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:

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

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.

Each preseparable abelian topological group is narrow.

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?