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 1.** Does every $\omega$-narrow topological group have countable discrete cellularity?
This problem can also be asked for uniform spaces. A uniform space $(X,\mathcal U)$ is *$\omega$-narrow* if for any entouage $U\in\mathcal U$ there exists a countable set $C\subset X$ such that for every $x\in X$ there exists $c\in C$ with $(c,x)\in U$. It can be shown that a uniform space is $\omega$-narrow if it has countable discrete cellularity. What about the converse?
>**Problem 2.** Does every $\omega$-narrow uniform space have countable discrete cellularity?
Dear Taras,
Sorry for my ignorance with uniform spaces. Do I understand correctly that the unit ball $B_X$ of a Banach space equipped with the weak topology is an example of $\omega$-narrow uniform space with respect to the natural uniformity that generates the weak topology?
Best, Vladimir
Taras Banakh
Dear Taras, my previous question about $B_X$ in weak topology (it seems to me that it is an $\omega$-narrow uniform space) is motivated by the following example.
Consider $X = \ell_1[0, 1]$, that is the Banach space of functions $x: [0,1] \to \mathbb R$ that have at most countable support and satisfy the condition
$$
\|x\| = \sum_{t \in [0, 1]} |f(t)| < \infty.
$$
The dual space is the space $X^* = \ell_\infty[0, 1]$ of all bounded functions on $[0, 1]$. For Every $t \in [0, 1]$ denote $\delta_t$ the evaluation functional, which maps each $x \in \ell_1[0, 1]$ to $x(t)$. Consider subsets $A_t = \{x \in B_X: \delta_t(x) > 9/10\}$. These subsets form a disjoint uncountable collection of relatively weakly open subsets of $B_X$. Let us demonstrate that this collection is a discrete family of subsets. Let $x \in B_X$ be an arbitrary element. If there is a $t \in [0, 1]$ such that $x(t) > 1/2$ then all the remaining values $x(s)$ must be smaller than 1/2, so the weak neighborhood $U = \{y \in B_X: \delta_t(y) > 1/2\}$ intersects $A_t$ and does not intersect all the remaining $A_s$. If $x(t) \le 1/2$ for all $t \in [0, 1]$, then pick a finite set $D \subset [0, 1]$ such that $\sum_{t \in [0, 1] \setminus D} x(t) < 1/10 $. Then $V = \{y \in B_X: \delta_t(y) \le 6/10, t \in D\} \cap \{y \in B_X: \sum_{t \in [0, 1] \setminus D} y(t) < 1/10 \}$ is a weak neighborhood of $x$ (it is generated by finitely many linear functionals), but it does not intersect any of $A_t$.
This demonstrates that for $X = \ell_1[0, 1]$ the topological subspace $B_X$ of $X$ equipped with the weak topology does not have countable discrete cellularity.
Best, Vladimir
Dear Vladimir, thank you for your answer, which nicely complements the answer of Tkachenko to the same question but posted on MathOverflow
https://mathoverflow.net/questions/325805/does-each-omega-narrow-topological-group-have-countable-discrete-cellularity
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