Does each $\omega$-narrow topological group have countable discrete cellularity?

A topological space $$X$$ is defined to have countable discrete cellularity if each discrete family of open subsets of $$X$$ is at most countable.

A family $$\mathcal F$$ of subsets of a topological space $$X$$ is called discrete if each point $$x\in X$$ has a neighborhood $$O_x\subset X$$ that intersects at most one set $$F\in\mathcal F$$.

It is easy to see that a Tychonoff space $$X$$ has countable discrete cellularity if and only if for any continuous map $$f:X\to M$$ to a metric space $$M$$ the image $$f(X)$$ is separable.

A topological group $$G$$ is $$\omega$$-narrow if for any neighborhood $$U$$ of the unit there exists a countable subset $$C\subset G$$ such that $$G=C\cdot U$$. By a classical theorem of Guran, a topological group is $$\omega$$-narrow if and only if $$G$$ is topologically isomorphic to a subgroup of a Tychonoff product of metrizable separable topological groups.

It is easy to see that a topological group is $$\omega$$-narrow if it has countable discrete cellularity. What about the converse?

Problem. Does every $$\omega$$-narrow topological group have countable discrete cellularity?

A general version of this problem for uniform spaces has negative version.

Example. Let $$X$$ be an uncountable discrete topological space endowed with the uniformity, generated by the base consisting of entourages $$E_F=\{(x,y)\in X\times X:\{x,y\}\cap F\ne\emptyset\;\Rightarrow\; x=y\}$$ where $$F$$ runs over finite subsets of $$X$$.

It is clear that $$X$$ does not have countable discrete cellularily and the uniformity $$\mathcal U$$ is totally bounded.

Such a totally bounded example with uncountable discrete cellularity cannot be constructed in the framework of topological groups because of the following known

Fact. Each totally bounded topological group has countable (discrete) cellularity.

Added in Edit. Mikhail Tkachenko informed me that Problem has a counterexample (described in my answer below). However, the group in this counterexample is essentially non-complete.

Problem 2. Let $$G$$ be an $$\omega$$-narrow Raikov-complete (Abelian) topological group. Has $$G$$ countable discrete cellularity?

Also the following related question of Tkachenko remains open.

Question. Is there an $$\omega$$-narrow topological group containing continuum many pairwise disjoint open sets?

Mikhail Tkachenko informed me that the problem has a counterexample, constructed in Example 8.2.1 of his book with Arhangelskii.

This example looks as follows. Consider the uncountable power $$C_2^{\omega_1}$$ of two-element cyclic group $$C_2=\{0,1\}$$, endowed with the $$\omega$$-box topology. In this topological group consider the subgroups $$G$$ (and $$H$$) consisting of functions $$\varphi:\omega_1\to C_2$$ whose support $$\operatorname{supp}(\varphi):=\varphi^{-1}(0)$$ is finite (and has even cardinality). It can be shown that

$$\bullet$$ the topological groups $$G$$ and $$H$$ are $$\omega$$-narrow;

$$\bullet$$ $$H$$ is dense in $$G$$;

$$\bullet$$ $$G$$ is Lindelof and hence has countable discrete cellularity;

$$\bullet$$ $$H$$ has uncountable discrete cellularity.

An example of a complete Abelian P-group which is $$\omega$$-narrow but has uncountable discrete cellularity was constructed by Tkachenko in this paper (see also another paper of Tkachenko for the proof of the equivalence of $$\mathbb R$$-factorizability and countable discrete cellularity for P-groups). This example answers Problem 2.

• Funny, I actually had a (too) quick look in that very book when I read the question and totally overlooked this. Mar 25 '19 at 15:06