[This document was typeset using Luca Trevisan’s LaTeX2WP. I will refer to result
n (or definition n) from last lecture as 3.n.]
A. The Galvin-Hajnal rank and an improvement of Theorem
3.1
Last lecture, I covered the first theorem of the Galvin-Hajnal paper and several corollaries. Recall that the result, Theorem 3.1,Qstates that if κ and λ are uncountable regular cardinals, and λ is κ-inaccessible,
then α<κ κα < ℵλ for any sequence (κα : α < λ)
Q
of cardinals such that α<β κα < ℵλ for all β < κ.
In particular (see, for example, Corollary 3.7), if cf(ξ) > ω and ℵξ is strong limit,
then 2ℵξ < ℵ(|ξ|cf(ξ) )+ .
The argument relied in the notion of an almost disjoint transversal. Assume that
κ is regular and uncountable, and recall that if A = (Aα : α < κ) is a sequence
of sets,
(A) = sup{|F| : F is an a.d.t. for A}. Here, F is an a.d.t. for A iff
Q then TQ
F ⊆
A := α<κ Aα and whenever f 6= g ∈ F, then {α < κ : f (α) = g(α)} is
bounded.
With
Q 3.1 was proved by showing that there is an a.d.t. for
Q κ, λ as above, Theorem
( β<α κβ : α < κ) of size α<κ κα , and then proving that, provided that |Aα | < ℵλ
for all α < κ, then T (A) < ℵλ .
In fact, the argument showed a bit more. Recall that if f : κ → ORD, then ℵf =
(ℵf (α) : α < κ). Then, for any f : κ → λ, T (ℵf ) < ℵλ .
The proof of this result was inductive, taking advantage of the well-foundedness of the
partial order <b,κ defined on κ ORD by f <b,κ g iff {α : f (α) ≥ g(α)} is bounded in κ.
That <b,κ is well-founded allows us to define a rank kf kb for each f : κ → ORD, and
we can argue by considering a counterexample of least possible rank to the statement
from the previous paragraph.
In fact, more precise results are possible. Galvin and Hajnal observed that replacing
the ideal of bounded sets with the nonstationary ideal (or, really, any normal ideal),
results in a quantitative improvement of Theorem 3.1.
Definition 1 For f, g : κ → ORD, let f <NSκ g iff {α < κ : f (α) ≥ g(α)} is
nonstationary. Similarly, define f ≤NSκ g iff {α : f (α) > g(α)} is nonstationary,
etc.
Since NSκ , the collection of nonstationary subsets of κ is a κ-complete (in fact, normal)
ideal, then <NSκ is well-founded, and kf k = kf kNSκ is defined for any f : κ → ORD.
k · k is usually called the Galvin-Hajnal rank of f.
1
Before stating the result, Theorem 8 below, we prove some preliminary lemmas about
the behavior of the Galvin-Hajnal rank.
Definition 2 Let κ be a regular uncountable cardinal. The canonical functions for κ
are defined inductively: A function f : κ → ORD is an α-th canonical function iff
1. A β-th canonical function fβ : κ → ORD exists for all β < α,
2. For all β < α, there is a β-th canonical function fβ for κ such that fβ <NSκ f,
and
3. For all h : κ → ORD, if for all β < α there is some β-th canonical function fβ
for κ such that fβ <NSκ h, then f ≤NSκ h.
Of course, the idea is that an α-canonical function f witnesses kf k = α and is
canonical just as ordinals are canonical among well-ordered sets. We prove this in a
series of lemmas.
Lemma 3 If κ is regular and uncountable, and f, g are α-th canonical functions for
κ, then f =NSκ g.
Proof: By definition, if f and g are α-th canonical functions for κ, then both
f ≤NSκ g and g ≤NSκ f hold, so f =NSκ g. By the lemma, we can simply talk about the α-th canonical function, if one exists.
Lemma 4 Let κ be an uncountable regular cardinal, let f : κ → ORD, and let α be
such that the α-th canonical function fα exists. Then kf k > α iff fα <NSκ f.
Proof: By induction, kfα k ≥ α, so if fα <NSκ f then α < kf k.
To prove the other implication, argue by induction, and assume that for all β < α
and all g : κ → ORD, if kgk > β then fβ <NSκ g where fβ is the β-th canonical
function.
Suppose that kf k > α. Then there is some f 0 <NSκ f such that kf 0 k ≥ α. It follows
that for any β < α, kf 0 k > β. By induction, fβ <NSκ f 0 . By definition, fα ≤NSκ f 0 .
