω ⊆ Σ such that
each of the projections δ 7−→ p̂δ ¹ η̄, δ 7−→ aδ ¹ η̄ and δ 7−→ bδ ¹ η̄ is constant
on Ω. Find γ < δ in Ω such that aγ 6= aδ ¹ γ, bγ 6= bδ ¹ γ and p̂γ and p̂δ are
isomorphic via the isomorphism that fixes their common restriction to the
η̄th levels of T (a) and T (b). It suffices to check that q = pγ ∪ pδ satisfies
∆(q(x), q(y)) ≥ ∆(x, y)
(I.60)
for every x ∈ dom(pγ ) and y ∈ dom(pδ ). If x and y correspond to each
other in the isomorphism between p̂γ and p̂δ , then either they belong to the
common part of pγ and pδ in which case (I.60) follows by the assumption
that pγ and pδ are strictly increasing level-preserving, or we have
∆(x, y) = ∆(aγ , aδ ) = ∆(bγ , bδ ) = ∆(pγ (x), pδ (y)).
(I.61)
If x is different from the copy ȳ ∈ dom(pγ ) of y ∈ dom(pδ ), then ∆(x, y) =
∆(x, ȳ) and ∆(pγ (x), pδ (y)) = ∆(pγ (x), pγ (y)), so we conclude that (I.60)
follows from the fact that pγ is strictly increasing level-preserving. This
finishes the proof.
a
6.11 Remark. It turns out that the ultrafilters of the form U(a) for a :
[ω1 ]2 −→ ω nontrivial and coherent are all Rudin-Keisler equivalent. The
proof of this fact can be found in [90].
6.12 Definition. The shift of a : [ω1 ]2 −→ ω is defined to be the mapping
a(1) : [ω1 ]2 −→ ω determined by
a(1) (α, β) = a(α + 1, β̂),
where β̂ = min{λ ∈ Λ : λ ≥ β}. The n-fold iteration of the shift operation
is defined recursively by the formula a(n+1) = (a(n) )(1) .
6. An ultrafilter from a coherent sequence
49
The following fact about this shift operation, together with Lemma 4.44,
can be used to show that Aronszajn trees are not well-quasi-ordered under
the quasi-ordering ≤ (see [90]).
6.13 Theorem. If a is nontrivial, coherent and orthogonal to Λ, then it
holds that T (a) > T (a(1) ).
Proof. By the assumption a ⊥ Λ, the subtree S of T (a) consisting of all
t’s that take the value 0 at every limit ordinal in their domains is an uncountable initial part of T (a). Note that every node of T (a) is equal to
the restriction of some finite change of a node from S. For a node t at
some limit level λ of S, let t(1) : λ −→ ω be determined by the formula
t(1) (α) = t(α + 1). Let S (1) = {t(1) : t ∈ S ¹ λ}. Every node of T (a(1) ) is
equal to the restriction of some finite change of a node of S (1) . Note also
that t 7−→ t(1) is a one-to-one map on S ¹ Λ and that its inverse naturally
extends to a strictly increasing map from T (a(1) ) into T . This shows that
T (a(1) ) ≤ T (a).
It remains to establish T (a) T (a(1) ). Note that if there is a strictly
increasing g : T (a) −→ T (a(1) ), then there is one that is moreover levelpreserving. For example f (t) = g(t) ¹ |t| is one such map. So let f :
T (a) −→ T (a(1) ) be a given level-preserving map. For a countable limit
ordinal ξ, set
(1)
Fξ = (supp(aξ ) ∩ Λ) ∪ {α < ξ : aξ (α) 6= f (aξ )(α)}.
(I.62)
Then Fξ is a finite set of ordinals < ξ, so by the Pressing Down Lemma
there is a stationary set Γ ⊆ Λ and F ⊆ ω1 such that Fξ = F for all
ξ ∈ Γ. Let α0 = max(F ) + 1. Shrinking Γ if necessary, assume that for
some s, t : α0 −→ 2,
aξ ¹ α0 = s and f (aξ ) ¹ α0 = t for all ξ ∈ Γ.
(I.63)
Choose ξ < η in Γ such that aξ 6= aη ¹ ξ and let δ = ∆(aξ , aη )(= min{α <
ξ : aξ (α) 6= aη (α)}). Then by (I.62) and (I.63), δ is not a limit ordinal and
(1)
δ − 1 = ∆(aξ , 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 finishes
the proof.
a
6.14 Corollary. If a is nontrivial, coherent and orthogonal to Λ + n for all
n < ω, then T (a(n) ) > T (a(m) ) whenever n < m < ω.
Proof. Note that if a is orthogonal to Λ + n for all n < ω, then so is every
of its finite shifts a(m) .
a
50
I. Coherent Sequences
(n)
(m)
6.15 Corollary. T (ρ3 ) > T (ρ3 ) whenever n < m < ω.
a
Somewhat unexpectedly, with very little extra assumptions we can say
much more about ≤ in the domain of coherent Aronszajn trees (see [90]).
6.16 Theorem. Under MAω1 , the family of coherent Aronszajn trees is
totally ordered under ≤.
Proof. Given two nontrivial coherent sequences a and b, we need to show
that either T (a) ≤ T (b) or T (b) ≤ T (a). It turns out that these two
alternatives correspond to the following two cases.
Case 1: There is uncountable Ω ⊆ ω1 such that ∆a is dominated by ∆b
on [Ω]2 . Let P be the poset of finite partial level-preserving functions p from
T (a) into T (b) that can be extended to strictly increasing level-preserving
maps from D̂p into T (b), where D̂p is the downward closure of the domain
Dp of p. Let Rp denote the range of p and R̂p its downward closure in T (b).
If P is a ccc poset, then applying MAω1 to P and the natural sequence
of dense open subsets of P would give us the desired strictly increasing
level-preserving map from T (a) into T (b). So let pξ (ξ < ω1 ) be a given
one-to-one sequence of members of P. We may assume that the pξ ’s form
a ∆-system with root r. Let Dξ = Dpξ \ Dr and Rξ = Rpξ \ Rr for ξ < ω1 .
For limit ξ fix g(ξ) < ξ and h(ξ) ∈ Ω \ ξ such that for every t ∈ Dξ with
height ≥ ξ:
t ≡ aξ ≡ ah(ξ)
pξ (t) ≡ bξ ≡ bh(ξ)
on [g(ξ), ξ),
on [g(ξ), ξ).
(I.65)
(I.66)
By the Pressing Down Lemma, there is a stationary Σ ⊆ ω1 such that g
is constantly equal to η0 on Σ and moreover that the projection maps of
Dξ ∪ {aξ , ah(ξ) } onto the η0 th level of T (a) and Rξ ∪ {bξ , bh(ξ) } onto the
η0 th level of T (b) are constant as long as ξ belongs to Σ. Moreover we
may assume for ξ 6= η, every node of Dξ ∪ {aξ , ah(ξ) } is incomparable to
every node of Dη ∪ {aη , ah(η) }, and similarly, every node of Rξ ∪ {bξ , bh(ξ) }
is incomparable to every node of Rη ∪ {bη , bh(η) }. Note that we may also
assume that every node of Dξ , for ξ ∈ Σ, has height ≥ ξ. So pick ξ 6= η in
Σ. Then for all s ∈ Dξ and t ∈ Dη :
∆(s, t) = ∆(ah(ξ) , ah(η) ) = ∆a (h(ξ), h(η)),
(I.67)
∆(pξ (s), pη (t)) = ∆(bh(ξ) , bh(η) ) = ∆b (h(ξ), h(η)).
(I.68)
Since {h(ξ), h(η)} belongs to [Ω]2 , we conclude that for all s ∈ Dξ and
t ∈ Dη , s and t split before pξ (s) and pη (t). So pξ ∪ pη extends to a strictly
increasing level-preserving function from the downward closure of Dpξ ∪Dpη
into the downward closure of Rpξ ∪ Rpη . This finishes the proof that P is a
ccc forcing relation.
7. The trace and the square-bracket operation
51
Case 2: For every uncountable Ω ⊆ ω1 there exists {α, β} in [Ω]2 such
that ∆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 . If Q is a ccc poset, then an application of
MAω1 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 that
Q is a ccc poset, let qξ (ξ < ω1 ) be a given one-to-one sequence of members
of Q. We may assume the qξ ’s form a ∆-system. Note that the root does
not contribute to the incompatibility of two conditions of the ∆-system and
so removing it, we assume that all qξ ’s are in fact pairwise disjoint. For a
limit ξ < ω1 , fix g(ξ) < ξ such that for all α ∈ qξ \ ξ:
aα ≡ aξ and bα ≡ bξ on [g(ξ), ξ).
(I.69)
Applying the Pressing Down Lemma and working as above, we find a stationary set Ω ⊆ ω1 such that for all ξ =
6 η in Ω, and α ∈ qξ , β ∈ qη :
∆a (α, β) = ∆a (ξ, η) and ∆b (α, β) = ∆b (ξ, η).
(I.70)
By the assumption of Case 2, we can find ξ 6= η in Ω such that ∆a (ξ, η) >
∆b (ξ, η). It follows that ∆a dominates ∆b on pξ ∪ pη . This completes the
proof that Q is a ccc poset and therefore the proof of Theorem 6.16.
a
6.17 Remark. While under MAω1 , the class of coherent Aronszajn trees is
totally ordered by ≤, Corollary 6.15 gives us that this chain of trees is not
well-ordered. This should be compared with an old result of Ohkuma [56]
that the class of all scattered trees is well-ordered by ≤ (see also [47]). It
turns out that the class of all Aronszajn trees is not totally ordered under
≤, i.e. there exist Aronszajn trees S and T such that S T and T S.
The reader is referred to [90] for more information on this and other related
results that we chose not to reproduce here.
7. The trace and the square-bracket operation
Recall the notion of a minimal walk from a countable ordinal β to a smaller
ordinal α along the fixed C-sequence Cξ (ξ < ω1 ) : β = β0 > β1 > · · · >
βn = α where βi+1 = min(Cβi \ α). Recall also the notion of a trace
Tr(α, β) = {β0 , β1 , · · · , βn },
the finite set of places visited in the minimal walk from β to α. The following
simple fact about the trace lies at the heart of all known definitions of
square-bracket operations not only on ω1 but also at higher cardinalities.
7.1 Lemma. For every uncountable subset Γ of ω1 the union of Tr(α, β)
for α < β in Γ contains a closed and unbounded subset of ω1 .
52
I. Coherent Sequences
Proof. It suffices to show that the union of traces contains every countable
limit ordinal δ such that sup(Γ ∩ δ) = δ. Pick an arbitrary β ∈ Γ \ δ and let
β = β0 > β1 > · · · > βk = δ
be the minimal walk from β to δ. Let γ < δ be an upper bound of all sets
of the form Cβi ∩ δ for i < k. By the choice of δ there is α ∈ Γ ∩ δ above
γ. Then the minimal walk from β to α starts as β0 > β1 > · · · > βk , so in
particular δ belongs to Tr(α, β).
a
We shall now see that it is possible to pick a single place [αβ] in Tr(α, β)
so that Lemma 7.1 remains valid with [αβ] in place of Tr(α, β). Recall that
by Lemma 1.14,
∆(α, β) = min{ξ ≤ α : ρ0 (ξ, α) 6= ρ0 (ξ, β)}
is a successor ordinal. We shall be interested in its predecessor,
7.2 Definition. ∆0 (α, β) = ∆(α, β) − 1.
Thus, if ξ = ∆0 (α, β), then ρ0 (ξ, α) = ρ0 (ξ, β) and so there is a natural isomorphism between Tr(ξ, α) and Tr(ξ, β). We shall define [αβ] by
comparing the three sets Tr(α, β), Tr(ξ, α) and Tr(ξ, β).
7.3 Definition. The square-bracket operation on ω1 is defined as follows:
[αβ] = min(Tr(α, β) ∩ Tr(∆0 (α, β), β))
= min(Tr(∆0 (α, β), β) \ α).
Next, recall the function ρ0 : [ω1 ]2 → ω <ω defined from the C-sequence
Cξ (ξ < ω1 ) and the corresponding tree T (ρ0 ). For γ < ω1 let (ρ0 )γ be the
fiber-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 of
the form [αβ] for some α < β in Γ contains a closed and unbounded subset
of ω1 .
Proof. For t ∈ T (ρ0 ) let
Γt = {γ ∈ Γ : (ρ0 )γ end-extends t}.
Let S be the collection of all t ∈ T (ρ0 ) for which Γt is uncountable. Clearly,
S is a downward closed uncountable subtree of T . The Lemma is proved
once we prove that every countable limit ordinal δ > 0 with the following
two properties can be represented as [αβ] for some α < β in Γ:
sup(Γt ∩ δ) = δ for every t ∈ S of length < δ,
(I.71)
7. The trace and the square-bracket operation
every t ∈ S of length < δ has two incomparable successors
in S both of length < δ.
53
(I.72)
Fix such a δ and choose β ∈ Γ \ δ such that (ρ0 )β ¹ δ ∈ S and consider
the minimal walk from β to δ:
β = β0 > β1 > · · · > βk = δ.
Let γ < δ be an upper bound of all sets of the form Cβi ∩ δ for i < k. Since
the restriction t = (ρ0 )β ¹ γ belongs to S, by (I.72) we can find one of its
end-extensions s in S which is incomparable with (ρ0 )β . It follows that for
α ∈ Γs , the ordinal ∆0 (α, β) has the fixed value
ξ = min{ξ < |s| : s(ξ) 6= ρ0 (ξ, β)} − 1.
Note that ξ ≥ γ, so the walk β → δ is a common initial part of walks β → ξ
and β → α for every α ∈ Γs ∩ δ. Hence if we choose α ∈ Γs ∩ δ above
min(Cδ \ ξ) (which we can by (I.71)), we get that the walks β → ξ and
β → α never meet after δ. In other words for any such α, the ordinal δ
is the minimum of Tr(α, β) ∩ Tr(ξ, β), or equivalently δ is the minimum of
Tr(ξ, β) \ α.
a
It should be clear that the above argument can easily be adjusted to
give us the following slightly more general fact about the square-bracket
operation.
7.5 Lemma. For every uncountable family A of pairwise disjoint finite
subsets of ω1 , all of the same size n, the set of all ordinals of the form
[a(1)b(1)] = [a(2)b(2)] = · · · = [a(n)b(n)] 25 for some a 6= b in A contains a
closed and unbounded subset of ω1 .
a
It turns out that the square-bracket operation can be used in constructions of various mathematical objects of complex behaviour where all known
previous constructions needed the Continuum Hypothesis or stronger enumeration principles. The usefulness of [··] in these constructions is based
on the fact that [··] reduces the quantification over uncountable subsets of
ω1 to the quantification over closed unbounded subsets of ω1 . For example
composing [··] with a unary operation ∗ : ω1 −→ ω1 which takes each of the
values stationary many times one gets the following fact about the mapping
c(α, β) = [αβ]∗ .
7.6 Theorem. There is a mapping c : [ω1 ]2 −→ ω1 which takes all the
values from ω1 on any square [Γ]2 of some uncountable subset Γ of ω1 . a
25 For a finite set x of ordinals of size n we use the notation x(1), x(2), . . . , x(n) or
x(0), x(1), . . . , x(n − 1), depending on the context, for the enumeration of x according to
the natural ordering on the ordinals.
54
I. Coherent Sequences
This result gives a definitive limitation on any form of a Ramsey Theorem
for the uncountable which tries to say something about the behaviors of
restrictions of colorings on an uncountable square. Recall that in Section
3.1 we have seen that there are mappings of this sort with complex behavior
on sets of the form
Γ0 ⊗ Γ1 = {{α0 , α1 } : α0 ∈ Γ0 , α1 ∈ Γ1 , α0 < α1 },
where Γ0 and Γ1 are two uncountable subsets of ω1 . It turns out that the
square bracket operation has a simple behavior on Γ0 × Γ1 for every pair
Γ0 and Γ1 of sufficiently thin uncountable subsets of ω1 . This can be made
precise as follows.
7.7 Lemma. For every uncountable family A of pairwise disjoint subsets
of ω1 , all of the same size n, there exist a partition I1 , . . . , Ik of {1, . . . , n},
an uncountable B ⊆ A, and an hb : n × n −→ ω1 for each b ∈ B such that
for 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 exist
a < b in B such that [a(p)b(q)] = δ for all p, q ∈ Ii and all i = 1, . . . , k
contains a closed and unbounded subset of ω1 .
Proof. Identifying an ordinal α with (ρ0 )α , we can think of A as a family
of n-tuples of nodes of the tree T (ρ0 ). For a limit ordinal δ < ω1 , choose aδ
in A such that every node from aδ has height ≥ δ. Let cδ be the projection
of aδ on the δth level of T (ρ0 ). Let h(δ) < δ be such that the projection of
cδ on the h(δ)th level has size equal to the size of cδ , i.e.
∆(s, t) < h(δ) for all s 6= t in cδ .
(I.73)
Find a stationary set Γ ⊆ ω1 such that h takes a constant value γ0 on Γ and
that the γ0 th projection takes the constant value on cδ (δ ∈ Γ). Fix δ < ε
in Γ and consider α ∈ aδ and β ∈ aε that project to different elements of
the γ0 th level of T (ρ0 ). Then ∆(α, β) is independent of α and β or δ and
ε as it is equal to ∆((ρ0 )α ¹ γ0 , (ρ0 )β ¹ γ0 ). So the trace Tr(∆0 (α, β), β)
depends only on β and the projection (ρ0 )α ¹ γ0 . Thus, going to a subset of
Γ, we may assume that its size depends only on the position of β in some
aε and that the minimal point of Tr(∆0 (α, β), β) \ α is the same for all α
in some aδ , for δ < ε, having a fixed projection on the γ0 th level of T (ρ0 ).
This is the content of the first part of the conclusion of Lemma 7.7. To see
the second part of the conclusion, we shrink Γ even further to assure that
the union of cδ (δ ∈ Γ) is an antichain of T (ρ0 ). Now note that for two
typical δ < ε in Γ, [αβ] = [α0 β 0 ] for all α, α0 ∈ aδ and β, β 0 ∈ aε with the
property that (ρ0 )α ¹ γ0 = (ρ0 )α0 ¹ γ0 = (ρ0 )β ¹ γ0 = (ρ0 )β 0 ¹ γ0 . So the
rest of the proof reduces to the proof of Lemma 7.5 above.
a
26 For a pair a and b of sets of ordinals, a < b denotes the fact that α < β for all α ∈ a
and β ∈ b.
27 By ∆, we denote the diagonal of k × k.
7. The trace and the square-bracket operation
55
7.8 Remark. To see the point of Lemma 7.7, consider the projection [αβ]∗
of the square-bracket operation generated by a mapping ∗ from ω1 onto ω
rather than ω1 . Then the sequence h∗b : n×n −→ ω (b ∈ B) of corresponding
compositions will have the same constant value h : n × n −→ ω modulo,
of course, shrinking B to some uncountable subset. Noting that the proof
of Lemma 7.7 applied to a typical A will actually result in the partition
Ii = {i} (i = 1, . . . , n), we conclude that if α and β are coming from two
different positions of some aδ and aε in B, then [αβ]∗ will depend on the
positions themselves rather than α and β.
7.9 Remark. It turns out that there is a natural variation of the squarebracket operation (see the mapping b of [78, p.287]), where, in Lemma 7.7,
we always have the finest possible partition Ii = {i} (i = 1, . . . , n) for every
uncountable family A of pairwise disjoint n-tuples of countable ordinals.
This property of the projection of the square-bracket operation has an interesting application in the Quadratic Form Theory, given by BaumgartnerSpinas [6] and Shelah-Spinas [67].
Note that the basic C-sequence Cα (α < ω1 ) which we have fixed at the
beginning of this chapter can be used to actually define a unary operation
∗ : ω1 −→ ω1 which takes each of the ordinals from ω1 stationarily many
times. So the projection [αβ]∗ can actually be defined in our basic structure
~ We are now at the point to see that our basic structure is actually
(ω1 , ω, C).
rigid.
7.10 Lemma. The algebraic structure (ω1 , [··], ∗) has no nontrivial automorphisms.
Proof. Let h be a given automorphism of (ω1 , [··], ∗). If the set Γ of fixed
points of h is uncountable, h must be the identity map. To see this, consider
a ξ < ω1 . By the property of the map c(α, β) = [αβ]∗ stated in Theorem
7.6 there exist γ < δ in Γ such that [γδ]∗ = ξ. Applying h to this equation
we get
h(ξ) = h([γδ]∗ ) = (h([γδ]))∗ = [h(γ)h(δ)]∗ = [γδ]∗ = ξ.
It follows that ∆ = {δ < ω1 : h(δ) 6= δ} is in particular uncountable.
Shrinking ∆ and replacing h by h−1 , if necessary, we may safely assume
that h(δ) > δ for all δ ∈ ∆. Consider a ξ < ω1 and let Sξ be the set of all
α < ω1 such that α∗ = ξ. By our choice of ∗ the set Sξ is stationary. By
Lemma 7.5 applied to the family A = {{δ, h(δ)} : δ ∈ ∆} we can find γ < δ
in ∆ such that [γδ] = [h(γ)h(δ)] belongs to Sξ , or in other words,
[γδ]∗ = [h(γ)h(δ)]∗ = ξ.
Since [h(γ)h(δ)]∗ = h([γδ]∗ ) we conclude that h(ξ) = ξ. Since ξ was an
arbitrary countable ordinal, this shows that h is the identity map.
a
56
I. Coherent Sequences
We give now an application of this rigidity result to a problem in Model
Theory about the quantifier Qx =’there exist uncountably many x’ and its
higher dimensional analogues Qn x1 · · · xn =’there exist an uncountable ncube many x1 , · · · , xn ’. By a result of Ebbinghaus and Flum [19] (see also
[57]) every model of every sentence of L(Q) has nontrivial automorphisms.
However we shall now see that this is no longer true about the quantifier
Q2 .
7.11 Example. The sentence φ will talk about one unary relation N , one
binary relation < and two binary functional symbols C and E. It is the
conjunction of the following seven sentences
(φ1 ) Qx
x = x,
(φ2 ) ¬Qx
N (x),
(φ3 ) < is a total ordering,
(φ4 ) E is a symmetric binary operation,
(φ5 ) ∀x < y
N (E(x, y)),
(φ6 ) ∀x < y < z
E(x, z) 6= E(y, z),
(φ7 ) ∀x∀n{N (n) → ¬Q2 uv[∃u0 < u∃v 0 < v(u0 6= v 0 ∧ E(u0 , u) = E(v 0 , v) =
n) ∧ ∀u0 < u ∀v 0 < v(E(u0 , u) = E(v 0 , v) = n → (C(u0 , v 0 ) 6= x ∨
C(u, v) 6= x))]}.
The model of φ that we have in mind is the model (ω1 , ω, <, c, e) where
c(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 = ρ̄1 or e = ρ̄). The
sentence φ7 is simply saying that for every ξ < ω1 and every uncountable
family A of pairwise disjoint unordered pairs of countable ordinals there
exist a 6= b in A such that
c(min a, min b) = c(max a, max b) = ξ.
This is a consequence of Lemma 7.5 and the fact that Sξ = {α : α∗ = ξ}
is a stationary subset of ω1 . These are the properties of [··] and ∗ which
we have used in the proof of Lemma 7.10 in order to prove that (ω1 , [··], ∗)
is a rigid structure. So a quite analogous proof will show that any model
(M, N, <, C, E) of φ must be rigid.
a
The crucial property of [··] stated in Lemma 7.5 can also be used to
provide a negative answer to the basis problem for uncountable graphs by
constructing a large family of pairwise orthogonal uncountable graphs.
7. The trace and the square-bracket operation
57
7.12 Definition. For a subset Γ of ω1 , let GΓ be the graph whose vertex-set
is ω1 and whose edge-set is equal to
{{α, β} : [αβ] ∈ Γ}.
7.13 Lemma. If the symmetric difference between Γ and ∆ is a stationary
subset of ω1 , then the corresponding graphs GΓ and G∆ are orthogonal to
each other, i.e. they do not contain uncountable isomorphic subgraphs. a
We have seen above that comparing [··] with a map π : ω1 −→ I where
I is some set of mathematical objects/requirements in such a way that
each object/requirement is given a stationary preimage, gives us a way to
meet each of these objects/requirements in the square of any uncountable
subset of ω1 . This observation is the basis of all known applications of the
square-bracket operation. A careful choice of I and π : ω1 −→ I gives us a
projection of the square-bracket operation that can be quite useful. So let
us illustrate this on yet another example.
7.14 Definition. Let H be the collection of all maps h : 2n −→ ω1 where
n is a positive integer denoted by n(h). Choose a mapping π : ω1 −→ H
which takes each value from H stationarily many times. Choose also a oneto-one sequence rα (α < ω1 ) of elements of the Cantor set 2ω . Note that
~
both these objects can actually be defined in our basic structure (ω1 , ω, C).
Consider the following projection of the square-bracket operation:
[[αβ]] = π([αβ])(rα ¹ n(π([αβ]))).
It is easily checked that the property of [··] stated in Lemma 7.5 corresponds to the following property of the projection [[αβ]]:
7.15 Lemma. For every uncountable family A of pairwise disjoint finite
subsets of ω1 , all of the same size n, and for every n-sequence ξ1 , . . . , ξn
of countable ordinals there exist a and b in A such that [[a(i)b(i)]] = ξi for
i = 1, . . . , n.
a
This projection of [··] leads to an interesting example of a Banach space
with ’few’ operators, which we will now describe.
7.16 Example. A nonseparable reflexive Banach space E with the property
that every bounded linear operator T : E −→ E can be expressed as T =
λI + S where λ is a scalar, I the identity operator of E, and S an operator
with separable range.
Let I = 3×[ω1 ]<ω and let us identify the index-set I with ω1 , i.e. pretend
that [[··]] takes its values in I rather than ω1 . Let [[··]]0 and [[··]]1 be the two
projections of [[··]].
58
I. Coherent Sequences
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 of
countable ordinals such that for all i < j < k:
max{αi , βi } < min{αj , βj },
(I.74)
[[αi αj ]]0 = [[βi βj ]]0 = 2,
(I.75)
[[αi αj ]]1 = [[βi βj ]]1 = {αl : l < i} ∪ {βl : l < i}.
(I.76)
The following properties of G, H and K should be clear:
G and H contain all the singletons, are closed under subsets
and they are 1-orthogonal to each other in the sense that
G ∩ H contains no doubleton.
G and H are both 2-orthogonal to the family of the unions
of members of K.
If K and L are two distinct members of K, then there are no
more than 5 ordinals α such that {α, β} ∈ K and {α, γ} ∈ L
for some β 6= γ.
For every sequence {αξ , βξ } (ξ < ω1 ) of pairwise disjoint
pairs of countable ordinals there exist arbitrarily large finite
sets Γ, ∆ ⊆ ω1 such that {αξ : ξ ∈ Γ} ∈ G, {βξ : ξ ∈ Γ} ∈ H
and {{αξ , βξ } : ξ ∈ ∆} ∈ K.
(I.77)
(I.78)
(I.79)
(I.80)
For a function x from ω1 into R, set
X
1
kxkH,2 = sup{(
x(α)2 ) 2 : H ∈ H},
α∈H
kxkK,2 = sup{(
X
1
(x(α) − x(β))2 ) 2 : K ∈ K}.
{α,β}∈K
Let k · k = max{k · k∞ , k · kH,2 , k · kK,2 } and define Ē2 = {x : kxk < ∞}. Let
1α be the characteristic function of {α}. Finally, let E2 be the closure of
the linear span of {1α : α ∈ ω1 } inside (Ē2 , k · k). The following facts about
the norm k · k are easy to establish using the properties of the families G, H
and K listed above
If x is supported by some G ∈ G, then kxk ≤ 2 · kxk∞ .
If x is supported by
[
(I.81)
K for some K in K, then kxk ≤ 10 · kxk∞ . (I.82)
7. The trace and the square-bracket operation
59
The role of the seminorm k·kH,2 is to ensure that every bounded operator T :
E2 −→ E2 can be expressed as D+S, where D is a diagonal operator relative
to the basis28 1α (α < ω1 ) and where S has separable range. Namely, if
this were not so, we could find a sequence {αξ , βξ } (ξ < ω1 ) of pairwise
disjoint pairs from ω1 such that T (1αξ )(βξ ) 6= 0 for all ξ < ω1 . Thinning
the sequence, we may assume that for some fixed r > 0 and all ξ < ω1 ,
|T (1αξ )(βξ )| ≥ r. By (I.80) for every n < ω we can find a subset Γ of ω1
of size n such that G = {αξ : ξ ∈ Γ} ∈ G and H = {βξ : ξ ∈ Γ} ∈ H.
P
√
1
Then, by (I.81), 2 · kT k ≥ kT (χG )k ≥ ( β∈H T (χG )(β)2 ) 2 ≥ r · n. Since
n is arbitrary, this contradicts the fact that T is bounded. The role of
k · kK,2 is to represent every bounded diagonal operator D(x)(ξ) = λξ x(ξ)
as λI + S for some scalar λ and operator S with separable range. This
reduces to the fact that the sequence λξ (ξ < ω1 ) is eventually constant. If
this were false, we would be able to extract a sequence {αξ , βξ } (ξ < ω1 )
of pairwise disjoint pairs of countable ordinals and an r > 0 such that
|λαξ − λβξ | ≥ r for all ξ < ω1 . By (I.80), for every n < ω, we can find a
subset ∆ of ω1 of size n such that K = {{αξ , βξ } : ξ ∈ ∆} belongs to K.
Then, by (I.82),
kD(χS K )k
X
1
≥ (
(D(χS K )(αξ ) − D(χS K )(βξ ))2 ) 2
10 · kDk ≥
ξ∈∆
= (
X
1
(λαξ − λβξ )2 ) 2
ξ∈∆
≥ r·
√
n.
Since n was arbitrary this contradicts the fact that D is a bounded operator.
Note that kxk ≤ 2kxk2 for all x ∈ `2 (ω1 ). It follows that `2 (ω1 ) ⊆ E2
and the inclusion is a bounded linear operator. Note also that `2 (ω1 ) is a
dense subset of E2 . Therefore E2 is a weak compactly generated space. For
example, W = {x ∈ `2 (ω1 ) : kxk2 ≤ 1} is a weakly compact subset of E2
and its linear span is dense in E2 . To get a reflexive example out of E2 one
uses an interpolation method of Davis, Figiel, Johnson and Pelczynski [11] as
follows. Let pn be the Minkowski functional of the set 2n W +2−n Ball(E2 ).29
Let
∞
X
1
E = {x ∈ E2 : kxkE = (
pn (x)2 ) 2 < ∞}.
n=0
28 Indeed it can be shown that 1 (α < ω ) is a ’transfinite basis’ of E in the sense
α
1
2
of [70]. So every vector x of E2 has a unique representation as Σα<ω1 x(α)1α and the
projection operators Pβ : E2 → E2 ¹ β (β < ω1 ) are uniformly bounded.
29 I.e. p (x) = inf{λ > 0 : x ∈ λB}, where B denotes this set.
n
60
I. Coherent Sequences
By [11, Lemma 1], E is a reflexive Banach space and `2 (ω1 ) ⊆ E ⊆ E2 are
continuous inclusions. Note that pn (x) < r iff x = y + z for some y ∈ E2
and z ∈ `2 (ω1 ) such that kyk < 2−n r and kzk2 < 2n r. The following two
facts about E correspond to the facts (I.81) and (I.82) and they are used in
the proof that every bounded operator T : E −→ E has the form λI + S in
a very similar way.
If x is the vector supported by some G ∈ G which can be
split into n blocks such that the k-th one has size 24k and
x takes the value 2−2k on each term of the k-th block, then
kxkE ≤ 2.
(I.83)
If {{αξ , βξ } : ξ ∈ ∆} is a member of K so that ∆ can be
4k
split into n blocks such that the k-th
S one ∆k has size 2
and if x is the vector supported by K for ξ ∈ ∆k , x(αξ ) =
2−2k and x(βξ ) = 2−2k , then kxkE ≤ 12.
(I.84)
We leave the checking of (I.83) and (I.84) as well as the corresponding
proof that E has ’few’ operators to the interested reader.
a
7.17 Remark. The above example is reproduced from Wark [94] who based
his example on a previous construction due to Shelah and Steprans [68]. The
reader is referred to these sources for more information.
We only mention yet another interesting application of the square-bracket
operation, given recently by Erdős, Jackson and Mauldin [21]:
7.18 Example. For every positive integer n there exist collections H and
X of hyperplanes and points of Rn , respectively, and a coloring P : H −→ ω
such that:
any n hyperplanes of distinct colors meet in at most one
point, and
(I.85)
there is no coloring Q : X −→ ω such that for every H ∈ H
there exist at most n − 1 points x in X ∩ H such that
Q(x) = P (H).
(I.86)
Finally we mention yet another projection of the square-bracket operation.
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 takes
each of the values stationarily many times. Define a new operation on ω1
by
|αβ| = π([αβ])(rα ¹ n(π([αβ])), rβ ¹ n(π([αβ]))).
(I.87)
7. The trace and the square-bracket operation
61
7.20 Lemma. For every positive integer n, every uncountable subset Γ of
ω1 and every symmetric n × n-matrix M of countable ordinals there is a
one-to-one φ : n −→ Γ such that |φ(i)φ(j)| = M (i, j) for i, j < n.
Proof. Let Γ be an uncountable subset of ω1 and let M be a given n × n
matrix on countable ordinals. Then for each α we can find tα in the αth level
of the tree T (ρ0 ) and a finite set Fα ⊆ Γ \ α of size n such that tα ⊆ (ρ0 )γ
for all γ ∈ Fα . Let nα be the minimal integer such that rγ ¹ nα 6= rδ ¹ nα
whenever γ 6= δ are members of Fα . For a limit ordinal α let ξ(α) < α be
an upper bound of all sets of the form Cγi ∩ α where γ ∈ Fα , i < k < ω
and γ = γ0 > . . . > γk = α is the minimal walk from γ to α along the Csequence fixed at the beginning of this chapter. Then there is a stationary
set ∆ ⊆ 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 in
(I.90)
some Fα for α ∈ ∆.
