A note on Factoring groups into dense subsets

1 Introduction

For a cardinal κ, a topological space X is called κ-resolvable if X can be partitioned into κ dense subsets [1]. In the case κ=2, these spaces were defined by Hewitt [4] as resolvable spaces. If X is not κ-resolvable, then X is called κ-irresolvable.

In topological groups, the intensive study of resolvability was initiated by the following remarkable theorem of Comfort and van Mill [2]: every countable non-discrete Abelian topological group G with finite subgroup B⁢(G) of elements of order 2 is 2-resolvable. In fact [11], every infinite Abelian group G with finite B⁢(G) can be partitioned into ω subsets dense in every non-discrete group topology on G. On the other hand, under Martin’s Axiom, the countable Boolean group G, G=B⁢(G) admits a maximal (hence, 2-irresolvable) group topology [5]. Every non-discrete ω-irresolvable topological group G contains an open countable Boolean subgroup provided that G is Abelian [6] or countable [10], but the existence of a non-discrete ω-irresolvable group topology on the countable Boolean group implies that there is a P-point in ω∗ (see [6]). Thus, in some models of ZFC (see [8]), every non-discrete Abelian or countable topological group is ω-resolvable. For a systematic exposition of resolvability in topological and left topological groups see [3, Chapter 13].

Recently, a new kind resolvability of groups was introduced in [7]. A group G provided with a topology 𝒯 is called box κ-resolvable if there is a factorization G=A⁢B such that |A|=κ and each subset aB is dense in 𝒯. If G is left topological (i.e. each left shift x↦g⁢x, g∈G is continuous), then this is equivalent to B being dense in 𝒯. We recall that a product AB of subsets of a group G is a factorization if G=A⁢B and the subsets {a⁢B:a∈A} are pairwise disjoint (equivalently, each g∈G has a unique representation g=a⁢b, a∈A, b∈B). For factorizations of groups into subsets see [9]. By [7, Theorem 1], if a topological group G contains an injective convergent sequence, then G is box ω-resolvable.

The aim of this note is to find some conditions under which an infinite group G of cardinality κ provided with a topology can be factorized into two dense subsets of cardinality κ. To this goal, we propose a new method of factorization based on filtrations of groups.

2 Theorem and question

We recall that the weight w⁢(X) of a topological space X is the minimal cardinality of bases of the topology of X.

We do not know whether or not this Theorem is true for κ=ℵ0 even if G is a topological group.

In Section 4, we give a positive answer in the following cases: each finitely generated subgroup of G is nowhere dense, the set {x2:x∈U} is infinite for each non-empty open subset of G, G is Abelian.

3 Proof

We begin with some general constructions of factorizations of a group G via filtrations of G.

Let G be a group with the identity e. Let κ be a cardinal. A family {Gα:α<κ} of subgroups of G is called a filtration if

1. G0={e}, G=⋃α<κGα,

2. G⁢α⊂Gβ for all α<β,

3. Gβ=⋃α<βGα for every limit ordinal β.

Every ordinal α<κ has the unique representation α=γ⁢(α)+n⁢(α), where γ⁢(α) is either a limit ordinal or 0 and n⁢(α)∈ω, ω={0,1,…}. We partition κ into two subsets

and

For each α∈E⁢(κ), we choose some system Lα of representatives of left cosets of Gα+1∖Gα by Gα so Gα+1∖Gα=Lα⁢Gα. For each α∈O⁢(κ), we choose some system Rα of representatives of right cosets of Gα+1∖Gα by Gα so we have Gα+1∖Gα=Gα⁢Rα.

We take an arbitrary element g∈G∖{e} and choose the smallest subgroup Gγ such that g∈Gγ. By (3), γ=α⁢(g)+1 so g∈Gα⁢(g)+1∖Gα⁢(g). If α⁢(g)∈E⁢(κ), we choose x0⁢(g)∈Lα⁢(g) and g0∈Gα⁢(g) such that g=x0⁢(g)⁢g0. If α⁢(g)∈O⁢(κ), we choose y0⁢(g)∈Rα⁢(g) and g0∈Gα⁢(g) such that g=g0⁢y0⁢(g). If g0=e, we stop. Otherwise we repeat the argument for g0 and so on. Since the set of ordinals less than κ is well ordered, after a finite number of steps we get the representation

where

If either {α0⁢(g),…,αλ⁢(g)⁢(g)}=∅ or {β0⁢(g),…,βρ⁢(g)⁢(g)}=∅, then we write g=yρ⁢(g)⁢…⁢y1⁢(g)⁢y0⁢(g) or g=x0⁢(g)⁢x1⁢(g)⁢…⁢xλ⁢(g)⁢(g). Thus, G=A⁢B where A is the set of all elements of the form x0⁢(g)⁢x1⁢(g)⁢…⁢xλ⁢(g) and B is the set of all elements of the form yρ⁢(g)⁢…⁢y1⁢(g)⁢y0⁢(g). To show that the product AB is a factorization of G, we assume that, besides (4), g has a representation

If g∈Gα+1∖Gα and α∈O⁢(κ), then z0⁢z1⁢…⁢zλ⁢tρ⁢…⁢t1∈Gα so t0=y0⁢(g). If α∈E⁢(κ), then z1⁢…⁢zλ⁢tρ⁢…⁢t1⁢t0∈Gα so z0=x0⁢(g). We replace g by g⁢t0-1 or by z0-1⁢g respectively and repeat the same arguments.