But then fα <NSκ f as wanted. Corollary 5 For any regular uncountable cardinal κ, if the α-th canonical function
fα for κ exists, then kfα k = α.
2
Proof: By induction on α, kfα k ≥ α. But kfα k > α is impossible by Lemma 4. It is easy to exhibit the first few canonical functions.
Lemma 6 Let κ be a regular uncountable cardinal.
1. The α-th canonical function fα exists for all α < κ+ .
2. If β < κ then kf k ≤ β iff {α : f (α) ≤ β} is stationary.
3. kf k ≤ κ iff {α : f (α) ≤ α} is stationary.
Proof: For α < κ we can set fα to be the function constantly equal to α, and we
can also set fκ to be the identity. It is straightforward from induction and Fodor’s
lemma that Definition 2 is satisfied. Items 2 and 3 follow now from Lemma 4.
In general, given κ < α < κ+ , choose a (nonstrictly) increasing sequence of ordinals
(iξ : ξ < κ) such that α = supξ (iξ + 1). This is possible, since cf(α) ≤ κ. Now set
fα (ξ) = supξ<α (fiξ (α) + 1) for all ξ < κ.
We argue by induction that Definition 2 is satisfied. For this, suppose that fα ≤
6 NSκ h.
Then {ξ < κ : h(ξ) < supρ<ξ (fiρ (ξ) + 1)} is stationary. By Fodor’s lemma, there is
some fixed ρ < κ such that {ξ < κ : h(ξ) ≤ fiρ (ξ)} is stationary. But then fiρ 6<NSκ h.
For κ regular uncountable and α ≥ κ+ , it may or may not be the case that the
κ+ -canonical function fκ+ exists.
For example, Galvin showed that the ω2 -th function f : ω1 → ω1 does not exist in L,
while in a model of Jech, Magidor, Mitchell, and Prikry, the α-th canonical function
f : ω1 → ω1 exists for all α. This statement is equiconsistent with the existence of a
measurable cardinal (the model is obtained by a construction that gives that NSω1 is
precipitous, and it is easy to see that the existence of fα for all α implies this). See
Thomas Jech, Menachem Magidor, William Mitchell, and Karel Prikry, Precipitous
Ideals, The Journal of Symbolic Logic, 45 (1) (Mar., 1980), 1–8.
If one is only interested in the existence of the α-th canonical function for ω1 for all
α < θ for some fixed θ, this does not require any large cardinals, and its consistency
was established in Thomas Jech, Saharon Shelah, A note on canonical functions,
Israel J. Math. 68 (3) (1989), 376–380.
Given an uncountable regular cardinal κ, forcing allows us to specify in which sense the
canonical functions for κ are, indeed, canonical: These are the functions f : κ → ORD
such that the ordinal [f ]G in the internal ultrapower V κ /G is independent of the NSκ generic filter G. This characterization is often taken in place of Definition 2.
Lemma 7 Let κ be an uncountable regular cardinal and let f : κ → ORD.
3
+

1. kf k < 
Y
|f (α)| .
f (α)6=0
2. If λ is regular uncountable and κ-inaccessible, and f : κ → λ, then kf k < λ.
Proof: Suppose g <NSκ f. Let
0
g (α) =
g(α) if g(α) < f (α),
0
otherwise.
Then g 0 =NSκ g so kg 0 k = kgk, g 0 (α) = 0 if f (α) = 0, and
Q if X = {α < κ : f (α) 6= 0},
0
to
X
is
in
the
set
product
then
the
restriction
of
g
α∈X f (α). There are at most
Q
0
since any ordinal below kf k is represented
α∈X |f (α)| many such functions g , and

+
Y
by one of them, it follows that kf k < 
|f (α)| , as wanted.
f (α)6=0
Item 2 follows immediately, since if f : κ → λ then f (α) < λ for all α, and δ =
supα f (α) < λ, by regularity. But then kf k < (δ κ )+ ≤ λ. Shelah has shown (by a more elaborate argument) that Lemma 7.2 still holds if one
weakens the κ-inaccessibility assumption on λ to the requirement that τ κ < λ for all
τ < λ of cofinality κ. This allows us to weaken the assumption in the same way in
the corollaries below.
We are now ready to state the second Galvin-Hajnal theorem.
For notational convenience, let’s write T (κ, δ) for T (A) where A = (Aα : α < κ) with
Aα = δ for all α < κ.
