# question about $‎\omega‎$‎-‎starcompact ‎space‎

‎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 $$Y=\{y\in \prod X_\alpha: |\{\alpha:y_\alpha\ne x_\alpha\}|\le\omega\}.$$

References

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

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