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


Definitions

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 $\operatorname{St}(K,\mathscr{U}) = \bigcup \{ U \in \mathscr{U} : K \cap U \neq \emptyset \}$.

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

  • This question does not appear to be about research level mathematics within the scope defined in the help center.
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ 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