Let h : 2m × 2m −→ ω1 be an arbitrary map subject to the requirement
that h : (si , sj ) = M (i, j) for all i, j < n. Let Σh = {δ < ω1 : π(δ) = h}.
By the choice of π the set Σh is stationary in ω1 . By the proof of the basic
property 7.4 of the square bracket operation we can find ν ∈ Σh , α ∈ ∆ ∩ ν
and β ∈ ∆ \ ν such that tα ¹ ξ = tβ ¹ ξ, Fα ⊆ ν and [αβ] = ν. We claim
that
[γδ] = ν for all γ ∈ Fα and δ ∈ Fβ .
(I.91)
To see this, first note that ∆0 (γ, δ) = ∆0 (α, β) = σ for all γ ∈ Fα and
δ ∈ Fβ . By the choice of ξ and the fact that σ ≥ ξ the walk from some
δ ∈ Fβ to σ goes through β, so for any γ ∈ Fα and δ ∈ Fβ we have
[γδ] = min(Tr(σ, δ) \ γ) = min(Tr(σ, β) \ α) = [αβ].
In fact the proof of Lemma 7.4 shows that we can find ν0 ∈ Σh , α0 ∈ ∆ ∩ ν
and an uncountable set ∆1 ⊆ ∆ \ ν0 such that Fα0 ⊆ ν0 , tα0 ¹ ξ = tβ ¹ ξ
and [α0 β] = 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 [α1 β] = ν1 for all
β ∈ ∆2 , and so on. Repeating this n times we get a sequence α0 , . . . , αn−1
such that for all i < j < n,
π([γδ]) = h for all γ ∈ Fαi and δ ∈ Fβ with β ∈ ∆j .
(I.93)
Define φ : n −→ ω1 by letting φ(i) be the ith member of Fαi . Going
back to the definition of the projection | · ·| of [··] one sees that indeed
|φ(i)φ(j)| = M (i, j) holds for all i < j < n.
a
62
I. Coherent Sequences
8. A square-bracket operation on a tree
In this section we try to show that the basic idea of the square-bracket
operation on ω1 can perhaps be more easily grasped by working on an
arbitrary special Aronszajn tree rather than T (ρ0 ). So let T = hT, <T i be
a fixed special Aronszajn tree and let a : T −→ ω be a fixed map witnessing
this, i.e. a mapping with the property that a−1 (n) is an antichain of T for
all n < ω. We shall assume that for every s, t ∈ T the greatest lower bound
s ∧ 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 = min{v ∈ Fa(s∧t) (t) : ht(v) ≥ ht(s)}.
(If ht(s) ≥ ht(t) we let [st]T = [ts]T .)
The following fact corresponds to Lemma 7.4 when T = T (ρ0 ).
8.1 Lemma. If X is an uncountable subset of T , the set of nodes of T of
the form [st]T for some s, t ∈ X intersects a closed and unbounded set of
levels of T .
a
We do not give a proof of this fact as it is almost identical to the proof of
Lemma 7.4 which deals with the special case T = T (ρ0 ). But one can go further and show that [··]T shares all the other properties of the square-bracket
operation [··] described in the previous section. Some of these properties,
however, are easier to visualize and prove in the general context. For example, consider the following fact which in the case T = T (ρ0 ) is the essence
of Lemma 7.20.
8.2 Lemma. Suppose A ⊆ T is an uncountable antichain and that for each
t ∈ A be given a finite set Ft of its successors. Then for every stationary
set Γ ⊆ ω1 there exists an arbitrarily large finite set B ⊆ A such that the
height of [xy]T belongs to Γ whenever x ∈ Fs and y ∈ Ft for some s =
6 t in
B.
a
Let us now examine in more detail the collection of graphs GΓ (Γ ⊆ ω1 )
of 7.12 but in the present more general context.
8.3 Definition. For Γ ⊆ ω1 , let
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 isomorphic uncountable subgraph whenever the symmetric difference between Γ
and ∆ is a stationary subset of ω1 , i.e. whenever they represent different
members of the quotient algebra P(ω1 )/N S. In particular, KΓ contains no
square [X]2 of an uncountable set X ⊆ T whenever Γ contains no closed
and unbounded subset of ω1 . The following fact is a sort of converse to this.
8.4 Lemma. If Γ contains a closed and unbounded subset of ω1 then there
is a proper forcing notion introducing an uncountable set X ⊆ T such that
[X]2 ⊆ KΓ .
Proof. The forcing notion P is defined to be the set of all pairs p = hXp , Np i
where
Np is a finite ∈-chain of countable elementary submodels
of Hω2 containing all the relevant objects.
(I.94)
Xp is a finite subset of T such that ht([st]T ) ∈ Γ for every
pair {s, t} of elements of Xp .
(I.95)
for all s 6= t in Xp there exist N ∈ Np such that s ∈ N iff
t∈
/ N.
(I.96)
for every s 6= t in Xp , N ∈ Np , if t ∈
/ N then the height of
min(Fa(s∧t) (t) \ N ) belongs to Γ.
(I.97)
The ordering of P is the coordinatewise inclusion. To show that P is a proper
forcing notion, let M be a given countable elementary submodel of some
large enough Hθ containing all the relevant objects and let p ∈ P ∩ M . We
shall show that
q = hXp , Np ∪ {M ∩ Hω2 }i
is an M -generic condition extending p. So let D ∈ M be a dense-open subset
of P and let r be a given extension of q. Extending r we may assume that
r ∈ D. Let p0 = r ∩ M . Note that p0 ∈ P ∩ M . Let m0 be the maximum of
the union of the following two finite sets of integers
{a(s ∧ t) : s, t ∈ Xr },
{a(t ¹ δ) : t ∈ Xr , δ = N ∩ ω1 for some N ∈ Np such that t ∈
/ N }.
Let k be the size of the projection of the set Xr \ M on the level TM ∩ω1
of the tree T . Let F be the family of all k-element subsets F of T for which
one can find r̄ ∈ D realizing the same type as r over p0 such that, if N is
the minimal model of Nr̄ \ Np0 then
F = {x ¹ (N ∩ ω1 ) : x ∈ Xr̄ \ Xp0 }.
64
I. Coherent Sequences
Since the projection of Xr̄ \ M onto TM ∩ω1 belongs to F, the family is a
rather large subset of [T ]k . It follows that the family E of all k-element
subsets E of T for which we can find δ < ω1 such that
FE = {F ∈ F : {x ¹ δ : x ∈ F } = E}
is uncountable, is also quite large. In particular, we can find an ordinal
α0 ∈ M ∩ ω1 above the heights of all nodes from the set
[
{M ∩ Fm0 (t) : t ∈ Xr }
and E1 ∈ E ∩ M such that
{x ¹ α0 : x ∈ E1 } = {x ¹ α0 : x ∈ Xr \ M }
and such that every node of Xr \ M is incomparable with every node of E1 .
Let
m+ = max{a(x ∧ y) : x ∈ Xr \ M, y ∈ E1 },
m− = min{a(x ∧ y) : x ∈ Xr \ M, y ∈ E1 and a(x ∧ y) > m0 }.
Let β0 be an ordinal of M ∩ ω1 above α0 which bounds the heights of all
nodes from the set
[
{M ∩ Fm+ (t) : t ∈ Xr \ M }.
By the definitions of E and F there is r̄ ∈ D∩M realizing the same type over
p0 as r such that if N is the minimal model of Nr̄ \ Np0 then N ∩ ω1 ≥ β0
and the projection
F = {x ¹ (N ∩ ω1 ) : x ∈ Xr̄ \ Xp0 }
dominates the set E1 . We claim that
q̄ = hXr̄ ∪ Xr , Nr̄ ∪ Nr i
is a condition of P. This will finish the proof of properness of P since clearly
q̄ extends both r̄ and r. Clearly, only (I.95) and (I.97) for q̄ require some
argumentation. We check first (I.97), so let s and t be a pair of distinct
members of Xr̄ ∪ Xr and let N ∈ Nr̄ ∪ Nr be a model such that t ∈
/ N.
Let m = a(s ∧ t). If t ∈ Xr̄ and if there is s0 ∈ Xr̄ such that s0 6= t and
s0 ∧ t = s ∧ t the conclusion that the height of min(Fm (t) \ N ) belongs to Γ
follows from the fact that r̄ satisfies (I.97). Similarly one considers the case
when t ∈ Xr \ M and when s0 ∧ t = s ∧ t for some s0 ∈ Xr , s0 6= t. For if
N belongs to Nr the conclusion follows from the fact that r satisfies (I.97)
and if N ∈ Nr̄ \ Nr then by the choice of r̄,
Fm (t) \ N = Fm (t) \ M,
8. A square-bracket operation on a tree
65
and so the conclusion that the height of min(Fm (t) \ N ) belongs to Γ follows
from the fact that r satisfies (I.97) for s0 and t from Xr and M ∩ Hω2 from
Nr . Let us now consider the case t ∈ Xr \ M , s ∈ Xr̄ \ Xr and s0 ∧ t 6= s ∧ t
for all s0 ∈ 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 0 ∩ω1 ) for every N 0 ∈ Nr .
Since every ordinal of the form N 0 ∩ ω1 (N 0 ∈ Nr ) belongs to Γ we are done
in case N belongs to Nr . So suppose N ∈ Nr̄ \ Nr . Then by the choice of
the bound β0 and the choice of r̄,
Fm (t) \ N = Fm (t) \ M
so the conclusion of (I.97) follows again from the fact that M ∩ Hω2 belongs
to Nr . Consider now the remaining case t ∈ Xr̄ \ Xr and s ∈ Xr \ Xr̄ . Note
that in this case the model N must belong to Nr so if s0 ∧ t = s ∧ t for
some s0 ∈ Xr̄ , s 6= t the conclusion follows from the fact that r̄ satisfies the
condition (I.97). So let us consider the subcase when s0 ∧ t 6= s ∧ t for all
s0 ∈ Xr̄ , s0 6= t. Thus, s ∧ t is the new splitting so by the choice of α0 and
E0 we have the inequalities:
m0 < m− ≤ m = a(s ∧ t).
Recall that r̄ was chosen to realize the same type over p0 as r so we in
particular have that Fm0 (t) and therefore its superset Fm (t) contains the
projection t ¹ (N ∩ ω1 ). It follows that min(Fm (t) \ N ) = t ¹ (N ∩ ω1 ), so
its height is indeed a member of Γ (as Γ ∈ N and Γ contains a closed and
unbounded subset of ω1 ).
It remains to check (I.95) for q̄. So let s and t be a given pair of distinct
members of Xr̄ ∪ Xr . We need to show that the height of [st]T belongs to Γ.
Clearly the only nontrivial case is when, say, s ∈ Xr̄ \ Xr and t ∈ Xr \ Xr̄ .
Suppose first that s ∧ t = s0 ∧ t for some s0 ∈ Xr , s0 6= t. Since M ∩ Hω2
belongs to Nr and since r satisfies (I.97), the height of min(Fa(s0 ∧t) (t) \ M )
belongs to Γ. Since the height of s is above the bound β0 for Fm+ (t) ∩ M ,
[st]T = min(Fa(s∧t) (t) \ M ) = min(Fa(s0 ∧t) (t) \ M ),
so we are done in this subcase. Suppose now that s ∧ t is a new splitting,
i.e. that
m0 < m = a(s ∧ t) ≤ m+ .
So again we know that in this subcase Fm (t) contains the projection t ¹
(M ∩ ω1 ). By our choice of the bound β0 and the condition r̄ we know
that this projection is the minimal point of Fm (t) whose height is bigger or
66
I. Coherent Sequences
equal to the height of s. By the definition of the square-bracket operation,
we have
[st]T = t ¹ (M ∩ ω1 ).
Since M ∩ ω1 belongs to Γ this finishes our checking that q̄ 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 Ẋ of the Xp ’s for
p belonging to the generic filter is included in KΓ . What is not so clear is
that P forces that the set Ẋ is uncountable as promised. To ensure this we
replace P with its restriction P(≤ r) to the condition
r = h{t}, {M ∩ Hω2 }i,
where M is an arbitrary countable elementary submodel of some large
enough Hθ containing all the relevant objects and where t is any node of
T of height at least M ∩ ω1 . Note that the above proof of properness of P
starting from p = h∅, ∅i shows that
q = h∅, {M ∩ Hω2 }i
is an M -generic condition of P so in particular its extension r is also M generic. It follows that if we let M [Ġ] be the name for the set formed by
interpreting all P-names belonging to M itself. Then r forces that Ẋ is
an element of the countable elementary submodel M [Ġ] of Hθ [Ġ] which
contains a point t of height above M [Ġ] ∩ ω1 . It follows that r forces Ẋ to
a
be uncountable. This finishes the proof of Lemma 8.4.
8.5 Corollary. The graph KΓ contains the square of some uncountable subset of T in some ω1 -preserving forcing extension if and only if Γ is a stationary subset of ω1 .
Proof. If Γ is disjoint from a closed and unbounded subset then in any ω1 preserving forcing extension its complement ∆ = ω1 \ Γ will be a stationary
subset of ω1 . So by the basic property 8.1 of the square-bracket operation
no such a forcing extension will contain an uncountable set X ⊆ T such
that [X]2 ⊆ KΓ . On the other hand, if Γ is a stationary subset of ω1 , going
first to some standard ω1 -preserving forcing extension in which Γ contains
a closed and unbounded subset of ω1 and then applying Lemma 8.4, we get
an ω1 -preserving forcing extension having an uncountable set X ⊆ T such
that [X]2 ⊆ KΓ .
a
8.6 Remark. Corollary 8.5 gives us a further indication of the extreme
complexity of the class of graphs on the vertex-set ω1 . It also bears some
relevance to the recent work of Woodin [98] who, working in his Pmax -forcing
extension, was able to associate a stationary subset of ω1 to any partition
of [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 sufficiently
generic filter to the forcing notion P = PΓ of Lemma 8.4 that would give us
an uncountable X ⊆ T such that [X]2 ⊆ KΓ . In other words, under a bit of
PFA or Woodin’s axiom (*), a set Γ ⊆ ω1 contains a closed and unbounded
subset 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 trees
is the question whether they contain any cofinal branch, a branch that intersects each level of the tree. The fundamental importance of this question
has already been realized in the work of Kurepa [45] and then later in the
works of Erdős and Tarski in their respective attempts to develop the theory
of partition calculus and large cardinals (see [23]). A tree T of height equal
to some regular cardinal θ may not have a cofinal branch for a very special
reason as the following definition indicates.
9.1 Definition. For a tree T = hT, <T i, a function f : T → T is regressive
if f (t) <T t for every t ∈ T that is not a minimal node of T. A tree T of
height θ is special if there is a regressive map f : T −→ T with the property
that the f -preimage of every point of T can be written as the union of < θ
antichains of T .
This definition in case θ = ω1 reduces indeed to the old definition of
special tree, a tree that can be decomposed into countably many antichains.
More generally we have the following:
9.2 Lemma. If θ is a successor cardinal then a tree T of height θ is special
if and only if T is the union of < θ antichains.
a
The new definition, however, seems to be the right notion of speciality
as it makes sense even if θ is a limit cardinal.
9.3 Definition. A tree T of height θ is Aronszajn if T has no cofinal
branches and if every level of T has size < θ.
Recall the well-known characterization of weakly compact cardinals due
to Tarski and his collaborators: a strongly inaccessible cardinal θ is weakly
compact if and only if there are no Aronszajn trees of height θ. We supplement this with the following:
9.4 Theorem. The following are equivalent for a strongly inaccessible cardinal θ:
(1) θ is Mahlo.
(2) there are no special Aronszajn trees of height θ.
68
I. Coherent Sequences
Proof. Suppose θ is a Mahlo cardinal and let T be a given tree of height
θ all of whose levels have size < θ. To show that T is not special let
f : T −→ T be a given regressive mapping. By our assumption of θ there
is an elementary submodel M of some large enough structure Hκ such that
T, f ∈ M and λ = M ∩ θ is a regular cardinal < θ. Note that T ¹ λ is a
subset of M and since this tree of height λ is clearly not special, there is
t ∈ T ¹ λ such that the preimage f −1 (t) is not the union of < λ antichains.
Using the elementarity of M we conclude that f −1 (t) is actually not the
union of < θ antichains.
The proof that (2) implies (1) uses the method of minimal walks in a
rather crucial way. So suppose to the contrary that our cardinal contains a
closed and unbounded subset C consisting of singular strong limit cardinals.
Using C, we choose a C-sequence Cα (α < θ) such that: Cα+1 = {α},
Cα = (ᾱ, α) for α limit such that ᾱ = sup(C ∩ α) < α but if α = sup(C ∩ α)
then take Cα such that:
tp(Cα ) = cf(α) < min(Cα ),
(I.98)
ξ = sup(Cα ∩ ξ) implies ξ ∈ C,
(I.99)
ξ ∈ Cα and ξ > sup(Cα ∩ ξ) imply ξ = η + 1 for some η ∈ C.
(I.100)
Given the C-sequence Cα (α < θ) we have the notion of minimal walk along
the sequence and various distance functions defined above. In this proof we
are particularly interested in the function ρ0 from [θ]2 into the set Qθ of all
finite sequences of ordinals from θ:
ρ0 (α, β) = htp(Cβ ∩ α)ia ρ0 (α, min(Cβ \ α))
where we stipulate that ρ0 (γ, γ) = 0 for all γ. We would like to show that
the tree
T (ρ0 ) = {(ρ0 )β ¹ α : α ≤ β < θ}
is a special Aronszajn tree of height θ. Note that the size of the αth level
(T (ρ0 ))α of T (ρ0 ) is controlled in the following way:
|(T (ρ0 ))α | ≤ |{Cβ ∩ α : α ≤ β < θ}| + |α| + ℵ0 .
(I.101)
So under the present assumption that θ is a strongly inaccessible cardinal,
all levels of T (ρ0 ) do indeed have size < θ. It remains to define the regressive
map
f : T (ρ0 ) −→ T (ρ0 )
that will witness speciality of T (ρ0 ). Note that it really suffices defining f
on all levels whose index belong to our club C of singular cardinals. So let
t = (ρ0 )β ¹ α be a given node of T such that α ∈ C and α ≤ β < θ. Note
that by our choice of the C-sequence every term of the finite sequence of
9. Special trees and Mahlo cardinals
69
ordinals ρ0 (α, β) is strictly smaller than α. So, if we let f (t) = t ¹ pρ0 (α, β)q,
where p·q is a standard coding of finite sequences of ordinals by ordinals,
we get a regressive map. To show that f is one-to-one on chains of T (ρ0 ),
which would be more than sufficient, suppose ti = (ρ0 )βi ¹ αi (i < 2) are
two nodes such that t0 $ t1 . Our choice of the C-sequence allows us to
deduce the following general fact about the corresponding ρ0 -function as in
the case θ = ω1 dealt with above in Lemma 1.14.
If α ≤ β ≤ γ, α is a limit ordinal, and if 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 that
ρ0 (α0 , β0 ) = ρ0 (α0 , β1 ). Now observe another fact about the ρ0 -function
whose proof is identical to that of θ = ω1 dealt with above in Lemma 1.13.
If α ≤ β ≤ γ then ρ0 (α, γ) <r ρ0 (β, γ).
(I.103)
Applying this to the triple α0 < α1 ≤ β1 we in particular have that
ρ0 (α0 , β1 ) 6= ρ0 (α1 , β1 ). Combining this with the above equality gives us
that ρ0 (α0 , β0 ) 6= ρ0 (α1 , β1 ) and therefore that f (t0 ) 6= f (t1 ).
a
This leads us to the following Ramsey-theoretic characterization of Mahlo
cardinals.
9.5 Theorem. A cardinal θ is a Mahlo cardinal if and only if every regressive map f defined on a cube [C]3 of a closed and unbounded subset of θ has
an infinite min-homogeneous set X ⊆ C.30
Proof. Again, the direct implication is simple so we leave it to the interested
reader. For the converse, note that the statement about regressive maps
easily implies that θ must be a strongly inaccessible cardinal. So assume
θ is a strongly inaccessible non-Mahlo cardinal. Let C be a closed and
unbounded subset of θ consisting of strong limit singular cardinals. We
shall find a regressive map on [C]3 without an infinite min-homogeneous
set. This will give us an opportunity to further expose the tree T (ρ0 ) and
the regressive map f : T (ρ0 ) ¹ C −→ T (ρ0 ) defined in the proof of Theorem
9.4. We have already observed that the length of every branching node t
must be of the form α + 1 for some α and in fact the following more precise
description is true:
If t is a branching node of T (ρ0 ) then its length is equal to
α + 1 for some α ∈ C.
Let
(I.104)
α− = max(C ∩ (α + 1)) and α+ = min(C \ (α + 1)).
30 Recall, that X is min-homogeneous for f if f (α, β, γ) = f (α0 , β 0 , γ 0 ) for every pair
α < β < γ and < α0 < β 0 < γ 0 of triples of elements of X such that α = α0 .
70
I. Coherent Sequences
Let β, γ > α + 1 be two ordinals such that (ρ0 )β and (ρ0 )γ split at t. Let
β = β0 > . . . > βk = α and γ = γ0 > . . . > γl = α be the walks from β to α
and γ to α respectively. From the fact that (ρ0 )β and (ρ0 )γ agree below α
we conclude that k = l and Cβi ∩ α = Cγi ∩ α for all i ≤ k. By the choice
of the C-sequence and the inequality α > α− we can deduce a sequence
of stronger agreements Cβi ∩ α+ = Cγi ∩ α+ for all i ≤ k. However, that
would mean that ρ0 (α + 1, β) = ρ0 (α + 1, γ), contrary to our assumption
that (ρ0 )β and (ρ0 )γ split at t. It follows that α = α− which was to be
shown.
Going back to the proof of Theorem 9.5, recall the regressive map f :
T (ρ0 ) ¹ C −→ T (ρ0 ) defined in 9.4:
f ((ρ0 )β ¹ α) = (ρ0 )β ¹ pρ0 (α, β)q.
We extend f to the whole tree by letting f (t) = t ¹ α where α is the maximal
element of C that is less than or equal to the length of t. Then we know
that f is one-to-one on any chain whose nodes are separated by C. From
f we shall derive three other regressive maps p : [C]3 −→ θ, q : [C]3 −→ ω
and 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(f n (β ∧ γ)), where n is the minimal integer such that f n (β ∧ γ) ⊆ (ρ0 )α .
q(α, β, γ) = min{n : f n (β ∧ γ) = ∅}.
r(α, β) = phρ0 (ξ, α), ρ0 (ξ, β)iq, where ξ = length(α ∧ β).
(I.105)
(I.106)
(I.107)
Now we claim that no infinite subset of C can be simultaneously minhomogeneous with respect to all three regressive maps. For, given an infinite
increasing sequence {αi } of elements of C using Ramsey’s theorem we may
assume that either
αi ∧ αj = αk ∧ αl for all i 6= j and k 6= l in ω, or
αi ∧ αi+1 $ αi+1 ∧ αi+2 for all i < ω.
(I.108)
(I.109)
Note that (I.108) would violate min-homogeneity with respect to the mapping r. If (I.109) holds and if the sequence is min-homogeneous with respect
to p and q another application of Ramsey’s theorem would give us an integer
m and an ordinal ξ such that for all i < j < k < ω:
m = min{n : f n (αj ∧ αk ) ⊆ (ρ0 )αi },
(I.110)
f m (αj ∧ αk ) = (ρ0 )α0 ¹ ξ(= t).
(I.111)
9. Special trees and Mahlo cardinals
71
By (I.104) the chain {αi ∧ αi+1 : i ≥ 1} is separated by C, so using the
basic property of f we conclude that m > 1. Then f −1 (t) includes the set
{f m−1 (αi ∧ αi+1 ) : i ≥ 1}. By (I.110) we have,
αi ∧ αi+1 $ f m−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, a
contradiction. This finishes the proof of Theorem 9.5.
a
Starting from the case n = 1 one can now easily deduce the following
characterization of n-Mahlo cardinals.
9.6 Theorem. The following are equivalent for an uncountable cardinal θ
and a positive integer n:
(1) θ is n-Mahlo.
(2) Every regressive map defined on [C]n+2 for some closed and unbounded
subset C of θ has an infinite min-homogeneous subset.
Proof. For a set X of ordinals and n ∈ ω, let W (X, n) denote the statement
that there exist f : [X]n+2 −→ ω and regressive g[X]n+3 −→ sup(X) such
that if H ⊆ X is min-homogeneous for g and f ”[H]n+2 = {k} for some
k ∈ ω, then |H| ≤ n + k + 5. Note that in the previous proof we have
established W (C, 0) for every closed unbounded subset C of some limit
ordinal θ such that C contains no inaccessible cardinals (the mapping q is
really a map from [C]2 into ω). The proof of the Theorem is finished once
we show that in general W (C, n) holds for every set of ordinals C of limit
order type that is closed in its supremum and which contains no n-Mahlo
cardinals. So let C be a given closed and unbounded subset of some limit
ordinal θ and suppose that C contains no n + 1-Mahlo cardinals. We need
to show that W (C, n + 1) holds. Note that for this it suffices to show by
induction that W (C ∩ δ, n) holds for all δ ∈ C since there is a natural
way to combine pairs of maps fδ : [C]n+2 −→ ω and gδ : [C]n+3 −→ θ
for δ ∈ C into maps f : [C]n+3 −→ ω and g : [C]n+4 −→ θ witnessing
W (C, n + 1). The successor steps of the induction follow from the easily
checked fact that W (X, n) implies W (Y, n) for every end-extension Y of X
of cardinality at most 2|sup(X)| . Assume now that δ is a limit point of C and
that W (C ∩γ, n) holds for all δ ∈ C ∩δ. Since δ is not an n+1-Mahlo cardinal
there is a closed and unbounded set D ⊆ C ∩ δ containing no n-Mahlo
cardinals. By the induction hypothesis, W (D, n) holds. Now, there is a
very natural way to combine maps fγ : [C]n+2 −→ ω and gγ : [C]n+3 −→ θ
for γ ∈ C ∩ δ witnessing W (C ∩ γ, n) and maps fδ : [D]n+2 −→ ω and
gδ : [D]n+3 −→ θ witnessing W (D, n) into maps Fδ : [C ∩ δ]n+2 −→ ω and
Gδ : [C ∩ δ]n+3 −→ θ witnessing W (C ∩ δ).
a
72
I. Coherent Sequences
9.7 Remark. Theorems 9.5 and 9.6 are due to Hajnal, Kanamori and Shelah [30] who were improving an earlier characterization of this sort due to
Schmerl [64]. Using Theorem 9.6 one can also prove the following characterization theorem essentially given in [64](see also [30]) which has some
interesting metamathematical applications (see [26] and [38]).
9.8 Theorem. The following are equivalent for an uncountable cardinal θ
and a positive integer n:
(1) θ is n-Mahlo.
(2) Every regressive map defined on [C]n+3 for some closed and unbounded
subset C of θ has a min-homogeneous set of size n + 5.
a
We finish this section by showing how the idea of proof of Theorem 9.4
leads us naturally to the following well-known fact, first established by Silver
(see [52]) when θ is a successor of a regular cardinal.
9.9 Theorem. If θ is a regular uncountable cardinal which is not Mahlo
in the constructible universe, then there is a constructible special Aronszajn
tree of height θ.
Proof. Working in L we choose a closed and unbounded subset C of θ
consisting of singular ordinals and a C-sequence Cα (α < θ) such that
Cα+1 = {α}, Cα = (ᾱ, α) when ᾱ = sup(C ∩ α) < α, while if α is a limit
point of C we take Cα to have the following properties:
ξ = sup(Cα ∩ ξ) implies ξ ∈ C
ξ > sup(Cα ∩ ξ) implies ξ = η + 1 for some η ∈ C.
(I.113)
(I.114)
We choose the C-sequence to also have the following crucial property:
|{Cα ∩ ξ : ξ ≤ α < θ}| ≤ |ξ| + ℵ0 for all ξ < θ.
(I.115)
It is clear then that the tree T (ρ0 ), where ρ0 is the ρ0 -function of Cα (α < θ),
is a constructible special Aronszajn tree of height θ.
a
We are also in a position to deduce the following well-known fact.
9.10 Theorem. The following are equivalent for a successor cardinal θ:
(a) there is a special Aronszajn tree of height θ.
(b) there is a C-sequence Cα (α < θ) such that tp(Cα ) ≤ θ− for all α and
such that {Cα ∩ ξ : α < θ} has size ≤ θ− for all ξ < θ.
10. The weight function on successor cardinals
73
Proof. If Cα (α < θ) is a C-sequence satisfying (b) and if ρ0 is the associated
ρ0 -function then T (ρ0 ) is a special Aronszajn tree of height θ. Suppose <T
is a special Aronszajn tree ordering on θ such that [θ− · α, θ− · (α + 1)) is its
αth 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 the
tree (θ, <T ). We choose a C-sequence Cα (α < θ) such that Cα+1 = {α},
Cα = (ᾱ, α) for α limit with the property that ᾱ = sup(C ∩ α) < α, but if α
is a limit point of C we take Cα more carefully as follows: Cα = {αξ : ξ < η}
where
αλ = sup{αξ : 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 subject to the requirement that f (αξ+1 ) > f (αζ+1 ) for all
ζ < ξ,
(I.118)
η is the limit ordinal ≤ θ− where the process stops, i.e.
sup{f (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 then
Cα ∩ γ = Cβ ∩ γ. From this one concludes that the C-sequence is locally
small, 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 of
height θ+ .
a
9.12 Corollary. In the constructible universe, special Aronszajn trees of
any regular uncountable non-Mahlo height exist.
a
9.13 Remark. In a large portion of the literature on this subject the notion
of a special Aronszajn tree of height equal to some successor cardinal θ+
is somewhat weaker, equivalent to the fact that the tree can be embedded
inside the tree {f : α −→ θ : α < θ & f is 1 − 1}. One would get our notion
of speciality by restricting the tree on successor ordinals losing thus the
frequently useful property of a tree that different nodes of the same limit
height have different sets of predecessors. The result 9.11 in this weaker
form is due to Specker [71], while the result 9.12 is essentially due to Jensen
[35].
10. The weight function on successor cardinals
In this section we assume that θ = κ+ and we fix a C-sequence Cα (α < κ+ )
such that
tp(Cα ) ≤ κ for all α < κ+ .
(I.120)
74
I. Coherent Sequences
Let ρ1 : [κ+ ]2 −→ κ be defined recursively by
ρ1 (α, β) = max{tp(Cβ ∩ α), ρ1 (α, min(Cβ \ α))}
where we stipulate that ρ1 (γ, γ) = 0 for all γ.
10.1 Lemma. |{ξ ≤ α : ρ1 (ξ, α) ≤ ν}| ≤ |ν| + ℵ0 for all α < κ+ and ν < κ.
Proof. Let ν + be the first infinite cardinal above the ordinal ν. The proof
of the conclusion is by induction on α. So let Γ ⊆ α be a given set of ordertype ν + . We need to find ξ ∈ Γ such that ρ1 (ξ, α) > ν. This will clearly be
true if there is ξ ∈ Γ such that tp(Cα ∩ ξ) > ν. So, we may assume that
tp(Cα ∩ ξ) ≤ ν for all ξ ∈ Γ.
(I.121)
Then there must be an ordinal α1 ∈ Cα such that
Γ1 = {ξ ∈ Γ : α1 = min(Cα \ ξ)}
has size ν + . By the inductive hypothesis there is ξ ∈ Γ1 such that (see
(I.121))
ρ1 (ξ, α1 ) > ν ≥ tp(Cα ∩ ξ).
(I.122)
It follows that
ρ1 (ξ, α) = max{tp(Cα ∩ ξ), ρ1 (ξ, α1 )} = ρ1 (ξ, α1 ) > ν.
This finishes the proof.
a
10.2 Lemma. If κ is regular, then {ξ ≤ α : ρ1 (ξ, α) 6= ρ1 (ξ, β)} has size
< κ for all α < β < κ+ .
Proof. The proof is by induction on α and β. Let Γ ⊆ α be a given set
of order-type κ. We need to find ξ ∈ Γ such that ρ1 (ξ, α) = ρ1 (ξ, β). Let
γ = sup(Γ) and
γ0 = max(Cβ ∩ γ), β0 = min(Cβ \ γ).
(I.123)
Note that by our assumption on κ and the C-sequence, these two ordinals
are well-defined and
γ0 < γ ≤ β0 < β.
(I.124)
By Lemma 10.1 and the inductive hypothesis there is ξ in Γ ∩ (γ0 , γ) such
that
ρ1 (ξ, α) = ρ1 (ξ, β0 ) > tp(Cβ ∩ γ).
(I.125)
It follows that Cβ ∩ γ = Cβ ∩ ξ and β0 = min(Cβ \ ξ), and so
ρ1 (ξ, β) = max{tp(Cβ ∩ ξ), ρ1 (ξ, β0 )} = ρ1 (ξ, β0 ) = ρ1 (ξ, α).
This completes the proof.
a
11. The number of steps
75
10.3 Remark. The assumption about the regularity of κ in Lemma 10.2 is
essential. For example, it can be seen (see [7, p.72]) that the conclusion of
this lemma fails if κ is a singular limit of supercompact cardinals.
10.4 Definition. Define ρ̄1 : [κ+ ]2 −→ κ by
ρ̄1 (α, β) = 2ρ1 (α,β) · (2 · tp{ξ ≤ α : ρ1 (ξ, β) = ρ1 (α, β)} + 1).
The following fact shows that this stretching of ρ1 keeps the basic coherence property stated in Lemma 10.2 but it gives us also the injectivity
property that could be quite usefull.
10.5 Lemma. If κ is a regular cardinal then
(a) ρ̄1 (α, γ) 6= ρ̄1 (β, γ) whenever α < β < γ < κ+ ,
(b) |{ξ ≤ α : ρ̄1 (ξ, α) 6= ρ̄1 (ξ, β)}| < κ whenever α < β < κ+ .
a
10.6 Remark. Note that Lemma 10.5 gives an alternative proof of Corollary 9.11 since under the assumption κ<κ = κ the tree T (ρ̄1 ) will have
levels of size at most κ. It should be noted that the coherent sequence
(ρ̄)α (α < κ+ ) of one-to-one mappings is an object of independent interest
which can be particularly useful in stepping-up combinatorial properties of
κ to κ+ . It is also an object that has interpretations in such areas as the
theory of Čech-Stone compactifications of discrete spaces (see, e.g. [95],
[10], [60], [44], [16]). We have already noted that if κ is singular then we
may no longer have the coherence property of Lemma 10.2. To get this
property, one needs to make some additional assumption on the C-sequence
Cα (α < κ+ ), an assumption about the coherence of the C-sequence. This
will be subject of some of the following chapters where we will concentrate
on the finer function ρ rather than ρ1 .
