INFINITE RAMSEY THEORY Contents 1. Ramsey's theorem 3 1.1. The arrow-notation 3 1.2. The ...

64 Pages·2009·0.39 MB·English

INFINITE RAMSEY THEORY
STEFAN GESCHKE
Contents
1. Ramsey’s theorem 3
1.1. The arrow-notation 3
1.2. The finite and infinite versions of Ramsey’s theorem 3
1.3. Limitations 6
1.4. Compactness: The infinite Ramsey theorem implies the
finite 6
1.5. Uncountable versions of Ramsey’s theorem 8
1.6. Weakly compact cardinals 12
2. Continuous Ramsey theory 13
2.1. Polish spaces 13
2.2. The Baire property 16
2.3. Galvin’s theorem 18
2.4. Covering the plane by functions 21
2.5. Homogeneity numbers and other cardinal invariants 24
2.6. The special role of 2ω and ωω 29
2.7. Constructing c 31
max
2.8. Continuous n-colorings 35
3. Open colorings 37
ˇ
4. Stone-Cech compactifications of discrete semigroups and
Ramsey theory 43
4.1. Compact right-topological semigroups 43
ˇ
4.2. Stone-Cech compactifications of discrete semigroups 44
4.3. The theorems of Hales-Jewett and van der Waerden 48
4.4. Hindman’s theorem 52
5. Metric Ramsey theory 55
5.1. Embedding sequences into R 55
5.2. Metric spaces with a set of non-zero distances bounded
from below and above 59
1
2 STEFAN GESCHKE
5.3. A Ramsey-type theorem for infinite metric spaces 60
5.4. A Ramsey-type theorem for complete metric spaces. 61
References 64
INFINITE RAMSEY THEORY 3
1. Ramsey’s theorem
1.1. The arrow-notation.
Definition 1.1. a) For a set X and a cardinal ν let
[X]ν = {Y ⊆ X :|Y|= ν},
[X]≤ν = {Y ⊆ X :|Y|≤ ν},
and
[X]<ν = {Y ⊆ X :|Y|< ν}.
b) For a natural number n and a cardinal µ, an n-coloring on X
with with µ colors is a function c : [X]n → µ. If µ is not specified
(and not clear from the context), we usually mean µ = 2. Also, if n
is not specified (and not clear from the context), we mean n = 2. In
particular, a coloring on X typically is a function c : [X]2 → 2 and an
n-coloring on X typically is a function c : [X]n → 2.
c) If c is an n-coloring on X with µ colors, then H ⊆ X is c-
homogeneous or just homogeneous if c is constant on [H]n. H is homo-
geneous of color i ∈ µ if c is constant on H with value i.
d)Letκandλbecardinals. Wewriteλ → (κ)n ifforeveryn-coloring
µ
on λ with µ colors there is a homogeneous set of size κ.
Lemma 1.2. Let κ(cid:48) ≤ κ, λ(cid:48) ≥ λ, and µ(cid:48) ≤ µ. Then λ → (κ)n implies
µ
λ(cid:48) → (κ(cid:48))n
µ(cid:48)
Proof. Easy. (cid:3)
Lemma 1.3. Let κ,λ,n,µ be as before and let n(cid:48) ≤ n. Then λ → (κ)n
µ
implies λ → (κ)n(cid:48)
µ
Proof. Let c(cid:48) be an n(cid:48)-coloring on λ with µ colors. We define an n-
coloring on λ as follows. Let {x ,...,x } ∈ [λ]n. We may assume that
1 n
x < ··· < x . Put c(x ,...,x ) = c(cid:48)(x ,...,x ).
1 n 1 n 1 n(cid:48)
Now, if H ⊆ λ is c-homogeneous, then H is c(cid:48)-homogeneous as well.
This shows the lemma. (cid:3)
1.2. The finite and infinite versions of Ramsey’s theorem.
Theorem 1.4 (Ramsey’s theorem, infinite version). For all n,m ∈ ω
with n,m > 0, ℵ → (ℵ )n.
0 0 m
4 STEFAN GESCHKE
Proof. We first observe that ℵ → (ℵ )n follows from ℵ → (ℵ )n by
0 0 m 0 0 2
induction on the number m of colors.
