0
$\begingroup$

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 $st(A,\mathcal{U})=\bigcup\{U\in\mathcal{U}:U\cap A\neq\emptyset\}=X$

$\endgroup$
0
1
$\begingroup$

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.

$\endgroup$
0

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.