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A topological space $X$ is said to be star compact if whenever $\mathscr{U}$ is an open cover of $X$, there is a compact subspace $K$ of $X$ such that $X = \operatorname{St}(K,\mathscr{U})$.

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 $X = \operatorname{St}(K,\mathscr{U})$.

We know that there are some topological spaces which are star countable but not star compact. However I don't know whether star compact implies star countable. Is there a topological space which is star compact but not star countable?

Added: $St(K, \mathscr{U})=\cup\{u\in \mathscr{U}: u \cap K \neq \emptyset\}$

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Could you provide a definition of $\operatorname{St}(K,\mathscr{U})$, please? –  user20266 Jan 1 '12 at 9:25
    
$St(K, \mathscr{U})=\cup\{u\in \mathscr{U}: u \cap K \neq \emptyset\}$ –  Paul Jan 1 '12 at 9:49
    
Such a space (if it exists) cannot be normal, because for $T_4$ spaces star compact is equivalent to star finite, so certainly star countable. –  Henno Brandsma Jan 1 '12 at 10:46
    
It also hasn't countable extent, and hence it can't be countably compact. However, every star compact space is pseducompact. So the space must be pseudocompact but not countably compact! –  Paul Jan 1 '12 at 10:58
    
The definition of $\operatorname{St}(K, \mathscr{U})$ added ad the end of your post does not say what you believe it says. You define a set $V\subseteq\mathscr{U}$ by $V=\{u\in \mathscr{U}: u \cap K \neq \emptyset\}$, then what is $\cup V$... I guess you mean something like $\operatorname{St}(K, \mathscr{U})=\bigcup\limits_{u\in V}u$. –  Did Jan 1 '12 at 12:28

1 Answer 1

up vote 2 down vote accepted

Let $X=\Big(\beta\omega_1\times(\omega_2+1)\Big)\setminus\Big((\beta\omega_1\setminus \omega_1)\times\{\omega_2\}\Big)$ as a subspace of $\beta\omega_1\times(\omega_2+1)$; I claim that $X$ is star compact.

Let $\mathscr{U}$ be an open cover of $X$. For each $\xi\in\omega_1$ there are $U_\xi\in\mathscr{U}$ and $\alpha_n\in\omega_2$ such that $$\langle \xi,\omega_2\rangle\in \{\xi\}\times(\alpha_n,\omega_2]\subseteq U_\xi\;.$$ Let $\alpha=\sup_\xi\alpha_\xi<\omega_2$, and let $K=\beta\omega_1\times\{\alpha+1\}$; $K$ is compact, and $$\omega_1\times \{\omega_2\}\subseteq \operatorname{st}(K,\mathscr{U})\;,$$ since $U_\xi\subseteq \operatorname{st}(K,\mathscr{U})$ for each $\xi\in\omega_1$. The ordinal space $\omega_2$ is countably compact, so $\beta\omega_1\times\omega_2$ is countably compact and therefore star finite, and there is a finite $F\subseteq \beta\omega_1\times\omega_2$ such that $\beta\omega_1\times\omega_2\subseteq\operatorname{st}(F,\mathscr{U})$. But then $K\cup F$ is compact, and $\operatorname{st}(K\cup F,\mathscr{U})=X$, as desired.

However, $X$ is not star countable. To see this, let $$\mathscr{U}=\{\beta\omega_1\times\omega_2\}\cup\Big\{\{\xi\}\times(\omega_2+1):\xi\in\omega_1\Big\}\;;$$ $\mathscr{U}$ is certainly an open cover of $X$, but if $C$ is any countable subset of $X$, we can choose $\xi\in\omega_1$ such that $C\cap\big(\{\xi\}\times (\omega_2+1)\big)=\varnothing$, and then $\langle \xi,\omega_2\rangle\notin\operatorname{st}(C,\mathscr{U})$.

This is a modification of Example 2.1 of Yan-Kui Song, On $\mathcal{K}$-Starcompact Spaces, Bull. Malays. Math. Soc. (2) 30(1) (2007), 59-64, which is available as a PDF here.

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