Theorem 8 (Galvin-Hajnal) Let κ be an uncountable regular cardinal, let (δα :
α < κ) be a (not necessarily strictly) increasing continuous sequence of cardinals, let
+ϕ(α)
for α < κ. If ∆ = 2κ supα<κ T (κ, δα ), then
ϕ : κ → ORD and set ψ(α) = δα
T (ψ) ≤ ∆+kϕk .
The proof of Theorem 8 is by induction on kϕk. When the δα are strictly increasing
and kϕk = 0, this is a result of Erdős, Hajnal, and Milner.
Corollary 9 If κ is uncountable regular and ϕ : κ → ORD, then T (ℵϕ ) ≤ (2κ )+kϕk .
Proof: In the notation of Theorem 8, take δα = ω for all α < κ, so ψ(α) = ℵϕ(α)
and T (ℵϕ ) = T (ψ). Note that ∆ = 2κ T (κ, ω) = 2κ . Notice that Theorem 3.1 follows immediately:
4
Corollary 10 Suppose that κ, λ are uncountable regular cardinals and λ is κ-inaccessible.
If f : κ → λ then T (ℵf ) < ℵλ .
Proof: By Corollary 9, T (ℵf ) ≤ (2κ )+kf k and by Lemma 7, kf k < λ. Corollary 11 Let κ, (δα )α<κ , ∆, ϕ be as in Theorem 8.
1. Let (τα : α < κ) be a sequence of cardinals such that ∀β < κ,
Q
Then α<κ τα ≤ ∆+kϕk .
Q
+ϕ(β)
α<β
τα ≤ δβ
.
2. Let (τα : α < κ) be an increasing sequence
of nonzero cardinals such that ∀β <
!κ
X
Q
+ϕ(β)
. Then
τα
≤ ∆+kϕk .
κ, α<β τα ≤ δβ
α<κ
3. Let (λα : α < κ) be a strictly increasing sequence of infinite cardinals. If
P
+ϕ(α)
λ = α<κ λα , ρ is a cardinal, and ∀α < κ, ρλα ≤ δα
, then ρλ ≤ ∆+kϕk .
Q
Proof: For 1, let Aα = β<α τβ and let F be an a.d.t. for (Aα : α < κ) of size
Q
+ϕ(α)
for all α, so |F| ≤ T (ψ) ≤ ∆+kϕk .
β<κ τβ . By assumption, |Aα | ≤ δα
!κ
X
For 2, recall from Corollary 8 in lecture II.2 that
τα
≤ ∆+kϕk .
α<κ
For 3, let τα = ρλα for α < κ. Then β<α τβ = ρ
and the result follows from item 1. Q
P
β<α
λβ
+ϕ(α)
≤ ρ λ α ≤ δα
for all α < κ,
Corollary 12 Let λ be a cardinal of uncountable cofinality κ, and suppose that λ is
κ-inaccessible. Let (λα : α < κ) be a strictly increasing and continuous sequence of
infinite cardinals cofinal in λ. Let ϕ : κ → ORD.
1. If ∀α < κ,
Q
+ϕ(α)
β<α
λβ ≤ λα
, then λκ ≤ λ+kϕk .
+ϕ(α)
2. If ∀α < κ, 2λα ≤ λα
, then 2λ ≤ λ+kϕk .
Q
+ϕ(α)
3. If 2τ ≤ λκ for all τ < λ and β<α λβ ≤ λα
for all α < κ, then 2λ ≤ λ+kϕk .
Proof:
In Corollary 11, take δα = τα = λα and ρ = 2. To prove 1, notice that
Q
κ
κ
κ
α λα = λ and ∆ = 2 supα T (κ, λα ) = supα λ = λ. Item 2 follows from Corollary
λ
11.3, and item 3 follows from the fact that 2 = λκ . Recall the very general statement of Silver’s theorem, Theorem 21 in lecture II.5:
5
Corollary 13 (Silver) Let λ be a κ-inaccessible cardinal of uncountable cofinality
κ. Let (λα : α < κ) be a strictly increasing sequence
of cardinals cofinal in λ. Suppose
Q
that there is some µ < κ such that {α < κ : β<α λβ ≤ λ+µ
α } is stationary in κ. Then
κ
+µ
λ ≤λ .
Proof: This follows from Corollary 12.1 and Lemma 6.2. For lack of time, I don’t include a full proof of Theorem 8 and instead present only
a sketch. To prove Theorem 8 two additional ingredients (whose proofs I skip) are
needed. One is the Erdős-Rado theorem, that we will prove in Chapter III.