11. The number of steps
The purpose of this section is to isolate a condition on C-sequences Cα (α <
θ) on regular uncountable cardinals θ as weak as possible subject to a requirement that the corresponding function
ρ2 (α, β) = ρ2 (α, min(Cβ \ α)) + 1
is in some sense nontrivial, i.e. far from being constant. Without doubt the
C-sequence Cα = α (α < θ) is the most trivial choice and the corresponding
ρ0 -function gives no information whatsoever about the cardinal θ. The
following notion of the triviality of a C-sequence on θ seems to be only
marginally different.
76
I. Coherent Sequences
11.1 Definition. A C-sequence Cα (α < θ) on a regular uncountable cardinal θ is trivial if there is a closed and unbounded set C ⊆ θ such that for
every α < θ there is β ≥ α with C ∩ α ⊆ Cβ .
11.2 Theorem. The following are equivalent for any C-sequence Cα (α <
θ) on a regular uncountable cardinal θ and the corresponding function ρ2 :
(i) Cα (α < θ) is nontrivial.
(ii) For every family A of θ pairwise disjoint finite subsets of θ and every
integer n there is a subfamily B of A of size θ such that ρ2 (α, β) > n
for all α ∈ a, β ∈ b and a 6= b in B.
Proof. Note that if a closed and unbounded set C ⊆ θ witnesses the triviality
of the C-sequence and if for α < θ we let β(α) ≥ α be the minimal β such
that C ∩ α ⊆ Cβ(α) , then any disjoint subfamily of A = {{α, β(α)} : α ∈ C}
of size θ violates (ii) for n = 1.
Conversely, assume we are given a nontrivial C-sequence Cα (α < θ).
Consider the following statement:
(iii) For every pair Γ and ∆ of unbounded subsets of θ and every integer n
there exist a tail subset Γ0 of Γ, an unbounded subset ∆0 of ∆ and a
regressive strictly increasing map f : ∆0 −→ θ such that ρ2 (α, β) > n
for all α ∈ Γ0 and β ∈ ∆0 with α < f (β).
To get (ii) from (iii) one assumes that the family A consists of finite sets
of some fixed size m and performs m2 successive applications of (iii): this
will give us an unbounded Ā ⊆ A and regressive strictly increasing maps
fij : πj [Ā] −→ θ (i, j < m), where πj is the projection mapping that takes
b to the jth member of b. We have then that for all a, b ∈ Ā 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 Ā once again.
Let us prove (iii) from (i) by induction on n. So suppose Γ and ∆ are
given two unbounded subsets of θ and that (iii) be true for n − 1. Let
Mξ (ξ < θ) be a continuous ∈-chain of submodels of Hθ+ containing all
relevant objects such that δξ = Mξ ∩ θ ∈ θ. Let
C = {ξ < θ : δξ = ξ}.
Then C is a closed and unbounded subset of θ, so by our assumption (i)
on Cα (α < θ) there is δ ∈ C such that C ∩ δ * Cβ for all β ≥ δ. Pick
an arbitrary β ∈ ∆ above δ. Then there is ξ ∈ C ∩ δ such that ξ ∈
/ Cβ .
Then ᾱ = sup(Cβ ∩ ξ) < ξ. Using the elementarity of the submodel Mξ
we conclude that for every α ∈ Γ \ ᾱ there is β(α) ∈ ∆ \ (α + 1) such that
sup(Cβ(α) ∩ α) = ᾱ. Applying the inductive hypothesis to the two sets
Γ \ ᾱ and {min(Cβ(α) \ α) : α ∈ Γ \ ᾱ}
11. The number of steps
77
we get a tail subset Γ0 of Γ \ ᾱ, an unbounded subset Γ1 of Γ \ ᾱ and a
strictly increasing map g : Γ1 −→ θ such that g(γ) < min(Cβ(γ) \ γ) and
ρ2 (α, min(Cβ(γ) \ γ)) > n − 1 for all α ∈ Γ0 , γ ∈ Γ1 with α < g(γ). (I.126)
Let ∆0 = {β(γ) : γ ∈ Γ1 } and f : ∆0 −→ θ be defined by
f (β(γ)) = min{γ, g(γ)}.
Then for α ∈ Γ0 and γ ∈ Γ1 with α < f (β(γ)) ≤ γ < β(γ) we have
ρ2 (α, β(γ))
This finishes the proof.
=
=
>
ρ2 (α, min(Cβ(γ) \ α)) + 1
ρ2 (α, min(Cβ(γ) \ γ)) + 1
(n − 1) + 1 = n.
a
11.3 Corollary. Suppose that Cα (α < θ) is a nontrivial C-sequence and
let T (ρ0 ) be the corresponding tree (see Section 9 above). Then every subset
of T (ρ0 ) of size θ contains an antichain of size θ.
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 assuming
without loss of generality that the set consists of successor nodes of T (ρ0 ).
Clearly, we may also assume that the set takes at most one point from a
given level of T (ρ0 ). Shrinking K further, we obtain that ρ2 is constant
on K. Let n be the constant value of ρ2 ¹ K. Applying 11.2(ii) to K
and n, we get K0 ⊆ K of size θ such that ρ2 (α, δ) > n for all {α, β}
and {γ, δ} from K0 with properties α < β, γ < δ and α < γ. Then
ρ0 (·, β) ¹ (α + 1) ({α, β} ∈ K0 ) is an antichain in T (ρ0 ).
a
11.4 Remark. It should be clear that nontrivial C-sequences exist on any
successor cardinal. Indeed, with very little extra work one can show that
nontrivial C-sequences exist for some inaccessible cardinals quite high in the
Mahlo-hierarchy. To show how close this is to the notion of weak compactness, we will give the following characterization of it which is of independent
interest. 31
11.5 Theorem. The following are equivalent for an inaccessible cardinal θ:
(i) θ is weakly compact.
(ii) For every C-sequence Cα (α < θ) there is a closed and unbounded set
C ⊆ θ such that for all α < θ there is β ≥ α such that Cβ ∩ α = C ∩ α.
31 It turns out that every C-sequence on θ being trivial is not quite as strong as the
weak compactness of θ. As pointed out to us by Donder and König, one can show this
using a model of Kunen [43, §3].
78
I. Coherent Sequences
Proof. To see the implication from (i) to (ii), let us assume that Cα (α < θ)
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 treeproperty of θ (see [37]) the tree T (ρ0 ) must contain a cofinal branch b. For
each α < θ fix β(α) ≥ α such that the restriction of (ρ0 )β(α) to α belongs to
b. For limit α < θ let γ(α) be the largest member γ appearing in the walk
β(α) = β0 (α) > β1 (α) > . . . > βk(α) (α) = α
along the C-sequence with the property that Cγ ∩ α is unbounded in α.
(Thus γ = α or γ = βk(α)−1 (α).) By the Pressing Down Lemma there is a
stationary set Γ ⊆ θ and an integer k, a sequence t in Qθ and an ordinal
ᾱ < θ such that for all α ∈ Γ:
k(α) = k and sup(Cβi (α) ∩ α) ≤ ᾱ for all i < k for which
(I.127)
this set is bounded in α.
tp(Cβi (α) ∩ α) = t(i) for all i < k for which this set is
(I.128)
bounded in α.
Shrinking Γ we may assume that either for all α ∈ Γ, γ(α) = α in which
case t is chosen to be a k-sequence, or for all α ∈ Γ, γ(α) = βk−1 (α) in
which case t is chosen to be a (k − 1)-sequence. It follows that
ρ0 (ξ, β(α)) = ta ρ0 (ξ, γ(α)) for all α ∈ Γ and ξ ∈ [ᾱ, α).
(I.129)
It follows that for α < α0 in Γ the mappings (ρ0 )γ(α) and (ρ0 )γ(α0 ) have the
same restrictions on the interval [ᾱ, α). So in particular we have that
Cγ(α) ∩ [ᾱ, α) = Cγ(α0 ) ∩ [ᾱ, α) whenever α < α0 in Γ.
(I.130)
Going to a stationary subset of Γ we may assume that all Cγ(α) (α ∈ Γ) have
the same intersection with ᾱ. 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 on
θ whose levels are intervals of θ or more precisely, there is a closed and
unbounded set ∆ of limit ordinals in θ such that if δξ (ξ < θ) is its increasing
enumeration then [δξ , δξ+1 ) is equal to the ξth level of T . Choose a Csequence Cα (α < θ) as follows. First of all let Cα+1 = {α} and Cα = (ᾱ, α)
if α is limit and ᾱ = 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 < θ
there is β(α) ≥ α with Cβ(α) ∩ α = C ∩ α. Let Γ be the set of all successor
ordinals 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. This
finishes the proof.
a
11. The number of steps
79
We have already remarked that every successor cardinal θ = κ+ admits
a nontrivial C-sequence Cα (α < θ). It suffices to take the Cα ’s to be all of
order-type ≤ κ. It turns out that for such a C-sequence the corresponding
ρ2 -function has a property that is considerably stronger than 11.2(ii).
11.6 Theorem. For every infinite cardinal κ there is a C-sequence on κ+
such that the corresponding ρ2 -function has the following unboundedness
property: for every family A of κ+ pairwise disjoint subsets of κ+ , all of
size < κ, and for every n < ω there exists B ⊆ A of size κ+ such that
ρ2 (α, β) > n whenever α ∈ a and β ∈ b for some a 6= b in B.
Proof. We shall show that this is true for every C-sequence Cα (α < κ+ )
with the property that tp(Cα ) ≤ κ for all α < κ+ . Let us first consider the
case when κ is a regular cardinal. The proof is by induction on n. Suppose
the conclusion is true for some integer n and let A be a given family of κ+
pairwise disjoint subsets of κ+ , all of size < κ+ . Let Mξ (ξ < κ+ ) be a
continuous ∈-chain of elementary submodels of Hκ++ such that for every
ξ the Mξ contains all the relevant objects and has the property that δξ =
Mξ ∩κ+ ∈ κ+ . Let C = {δξ : ξ < κ+ }. Choose ξ = δξ ∈ C of cofinality κ and
b ∈ A such that β > ξ for all β ∈ b. Then γ = sup{max(Cβ ∩ξ) : β ∈ b} < ξ.
By the elementarity of Mξ we conclude that for every η ∈ (γ, κ+ ) there is bη
in A above η with γ = sup{max(Cβ ∩ η) : β ∈ bη }. Consider the following
family of subsets of κ+ : âη = bη ∪ {min(Cβ \ η) : β ∈ bη } (η ∈ (γ, κ+ )).
By the inductive hypothesis there is an unbounded Γ ⊆ κ+ such that for
all η < ζ in Γ, bη ⊆ ζ and ρ2 (α, β) > n for all α ∈ âη and β ∈ âζ . Let
B = {bη : η ∈ Γ} and let bη and bζ for η < ζ be two given members of B.
Then ρ2 (α, β) = ρ2 (α, min(Cβ \ α)) + 1 ≥ n + 1 for all α ∈ bη and β ∈ bζ .
This completes the inductive step and therefore the proof of the Theorem
when κ is a regular cardinal.
If κ is a singular cardinal, going to a subfamily of A, we may assume that
there is a regular cardinal λ < κ and an unbounded set Γ ⊆ κ+ with the
property that A can be enumerated as aη (η ∈ Γ) such that for all η ∈ Γ:
|aη | < λ,
(I.131)
β > η for all β ∈ aη ,
(I.132)
|Cβ ∩ η| < λ for all β ∈ aη .
(I.133)
Choose ξ = δξ ∈ C of cofinality λ and η̂ ∈ Γ above ξ and proceed as in
the previous case. This completes the inductive step and the proof of the
theorem 11.6.
a
Note that the combination of the ideas from the proofs of Theorem 11.2
and 11.6 gives us the following variation.
80
I. Coherent Sequences
11.7 Theorem. Suppose that a regular uncountable cardinal θ supports a
nontrivial C-sequence and let ρ2 be the associated function. Then for every
integer n and every pair of θ-sized families A0 and A1 , where the members
of A0 are pairwise disjoint bounded subsets of θ and the members of A1 are
pairwise disjoint finite subsets of θ, there exist B0 ⊆ A0 and B1 ⊆ A1 of
size θ such that ρ2 (α, β) > n whenever α ∈ a and β ∈ b for some a ∈ B0
and b ∈ B1 such that sup(a) < min(b).
a
An interesting application of this result has recently been given by Gruenhage [28]. To state his result, given a map from a space X onto a space
Y , let ](f ) be the order-type of the shortestT
chain C of closed subsets of X
such that f 00 H = Y for all H ∈ C, while f 00 ( C) 6= Y . If such a chain does
not exist, set ](f ) = ∞.
11.8 Theorem. Let f : X −→ Y be a given closed surjection. Assume
that the space X has the property that every open cover of X has a point
countable open refinement and that the space Y has the property that every
open subspace of Y either contains an isolated point or a sequence of open
sets whose intersection is not open. Then ](f ) does not support a nontrivial
C-sequence.
a
This shows that under some very mild assumption on a pair of spaces X
and Y , if their sizes are smaller than the first regular uncountable cardinal
θ not supporting a nontrivial C-sequence, then for every closed map f from
X onto Y , there is a closed subspace H of X on which f is irreducible, i.e.
f 00 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 only
if it is coherent, i.e. it has the property that Cα = Cβ ∩ α whenever α is a
limit point of Cβ .
Note that the nontriviality conditions appearing in Definition 11.1 and
Theorem 11.5 coincide in the realm of square-sequences:
12.2 Lemma. A square-sequence Cα (α < θ) is trivial if and only if there
is a closed and unbounded subset C of θ such that Cα = C ∩ α whenever α
is a limit point of C.
a
To a given square-sequence Cα (α < θ) one naturally associates a tree
ordering <2 on θ as follows:
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 that
the tree (θ, <2 ) has a chain of size θ. In fact, one can characterize the tree
orderings <T on θ for which there exists a square sequence Cα (α < θ) such
that 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 β
whenever cf(β) > ω and
(iii) minimal as well as successor nodes of the tree <T on θ are ordinals
of cofinality ω.
Proof. For each ordinal α < θ of countable cofinality we fix a subset Sα ⊆ α
of order-type ω cofinal with α. Given a tree ordering <T on θ with properties
(i)-(iii), for a limit ordinal β < θ let Pβ+ be the set of all successor nodes from
Pβ ∪ {β} including the minimal one. For α ∈ Pβ+ let α− be its immediate
predecessor in Pβ . Finally, set
Cβ = Pβ ∪
[
{Sα ∩ [α− , α) : α ∈ Pβ+ }.
It is easily checked that this defines a square-sequence Cβ (β < θ) with the
property 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 that
the exact analogue of this result is true for any cofinality κ < θ rather than
ω.
An important result about square sequences is the following result that
can be deduced on the basis of the well-known construction of square sequences in the constructible universe due to Jensen [35] (see 1.10 of [78]).
12.5 Theorem. If a regular uncountable cardinal θ is not weakly compact
in the constructible subuniverse then there is a nontrivial square sequence
on θ which is moreover constructible.
a
12.6 Corollary. If a regular uncountable cardinal θ is not weakly compact
in the constructible subuniverse then there is a constructible Aronszajn tree
on θ.
82
I. Coherent Sequences
Proof. Let Cα (α < θ) be a fixed nontrivial square sequence which is constructible. Changing the Cα ’s a bit, we may assume that if β is a limit
ordinal with α = min Cβ or if α ∈ Cβ but sup(Cβ ∩α) < α then α must be a
successor ordinal in θ. Consider the corresponding function ρ0 : [θ]2 −→ Qθ
ρ0 (α, β) = tp(Cβ ∩ α)a ρ0 (α, min(Cβ \ α)),
where ρ0 (γ, γ) = ∅ for all γ < θ. Consider the tree
T (ρ0 ) = {(ρ0 )β ¹ α : α ≤ β < θ}.
Clearly T (ρ0 ) is constructible. By (I.101) the αth level of T (ρ0 ) is bounded
by the size of the set {Cβ ∩ α : β ≥ α}. Since the intersection of the form
Cβ ∩ α is determined by its maximal limit point modulo a finite subset of
α, we conclude that the αth level of T (ρ0 ) has size ≤ |α| + ℵ0 . Since the
sequence Cα (α < θ) is nontrivial, the proof of Theorem 11.5 shows that
T (ρ0 ) has no cofinal branches.
a
12.7 Lemma. Suppose Cα (α < θ) is a square sequence on θ, <2 the
associated tree ordering on θ and T (ρ0 ) = {(ρ0 )β ¹ α : α ≤ β < θ} where
ρ0 : [θ]2 −→ Qθ is the associated ρ0 -function. Then α 7−→ (ρ0 )α is a strictly
increasing 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 and
β → ξ for ξ < α get the same code ρ0 (ξ, α) = ρ0 (ξ, β).
a
The purpose of this section, however, is to analyze a family of ρ-functions
associated with a square sequence Cα (α < θ) on some regular uncountable
cardinal θ, both fixed from now on. Recall that an ordinal α divides an
ordinal γ if there is β such that γ = α · β, i.e. γ can be written as the union
of an increasing β-sequence of intervals of type α. Let κ ≤ θ be a fixed
infinite regular cardinal. Let Λκ : [θ]2 −→ θ be defined by
Λκ (α, β) = max{ξ ∈ Cβ ∩ (α + 1) : κ divides tp(Cβ ∩ ξ)}.
(I.136)
Finally we are ready to define the main object of study in this section:
ρκ : [θ]2 −→ κ
(I.137)
defined recursively by
ρκ (α, β)
=
sup{tp(Cβ ∩ [Λκ (α, β), α)), ρκ (α, min(Cβ \ α)),
ρκ (ξ, α) : ξ ∈ Cβ ∩ [Λκ (α, β), α)},
where we stipulate that ρκ (γ, γ) = 0 for all γ.
The following consequence of the coherence property of Cα (α < θ) will
be quite useful.
12. Square sequences
83
12.8 Lemma. If α is a limit point of Cβ then ρκ (ξ, α) = ρκ (ξ, β) for all
ξ < α.
a
Note that ρκ is something that corresponds to the function ρ : [ω1 ]2 −→ ω
considered in Definition 4.1 (see also Section 14) and that the ρκ ’s are simply
various local versions of the key definition. We shall show that they all have
the crucial subadditive properties.
12.9 Lemma. If α < β < γ < θ then
(a) ρκ (α, γ) ≤ max{ρκ (α, β), ρκ (β, γ)},
(b) ρκ (α, β) ≤ max{ρκ (α, γ), ρκ (β, γ)}.
Proof. The proof is by induction an α, β and γ. We first prove the inductive step for (a). Let ν = max{ρκ (α, β), ρκ (β, γ)}. We need to show that
ρκ (α, γ) ≤ ν. To this end, let
γα = min(Cγ \ α) and γβ = min(Cγ \ β).
(I.138)
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 we
conclude 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 = γ̄. By the inductive hypothesis (a) for the
triple α < β ≤ γ̄ we get
ρκ (α, γ̄) ≤ max{ρκ (α, β), ρκ (β, γ̄)} ≤ ν,
(I.140)
since ρκ (β, γ̄) ≤ ρκ (β, γ). Note that [λ, α)∩Cγ is an initial part of [λ, β)∩Cγ
so its order-type is bounded by ρκ (β, γ) and therefore by ν. Consider a ξ ∈
[λ, α) ∩ Cγ . Applying the inductive hypothesis (b) for the triple ξ < α < β
we conclude that
ρκ (ξ, α) ≤ max{ρκ (ξ, β), ρκ (α, β)} ≤ ν,
(I.141)
since ρκ (ξ, β) ≤ ρκ (β, γ) by the definition of ρκ (β, γ). This shows that all
ordinals appearing in the formula for ρκ (α, γ) are ≤ ν and so ρκ (α, γ) ≤ ν.
Subcase 2a .2: γα < γβ . Then γα belongs to Cγ ∩ [λ, β) and therefore
ρκ (γα , β) ≤ ρκ (β, γ) ≤ ν. Applying the inductive hypothesis (b) for α ≤
γα < β gives us
ρκ (α, γα ) ≤ max{ρκ (α, β), ρκ (γα , β)} ≤ ν.
(I.142)
84
I. Coherent Sequences
Given ξ ∈ Cγ ∩ [λ, α) using the inductive hypothesis (b) for ξ < α < β we
get that
ρκ (ξ, α) ≤ max{ρκ (ξ, β), ρκ (α, β)} ≤ ν,
(I.143)
since ρκ (ξ, β) ≤ ρκ (β, γ). Since tp(Cγ ∩[λ, α)) ≤ tp(Cγ ∩[λ, β)) ≤ ρκ (β, γ) ≤
ν we see again that all the ordinals appearing in the formula for ρκ (α, γ)
are ≤ ν.
Let us now concentrate on the inductive step for (b). This time let
ν = max{ρκ (α, γ), ρκ (β, γ)} and let γα and γβ be as in (I.138). We need to
show that ρκ (α, β) ≤ ν.
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 we
get
ρκ (α, β) ≤ max{ρκ (α, Λκ (β, γ)), ρκ (Λκ (β, γ), β)} ≤ ν.
(I.145)
Case 2b : α ≥ Λκ (β, γ).
Subcase 2b .1: γα = γβ = γ̄. Then ρκ (α, γ̄) ≤ ρκ (α, γ) ≤ ν and
ρκ (β, γ̄) ≤ ρκ (β, γ) ≤ ν. Applying the inductive hypothesis (b) for α <
β < γ̄ we get
ρκ (α, β) ≤ max{ρκ (α, γ̄), ρκ (β, γ̄)} ≤ ν.
(I.146)
Subcase 2b .2: γα < γβ . Then γα ∈ Cγ ∩ [λ, β) and so ρκ (γα , β) appears
in the formula for ρκ (β, γ) and so we conclude that ρκ (γα , β) ≤ ρκ (β, γ) ≤ ν.
Similarly, ρκ (α, γα ) ≤ ρκ (α, γ) ≤ ν. Applying the inductive hypothesis (a)
for the triple α ≤ γα < β we get
ρκ (α, β) ≤ max{ρκ (α, γα ), ρκ (γα , β)} ≤ ν.
This finishes the proof.
(I.147)
a
The following is an immediate consequence of the fact that the definition
of ρκ is closely tied to the notion of a walk along the fixed square sequence.
12.10 Lemma. ρκ (α, γ) ≥ ρκ (α, β) whenever α ≤ β ≤ γ and β belongs to
the trace of the walk from γ to α.
a
12.11 Lemma. Suppose β ≤ γ < θ and that β is a limit ordinal > 0. Then
ρκ (α, γ) ≥ ρκ (α, β) for coboundedly many α < β.
Proof. Let γ = γ0 > γ1 > . . . > γn−1 > γn = β be the trace of the walk
from γ to β. Let γ̄ = γn−1 if β is a limit point of Cγn−1 , otherwise let γ̄ = β.
Note that by Lemma 12.8, in any case we have that
ρκ (α, β) = ρκ (α, γ̄) for all α < β.
(I.148)
12. Square sequences
85
Let β̄ < β be an upper bound of all Cγi ∩ β (i < n) which are bounded in
β. Then γ̄ is a member of the trace of any walk from γ to some ordinal α
in the interval [β̄, β). Applying Lemma 12.10 to this fact gives us
ρκ (α, γ) ≥ ρκ (α, γ̄) for all α ∈ [β̄, β).
(I.149)
Since ρκ (α, γ̄) = ρκ (α, β) for all α < β (see (I.148)), this gives us the
a
conclusion of the Lemma.
12.12 Lemma. The set Pνκ (β) = {ξ < β : ρκ (ξ, β) ≤ ν} is a closed subset
of β for every β < θ and ν < κ.
Proof. The proof is by induction on β. So let α < β be a limit point
of Pνκ (β). Let β1 = min(Cβ \ α) and let λ = Λκ (α, β). If λ = α then
ρκ (α, β) = 0 and therefore α ∈ Pνκ (β). So we may assume that λ < α.
Then
Λκ (ξ, β) = λ for every ξ ∈ [λ, α).
(I.150)
Case 1: ᾱ = sup(Cβ ∩ α) < α. Then β1 = min(Cβ \ ξ) for every
ξ ∈ Pνκ (β) ∩ (ᾱ, α). It follows that
Pνκ (β) ∩ (ᾱ, α) ⊆ Pνκ (β1 ),
(I.151)
so α is also a limit point of Pνκ (β1 ). By the inductive hypothesis α ∈ Pνκ (β1 ),
i.e. ρκ (α, β1 ) ≤ ν. Since Λκ (ξ, β) = λ for all ξ ∈ [λ, α), we get
tp(Cβ ∩ [λ, α)) = sup{tp(Cβ ∩ [λ, ξ)) : ξ ∈ Pνκ (β) ∩ [λ, α)} ≤ ν.
(I.152)
Consider η ∈ Cβ ∩ [λ, α). Then η ∈ Cβ ∩ [λ, ξ) for some ξ ∈ Pνκ (β) ∩ [ᾱ, α)
so by the definition of ρκ (ξ, β) we conclude that ρκ (η, ξ) ≤ ρκ (ξ, β) ≤ ν. By
(I.151) we also have ρκ (ξ, β1 ) ≤ ν so by 12.9(a), ρκ (η, β1 ) ≤ ν. Applying
12.9(b) to η < α < β1 , we conclude that
ρκ (η, α) ≤ max{ρκ (η, β1 ), ρκ (α, β1 )} ≤ ν.
(I.153)
This shows that all the ordinals appearing in the formula for ρκ (α, β) are
≤ ν and therefore that ρκ (α, β) ≤ ν.
Case 2: α is a limit point of Cβ . As in the previous case, to show that
ρκ (α, β) ≤ ν we need to have tp(Cβ ∩ [λ, α)), ρκ (α, β1 ) and ρκ (η, α) with
η ∈ Cβ ∩ [λ, α) all ≤ ν. Note that β1 = α so ρκ (α, β1 ) = 0. Similarly as in
the first case we conclude tp(Cβ ∩ [λ, α)) ≤ ν. By Lemma 12.8,
ρκ (ξ, α) = ρκ (ξ, β) ≤ ν for all ξ ∈ Pνκ (β) ∩ α.
(I.154)
Pνκ (β)
Consider an η ∈ Cβ ∩ [λ, α). Since α is a limit point of
there is ξ
from this set above η. By (I.150) we have that η ∈ Cβ ∩ [Λκ (ξ, β), ξ) so
ρκ (η, ξ) ≤ ρκ (ξ, β) ≤ ν. Applying 12.9(a) to η < ξ < α we get
ρκ (η, α) ≤ max{ρκ (η, ξ), ρκ (ξ, α)} ≤ ν.
This finishes the proof.
(I.155)
a
86
I. Coherent Sequences
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 ⊆<κµ whenever ν < µ < κ,
S
(3) ∈¹ θ = ν<κ <κν .
Proof. This follows immediately from Lemma 12.9.
a
Recall the notion of a special tree of height θ from Section 9, a tree T for
which one can find a T -regressive map f : T −→ T with the property that
the preimage of any point is the union of < θ antichains. By a tree on θ
we mean a tree of the form (θ, <T ) with the property that α <T β implies
α < β.
12.14 Lemma. If a tree T on θ is special, then there is an ordinal-regressive
map f : θ −→ θ and a closed and unbounded set C ⊆ θ such that f is oneto-one on all chains separated by C.
Proof. Let g : θ −→ θ be a T -regressive map such that for each ξ < θ the
preimage 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 < θ
with the property that λξ < α for all ξ < α. Let p., .q denote the Gödel
pairing function which in particular has the property that pα, βq < δ for
δ ∈ C and α, β, δ. Define ordinal-regressive map f : θ −→ θ by letting
f (α) = max(C ∪ α) if α ∈
/ C and for α ∈ C, let f (α) = pg(α), γq where
γ < λg(α) (< α) is minimal ordinal with property that α ∈ Aγ (g(α)). Then
f is as required.
a
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 then
there must be ν < κ such that (θ, <κν ) is a nonspecial tree on θ.
Proof. Suppose to the contrary that all trees are special. By Lemma 12.14
we may choose ordinal-regressive maps fν : θ −→ θ for all ν < κ and a
single closed and unbounded set C ⊆ θ such that each of the maps fν is
one-to-one on <κν -chains separated by C. Using the Pressing Down Lemma
we find a stationary set Γ of cofinality κ+ ordinals < θ and λ < θ such that
fν (γ) < λ for all γ ∈ Γ and ν < κ. If |λ|+ < θ, let ∆ = λ, Γ = Γ0 and if
|λ|+ = θ, represent λ as the increasing union of a sequence ∆ξ (ξ < cf(|λ|))
12. Square sequences
87
of sets of size < |λ|. Since κ 6= cf(|λ|) there is a ξ¯ < cf(|λ|) and a stationary
Γ0 ⊆ Γ such that for all γ ∈ Γ0 , fν (γ) ∈ ∆ξ̄ for κ many ν < κ. Let ∆ = ∆ξ̄ .
This gives us subsets ∆ and Γ0 of θ such that
|∆|+ < θ and Γ0 is stationary in θ,
Σγ = {ν < κ : fν (γ) ∈ ∆} is unbounded in κ for all γ ∈ Γ0 .
(I.157)
(I.158)
Let θ̄ = κ+ ·|∆|+ . Then θ̄ < θ and so we can find β ∈ Γ0 such that Γ0 ∩C ∩β
has size θ̄. Then there will be ν0 < κ and Γ1 ⊆ Γ0 ∩ C ∩ β of size θ̄ such
that ρκ (α, β) ≤ ν0 for all α ∈ Γ1 . By (I.158) we can find Γ2 ⊆ Γ1 of size
θ̄ and ν1 ≥ ν0 such that fν1 (α) ∈ ∆ for all α ∈ Γ2 . Note that Γ2 is a
<κν1 -chain separated by C, so fν1 is one-to-one on Γ2 . However, this gives
us the desired contradiction since the set ∆, in which fν1 embeds Γ2 has
size smaller than the size of Γ2 . This finishes the proof.
a
It is now natural to ask the following question: under which assumption
on the square sequence Cα (α < θ) can we conclude that neither of the trees
(θ, <κν ) 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, <κν ) 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 belongs
to Γκ . Pick α ∈ B ∩ β such that tp(Cβ ∩ α) > ν. By definition of ρκ (α, β)
we have that ρκ (α, β) ≥ tp(Cβ ∩ α) > ν since clearly Λκ (α, β) = 0. This
contradicts the fact that α <κν β and finishes the proof.
a
12.17 Definition. A square sequence on θ is special if the corresponding
tree (θ, <2 ) is special, i.e. there is a <2 -regressive map f : θ −→ θ with the
property that the f -preimage of every ξ < θ is the union of < θ antichains
of (θ, <2 ).
12.18 Theorem. Suppose κ < θ are regular cardinals such that θ is not
a successor of a cardinal of cofinality κ. Then to every square sequence
Cα (α < θ) for which there exist stationarily many α such that tp Cα = κ,
one can associate a sequence Cαν (α < θ, ν < κ) such that:
(i) Cαν ⊆ Cαµ for all α and ν ≤ µ,
(ii) α =
S
ν<κ
Cαν for all limit α,
(iii) Cαν (α < θ) is a nonspecial (and nontrivial) square sequence on θ for
all ν < κ.
88
I. Coherent Sequences
Proof. Fix ν < κ and define Cαν by induction on α < θ. So suppose β is a
limit ordinal < θ and that Cαν is defined for all α < β. If Pνκ (β) is bounded
in β, let β̄ be the maximal limit point of Pνκ (β) (β̄ = 0 if the set has no
limit points) and let
Cβν = Cβ̄ν ∪ Pνκ (β) ∪ (Cβ ∩ [max(Pνκ (β)), β)).
If Pνκ (β) is unbounded in β, let
[
Cβν = Pνκ (β) ∪ {Cαν : α ∈ Pνκ (β) & α = sup(Pνκ (β) ∩ α)}.
(I.159)
(I.160)
By Lemmas 12.9 and 12.12, Cβν (β < θ) is well defined and it forms a
square sequence on θ. The properties (i) and (ii) are also immediate. To see
that for each ν < κ the sequence Cβν (β < θ) is nontrivial, one uses Lemma
12.16 and the fact that if α is a limit point of Cβν occupying a place in Cβν
that is divisible by κ, then α <κν β. By Lemma 12.15, or rather its proof,
we conclude that there is ν̄ < κ such that Cβν (β < θ) is nonspecial for all
ν ≥ ν̄. This finishes the proof.
a
12.19 Lemma. For every pair of regular cardinals κ < θ, every special
square sequence Cα (α < θ) can be refined to a special square sequence
C̄α (α < θ) with the property that tp C̄α = κ for stationarily many α < θ.
Proof. Let <2 be the tree ordering associated with Cα (α < θ) and let
f : θ −→ θ be a <2 -regressive map such that f −1 (ξ) is the union of < θ
antichains of (θ, <2 ) for all ξ < θ.
Case 1: κ = ω. By the Pressing Down Lemma there is a stationary set
Γ of cofinality ω ordinals < θ on which f takes some constant value. Thus
Γ can be decomposed into < θ antichains of (θ, <2 ) and so Γ contains a
stationary subset Γ0 of pairwise <2 -incomparable ordinals. For α ∈ Γ0 let
C̄α be any ω-sequence of successor ordinals converging to α. If α is a limit
ordinal not in Γ0 but there is (a unique!) ᾱ <2 α in Γ0 , let C̄α = Cα \ ᾱ. In
all other cases, let C̄α = Cα . It is easily seen that C̄α (α < θ) is as required.
