On a topological Ramsey Theorem

We introduce natural strengthenings of sequential compactness called the $r$-Ramsey property for each natural number $r\geq 1$. We prove that metrizable compact spaces are $r$-Ramsey for all $r$ and give examples of compact spaces that are $r$-Ramsey but not $r+1$-Ramsey for each $r\geq 1$ (assuming CH for all $r>1$


Introduction
Let K be a compact space and let r be a positive integer. Following [2], we say that a function f : [S] r → K converges to p ∈ K if for every neighborhood U of p there is a finite set F such that f [S \ F ] r ⊆ U. Once this happens for some p, we say that f is convergent.
This notion, for r = 2 was introduced in [2] where the special case of our Theorem 2.1 was stated and proved. Their main motivation was to obtain idempotents in compact semigroups K as limits of certain functions f : [ω] 2 → K. We show that more general notion of a space satisfying the r-Ramsey property (Definition 2.2) given below is a quite natural strengthening of sequential compactness and the main motivation of this paper is to prove some basic facts about this class of spaces and describe some examples showing that r-Ramsey can be strictly weaker than (r + 1)-Ramsey.
Our topological terminology is standard and basic definitions and notions can be found in [4]. Set theoretic notation and terminology including some background on Ramsey's Theorem can be found in [6]. And for a more detailed analysis of almost disjoint families, Ψ spaces and the ideals FIN n , we refer the reader to [5].
2 A sequential Ramsey theorem Theorem 2.1. Let (K, ̺) be a compact metric space and let f : [N] r → K be an arbitrary function, where r > 0. Then there exists an infinite set B ⊆ N such that f ↾ [B] r converges to some element of K.
Proof. For each n ∈ ω choose a finite cover U n of K consisting of balls of radius 2 −n . Then U n induces a finite coloring of [N] r . Inductively, choose infinite sets N ⊇ A 0 ⊇ A 1 ⊇ · · · such that A n is monochromatic for the coloring induced by U n (here we have used the classical Ramsey's theorem). Let B be any infinite set such that B \ A n is finite for every n ∈ ω. By compactness, f ↾ [A] r is convergent.
The result above motivates the following definition: Definition 2.2. Let X be a topological space and let r ∈ N be positive. We shall say that X has the r-Ramsey property (or X is an r-Ramsey space) if for every function f : [N] r → X there exist p ∈ X and an infinite set B ⊆ N such that f ↾ [B] r converges to p. We shall say that X has the Ramsey property if it has the r-Ramsey property for every positive r ∈ N. We will say that the set B is a convergent subsequence of f .
Note that the 1-Ramsey property is just the sequential compactness. Recall that a topological space X is sequentially compact if every sequence in X has a convergent subsequence. Proposition 2.3. Every space with the r-Ramsey property has the (r − 1)-Ramsey property, whenever r > 1. In particular, every space with the r-Ramsey property for some r > 0 is sequentially compact.
Proof. Assume X has the r-Ramsey property and fix g : It is useful to note that if f : [S] r → X converges in a sequentially compact X, then f has a somewhat nice canonical subsequence. Note that the fact that g, as defined in the definition, is convergent and converges to x follows from the convergence of f to x.
Lemma 2.5. If X is sequentially compact, then any convergent f : [S] r → X has an r-nice convergent subsequence.
Proof. By induction on r. If r = 1 then this follows since X is sequentially compact. Fix r > 1 and f : [S] r → X convergent to some point x ∈ X. Enumerate [S] r−1 as Now take T to be a pseudo-intersection of the S n 's and note that g : [T ] r−1 → X defined by g(s) = x s is defined and, since f converges to x, so g also converges to x. And so by our induction hypothesis, g has an (r − 1)-nice subsequence and so the lemma is proven.
Note that it is easy to prove by induction that the closure of the image of an r-nice convergent f is countable and so Corollary 2.6. If f : [ω] r → X and X has the r-Ramsey property, then f has a convergent subsequence f ↾ [T ] r → X such that the closure of f ′′ [T ] r is countable.