Assume ℵ → (ℵ )n and ℵ → (ℵ )n. Let c be an n-coloring on ω
0 0 m 0 0 2
with m+1 colors. Define an n-coloring c(cid:48) on ω with 2 colors as follows:
Let

0, if c(x ,...,x ) < m
c(cid:48)(x ,...,x ) = 1 n
1 n
1, if c(x ,...,x ) = m.
1 n
Now, by ℵ → (ℵ )n there is an infinite c(cid:48)-homogeneous set H ⊆ ω. If
0 0 2
c(cid:48) has the value 1 on [H]n, then H is c-homogeneous of color m. If the
value of c(cid:48) is 0 on all n-element subsets of H, then c assumes only m
different colors on [H]n and the existence of an infinite c-homogeneous
subset of H follows from ℵ → (ℵ )n. This shows ℵ → (ℵ )n .
0 0 m 0 0 m+1
Hence we may restrict our attention to the case m = 2. We show
ℵ → (ℵ )n by induction on n. Note that for n = 1 the statement is
0 0 2
just the familiar Pigeon Hole Principle. Now assume ℵ → (ℵ )n. We
0 0 m
will show ℵ → (ℵ )n+1.
0 0 m
Let c be an (n+1)-coloring on ω with two colors. We define a strictly
increasing sequence (a ) of natural numbers, a decreasing sequence
k k∈ω
(A ) of infinite subsets of ω and a sequence (i ) of elements of 2.
k k∈ω k k∈ω
Let a = 0 and A = ω. Suppose we have already defined a and A .
0 0 k k
Let c : [A \(a +1)]n → 2 be defined by
k+1 k k
c (x ,...,x ) = c(a ,x ,...,x ).
k+1 1 n k 1 n
Using the induction hypothesis ℵ → (ℵ )n, there is an infinite c -
0 0 2 k+1
homogeneous set A ⊆ A \ (a + 1) of some color i . Let a be
k+1 k k k k+1
the least element of A . The finishes the recursive construction of
k+1
the three sequences.
Let A = {a : k ∈ ω}. Given k ,...,k ∈ ω with k < ··· < k we
k 0 n 0 n
have
c(a ,...,a ) = c (a ,...,a ) = i .
k0 kn k0+1 k1 kn k0
In other words, the color of the (n+1)-element set {a ,...,a } only
k0 kn
depends on its smallest element. Let C ⊆ ω be an infinite set such that
for all k ∈ C, i is the same i ∈ 2. Let H = {a : k ∈ C}. Now H is
k k
infinite and c-homogeneous of color i. (cid:3)
Exercise 1.5. Show that every sequence (x ) of real numbers has
n n∈ω
an infinite subsequence that is decreasing or increasing.
INFINITE RAMSEY THEORY 5
Theorem 1.6 (Ramsey’s theorem, finite version). For all n,m,k ∈ ω
with m > 0 there is (cid:96) ∈ ω such that (cid:96) → (k)n.
m
It is convenient to introduce Ramsey numbers for the proof of this
theorem.
Definition 1.7. Let n > 0. For each k ∈ ω let R(k;n) denote the least
cardinal (cid:96) such that (cid:96) → (k)n. Note that Theorem 1.4 guarantees the
2
existence of such a cardinal (cid:96).
Proof of Theorem 1.6. An inductive argument similar to the one used
in the proof of the infinite version of Ramsey’s theorem shows that we
may restrict our attention to the case m = 2. Also, it is easily seen
that 2k → (k)1.
2
Now assume that for some n > 0 and all k ∈ ω, R(k;n) is finite. Fix
k ∈ ω. We show that there is (cid:96) ∈ ω with (cid:96) → (k)n+1.
2
We start by defining a sequence ((cid:96) ) of natural numbers. Let
j j∈ω
(cid:96) = 1. Suppose (cid:96) has been chosen for some j ∈ ω. Let
0 j
(cid:96) = R((cid:96) ;n)+1.
j+1 j
This finishes the definition of the (cid:96) . Now let (cid:96) = (cid:96) .
j 2k
Claim 1.8. (cid:96) → (k)n+1
2
To show the claim, let c : [(cid:96)]n+1 → 2. We define an increasing
sequence (a ) of elements of (cid:96), a decreasing sequence (A ) of
j j<2k j j<2k
subsets of (cid:96) and a sequence (i ) of colors in 2.
j j<2k
Let A = (cid:96). Suppose for some j < 2k we have chosen A ⊆ (cid:96) of size
0 j
(cid:96) . Let a be the least element of A . Define
2k−j j j
c : [A \{a }]n → 2
j j j
by c (x ,...,x ) = c(a ,x ,...,x ). The set A \{a } is of size
j 1 n j 1 n j j
(cid:96) −1 = R((cid:96) ;n).