Case 2: κ > ω. By Lemma 12.14 we may find a closed and unbounded
subset C of θ and an ordinal-regressive map g : C −→ θ which is one-to-one
on <2 -chains included in C (and which has the property that for α ∈ C
g(α) is the Gödel pairing of f (α) and the index of the antichain which contains α in the fixed antichain-decomposition of the preimage f −1 (f (α)).By
the argument from Case 1 there is a stationary set Γ of limit points of C
consisting of cofinality κ ordinals < θ such that Γ is an antichain of (θ, <2 ).
Given α ∈ Γ, let h(α) be the minimal ordinal λ < θ such that g(ξ) < λ
for unboundedly many limit points ξ of Cα that also belong to C. Since g
is regressive and since κ, the cofinality of α, is uncountable, the Pressing
Down Lemma gives us that h(α) < α. Choose a stationary set Γ0 ⊆ Γ such
that h takes some constant value λ on Γ0 . Note that since g is one-to one
12. Square sequences
89
on <2 -chains included in C the constant value λ cannot be a successor ordinal. It follows that λ must be a limit ordinal of cofinality equal to κ. Let
λν (ν < κ) be a strictly increasing sequence which converges to λ.
We first define C̄α for α ∈ Γ0 as a strictly increasing continuous sequence
cν (α) (ν < κ) of limit points of Cα that also belong to C and satisfy the
following requirements at each ν < κ:
cµ (α) = supν<µ cν (α) for µ limit
(I.161)
cν+1 (α) > cν (α) is minimal subject to being a limit point
of Cα , belonging to C, and being > γ for every limit point
γ of Cα such that γ ∈ C and g(γ) < λν+1 .
(I.162)
If α is a limit point of C̄β for some β ∈ Γ0 , set C̄α = C̄β ∩ α. Note that by
the canonicity of the definition of cν (β) (ν < κ) for β ∈ Γ0 , we would get
the same result, as C̄β ∩ α = C̄β̄ ∩ α for any other β̄ ∈ Γ0 for which α is a
limit point of C̄β̄ .
If α is a limit ordinal which is not a limit point of any C̄β for β ∈ Γ0 and
there is (a unique!) ᾱ <2 α in Γ0 , let C̄α = Cα \ ᾱ.
If α is a limit ordinal such that α <2 ᾱ for some ᾱ ∈ Γ0 but α is not a
limit point on any C̄β for β ∈ Γ0 , let C̄α = Cα \ max(C̄ᾱ ∩ α). Note again
that there is no ambiguity in this definition, as
C̄β ∩ α = C̄γ ∩ α
holds for every β and γ in Γ0 such that α <2 β and α <2 γ. In all other
cases set C̄α = Cα . It should be clear that C̄α (α < θ) is a nontrivial square
sequence satisfying the conclusion of Lemma 12.19.
a
Finally we can state the main result of this Section which follows from
Theorem 12.18 and Lemma 12.19.
12.20 Theorem. If regular uncountable cardinal θ 6= ω1 carries a nontrivial
square sequence then it also carries a nontrivial square-sequence which is
moreover nonspecial.
a
12.21 Corollary. If a regular uncountable cardinal θ 6= ω1 is not weakly
compact in the constructible subuniverse then there is a nonspecial Aronszajn tree of height θ.
Proof. By Theorem 12.5, θ carries a nontrivial square sequence Cα (α < θ).
By Theorem 12.20 we may assume that the sequence is moreover nonspecial.
Let ρ0 be the associated ρ0 -function and consider the tree T (ρ0 ). As in
Corollary 12.6 we conclude that T (ρ0 ) is an Aronszajn tree of height θ. By
Lemma 12.7 there is a strictly increasing map from (θ, <2 ) into T (ρ0 ), so
T (ρ0 ) must be nonspecial.
a
90
I. Coherent Sequences
12.22 Remark. The assumption θ 6= ω1 in Theorem 12.20 is essential as
there is always a nontrivial square sequence on ω1 but it is possible to have
a situation where all Aronszajn trees on ω1 are special. For example MAω1
implies this. In [48], Laver and Shelah have shown that any model with a
weakly compact cardinal admits a forcing extension satisfying CH and the
statement that all Aronszajn trees on ω2 are special. A well-known open
problem in this area asks whether one can have GCH rather than CH in a
model where all Aronszajn trees on ω2 are special.
13. The full lower trace of a square sequence
In this section θ is a regular uncountable cardinal and Cα (α < θ) is a
nontrivial 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 formula, describing a mapping F : [θ]2 −→ [θ]<ω :
[
F (α, β) = F (α, min(Cβ \ α)) ∪ {F (ξ, α) : ξ ∈ Cβ ∩ [Λ(α, β), α)},
where F (γ, γ) = {γ} for all γ.
13.1 Lemma. For all α ≤ β ≤ γ,
(a) F (α, γ) ⊆ F (α, β) ∪ F (β, γ),
(b) F (α, β) ⊆ F (α, γ) ∪ F (β, γ).
Proof. The proof of the Lemma is by simultaneous induction. We first check
(a). Let γα = min(Cγ \ α), γβ = min(Cγ \ β) and λ = Λ(β, γ). If α < λ
then Cλ = Cγ ∩ λ and therefore F (α, λ) = F (α, γ). Applying the inductive
hypothesis (b) for α < λ ≤ β we get the required conclusion
F (α, γ) = F (α, λ) ⊆ F (α, β) ∪ F (λ, β) ⊆ F (α, β) ∪ F (β, γ).
(I.163)
So we may assume that λ ≤ α. Then Λ(α, γ) = λ. If γα < β, applying the
inductive hypothesis (b) for α ≤ γα < β, we have
F (α, γα ) ⊆ F (α, β) ∪ F (γα , β) ⊆ F (α, β) ∪ F (β, γ),
(I.164)
since γα ∈ Cγ ∩ [Λ(β, γ), β). If γα ≥ β then γα = γβ , and applying the
inductive hypothesis (a) for α < β ≤ γα = γβ , we have
F (α, γα ) ⊆ F (α, β) ∪ F (β, γβ ) ⊆ F (α, β) ∪ F (β, γ).
(I.165)
13. The full lower trace of a square sequence
91
Consider now ξ ∈ Cγ ∩[λ, α). By the inductive hypothesis (b) for ξ < α ≤ β
we have
F (ξ, α) ⊆ F (α, β) ∪ F (ξ, β) ⊆ F (α, β) ∪ F (β, γ),
(I.166)
since ξ ∈ Cγ ∩ [Λ(β, γ), β). It follows that all the factors of F (α, γ) are
included in the union F (α, β) ∪ F (β, γ), so this completes the checking of
(a).
To prove (b) define γα , γβ and λ as above and consider first the case
λ ≥ α. Then Cλ = Cγ ∩ λ and therefore F (α, λ) = F (α, γ). Applying the
inductive hypothesis (a) for α < λ ≤ β we get
F (α, β) ⊆ F (α, λ) ∪ F (λ, β) ⊆ F (α, γ) ∪ F (β, γ).
(I.167)
So we may assume that λ < α in which case we also know that Λ(α, γ) = λ.
Consider first the case γα < β. Applying the inductive hypothesis (a) for
α ≤ γα < β we have
F (α, β) ⊆ F (α, γα ) ∪ F (γα , β) ⊆ F (α, γ) ∪ F (β, γ),
(I.168)
since γα ∈ Cγ ∩ [Λ(β, γ), β). Suppose now that γα ≥ β. Then γα = γβ and
applying the inductive hypothesis (b) for the triple α ≤ β ≤ γα = γβ we
have
F (α, β) ⊆ F (α, γα ) ∪ F (β, γβ ) ⊆ F (α, γ) ∪ F (β, γ).
This completes the proof.
(I.169)
a
13.2 Lemma. For all α ≤ β ≤ γ,
(a) ρ0 (α, β) = ρ0 (min(F (β, γ) \ α), β)a ρ0 (α, min(F (β, γ) \ α)),
(b) ρ0 (α, γ) = ρ0 (min(F (β, γ) \ α), γ)a ρ0 (α, min(F (β, γ) \ α)).
Proof. The proof is by induction on α, β and γ. Let
λ = Λ(β, γ), γ1 = min(Cγ \ α) and α1 = min(F (β, γ) \ α).
Pick ξ ∈ {min(Cγ \ β)} ∪ (Cγ ∩ [λ, β)) such that α1 ∈ F (ξ, β). Then α1 =
min(F (ξ, β) \ α), so applying the inductive hypothesis (b) for α ≤ ξ < β we
get
ρ0 (α, β) = ρ0 (α1 , β)a ρ0 (α, α1 )
(I.170)
which checks (a) for α ≤ β ≤ γ. Suppose first that λ > α. Note that
F (λ, β) is a subset of F (β, γ) so α2 = min(F (λ, β) \ α) ≥ α1 . Applying the
inductive hypothesis for α ≤ λ ≤ β we get
ρ0 (α, β) = ρ0 (α2 , β)a ρ0 (α, α2 ),
(I.171)
92
I. Coherent Sequences
ρ0 (α, λ) = ρ0 (α2 , λ)a ρ0 (α, α2 ).
(I.172)
Combining (I.170) and (I.171) we infer that ρ0 (α, α1 ) is a tail of ρ0 (α, α2 ),
so (I.172) can be written as
ρ0 (α, λ) = ρ0 (α2 , λ)a ρ0 (α1 , α2 )a ρ0 (α, α1 ) = ρ0 (α1 , λ)a ρ0 (α, α1 ). (I.173)
Since λ is a limit point of Cγ , we have ρ0 (α, γ) = ρ0 (α, λ), so (I.173) gives
us the conclusion (b) of Lemma 13.2.
Consider now the case λ ≤ α. Then γ1 is either equal to min(Cγ \ β) or
it belongs to [λ, β) ∩ Cγ . In any case we have that F {γ1 , β} ⊆ F (β, γ), so
γ2 = min(F {γ1 , β} \ α) ≥ α1 .
(I.174)
Applying the inductive hypothesis for α, γ1 and β we get
ρ0 (α, β) = ρ0 (γ2 , β)a ρ0 (α, γ2 ),
(I.175)
ρ0 (α, γ1 ) = ρ0 (γ2 , γ1 )a ρ0 (α, γ2 ).
(I.176)
Combining (I.170) and (I.175) we see that ρ0 (α, α1 ) is a tail of ρ0 (α, γ2 ), so
(I.176) can be rewritten as
ρ0 (α, γ1 ) = ρ0 (γ2 , γ1 )a ρ0 (α1 , γ2 )a ρ0 (α, α1 ) = ρ0 (α1 , γ1 )a ρ0 (α, α1 ).
(I.177)
Since ρ0 (α, γ) = tp(Cγ ∩ α)a ρ0 (α, γ1 ), (I.177) gives us the conclusion (b) of
Lemma 13.2. This finishes the proof.
a
Recall the function ρ2 : [θ]2 −→ ω which counts the number of steps in
the walk along the fixed C-sequence Cα (α < θ) which in this Section is
assumed to be moreover a square sequence:
ρ2 (α, β) = ρ2 (α, min(Cβ \ α)) + 1,
where we let ρ2 (γ, γ) = 0 for all γ. Thus ρ2 (α, β) + 1 is simply equal to the
cardinality of the trace Tr(α, β) of the minimal walk from β to α.
13.3 Lemma. supξ<α |ρ2 (ξ, α) − ρ2 (ξ, β)| < ∞ for all α < β < θ.
Proof. By Lemma 13.2,
supξ<α |ρ2 (ξ, α) − ρ2 (ξ, β)| ≤ supξ∈F (α,β) |ρ2 (ξ, α) − ρ2 (ξ, β)|.
a
13.4 Definition. Set I to be the set of all countable Γ ⊆ θ such that
supξ∈∆ ρ2 (ξ, α) = ∞ for all α < θ and infinite ∆ ⊆ Γ ∩ α.
13.5 Lemma. I is a P-ideal of countable subsets of θ.
13. The full lower trace of a square sequence
93
Proof. Let Γn (n < ω) be a given sequence of members of I and fix β < θ
such that Γn ⊆ β for all n. For n < ω set Γ∗n = {ξ ∈ Γn : ρ2 (ξ,S
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.
a
13.6 Theorem. The P-ideal dichotomy implies that a nontrivial square
sequence can exist only on θ = ω1 .
Proof. Applying the P-ideal dichotomy on I from 13.4 we get the two alternatives (see 4.45):
there is an uncountable ∆ ⊆ θ such that [∆]ω ⊆ I, or
(I.178)
S
there is a decomposition θ = n<ω Σn such that Σn ⊥I for
(I.179)
all n.
By Lemma 13.3, if (I.178) holds, then ∆ ∩ α must be countable for all α < θ
and so the cofinality of θ must be equal to ω1 . Since we are working only
with regular uncountable cardinals, we see that (I.178) gives us that θ = ω1
must hold. Suppose now (I.179) holds and pick k < ω such that Σk is
unbounded in θ. Since Σk ⊥I we have that (ρ2 )α is bounded on Σk ∩ α
for all α < θ. So there is an unbounded set Γ ⊆ θ and an integer n such
that for each α ∈ Γ the restriction of (ρ2 )α on Σk ∩ α is bounded by n. By
Theorem 11.2 we conclude that the square sequence Cα (α < θ) we started
with must be trivial.
a
13.7 Definition. By Sθ we denote the sequential fan with θ edges, i.e. the
space on (θ × ω) ∪ {∗} with ∗ as the only nonisolated point, while a typical
neighborhood of ∗ has the form Uf = {(α, n) : n ≥ f (α)} ∪ {∗} where
f : θ −→ ω.
13.8 Theorem. If there is a nontrivial square sequence on θ then the square
of 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 of
steps function ρ2 : [θ]2 −→ ω we define d : [θ]2 −→ ω by letting
d(α, β) = sup |ρ2 (ξ, α) − ρ2 (ξ, β)|.
ξ≤α
13.10 Lemma. For all α ≤ β ≤ γ,
(a) d(α, β) ≥ ρ2 (α, β),
32 The tightness of a point x in a space X is equal to θ if θ is the minimal cardinal such
that, if a set W ⊆ X \ {x} accumulates to x, then there is a subset of W of size ≤ θ that
accumulates to x.
94
I. Coherent Sequences
(b) d(α, γ) ≤ d(α, β) + d(β, γ),
(c) d(α, β) ≤ d(α, γ) + d(β, γ).
Proof. The conclusion (a) follows from the fact that we allow ξ = α in the
definition of d(α, β). The conclusions (b) and (c) are consequences of the
triangle inequalities of the `∞ -norm and the fact that in both inequalities
we have that the domain of functions on the left-hand side is included in
the domain of functions on the right-hand side.
a
13.11 Definition. For γ ≤ θ, let
Wγ = {((α, d(α, β)), (β, d(α, β))) : α < β < γ}.
The following Lemma establishes that the tightness of the point (∗, ∗) of
Sθ2 is equal to θ, giving us the proof of Theorem 13.8.
13.12 Lemma. (∗, ∗) ∈ W̄θ but (∗, ∗) ∈
/ W̄γ for all γ < θ.
Proof. To see that Wθ accumulates to (∗, ∗), let Uf2 be a given neighborhood
of (∗, ∗). Fix an unbounded set Γ ⊆ θ on which f is constant. By Theorem
11.2 and Lemma 13.10(a) there exist α < β in Γ such that d(α, β) ≥ f (α) =
f (β). Then ((α, d(α, β)), (β, d(α, β))) belongs to the intersection Wθ ∩ Uf2 .
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 ∩ Ug2 is nonempty and choose ((α, d(α, β)), (β, d(α, β))) from
this set. Then
d(α, β) ≥ 2d(α, γ) + 1 and d(α, β) ≥ 2d(β, γ) + 1.
(I.180)
It follows that d(α, β) ≥ d(α, γ) + d(β, γ) + 1 contradicting Lemma 13.10(c).
a
Since θ = ω1 admits a nontrivial square sequence, Theorem 13.8 leads to
the following result of Gruenhage and Tanaka [29].
13.13 Corollary. The square of the sequential fan with ω1 edges is not
countably tight.
a
13.14 Question. What is the tightness of the square of the sequential fan
with ω2 edges?
13.15 Corollary. If a regular uncountable cardinal θ is not weakly compact
in the constructible subuniverse then the square of the sequential fan with θ
edges has tightness equal to θ.
a
13. The full lower trace of a square sequence
95
The way we have proved that Sθ2 has tightness equal to θ is of independent interest. We have found a point (∗, ∗) of Sθ2 and a set Wθ of isolated
points which accumulates to (∗, ∗) but has no other accumulation points
and moreover, no subset of Wθ of size < θ has accumulation points at all.
This is interesting in view of the following result of Dow and Watson [17],
where C denotes the category of spaces that contain the converging sequence
ω +1 and is closed under finite products, quotients and arbitrary topological
sums.
13.16 Theorem. If for every regular cardinal θ there is a space X in C
which has a set of isolated points of size θ witnessing tightness θ of its
unique accumulation point in X, then every space belongs to C.
a
13.17 Corollary. If every regular uncountable cardinal supports a nontrivial square sequence then every topological space X can be obtained from
the converging sequence after finitely many steps of taking (finite) products,
quotients, and arbitrary topological sums.
Proof. This follows from combining (the proof of) Theorem 13.8 and Theorem 13.16.
a
If κ is a strongly compact cardinal then the tightness of the product of
every pair of spaces of tightness < κ is < κ and so C contains no space
of tightness κ or larger. It has been shown in [17] that if the Lebesque
measure extends to a countably additive measure on all sets of reals then
there even is a countable space which does not belong to C. On the other
hand, by Theorem 12.5 and Corollary 13.17, if no regular cardinal above ω1
is weakly compact in the constructible subuniverse, then every space can be
generated by the converging sequence by taking finite products, quotients
and arbitrary topological sums.
13.18 Remark. We have seen that the case θ = ω1 is quite special when one
considers the problem of existence of various nontrivial square sequences on
θ. It should be noted that a similar result about the problem of the tightness
of Sθ2 is not available. In particular, it is not known whether the P-ideal
dichotomy or a similar consistent hypothesis of set theory implies that the
tightness of the square of, say, Sω2 is smaller than ω2 . It is interesting
that considerably more is known about the dual question, the question of
initial compactness of the Tychonoff cube Nθ . For example, if one defines
Bαβ = {f ∈ Nθ : f (α), f (β) ≤ d(α, β)} (α < β < θ) one gets an open
cover of Nθ without a subcover of size < θ. However, for small θ such
as θ = ω2 one is able to find such a cover of Nθ without any additional
set-theoretic assumption and in particular without the assumption that θ
carries a nontrivial square sequence.
96
I. Coherent Sequences
14. Special square sequences
The following well-known result of Jensen [35] supplements the corresponding result for weakly compact cardinals listed above as Theorem 12.5.
14.1 Theorem. If a regular uncountable cardinal θ is not Mahlo in the
constructible subuniverse then there is a special square sequence on θ which
is moreover constructible.
a
Today we know many more inner models with sufficient amount of fine
structure necessary for building special square sequences. So the existence
of special square sequences, especially at successors of strong-limit singular
cardinals, is tied to the existence of some other large cardinal axioms. The
reader is referred to the relevant chapters of this Handbook for the specific
information. In this section we give the combinatorial analysis of walks
along special square sequences and the corresponding distance functions.
Let us start by restating some results of Section 12.
14.2 Theorem. Suppose κ < θ are regular cardinals and that θ carries a
special square sequence. Then there exist Cαν (α < θ, ν < κ) such that:
(1) Cαν ⊆ Cαµ for all α and ν < µ,
S
(2) α = ν<κ Cα for all limit α,
(3) Cαν (α < θ) is a nontrivial square sequence on θ for all ν < κ.
Moreover, if θ is not a successor of a cardinal of cofinality κ then each of
the square sequences can be chosen to be nonspecial.
a
14.3 Theorem. Suppose κ < θ are regular cardinals and that θ carries a
special square sequence. Then there exist <ν (ν < κ) such that:
(i) <ν is a closed tree ordering of θ for each ν < κ,
S
(ii) ∈¹ θ = ν<κ <ν ,
(iii) no tree (θ, <ν ) has a chain of size θ.
a
14.4 Lemma. The following are equivalent when θ is a successor of some
cardinal κ:
(1) there is a special square sequence on θ,
(2) there is a square sequence Cα (α < θ) such that tp(Cα ) ≤ κ for all
α < θ.
14. Special square sequences
97
Proof. Let Dα (α < κ+ ) be a given special square sequence. By Lemma
9.2 the corresponding tree (κ+ , <2 ) can be decomposed into κ antichains so
let f : κ+ −→ κ be a fixed map such that f −1 (ξ) is a <2 -antichain for all
ξ < κ. Let α < κ+ be a given limit ordinal. If Dα has a maximal limit point
ᾱ < α, let Cα = Dα \ ᾱ. Suppose now that {ξ : ξ <2 α} is unbounded in
α and define a strictly increasing continuous sequence cα (ξ) (ξ < ν(α)) of
its elements as follows. Let cα (0) = min{ξ : ξ <2 α}, cα (η) = supξ<η cα (ξ)
for η limit, and cα (ξ + 1) is the minimal <2 -predecessor γ of α such that
γ > 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 easily
checked that this gives a square sequence Cα (α < κ+ ) with the property
that tp(Cα ) ≤ κ for all α < κ+ .
a
Square sequences Cα (α < κ+ ) that have the property tp(Cα ) ≤ κ for
all α < κ+ are usually called ¤κ -sequences. So let Cα (α < κ+ ) be a
¤κ -sequence fixed from now on. Let
Λ(α, β) = maximal limit point of Cβ ∩ (α + 1)
(I.181)
when such a limit point exists; otherwise Λ(α, β) = 0. The purpose of this
section is to analyze the following distance function:
ρ : [κ+ ]2 −→ κ defined recursively by
ρ(α, β) =
(I.182)
max{tp(Cβ ∩ α), ρ(α, min(Cβ \ α)),
ρ(ξ, α) : ξ ∈ Cβ ∩ [Λ(α, β), α)},
where we stipulate that ρ(γ, γ) = 0 for all γ. Clearly ρ(α, β) ≥ ρ1 (α, β), so
by Lemma 10.1 we have
14.5 Lemma. |{ξ ≤ α : ρ(ξ, α) ≤ ν}| ≤ |ν| + ℵ0 for all α < κ+ and ν < κ.
a
The following two crucial subadditive properties of ρ have proofs that
are almost identical to the proof of the corresponding properties of, say, the
function ρω discussed above in Section 12.
14.6 Lemma. For all α ≤ β ≤ γ,
(a) ρ(α, γ) ≤ max{ρ(α, β), ρ(β, γ)},
(b) ρ(α, β) ≤ max{ρ(α, γ), ρ(β, γ)}.
a
The following immediate fact will also be quite useful.
98
I. Coherent Sequences
14.7 Lemma. If α is a limit point of Cβ , then ρ(ξ, α) = ρ(ξ, β) for all
ξ < α.
a
The following as well is an immediate consequence of the fact that the
definition of ρ is closely tied to the notion of a minimal walk along the
square sequence.
14.8 Lemma. ρ(α, γ) ≥ max{ρ(α, β), ρ(β, γ)} whenever α ≤ β ≤ γ and β
belongs to the trace of the walk from γ to α.
a
Using Lemmas 14.7 and 14.8 one proves the following fact exactly as in
the case of ρκ of Section 12 (the proof of Lemma 12.11).
14.9 Lemma. If 0 < β ≤ γ and if β is a limit ordinal, then there is β̄ < β
such that ρ(α, γ) ≥ ρ(α, β) for all α in the interval [β̄, β).
a
The proof of the following fact is also completely analogous to the proof
of the corresponding fact for the local version ρκ considered above in Section
12 (the proof of Lemma 12.12).
14.10 Lemma. Pν (γ) = {β < γ : ρ(β, γ) ≤ ν} is a closed subset of γ for
all γ < κ+ and ν < κ.
a
The discussion of ρ : [κ+ ]2 −→ κ now splits naturally into two cases
depending on whether κ is a regular or a singular cardinal (with the case
cf(κ) = ω of special importance). This is done in the following two sections.
15. Successors of regular cardinals
In this section κ is a fixed regular cardinal, Cα (α < κ+ ) a fixed ¤κ -sequence
and ρ : [κ+ ]2 −→ κ the corresponding ρ-function.
For ν < κ and α < β < κ+ set α <ν β if and only if
ρ(α, β) ≤ ν.
(I.183)
The following is an immediate consequence of the analysis of ρ given in
the previous section.
15.1 Lemma.
(a) <ν is a closed tree ordering of κ+ of height ≤ κ for all ν < κ,
(b) <ν ⊆<µ whenever ν ≤ µ < κ,
S
(c) ∈¹ κ+ = ν<κ <ν .
a
15. Successors of regular cardinals
99
15.2 Remark. Recall that by Lemma 14.3 for every regular cardinal λ ≤ κ
the ∈-relation of κ+ can also be written as an increasing union of a λsequence <ν (ν < λ) of closed tree orderings. When λ < κ, however, we
can no longer insist that the trees (κ+ , <ν ) have heights ≤ κ. Using an
additional set-theoretic assumption (compatible with the existence of a ¤κ sequence) one can strengthen (a) of Lemma 15.1 and have that the height of
(κ+ , <ν ) is < κ for all ν < κ. Without additional set-theoretic assumptions,
these trees, however, do have some properties of smallness not covered by
Lemma 15.1.
15.3 Lemma. If κ > ω, then no tree (κ+ , <ν ) has a branch of size κ.
Proof. Suppose towards a contradiction that some tree (κ+ , <ν ) does have
a branch of size κ and let B be one such fixed branch (maximal chain). By
Lemmas 14.5 and 14.10, if γ = sup(B) then B is a closed and unbounded
subset of γ of order-type κ. Since κ is regular and uncountable, Cγ ∩ B
is unbounded in Cγ , so in particular we can find α ∈ Cγ ∩ B such that
tp(Cγ ∩ α) > ν. Reading off the definition of ρ(α, γ) we conclude that
ρ(α, β) = tp(Cγ ∩ α) > ν. Similarly we can find a β > α belonging to the
intersection of lim(Cγ ) and B. Then Cβ = Cγ ∩ β so α ∈ Cβ and therefore
ρ(α, β) = tp(Cβ ∩ α) > ν contradicting the fact that α <ν β.
a
15.4 Lemma. If κ > ω, then no tree (κ+ , <ν ) has a Souslin subtree of
height κ.
Proof. Forcing with a Souslin subtree of (κ+ , <ν ) of height κ would produce
an ordinal γ of cofinality κ and a closed and unbounded subset B of Cγ
forming a chain of the tree (κ+ , <ν ). It is well-known that in this case
B would contain a ground model subset of size κ contradicting Lemma
15.3.
a
15.5 Lemma. If κ > ω then for every ν < κ and every family A of κ
pairwise disjoint finite subsets of κ+ there exists A0 ⊆ A of size κ such that
for all a 6= b in A0 and all α ∈ a, β ∈ b we have ρ(α, β) > ν.
Proof. We may assume that for some n and all a ∈ A we have |a| = n.
Let a(0), . . . , a(n − 1) enumerate a given element a of A increasingly. By
Lemma 15.4, shrinking A, we may assume that a(i) (a ∈ A) is an antichain
of (κ+ , <ν ) for all i < n. Going to a subfamily of A of equal size we may
assume to have a well-ordering <w of A with the property that if a <w b
then no node from a is above a node from b in the tree ordering <ν . Define
f : [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 minimal
pair 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 0
on [A0 ]2 or there exist (i, j) ∈ n × n and an infinite A1 ⊆ A such that
100
I. Coherent Sequences
f is constantly equal to (i, j) on [A1 ]2 . The first alternative is what we
want, so let us see that the second one is impossible. Otherwise, choose
a <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 about
A. This completes the proof.
a
The unboundedness property of Lemma 15.5 can be quite useful in designing forcing notions satisfying good chain conditions. Having such applications in mind, we shall now work on refining further this kind of unboundedness property of the ρ-function.
15.6 Lemma. Suppose κ > 0, let γ < κ+ and let {αξ , βξ } (ξ < κ) be a sequence of pairwise disjoint elements of [κ+ ]≤2 . Then there is an unbounded
set Γ ⊆ κ such that ρ{αξ , βη } ≥ min{ρ{αξ , γ}, ρ{βη , γ}} for all ξ 6= η in
Γ.33
Proof. Clearly we may assume that the sequences αξ (ξ < κ) and βξ (ξ < κ)
are strictly increasing. For definiteness we assume that αξ ≤ βξ for all ξ.
The case αξ > βξ for all ξ is considered similarly. Let
α = sup αξ and β = sup βξ .
(I.184)
Then α and β are ordinals of cofinality κ, so the order-types of Cα and Cβ
are both equal to κ. It will be convenient to assume that the two sequences
{αξ } and {βξ } are actually indexed by Cα rather than κ. It is then clear
we may assume that
βξ ≥ αξ ≥ ξ for all ξ ∈ Cα .
(I.185)
Case 1: α = β. By Lemma 14.9, for each limit point ν of Cα there is
f (ν) < ν in Cα such that
ρ(ξ, αν ), ρ(ξ, βν ) ≥ ρ(ξ, ν) = ρ(ξ, α) for all ξ ∈ [f (ν), ν).
(I.186)
Fix a stationary Γ ⊆ lim(Cα ) such that f is constant on Γ. Then
ρ(αξ , βη ) ≥ ρ(αξ , α) for all ξ < η in Γ, and
(I.187)
ρ(βη , αξ ) ≥ ρ(βη , α) for all η < ξ in Γ.
(I.188)
a
Subcase 1 : γ ≥ α. Using the two subadditive properties of ρ in Lemma
14.6 one easily concludes that ρ(ξ, α) = ρ(ξ, γ) for any ξ < α such that
ρ(ξ, α) > ρ(α, γ). So, going to a tail of Γ we may assume that
ρ(αξ , α) = ρ(αξ , γ) and ρ(βξ , α) = ρ(βξ , γ) for all ξ ∈ Γ.
(I.189)
33 Here, and everywhere else later in this Chapter, the convention is that, ρ{α, β} is
meant to be equal to ρ(α, β) if α < β, equal to ρ(β, α) if β < α, and equal to 0 if α = β.
15. Successors of regular cardinals
101
Consider ξ < η in Γ. Combining (I.187) and the first equality of (I.189)
gives us
ρ(αξ , βη ) ≥ ρ(αξ , α) = ρ(αξ , γ),
which is sufficient for the conclusion of Lemma 15.6. Suppose ξ > η are
chosen from Γ in this order. Combining (I.188) and the second equality of
(I.189) gives us
ρ(βη , αξ ) ≥ ρ(βη , α) = ρ(βη , γ),
which is also sufficient for the conclusion of Lemma 15.6.
Subcase 1b : γ < α. Using Lemma 14.5 and going to a tail of the set Γ
we may assume that Γ lies above γ and that
ρ(αξ , α), ρ(βξ , α) > ρ(γ, α) for all ξ ∈ Γ.
(I.190)
Applying the subadditive property 14.6(b) of ρ we get for all ξ ∈ Γ:
ρ(γ, αξ ) ≤ max{ρ(γ, α), ρ(αξ , α)} = ρ(αξ , α), and
ρ(γ, βξ ) ≤ max{ρ(γ, α), ρ(βξ , α)} = ρ(βξ , α).
(I.191)
(I.192)
If ξ < η are chosen from Γ in this order then (I.187) and (I.191) give us
ρ(αξ , βη ) ≥ ρ(αξ , α) ≥ ρ(γ, αξ ),
which is sufficient for the conclusion of Lemma 15.6. If ξ > η are chosen
from Γ in this order, then combining (I.188) and (I.192) give us
ρ(βη , αξ ) ≥ ρ(βη , α) ≥ ρ(γ, βη ),
which is again sufficient for the conclusion of Lemma 15.6.
Case 2: α < β. We may assume all βξ > α for all ξ and working as in
the Case 1 we find a stationary set Γ of limit points of Cα such that
ρ(αξ , βη ) ≥ ρ(αξ , α) for all ξ < η in Γ.
(I.193)
By Lemma 14.5 for each η ∈ Γ there is g(η) ∈ Cα such that ρ(ξ, α) >
ρ(α, βη ) for all ξ in the interval [g(η), α). Using the two subadditive properties of ρ from Lemma 14.6, one concludes that ρ(ξ, α) = ρ(α, βη ) for all
ξ ∈ [g(η), α). Intersecting Γ with the closed and unbounded subset of Cα
of all ordinals that are closed under the mapping g we may assume
ρ(αξ , βη ) = ρ(αξ , α) for all ξ > η in Γ.
(I.194)
Subcase 2a : γ < α. Applying Lemma 14.5 again and going to a tail of
Γ we may assume that Γ lies above γ and
ρ(αξ , α) > ρ(γ, α) for all ξ ∈ Γ.
(I.195)
102
I. Coherent Sequences
Applying the subadditive properties of ρ we get
ρ(γ, αξ ) ≤ max{ρ(γ, α), ρ(αξ , α)} = ρ(αξ , α) for all ξ ∈ Γ.
(I.196)
If ξ < η are chosen from Γ in this order, then combining the inequalities
(I.193) and (I.196) we get
ρ(αξ , βη ) ≥ ρ(αξ , α) ≥ ρ(γ, αξ ),
which is sufficient to give us the conclusion of Lemma 15.6. If ξ > η are
chosen from Γ in this order, then combining (I.194) and (I.196) we get
ρ(αξ , βη ) = ρ(αξ , α) ≥ ρ(γ, αξ ),
which is again sufficient for the conclusion of Lemma 15.6.
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 are
chosen from Γ in this order, then combining the equalities (I.194) and (I.197)
we have
ρ(αξ , βη ) = ρ(αξ , α) = ρ(αξ , γ),
which is again sufficient to give us the conclusion of Lemma 15.6. This
finishes the proof.
a
15.7 Lemma. Suppose κ is λ-inaccessible 34 for some λ < κ and that A is
a family of size κ of subsets of κ+ , all of size < λ. Then for every ordinal
ν < κ there is a subfamily B of A of size κ such that for all a and b in B:
(a) ρ{α, β} > ν for all α ∈ a \ b and β ∈ b \ a.
