HINDMAN SPACES

MENACHEM KOJMAN

Abstract. A topological space X is Hindman if for every sequence (x_{n})_{n} in X there exists an infinite D ⊆ ℕ so that the sequence (x_{n})_{n∈F S(D)}, indexed by all finite sums over D, is IP-converging in X. Not all sequentially compact spaces are Hindman. The product of two Hindman spaces is Hindman.

Furstenberg and Weiss [4] proved that all compact metric spaces are Hindman. We show that every Hausdorff space X that satisfies the following condition (∗) The colsure of every countable set in X is compact and first-countable is Hindman.

Consequently, there exist nonmetrizable and non-compact Hindman spaces. The following are two particulal consequences of the main result: every sequence of ordinal numbers and every bounded sequence of monotone (not necessarily continuous) real functions on [0, 1] have IP-converging subsequences.

1. Introduction

Let us start by recalling Hindman’s finite-sum Theorem about partitions of ℕ. Let ℕ = {1, 2, … } denote the set of natural numbers, with the convention that n = {0, 1, …, n − 1}.

Definition 1. A set A of natural numbers is called an IP set if there exists an infinite set D ⊆ ℕ so that F S(D) ⊆ A, where F S(D) = {∑ α : α ⊆ D and α ≠ ∅ is finite}.

Theorem 2 (Hindman [8]). If the integers are partitioned into finitely many parts, then one of the parts is IP. In fact, if an IP set is partitioned into finitely many parts, then one of the parts is IP.

In this paper I examine which topological spaces satisfy a topological generalization of Hindman’s theorem. We shall call those spaces Hindman spaces.

The results show that in addition to the compact metric spaces, which were shown to be Hindman in [5], Theorem 8.14, there are plenty of Hindman spaces which are neither compact nor metrizable.

2. Hindman spaces

Hindman’s Theorem can be formulated topologically as follows: every sequence (x_{n}) in a finite (Hausdorff) space X has a converging subsequence whose set of indices is IP. It is tempting to define a Hindman space as a Hausdorff space X that satisfies this property, but that definition would not get us very far. The blame lies on the distributive law.

1991 Mathematics Subject Classification. Primary: 05C55, 54F65; Secondary 04A20, 11P99, 26A40.

Key words and phrases. Hindman’s Theorem, converging sequence, compactification, finite sums, nonmetrizable topological spaces.

Theorem 3. If a T_{1} space X satisfies that for every sequence (x_{n})_{n∈ℕ} there exists a converging subsequence (x_{n})_{n∈A} for some IP set A, then X is finite.

Proof. For every n ∈ ℕ let i(n) = max{i : 2^{i}|n}. Note that i(n_{1} + n_{2}) > i(n_{1}) if i(n_{1}) = i(n_{2}) and that i(n_{1} + n_{2}) = i(n_{1}) if i(n_{1}) < i(n_{2})

Suppose X satisfies the condition of the theorem and that X is infinite. Necessarily, X contains a converging injective sequence (x_{n})_{n∈ℕ} with m < n ⇒ x_{m} ≠ x_{n}. Define a new sequence (y_{n})_{n} in X by y_{n} := x_{i(n)}

Suppose now that D ⊆ ℕ is infinite and A = F S(D) satisfies that (y_{n})_{n∈A} is converging. Assume first that {y_{n} : n ∈ A} is finite. Then, since X is T_{1}, (y_{n})_{n∈A} is eventually constant. Pick n_{1} < n_{2} in D with y_{n_{1}} = y_{n_{2}} = y_{n_{1}+n_{2}}. Then i(n_{1}) = i(n_{2}) = i(n_{1} + n_{2}), a contradiction.

Thus one must have that {y_{n} : n ∈ A} is infinite. In this case we may assume, by thinning D out, that i(n_{1}) < i(n_{2}) for all n_{1} < n_{2} in D. But then y_{n_{1}+n_{2}} = y_{n_{1}} for all n_{1} < n_{2} in D, and (y_{n})_{n∈A} cannot converge.

To obtain a meaningful topological definition of a Hindman space we use the notion of IP-convergence, introduced by Furstenberg and Weiss in [4] (and presented in detail also in [5], Chapter 8).

Definition 4. 1. Suppose that D ⊆ ℕ is infinite. An IP-sequence in a topological space is a sequence indexed by F S(D) for some infinite D ⊆ ℕ. An IP-sequence (x_{n})_{n∈F S(D)} in a topological space X IP-converges to a point x ∈ X if for every neighborhood u of x there exists m ∈ ℕ so that {x_{n} : n ∈ F S(D \ m)} ⊆ u.