2k−j 2k−j−1
It follows that there is a c -homogeneous set A ⊆ A \{a } of some
j j+1 j j
color i ∈ 2 of size (cid:96) = (cid:96) . This concludes the definition of
j 2k−j−1 2k−(j+1)
the three sequences.
Now, given j < ··· < j < 2k, we have
0 n
c(a ,...,a ) = c (a ,...,a ) = i .
j0 jn j0 j1 jn j0
6 STEFAN GESCHKE
In particular, the color of {a ,...,a } only depends on j . Choose a
j0 j1 0
setC ⊆ 2k ofsizek suchthatforsomefixedi ∈ 2andallj ∈ C wehave
i = i. Let H = {a : j ∈ C}. H is of size k and c-homogeneous. (cid:3)
j j
Exercise 1.9. Show that for k ∈ ω we have 9k → (k)2.
2
Hint: Go through the proof of Theorem 1.6 and do some explicit
computations.
1.3. Limitations.
Theorem 1.10. a) 2ℵ0 (cid:54)→ (3)2
ℵ0
b) 2ℵ0 (cid:54)→ (ℵ )2 (Sierpin´ski)
1 2
Proof. a) Consider the set 2ω of all sequences of zeroes and ones of
length ω. Given {x,y} ∈ [2ω]2 let c(x,y) be the least n ∈ ω such that
x(n) (cid:54)= y(n). It is easily checked that there is now c-homogeneous set
of size three.
b) Let < denote the usual linear order in R. Choose a well-ordering
≺ of R. Define c : [R]2 → 2 by

0, if < and ≺ agree on {x,y}
c(x,y) =
1, otherwise.
Now every homogeneous set of color 0 is an increasing well-ordered
subset of R and every homogeneous set of color 1 is a reversely well-
ordered subset of R. But since not uncountable ordinal order-embeds
into R, the c-homogeneous sets cannot be uncountable. (cid:3)
Exercise 1.11. Show that for n ∈ ω with n > 1, 2n (cid:54)→ (3)2.
n
Exercise 1.12. Sierpin´ski’s example relies on the fact that ω , the first
1
uncountable ordinal, does not order-embed into R. Give a proof of that
fact.
Hint: R has a countable dense subset.
1.4. Compactness: The infinite Ramsey theorem implies the
finite.
Definition 1.13. A tree is a partial order (T,<) such that for all t ∈ T
the set {s ∈ T : s < t} is well-ordered by <. If (T,<) is a tree and
t ∈ T, then the height ht(t) is the ordertype of {s ∈ T : s < t}. Given
INFINITE RAMSEY THEORY 7
an ordinal α, the α-th level of T is the set
Lev (T) = {t ∈ T : ht(t) = α}.
α
The height ht(T) of T is the least α with Lev (T) = ∅. For t ∈ T let
α
succ (t) = {s ∈ T : t < s∧ht(s) = ht(t)+1}
T
denote the set of immediate successors of t. T is finitely branching if
succ (t) is finite for all t ∈ T. T is rooted if it has a unique minimal
T
element, i.e., if Lev (T) is a singleton. A branch of T is a maximal
0
linearly ordered subset of T.
Theorem 1.14 (Ko¨nig’s lemma). Every infinite, finitely branching,
rooted tree has an infinite branch.
Proof. Let T be an infinite, finitely branching, rooted tree. We choose
a strictly increasing infinite chain (t ) in T. By Zorn’s lemma, this
n n∈ω
chain extends to an infinite branch of T.