(b) ρ{α, β} ≥ min{ρ{α, γ}, ρ{β, γ}} for all α ∈ a \ b, β ∈ b \ a and
γ ∈ a ∩ b.
Proof. Consider first the case λ = ω. Going to a subfamily, assume A forms
a ∆-system with root r and that for some fixed integer n ≥ 1 and all a ∈ A
we have |a \ r| = n. By Lemma 15.5, going to a subfamily we may assume
that A satisfies (a). For a ∈ A let a(0), . . . , a(n − 1) be the increasing
34 I.e.
ν τ < κ for all ν < κ and τ < λ.
15. Successors of regular cardinals
103
enumeration of a \ r. Going to a subfamily, we may assume that A can
be enumerated as aξ (ξ < κ) in such a way that aξ (i) (ξ < κ) is strictly
increasing for all i < n. By n2 · |r| successive applications of Lemma 15.6
we find a single unbounded set Γ ⊆ κ such that for all γ ∈ r, (i, j) ∈ n × n
and ξ 6= η in Γ:
ρ{aξ (i), aη (j)} ≥ min{ρ{aξ (i), γ}, ρ{aη (j), γ}}.
(I.198)
This gives us the conclusion (b) of the Lemma.
Let us now consider the general case. Using the λ-inaccessibility of κ
we may assume that A forms a ∆-system with root r and that members
of A have some fixed order-type, i.e. that for some fixed ordinal µ < λ
we can enumerate each a \ r for a ∈ A in increasing order as a(i) (i < µ).
Using λ-inaccessibility again and refining A we may assume that there is an
enumeration aξ (ξ < κ) of A such that aξ (i) (ξ < κ) is a strictly increasing
sequence for all i < µ. For i < µ, set
δ(i) = supξ aξ (i).
Then cf(δ(i)) = κ and therefore tp Cδ(i) = κ for all i < µ. Going to a
smaller cardinal we may assume that λ is a regular cardinal < κ. Applying
Lemma 14.9 simultaneously µ times gives us a regressive map f : κ −→ κ
such that for every η < κ of cofinality λ, all ξ ∈ [cδ(i) (f (η)), cδ(i) (η)) and
i < µ:
ρ(ξ, aη (i)) ≥ ρ(ξ, cδ(i) (η)) = ρ(ξ, δ(i)).
(I.199)
Here cδ(i) (ξ) (ξ < κ) refers to the increasing enumeration of the set Cδ(i) for
each i < µ. Pick a stationary set Γ of cofinality λ ordinals < κ such that f
is constant on Γ. It follows that for all i, j < µ, if δ(i) = δ(j) then
ρ{aξ (i), aη (j)} ≥ ρ(aξ (i), δ(i)) for ξ < η in Γ,
(I.200)
ρ{aη (j), aξ (i)} ≥ ρ(aη (j), δ(i)) for ξ > η in Γ.
(I.201)
On the other hand, if i, j < µ such that δ(i) < δ(j) we have
ρ{aξ (i), aη (j)} ≥ ρ(aξ (i), δ(i)) for all ξ < η in Γ.
(I.202)
Intersecting Γ with a closed and unbounded subset of κ we also assume that
aξ (i) ≥ cδ(i) (ξ) for all i < µ and ξ ∈ Γ.
aξ (i) ≤ aξ (j) for all ξ ∈ Γ and i, j < µ such that δ(i) ≤ δ(j).
(I.203)
(I.204)
Since ρ(α, δ(i)) ≥ tp(Cδ(i) ∩ α) for every i < µ and α < δ(i), combining
(I.200)-(I.204) we see that
ρ{aξ (i), aη (j)} ≥ min{ξ, η} for all i, j < µ and ξ 6= η in Γ.
(I.205)
104
I. Coherent Sequences
So replacing Γ with Γ \ (ν + 1) gives us the conclusion (a) of Lemma 15.7.
Suppose now we are given two coordinates i, j < µ and an ordinal γ in
the root r. Let
αξ = aξ (i) and βξ = aξ (j)
for ξ ∈ Γ. Note that (I.200) and (I.201) give us the conditions (I.187)
and (I.188) of the Case 1 (i.e. δ(i) = α = β = δ(j)) in the proof of
Lemma 15.6 while (I.202) gives us the condition (I.193) of the Case 2 (i.e.
δ(i) = α < β = δ(j)) in the proof of Lemma 15.6. Observe that in this
proof of Lemma 15.6, once the set Γ was obtained by a single application of
the Pressing Down Lemma to finally give us (I.187) and (I.188) of Case 1 or
(I.193) of Case 2, the rest of the refinements of Γ all lie in the restriction of
the closed and unbounded filter to Γ. In other words, we can now proceed
as in the case λ = ω and successively apply this refining procedure for each
γ ∈ r and (i, j) ∈ µ × µ obtaining a set Γ0 ⊆ Γ with the property that Γ \ Γ0
is nonstationary and
ρ{aξ (i), aη (j)} ≥ min{ρ{aξ (i), γ}, ρ{aη (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, the
proof is completed.
a
15.8 Definition. D : [κ+ ]2 −→ [κ+ ]<κ is defined by
D(α, β) = {ξ ≤ α : ρ(ξ, α) ≤ ρ(α, β)}.
(Note that D(α, β) = {ξ ≤ α : ρ(ξ, β) ≤ ρ(α, β)}, so we could take the
formula
D{α, β} = {ξ ≤ min{α, β} : ρ(ξ, α) ≤ ρ{α, β}}
as our definition of D{α, β} when there is no any implicit assumption about
the ordering between α and β as there is whenever we write D(α, β). )
15.9 Lemma. If κ is λ-inaccessible for some λ < κ, then for every family
A of size κ of subsets of κ+ , all of size < λ, there exists B ⊆ A of size κ
such that for all a and b in B and all α ∈ a \ b, β ∈ b \ a and γ ∈ a ∩ b:
(a) α, β > γ
(b)
β>γ
(c)
α>γ
(d) γ > α, β
implies
implies
implies
implies
D{α, γ} ∪ D{β, γ} ⊆ D{α, β},
D{α, γ} ⊆ D{α, β},
D{β, γ} ⊆ D{α, β},
D{α, γ} ⊆ D{α, β} or D{β, γ} ⊆ D{α, β}.
Proof. Choose B ⊆ A of size κ satisfying the conclusion (b) of Lemma 15.7.
Pick a 6= b in B and consider α ∈ a \ b, β ∈ b \ a and γ ∈ a ∩ b. By the
conclusion of 15.7(b), we have
ρ{α, β} ≥ min{ρ{α, γ}, ρ{β, γ}}.
(I.207)
15. Successors of regular cardinals
105
a. Suppose α, β > γ. Note that in this case a single inequality ρ(γ, α) ≤
ρ{α, β} or ρ(γ, β) ≤ ρ{α, β} given to us by (I.207) implies that we actually
have both inequalities simultaneously holding. The subadditivity of ρ gives
us ρ(ξ, α) ≤ ρ{α, β}, or equivalently ρ(ξ, β) ≤ ρ{α, β} for any ξ ≤ γ with
ρ(ξ, γ) ≤ ρ(γ, α) or ρ(ξ, γ) ≤ ρ(γ, β). This is exactly the conclusion of
15.9(a).
b. Suppose that β > γ > α. Using the subadditivity of ρ we see that
in both cases given to us by (I.207) we have that ρ(α, γ) ≤ ρ{α, β}. So the
inclusion D{α, γ} ⊆ D{α, β} follows immediately.
c. Suppose that α > γ > β. The conclusion D{β, γ} ⊆ D{α, β} follows
from the previous case by symmetry.
d. Suppose that γ > α, β. Then ρ(α, γ) ≤ ρ{α, β} gives D{α, γ} ⊆
D{α, β} while ρ(β, γ) ≤ ρ{α, β} gives us D{β, γ} ⊆ D{α, β}.
This completes the proof.
a
15.10 Remark. Note that min{x, y} ∈ D{x, y} for every {x, y} ∈ [κ+ ]2 , so
the conclusion (a) of Lemma 15.9 in particular means that γ < min{α, β}
implies γ ∈ D{α, β}. In applications, one usually needs this consequence of
15.9(a) rather than 15.9(a) itself.
15.11 Definition. The ∆-function of some family F of subsets of some
ordinal κ (respectively, a family of functions with domain κ) is the function
∆ : [F]2 −→ κ defined by
∆(f, g) = min(f 4 g),
(respectively, ∆(f, g) = min{ξ : f (ξ) 6= g(ξ)}).
Note the following property of ∆:
15.12 Lemma. ∆(f, g) ≥ min{∆(f, h), ∆(g, h)} for all {f, g, h} ∈ [F]3 .
a
15.13 Remark. This property can be very useful when transferring objects
that live on κ to objects on F. This is especially interesting when F is of
size larger than κ while all of its restrictions F ¹ ν = {f ∩ν : f ∈ F} (ν < κ)
have size < κ, i.e. when F is a Kurepa family (see for example [12]). We shall
now see that it is possible to have a Kurepa family F = {fα : α < κ+ } whose
∆-function is dominated by ρ, i.e. ∆(fα , fβ ) ≤ ρ(α, β) for all α < β < κ+ .
15.14 Theorem. If ¤κ holds and κ is λ-inaccessible then there is a λ-closed
κ-cc forcing notion P that introduces a Kurepa family on κ.
Proof. Going to a larger cardinal, we may assume that λ is regular. Put p
in P, if p is a one-to-one function from a subset of κ+ of size < λ into the
family of all subsets of κ of size < λ such that for all α and β in dom(p):
p(α) ∩ p(β) is an initial part of p(α) and of p(β),
(I.208)
106
I. Coherent Sequences
∆(p(α), p(β)) ≤ ρ(α, β) provided that α 6= β.
(I.209)
Let p ≤ q whenever dom(p) ⊇ dom(q) and p(α) ⊇ q(α) for all α ∈ dom(q).
Clearly P is a λ-closed forcing notion. To see that it satisfies the κ-chain
condition, let A be a given subset of P of size κ. By the assumption that κ
is λ-inaccessible, going to a subfamily of A of size κ, we may assume that
A forms a ∆-system, or Smore precisely, that dom(p) (p ∈ A) forms a ∆system with root d, that rng(p) (p ∈ A) forms a ∆-system with root c and
that any two members of A generate isomorphic structures via the natural
isomorphism that fixes the roots. By Lemma 15.7 there exists B ⊆ A of size
κ such that for all p, q ∈ B, all α ∈ dom(p)\dom(q), all β ∈ dom(q)\dom(p)
and all γ ∈ dom(p) ∩ dom(q)(= d):
ρ{α, β} > max(c),
(I.210)
ρ{α, β} ≥ min{ρ{α, γ}, ρ{β, γ}}.
(I.211)
S
Suppose
p, q ∈ B and that a = ( rng(p)) \ c lies entirely below b =
S
( rng(q)) \ c. Define r : dom(p) ∪ dom(q) −→ [κ]<λ as follows. For γ ∈ d,
let r(γ) = p(γ) ∪ q(γ). If β ∈ dom(q) \ dom(p) and if there is γ ∈ d such
that ∆(q(β), q(γ)) ∈ b, the γ must be unique. Put r(β) = p(γ) ∪ q(β) in
this case. If such γ does not exist, put r(β) = q(β) ∪ {max(c) + 1}. For
α ∈ dom(p) \ dom(q), put r(α) = p(α). Note that this choice of r does not
change the behaviour of ∆ on pairs coming from [dom(p)]2 or [dom(q)]2 , so
in checking that r satisfies (I.208) it suffices to assume α ∈ dom(p) \ dom(q)
and β ∈ dom(q) \ dom(p). If r(β) was defined as p(γ) ∪ q(β) for some γ ∈ d
we have that ∆(r(α), r(β)) = ∆(p(α), p(γ)) and that ∆(q(β), q(γ)) ∈ b. It
follows that
∆(r(α), r(β))
= min{∆(p(α), p(γ)), ∆(q(β), q(γ))}
≤ min{ρ{α, γ}, ρ{β, γ}}.
(I.212)
Applying (I.211) we get the desired inequality ∆(r(α), r(β)) ≤ ρ{α, β}. If
r(β) was chosen to be q(β)∪{max(c)+1}. Then ∆(r(α), r(β)) ≤ max(c)+1
and this is ≤ ρ{α, β} by (I.210).
Let Ġ be the generic filter of P and let Ḟ = {f˙α : α < κ+ }, where f˙α is
the union of p(α) (p ∈ Ġ). A simple genericity argument shows that each f˙α
intersects every interval if the form [ν, ν +λ) (ν < κ). It follows (see (I.208))
that Ḟ ¹ ν has size at most λ for all ν < κ. This finishes the proof.
a
15.15 Corollary. If ¤ω1 holds, and so in particular if ω2 is not a Mahlo
cardinal in the constructible universe, then there is a property K poset, forcing the Kurepa hypothesis.
a
15. Successors of regular cardinals
107
15.16 Remark. This is a variation on a result of Jensen who proved that
under ¤ω1 there is a ccc poset forcing the Kurepa hypothesis. It should be
noted that Jensen also proved (see [36]) that in the Levy collapse of a Mahlo
cardinal to ω2 there is no ccc poset forcing the Kurepa hypothesis. It should
also be noted that Veličković [91] was the first to use ρ to reprove Jensen’s
result. He has also shown (see [91]) that ρ easily yields an example of a
function with the ’∆-property’ in the sense of Baumgartner and Shelah [5].
Recall that a function with the ∆-property in the sense of [5] is a function
that satisfies some of the properties of the function D of Lemma 15.9 and
a function that forms a ground for the well-known forcing construction of
a locally compact scattered topology on ω2 all of whose Cantor-Bendixson
ranks are countable. We now use ρ to improve the chain-condition of the
Baumgartner-Shelah forcing.
15.17 Theorem. If ¤ω1 holds then there is a property K forcing notion that
introduces a locally compact scattered topology on ω2 all of whose CantorBendixson ranks are countable.
Proof. Let P be the set of all p = hDp , ≤p , Mp i where Dp is a finite subset of
ω2 , where ≤p is a partial ordering of Dp compatible with the well-ordering
and Mp : [Dp ]2 −→ [ω2 ]<ω has the following properties:
Mp {α, β} ⊆ D{α, β} ∩ Dp ,
Mp {α, β} = {α} if α ≤p β and Mp {α, β} = {β} if β ≤p α,
γ ≤p α, β for all γ ∈ Mp {α, β},
for every δ ≤p α, β there is γ ∈ Mp {α, β} such that δ ≤p γ.
(I.213)
(I.214)
(I.215)
(I.216)
We let p ≤ q if and only if Dp ⊇ Dq , ≤p ¹ Dq =≤q and Mp ¹ [Dq ]2 =
Mq . To verify that P satisfies Knaster’s chain condition, let A be a given
uncountable subset of P. Shrinking A we may assume that A forms a
∆-system with root R and that the conditions of A generate isomorphic
structures over R. By Lemma 15.9 there is an uncountable subset B of
A such that for all p and q in B and all α ∈ Dp \ Dq , β ∈ Dq \ Dp and
γ ∈ R = Dp ∩ Dq :
α, β > γ
β>γ
α>γ
implies
implies
implies
D{α, γ} ∪ D{β, γ} ⊆ D{α, β},
D{α, γ} ⊆ D{α, β},
D{β, γ} ⊆ D{α, β}.
(I.217)
(I.218)
(I.219)
Consider two conditions p and q from B. Let r = hDr , ≤r , Mr i be defined
as follows: Dr = Dp ∪ Dq and α ≤r β if and only if α ≤p β, or α ≤q β, or
there is γ ∈ R such that α ≤p γ ≤q β or α ≤q γ ≤p β. It is easily checked
that ≤r is a partial ordering on Dr compatible with the well-ordering such
108
I. Coherent Sequences
that ≤r ¹ Dp =≤p and ≤r ¹ Dq =≤q . Define Mr : [Dr ]2 −→ [ω2 ]<ω by
letting Mr ¹ [Dp ]2 = Mp , Mr ¹ [Dq ]2 = Mq , Mr {α, β} = {α} if α ≤r β,
Mr {α, β} = {β} if β ≤r α and if α ∈ Dp \ R and β ∈ Dq \ R are ≤r incomparable, set
Mr {α, β} = {γ ∈ D{α, β} ∩ Dr : γ ≤r α, β}.
(I.220)
Only nontrivial to check is that a so defined r satisfies (I.216). So let
δ ≤r α, β be given. If α, β ∈ Dp \ Dq and δ ∈ Dq \ Dp then there exist
γα , γβ ∈ R such that δ ≤q γα ≤p α and δ ≤q γβ ≤p β. By (I.216) for q
there exists γ ∈ Mq {γα , γβ } such that δ ≤q γ. Since Mq {γα , γβ } ⊆ R we
conclude that γ ∈ R. Thus γ ≤p α, β, so by (I.216) there is γ̄ ∈ Mp {α, β}
such that γ ≤p γ̄ and therefore δ ≤r γ̄. The case α ∈ R, β ∈ Dp \ D and
δ ∈ Dq \ Dp is quite similar.
Consider now the case when α ∈ Dp \ Dq , β ∈ Dq \ Dp and they are
≤r -incomparable, i.e. Mr {α, β} is given by (I.220). For symmetry we may
consider only the subcase δ ≤r α, β with δ ∈ Dp . Then δ ≤p α and
δ ≤p γ ≤q β for some γ ∈ R. By (I.216) for p there is γ̄ ∈ Mp {α, γ} such
that δ ≤p γ̄. It remains to show that γ̄ was put in Mr {α, β} in (I.220), i.e.
that γ̄ belongs to D{α, β}. Note that since ≤r agrees with the well-ordering,
we have that β > γ, so by (I.218) we have that D{α, γ} ⊆ D{α, β}. By
(I.213) for p we have that γ̄ ∈ Mp {α, γ} ⊆ D{α, γ} so we can conclude that
γ̄ ∈ D{α, β}. The rest of the cases are symmetric to the ones above.
Let Ġ be a generic filter of P and let ≤Ġ be the union of ≤p (p ∈ Ġ). For
α < ω2 set Ḃα = {ξ : ξ ≤Ġ α}. Let τ be the topology on ω2 generated by
Ḃα (α < ω2 ) as a clopen subbasis. It is easily checked that τ is a Hausdorff
locally compact scattered topology on ω2 . A simple density argument and
induction on α shows that the interval [ωα, ωα + ω) is the αth CantorBendixson rank of the space (ω2 , τ ).
a
15.18 Definition. A function f : [ω2 ]2 −→ [ω2 ]≤ω has property ∆ if for
every uncountable set A of finite subsets of ω2 there exist a and b in A such
that for all α ∈ a \ b, β ∈ b \ a and γ ∈ a ∩ b:
(a) α, β > γ
(b)
β>γ
(c)
α>γ
implies
implies
implies
γ ∈ f {α, β},
f {α, γ} ⊆ f {α, β},
f {β, γ} ⊆ f {α, β}.
15.19 Remark. This definition is due to Baumgartner and Shelah [5] who
have used it in their well-known forcing construction. It should be noted
that they were also able to force a function with the property ∆ using a
σ-closed ω2 -cc poset. As shown above (a fact first checked in [5, p.129]),
the function
D{α, β} = {ξ ≤ min{α, β} : ρ(ξ, α) ≤ ρ{α, β}}
16. Successors of singular cardinals
109
has property ∆. However, Lemma 15.9 shows that the function D has many
other properties that are of independent interest and that are likely to be
needed in other forcing constructions of this sort. The reader is referred to
papers of Koszmider [41] and Rabus [58] for further work in this area.
16. Successors of singular cardinals
In the previous section we have witnessed the fact that the function ρ :
[κ+ ]2 −→ κ defined from a ¤κ -sequence Cα (α < κ+ ) can be a quite useful
tool in stepping-up objects from κ to κ+ . In this section we analyse the
stepping-up power of ρ under the assumption that κ is a singular cardinal
of cofinality ω. So let κn (n < ω) be a strictly increasing sequence of regular
cardinals converging to κ fixed from now on. This immediately gives rise to
a rather striking tree decomposition <n (n < ω) of the ∈-relation on κ+ :
α <n β if and only if ρ(α, β) ≤ κn .
(I.221)
16.1 Lemma.
S
(1) ∈¹ κ+ = n<ω <n ,
(2) <n ⊆<n+1 ,
(3) (κ+ , <n ) is a tree of height ≤ κ+
n.
a
16.2 Definition. For α < κ+ and n < ω set
Fn (α) = {ξ ≤ α : ρ(ξ, α) ≤ κn },
fα (n) = tp(Fn (α)).
Let L = {fα : α < κ+ }, considered as a linearly ordered set with the
lexicographical ordering. Since L is a subset of ω κ, it has an order-dense
subset of size κ, so in particular it contains no well-ordered subset of size
κ+ . The following result shows that, however, every subset of L of smaller
size is the union of countably many well-ordered subsets.
16.3 Lemma. For each β < κ+ , Lβ = {fα : α < β} can be decomposed
into countably many well-ordered subsets.
Proof. Let Lβn = {fα : α ∈ Fn (β)} for n < ω. Note that the projection
f 7−→ f ¹ (n + 1) is one-to-one on Lβn so each Lβn is lexicographically
well-ordered.
a
110
I. Coherent Sequences
16.4 Remark. Note that K = {{(n, fα (n)) : n < ω} : α < κ+ } is a family
of countable subsets of ω × κ which has the property that K ¹ X = {K ∩ X :
K ∈ K} has size ≤ |X| + ℵ0 for every X ⊆ ω × κ of size < κ. We now
give an application of the existence of such a family of countable sets, due
to Fremlin [25]. This requires a few standard notions from measure theory
and real analysis.
16.5 Definition. A Radon measure space is a Hausdorff topological space
X with a σ-finite measure µ defined on a field of subsets of X containing
open sets with the property that the measure of every set is equal to the
supremum of measures of its compact subsets. When µ(X) = 1 then (X, µ)
is called Radon probability space. Finally recall that the Maharam type of
(X, µ) is the least cardinality of a subset of the measure algebra of µ which
completely generates the algebra.
16.6 Definition. A directed set D is Tukey reducible to a directed set E,
written as D ≤ E, if there is f : D −→ E which maps unbounded sets to
unbounded sets. In case D and E are ideals of subsets, this is equivalent to
saying that there is q : E −→ D which is monotone and has cofinal range.
We say that D and E are Tukey equivalent, D ≡ E, whenever D ≤ E and
E ≤ D.
16.7 Theorem. If there is a locally countable subset of [κ]ω of size cf[κ]ω
then the null ideal of every atomless Radon probability space of Maharam
type κ is Tukey equivalent to the product N × [κ]ω where N is the ideal of
measure zero subsets of the unit interval.
Proof. We shall give an argument for the case X = {0, 1}κ and µ the Haar
measure of X, referring the reader to the general result from [25] that the
null ideal of any atomless probability space of Maharam type κ is Tukey
equivalent to the null ideal Nκ of {0, 1}κ . The only nontrivial direction is
to produce a Tukey map from Nκ into Nω × [κ]ω . Let K ⊆ [κ]ω of size
cf[κ]ω with the property that K ¹ Γ = {a ∩ Γ : a ∈ K} is countable for
all countable Γ ⊆ κ. Let h : K −→ [κ]ω be a mapping whose range is
cofinal in [κ]ω when considered as a directed set ordered by the inclusion.
Let C = {a ∪ h(a) : a ∈ K}. Then C is a cofinal subset of [κ]ω with the
property that C(b) = {a ∈ C : a ⊆ b} is countable for every b ∈ [κ]ω . For
c ∈ C fix a bijection ec : c −→ ω. This gives us a way to define a function
πc mapping cylinders of {0, 1}κ with supports included in c into 2ω in the
natural way. Let Nκcl be the family of all null subsets of {0, 1}κ that are
countably supported cylinders of {0, 1}κ . Clearly, Nκcl is cofinal in Nκ , so
it suffices to produce a Tukey map f : Nκcl −→ Nω × [κ]ω . Given N ∈ Nκcl ,
find c(N ) ∈ C such that supp(N ) ⊆ c(N ) and put
f (N ) = (πc(N ) 00 N, c(N )).
16. Successors of singular cardinals
111
It is straightforward to check that f maps unbounded subsets of Nκcl to
unbounded subsets of the product Nω × [κ]ω .
a
16.8 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 16.4
the hypothesis of Theorem 16.7 is satisfied for all κ if one assumes ¤κ and
cf[κ]ω = κ+ for every κ of uncountable cofinality. Note that the transfer
from a locally countable K to a cofinal C = {a ∪ h(a) : a ∈ K} in the above
proof does not preserve local countability. To obtain a locally countable
cofinal C ⊆ [κ]ω one needs to work a bit harder along the same lines.
16.9 Definition. If a family K ⊆ [S]ω is at the same time locally countable
and cofinal in [S]ω then we call it a cofinal Kurepa family (cofinal K-family
in short). Two cofinal K-families H and K are compatible if H ∩ K ∈ H ∩ K
for all H ∈ H and K ∈ K. We say that K extends H if they are compatible
and if H ⊆ K.
16.10 Remark. Note that the size of any cofinal K-family K on a set S
is equal to the cofinality of [S]ω . Note also that for every X ⊆ S there is
Y ⊇ X of size cf[X]ω such that K ∩ Y ∈ K for all K ∈ K.
16.11 Definition. Define CK(θ) to be the statement that every sequence
Kn (n < ω) of comparable cofinal K-families with domains included in
θ which are closed under ∪, ∩ and \ can be extended to a single cofinal
K-family on θ, which is also closed under these three operations.
16.12 Lemma. CK(ω1 ) is true and if CK(θ) is true for some θ such that
cf[θ]ω = θ then CK(θ+ ) is also true.
Proof. The easy proof of CK(ω1 ) is left to the reader.
Suppose CK(θ) and let Kn (n < ω) be a given sequence of compatible
cofinal K-families as in the hypothesis of CK(θ+ ). By Remark 16.10 there
is a strictly increasing sequence δξ (ξ < θ+ ) of ordinals < θ such that
Kn ¹ δξ ⊆ Kn for all ξ < θ+ and n < ω. Recursively on ξ < θ+ we
construct a chain Hξ (ξ < θ+ ) of cofinal K-families as follows. If ξ = 0
or ξ = η + 1 for some η, using CK(δξ ) we can find a cofinal K-family
Hξ on δξ extending Hξ−1 and K
Sn ¹ δξ (n < ω). If ξ has uncountable
cofinality then the union of H̄ξ = 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 H̄ξ and Kn ¹ δξ (n < ω). If ξ has countable cofinality, pick a
sequence {ξn } converging to ξ and use CK(δξ ) to find a cofinal K-family
Hξ on δξ extending
S Hξn (n < ω) and Kn (n < ω). When the recursion is
done, set H = ξ<θ+ Hξ . Then H is a cofinal K-family on θ+ extending
Kn (n < ω).
a
16.13 Corollary. For each n < ω there is a cofinal Kurepa family on ωn .
a
112
I. Coherent Sequences
16.14 Definition. Let κ be a cardinal of cofinality ω. A Jensen matrix on
κ+ is a matrix Jαn (α < κ+ , n < ω) of subsets of κ+ with the following
properties, where κn (n < ω) is some fixed increasing sequence of cardinals
converging to κ:
(1) |Jαn | ≤ κn for all α < κ+ and n < ω,
(2) for all α < β and n < ω there is m < ω such that Jαn ⊆ Jβm ,
S
S
S
(3) n<ω [Jβn ]ω = α<β n<ω [Jαn ]ω whenever cf(β) > ω,
S
S
(4) [κ+ ]ω = α<κ+ n<ω [Jαn ]ω .
16.15 Remark. The notion of a Jensen matrix is the combinatorial essence
behind Silver’s proof of Jensen’s model-theoretic two-cardinal transfer theorem in the constructible universe (see [35, appendix]), so the matrix could
equally well be called ’Silver matrix’. It has been implicitly or explicitly
used on several places in the literature. The reader is referred to the paper
of Foreman and Magidor [24] which gives a quite complete discussion of this
notion and its occurrence in the literature.
16.16 Lemma. Suppose some cardinal κ of countable cofinality carries a
Jensen matrix Jαn (α < κ+ , n < ω) relative to some sequence of cardinals
κn (n < ω) that converge to κ. If CK(κn ) holds for all n < ω then CK(κ+ )
is also true.
Proof. Let Kn (n < ω) be a given sequence of compatible cofinal K-families
with domains included in κ+ . Given Jαn , there is a natural continuous
chain Jξαn (ξ < ω1 ) of subsets of κ+ ofSsize ≤ κn such that J0αn = Jαn
and Jξ+1
equal to the union of all K ∈ n<ω Kn which intersect Jξαn . Let
αn
S
∗
Jαn = ξ<ω1 Jξαn . It is easily seen that J∗αn (α < κ+ , n < ω) is also a Jensen
matrix. By recursion on α and n we define a sequence Hαn (α < κ+ , n < ω)
of compatible cofinal K-families as follows. If α = β + 1 or α = 0 and
n < ω using CK(κn ) we can find a cofinal K-family Hαn with domain 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 converging
to α. Using CK(κn ) we can find a cofinal K-family Hαn which extends
Hαm (m < n), Km ¹ J∗αn (m < ω) and each of the families Hαi k (i < ω, k <
ω and J∗αi k ⊆ J∗αn ). Finally suppose that cf(α) > ω. For n < ω, set
Hαn = [J∗αn ]ω ∩ (
[ [
Hξm ).
ξ<α m<ω
Using the properties of the Jensen matrix (especially (3)) as well as the
compatibility of Hξm (ξ < α, m < ω) one easily checks that Hαn is a cofinal
K-family with domain J∗αn which extends each member of Hαm (m < n)
16. Successors of singular cardinals
113
and Km ¹ J∗αn (m < ω) and which is compatible with all of the previously
constructed families Hξm (ξ < α, m < ω). When the recursion is done we
set
[ [
H=
Hαn .
α<κ+ n<ω
J∗αn
Using the property (4) of
(α < κ+ , n < ω), it follows easily that H is a
+
cofinal K-family on κ extending Kn (n < ω).
a
16.17 Theorem. If a Jensen matrix exists on any successor of a cardinal
of cofinality ω, then a cofinal Kurepa family exists on any domain.
a
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 properties
not captured by the definition of a Jensen matrix. If one additionally has
a 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 =
(aα ∪ {α}) (β < κ+ , n < ω).
(I.223)
α<n β
(Recall that <n is the tree ordering on κ+ defined by the formula α <n β iff
ρ(α, β) ≤ κn where κn is a fixed increasing sequence of cardinals converging
to κ.) The matrix Mβn (β < κ+ , n < ω) has properties not captured by
Definition 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 = sup{α : α <n β} then Mβn =
S
α<n β
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 have
started with a locally countable K = {aα : α < κ+ }.
a
16.19 Remark. One can think of the matrix M = {Mβn : β < κ+ , n < ω}
as a version of a ’morass’ for the singular cardinal κ (see [92]). It would be
interesting to see how far one can go in this analogy. We give a few applications just to illustrate the usefulness of the families we have constructed
so far.
114
I. Coherent Sequences
16.20 Definition. A Bernstein decomposition of a topological space X is
a function f : X −→ 2N with the property that f takes all the values from
2N on any subset of X homeomorphic to the Cantor set.
16.21 Remark. The classical construction of Bernstein [8] can be interpreted by saying that every space of size at most continuum admits a Bernstein decomposition. For larger spaces one must assume Hausdorff’s separation axiom, a result of Nešetril and Rődl (see [55]). In this context
Malykhin was able to extend Bernstein’s result to all spaces of size < c+ω
(see [50]). To extend this to all Hausdorff spaces, some use of square sequences seems natural. In fact, the first Bernstein decompositions of an
arbitrary Hausdorff space have been constructed using ¤κ and κω = κ+
for every κ > c of cofinality ω by Weiss [96] and Wolfsdorf [97]. We shall
now see that cofinal K-families are quite natural tools in constructions of
Bernstein decompositions.
16.22 Theorem. Suppose every regular θ > c supports a cofinal Kurepa
family of size θ. Then every Hausdorff space admits a Bernstein decomposition.
Proof. Let us say that a subset S of a topological space X is sequentially
closed in X if it contains all limit points of sequences {xn } ⊆ S that converge
in X. We say S is σ-sequentially closed in X if it can be decomposed
into countably many sequentially closed sets. For a given cardinal θ, let
B(θ) denote the statement that for every Hausdorff space X of size ≤ θ,
every σ-sequentially closed subset S of X and every Bernstein decomposition
f : S −→ 2N there is a Bernstein decomposition g : X −→ 2N extending
f . Using the fact that a Hausdorff space of size ≤ c has at most c copies
of the Cantor set and following the idea of Bernstein’s original proof, one
sees that B(c) holds. Note also that by our assumption θℵ0 = θ for every
cardinal θ ≥ c of uncountable cofinality. So if we have B(θ) for such θ
one proves B(θ+ ) by constructing a chain fξ (ξ < θ+ ) of partial Bernstein
decompositions on any Hausdorff topology τ on θ+ using the fact that the
set C of all δ < θ+ which are σ-sequentially closed in (θ+ , τ ) is closed and
unbounded in θ+ . Thus if δξ (ξ < θ+ ) enumerates C increasingly we will
have gξ : δξ −→ 2N and gξ compatible with the given partial Bernstein
decomposition f of a given σ-sequentially closed set S ⊆ θ+ . Note that
S ∩ δξ will be σ-sequentially closed, so the use of B(δξ ) to get fξ will be
possible. This inductive procedure encounters a crucial difficulty only when
θ = κ+ for some κ > c of cofinality ω. This is where a cofinal K-family K
on κ+ is useful. Let τ be a given Hausdorff topology on θ+ and for K ∈ K
let K 0 be the collection of all limits of τ -convergent sequences of elements
of K. Note that if a set Γ ⊆ κ+ is K-closed in the sense that K ¹ Γ ⊆ K
(i.e. K ∩ Γ ∈ K for all K ∈ K) then Γ0 will be equal to the union of all
K 0 (K ∈ K ¹ Γ). So if |Γ| > c is a cardinal of uncountable cofinality then
16. Successors of singular cardinals
115
|Γ0 | = |Γ|. It follows that every δ < θ+ with the property that δ = lim δn for
some sequence δn (n < ω) of ordinals ≤ δ with the property that K ¹ δn ⊆ K
for all n can be decomposed into countably many sets Γδn (n < ω) such that
for each n, the set Γδn is σ-sequentially closed in τ and |Γδn | is a cardinal
> c of uncountable cofinality. However, note that the set C of all such
δ < θ+ is closed and unbounded in θ+ so as above we use our inductive
hypothesis B(θ) (θ < κ) to construct a chain gξ : δξ −→ 2N (ξ < θ+ ) of
Bernstein decompositions. This completes the proof.
a
It is interesting that various less pathological classes of spaces admit a
local version of Theorem 16.22.