Spaces with the Ramsey property
It is clear that the class of all topological spaces with the Ramsey property is stable under closed subspaces and continuous images. The same applies to the r-Ramsey property. In order to see that there are arbitrarily large spaces with the Ramsey property, consider the Σ-product Note that the closure of every countable subset of Σ(κ) is compact and metrizable, therefore Σ(κ) has the Ramsey property. More generally, all monolithic countably compact spaces have the Ramsey property (recall that a space X is monolithic if the closure of every countable subset of X is second countable).
Recall that the unboundedness number b is the minimal cardinality of a family F ⊆ ω ω which is unbounded with respect to the almost domination < * , where f < * g if f (n) < g(n) for all but finitely many n ∈ ω.
Theorem 3.1. Every sequentially compact space of character < b has the Ramsey property.
Proof. We use induction on r. Sequential compactness implies that the theorem is true for r = 1. Suppose r > 1 and the statement is true for We construct inductively a sequence a 0 < a 1 < · · · in N, a sequence {p n } n>0 ⊆ X, and a decreasing sequence of infinite sets Suppose n > 0 and a i , A i , and {p i } i<n−1 have been constructed for i < n. Using the theorem for r − 1, we find p n−1 ∈ X and an infinite set A n ⊆ A n−1 such that f a n−1 ↾ [A n ] r−1 converges to p n−1 . We set a n = min(A n \ (a n−1 + 1)). Now let M ∈ [N] ω be such that {p n } n∈M is convergent to some p ∈ X (here we have used the sequential compactness of X). Let Re-enumerating A n and p n , we may assume that M = N.
Note that the set B has the following property: r is such that min s = a n with n m(U) and min(s \ {a n }) ϕ U (n) (the last fact follows from (ii), because s \ {a n } ⊆ A n+1 ). Define ϕ U (n) arbitrarily for n < m(U).
Let U (p) be a fixed base at p such that |U (p)| < b. Then {ϕ U } U ∈U (p) ⊆ ω ω has cardinality < b, therefore we can find a strictly increasing function ψ ∈ B ω such that ϕ U < * ψ for every U ∈ U (p). Now let C = {ψ(n) : n ∈ ω}. We claim that C is as required.
Fix U ∈ U (p) and fix s ∈ [C] r such that min s = a k , where k > m(U) and ψ(n) > ϕ U (n) for every n k.
This shows that f ↾ [B] r converges to p.
Corollary 3.2. Let X be a sequentially compact space in which the closure of every countable set is first countable. Then X has the Ramsey property. Concerning products, it is known that the product of countably many sequentially compact spaces is sequentially compact. The same proof also applies to show Theorem 3.4. The product of countably many r-Ramsey spaces is r-Ramsey.
Proof. Suppose we are given r-Ramsey spaces X i and The splitting number, s can be characterized as the minimal κ such that 2 κ is not sequentially compact (see [3]). The analogous cardinal characteristic of the continuum that characterizes the r-Ramsey property in Cantor cubes is par and was introduced by Blass in [1]: par n denotes the smallest cardinal κ such that there is a family of partitions of [ω] n → 2 of size κ such that no infinite set is almost homogeneous for all of them simultaneously.
First notice that par 1 is just the splitting number s. Also, note that if we consider partitions into some finite number of pieces k, instead of 2 pieces, we obtain the same cardinal. And, moreover, for all n ≥ 2 we have that par n = par 2 , in fact: Theorem 3.6 ([1]). For each n ≥ 2, par n = min{b, s}. Now we prove that the minimal κ such that 2 κ is not r-Ramsey is precisely par 2 .
Proof. Fix κ < par 2 and fix f : [ω] r → 2 κ . For each α < κ let f α = π α · f . By the definition of κ < par 2 = par r , there is B ⊆ ω and for each α an i α such that for every α < κ there is a finite set F such that f α is constant with value i α on [B \ F ] r . This just means that f : [B] r → 2 κ converges to (i α ) α∈κ as required.
To complete the proof, to see that 2 κ is not 2-Ramsey for κ = par 2 , fix a family {f α : α < κ} so that no B ⊆ ω is almost homogeneous for all functions f α . Taking the product function f = f α we have, as above, that for no B can f : [B] 2 → 2 κ converge.

Examples
In this section we give some examples of spaces with the k-Ramsey Property that do not have the (k + 1)-Ramsey Property. The first example of a 1-Ramsey (i.e., sequentially compact) not 2-Ramsey is in ZFC, but for larger k we assume CH. All of the examples are of the form K(A) = α(Ψ(A )), the Alexandrov-Urysohn compactum formed by taking the one-point compactification of the Ψ-space determined by an almost disjoint family A of infinite subsets of ω. If A is a maximal almost disjoint (mad) family, then K(A ) is not even 2-Ramsey. We now describe, for each r > 1 an r-Ramsey not (r + 1)-Ramsey compact space of the form K(A ). We state a few lemmas about these properties in these types of spaces. Proof. The property is clearly necessary. For sufficiency, note that for any f : [ω] r → K(C ) one can find B such that either f ′′ [B] r ⊆ C (in which case, since any infinite subset converges, one can easily find B ′ ⊆ B witnessing r-Ramsey) or f ′′ [B] r ⊆ I, as required.
Proof. If we can find such a B, then the closure of the subset f ′′ [B] r in K(A ) has countable character at all points of A and character less than b at ∞ and so by Theorem 3.1 this subspace is r-Ramsey and so we can find B ′ ⊆ B on which f converges.
We now turn to a construction of an r-Ramsey example of an Alexandrov-Urysohn compactum that is not (r + 1)-Ramsey.
We first need some basic properties of the Fubini product of the ideal FIN. Recall, FIN is the ideal of finite subsets of ω and for each n > 1, FIN n is the ideal on ω n defined recursively by The following lemma is easily proved using the definition.
Lemma 4.4. For any X ⊆ ω n , if X ∈ FIN n then there is a T ⊆ X such that T forms an everywhere ω-splitting tree. I.e., letting T i = {x ↾ i : x ∈ T }, for each i < n and each s ∈ T i , {t ∈ T i+1 : t ↾ i = s} is infinite.