Now we are ready to prove the Theorem. Let {Uα:α<κ} be a κ-sequence of non-empty open sets such that each non-empty U∈𝒯 contains some Uα. Since |Uα|=κ for every α<κ, we can construct inductively a filtration {Gα:α<κ}, |Gα|=max⁡{ℵ0,|α|} such that for each α∈E⁢(κ) (resp. α∈O⁢(κ)) there is a system Lα (resp. Rα) of representatives of left (resp. right ) cosets of Gα+1∖Gα by Gα such that Lα∩Uγ≠∅ (resp. Rα∩Uγ≠∅) for each γ≤α. Then the subsets A,B of the above factorization of G are dense in 𝒯 because Lα⊂A, Rβ⊂B for each α∈E⁢(κ), β∈O⁢(κ).

4 Comments

1. Analyzing the proof, we see that the Theorem holds under the weaker condition: G has a family ℱ of subsets such that |ℱ|=κ, |F|=κ for each F∈ℱ and, for every non-empty U∈𝒯, there is F∈ℱ such that F⊆U.

If κ=ℵ0 but each finitely generating subgroup of G is nowhere dense, we can choose a family {Gn:n∈ω} such that the corresponding A,B are dense. Thus, we get a positive answer to the Question if each finitely generated subgroup H of G is nowhere dense (equivalently the closure of H is not open).

2. Let G be a group and let A,B be subsets of G. We say that the product AB is a partial factorization if the subsets {a⁢B:a∈A} are pairwise disjoint (equivalently, {A⁢b:b∈B} are pairwise disjoint).

We assume that AB is a partial factorization of G into finite subsets and that X is an infinite subset of G. Then the following statements are easily verified

1. there is x∈X such that x∉B and A⁢(B∪{x}) is a partial factorization;

2. if the set {x2:x∈X} is infinite, then there is an element x∈X such that (A∪{x,x-1})⁢B is a partial factorization.

3. Let G be a non-discrete Hausdorff topological group, let AB be a partial factorization of G into finite subsets, A=A-1, e∈A∩B and g∉A⁢B. Then

1. there is a neighbourhood V of e such that, for U=V∖{e} and for any x∈U, the product (A∪{x,x-1})⁢(B∪{x-1⁢g}) is a partial factorization (so g∈(A∪{x,x-1})⁢(B∪{x-1⁢g})).

It suffices to choose V so that V=V-1 and

We use A=A-1 only in U⁢B∩A⁢U⁢g=∅.

4. Let G be countable non-discrete Hausdorff topological group such that the set {x2:x∈U} is infinite for every non-empty open subset U of G. We enumerate G={gn:n∈ω}, g0=e and choose a countable base {Un:n∈ω} for non-empty open sets. We put A0={e}, B0={e} and use (5), (6), (7) to choose inductively two sequences (An)n∈ω and (Bn)n∈ω of finite subsets of G such that for every n∈ω, An⊂An+1, Bn⊆Bn+1, An=An-1, An⁢Bn is a partial factorization, gn∈An⁢Bn, An∩Un≠∅, Bn∩Un≠∅. We put

and note that AB is a factorization of G into dense subsets.

5. Let G be a countable Abelian non-discrete Hausdorff topological group of countable weight. We suppose that G contains a non-discrete finitely generated subgroup H. Given any non-empty open subset U of G, we choose a neighborhood X of e in H and g∈S such that X⁢g⊂U. Since H is finitely generated, the set {x2:x∈X} is infinite so we can apply comment 4. If each finitely generated subgroup of G is discrete then, to answer the Question, we use comment 1.

6. Let G be a countable group endowed with a topology 𝒯 of countable weight such that U is infinite for every U∈𝒯. Applying the inductive construction from comment 5 to An⁢Bn and Bn+1-1⁢An-1, we get a partial factorization of G into two dense subsets.

7. Let G be a group satisfying the assumption of the Theorem and let γ be an infinite cardinal, γ<κ. We take a subgroup A of cardinality γ and choose inductively a dense set B of representatives of right cosets of G by A. Then we get a factorization G=A⁢B. In particular, if G is left topological, then G is box γ-resolvable.