16.23 Theorem. A sigma-product 35 of the unit interval admits a Bernstein decomposition if its index set carries a cofinal Kurepa family. So in
particular, any metric space that carries a cofinal Kurepa family admits a
Bernstein decomposition.
Proof. So let K be a fixed well-founded cofinal K-family on Γ and let <w
be a well-ordering of K compatible with ⊆. For K ∈ K fix a Bernstein
decomposition 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 -minimal
K ∈ K such that supp(x) ⊆ K. It is easily checked that f is a Bernstein
decomposition of the sigma-product.
a
16.24 Definition. Recall the notion of a coherent family of partial functions
indexed by some ideal I, a family of the form fa : a −→ ω (a ∈ I) with the
property 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 strong
influence on such families provided I is a P-ideal of countable subsets of
some set Γ.
16.25 Theorem. Assuming the P-ideal dichotomy, for every coherent family of functions fa : a −→ ω (a ∈ I) indexed by some P-ideal I of countable
subsets of some set Γ, either
1. There is an uncountable ∆ ⊆ Γ such that fa ¹ ∆ is finite-to-one for
all a ∈ I, or
2. There is g : Γ −→ ω such that g ¹ a =∗ fa for all a ∈ I.
Proof. Let L be the family of all countable subsets b of Γ for which one can
find an a in I such that b \ a is finite and fa is finite-to-one on b. To see
that L is a P-ideal, let {bn } be a given sequence of members of L and for
35 Recall that a sigma-product is the subspace of some Tychonoff cube [0, 1]Γ consisting
of all mappings x such that supp(x) = {ξ ∈ Γ : x(ξ) 6= 0} is countable. The set Γ is its
index-set .
116
I. Coherent Sequences
each n fix a member an of I such that fan is finite-to-one on bn . Since I is
a P-ideal, we can find a ∈ I such that an \ a is finite for all n. Note that
for all n, bn \ a is finite and that fa is finite-to-one on bn . For n < ω, let
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 the
union of the b∗n ’s, we get a subset of a which almost includes each bn and
on which fa is finite-to-one. It follows that b belongs to L. This finishes
the proof that L is a P-ideal. Applying the P-ideal dichotomy to L, we get
the two alternatives that translate into the alternatives 1 and 2 of Theorem
16.25.
a
This leads to the natural question whether for any set Γ one can construct
a family {fa : a −→ ω} of finite-to-one mappings indexed by [Γ]ω . This
question was answered by Koszmider [40] using the notion of a Jensen matrix
discussed above. We shall present Koszmider’s result using the notion of a
cofinal Kurepa family instead.
16.26 Theorem. If Γ carries a cofinal Kurepa family then there is a coherent 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 be
a well-ordering of K compatible with ⊆. It suffices to produce a coherent
family of finite-to-one mappings indexed by K. This is done by induction
on <w . Suppose K ∈ K and fH : H −→ ω is determined for all H ∈ K
with 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 is
n < ω such that H ∩ K =∗ Hn ∩ K. So it suffices to construct a finite-toone fK : K −→ ω which coheres with each fHn (n < ω), a straightforward
task.
a
16.27 Corollary. For every nonnegative integer n there is a coherent family
fa : 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 of
one-to-one mappings fa : a −→ ω (a ∈ [c+ ]ω ). We finish this section with a
typical application of coherent families of finite-to-one mappings discovered
by Scheepers [61].
16.29 Theorem. If there is a coherent family fa : a −→ ω (a ∈ [Γ]ω ) of
finite-to-one mappings, then there is F : [[Γ]ω ]2 −→ [Γ]<ω with the property
that for every strictly ⊆-increasing sequence an (n < ω) of countable subsets
of Γ, the union of F (an , an+1 ) (n < ω) covers the union of an (n < ω).
17. The oscillation mapping
117
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 a
proper subset of b. Choose Φ : ω ω −→ ω ω with the property that x <∗ y
implies Φ(y) <∗ Φ(x), where <∗ is the ordering of eventual dominance on
ω ω (i.e. x <∗ y if x(n) < y(n) for all but finitely many n’s). Define another
family 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 Γ
¯ (n̄ ≤ n < ω)
and ξ¯ belongs to some an̄ then the sequence of integers gan (ξ)
¯
¯ i.e.
must have some place n ≥ n̄ with the property that gan (ξ) < gan+1 (ξ),
a place n ≥ n̄ such that ξ ∈ F (an , an+1 ).
a
16.30 Remark. Note that if κ is a singular cardinal of cofinality ω with the
property that cf[θ]ω < κ for all θ < κ, then the existence of a cofinal Kurepa
family on κ+ implies the existence of a Jensen matrix on κ+ . So these two
notions appear to be quite close to each other. The three basic properties
of the function ρ : [κ+ ] −→ κ (14.5 and 14.6(a),(b)) seem much stronger
in view of the fact that the linear ordering as in 16.3 cannot exist for κ
above a supercompact cardinal and the fact that Foreman and Magidor [24]
have produced a model with a supercompact cardinal that carries a Jensen
matrix on any successor of a singular cardinal of cofinality ω. Note that
Chang’s conjecture of the form (κ+ , κ) ³ (ω1 , ω) for some singular κ of
cofinality ω implies that every locally countable family K ⊆ [κ]ω must have
size ≤ κ. So, one of the models of set theory that has no cofinal K-family on,
say ℵω+1 , is the model of Levinski, Magidor and Shelah [49]. It seems still
unknown whether the conclusion of Theorem 16.29 can be proved without
additional set-theoretic assumptions.
17. The oscillation mapping
In what follows, θ will be a fixed regular infinite cardinal.
osc : P(θ)2 −→ Card
(I.226)
is defined by
osc(x, y) = |x \ (sup(x ∩ y) + 1)/ ∼ |,
where ∼ is the equivalence relation on x \ (sup(x ∩ y) + 1) defined by letting
α ∼ β iff the closed interval determined by α and β contains no point from
118
I. Coherent Sequences
y. So, if x and y are disjoint, osc(x, y) is simply the number of convex pieces
the set x is split by the set y. The oscillation mapping has proven to be a
useful device in various schemes for coding information. It usefulness in a
given context depend very much of the corresponding ’oscillation theory’,
a set of definitions and lemmas that disclose when it is possible to achieve
a given number as oscillation between two sets x and y in a given family
X . The following definition reveals the notion of largeness relevant to the
oscillation theory that we develop in this section.
17.1 Definition. A family X ⊆ P(θ) is unbounded if for every closed and
unbounded subset C of θ there exist x ∈ X and an increasing sequence
{δn : n < ω} ⊆ C such that sup(x ∩ δn ) < δn and [δn , δn+1 ) ∩ x 6= ∅ for all
n < ω.
This notion of unboundedness has proven to be the key behind a number
of results asserting the complex behaviour of the oscillation mapping on X 2 .
The case θ = ω seems to contain the deeper part of the oscillation theory
known so far (see [79],[80, §1] and [88]), though in this section we shall only
consider the case θ > ω. We shall also restrict ourselves to the family K(θ)
of all closed bounded subsets of θ rather than the whole power-set of θ. The
following is the basic result about the behaviour of the oscillation mapping
in this context.
17.2 Lemma. If X is an unbounded subfamily of K(θ) then for every positive integer n there exist x and y in X such that osc(x, y) = n.
Proof. Choose an ∈-chain M of length θ of elementary submodels M of some
large enough structure of the form Hλ such that X ∈ M and M ∩ θ ∈ θ.
Applying the fact that X is unbounded with respect to the closed and
unbounded set {M ∩ θ : M ∈ M} we can find y ∈ X such that:
δi = Mi ∩ θ ∈
/ y for all i < n,
(I.227)
δ0 ∩ y 6= ∅ and y \ δn−1 6= ∅,
(I.228)
[δi−1 , δi ) ∩ y 6= ∅ for all 0 < i < n.
(I.229)
J0 = [0, sup(y ∩ δ0 )],
(I.230)
Ji = [δi−1 , max(y ∩ δi )] for 0 < i < n,
(I.231)
Jn = [δn−1 , max(y)].
(I.232)
Let
Let Fn be the collection of all increasing sequences I0 < I1 < . . . < In of
closed intervals of θ such that I0 = J0 and such that there is x = xI in X
such that
x ⊆ I0 ∪ I1 ∪ . . . ∪ In ,
(I.233)
17. The oscillation mapping
119
max(Ii ) ∈ x for all i ≤ n.
(I.234)
Clearly, Fn ∈ M0 and hJ0 , J1 , . . . , Jn i ∈ Fn . Let Fn−1 be the collection
of all hI0 , . . . , In−1 i such that for every α < θ there exists an interval I
of θ above α such that hI0 , . . . , In−1 , Ii ∈ Fn . Note that Fn−1 ∈ M0 and
that using the elementarity of Mn−1 one proves that hJ0 , . . . , Jn−1 i ∈ Fn−1 .
Let Fn−2 be the collection of all hI0 , . . . , In−2 i such that for every α < θ
there exists interval I above α such that hI0 , . . . , In−2 , Ii ∈ Fn−1 . Then
again Fn−2 ∈ M0 and using the elementarity of Mn−2 one shows that
the restriction hJ0 , . . . , Jn−2 i belongs to Fn−2 . Continuing in this way we
construct families Fn , Fn−1 , . . . , F0 such that:
Fi ∈ M0 for all i ≤ n,
(I.235)
F0 = {J0 }.
(I.236)
hI0 , . . . , In−i i ∈ Fn−i implies that for all α < θ there exists
an interval I above α such that hI0 , . . . , In−i , Ii ∈ Fn−i+1 .
(I.237)
Recursively in i ≤ n we choose intervals I0 < I1 < . . . < In such that:
hI0 , . . . , Ii i ∈ Fi for all i ≤ n,
(I.238)
Ji−1 < Ii ∈ Mi−1 for 0 < i ≤ n.
(I.239)
Clearly, there is no problem in choosing the sequence using the elementarity of the submodels Mi−1 and the condition (I.237). By the definition
hI0 , . . . In i ∈ Fn means that there is x ∈ X satisfying (I.233) and (I.234).
It follows that max(x ∩ y) = max(I0 ) and that
x ∩ Ii 6= ∅ (0 < i ≤ n)
(I.240)
are the convex pieces of x \ (max(x ∩ y) + 1) to which this set is split by y.
It follows that osc(x, y) = n. This finishes the proof.
a
Lemma 17.2 also has a rectangular form.
17.3 Lemma. If X and Y are two unbounded subfamilies of K(θ) then for
all but finitely many positive integers n there exist x ∈ X and y ∈ Y such
that osc(x, y) = n.
a
Recall the notion of a nontrivial C-sequence Cα (α < θ) on θ from Section
11, a C-sequence with the property that for every closed and unbounded
subset C of θ there is a limit point δ of C such that C ∩ δ * Cα for all
α < θ.
17.4 Definition. For a subset D of θ let lim(D) denote the set of all α < θ
with the property that α = sup(D ∩ α). A subsequence Cα (α ∈ Γ) of some
C-sequence Cα (α < θ) is stationary if the union of all lim(Cα ) (α ∈ Γ) is a
stationary subset of θ.
120
I. Coherent Sequences
17.5 Lemma. A stationary subsequence of a nontrivial C-sequence on θ is
an unbounded family of subsets of θ.
Proof. Let Cα (α ∈ Γ) be a given stationary subsequence of a nontrivial
C-sequence on θ. Let C be a given closed and unbounded subset of θ. Let
∆ be the union of all lim(Cα ) (α ∈ Γ). Then ∆ is a stationary subset of θ.
For ξ ∈ ∆ choose αξ ∈ Γ such that ξ ∈ lim(Cαξ ). Applying the assumption
that Cα (α ∈ Γ) is a nontrivial C-subsequence, we can find ξ ∈ ∆ ∩ lim(C)
such that
C ∩ [η, ξ) * Cαξ for all η < ξ.
(I.241)
If such ξ cannot be found using the stationarity of the set ∆ ∩ lim(C) we
would be able to use the Pressing Down Lemma on the regressive mapping
that would give us an η < ξ violating (I.241) and get that a tail of C
trivializes Cα (α ∈ Γ). Using (I.241) and the fact that Cαξ ∩ ξ is unbounded
in ξ we can find a strictly increasing sequence δn (n < ω) of elements of
(C ∩ ξ) \ Cαξ such that [δn , δn+1 ) ∩ Cαξ 6= ∅ for all n. So the set Cαξ satisfies
the conclusion of Definition 17.1 for the given closed and unbounded set
C.
a
Recall that Qθ denotes the set of all finite sequences of ordinals < θ and
that we consider it ordered by the right lexicographical ordering. We need
the following two further orderings on Qθ :
s v t if and only if s is an initial segment of t.
s < t if and only if s is a proper initial part of t.
(I.242)
(I.243)
17.6 Definition. Given a C-sequence Cα (α < θ) we can define an action
(α, t) 7−→ αt of Qθ on θ recursively on the ordering v of Qθ as follows:
α∅ = α,
(I.244)
αhξi = the ξth member of Cα if ξ < tp(Cα ); otherwise αhξi = α, (I.245)
αta hξi = (αt )hξi .
(I.246)
17.7 Remark. Note that if ρ0 (α, β) = t for some α < β < θ then βt = α.
In fact, if β = β0 > . . . > βn = α is the walk from β to α along the
C-sequence, each member of the trace Tr(α, β) = {β0 , β1 , . . . , βn } has the
form βs where s is the uniquely determined initial part of t. Note, however,
that in general βt = α does not imply that ρ0 (α, β) = t.
17.8 Notation. Given a C-sequence Cα (α < θ) on θ we shall use osc(α, β)
to denote osc(Cα , Cβ ).
17.9 Theorem. If Cα (α < θ) is a nontrivial C-sequence on θ, then for
every unbounded set Γ ⊆ θ and positive integer n there exist α < β in Γ and
t v ρ0 (α, β) such that
17. The oscillation mapping
121
(a) osc(αt , βt ) = n,
(b) osc(αs , βs ) = 1 for all s < t.
Proof. Choose an elementary submodel M of some large enough structure
of the form Hλ such that δ̄ = M ∩ θ ∈ θ and M contains all the relevant
objects. Choose β̄ ∈ Γ above δ̄ and let
β̄ = β̄0 > β̄1 > . . . > β̄k = δ̄
be the minimal walk from β̄ to δ̄ along the C-sequence. Let k̄ = k − 1 if
Cβ̄k−1 ∩ δ̄ is unbounded; otherwise k̄ = k. Recall our implicit assumption
about essentially all C-sequences that we consider here: if γ is a limit ordinal
then a ξ ∈ Cγ occupying a successor place in Cγ must be a successor ordinal.
In this case, since δ is a limit ordinal and since δ̄ ∈ Cβ̄k−1 , either it must
be that β̄k−1 is a limit ordinal and δ̄ = sup(Cβ̄k−1 ∩ δ̄) or β̄k−1 = δ̄ + 1.It
follows that β̄k̄ is a limit ordinal. Set t = ρ0 (β̄k̄ , β̄). Let
∅ = t ¹ 0, t ¹ 1, . . . , t ¹ k̄ = t
be our notation for all the restrictions of the sequence t. Note that β̄i = β̄t¹i
for i ≤ k. Let Γ0 be the set of all β ∈ Γ such that:
β = β∅ > βt¹1 > . . . > βt ,
(I.247)
sup(Cβt¹i ∩ βt ) = sup(Cβ̄i ∩ β̄k̄ ) for all i < k̄.
(I.248)
Note that Γ0 ∈ M and β̄ ∈ Γ0 . Let ∆ = {βt : β ∈ Γ0 }. Then ∆ ∈ M
and β̄k̄ ∈ ∆. By the very choice of k̄, δ̄ belongs to lim(Cβ̄k̄ ), so using the
elementarity of M one concludes that Cδ (δ ∈ ∆) is a stationary subsequence
of our original sequence. Shrinking ∆ a bit if necessary, we may assume that
for every γ ∈ ∆ an α ∈ Γ0 with αt = γ can be found below the next member
of ∆ above γ. By Lemma 17.5, Cδ (δ ∈ ∆) is an unbounded family of subsets
of θ and so we can apply the basic oscillation Lemma 17.2 and find γ < δ
such that osc(αt , βt ) = n. Using the fact α < δ and the fact that α and β
satisfy (I.248) we conclude that osc(αt¹i , βt¹i ) = 1 for all i < k̄. This finishes
the proof.
a
17.10 Corollary. Suppose a regular uncountable cardinal θ carries a nontrivial C-sequence. Then there is f : [θ]2 −→ ω which takes all the values
from ω on any set of the form [Γ]2 for an unbounded subset of θ.
Proof. Given α < β < θ, if there is t v ρ0 (α, β) satisfying 17.9(a),(b) put
f (α, β) = osc(αt , βt ) − 2
otherwise put f (α, β) = 0.
(I.249)
a
122
I. Coherent Sequences
17.11 Remark. The class of all regular cardinals θ that carry a nontrivial
C-sequence is quite extensive. It includes not only all successor cardinals
but also some inaccessible as well as hyperinaccessible cardinals such as for
example, first inaccessible or first Mahlo cardinals. In view of the wellknown Ramsey-theoretic characterization of weak compactness, Corollary
17.10 leads us to the following natural question.
17.12 Question. Can the weak compactness of a strong limit regular
uncountable cardinal be characterized by the fact that for every f : [θ]2 −→
ω there exists an unbounded set Γ ⊆ θ such that f 00 [Γ]2 6= ω? This is true
when ω is replaced by 2, but can any other number beside 2 be used in this
characterization?
18. The square-bracket operation
In this Section we show that the basic idea of the square-bracket operation
on ω1 introduced in Definition 7.3 extends to a general setting on an arbitrary uncountable regular cardinal θ that carries a nontrivial C-sequence
Cα (α < θ). The basic idea is based on the oscillation map defined in the
previous section and, in particular, on the property of this map exposed in
Theorem 17.9: for α < β < θ we set
[αβ] = βt , where t v ρ0 (α, β) is such that osc(αt , βt ) ≥ 2
but osc(αs , βs ) = 1 for all s < t; if such t does not exist, we
let [αβ] = α.
(I.250)
Thus, [αβ] is the first place visited by β on its walk to α where a nontrivial
oscillation with the corresponding step of α occurs. What Theorem 17.9
is telling us is that the nontrivial oscillation indeed happens most of the
time. Results that would say that the set of values {[αβ] : {α, β} ∈ [Γ]2 }
is in some sense large no matter how small the unbounded set Γ ⊆ θ is,
would correspond to the results of Lemmas 7.4-7.5 about the square-bracket
operation on ω1 . It turns out that this is indeed possible and to describe it
we need the following definition.
18.1 Definition. A C-sequence Cα (α < θ) on θ avoids a given subset ∆
of θ if Cα ∩ ∆ = ∅ for all limit ordinals α < θ.
18.2 Lemma. Suppose Cα (α < θ) is a given C-sequence on θ that avoids
a set ∆ ⊆ θ. Then for every unbounded set Γ ⊆ θ, the set of elements of ∆
not of the form [αβ] for some α < β in Γ is nonstationary in θ.
Proof. Let Ω be a given stationary subset of ∆. We need to find α < β in
Γ such that [αβ] ∈ Ω. For a limit δ ∈ Ω fix a β = β(δ) ∈ Γ above δ and let
β = β0 > β1 > . . . > βk(δ) = δ
18. The square-bracket operation
123
be the walk from β to δ along the sequence Cα (α < θ). Since δ ∈
/ Cα for any
limit α and since Cα+1 = {α} for all α, we must have that βk(δ)−1 = δ + 1.
In particular,
Cβi ∩ δ is bounded in δ for all i < k(δ).
(I.251)
Applying the Pressing Down Lemma gives us a stationary set Ω0 ⊆ Ω, an
integer k, ordinals ξi (i < k), and a sequence t ∈ Qθ such that for all δ ∈ Ω0 :
k(δ) = k and ρ0 (δ, β(δ)) = t,
(I.252)
max(Cβ(δ)i ∩ δ) = ξi for all i < k.
(I.253)
Intersecting Ω0 with a closed and unbounded subset of θ, we may assume
for every δ ∈ Ω0 that β(δ) is smaller than the next member of Ω0 above δ.
Note that any C-sequence on a regular uncountable cardinal that avoids a
stationary subset of the cardinal, in particular, must be nontrivial. Thus
Cδ (δ ∈ Ω0 ) is a stationary subsequence of a nontrivial C-sequence, and
so by Lemma 17.5, unbounded as a family of subsets of θ. By the basic
oscillation Lemma 17.2, we can find γ < δ in Ω0 such that osc(Cγ , Cδ ) = 2.
Then by the choice of Ω, if α = β(γ) and β = β(δ) then γ = αt , δ = βt
and α < δ. Comparing this with (I.252) and (I.253) we conclude that
[αβ] = βt = δ ∈ Ω. This finishes the proof.
a
It is clear that the argument gives the following more general result.
18.3 Lemma. Suppose Cα (α < θ) avoids ∆ ⊆ θ and let A be a family
of size θ consisting of pairwise disjoint finite sets, all of some fixed size
n. Then the set of all elements of ∆ that are not of the form [a(1)b(1)] =
[a(2)b(2)] = . . . = [a(n)b(n)] for some a 6= b in A is nonstationary in θ. a
Since [αβ] belongs to the trace Tr(α, β) of the walk from β to α it is not
surprising that [··] strongly depends on the behaviour of Tr. For example
the following should be clear from the proof of Lemma 18.2.
18.4 Lemma. The following are equivalent for a given C-sequence and a
set Ω:
(a) Ω \ {[αβ] : {α, β} ∈ [Γ]2 } is nonstationary in θ,
S
(b) Ω \ {Tr(α, β) : {α, β} ∈ [Γ]2 } is nonstationary in θ.
a
This fact suggests the following definition.
18.5 Definition. The trace filter of a given C-sequence
Cα (α < θ) is the
S
normal filter on θ generated by sets of the form {Tr(α, β) : {α, β} ∈ [Γ]2 }
where Γ is an unbounded subset of θ.
124
I. Coherent Sequences
18.6 Remark. Having a proper (i.e. 6= P(θ)) trace filter is a strengthening
of the nontriviality requirement on a given C-sequence Cα (α < θ). For
example, if a C-sequence avoids a stationary set Ω ⊆ θ, then its trace filter
is nontrivial and in fact no stationary subset of Γ is a member of it. Note the
following analogue of Lemma 18.4: the trace filter of a given C-sequence
is the normal filter generated by sets of the form {[αβ] : {α, β} ∈ [Γ]2 }
where Γ is an unbounded subset of θ. So to obtain the analogues of the
results of Section 7 about the square-bracket operation on ω1 one needs a
C-sequence Cα (α < θ) on θ whose trace filter is not only nontrivial but
also not θ-saturated, i.e. it allows a family of θ pairwise disjoint positive
sets. It turns out that the hypothesis of Lemma 18.2 is sufficient for both
of these conclusions.
18.7 Lemma. If a C-sequence on θ avoids a stationary subset of θ, then
there exist θ pairwise disjoint subsets of θ that are positive with respect to
its trace filter. 36
Proof. This follows from the well-known fact (see [37]) that if there is a
normal, nontrivial and θ-saturated filter on θ, then for every stationary
Ω ⊆ θ there exists λ < θ such that Ω ∩ λ is stationary in λ (and the fact
that the stationary set which is avoided by the C-sequence does not reflect
in this way).
a
18.8 Corollary. If a regular cardinal θ admits a nonreflecting stationary
subset then there is c : [θ]2 −→ θ which takes all the values from θ on any
set of the form [Γ]2 for some unbounded set Γ ⊆ θ.
a
To get such a c, one composes the square-bracket operation of some Csequence, that avoids a stationary subset of θ, with a mapping ∗ : θ −→ θ
with the property that the ∗-preimage of each point from θ is positive with
respect to the trace filter of the square sequence. In other words c is equal
to the composition of [··] and ∗, i.e. c(α, β) = [αβ]∗ . Note that, as in Section
7, the property of the square-bracket operation from Lemma 18.3 leads to
the following rigidity result which corresponds to Lemma 7.10.
18.9 Lemma. The algebraic structure (θ, [··], ∗) has no nontrivial automorphisms.
a
18.10 Remark. Note that every θ which is a successor of a regular cardinal
κ admits a nonreflecting stationary set. For example Ω = {δ < θ : cf δ = κ}
is such a set. Thus any C-sequence on θ that avoids Ω leads to a square
bracket operation which allows analogues of all the results from Section 7
about the square-bracket operation on ω1 . The reader is urged to examine
36 A subset A of the domain of some filter F is positive with respect to F if A ∩ F 6= ∅
for every F ∈ F .
18. The square-bracket operation
125
these analogues. We finish this section by giving a projection of the squarebracket operation (the analogue of 7.14) which has been used by Shelah and
Steprans [69] in order to construct a family of nonisomorphic extraspecial
p-groups that contain no large abelian subgroups.
Suppose now that θ = κ+ for some regular cardinal κ and fix a C-sequence
Cα (α < κ+ ) on κ+ such that tp(Cα ) ≤ κ for all α, or equivalently, such
that Cα (α < κ+ ) avoids the set Ωκ = {δ < κ+ : cf δ = κ}. Let [··] be
the corresponding square-bracket operation. Let λ be the minimal cardinal
such that 2λ ≥ κ+ . Choose a sequence rξ (ξ < κ+ ) of distinct subsets of
λ. Let H be the collection of all maps h : P(D(h)) −→ κ+ where D(h) is
a finite subset of λ. Let π : κ+ −→ H be a map with the property that
π −1 (h) ∩ Ωκ is stationary for all h ∈ H. Finally, define an operation [[··]] on
κ+ as follows:
[[αβ]] = π([αβ])(rα ∩ D(π([αβ]))).
(I.254)
The following is a simple consequence of the property 18.3 of the squarebracket operation.
18.11 Lemma. For every family A of size κ+ consisting of pairwise disjoint
finite subsets of κ+ all of some fixed size n and every sequence ξ0 , . . . , ξn−1
of ordinals < κ+ there exist a 6= b in A such that [[a(i)b(i)]] = ξi for all
i < n.
a
18.12 Remark. Shelah and Steprans consider the function Φ : [κ+ ]2 −→ 2
defined by Φ(α, β) = 0 iff [[αβ]] = 0 in [69] which they then incorporate
into a general construction, due to Ehrenfeucht and Faber, to obtain an
extraspecial p-group GΓ associated to essentially every unbounded subset
Γ of κ+ . They show that GΓ contains no abelian subgroup of size κ+ using
not only Lemma 18.11 but also the behaviour of [[a(i)b(j)]] for i 6= j < n,
a 6= b ∈ A that follows from the behaviour of the square-bracket operation,
exposed above in Lemma 7.7.
For sufficiently large cardinals θ we have the following variation on the
theme first encountered above in Theorem 11.2.
18.13 Theorem. Suppose θ is bigger than the continuum and carries a Csequence avoiding a stationary set Γ of cofinality > ω ordinals in θ. Let A
be a family of θ pairwise disjoint finite subsets of θ, all of some fixed size n.
Then for every stationary Γ0 ⊆ Γ there exist s, t ∈ ω n and a positive integer
k 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,
37 Recall that if a and b are two sets of ordinals, then the notation a < b means that
max(a) < min(b).
126
I. Coherent Sequences
(2) ρ0 (a(i), b(j)) = ρ0 (δ0 , b(j))a ρ0 (δ1 , δ0 )a . . . a ρ0 (δl , δl−1 )a ρ0 (a(i), δl )
for all i, j < n,
(3) ρ2 (δ0 , b(j)) = tj and ρ2 (a(i), δl ) = si for all i, j < n.
Proof. For ∆ ⊆ Γ and δ < θ set
\
Sδ (A) =
{hρ2 (a(i), δ) : i < ni : a ∈ A ¹ [γ, δ)},
(I.255)
γ<δ
Tδ (∆) =
\
{ρ2 (α, δ) : α ∈ ∆ ∩ [γ, δ)}.
(I.256)
γ<δ
Note that Sδ (A) and Tδ (∆) range over subsets of ω n and ω, respectively, so
if δ has uncountable cofinality and if A and ∆ are unbounded in δ then both
of these sets, Sδ (A) and Tδ (∆), are nonempty. Let Γ0 be a given stationary
subset of Γ. Since θ has size bigger than the continuum, starting from Γ0
one can
T find an infinite sequence Γ0 ⊇ Γ1 ⊇ . . . of subsets of Γ such that
Γω = i<ω Γi is stationary in θ and
Sδ (A) = Sε (A) 6= ∅ for all δ, ε ∈ Γ1 ,
Tδ (Γi ) = Tε (Γi ) 6= ∅ for all δ, ε ∈ Γi+1 and i < ω.
(I.257)
(I.258)
Pick δ ∈ Γω such that Γω ∩ δ is unbounded in δ and choose b ∈ A above δ.
Let tj = ρ2 (δ, bj ) for j < n. Now let s be an arbitrary member of Sδ (A) and
k an arbitrary member of Tδ (Γω ). To check that these objects satisfy the
conclusion of the Theorem, let l < ω be given. Let δ0 = δ and let γ0 < δ0
be an upper bound for all sets of the form Cξ ∩ δ0 (ξ ∈ Tr(δ0 , b(j)), ξ 6=
δ0 , j < n). Then the walk from any b(j) to any α ∈ [γ0 , δ0 ) must pass
through δ0 . Since k ∈ Tδ (Γω ) ⊆ Tδ (Γl ) we can find δ1 ∈ Γl ∩ (γ0 , δ0 ) such
that ρ2 (δ1 , δ0 ) = k. Choose γ1 ∈ (γ0 , δ1 ) such that the walk from δ0 to any
α ∈ [γ1 , δ1 ) must pass through δ1 . Since k ∈ Tδ (Γω ) ⊆ Tδ (Γl−1 ) = Tδ1 (Γl−1 )
there exists δ2 ∈ Γl−1 ∩(γ1 , δ1 ) such that ρ2 (δ2 , δ1 ) = k. Choose γ2 ∈ (γ1 , δ2 )
such that the walk from δ1 to any α ∈ [γ2 , δ2 ) must pass through δ2 , and so
on. Proceeding in this way, we find δ = δ0 > δ1 > . . . > δl > γl > γl−1 >
. . . > γ0 such that:
δi ∈ Γl−i+1 for all 0 < i ≤ l,
(I.259)
ρ2 (δi+1 , δi ) = k for all i < l,
for every i < l the walk from δi to some α ∈ [γi+1 , δi+1 )
passes through δi+1 .
(I.260)
(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) first
passes through all the δi (i ≤ l), so ρ0 (a(i), b(j)) is simply equal to the
concatenation of ρ0 (δ0 , b(j)), ρ0 (δ1 , δ0 ), . . ., ρ0 (δl , δl−1 ), ρ0 (a(i), δl ). This
together with (I.260) gives us the three conclusions of Theorem 18.13. a
18. The square-bracket operation
127
From now on θ is assumed to be a fixed cardinal satisfying the hypotheses
of Theorem 18.13. It turns out that Theorem 18.13 gives us a way to define
another square-bracket operation which has complex behavior not only on
squares of unbounded subsets of θ but also on rectangles formed by two
unbounded subsets of θ. To define this new operation we choose a mapping
h : ω −→ ω such that:
for every k, m, n, p < ω and s ∈ ω n there exist l < ω such
that 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 where
t = ρ0 (α, β) ¹ h(ρ2 (α, β)).
Thus, [αβ]h is the h(ρ2 (α, β))th place that β visits on its walk to α. It is
clear that Theorem 18.13 and the choice of h in (I.262) give us the following
conclusion.
18.15 Lemma. Let A be a family of θ pairwise disjoint finite subsets of θ,
all of some fixed size n, and let Ω be an unbounded subset of θ. Then almost
every 38 δ ∈ Γ has the form [a(0)β]h = [a(1)β]h = . . . = [a(n − 1)β]h for
some a ∈ A, β ∈ Ω, a < β.
a
In fact, one can get a projection of this square-bracket operation with
seemingly even more complex behavior. Keeping the notation of Theorem
18.13, pick a function ξ 7−→ ξ ∗ from θ to ω such that {ξ ∈ Γ : ξ ∗ = n} is
stationary for all n. This gives us a way to consider the following projection
of the trace function Tr∗ : [θ]2 −→ ω <ω :
Tr∗ (α, β) = hmin(Cβ \ α)∗ ia Tr∗ (α, min(Cβ \ α)),
(I.263)
where we stipulate that Tr∗ (γ, γ) = hγ ∗ i for all γ. It is clear that the proof
of Theorem 18.13 allows us to add the following conclusions:
18.13∗ Theorem. Under the hypothesis of Theorem 18.13, its conclusion
can be extended by adding the following two new statements:
(4) Tr∗ (δ1 , δ0 ) = . . . = Tr∗ (δl , δl−1 ),
(5) The maximal term of the sequence Tr∗ (δ1 , δ0 ) = . . . = Tr∗ (δl , δl−1 ) is
bigger than the maximal term of any of the sequences Tr∗ (δ0 , b(j)) or
Tr∗ (a(i), δl ) for i, j < n.
a
18.16 Definition. For α < β < θ, let [αβ]∗ = βt for t the minimal initial
part of ρ0 (α, β) such that βt∗ = max(Tr∗ (α, β)).
38 Here
’almost every’ is to be interpreted by ’all except a nonstationary set’.