Let t be the root of T, i.e., the unique minimal element. Suppose
0
for some n ∈ ω, t has been defined such that
n
T(t ) = {s ∈ T : s < t ∨s = t ∨s > t }
n n n n
(cid:83)
isinfinite. Sincesucc (t )isfiniteandT(t ) = {T(s) : s ∈ succ (t )},
T n n T n
there is t ∈ succ (t ) such that T(t ) is infinite. This finishes the
n+1 T n n+1
construction of the sequence (t ) . (cid:3)
n n∈ω
Exercise 1.15. 2<ω is the tree of all finite sequences of zeroes and
ones. The order on 2<ω is proper set-theoretic inclusion. Let S ⊆ 2<ω
be such that for every x ∈ 2ω there is s ∈ S such that s is an initial
segment of x, i.e., s ⊆ x. Show that there is a finite set S ⊆ S such
0
that for all x ∈ 2ω there is s ∈ S with s ⊆ x.
0
Hint: Suppose this fails for some S ⊆ 2<ω and consider the collection
of those t ∈ 2<ω that don’t have an initial segment in S. Use Ko¨nig’s
lemma.
For topologists: This exercise essentially asks you to use Ko¨nig’s
lemma to show the compactness of the Hausorff space 2ω.
We now give a proof of the finite version of Ramsey’s theorem from
the infinite.
8 STEFAN GESCHKE
Alternative proof of Theorem 1.6. Suppose for some k,m,n ∈ ω with
m,n > 0 there is no (cid:96) ∈ ω such that (cid:96) → (k)n. Then for each (cid:96) ∈ ω,
m
the set
C = {c : c is an n-coloring on (cid:96) with m colors without
(cid:96)
a c-homogeneous set of size k}.
is finite and nonempty.
Let T = (cid:83) C . T, ordered by proper set-theoretic inclusion (cid:40), is
(cid:96)∈ω (cid:96)
an infinite tree and for all (cid:96) ∈ ω there is (cid:96)(cid:48) ∈ ω such that Lev (T) = C .
(cid:96) (cid:96)(cid:48)
It follows that T is finitely branching. The lowest level Lev (T) has
0
exactly one element, the empty coloring.
Hence K¨onig’s lemma applies to T and therefore T has an infinite
(cid:83)
branch {c : (cid:96) ∈ ω}. Now c = c is an n-coloring on ω with m
(cid:96) (cid:96)∈ω (cid:96)
colors without a homogeneous set even of size k. But this constradicts
ℵ → (ℵ )n. (cid:3)
0 0 m
1.5. Uncountable versions of Ramsey’s theorem. In Lemma 1.10
we observed that the natural generalization of the infinite version of
Ramsey’s theorem to the uncountable fails. However, there is an un-
countable version of Ramsey’s theorem if you choose the cardinal on
the left hand side of the arrow relation sufficiently large.
Definition 1.16. For a cardinal κ let (cid:105) (κ) = κ. If (cid:105) (κ) has been
0 α
defined for some ordinal α, let (cid:105) (κ) = 2(cid:105)α(κ). If α is a limit ordinal
α+1
and (cid:105) (κ) has been defined for all β < α, let
β
(cid:105) (κ) = sup{(cid:105) (κ) : β < α}.
α β
By (cid:105) we denote (cid:105) (ℵ ).
α α 0
Theorem 1.17 (Erdo˝s-Rado). For all n ∈ ω and every infinite cardinal
κ,
((cid:105) (κ))+ → (κ+)n+1.
n κ
In particular, (2κ)+ → (κ+)2 and therefore (2ℵ0)+ → (ℵ )2 .
κ 1 ℵ0
Proof. First, we consider the case n = 0. We have to show
κ+ → (κ+)1.
κ
But this is just an instance of the pigeon hole principle: If κ+ is parti-
tioned into κ classes, one of the classes has to be of size κ+.
