Igor Protasov and Serhii Slobodianiuk

A note on Factoring groups into dense subsets

Published online: May 18, 2016

Abstract

Let G be a group of cardinality κ > 0 endowed with a topology 𝒯 such that | U | = κ for every non-empty U 𝒯 and 𝒯 has a base of cardinality κ. We prove that G can be factorized G = A B (i.e. each g G has a unique representation g = a b , a A , b B ) into dense subsets A, B, | A | = | B | = κ . We do not know if this statement holds for κ = 0 even if G is a topological group.

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.

Theorem

Let G be an infinite group of cardinality κ , κ > 0 , endowed with a topology T such that w ( G , T ) κ and | U | = κ for each non-empty U T . Then there is a factorization G = A B into dense subsets A , B , | A | = | B | = κ .

We do not know whether or not this Theorem is true for κ = 0 even if G is a topological group.

Question

Let G be a non-discrete countable Hausdorff topological group G of countable weight. Can G be factorized G = A B into two countable dense subsets?

In Section 4, we give a positive answer in the following cases: each finitely generated subgroup of G is nowhere dense, the set { x 2 : 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 { G α : α < κ } of subgroups of G is called a filtration if

  1. (1)

    G 0 = { e } , G = α < κ G α ,

  2. (2)

    G α G β for all α < β ,

  3. (3)

    G β = α < β G α 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

E ( κ ) = { α < κ : n ( α ) is even }

and

O ( κ ) = { α < κ : n ( α ) is odd } .

For each α E ( κ ) , we choose some system L α of representatives of left cosets of G α + 1 G α by G α so G α + 1 G α = L α G α . For each α O ( κ ) , we choose some system R α of representatives of right cosets of G α + 1 G α by G α so we have G α + 1 G α = G α R α .

We take an arbitrary element g G { e } and choose the smallest subgroup G γ such that g G γ . By (3), γ = α ( g ) + 1 so g G α ( g ) + 1 G α ( g ) . If α ( g ) E ( κ ) , we choose x 0 ( g ) L α ( g ) and g 0 G α ( g ) such that g = x 0 ( g ) g 0 . If α ( g ) O ( κ ) , we choose y 0 ( g ) R α ( g ) and g 0 G α ( g ) such that g = g 0 y 0 ( g ) . If g 0 = e , we stop. Otherwise we repeat the argument for g 0 and so on. Since the set of ordinals less than κ is well ordered, after a finite number of steps we get the representation

\text{(4)} g = x 0 ( g ) x 1 ( g ) x λ ( g ) ( g ) y ρ ( g ) y 1 ( g ) y 0 ( g ) ,

where

x i L α i ( g ) , α 0 ( g ) > α 1 ( g ) > > α λ ( g ) ( g ) , y i R β i ( g ) , β 0 ( g ) > β 1 ( g ) > > β ρ ( g ) ( g ) .

If either { α 0 ( g ) , , α λ ( g ) ( g ) } = or { β 0 ( g ) , , β ρ ( g ) ( g ) } = , then we write g = y ρ ( g ) y 1 ( g ) y 0 ( g ) or g = x 0 ( g ) x 1 ( g ) x λ ( g ) ( g ) . Thus, G = A B where A is the set of all elements of the form x 0 ( g ) x 1 ( g ) x λ ( g ) and B is the set of all elements of the form y ρ ( g ) y 1 ( g ) y 0 ( g ) . To show that the product AB is a factorization of G, we assume that, besides ( 4 ) , g has a representation

g = z 0 z 1 z λ t ρ t 1 t 0 .

If g G α + 1 G α and α O ( κ ) , then z 0 z 1 z λ t ρ t 1 G α so t 0 = y 0 ( g ) . If α E ( κ ) , then z 1 z λ t ρ t 1 t 0 G α so z 0 = x 0 ( g ) . We replace g by g t 0 - 1 or by z 0 - 1 g respectively and repeat the same arguments.

Now we are ready to prove the Theorem. Let { U α : α < κ } be a κ -sequence of non-empty open sets such that each non-empty U 𝒯 contains some U α . Since | U α | = κ for every α < κ , we can construct inductively a filtration { G α : α < κ } , | G α | = max { 0 , | α | } such that for each α E ( κ ) (resp. α O ( κ ) ) there is a system L α (resp. R α ) of representatives of left (resp. right ) cosets of G α + 1 G α by G α such that L α U γ (resp. R α U γ ) for each γ α . Then the subsets A , B of the above factorization of G are dense in 𝒯 because L α A , R β 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 of subsets such that | | = κ , | F | = κ for each F and, for every non-empty U 𝒯 , there is F such that F U .

If κ = 0 but each finitely generating subgroup of G is nowhere dense, we can choose a family { G n : 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. (5)

    there is x X such that x B and A ( B { x } ) is a partial factorization;

  2. (6)

    if the set { x 2 : 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. (7)

    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

A U g A B = , U B ( A B A U g ) = , U 2 g A B = , U A = .

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 { x 2 : x U } is infinite for every non-empty open subset U of G. We enumerate G = { g n : n ω } , g 0 = e and choose a countable base { U n : n ω } for non-empty open sets. We put A 0 = { e } , B 0 = { e } and use ( 5 ) , ( 6 ) , ( 7 ) to choose inductively two sequences ( A n ) n ω and ( B n ) n ω of finite subsets of G such that for every n ω , A n A n + 1 , B n B n + 1 , A n = A n - 1 , A n B n is a partial factorization, g n A n B n , A n U n , B n U n . We put

A = n ω A n , B = n ω B n

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 { x 2 : 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 A n B n and B n + 1 - 1 A n - 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.

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Received: 2016-2-14
Revised: 2016-4-29
Published Online: 2016-5-18
Published in Print: 2017-1-1

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