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

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

‎‎‎$ ‎st‎^{n+1}‎‎ (K, \mathscr{U}‎) =‎ ‎\bigcup ‎ \{ U‎ ‎\in‎ \mathscr{‎U} : U ‎\cap st‎^{n}‎‎‎(K, \mathscr{U}‎) ‎\neq ‎\emptyset \}‎‎‎$

Definition: ‎A space ‎‎$‎X‎$‎ is said to be ‎$ ‎\omega‎ $‎-starcompact if for every open cover ‎$ ‎‎\mathscr{‎U}‎‎‎ $‎ of ‎‎$‎X‎$‎, there is some ‎‎$‎n ‎\in ‎\mathbf{N‎^{+}‎}‎‎$‎ and some finite subset ‎‎$‎B‎$‎ of ‎‎$‎X‎$‎ such that ‎‎$‎st‎^{‎n‎}‎( B, \mathscr{‎U}‎‎) = ‎X‎$‎.‎ ‎

Definition: ‎A Hausdorff topological space ‎‎$‎(X,‎\tau‎)‎‎$‎ is called ‎H‎‎-closed or absolutely closed if it is closed in any Hausdorff space, which contains ‎‎$‎X‎$‎ as a ‎subspace.‎ ‎‎

‎Theorem:‎ Let $X$ be a Hausdorff space. $X$ is H-closed if and only if every open cover $\mathcal{C} $ of $X$ contains a finite subsystem $\mathcal{D}$ such that $\bigcup \{‎‎\overline{‎D‎}‎ : D \in \mathcal{D} \}=X$, i.e., the closures of the sets from $\mathcal{D}$ cover $X$. ‎‎ ‎

According to the above ‎theorem ‎, I concluded that each ‎H-closed ‎space is $ ‎\omega‎ $‎-‎starcompact‎.

Is it possible to conclude that any $ ‎\omega‎ $‎-‎starcompact ‎space‎ is ‎H-‎closed? Or in the Hausdorff ‎space‎‎‎‎ is any $ ‎\omega‎ $‎-‎starcompact‎ space ‎H-‎closed?‎


No. It is well-known (see, for instance, [Mat, Theorem 3]) that a Hausdorff space $X$ is countably compact iff for every open cover $\mathcal U$ there exists a finite subset $F$ of $X$ such that $St(F,\mathcal U)=X$. In particular, each countably compact space is $\omega$-starcompact. So any proper dense countably compact space $X$ of a Hausdorff compact space $Y$ is a counterexample. For instance, we can put $Y=[0,\omega_1]$ endowed with the order topology and $X=[0,\omega_1)$ or $Y$ be a $\Sigma$-product of an uncountable family of compact spaces with at least two points each. I recall that a space $Y$ is $\Sigma$-product of a family $\{X_\alpha:\alpha\in A\}$ if there exists a point $x\in\prod X_\alpha$ such that $Y=\{y\in \prod X_\alpha: |\{\alpha:y_\alpha\ne x_\alpha\}|\le\omega\}.$


[Mat] M. Matveev, A Survey on Star Covering Properties.

  • $\begingroup$ I can not downlode the reference, is it an article? $\endgroup$ – Adrina Jul 10 '19 at 18:28
  • $\begingroup$ @adin Yes, it is an article, available, I guess, in several formats. I have just downloaded its PostScript version. $\endgroup$ – Alex Ravsky Jul 10 '19 at 19:12
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    $\begingroup$ @adin my answer has a downloadable version of the original paper. $\endgroup$ – Henno Brandsma Jul 10 '19 at 21:11

If $X$ is $H$-closed and $\mathcal{U}$ is an open cover, subset of $\mathcal{U}$, say $\mathcal{U}'$, such that $\bigcup \{\overline{U}: U \in \mathcal{U}'\} = \overline{\bigcup \mathcal{U}'} = X$ and if we pick one $x$ from each $U \in \mathcal{U}'$ we get a finite subset $F$ of $X$ such that $\operatorname{st}(F,\mathcal{U})\supseteq \bigcup\mathcal{U}'$ so $\operatorname{st}(F,\mathcal{U})$ is dense and so $\operatorname{st}^2(F,\mathcal{U})= \operatorname{st}(\operatorname{st}(F,\mathcal{U}),\mathcal{U})= X$ and $X$ is star-2-compact, as this is called and so certainly $\omega$-starcompact.

The reverse is far from true, star-1-compact is equivalent to countable compactness for Hausdorff spaces van Douwen et al.'s paper). So Alex Ravsky's examples are $\omega$-starcompact Hausdorff (even Tychonoff) but not $H$-compact.


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