# Is each preseparable topological group narrow?

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?