Now we need the following
Proof. By induction on n. When n = 1 we can find B such that f ′′ B is either contained in a column {k} × ω or is a partial function. In either case f ′′ B ∈ FIN 2 .
For the inductive step, fix f : [ω] n → ω n+1 and fix k 0 ∈ ω arbitrary. Define  There is i such that x(i) = ω. In this case, there is some N such that And this latter set is not in FIN n+1 . Now fix r > 1 and we will need to assume CH. We build an almost disjoint family A on ω r+1 so that for the function G : [ω] r+1 → ω r+1 defined by G(x) = x(0), ..., x(r) where x = {x(0), ..., x(r)} is the increasing enumeration of x, G will have no convergent subsequence B.
To simplify some notation for any B, let Lemma 4.6. Suppose A is an almost disjoint family on ω r+1 . For the function G defined as above, B is a convergent subsequence for G with limit ∞ in K(A ) if and only if for every a ∈ A there is an n such that a ∩ (B \ n) ↑r+1 = ∅.
Proof. Directly from the definitions.
Theorem 4.7. Assume CH. For each r > 1 there is an almost disjoint family A on ω such that K(A ) is r-Ramsey but not (r + 1)-Ramsey.
Proof. We will construct A = {a α : α < ω 1 } an almost disjoint family on ω r+1 by defining each a α by recursion on α. We start by letting {a n : n ∈ ω} be an enumeration of the disjoint family {{k} × ω : k ∈ ω r } Note that each a n ∈ FIN n+1 and any x that is almost disjoint from all the a n is also in FIN n+1 .
We enumerate as {B α : α < ω 1 } all infinite subsets of ω and fix an enumeration To define a α and X α , note first that the following family of sets Therefore the family {(ω n+1 ) \ H : H ∈ H } has an infinite pseudo-intersection. Let a α be any such pseudo-intersection and it directly follows that a α satisfies the inductive hypotheses (2)-(4). So we need only define X α to satisfy (1). But that there is such an X α follows from Lemma 4.5.
This completes the construction of A = {a α : α < ω 1 }. To see that it is r-Ramsey, by Lemma 4.2 we need only consider functions f : [ω] r → ω r+1 , and each such f appears as an f α . For each α by inductive hypothesis (3) we have that is countable. And so by Lemma 4.3, it follows that K(A ) is r-Ramsey. On the other hand, to see that G has no convergent subsequence, we note that by inductive hypothesis (4) we have, by Lemma 4.6, that no B α is a convergent subsequence for G and so K(A ) is not (r + 1)-Ramsey.

Questions
We finish with the following questions: Question 5.1. Does there exist a space with the r-Ramsey property and without the (r + 1)-property assuming only ZFC? Here r > 1.
Note that all our examples of r-Ramsey and not (r + 1)-Ramsey spaces (which required CH) are Fréchet-Urysohn but the ZFC example that was not 2-Ramsey, being a K(A ) where A is mad, is not Fréchet. However, a similar example could be constructed from a completely separable mad family. The main idea is to start with a completely separable mad family on ω × ω as in the construction of Example 4.1. Then, deleting an infinite set from each A ∈ A other than the fixed columns {n}×ω will still give an example that fails to be 2-Ramsey. But it will be Fréchet since it is nowhere mad ((see [5] for the definitions of completely separable and nowhere mad). So we have Example 5.2. Assuming the existence of a completely separable mad family, there is an A such that K(A ) is Fréchet and not 2-Ramsey.
Although the existence of a completely separable mad family is a relatively weak one (e.g., it follows from c < ℵ ω or s ≤ a), we ask Question 5.3. Does there exist a ZFC example of a Fréchet-Urysohn compact space without the Ramsey property?
We know that a product of any countable family of r-Ramsey spaces is r-Ramsey, and we have characterized when 2 κ is r-Ramsey. Moreover, h is the minimal κ such that a product of fewer than κ many sequentially compact spaces is sequentially compact [7], and we conjecture the same holds for r-Ramsey.
Indeed, the proof that the product of fewer than h sequentially compact spaces is sequentially compact also shows the same for r-Ramsey, but the family of sequentially compact spaces whose product is not sequentially compact given in [7] are, in fact, Alexandrov-Urysohn compacta that we have seen are not even 2-Ramsey. So, if µ is the minimal cardinal for productivity of the class of 2-Ramsey spaces, then h ≤ µ and we conjecture that µ = h.
Question 5.4. Characterize the minimal cardinal κ satisfying the product of fewer than κ many r-Ramsey spaces is always r-Ramsey.