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


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 star countable if whenever $\mathscr{U}$ is an open cover of $X$, there is a countable subspace $K$ of $X$ such that $X = \operatorname{St}(K,\mathscr{U})$, where $\operatorname{St}(K,\mathscr{U}) = \bigcup \{ U \in \mathscr{U} : K \cap U \neq \emptyset \}$.


closed as off-topic by Todd Trimble Apr 7 '17 at 22:18

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  • $\begingroup$ Is this related to your earlier question on countable tighness etc.? A semi-startifiable space obviously has countable pseudocharacter and if Lindelöf (pretty close to star countable) it would be hereditarily separable,and so countably tight. The other ound would then imply $|X| \le \mathfrak{c}$ So you need a non-separable space. $\endgroup$ – Henno Brandsma Apr 7 '17 at 20:20
  • $\begingroup$ I'm voting to close this question as "off-topic" because it is a duplicate of this question: mathoverflow.net/questions/132209/… (Technically, that's a site violation. Please don't repeat questions.) $\endgroup$ – Todd Trimble Apr 7 '17 at 22:18
  • $\begingroup$ This question (and its older duplicate) is answered in (mathoverflow.net/questions/132209/…). A counterexample if the Katetov extension of $\omega$. $\endgroup$ – Taras Banakh Apr 15 '17 at 12:58