A question on semi-stratifiable space

Question

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A space $X$ is callled semi-stratifiable space if it has a $g$-function such that: for any point $x$ of $X$ and a sequence $\{x_n\}$ of $X$ if $x \in g(n,x_n)$, then $x_n \to x$.

Note that every Moore space is semi-stratifiable. We know the cardinality of a star countable Moore space is not greater than $\mathfrak c$.

A topological space $X$ is said to be star countable if whenever $\mathscr{U}$ is an open cover of $X$, there is a countable subspace $K$ of $X$ such that $X = \operatorname{St}(K,\mathscr{U})$.

Is there a star countable semi-stratifiable space $X$ with $|X|> \mathfrak c$?

Thanks for your help.

Answer 1

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 and hence star-countable. On the other hand, $\kappa\omega$ has cardinality $2^{\mathfrak c}>\mathfrak c$. Also 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.

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. This answers question Is every semi-stratifiable space $\omega$-monolithic?