Let G be a group of cardinality endowed with a topology such that for every non-empty and has a base of cardinality κ. We prove that G can be factorized (i.e. each has a unique representation , , ) into dense subsets A, B, . We do not know if this statement holds for even if G is a topological group.
For a cardinal , a topological space X is called -resolvable if X can be partitioned into dense subsets . In the case , these spaces were defined by Hewitt  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 : every countable non-discrete Abelian topological group G with finite subgroup of elements of order 2 is 2-resolvable. In fact , every infinite Abelian group G with finite 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, admits a maximal (hence, 2-irresolvable) group topology . Every non-discrete ω-irresolvable topological group G contains an open countable Boolean subgroup provided that G is Abelian  or countable , but the existence of a non-discrete ω-irresolvable group topology on the countable Boolean group implies that there is a P-point in (see ). Thus, in some models of ZFC (see ), 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 . A group G provided with a topology is called box -resolvable if there is a factorization such that and each subset aB is dense in . If G is left topological (i.e. each left shift , is continuous), then this is equivalent to B being dense in . We recall that a product AB of subsets of a group G is a factorization if and the subsets are pairwise disjoint (equivalently, each has a unique representation , , ). For factorizations of groups into subsets see . By [7, Theorem 1], if a topological group G contains an injective convergent sequence, then G is box -resolvable.
The aim of this note is to find some conditions under which an infinite group G of cardinality provided with a topology can be factorized into two dense subsets of cardinality . To this goal, we propose a new method of factorization based on filtrations of groups.
We recall that the weight of a topological space X is the minimal cardinality of bases of the topology of X.
Let G be an infinite group of cardinality , , endowed with a topology such that and for each non-empty . Then there is a factorization into dense subsets , .
We do not know whether or not this Theorem is true for even if G is a topological group.
Let G be a non-discrete countable Hausdorff topological group G of countable weight. Can G be factorized 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 is infinite for each non-empty open subset of G, G is Abelian.
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 be a cardinal. A family of subgroups of G is called a filtration if
for all ,
for every limit ordinal β.
Every ordinal has the unique representation , where is either a limit ordinal or 0 and , . We partition into two subsets
For each , we choose some system of representatives of left cosets of by so . For each , we choose some system of representatives of right cosets of by so we have .
We take an arbitrary element and choose the smallest subgroup such that . By (3), so . If , we choose and such that . If , we choose and such that . If , we stop. Otherwise we repeat the argument for and so on. Since the set of ordinals less than is well ordered, after a finite number of steps we get the representation
If either or , then we write or . Thus, where A is the set of all elements of the form and B is the set of all elements of the form . To show that the product AB is a factorization of G, we assume that, besides , g has a representation
If and , then so . If , then so . We replace g by or by respectively and repeat the same arguments.
Now we are ready to prove the Theorem. Let be a -sequence of non-empty open sets such that each non-empty contains some . Since for every , we can construct inductively a filtration , such that for each (resp. ) there is a system (resp. ) of representatives of left (resp. right ) cosets of by such that (resp. ) for each . Then the subsets of the above factorization of G are dense in because , for each , .
1. Analyzing the proof, we see that the Theorem holds under the weaker condition: G has a family of subsets such that , for each and, for every non-empty , there is such that .
If but each finitely generating subgroup of G is nowhere dense, we can choose a family such that the corresponding 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 be subsets of G. We say that the product AB is a partial factorization if the subsets are pairwise disjoint (equivalently, 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
there is such that and is a partial factorization;
if the set is infinite, then there is an element such that 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, , and . Then
there is a neighbourhood V of e such that, for and for any , the product is a partial factorization (so ).
It suffices to choose V so that and
We use only in .
4. Let G be countable non-discrete Hausdorff topological group such that the set is infinite for every non-empty open subset U of G. We enumerate , and choose a countable base for non-empty open sets. We put , and use , , to choose inductively two sequences and of finite subsets of G such that for every , , , , is a partial factorization, , , . 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 such that . Since H is finitely generated, the set 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 of countable weight such that U is infinite for every . Applying the inductive construction from comment 5 to and , 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, . 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 . In particular, if G is left topological, then G is box γ-resolvable.
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