2. A topological space X is Hindman if for every sequence (x_{n}) in X there exists an infinite set D ⊆ ℕ so that (x_{n})_{n∈F S(D)} IP-converges to some x ∈ X.

Suppose (x_{n})_{n∈F S(D)} is IP-converging in X. For every infinite D_{1} ⊆ D, (x_{n})_{n∈F S(D_{1})} is clearly also IP-converging to x. By thinning any infinite D out and enumerating it increasingly as {d_{1}, d_{2}, … }, we may assume that for every n ∈ F S(D) there is a unique finite α ⊆ ℕ for which n = ∑_{i∈α} d_{i} and denote that unique α by α_{D}(n). Call an infinite D ⊆ ℕ for which α_{D}(n) is defined for all n ∈ F S(D) a sparse subset of ℕ.

Let E = {2^{i} : i ∈ ℕ}. Then F S(E) = ℕ and α_{E}(n) is defined for all n, and is exactly the set of i-s for which the i-th digit in the binary expansion of n is equal to 1.

Let x_{n} = n for all n. The IP-sequence (x_{n})_{n∈F S(E)} IP-converges to 0 in the 2-adic topology on ℕ∪{0}, (that is, in the metric topology generated by the metric d(m, n) = 1/ max{i : i|(m − n)} for m ≠ n). Let D_{1} = {3^{i} : i ∈ ℕ}. Now F S(D_{1}) ⊆ F S(E) = ℕ but (x_{n})_{n∈F S(D_{1})} does not IP-converge to 0. The possibility just sketched, of a non-IP-converging IP-sequence contained in an IP-converging IPsequence, needs to be handled in proving that the product of two Hindman spaces is Hindman.

Definition 5. Suppose D is sparse. D_{1} ⊆ F S(D) is normal in F S(D) if D_{1} is sparse and for all n_{1}, n_{2} ∈ D_{1}, n_{1} ≠ n_{2} ⇒ α_{D}(n_{1}) ∩ α_{D}(n_{2}) = ∅.

Fact 6. If D_{1} is normal in F S(D) then:

1. F S(D_{1}) ⊆ F S(D)

2. Whenever (x_{n})_{n∈F S(D)} IP-converges to x in some space X also (x_{n})_{n∈F S(D_{1})} IP-converges to x.

Proof. The first part is clear. For the second part, let u be an arbitrary neighborhood of x and fix m such that x ∈ F S(D \ m) ⇒ x_{n} ∈ u. Since {a_{D}(n) : n ∈ D_{1}} is a disjoint family, there is some m′ so that n ∈ D_{1} \m′ ⇒ {d_{i} : i ∈ α_{D}(n)} > m.

Lemma 7. Suppose that D is sparse. Then there exists D_{1} ⊆ F S(D) which is normal both in F S(E) and in F S(D).

Proof. For finite sets of integers α_{1}, α_{2} we write α_{1} < α_{2} to mean max α_{1} < min α_{2}. By induction on i find a sequence of integers n_{i} ∈ F S(D) so that i < j ⇒ α_{D}(n_{i}) < α_{D}(n_{j}) ∧ α_{E}(n_{i}) < α_{E}(n_{j}). Let n_{1} = min D. Suppose n_{i} has been chosen. Choose some m_{i} so that 2^{m_{i}} > ∑_{ℓ≤i} n_{ℓ} and fix T ⊆ D \ 2^{m_{i}} with |T | = 2^{m_{i}}. It is easy to see that for some nonempty subset T_{1} ⊆ T the sum ∑ T_{1} is divisible by 2^{m_{i}}. Let n_{i+1} = ∑ T_{1}. It is clear that α_{D}(n_{i+1}) > α_{D}(n_{i}) and, since 2^{m_{i}}|n_{i+1}, α_{E}(n_{i+1}) > α_{E}(n_{i}). Let D_{1} = {n_{i} : i ∈ ℕ}.

Lemma 8. If X is Hindman then for every sparse D ⊆ ℕ and an IP-sequence (x_{n})_{n∈F S(D)} there exists D_{1} normal in F S(D) so that (x_{n})_{n∈F S(D_{1})} IP-converges to some x ∈ X.

