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A topological group $G$ is defined to be

$\bullet$ precompact if for any neighborhood $U\subseteq G$ of the unit there exists a finite subset $F\subseteq G$ such that $G=UF$;

$\bullet$ narrow if for any neighborhood $U\subseteq G$ of the unit there exists a countable subset $S\subseteq G$ such that $G=US$;

$\bullet$ separable if there exists a countable subset $S\subseteq G$ such that for any neighborhood $U\subseteq G$ of the unit we have $G=SU$;

$\bullet$ preseparable if there exists a countable subset $S\subseteq G$ such that for any neighborhood $U\subseteq G$ of the unit there exists a finite subset $F\subseteq G$ such that $G=SUF$.

Let us observe the following facts concerning those concepts:

  1. A topological group is preseparable if it is precompact or separable.

  2. More generally, a topological group $G$ is preseparable if $G$ contains a separable closed normal subgroup $H$ whose quotient group $G/H$ is precompact. In the latter case the group $G$ is also narrow.

  3. Each preseparable abelian topological group is narrow.

  4. For any cardinal $\kappa>\mathfrak c$, the Tychonoff power $\mathbb R^\kappa$ is an example of a narrow abelian topological group which is not preseparable.

Problem. Is each preseparable topological group narrow?

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1 Answer 1

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Jan Pachl has informed me that the answer to this problem is affirmative and can be derived from the following helpful fact, proved in Lemma 3.31 of his book "Uniform spaces and measures". I also remember that a similar theorem was proved in the book "Topologies on groups determined by sequences" by I.Protasov and E.Zelenyuk.

Theorem. If a group $G$ is written as $G=\bigcup_{i=1}^nU_iA$ for some sets $U_1,\dots,U_n,A\subset G$, then $G=U_i^{-1}U_iB$ for some $i\in\{1,\dots,n\}$ and some set $B\subseteq G$ of cardinality $|B|\le f(n,|A|)$, where the function $f(n,\kappa)$ is defined by the recursive formula: $f(1,\kappa)=\kappa$ and $f(n,\kappa)=f(n-1,\kappa+\kappa^2)$. In particular, $f(n,\kappa)=\kappa$ for any infinite cardinal $\kappa$ and any $n\in\mathbb N$.

Proof. The proof is by induction on $n$. For $n=1$ it is trivial. Assume that the theorem is proved for all $k<n$. Write $G$ as $G=\bigcup_{i=1}^nU_iA$ for some sets $U_1,\dots,U_n,A\subset G$. If $U_n^{-1}U_nA=G$, then we are done. If $U_n^{-1}U_nA\ne G$, then we can choose a point $x\in G\setminus U_n^{-1}U_nA$ and conclude that $U_nx\cap U_nA=\emptyset$ and hence $U_nx\subset \bigcup_{i=1}^{n-1}U_iA$. Then $U_nA\subset \bigcup_{i=1}^{n-1}U_iAx^{-1}A$ and $G=\bigcup_{i=1}^{n-1}U_i(A\cup Ax^{-1}A)$. By the induction hypothesis, there exists $i\in\{1,\dots,n-1\}$ and a set $B\subset G$ of cardinality $|B|\le f(n-1,|A\cup Ax^{-1}A|)\le f(n-1,|A|+|A|^2)=f(n,|A|)$ such that $G=U_i^{-1}U_iB$.

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