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.

  • $\begingroup$ 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

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?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.