Is every [semi-stratifiable](https://topospaces.subwiki.org/wiki/Semi-stratifiable_space) space $\omega$-monolithic?




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Definitions
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A topological space $(X,\tau)$ is called **semi-stratifiable** if there exists a function $g:\omega\times X\to\tau$ such that:


1. for any point $x$ of $X$ holds $\{x\}=\bigcap\_{n\in\omega} g(n,x)$;
2. for any point $x$ of $X$ and a sequence $\{x\_n\}$ of $X$, if $x \in g(n,x\_n)$ for each $n$, then $x\_n \to x$.


A topological space $X$ is said to be **$\omega$-monolithic** if $nw(\overline{A}) \le \omega$ for any subset $A \subset X$ with $|A| \le \omega$.


$nw(X)$ denotes the [cardinal function](https://en.wikipedia.org/wiki/Cardinal_function#Cardinal_functions_in_topology)
 called [network weight](https://en.wikipedia.org/wiki/Base_(topology)#Weight_and_character), which is minimal cardinality of a network
$$nw(X)=\min\{|\mathcal N|: \mathcal N \text{ is a net for } X\}+\omega.$$


If not. What if $X$ is semi-metric space?


Note that $X$ is semi-metrisable iff $X$ is first countable and semi-stratifiable;



Is every semi-stratifiable space $\omega$-monolithic?
As a counterexample to this question we can consider the Katetov extension $\kappa\omega$ of the discrete space of all finite ordinals $\omega$. 


By definition, $\kappa\omega$ is the space of all ultrafilters on $\omega$ with the topology in which a neighborhood base of an ultrafilter $\mathcal U$ consists of the sets $\{\mathcal U\}\cup U$ where $U\in\mathcal U$. Here we identify $\omega$ with the set of principal ultrafilters on $\omega$. So, $\kappa\omega=\omega\cup\omega^\*$ where $\omega^\*$ is the set of free ultrafilters on $\omega$. The space $\kappa\omega$ is separable but has cardinality $2^{\mathfrak c}>\mathfrak c$. Since the subspace $\omega^\*$ of free ultrafilters is discrete and uncountable, the separable space $\kappa\omega$ has uncountable network weight, so is not $\omega$-monolithic.


On the other hand, the space $\kappa\omega$ is semi-stratifiable. This is witnessed by the function $g$ defined by $g(n,\mathcal U)=\{\mathcal U\}$ if the ultrafilter $\mathcal U$ is principal and $g(n,\mathcal U)=\{\mathcal U\}\cup(\omega\setminus n)$ if $\mathcal U$ is free. 


This example also answers [A question on semi-stratifiable space](https://mathoverflow.net/questions/132209/a-question-on-semi-stratifiable-space).


