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?



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> **Problem.** Does every $\omega$-narrow topological group have countable discrete cellularity?
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A general version of this problem for uniform spaces has negative version.



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> **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 $X$. 
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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



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



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> **Problem 2.** Let $G$ be an $\omega$-narrow Raikov-complete (Abelian) topological group. Has $G$ countable discrete cellularity? 
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Also the following related question of Tkachenko remains open.



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> **Question.** Is there an $\omega$-narrow topological group containing continuum many pairwise disjoint open sets?
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Does each $\omega$-narrow topological group have countable discrete cellularity?
Mikhail Tkachenko informed me that the problem has a counterexample, constructed in [Example 8.2.1](https://books.google.com/books?id=hIEnzrOBbW0C&pg=PA526) of [his book](https://books.google.com/books/about/Topological_Groups_and_Related_Structure.html?id=hIEnzrOBbW0C&redir_esc=y) 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.




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An example of a complete Abelian P-group which is $\omega$-narrow but has uncountable discrete cellularity was constructed by Tkachenko in [this paper](https://www.emis.de/journals/CMUC/cmuc0103/abs/tkachen.htm) (see also [another paper](https://doi.org/10.1016/S0166-8641(03)00217-7) of Tkachenko for the proof of the equivalence of $\mathbb R$-factorizability and countable discrete cellularity for P-groups). This example answers Problem 2. 


