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This question is also posted here.

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.

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  • $\begingroup$ Could you kindly tell us what a $g$-function is? $\endgroup$ – Joel David Hamkins May 29 '13 at 11:32
  • $\begingroup$ $g: \mathbb N \times X \to \tau_X$ is a $g$-function of $X$ if for any $x$ and $n \in \mathbb N$, $x \in g(n+1,x) \subset g(n,x)$. $\endgroup$ – Paul May 29 '13 at 11:40
  • $\begingroup$ It seems that you want to impose some separation axiom, since otherwise the indiscrete space (of any cardinality) would seem to be trivially semi-stratifiable and star-countable. $\endgroup$ – Joel David Hamkins May 29 '13 at 15:00
  • $\begingroup$ It's customary in generalised metric spaces to assume at least $T_3$ ($T_1$ plus regular). This is also customary for stratifiable and semi-stratifiable spaces, AFAIK. $\endgroup$ – Henno Brandsma May 29 '13 at 18:12
  • $\begingroup$ @Joel: maybe I should mentioned it. $\endgroup$ – Paul May 30 '13 at 0:03
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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?

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