Definition 14 Let κ, λ, ρ be cardinals and let n < ω. Given a set X, recall that
[X]n = {Y ⊆ X : |Y | = n}. The statement κ → (λ)nρ means that whenever f : [κ]n →
ρ, there is Y ⊆ κ, |Y | = λ, such that the restriction of f to [Y ]n is constant. If this
is not the case, we write κ 6→ (λ)nρ .
It is customary in the field to call the functions f as above colorings, and to refer to
the sets Y as homogeneous with respect to f.
This “arrow notation” turns to be very useful. For example, the infinitary version of
Ramsey’s theorem is simply ω → (ω)22 . A well known example due to Sierpinski shows
that c 6→ (ω1 )22 or, more generally, 2κ 6→ (κ+ )22 . This will be discussed in Chapter 3.
Theorem 15 (Erdős-Rado) For any infinite cardinal κ, (2κ )+ → (κ+ )2κ . The second ingredient is a result of Hajnal on set-mappings.
Definition 16 Let E be a set. A set-mapping on E is a function f : E → P(E)
such that ∀x ∈ E (x ∈
/ f (x)). Given such f, E, we say that H ⊆ E is free with respect
to f iff ∀x ∈ H (H ∩ f (x) = ∅).
For example, the function f : κ → P(κ) given by f (α) = {β < κ : β ∈ α} = α
is a set-mapping on κ, |f (α)| < κ for all α, and any free H ⊆ κ is a singleton.
Ruziewicz conjectured in 1936 that if one strengthens the cardinality restriction on f
to |f (α)| < λ for all α < κ, where λ < κ is some fixed cardinal, then there is now a
free H of size κ. This was shown by Hajnal in 1961.
Theorem 17 (Hajnal) Let κ and λ be cardinals with λ < κ and κ infinite. Let f
be a set-mapping on κ such that |f (α)| < λ for all α < κ. Then there is an H ⊆ κ of
size κ that is free with respect to f. 6
Proof of Theorem 8: Proceed by induction on kϕk.
Suppose first that kϕk = 0.
Let F be an a.d.t. for ψ. By Lemma 6.2, {α < κ : ϕ(α) = 0} is stationary. Let
X0 = {α : ϕ(α) = 0 and α is limit}, so X0 is also stationary. Note that f (α) < δα for
all f ∈ F and α ∈ X0 and there is therefore a βf (α) < α such that f (α) < δβf (α) . (It
is here that the continuity of the sequence (δα : α < κ) is used.)
By Fodor’s lemma, for each f ∈ F there is a stationary set Xf ⊆ X0 and an ordinal
βf < κ such that f (α) < δβf for any α ∈ Xf .SFor fixed sets X, β, let FX,β = {f ∈ F :
Xf = X and βf = β}, and notice that F = X,β FX,β .
Since there are only 2κ such pairs (X, β) and |FX,β | ≤ T (κ, δβ ) ≤ ∆, it follows that
|F| ≤ 2κ ∆ = ∆, and we are done.
Suppose now that kϕk = ν > 0 and that the result holds for all ϕ : κ → ORD and
corresponding ψ such that kϕk < ν.
Let F be an a.d.t. for ψ. For f ∈ F and α < κ let ϕf (α) be least such that |f (α)| ≤
ϕ (α)
+ϕ (α)
δα f , and let ψf (α) = δα f . Notice that ϕf <NSκ ϕ, since ϕf (α) ≥ ϕ(α) only if
ϕ(α) = 0, but the set of these ordinals α is nonstationary since kϕk > 0. It follows
that kϕf k < ν.
S
This shows that F = µ<ν Fµ , where Fµ = {f ∈ F : kϕf k = µ}, and therefore
|F| ≤ |ν| supµ<ν |Fµ |, and we are done if we show that Fµ ≤ ∆+ν for each µ < ν.
In fact, |Fµ | ≤ ∆µ+1 for all µ < ν. To see this, fix such a µ and begin by defining a
set-mapping H on Fµ by
Y
H(f ) = {g ∈ Fµ : g 6= f and g ∈
(f (α) + 1)}.
α<κ
ϕ (α)
Clearly, H(f ) is an a.d.t. for f +1. Note that for all α < κ, |f (α)+1| ≤ δα f = ψf (α),
so |H(f )| ≤ T (ψf ). Notice now that the hypotheses of the theorem apply to ϕf and
ψf in place of ϕ and ψ. Since kϕf k = µ < ν, by the induction hypothesis we have
that T (ψf ) ≤ ∆+µ .