128
I. Coherent Sequences
Thus [αβ]∗ is the first place in the walk from β to α where the function ∗
reaches its maximum among all other places visited during the walk. Note
that combining the conclusions (1)-(5) of Theorem 18.13(∗) we get:
18.13∗∗ Theorem. Under the hypothesis of Theorem 18.13, its conclusion
can be extended by adding the following:
(6) [a(i)b(j)]∗ = [δ1 δ0 ]∗ for all i, j < n.
a
Having in mind the property of [··]h stated in Lemma 18.15, the following
variation is now quite natural.
18.17 Definition. [αβ]∗h = [α[αβ]∗ ]h for α < β < θ.
Using Theorem 18.13(∗∗) (1)-(6) one easily gets the following conclusion.
18.18 Lemma. Let A be a family of θ pairwise disjoint finite subsets of θ,
all of some fixed size n. Then for all but nonstationarily many δ ∈ Γ one
can find a < b in A such that [a(i)b(j)]∗h = δ for all i, j < n.
a
18.19 Remark. Composing [··]∗h with a mapping π : θ −→ θ with the
property that π −1 (ξ) ∩ Γ is stationary for all ξ < θ, one gets a projection
of [··]∗h for which the conclusion of Lemma 18.18 is true for all δ < κ.
Assuming that θ is moreover a successor of a regular cardinal κ (of size at
least continuum), in which case Γ can be taken to be {δ < κ+ : cf δ = κ},
and proceeding as in 18.11 above we get a projection [[··]]∗h with the following
property:
18.20 Lemma. For every family A of pairwise disjoint finite subsets of κ+
all of some fixed size n and for every q : n × n −→ κ+ there exist a < b in
A such that [[a(i)b(j)]]∗h = q(i, j) for all i, j < n.
a
18.21 Remark. The first example of a cardinal with such a complex binary operation was given by the author [79] using the oscillation mapping
described above in Section 17. It was the cardinal b, the minimal cardinality
of an unbounded subset of ω ω under the ordering of eventual dominance.
The oscillation mapping restricted to some well-ordered unbounded subset
W of ω ω is perhaps still the most interesting example of this kind due to
the fact that its properties are preserved in forcing extensions that do not
change the unboundedness of W (although they can collapse cardinals and
therefore destroy the properties of the square-bracket operations on them).
This absoluteness of osc is the key feature behind its applications in various
coding procedures (see, e.g. [85]).
18.22 Theorem. For every regular cardinal κ of size at least the continuum, the κ+ -chain condition is not productive, i.e., there exist two partially
ordered sets P0 and P1 satisfying the κ+ -chain condition but their product
P0 × P1 fails to have this property.
19. Unbounded functions on successors of regular cardinals
129
Proof. Fix two disjoint stationary subsets Γ0 and Γ1 of {δ < κ+ : cf δ = κ}.
Let Pi be the collection of all finite subsets p of κ+ with the property that
[αβ]∗h ∈ Γi for all α < β in p. By Lemma 18.18, P0 and P1 are κ+ -cc posets.
Their product P0 × P1 , however, contains a family h{α}, {α}i (α < κ+ ) of
pairwise incomparable conditions.
a
18.23 Remark. Theorem 18.22 is due to Shelah [65] who proved it using
similar methods. The first ZFC-examples of non-productiveness of the κ+ chain condition were given by the author in [77] using what is today known
under the name ’pcf theory’. After the full development of pcf theory it
became apparent that the basic construction from [77] applies to every successor of a singular cardinal [66]. A quite different class of cardinals θ with
θ-cc non-productive was given by the author in [76]. For example, θ = cf c
is one of these cardinals. For an overview of recent advances in this area,
the reader is referred to [53]. The following problem seems still open:
18.24 Question. Suppose that θ is a regular strong limit cardinal and the
θ-chain condition is productive. Is θ necessarily a weakly compact cardinal?
19. Unbounded functions on successors of regular cardinals
In this section κ is a regular cardinal and Cα (α < κ+ ) is a fixed sequence
with tp(Cα ) ≤ κ for all α < κ+ . Define ρ∗ : [κ+ ]2 −→ κ by
ρ∗ (α, β) =
sup{tp(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) ≤ ν}
has size no more than |ν| + ℵ0 .
a
The proof of the following subadditivity properties of ρ∗ is very similar
to the proof of the corresponding fact for the function ρ from Section 14.
19.2 Lemma. For all α ≤ β ≤ γ,
(a) ρ∗ (α, γ) ≤ max{ρ∗ (α, β), ρ∗ (β, γ)},
(b) ρ∗ (α, β) ≤ max{ρ∗ (α, γ), ρ∗ (β, γ)}.
a
130
I. Coherent Sequences
We mention a typical application of this function to the problem of existance of partial square sequences which, for example, have some applications
in the pcf theory (see [9]).
19.3 Theorem. For every regular uncountable cardinal λ < κ and stationary Γ ⊆ {δ < κ+ : cf(δ) = λ}, there is a stationary set Σ ⊆ Γ and a
sequence Cα (α ∈ Σ) such that:
(1) Cα is a closed and unbounded subset of α,
(2) Cα ∩ ξ = Cβ ∩ ξ for every ξ ∈ Cα ∩ Cβ .
Proof. For each δ ∈ Γ, choose ν = ν(δ) < κ such that the set P<ν (δ) =
{ξ < δ : ρ∗ (ξ, δ) < ν} is unbounded in δ and closed under taking suprema
of sequences of size < λ. Then there is ν̄, µ̄ < κ and stationary Σ ⊆ Γ such
that ν(δ) = ν̄ and tp(P<ν̄ (δ)) = µ̄ for all δ ∈ Σ. Let C be a fixed closed
and unbounded subset of µ̄ of order-type λ. Finally, for δ ∈ Γ set
Cδ = {α ∈ P<ν̄ (δ) : tp(P<ν̄ (α)) ∈ C}.
Using Lemma 19.2, one easily checks that Cα (α ∈ Σ) satisfies the conditions
(1) and (2).
a
Another application concerns the fact exposed above in Section 16, that
the inequalities 19.2(a),(b) are particularly useful, when κ has cofinality
ω. Also consider the well-known phenomenon first discovered by K.Prikry
(see [37]), that in some cases, the cofinality of a regular cardinal κ can be
changed to ω, while preserving all cardinals.
19.4 Theorem. In any cardinal-preserving extension of the universe, which
has no new bounded subsets of κ, but in which κ has a cofinal ω-sequence
diagonalizing the filter of closed and unbounded subsets of κ restricted to
the ordinals of cofinality > ω, there is a sequence Cαn (α ∈ lim (κ+ ), n < ω)
such that for all α < β in lim(κ+ ):
(1) Cαn is a closed subset of α for all n,
(2) Cαn ⊆ Cαm , whenever n ≤ m,
S
(3) α = n<ω Cαn ,
(4) α ∈ lim(Cβn ) implies Cαn = Cβn ∩ α.
Proof. For α < κ+ , let Dα be the collection of all ν < κ for which P<ν (α)
is σ-closed, i.e., closed under supremums of all of its countable bounded
subsets. Clearly, Dα contains a closed unbounded subset of κ, restricted to
cofinality > ω ordinals. Note that
ν ∈ Dβ and ρ∗ (α, β) < ν imply that ν ∈ Dα .
(I.265)
19. Unbounded functions on successors of regular cardinals
131
In the extended universe, pick a strictly increasing sequence νn (n < ω),
which converges to κ and has the property that for each α < κ+ , there is
n < ω such that νm ∈ Dα for all m ≥ n. Let n(α) be the minimal such
integer n.
Fix α ∈ lim(κ+ ) and n < ω. If there is γ ≥ α such that n ≥ n(γ) and
α ∈ lim(P<νn (γ))
(I.266)
let γ(α) be the minimal such γ and let
Cαn = P<νn (γ(α)) ∩ α.
(I.267)
If such a γ ≥ α does not exist, let Cαn = ∅ for n < n(α) and let
Cαn = P<νn (α) ∩ α.
(I.268)
for n ≥ n(α). Clearly, this choice of Cαn ’s satisfies (1),(2) and (3). To verify
(4), let α be a limit point of some Cβn .
Suppose first that cf(α) = ω. If Cβn was defined according to the
case I.267 so is Cαn . Since P<νn (γ(β)) is σ-closed we conclude that α ∈
P<νn (γ(β)) which by 19.2(a),(b) means that P<ν (α) = P<ν (γ(β)) ∩ α for
all ν ≥ νn and therefore Dγ(β) \ νn ⊆ Dα . It follows that n(α) ≥ n and
therefore that γ(α) = α. Hence, Cαn = Cβn ∩ α. If Cβn was defined according to the case I.268 then our assumption α ∈ lim(Cβn ) and the fact
that n ≥ n(β) means that Cαn was defined according to I.267. Similarly as
above we conclude that actually α ∈ P<νn (β) which gives us n ≥ n(α) and
γ(α) = α. It follows that Cαn = Cβn ∩ α in this case as well.
Suppose now that cf(α) > ω. If Cβn was defined according to I.267 then
so is Cαn and so Pνn (γ(α)) ∩ α and Pνn (γ(β)) ∩ α are two σ-closed and
unbounded subsets of α. So their intersection is unbounded in α which by
19.2(a),(b) gives us that Pνn (γ(α)) ∩ α = Pνn (γ(β)) ∩ α. It follows that
Cαn = Cβn ∩ α in this case. Suppose now that Cβn was defined according
to I.268. Then γ = β satisfies n ≥ n(γ) and I.266. So Cαn was defined
according to the case I.267. It follows that Pνn (γ(α)) ∩ α and Pνn (β) ∩ α are
two σ-closed and unbounded subsets of α so their intersection is unbounded
in α. Applying 19.2(a),(b) we get that Pνn (γ(α))∩α = Pνn (β)∩α. It follows
that Cαn = Cβn ∩ α in this case as well.
a
19.5 Remark. The combinatorial principle appearing in the statement of
Theorem 19.4 is a member of a family of square principles that has been
studied systematically by Schimmerling and others (see e.g. [62]). It is
definitely a principle sufficient for all of the applications of ¤κ appearing in
Section 16 above.
19.6 Definition. A function f : [κ+ ]2 −→ κ is unbounded if f 00 [Γ]2 is
unbounded in κ for every Γ ⊆ κ+ of size κ. We shall say that such an f is
132
I. Coherent Sequences
strongly unbounded if for every family A of size κ+ , consisting of pairwise
disjoint finite subsets of κ+ , and every ν < κ there exists A0 ⊆ A of size κ
such that f (α, β) > ν for all α ∈ a, β ∈ b and a 6= b in A0 .
19.7 Lemma. If f : [κ+ ]2 −→ κ is unbounded and subadditive (i.e. it
satisfies the two inequalities 19.2(a),(b)), then f is strongly unbounded.
Proof. For α < β < κ+ , set α <ν β if and only if f (α, β) ≤ ν. Then
our assumption about f satisfying 19.2(a) and (b) reduces to the fact that
each <ν is a tree ordering on κ+ compatible with the usual ordering on
κ+ . Note that the unboundedness property of f is preserved by any forcing
notion satisfying the κ-chain condition, so in particular no tree (κ+ , <ν ) can
contain a Souslin subtree of height κ. In the proof of Lemma 15.5 above
we have seen that this property of (κ+ , <ν ) alone is sufficient to conclude
that every family A of κ many pairwise disjoint subsets of κ+ contains a
subfamily A0 of size κ such that for every a 6= b in A0 every α ∈ a is
<ν -incomparable to every β ∈ b, which is exactly the conclusion of f being
strongly unbounded.
a
19.8 Lemma. The following are equivalent:
(1) There is a structure (κ+ , κ, <, Rn )n<ω with no elementary substructure B of size κ such that B ∩ κ is bounded in κ.
(2) There is an unbounded function f : [κ+ ]2 −→ κ,
(3) There is a strongly unbounded, subadditive function f : [κ+ ]2 −→ κ.
Proof. Only the implication from (2) to (3) needs some argument. So let
f : [κ+ ]2 −→ κ be a given unbounded function. Increasing f we may assume
that each of its sections fα : α −→ κ is one-to-one. For a set Γ ⊆ κ+ , let
the f -closure of Γ, CLf (Γ), be the minimal ∆ ⊇ Γ with the property that
f 00 [∆]2 ⊆ ∆ and fα−1 (ξ) ∈ ∆, whenever α ∈ ∆ and ξ < sup(∆ ∩ κ). Define
e : [κ+ ]2 −→ κ by letting
e(α, β) = sup(κ ∩ CLf (ρ∗ (α, β) ∪ Pρ∗ (α,β) (α))).
(I.269)
We show the two subadditive properties for e:
e(α, γ) ≤ max{e(α, β), e(β, γ)} whenever α ≤ β ≤ γ.
(I.270)
By 19.2(a) either ρ∗ (α, γ) ≤ ρ∗ (α, β) in which case Pρ∗ (α,γ) (α) ⊆ Pρ∗ (α,β) (α)
holds, or ρ∗ (α, γ) ≤ ρ∗ (β, γ) when we have Pρ∗ (α,γ) (α) ⊆ Pρ∗ (β,γ) (β). By
the definition of e, the first case gives us the inequality e(α, γ) ≤ e(α, β)
and the second gives us e(α, γ) ≤ e(β, γ).
e(α, β) ≤ max{e(α, γ), e(β, γ)} whenever α ≤ β ≤ γ.
(I.271)
19. Unbounded functions on successors of regular cardinals
133
By 19.2(b) either ρ∗ (α, β) ≤ ρ∗ (α, γ) in which case Pρ∗ (α,β) (α) ⊆ Pρ∗ (α,γ) (α)
holds, or ρ∗ (α, β) ≤ ρ∗ (β, γ) when we have Pρ∗ (α,β) (α) ⊆ Pρ∗ (β,γ) (β). In
the first case we get e(α, β) ≤ e(α, γ) and in the second e(α, β) ≤ e(β, γ).
By Lemma 19.7, to get the full conclusion of (3) it suffices to show that e
is unbounded. So let Γ be a given subset of κ+ of order-type κ and let ν < κ
be a given ordinal. Since e(α, β) ≥ ρ∗ (α, β) for all α < β < κ+ we may
assume that ρ∗ is bounded on Γ, or more precisely, that for some ν̄ ≤ ν and
all α < β in Γ, ρ∗ (α, β) ≤ ν̄. For α ∈ Γ, let β(α) be the minimal point of Γ
above α. Then there is µ ≤ ν̄ and ∆ ⊆ Γ of size κ such that ρ∗ (α, β(α)) = µ
for all α ∈ ∆. Note that the sequence of sets Pµ (α) (α ∈ ∆) forms a chain
under end-extension. Let Γ∗ be the union of this sequence of sets. Since f
is unbounded, there exist ξ < η in Γ∗ such that f (ξ, η) > ν. Pick α ∈ ∆
above η. Then ξ, η ∈ Pµ (α) and therefore e(α, β(α)) ≥ f (ξ, η) > ν. This
finishes the proof.
a
19.9 Remark. Recall that Chang’s conjecture is the model-theoretic transfer principle asserting that every structure of the form (ω2 , ω1 , <, . . .) with
a countable signature has an uncountable elementary submodel B with the
property that B ∩ ω1 is countable. This principle shows up in many considerations including the first two uncountable cardinals ω1 and ω2 . For
example, it is known that it is preserved by ccc forcing extensions, that it
holds in the Silver collapse of an ω1 -Erdős cardinal, and that it in turn implies that ω2 is an ω1 -Erdős 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 the
statement that there exists e : [ω2 ]2 −→ ω1 such that:
(a) e(α, γ) ≤ max{e(α, β), e(β, γ)} whenever α ≤ β ≤ γ,
(b) e(α, β) ≤ max{e(α, γ), e(β, γ)} whenever α ≤ β ≤ γ,
(c) For every uncountable family A of pairwise disjoint finite subsets of
ω2 and every ν < ω1 there exists an uncountable A0 ⊆ A such that
e(α, β) > ν whenever α ∈ a and β ∈ b for every a 6= b ∈ A0 .
a
19.11 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 by
De {α, β} = {ξ ≤ min{α, β} : e(ξ, α) ≤ e{α, β}} satisfies the weak form of
Definition 15.18, where only the conclusion (a) is kept. The full form of
Definition 15.18, however, cannot be achieved assuming only the negation
of Chang’s conjecture. This shows that the function ρ, based on a ¤ω1 sequence, is a considerably deeper object than an e : [ω2 ]2 −→ ω1 satisfying
19.10(a),(b),(c).
134
I. Coherent Sequences
Recall that the successor of the continuum is characterized as the minimal cardinal θ with the property that every f : [θ]2 −→ ω is constant on
the square of some infinite set. We shall now see that in slightly weakening the partition property by replacing squares by rectangles one gets a
characterization of a quite different sort. To see this, let us use the arrow
notation
µ ¶1,1
µ ¶
ω
θ
−→
ω ω
θ
to succinctly express the statement that for every map f : θ × θ −→ ω,
there exist infinite sets A, B ⊆ θ such that f is constant on their product.
Let θ2 be the minimal θ which fails to satisfy this property. Note that
ω1 < θ2 ≤ c+ . The following result shows that θ2 can have the minimal
possible value ω2 , as well as that θ2 can be considerably smaller than the
continuum.
19.12 Theorem. Chang’s conjecture is equivalent to the statement that
µ ¶
µ ¶1,1
ω2
ω
−→
holds in every ccc forcing extension.
ω2
ω ω
Proof. We have already noted that Chang’s conjecture is preserved by ccc
forcing extensions, so in order to prove the direct implication, it suffices
to show that Chang’s conjecture implies that every f : ω2 × ω2 −→ ω is
constant on the product of two infinite subsets of ω2 . Given such an f ,
using Chang’s conjecture it is possible to find an elementary submodel M
of Hω3 such that f ∈ M , δ = M ∩ ω1 ∈ ω1 and Γ = M ∩ ω2 is uncountable.
Choose i ∈ ω and uncountable ∆ ⊆ Γ such that f (δ, β) = i for all β ∈ ∆.
Choose α0 < δ such that ∆0 = {β ∈ ∆ : f (α0 , β) = i} is uncountable.
Such α0 exists by elementarity of M . Pick β0 ∈ ∆0 above ω1 . Note that
by the elementarity of M , for every β ∈ ∆0 there exist unboundedly many
α < δ such that f (α, β0 ) = f (α, β) = i. So we can choose α1 < δ above
α0 such that the set ∆1 = {β ∈ ∆0 : f (α1 , β) = i} is uncountable. Then
again for every β ∈ ∆1 there exist unboundedly many α < δ such that
f (α, β0 ) = f (α, β1 ) = f (α, β) = i, and so on. Continuing in this way,
we build two increasing sequences {αn } ⊆ δ and {βn } ⊆ Γ \ ω1 such that
f (αn , βm ) = i for all n, m < ω.
Assume now that Chang’s conjecture fails and fix e : [ω2 ]2 −→ ω as
in Corollary 19.10. We shall describe a ccc forcing notion P which forces
an f : ω2 × ω2 −→ ω which is not constant on the product of any two
infinite subsets of ω2 . The conditions of P are simply maps of the form
p : [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 and
have the property that e(ξ, α) > e(α, β).
(I.272)
19. Unbounded functions on successors of regular cardinals
135
We order P by letting p extend q if and only if p extends q as a function
and
p(ξ, α) 6= p(ξ, β) whenever α < β belong to Dq and ξ belongs to Dp \ Dq .
(I.273)
Let A be a given uncountable subset of P. Refining A, we may assume that
Dp (p ∈ A) forms a ∆-system with root D and that Fp = e00 [Dp ]2 (p ∈ A)
form an increasing ∆-system on ω1 with root F . Moreover, we may assume
that for every p and q in A, the finite structures they generate together with
e are isomorphic via the isomorphism that fixes D and F . By 19.10(c) there
is uncountable A0 ⊆ A such that for all p, q ∈ A0 ,
e(α, β) > max(F ) whenever α ∈ Dp \ D and β ∈ Dq \ D.
(I.274)
We claim that every two conditions p and q from A0 are compatible. To see
this, extend p ∪ q to a mapping r : [Dp ∪ Dq ]2 −→ ω which is one-to-one on
new pairs, avoiding the old values. To see that r is indeed a member of P we
need to check (I.272) for r. So let ξ < α < β be three given ordinals from
Dr such that e(ξ, α) > e(α, β). We have to prove that r(ξ, α) 6= r(ξ, β).
By the choice of r we may assume that {ξ, α} and {ξ, β} are old pairs, i.e.
that they belong to [Dp ]2 ∪ [Dq ]2 . Since p and q both satisfy (I.272), we
may consider only the case when (say) α ∈ Dp \ Dq , β ∈ Dq \ Dp and
ξ ∈ D. Recall that e(ξ, α) > e(α, β) and the subadditive properties of e
(see 19.10(a),(b)) give us the equality e(ξ, α) = e(ξ, β), i.e. e(ξ, α) belongs
to F contradicting (I.274). To see that r extends p and q, by symmetry it
suffices to check that r extends, say, p. So suppose ξ < α < β are given
such that α < β belong to Dp and ξ belongs to Dq \ Dp . We need to show
that r(ξ, α) 6= r(ξ, β). By the choice on new pairs, this is automatic if one
of the two pairs is new. So we may consider the case when both α and
β are in D. Since r(ξ, α) = q(ξ, α) and r(ξ, β) = q(ξ, β) in this case the
inequality r(ξ, α) 6= r(ξ, β) would follow from (I.272) for q if we show that
e(ξ, α) > e(α, β). Otherwise e(ξ, α) would belong to the root F and by the
fact that all conditions are isomorphic, e(ξ(s), α) ∈ F holds for all s ∈ A0 ,
where ξ(s) ∈ Ds (s ∈ A0 ) are the copies of ξ. This of course contradicts the
property 19.10(c) of e since ξ(s) 6= ξ(t) when s 6= t.
Forcing with P gives us a mapping g : [ω2 ]2 −→ ω with the property that
{ξ < α : g(ξ, α) = g(ξ, β)} is finite for all α < β < ω2 . Define f : ω2 ×ω2 −→
ω by letting f (α, β) = 2g{α, β} + 2 when α < β, f (α, β) = 2g{α, β} + 1
when α > β, and f (α, β) = 0 when α = β. Then f is not constant on any
product of two infinite sets.
a
19.13 Remark. The relative size of θ2 (or its higher-dimensional analogues
θ3 , θ4 , . . .) in comparison to the sequence of cardinals ω2 , ω3 , ω4 , . . . is of
considerable interest, both in Set Theory and Model Theory (see e.g. [63],
136
I. Coherent Sequences
[82], [87]). On the other hand, even the following most simple questions,
left open by Theorem 19.12, are still unanswered.
19.14 Question. Can one prove any of the bounds like θ2 ≤ ω3 , θ3 ≤ ω4 ,
θ4 ≤ ω5 , etc. without appealing to additional axioms?
Note that by Corollary 19.10, Chang’s conjecture is equivalent to the
statement that within every decomposition of the usual ordering on ω2 as
an increasing chain of tree orderings, one of the trees has an uncountable
chain. Is it possible to have decompositions of ∈¹ ω2 into an increasing ω1 chain of tree orderings of countable heights? It turns out that the answer to
this question is equivalent to a different well-known combinatorial statement
about ω2 rather than Chang’s conjecture itself. Recall that f : [κ+ ]2 −→ κ
is transitive if
f (α, γ) ≤ max{f (α, β), f (β, γ)} whenever α ≤ β ≤ γ.
(I.275)
Given a transitive map f : [κ+ ]2 −→ κ, one defines ρf : [κ+ ]2 −→ κ
recursively on α ≤ β < κ as follows:
ρf (α, β) is equal to the supremum of:
1. f (min(Cβ \ α), β),
2. tp(Cβ ∩ α),
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 corresponding
ρf : [κ+ ]2 −→ κ has the following properties:
(a) ρf (α, γ) ≤ max{ρf (α, β), ρf (β, γ)} whenever α ≤ β ≤ γ,
(b) ρf (α, β) ≤ max{ρf (α, γ), ρf (β, γ)} whenever α ≤ β ≤ γ,
(c) |{ξ ≤ α : ρf (ξ, α) ≤ ν}| ≤ |ν| + ℵ0 for ν < κ and α < κ+ ,
(d) ρf (α, β) ≥ f (α, β) for all α < β < κ+ .
Proof. Proofs of (a)-(c) are almost identical to the corresponding proofs
for the function ρ itself and so we avoid repetition. The inequality (d) is
deduced from (I.276) by using the transitivity of f as follows:
ρf (α, β) ≥ max f (βi+1 , βi ) ≥ f (α, β),
i<n
where β = β0 > β1 > . . . > βn−1 > βn = β is the walk from β to α along
the given C-sequence.
a
19.16 Remark. Transitive maps are frequently used combinatorial objects,
especially when one works with quotient structures. Adding the extra subadditivity condition 19.15(b), one obtains a considerably more subtle object
which is much less understood.
19. Unbounded functions on successors of regular cardinals
137
19.17 Definition. Let fα : κ −→ κ (α < κ+ ) be a given sequence of
functions such that fα <∗ fβ whenever α < β (i.e. {ν < κ : fα (ν) ≥ fβ (ν)}
is bounded in κ whenever α < β). Then the corresponding transitive map
f : [κ+ ]2 −→ κ is defined by
f (α, β) = min{µ < κ : fα (ν) < fβ (ν) for all ν ≥ µ}.
(I.277)
Let ρf be the corresponding ρ-function that dominates this particular f
and for ν < κ let <fν 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 of
everywhere dominance. Then for every ν < κ the tree (κ+ , <fν ) 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 sequence
fα (α < κ+ ) in the ordering <∗ of eventual dominance, then the new
sequence f¯α = min{fα , g} (α < κ+ ) is still strictly <∗ -increasing but now
bounded by g even in the ordering of everywhere dominance. So this proves
the following result of Galvin (see [34],[59]).
19.19 Corollary. The following two conditions are equivalent for every
regular cardinal κ.
(1) There is a sequence fα : κ −→ κ (α < κ+ ) which is strictly increasing
and bounded in the ordering of eventual dominance.
(2) The usual order-relation of κ+ can be decomposed into an increasing
κ-sequence of tree orderings of heights < κ.
a
19.20 Remark. The assertion that every strictly <∗ -increasing κ+ -sequence of functions from κ to κ is <∗ -unbounded is strictly weaker than Chang’s
conjecture and in the literature it is usually referred to as weak Chang’s
conjecture. This statement still has considerable large cardinal strength
(see [14]). Also note the following consequence of Corollary 19.19 which can
be deduced from Lemmas 19.1 and 19.2 above as well.
19.21 Corollary. If κ is a regular limit cardinal (e.g. κ = ω), then the
usual order-relation of κ+ can be decomposed into an increasing κ-sequence
of tree orderings of heights < κ.
a
138
I. Coherent Sequences
20. Higher dimensions
The reader must have noticed already that in this Chapter so far, we have
only considered functions of the form f : [θ]2 −→ I or equivalently sequences fα : α −→ I (α < θ) of one-place functions. To obtain analogous results about functions defined on higher-dimensional cubes [θ]n one
usually develops some form of stepping-up procedure that lifts a function
of the form f : [θ]n −→ I to a function of the form g : [θ+ ]n+1 −→ I.
The basic idea seems quite simple. One starts with a coherent sequence
eα : α −→ θ (α < θ+ ) of one-to-one mappings and wishes to define
g : [θ+ ]n+1 −→ I as follows:
g(α0 , α1 , . . . , αn ) = f (e(α0 , αn ), . . . , e(αn−1 , αn )).
(I.279)
In other words, we use eαn to send {α0 , . . . , αn−1 } to the domain of f and
then apply f to the resulting n-tuple. The problem with such a simpleminded definition is that for a typical subset Γ of θ+ , the sequence of restrictions eδ ¹ (Γ∩δ) (δ ∈ Γ) may not cohere, so we cannot produce a subset
of θ that would correspond to Γ and on which we would like to apply some
property of f . It turns out that the definition (I.279) is basically correct except 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-sequence
Cα (α < θ+ ) on θ+ . The following result shows that if the C-sequence is
carefully chosen, the function τ will serve as a stepping-up tool.
20.1 Lemma. Suppose ρ0 and τ are based on some ¤θ -sequence Cα (α <
θ+ ) and let κ be a regular uncountable cardinal ≤ θ. Then every set Γ ⊆ θ+
of order-type κ contains a cofinal subset ∆ such that, if ε = sup(Γ) =
sup(∆), then ρ0 (ξ, ε) = ρ0 (ξ, τ (α, β, γ)) for all ξ < α < β < γ in ∆.
Proof. For δ ∈ lim(Cε ), let ξδ be the minimal element of Γ above δ. Let εδ
be the maximal point of the trace Tr(δ, ξδ ) of the walk from ξδ to δ along
the ¤θ -sequence Cα (α < θ+ ). Thus εδ = δ or εδ = min(Tr(δ, ξδ ) \ (δ + 1))
and Cξ ∩ δ is bounded in δ for all ξ ∈ Tr(δ, ξδ ) strictly above εδ . Let f (δ)
be a member of Cε ∩ δ which bounds all these sets. By the Pressing Down
Lemma there is a stationary set Σ ⊆ lim(Cε ) and δ̄ ∈ Cε such that f (δ) ≤ δ̄
for all δ ∈ Σ. Shrinking Σ we may assume that ξγ < δ whenever γ < δ are
members of Σ. Note that
εγ = min(Tr(ξα , ξγ ) ∩ Tr(ξβ , ξγ )) for every α < β < γ in Σ.
(I.281)
In other words, τ (ξα , ξβ , ξγ ) = εγ for all α < β < γ in Σ. By the definition
of εδ for δ ∈ lim(Cε ), δ belongs to Cεδ and so by the implicit assumption
20. Higher dimensions
139
that nonlimit points of Cεδ are successor ordinals, we conclude that δ is a
limit point of Cεδ . Thus the section (ρ0 )δ of the function ρ0 is an initial
part of both the section (ρ0 )ε and (ρ0 )εδ , which is exactly what is needed
for the conclusion of Lemma 20.1.
a
Recall that for a given C-sequence Cα (α < θ+ ) such that tp(Cα ) ≤ θ
for all α < θ+ , the range of ρ0 is the collection of all finite sequences of
ordinals < θ. It will be convenient to identify Qθ with θ via, for example,
Gödel’s coding of sequences of ordinals by ordinals. This identification gives
us a well-ordering <w of Qθ of length θ . We use this identification to liftup of an arbitrary map f : [θ]n −→ I (really, f : [Qθ ]n −→ I) to a map
g : [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 particular
example.
20.2 Theorem. Suppose θ is an arbitrary cardinal for which ¤θ holds.
Suppose further that for some regular κ > ω and integer n ≥ 2 there is a
map f : [θ]n −→ [[θ]<κ ]<κ such that:
(1) A ⊆ min(a) for all a ∈ [θ]n and A ∈ f (a).
(2) For all ν < κ and Γ ⊆ θ of size κ there exist a ∈ [Γ]n and A ∈ f (a)
such that tp(A) ≥ ν and A ⊆ Γ.
Then θ+ and κ satisfy the same combinatorial property, but with n + 1 in
place of n.
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 )−1
ε (f (ρ0 (α0 , ε), . . . , ρ0 (αn−1 , ε))),
(I.283)
where ε = τ (αn−2 , αn−1 , αn ), and where ρ0 and τ are based on some fixed
¤θ -sequence. Note that the transformation does not necessarily preserve
(1), so we intersect each member of a given g(a) with min(a) in order to
satisfy this condition. To check (2), let Γ ⊆ θ be a given set of size κ. By
Lemma 20.1, shrinking Γ we may assume that Γ has order-type κ and that
if ε = sup(Γ), then
ρ0 (ξ, ε) = ρ0 (ξ, τ (α, β, γ)) for all α < β < γ in Γ.
(I.284)
It follows that g restricted to [Γ]n+1 satisfies the formula
g(α0 , . . . , αn−1 , αn ) = (ρ0 )−1
ε (f (ρ0 (α0 , ε), . . . , ρ0 (αn−1 , ε))).
(I.285)
140
I. Coherent Sequences
Shrinking Γ further we assume that the mapping (ρ0 )ε , as a map from
the ordered set (θ+ , ∈) into the ordered set (Qθ , <w ) is strictly increasing
when restricted to Γ. Given an ordinal ν < κ, we apply (2) for f to the
set ∆ = {ρ0 (α, ε) : α ∈ Γ} and find a ∈ [∆]n and A ∈ f (a) such that
tp(A) ≥ ν and A ⊆ ∆. Let {α0 , . . . , αn−1 } be the increasing enumeration
−1
of the preimage (ρ0 )ε (a) and pick αn ∈ Γ above αn−1 . Let B be the
−1
preimage (ρ0 )ε (A). Then B ∈ g(α0 , . . . , αn−1 , αn ), tp(B) ≥ ν and B ⊆ Γ.
This finishes the proof.
a
20.3 Remark. It should be noted that Velleman [93] was the first to attempt to step-up the combinatorial property appearing in Theorem 20.2
using his version of the gap-2 morass. Unfortunately the stepping up procedure of [93] worked only up to the value n = 3 so it remains unclear if the
higher-gap morass could be used to reach all the other values of n.
If we apply this stepping-up procedure to the projection [[··]] of the squarebracket operation defined in 7.14, one obtains analogues of families G, H
and K of Example 7.16 for ω2 instead of ω1 . To make this precise, we
first fix a ¤ω1 -sequence Cα (α < ω2 ) and consider the corresponding maps
ρ0 : [ω2 ]2 −→ ω1<ω and ρ : [ω2 ]2 −→ ω1 . Let ρ̄ : [ω2 ]2 −→ ω1 be defined by
ρ̄(α, β) = 2ρ(α,β) · (2 · tp{ξ ≤ α : ρ(ξ, α) ≤ ρ(α, β)} + 1).
Then for α < ω2 , the mapping ρ̄α : α −→ ω1 that sends ξ to ρ̄(ξ, α) is
1-1 and we have the following coherence property:
ρ̄α = ρ̄β ¹ α whenever α ∈ lim(Cβ ).