Proof. Suppose (x_{n})_{n∈F S(D)} is given in some Hindman space X where D is sparse. Let {d_{i} : i ∈ ℕ} be the increasing enumeration of D. Define g : ℕ → F S(D) by g(∑_{i∈α} 2^{i}) = ∑_{i∈α} d_{i} for all nonempty finite α ⊆ ℕ. Since X is Hindman, there exists a sparse F ⊆ ℕ so that (x_{g(n)})_{n∈F S(F)} IP-converges to some x ∈ X. Using the previous Lemma, we may assume that F is normal in F S(E). This implies that D_{1} := g[F] is normal in F S(D) and that (x_{n})_{n∈F S(D_{1})} IP-converges to x.

Fact 9. The product of two Hindman spaces is Hindman.

Proof. Suppose that Z = X × Y where X, Y are Hindman and suppose that (x_{n}, y_{n})_{n} is a sequence in Z. Since X is Hindman, one can fix a sparse D ⊆ ℕ so that (x_{n})_{n∈F S(D)} IP-converges to some x ∈ X. By Lemma 8, there is some D_{1} normal in F S(D) so that (y_{n})_{n∈F S(D)} IP-converges to some y ∈ Y. Now (x_{n}, y_{n})_{n∈F S(D_{1})} IP-converges to (x, y) in Z.

Every Hindman space is surely sequentially compact. The following shows that being Hindman is strictly stronger than being sequentially compact:

Theorem 10. There exists a Hausdorff, compact, sequentially compact, separable space which is first-countable at all points but one, which is not Hindman.

Proof. Let F be a maximal element in {A : A is an infinite family of infinite almost disjoint subsets of ℕ and every member of A is not IP}. Then F is in fact a maximal almost disjoint family (suppose to the contrary that B ⊆ ℕ is infinite and almost disjoint to every member in A. By thinning out, B can be made non-IP.)

For every A ∈ F let p_{A} ∉ ℕ be a distinct point. Define a topology T on Y = ℕ ∪ {p_{A} : A ∈ F} by requiring that B ∈ T if and only if for every p_{A} ∈ B the set A \ B is finite. T is a Hausdorff topology on Y with respect to which ℕ is a discrete subspace. Since {p_{A}} ∪ A is a compact neighborhood of p_{A} for every A ∈ F, T is locally compact.

Let X = Y ∪ {p} be the one-point compactification of Y. Let us check that X is sequentially compact. Let (x_{n}) be an arbitrary sequence in X. If the range of the sequence is finite, there is little to prove. If the range of the sequence is infinite, assume, be passing to a subsequence, that the sequence is injective and that its range is contained either in ℕ or in {p_{A} : A ∈ F}. In the first case, the maximality of F guarantees the existence of some A ∈ F and a subsequence (x_{n_{k}}) whose range is contained in A; this subsequence converges to p_{A}. In the other case, the sequence itself converges to p, since the intesection of its range with every compace subset of Y is finite.

To see that X is not Hindman, let (x_{n})_{n} in X be defined by x_{n} = n. For each p_{A}, {p_{A}} ∪ A is a neighborhood of p_{A} which does not contain any IP subsets of ℕ. Therefore, for no D ⊆ ℕ can (x_{n})_{n∈F S(D)} IP-converge to p_{A}. To see that for no D can (x_{n})_{n∈F S(D)} IP-converge to p, let D be arbitrary, and consider the family of sets {F S(D \ m) : m ∈ ℕ}. Every finite intersection of members from this (countable) family is infinite, hence there exists an infinite pseudo intersection of the family, namely an infinite set Z so that for every m ∈ ℕ satisfies Z \F S(D\m) is finite. By maximality of F, there is some member A ∈ F for which A ∩ Z is infinite. The set X \ ({p_{A}} ∪ A) is an open neighborhood of p, but for every m ∈ ℕ, the intersection F S(D \ m) ∩ A is infinite; therefore (x_{n})_{n∈F S(D)} does not IP-converge to p.

Theorem 8.14 in [5], due to Furstenberg and Weiss [4], implies that compact metric spaces are Hindman. The following Theorem extends this result, using the technology of idempotents in (βℕ, +). βℕ is the set of all ultrafilters over ℕ and the operation of addition in ℕ can be extended to βℕ (turning it to a left continuous and compact semi-group) via the formula

A ∈ U + V ⟺ {n : A − n ∈ U} ∈ V (1)

For a good and recent exposition of this material see [7] (with the caution that they reverse the operation from the way it is described in (1)). An ultrafilter U ∈ βℕ is an idempotent if U + U = U. The Galvin-Glazer proof of Hindman’s Theorem by using an idempotent in (βℕ, +) is presented also in [6] and in [2].