Let λ = ∆+µ+1 , and ρ = ∆+µ+2 . If, towards a contradiction, |Fµ | ≥ ρ, then by
Theorem 17 there is a set F 0 ⊆ Fµ of size ρ that is free with respect to H.
Notice that ρ > ∆ ≥ 2κ , and therefore (since H is free) we can inductively find a
sequence (fξ : ξ < (2κ )+ ) with each fξ ∈ F 0 and fξ ∈
/ H(fη ) whenever ξ < η < (2κ )+ .
By definition of H, it follows that for all such ξ < η there is an α = α(ξ, η) < κ such
that fξ (α) > fη (α).
By the Erdős-Rado Theorem 15 there is an infinite set A ⊆ (2κ )+ homogeneous for
the coloring α : [(2κ )+ ]2 → κ. Let γ < κ be the constant value that the map α takes
on [A]2 . If ξ < η ∈ A, then fξ (γ) > fη (γ), and therefore there is an infinite decreasing
sequence of ordinals, contradiction. 7
I conclude with an application not mentioned in lecture.
Definition 18 A structure M = (M, R, . . . ) in a countable language with a distinguished unary relation (interpreted in M by R ⊆ M ) is said to be of type (κ, λ) iff
|M | = κ and R = λ.
Chang’s conjecture is the statement that whenever M is of type (ω2 , ω1 ), there is an
elementary substructure N ≺ M of type (ω1 , ω).
Chang’s conjecture is consistent. It holds under MM and can be forced from a Ramsey
cardinal. A large cardinal of comparable strength is required, since Silver showed that
Chang’s conjecture implies the existence of 0] .
By arguing about Skolem functions, one can show without too much effort that
Chang’s conjecture is equivalent to ω2 → [ω1 ]<ω
ω1 ,ω , the statement that whenever
<ω
f : [ω2 ]
→ ω1 , there is a subset A of ω2 of order type ω1 such that f ”[A]<ω is
countable.
Theorem 19 (Magidor) Chang’s conjecture implies that if ℵω1 is strong limit, then
2ℵω1 < ℵω2 .
Notice that from Corollary 3.7 from last lecture without assuming Chang’s conjecture
we have that if ℵω1 is strong limit then 2ℵω1 < ℵ(2ℵ1 )+ . It is open whether Theorem
19 holds without the additional assumption.
Proof: The following argument is due to Galvin.
By say, Corollary 12, it is enough to show that kf k < ω2 for all f : ω1 → ω1 .
To prove this, let f : ω1 → ORD be such that kf k ≥ ω2 . Using Lemma 6, for
β < ω2 let fβ be the β-th canonical function for ω1 , so fβ <NSκ fγ <NSκ f for all
β < γ < ω2 . Let Cβ,γ be a club witnessing this, so for all α < ω1 , if α ∈ Cβ,γ then
fβ (α) < fγ (α) < f (α).
T
Define h : [ω2 ]<ω → ω1 as follows: If x ⊆ ω2 is finite, let h(x) = min {Cβ,γ : β, γ ∈
x, β < γ}, where we take the minimum of an empty intersection to be zero.
By Chang’s conjecture, there is an A ⊆ ω2 of order type ω1 such that α = sup h00 [A]<ω <
ω1 .
We claim that f (α) ≥ ω1 , which concludes the proof.
To see the claim, notice first that α ∈ Cβ,γ whenever β, γ ∈ A and β < γ. Suppose
otherwise, and fix β1 < γ1 counterexamples. Then α1 := sup(Cβ1 ,γ1 ∩ (α + 1) =
max(Cβ1 ,γ1 ∩ α < α. By definition of α, this means that there must be some finite
x ⊂ A such that h(x) > α1 and therefore, by definition of h and α1 , it follows that
h(x ∪ {β1 , γ1 }) > α, contradiction.
8
But the claim concludes the proof, since then fβ (α) < fγ (α) < f (α) for all β < γ in
A, and therefore, since |A| = ω1 , we must have f (α) ≥ ω1 . Besides the original Galvin-Hajnal paper, a good reference for this lecture is the
book Paul Erdős, András Hajnal, Attila Máté, Richard Rado, Combinatorial set
theory: Partition relations for cardinals, North-Holland (1984).
9