(I.286)
20.4 Definition. The ternary operation [[[··]]] : [ω2 ]3 −→ ω2 is defined as
follows:
[[[αβγ]]] = (ρ̄τ (α,β,γ) )−1 [[ρ̄τ (α,β,γ) (α) ρ̄τ (α,β,γ) (β)]].
It should be clear that the stepping-up method lifts the property 7.15 of
[[··]] to the following property of the ternary operation [[[··]]]:
20.5 Lemma. Let A be a given uncountable family of pairwise disjoint
subsets of ω2 , all of some fixed size n. Let ξ1 , . . . , ξn be a sequence of ordinals
< ω2 such that for all i = 1, . . . , n and all but countably many a ∈ A we have
that ξi < a(i). Then there exist a,b and c in A such that [[[a(i)b(i)c(i)]]] = ξi
for i = 1, . . . , n.
a
As 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 two
projections of [[[··]]]. Set
G = {G ∈ [ω2 ]<ω : [[[αβγ]]]0 = 0 for all {α, β, γ} ∈ [G]3 },
20. Higher dimensions
141
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 subsets
of ω2 there is an arbitrarily large finite subset B of A with {b(0) : b ∈ B} ∈ G
and {b(1) : b ∈ B} ∈ H.
a
Recall that the parameter in all our definitions so far is a fixed ¤ω1 sequence, so we also have the corresponding ρ-function at our disposal. We
use it to define an analogue of the family K of Example 7.16 as follows. In
this case, let K be the collection of all finite sets {ai : i < n} of elements of
[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,
[[[ai (0)aj (0)ak (0)]]]1 = [[[ai (1)aj (1)ak (1)]]]1 =
[
al .
(I.288)
(I.289)
l<i
The following property of K is a consequence of Lemma 20.5 and the
subadditivity 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 there
exist ξ0 < ξ1 < . . . < ξn−1 < ω1 such that {aξi : i < n} ∈ K.
a
The particular definition of K is made in order to have the following
property of K which is analogous to (I.79) of Section 7:
If K and L are two distinct members of K, then there are no
more than 9 ordinals α such that {α, β} ∈ K and {α, γ} ∈ L
for some β 6= γ.
(I.290)
The definitions of G, H and K immediately lead to the following analogues
of (I.77) and (I.78):
G and H contain all singletons, are closed under subsets
and are 3-orthogonal to each other.
(I.291)
G and H are both 5-orthogonal to the family of all unions
of members of K.
(I.292)
So we can proceed as in Example 7.16 and for a function x from ω2 into
R, define
142
I. Coherent Sequences
kxkH,2 = sup{(
X
1
x(α)2 ) 2 : H ∈ H},
α∈H
kxkK,2 = sup{(
X
1
(x(α) − x(β))2 ) 2 : K ∈ K}.
{α,β}∈K
Let k·k = max{k·k∞ , k·kH,2 , k·kK,2 }, let Ē2 = {x : kxk < ∞}, and let E2
be the closure of the linear span of {1α : α ∈ ω2 } inside (Ē2 , k · k). Then as
in Example 7.16 one shows, using the properties of G, H and K listed above
in Lemmas 20.6 and 20.7 and in (I.287-I.292), that E2 is a Banach space
with {1α : α ∈ ω2 } as its transitive basis and with the property that every
bounded linear operator T : E2 −→ E2 can be written as S + D, where S is
an operator with separable range and D is a diagonal operator relative to
the basis {1α : α ∈ ω2 } with only countably many changes of the constants.
The new feature here is the use of (I.287) in checking that the projections
on countable sets of the form {α < β : ρ(α, β) < ν} (α < ω2 , ν < ω1 ) are
(uniformly) bounded. This is needed among other things in showing, using
Lemma 20.6, that T can be written as a sum of one such projection and a
diagonal operator relative to {1α : α ∈ ω2 }.
Using the interpolation method of [11], one can turn E2 into a reflexive
example. Thus, we have the following:
20.8 Example. Assuming ¤ω1 , there is a reflexive Banach space E with a
transitive basis of type ω2 with the property that every bounded operator
T : E −→ E can be written as a sum of an operator with a separable range
and a diagonal operator (relative to the basis) with only countably many
changes of constants.
a
20.9 Remark. In [42], P.Koszmider has shown that such a space cannot
be constructed on the basis of the usual axioms of set theory. We refer
the reader to that paper for more details about these kinds of examples of
Banach spaces.
For the rest of this section we shall examine the stepping-up method with
less restrictions on the given C-sequence Cα (α < θ+ ) on which it is based.
20.10 Theorem. The following are equivalent for a regular cardinal θ such
that log θ+ = θ. 39
(1) There is a substructure of the form (θ++ , θ+ , <, . . .) with no substructure B of size θ+ with B ∩ θ+ of size θ.
(2) There is f : [θ++ ]3 −→ θ+ which takes all the possible values on the
cube of any subset Γ of θ++ of size θ+ .
39 log
κ = min{λ : 2λ ≥ κ}.
20. Higher dimensions
143
Proof. To prove the nontrivial direction from (1) to (2), we use Lemma 19.8
and choose a strongly unbounded and subadditive e : [θ++ ]2 −→ θ+ . We
also choose a C-sequence Cα (α < θ+ ) such that tp(Cα ) ≤ θ for all α < θ+
and consider the corresponding function ρ∗ : [θ+ ]2 −→ θ defined above in
I.264. Finally we choose a one-to-one sequence rα (α < θ++ ) of elements of
+
{0, 1}θ and consider the corresponding function ∆ : [θ++ ]2 −→ θ+ :
∆(α, β) = ∆(rα , rβ ) = min{ν : rα (ν) 6= rβ (ν)}.
(I.293)
The definition of f : [θ++ ]3 −→ θ is given according to the following two
rules applied to a given triple x = {α, β, γ} ∈ [θ++ ]3 (α < β < γ):
Rule 1: If ∆(rα , rβ ) < ∆(rβ , rγ ) and rα <lex rβ <lex rγ or rα >lex
rβ >lex rγ , let
f (α, β, γ) = min(Pν (∆(β, γ)) \ ∆(α, β)),
where ν = ρ∗ (min{ξ ≤ ∆(α, β) : ρ∗ (ξ, ∆(α, β)) 6= ρ∗ (ξ, ∆(β, γ))}, ∆(β, γ)).
Rule 2: If α ∈ x is such that rα is lexicographically between the other
two rξ ’s for ξ ∈ x, if β ∈ x \ {α} is such that ∆(rα , rβ ) > ∆(rα , rγ ), where
γ is the remaining element of x and if x does not fall under Rule 1, let
f (α, β, γ) = min(Pν (e(β, γ)) \ e(α, β)),
where ν = ρ∗ {∆(α, β), e(β, γ)}.
The proof of the Theorem is finished once we show the following: for
every stationary set Σ of cofinality θ ordinals < θ+ and every Γ ⊆ θ++ of
size θ+ there exist α < β < γ in Γ such that f (α, β, γ) ∈ Σ. Clearly we may
assume that Γ has order-type θ+ . Let RΓ = {rα : α ∈ Γ}.
Case 1: RΓ contains a subset of size θ+ which is lexicographically wellordered or conversely well-ordered. Going to a subset of Γ, we may assume
that
∆(rα , rβ ) = ∆(rα , rγ ) for all α < β < γ in Γ.
(I.294)
It follows that Rule 1 applies in the definition of f (α, β, γ) for any triple
α < 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 range
of [··] on [∆]2 , where [··] is the square-bracket operation on θ+ defined as
follows:
[αβ] = min(Pν (β) \ α),
(I.295)
where ν = ρ∗ (min{ξ ≤ α : ρ∗ (ξ, α) 6= ρ∗ (ξ, β)}, β). Note that this is the
exact analogue of the square-bracket operation that has been analysed above
in Section 7. So exactly as in Section 7, one establishes the following fact.
For ∆ ⊆ θ+ unbounded, the set {[αβ] : {α, β} ∈ [∆]2 }
contains almost all ordinals < θ+ of cofinality θ.
(I.296)
144
I. Coherent Sequences
Case 2: RΓ contains no subsets of size θ+ which are lexicographically
+
well-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 of
+
{0, 1}<θ of size θ and height λ < θ+ , or a downwards closed Aronszajn
+
subtree of {0, 1}<θ of height θ+ . For β ∈ Γ, let
eβ 00 Γ+ = {e{β, γ} : γ ∈ Γ, rβ <lex rγ }.
(I.297)
Case 2.1: There exist θ+ many β ∈ Γ such that the set eβ 00 Γ+ is bounded
for all β ∈ Γ. Choose an elementary submodel M of some large enough
structure of the form Hκ such that δ = M ∩ θ+ ∈ Σ and M contains all
the relevant objects. By the elementarity of M , strong unboundedness of
e and the fact that T contains no θ+ -chain, there exist β and γ in Γ \ M
such that rγ <lex rβ , ε = e{β, γ} ≥ δ and ∆(rβ , rγ ) < δ. Let ν̄ = ρ∗ (δ, ε).
Note that there exist θ many ξ > ∆(rβ , rγ ) such that (rβ ¹ ξ)a 0 belongs to
M and rβ (ξ) = 1. This is clear in the case when T is an Aronszajn tree,
by the elementarity of M and the fact that cf(δ) = θ. However, this is also
true in case T has height λ < θ+ and has size θ, since from our assumption
log θ+ = θ the ordinal λ must have cofinality exactly θ. So by the property
19.1 of ρ∗ there is one such ξ such that ν = ρ∗ (ξ, ε) ≥ ν̄. Let t = (rβ ¹ ξ)a 0.
Then t ∈ T ∩ M and so Γt is a subset of Γ of size θ+ which belongs to M .
Using the strong unboundedness of e and the elementarity of M , there must
be an α in Γt ∩ M such that e(α, β) > sup(Pν (ε) ∩ δ). Since e(α, β) belongs
to eα 00 Γ+ , a set which, by the assumption of Case 2.1, is bounded in θ+ , we
conclude that e(α, β) < δ. It follows that f (α, β, γ) is defined according to
Rule 2, and therefore
f (α, β, γ) = min(Pν (ε) \ e(α, β)) = δ.
(I.298)
Case 2.2: For every β in some tail of Γ the set eβ 00 Γ+ is unbounded in
θ . Going to a tail, we assume that this is true for all β ∈ Γ. Let M and
δ = M ∩ θ+ ∈ Σ be chosen as before. If T is Aronszajn, using elementarity
of M , there must be β ∈ Γ ∩ M and δ̄ < δ of cofinality θ such that there
exist unboundedly many ξ below δ̄ such that the set
+
Ωξ = {e{β, χ} : χ ∈ Γ, rβ <lex rχ , ∆(rβ , rχ ) = ξ}
(I.299)
is unbounded in θ+ . If T is a tree of height λ < θ+ , δ̄ = λ will satisfy this
for 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 ) < δ̄. Let
ν̄ = ρ∗ (δ, ε). By the property 19.1 of ρ∗ , there is µ̄ < δ̄ such that ρ∗ (ξ, ε) ≥ ν̄
for all ξ ∈ [µ̄, δ̄). Choose ξ ∈ [µ̄, δ̄) above ∆(rβ , rγ ) such that the set Ωξ is
unbounded in θ+ . Let ν = ρ∗ (ξ, ε). Then ν ≥ ν̄. The set Pν (ε) ∩ δ being
of size < θ is bounded in δ. Since Ωξ ∈ M , there is α ∈ Γ ∩ M above β
20. Higher dimensions
145
such that ∆(rβ , rα ) = ξ, rβ <lex rα and e(β, α) > sup(Pν (ε) ∩ δ). Note that
rα <lex rγ , so the definition of f {α, β, γ} obeys Rule 2. Applying this rule,
we get
f {α, β, γ} = min(Pν (ε) \ e(β, α)) = δ.
(I.300)
This finishes the proof of Theorem 20.10.
a
20.11 Theorem. If θ is a regular strong limit cardinal carrying a nonreflecting stationary set, then there is f : [θ+ ]3 −→ θ which takes all the values
from 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 us
only indicate the adjustments. By Corollary 19.21 and Lemma 19.7, we
can choose a strongly unbounded subadditive map e : [θ+ ]2 −→ θ. By
the assumption about θ we can choose a C-sequence Cα (α < θ) avoiding
a 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 sequence
rα (α < θ+ ) of elements of {0, 1}θ and consider the corresponding function
∆ : [θ+ ]2 −→ θ:
∆(α, β) = ∆(rα , rβ ) = min{ν < θ : rα (ν) 6= rβ (ν)}.
(I.301)
The definition of f : [θ+ ]3 −→ θ is given according to the following rules,
applied to a given x ∈ [θ+ ]3 .
Rule 1: If x = {α < β < γ}, ∆(rα , rβ ) < ∆(rβ , rγ ) and rα <lex rβ <lex
rγ , or rα >lex rβ >lex rγ , let
f {α, β, γ} = [∆(α, β)∆(β, γ)].
Rule 2: If α ∈ x is such that rα is lexicographically between the other
two rξ ’s for ξ ∈ x, if β ∈ x \ {α} is such that ∆(rα , rβ ) > ∆(rα , rγ ), where
γ is the remaining element of x, and they do not satisfy the conditions of
Rule 1, set
f {α, β, γ} = min(Tr(∆(α, β), e{β, γ}) \ e{α, β}),
i.e. f {α, β, γ} is the minimal point on the trace of the walk from e{β, γ}
to ∆(α, β) above the ordinal e{α, β}; if such a point does not exist, set
f {α, β, γ} = 0.
The proof of the Theorem is finished once we show that for every stationary Ω ⊆ Σ and every Γ ⊆ θ+ of size θ, there exist x ∈ [Γ]3 such that
f (x) ∈ Ω. This is done, as in the proof of Theorem 20.10, by considering
the two cases. Case 1 is reduced to the property 18.2 of the square-bracket
operation. The treatment of Case 2 is similar and a bit simpler than in the
proof of Theorem 20.10, since the case that T is of height < θ is impossible
due to our assumption that θ is a strong limit cardinal. This finishes the
proof.
a
146
I. Coherent Sequences
Since log ω1 = ω, the following is an immediate consequence of Theorem
20.10.
20.12 Theorem. Chang’s conjecture is equivalent to the statement that for
every f : [ω2 ]3 −→ ω1 there is uncountable Γ ⊆ ω2 such that f 00 [Γ]3 6= ω1 .
a
20.13 Remark. Since this same statement is stronger for functions from
higher dimensional cubes [ω2 ]n into ω1 the Theorem 20.12 shows that they
are all equivalent to Chang’s conjecture. Note also that n = 3 is the minimal
dimension for which this equivalence holds, since the case n = 2 follows from
the Continuum Hypothesis, which has no relationship to Chang’s conjecture.
For the rest of this section we examine the stepping-up procedure without
the assumption that some form of Chang’s conjecture is false. So let θ be
a given regular uncountable cardinal and let Cα (α < θ+ ) be a fixed Csequence such that tp(Cα ) ≤ κ for all α < θ+ . Let ρ∗ : [θ+ ]2 −→ θ be
the ρ∗ -function defined above in (I.264). Recall that, in case Cα (α < θ+ )
is a ¤θ -sequence, the key to our stepping-up procedure was the function
τ : [θ+ ]3 −→ θ+ defined by the formula (I.280). Without the assumption of
Cα (α < θ+ ) being a ¤θ -sequence, the following related function turns out
to be a good substitute: χ : [θ+ ]3 −→ ω defined by
χ(α, β, γ) = |ρ0 (α, γ) ∩ ρ0 (β, γ)|.
(I.302)
Thus χ(α, β, γ) is equal to the length of the common part of the walks
γ → α and γ → β.
20.14 Definition. A subset Γ of θ+ is stable if χ is bounded on [Γ]3 .
20.15 Lemma. Suppose that Γ is a stable subset of θ+ of size θ. Then
{ρ∗ (α, β) : {α, β} ∈ [Ω]2 } is unbounded in θ for every Ω ⊆ Γ of size θ.
Proof. Suppose ρ∗ is bounded by ν on [Ω]2 for some Ω ⊆ θ+ of size θ. We
need to show that χ is unbounded on [Ω]3 . Therefore we may assume that
Ω is of order-type θ and let λ = sup(Ω). Construct a sequence Ω = Ω0 ⊇
Ω1 ⊇ . . . of subsets of Ω of order-type θ such that:
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 of
some large enough structure of the form Hκ such that M ∩ θ ∈ θ and M
contains all the relevant objects. Now choose another elementary submodel
N of Hκ which contains M as well as all the other relevant objects such
that N ∩ θ ∈ θ. Let λN = sup(N ∩ λ). Choose γ ∈ Ωn above λN . Let
γ = γ0 > . . . > γk = λN be the walk from γ to λN . Let γ̄ be the γi with
20. Higher dimensions
147
the smallest index such that Cγi ∩ λN is unbounded in λN . Then γ̄ 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 is
bounded in λN .
(I.304)
Then {γi : i < k} \ γ̄ ⊆ Tr(α, γ) for all α ∈ N ∩ Ωn above λ0 . So if we
choose α < β in N ∩ Ω0 such that Cγ̄ ∩ [α, β) 6= ∅, then
Tr(α, γ) ∩ Tr(β, γ) = {γi : i ≤ k} \ γ̄.
(I.305)
From this and our assumption χ(α, β, γ) ≥ n we conclude that this set has
size at least n. For M ∈ M, let λM = sup(M ∩λ) and C = {λM : M ∈ M}.
Then C is a closed and unbounded subset of λ of order-type θ. By our
assumption that ρ∗ is bounded by ν on [Ω]2 , we get:
tp(Cγ̄ ∩ λN ) ≤
≤
≤
sup{tp(Cγ̄ ∩ α) : α ∈ Ω ∩ [λ0 , λN )}
sup{ρ∗ (α, γ) : α ∈ Ω ∩ [λ0 , λN )}
ν.
(I.306)
Since C ∩ [λ0 , λN ) has order-type > ν there must be M ∈ M ∩ N such
that λM ∈
/ Cγ̄ and λM > λ0 . Thus, we have an elementary submodel M
of Hκ which contains all the relevant objects and a γ ∈ Ωn above λM such
that the walk γ → λM contains all the points from the set {γi : i < k} \ γ̄,
but includes at least the point min(Cγ̄ \ λM ) which is strictly above λM .
Hence, if γ = γ0 > γ1 > . . . > γl = λM is the trace of γ → λM and if γ̂
is the γi (i ≤ l) with the smallest index, subject to the requirement that
Cγ̂ ∩ λM is unbounded in λM , then {γi : i < l} \ γ̂ has size at least n + 1.
Let λ̂0 ∈ λ ∩ M be such that:
λ̂0 > max(Cγi ∩ λM ) for all i < l for which Cγi ∩ λM is
bounded in λM .
(I.307)
Then {γi : i < l} \ γ̄ ⊆ Tr(α, γ) for all α ∈ M ∩ Ωn above λ̂0 . Using the
elementarity of M we conclude that for every δ < λ there exist ε ∈ Ωn \ δ
and a set aε ⊆ [δ, ε) of size at least n + 1 such that aε ⊆ Tr(α, ε) for all
α ∈ Ωn ∩ [λ̂0 , δ). Hence we can choose a closed and unbounded set D of
[λ̂0 , 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 ∈ [λ̂0 , δ),
(I.308)
ε(δ) < min(D \ (δ + 1)).
(I.309)
148
I. Coherent Sequences
Finally let Ωn+1 = {ε(δ) : δ ∈ D}. Then
aε(γ) ⊆ Tr(ε(α), ε(γ)) ∩ Tr(ε(β), ε(γ)) for all α < β < γ in ∆.
(I.310)
This gives us that χ(α, β, γ) ≥ n + 1 for every triple α < β < γ in Ωn+1 ,
finishing the inductive step and therefore the proof of Lemma 20.15.
a
20.16 Definition. The 3-dimensional version of the oscillation mapping,
osc : [θ+ ]3 −→ ω, is defined on the basis of the 2-dimensional version of
Section 17 as follows
osc(α, β, γ) = osc(Cβs \ α, Cγt \ α),
where s = ρ0 (α, β) ¹ χ(α, β, γ) and t = ρ0 (α, γ) ¹ χ(α, β, γ).
In other words, we let n be the length of the common part of the two
walks γ → α and γ → β. Then we consider the walks γ = γ0 > . . . > γk = α
and β = β0 > . . . > βl = α from γ to α and β to α respectively; if both k
and l are bigger than n (and note that k ≥ n), i.e. if γn and βn are both
defined, we let osc(α, β, γ) be equal to the oscillation of the two sets Cβn \ α
and Cγn \ α. If not, we let osc(α, β, γ) = 0.
20.17 Lemma. Suppose that Γ is a subset of θ+ of size κ, a regular uncountable cardinal, and that every subset of Γ of size κ is unstable. Then for
every positive integer n, there exist α < β < γ in Γ such that osc(α, β, γ) =
n.
Proof. We may assume that Γ is actually of order-type κ and let λ = sup(Γ).
Let M be an ∈-chain of length κ of elementary submodels M of Hθ++ such
that M ∩ κ ∈ κ and M contains all the relevant objects. For M ∈ M, let
λM = sup(λ ∩ M ). Now choose another elementary submodel N of Hθ++
containing M as well as all the other relevant objects with N ∩ κ ∈ κ. Let
λN = sup(λ ∩ N ) and choose γ ∈ Γ above λN and let γ = γ0 > . . . > γk =
λN be the walk from γ to λN . Let γ̄ and λ0 be chosen as above in (I.304).
Then γ̄ = γk or γ̄ = γk−1 and
Tr(α, γ) ∩ Tr(β, γ) = {γi : i ≤ k} \ γ̄ for all α < β in
Γ ∩ [λ0 , λN ) such that Cγ̄ ∩ [α, β) 6= ∅.
(I.311)
So if a tail of the set {λM : M ∈ M ∩ N } is included in Cγ̄ , then using the
elementarity of N , we would be able to construct a cofinal subset Γ0 ⊆ Γ
such that χ is constantly equal to k̄ on [Γ0 ]3 , where k̄ is the place of γ̄ in
the sequence γ0 , . . . , γk . This, of course, would contradict our assumption
that Γ contains no large stable subsets. It follows that the set of all λN for
M ∈ M ∩ N which do not belong to Cγ̄ is cofinal in λN . So we can pick
M1 ∈ M2 ∈ . . . ∈ Mn+1 in M ∩ N such that, if λi = λMi (1 ≤ i ≤ n + 1),
then:
λi ∈ [λ0 , λ) \ Cγ̄ for all 1 ≤ i ≤ n + 1,
(I.312)
20. Higher dimensions
149
[λi , λi+1 ) ∩ Cγ̄ 6= ∅ for all 1 ≤ i ≤ n.
(I.313)
Fix α ∈ Γ ∩ M1 above max(Cγ̄ ∩ λ1 ). For 1 ≤ i ≤ n, pick λ̄i ∈ [λi , λi+1 ) ∩
Mi+1 such that λ̄i ≥ max(Cγ̄ ∩ λi+1 ). For 1 ≤ i ≤ n, let I¯i = [λi , λ̄i ], and
let I¯n+1 = [λn , γ]. Now set F to be the collection of all increasing sequences
I1 < I2 < . . . < In+1 of closed intervals of ordinals < λ with the property
that there is β = β( I~ ) in Γ ∩ In+1 such that, if β = β0 > . . . > βl = α is
the walk from β to α, then:
l > k̄ and βk̄ = min(Tr(α, β) ∩ In+1 ),
Cβk̄ \ α ⊆
n+1
[
Ii ,
(I.314)
(I.315)
i=1
Cβk̄ ∩ Ii 6= ∅ for all 1 ≤ n ≤ n + 1.
(I.316)
Clearly, F ∈ M1 and hI¯1 , . . . , I¯n+1 i ∈ F , as γ is a witness of this. So
working as in the proof of Lemma 17.2, we can find F0 ⊆ F in M1 such
that, if some sequence J~ of length ≤ n (including the case J~ = ∅) endextends to a member of F0 , then for every ξ < λ there is a closed interval
K included in [ξ, λ) such that the concatenation J~a K also end-extends to
a sequence in F0 . Working inductively on 1 ≤ i ≤ n + 1 we can select a
sequence I1 , I2 , . . . , In such that:
I1 , . . . , Ii ∈ Mi+1 for 1 ≤ i ≤ n,
(I.317)
Ii ⊆ [λ̄i , λi+1 ) for 1 ≤ i ≤ n.
(I.318)
hI1 , . . . , Ii i end-extends to a member of F0 .
(I.319)
Clearly, there is no problem in choosing these objects, using the elementarity
of the models Mi (1 ≤ i ≤ n + 1) and the splitting property of F0 . Working
in Mn+1 , we find an interval In+1 such that hI0 , . . . , In , In+1 i belongs to
F0 . Let β = β(hI0 , . . . , In+1 i). Since Cγ̄ intersects the interval [λ1 , λ̄1 )
it separates α from β, so by (I.311) we get that τ (α, β, γ) = k̄, hence by
definition
osc(α, β, γ) = osc(Cβk̄ \ α, Cγk̄ \ α).
(I.320)
Recall that γk̄ = γ̄, so referring to (I.313),(I.315) and (I.316) we conclude
that I1 ∩Cβk̄ , . . . , In−1 ∩Cβk̄ and (In ∪In+1 )∩Cβk̄ are the n convex pieces in
which the set Cγ̄ \ α splits the set Cβk̄ \ α. Hence osc(Cβk̄ \ α, Cγk̄ \ α) = n.
This completes the proof of Lemma 20.17.
a
Applying the last two Lemmas to the subsets of θ+ of size θ, we get an
interesting dichotomy:
20.18 Lemma. Every Γ ⊆ θ+ of size θ can be refined to a subset Ω of size
θ such that either:
150
I. Coherent Sequences
(1) ρ∗ is unbounded and therefore strongly unbounded on Ω, or
(2) the oscillation mapping takes all its possible values on the cube of Ω.
a
We finish the section with a typical application of this dichotomy.
20.19 Theorem. Suppose θ is a regular cardinal such that log θ+ = θ.
Then there is f : [θ++ ]3 −→ ω which takes all the values from ω on the cube
of any subset of θ++ of size θ+ .
Proof. We choose two C-sequences Cα (α < θ+ ) and Cα+ (α < θ++ ) on θ+
and θ++ respectively, such that tp(Cα ) ≤ θ for all α < θ+ and tp(Cα+ ) ≤
θ+ for all α < θ++ . Let ρ∗ : [θ+ ]2 −→ θ and ρ∗+ : [θ++ ]2 −→ θ+ be
the corresponding ρ∗ -functions defined above in I.264. Also choose a one+
to-one sequence rα (α < θ++ ) of elements of {0, 1}θ and consider the
corresponding function ∆ : [θ++ ]2 −→ θ+ defined in (I.293). We define
f : [θ++ ]3 −→ θ+ according to the following two cases for a given triple
α < β < γ of elements of θ++ .
Case 1: (Cβs ∩ Cγt ) \ α 6= ∅, where s = ρ0 (α, β) ¹ χ(α, β, γ) and t =
ρ0 (α, γ) ¹ χ(α, β, γ) of course assuming that ρ0 (α, β) has length at least
χ(α, β, γ).
Rule 1: If ∆(rα , rβ ) < ∆(rβ , rγ ) and rα <lex rβ <lex rγ or rα >lex
rβ >lex rγ , set
f (α, β, γ) = min(Pν (∆(β, γ)) \ ∆(α, β)),
where ν = ρ∗ (min{ξ ≤ ∆(α, β) : ρ∗ (ξ, ∆(α, β)) 6= ρ∗ (ξ, ∆(β, γ))}, ∆(β, γ)).
Rule 2: If ᾱ ∈ {α, β, γ} is such that rᾱ is lexicographically between
the other two rξ ’s for ξ ∈ {α, β, γ}, if β̄ ∈ {α, β, γ} \ {ᾱ} is such that
∆(rᾱ , rβ̄ ) > ∆(rᾱ , rγ̄ ), where γ̄ is the remaining member of {α, β, γ}, and
if {α, β, γ} does not fall under Rule 1, let
f (α, β, γ) = min(Pν (ρ∗+ {β̄, γ̄}) \ ρ∗+ (α, β)),
where ν = ρ∗ {∆(α, β), ρ∗+ (β, γ)}.
Case 2: (Cβs ∩ Cγt ) \ α = ∅, where s = ρ0 (α, β) ¹ χ(α, β, γ) and t =
ρ0 (α, γ) ¹ χ(α, β, γ) of course assuming that ρ0 (α, β) has length at least
χ(α, β, γ). Let
f (α, β, γ) = osc(α, β, γ).
If a given triple α < β < γ does not fall into one of these two cases, let
f (α, β, γ) = 0.
The proof of Theorem 20.19 is finished if we show that for every Γ ⊆ θ++
of size θ+ , the image f 00 [Γ]3 either contains all positive integers or almost
all ordinals < θ+ of cofinality θ.
20. Higher dimensions
151
So let Γ be a given subset of θ++ of order-type θ+ and let λ = sup(Γ).
By the proof of Lemma 20.15, if Γ is stable, then it must contain a (stable)
subset Ω such that for some n < ω and all α < β < γ in Ω:
χ(α, β, γ) = n,
(Cβs ∩ Cγt ) \ α 6= ∅ where s = ρ0 (α, β) ¹ n and t = ρ0 (α, γ) ¹ n.
(I.321)
(I.322)
In other words, Case 1 of the definition of f (α, β, γ) applies to any triple
α < β < γ from Ω. Note that in this case, the definition follows exactly the
definition of the analogous function in the proof of Theorem 20.10 with ρ∗+
in the role of the strongly unbounded subadditive function e : [θ++ ]2 −→ θ+ .
By Lemma 20.15, ρ∗+ is strongly unbounded on subsets of Ω, so by the proof
of Theorem 20.10 we conclude that in this case the image f 00 [Ω]3 contains
almost all ordinals < θ+ of cofinality θ.
Suppose now that no subset of Γ of size θ+ is stable. By Lemma 20.17 (in
fact, its proof) for every integer n ≥ 1 there exist α < β < γ in Γ such that
for s = ρ0 (α, β) ¹ χ(α, β, γ) and t = ρ0 (α, γ) ¹ χ(α, β, γ), where ρ0 (α, β)
has length > χ(α, β, γ):
osc(α, β, γ) = n,
(I.323)
(Cβs ∩ Cγt ) \ α = ∅.
This finishes the proof.
(I.324)
a
20.20 Corollary. There is f : [ω2 ]3 −→ ω which takes all the values on the
cube of any uncountable subset of ω2 .
a
20.21 Remark. Note that the dimension 3 in this Corollary cannot be
lowered to 2 as long as one does not use some additional axioms to construct
such f . Note also that the range ω cannot be replaced by a set of bigger
size, as this would contradict Chang’s conjecture. We have seen above that
Chang’s conjecture is equivalent to the statement that for every f : [ω2 ]3 −→
ω1 there is an uncountable set Γ ⊆ ω2 such that f 00 [Γ]3 6= ω1 . Is there a
similar reformulation of the Continuum Hypothesis? More precisely, one
can ask the following question.
20.22 Question. Is CH equivalent to the statement that for every
f : [ω2 ]2 −→ ω there exists an uncountable Γ ⊆ ω1 with f 00 [Γ]2 6= ω?
152
I. Coherent Sequences
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Index
<c , 6
<p , 29
<r , 6
<x , 29
<ρ̄ , 31
D, 104
F, 5
Fn (α), 109
G 7−→ G∗ , 15
Gα , 8
KΓ , 62
Sθ , 93
T (ρ0 ), 8
Wγ , 94
[αβ], 52
[αβ]∗ , 127
[αβ]∗h , 128
[··]h , 127
∆, 105
∆-function, 105
Λ, 38
Λ + n, 38
αt , 120
ρ̄, 29
ρ̄1 , 75
χV , 13
≤, 46
F-Cohen, 32
F-Souslin, 32
GΓ , 57
I, 92
Uf , 93
| αβ |, 60
ρ, 26
ρ0 , 4
ρ1 , 9
ρ2 , 36
ρ3 , 37
ρa , 33
σ-sequentially closed, 114
σ(α, β), 52
τe , 33
[[αβ]], 57
e 1 ), 12
A(ρ
Te(ρ1 ), 12
Vet , 13
a(n) , 48
d, 93
fα (n), 109
CK(θ), 111
lim(D), 119
osc, 148
action, 120
Aronszajn tree, 8, 44, 67
avoid ∆, 122
Bernstein decomposition, 114
Chang’s conjecture, 133
clique, 15
cofinal branch, 67
coherent C-sequence, 80
coherent family, 115
coherent function sequence, 36
coherent mapping, 44
coherent tree, 44
corresponding transitive map, 137
division on ordinals, 82
160
INDEX
161
Eberlein compactum, 13
full code, 4
full lower trace, 5
Hausdorff gap, 34, 44
irreducible, 80
Jensen matrix, 112
Kurepa
Kurepa
Kurepa
Kurepa
family,
family,
family,
family,
105
cofinal, 111
compatibility, 111
extendibility, 111
last step, 37
lower trace, 4
Maharam type, 110
maximal weight, 9
metric, 28
minimal walk, 3, 51
nontrivial, 36, 44
nowhere dense, 44
number of steps, 36
orthogonal, 38, 44
P-ideal, 38
P-ideal dichotomy, 38
positive with respect to a filter, 124
property ∆, 108
property C, 14
property C 00 , 14
Radon measure space, 110
Radon probability space, 110
right lexicographical ordering, 6
sequential fan, 93
sequentially closed, 114
shift, 48
sigma-product, 115
sigma-product, index-set, 115
Souslin, 32
special Aronszajn tree, 8
special square sequence, 87
special tree, 67
square-bracket operation, 52
square-sequence, 80
stable, 146
stationary subsequence, 119
step, 3
stepping-up, 138
strong measure zero, 14
strongly unbounded function, 132
subadditivity, 28
support, 44
tightness, 93
trace, 4, 51
trace filter, 123
transitive map, 33, 136
tree on θ, 86
trivial C-sequence, 76
Tukey equivalent, 110
Tukey reducible, 110
ultra-metric, 28
ultra-subadditivity, 28
unbounded family, 118
unbounded function, 131
upper trace, 4
walk, 3
weak Chang’s conjecture, 137
weight, 3

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