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

Star compactness is stronger than pseudocompactness and weaker than countable compactness.

Is every regular star compact metaLindelof space compact?

Thanks for your help.

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$\newcommand{\st}{\operatorname{st}}$You can also derive the result as a corollary of the result that a $T_1$, star-Lindelöf, metaLindelöf space is Lindelöf, which can be proved using the same ideas, but perhaps even more easily.

Suppose that $X$ is $T_1$, star-Lindelöf and metaLindelöf. Let $\mathscr{U}$ be an open cover of $X$; $\mathscr{U}$ has a point-countable open refinement $\mathscr{V}$, and there is a Lindelöf $A\subseteq X$ such that $\st(A,\mathscr{V})=X$. Recursively construct a set $A_0=\{x_\xi:\xi<\alpha\}\subseteq A$ such that $\bigcup_{\xi<\alpha}\st(x_\xi,\mathscr{V})=X$ and $x_\eta\notin\st(x_\xi,\mathscr{V})$ whenever $\xi<\eta<\alpha$. For $\xi<\alpha$ let $W_\xi=\st(x_\xi,\mathscr{V})$; $W_\xi\cap A_0=\{x_\xi\}$, so $A_0$ is a closed, discrete subset of $X$. But then $A_0$ is a closed subset of the Lindelöf subspace $A$, so $A_0$ is Lindelöf and must therefore be countable, $\{V\in\mathscr{V}:V\cap A_0\ne\varnothing\}$ is a countable subcover of $\mathscr{V}$, and $X$ is Lindelöf.

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Here I give a proof for the question. However I'm not sure that it is right. Could you help me ? Thanks for your any comment.

Proof: Let $X$ be a star compact metaLindelöf space. To prove that $X$ is compact it suffices to show that $X$ is Lindelöf since every regular star compact Lindelöf space is compact.

Suppose not. Then there exists an open cover $\mathcal U$ of $X$ such that $\mathcal U$ contains no countable subcover. Since $X$ is metaLindelöf, it follows that $\mathcal U$ has an point-countable open refinement $\mathcal U'$. For $\mathcal U'$ as an open cover of $X$ there exists a closed discrete set $D$ of $X$ such that $\operatorname{St}(D, \mathcal U') = X$. To see it, choose inductively a point $x_\alpha \notin \operatorname{St}(\{x_\beta: \beta < \alpha\}, \mathcal U')$. If $\lambda$ is the first ordinal for which this choice is impossible, then $D=\{x_\alpha: \alpha < \lambda \}$ is a closed discrete subset of $X$ and $\operatorname{St}(D, \mathcal U') = X$. It is evident that $|D| > \omega$ otherwise $\mathcal U$ would contain countable subcover.

Let $\mathcal W=\{\operatorname{St}(x_\alpha, \mathcal U'): x_\alpha \in D\}$. Clearly, $\mathcal W$ is a new point-countable open cover of $X$ the cardinality of which is greater than $\omega$. Since $X$ is star compact, it follows that there is a compact subset $K$ of $X$ such that $\operatorname{St}(K, \mathcal W) = X$. It is not difficult to see that, for each $x_\alpha \in D$, $K \cap \operatorname{St}(x_\alpha, \mathcal U') \not= \emptyset$. Pick a point $y_\alpha \in K \cap \operatorname{St}(x_\alpha, \mathcal U')$ and let $K'=\{y_\alpha: \alpha < \lambda\} \subset K$. It is clear that $|K'| > \omega$ otherwise we can conclude that $|\mathcal W| \le \omega$, which contradicts that $|\mathcal W| > \omega$.

Finally we prove that $K'$ is a closed discrete subset of $K$. To see it, it suffices to show that for any $z \in X \setminus K'$, there is a neighborhood $U$ of $z$ such that $U \cap K'=\emptyset$. In fact, since $\mathcal W$ covers $X$, there exists a point $x_\alpha \in D$ such that $z \in \operatorname{St}(x_\alpha, \mathcal U')$. Choose an open set $U$ such that $y_\alpha \notin U$ and $U \subset \operatorname{St}(x_\alpha, \mathcal U')$. Clearly, $U \cap K' = \emptyset$.

Now we can conclude that $K'$ is an infinite closed discrete subset of $K$. This contradicts that $K$ is compact. This completes the proof.

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  • $\begingroup$ I think that you need to choose $K'$ a little more carefully in order to ensure that it’s discrete: use $\mathscr{W}$ to choose $K'\subseteq K$ the same way that you used $\mathscr{U}'$ to choose $D\subseteq X$. Then for each $y\in K'$ you can choose some $\alpha(y)<\lambda$ such that $y\in\operatorname{St}(x_{\alpha(y)},\mathscr{U}')$ and know that $K'\cap\operatorname{St}(x_{\alpha(y)},\mathscr{U}')=\{y\}$. ... $\endgroup$ – Brian M. Scott Dec 5 '13 at 18:46
  • $\begingroup$ ... And if $z\in X\setminus K'$, there is a $y\in K'$ such that $z\in\operatorname{St}(x_{\alpha(y)},\mathscr{U}')$, and $\big(\operatorname{St}(x_{\alpha(y)},\mathscr{U}')\big)\setminus\{y\}$ is then a nbhd of $z$ disjoint from $K'$. $\endgroup$ – Brian M. Scott Dec 5 '13 at 18:47
  • $\begingroup$ @BrianM.Scott: Thanks for your nice comment. Do you think the proof is OKay? By the same method, can we prove that every star lindelof metalindelof is lindelof since every lindelof space cannot contain a closed discrete subset the cardinality of which is greater than $\omega$? $\endgroup$ – Paul Dec 6 '13 at 0:50
  • $\begingroup$ With the change at the end that I suggested it would be okay. Or you can get it as a corollary of the result that you suggested, that a star-Lindelöf, metaLindelöf space is Lindelöf; I’ve posted a short proof of that, using basically the same techniques. $\endgroup$ – Brian M. Scott Dec 6 '13 at 18:53

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