star compact Hausdorff is countably compact

Question

I have read in some papers stated that every star compact Hausdorff is countably compact. I don't know how to prove it.

Note: A space $X$ is called star compact if for any open cover $\mathcal{U}$ of $X$, there exists finite subset $A\subseteq X$ such that $st(A,\mathcal{U})=\bigcup\{U\in\mathcal{U}:U\cap A\neq\emptyset\}=X$

Answer 1

This is theorem 2.1.5 in this paper, Star covering properties, by E.K. van Douwen, G.M. Reed, A.W. Roscoe, and I.J. Tree, Topology and its Applications $39$ ($1991$), $71$-$103$.

Read the definitions at the start and you will see this is exactly what you need.

Added: Here is the proof, slightly revised and with one minor error corrected.

If $X$ is Hausdorff but not countably compact, there is an infinite closed discrete $D\subseteq X$; say $D=\{x_n:n\in\Bbb N\}$. For each $n\in\Bbb N$ there is an open $U_n$ such that $U_n\cap D=\{x_n\}$. For each $m\in\Bbb N$ let $D_m=\{x_n\in D:2^m\le n<2^{m+1}\}$; clearly $|D_m|=2^m$. Since $D_m$ is finite, the Hausdorffness of $X$ ensures that the points $x_n\in Y_m$ have pairwise disjoint open nbhds $V_n$, respectively, and we may further assume that $V_n\subseteq U_n$ for each $n\in\Bbb N$.

Now let $\mathscr{V}_m=\{V_n:x_n\in D_m\}$, and let $\mathscr{V}=\{X\setminus D\}\cup\bigcup_{m\in\Bbb N}\mathscr{V}_m$; clearly $\mathscr{V}$ is an open cover of $X$. Let $F\subseteq X$ be finite, and let $m=|F|$. Then $m<2^m=|D_m|=|\mathscr{V}_m|$, and each member of the pairwise disjoint family $\mathscr{V}_m$ can contain at most one point of $F$, so there is some $n$ such that $x_n\in D_m$ and $V_n\cap F=\varnothing$. But then $x_n\notin\operatorname{st}(F,\mathscr{V})$, so $X$ is not star compact.