INFINITE RAMSEY THEORY 9
Now let n > 0 and assume
((cid:105) (κ))+ → (κ+)n.
n−1 κ
Let c : [λ]n+1 → κ, where λ = ((cid:105) (κ))+. For each a ∈ λ let the coloring
n
c : [λ]n → κ be defined by
a
c (x ,...,x ) = c(x ,...,x ,a).
a 0 n−1 0 n−1
Claim 1.18. There is a set A ⊆ λ of size (cid:105) (κ) such that for every
n
set B ⊆ A of size (cid:105) (κ) and every b ∈ λ\B there is a ∈ A\B such
n−1
that c and c agree on [B]n.
a b
To show the claim, we construct an increasing chain (A )
α α<((cid:105)n−1(κ))+
of subsets of λ of size (cid:105) (κ). We start with an arbitrary set A ⊆ λ
n 0
of size (cid:105) (κ). Suppose we have chosen A for some α < ((cid:105) (κ))+.
n α n−1
Observe that there are
(cid:105) (κ)(cid:105)n−1(κ) = 2(cid:105)n−1(κ) = (cid:105) (κ)
n n
subsets of A of size (cid:105) (κ). Given a set B ⊆ A of size (cid:105) (κ),
α n−1 α n−1
there are
((cid:105) (κ))κ ≤ (cid:105) (κ)
n−1 n
functions from [B]n to κ.
Choose A ⊆ λ such that A ⊆ A , |A |= (cid:105) (κ) and such
α+1 α α+1 α+1 n
that for every B ⊆ A of size (cid:105) (κ) and every b ∈ λ \ B there is
α n−1
a ∈ A such that c and c agree on [B]n. This is possible since there
α+1 a b
are not too many B ⊆ A of size (cid:105) (κ) and functions c (cid:22) [B]n.
α n−1 b
If α < ((cid:105) (κ))+ is a limit ordinal, let
n−1
(cid:91)
A = {A : β < α}.
α β
This finishes the construction of the sequence (A ) .
α α<((cid:105)n−1(κ))+
Let A = (cid:83){A : α < ((cid:105) (κ))+}. Now, whenever B ⊆ A is of size
α n−1
(cid:105) (κ), then there is α < ((cid:105) (κ))+ such that B ⊆ A . If b ∈ λ\B,
n−1 n−1 α
then by the choice of A , there is a ∈ A ⊆ A such that c and c
α+1 α+1 a b
agree on [B]n. This shows that A works for the claim.
Continuing the proof of the Erd˝os-Rado theorem, let A ⊆ λ be as
in the claim. Choose a ∈ λ\A. Recursively, we construct a sequence
(x ) of pairwise distinct elements of A such that for all
α α<((cid:105)n−1(κ))+
10 STEFAN GESCHKE
α < ((cid:105) (κ))+, c agrees with c on [{x : β < α}]n. This is possible
n−1 xα a β
by the choice of A.
Now let X = {x : β < ((cid:105) (κ))+}. Define d : [X]n → κ by letting
β n−1
d(x ,...,x ) = c (x ,...,x ) = c(a,x ,...,x ).
α0 αn−1 a α0 αn−1 α1 αn
By the choice of the x , for all α < ··· < α < ((cid:105) (κ))+,
α 0 n n−1
(1) c(x ,...,x ) = c (x ,...,x )
α0 αn xαn α0 αn−1
= c (x ,...,x ) = d(x ,...,x ).
a α0 αn−1 α0 αn−1
By ((cid:105) (κ))+ → (κ+)n, there is a d-homogeneous set H ⊆ X of size
n−1 κ
κ+. H is in fact c-homogeneous by equation (1). (cid:3)
Exercise 1.19. Let κ be an infinite cardinal a consider the set 2κ
of all sequences of zeroes and ones of length κ. Let < denote the
lex
lexicographic order on 2κ, i.e., let x < y if x (cid:54)= y and for the smallest
lex
α < κ with x(α) (cid:54)= y(α) we have x(α) < y(α).
Show that 2κ does not contain strictly increasing or decreasing se-
quences of length κ+.
Instructions: Assume 2κ contains a strictly increasing sequence of
lengthκ+. Letγ ≤ κbetheminimalordinalsuchthatthereisastrictly
increasing sequence (x ) in 2γ (with respect to the lexicographic
α α<κ+
ordering on 2γ). For each α < κ+ let ξ < γ be the unique ordinal with
α
x (cid:22) ξ = x (cid:22) ξ and x (ξ ) = 0 and x (ξ ) = 1.
α α α+1 α α α α+1 α
How many possibilities are there for ξ ? Can γ really be minimal?
α
Arrive at a contradiction.
Exercise 1.20. Show that for every infinite cardinal κ, 2κ (cid:54)→ (κ+)2.
2
Hint: Generalize Theorem 1.10 b) using Exercise 1.19.