Theorem 11. Suppose that a Hausdorff space X satisfies the following condition:

(∗) The closure of every countable set in X is compact and first-countable.

Then X is Hindman.

Proof. Suppose (x_{n}) is given in X. Let D = cl_{X}{x_{n} : n ∈ ℕ}. Fix an idempotent U ∈ βℕ and let U′ be the ultrafilter over D which is defined by A ∈ U′ ⟺ {n : x_{n} ∈ A} ∈ U. Since D is compact, we can find x ∈ D so that every open neighborhood of x (in D) belongs to U′. Fix a decreasing neighborhood base (u_{n})_{n} at x with u_{1} = D, and denote ū_{n} = {k ∈ ℕ : x_{k} ∈ u_{n}}. We shall find an infinite {m_{n} : n ∈ ℕ} ⊆ ℕ so that for every finite nonempty α ⊆ ℕ, ∑_{n∈α} m_{n} ∈ ū_{min α}. This clearly implies IP-convergence of (x_{m})_{m∈F S({m_{n}:n∈ℕ})} to x.

By induction on n define a set A_{n} ⊆ ℕ and a number m_{n} that satisfy for all n:

1. A_{n} ∈ U

2. A_{n} ⊆ ū_{n}.

3. m_{n} ∈ A_{n} and m_{n} > ∑_{i<n} m_{i}.

4. k ∈ A_{n+1} ⇒ {k, k + m_{n}} ⊆ A_{n}.

Let A_{1} = A. Suppose that A_{n} and m_{1}, … m_{n−1} are defined and A_{n} satisfies conditions 1. and 2. We shall now define m_{n} that satisfies 3. and then define A_{n+1} so that 4. is satisfied and so that 1. and 2. hold for n + 1.

A_{n} ∈ U and U is an idempotent (hence nonprincipal). So B_{n+1} = {k : A_{n} − k ∈ U} ∩ A_{n} ∈ U. Fix m_{n} ∈ B_{n+1} which satisfies m_{n} > ∑_{i<n} m_{i}. Condition 3 is now satisfied. Let A_{n+1} = A_{n} ∩ (A_{n} − m_{n}) ∩ ū_{n+1}. Now condition 4 is satisfied. Since A_{n+1} is defined as the intersection of three sets in U, A_{n+1} ∈ U, so condition 1 holds for n+1. Since A_{n+1} ⊆ ū_{n+1}, condition 2 for n+1 holds too. That completes the inductive definition.

To show that for each nonempty finite set α ⊆ N, it holds that ∑_{n∈α} m_{n} ∈ ū_{min α}, we argue by induction on |α| that ∑_{n∈α} m_{n} ∈ A_{n}, which suffices, since A_{n} ⊆ ū_{n} by condition 2. If α = {n} is a singleton, then ∑_{n∈α} m_{n} = m_{n} ∈ A_{n} by condition 3. Suppose α = {m_{n_{1}}, m_{n_{2}}, …, m_{n_{t}}}. By the induction hypothesis, k := ∑_{i=2}^{t} m_{n_{i}} ∈ A_{n_{2}}. But A_{n_{2}} ⊆ A_{m_{1}+1} (since the A_{n}’s are decreasing by 4.), so by condition 4, m_{n_{1}} + k ∈ A_{m_{1}}.

Every space which which satisfies condition (∗) is Hindman. The class of spaces which satisfy (∗) includes all compact metric spaces, every successor ordinal and every limit ordinal of uncountable cofinality, with the order topology. In particular,

Corollary 12. For every sequence (α_{n}) of ordinal numbers there is an IP-converging sequence (α_{n})_{n∈F S(D)} for some infinite D ⊆ ℕ.

The space ω_{2} with the order topology is an example of a space which satisfies (∗) which is neither first-countable nor compact (although all of its separable subspaces are metrizable). Helly’s space of all monotone functions from [0, 1] to [0, 1] with the topology induced from the product topology on [0, 1]^{[0,1]} (see [12]) is an example of a separable, compact and first-countable space, which, therefore, satisfies (∗), which is not metrizable. The instance of Theorem 11 for this particular space gives:

Corollary 13. Suppose that f_{n} : [0, 1] → [0, 1] is monotone for each n. Then there is a monotone function f : [0, 1] → [0, 1] and an infinite D ⊆ ℕ so that (f_{n})_{n∈F S(D)} IP-converges pointwise to f.

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