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

We know:

1: A Hausdorff space X is H-closed if and only if every open filter in X has a cluster point.

2: Every ultrafilter in the family of all open subsets of X converges.

‎Can ‎any‎one ‎help ‎me ‎to ‎prove the statement below:‎

‎ 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 $$X$$.

• This question seems relevant: math.stackexchange.com/questions/2202773/h-closed-compact, see in particular at.yorku.ca/p/a/c/a/15.pdf Jul 23 '19 at 20:34
• I have given an A for one direction, And it is easy to show that if X is Hausdorff, regular, and H-closed then X is compact. But I do not as yet have a general answer for the other direction. Jul 24 '19 at 20:11
• Got the other half. Added it to my A. Jul 24 '19 at 21:05

PART ONE. let $$X$$ be a subspace of $$Y$$ where $$Y$$ is Hausdorff and $$X$$ is not closed in $$Y.$$

Notation: $$Cl_Y$$ and $$Cl_X$$ denote, respectively, closure in $$Y$$ and closure in $$X.$$

Take $$y\in Cl_Y(X)$$ with $$y\not \in X.$$ For each $$x\in X$$ let $$U_x, V_x$$ be disjoint open subsets of $$Y$$ with $$x\in U_x$$ and $$y\in V_x.$$ Then in the space $$X,$$ the family $$C=\{X\cap U_x: x\in X\}$$ is an open cover of $$X.$$

Suppose $$D$$ is a finite subset of $$C$$ such that $$\cup \{Cl_X(d):d\in D\}=X.$$ Let $$E$$ be a finite subset of $$X$$ such that $$D=\{X\cap U_x: x\in E\}.$$

Now $$E$$ is not empty ... (otherwise $$X$$ is empty, but $$X$$ is not closed in $$Y$$)... but it is finite, so $$V=\cap_{x\in E}V_x$$ is open in the space $$Y,$$ and $$y\in V.$$

Now $$V$$ is open in $$Y$$ and disjoint from $$\cup_{x\in E}U_x,$$ so (using the finiteness of $$E$$ again ) we have$$\emptyset=V\cap Cl_Y(\,\cup_{x\in E} \,U_x\,)=$$ $$= V\cap (\,\cup_{x\in E}\, Cl_Y ( U_x)\,) \supset$$ $$\supset V\cap (\,\cup_{x\in E}\,Cl_X(X\cap U_x)\,)=V\cap X.$$ Bur $$y\in V$$ and $$V$$ is open in $$Y$$, so $$y \not \in Cl_Y(X),$$ a contradiction.

So no such $$D$$ exists.

PART TWO. In Part One we showed that if $$X$$ is Hausdorff and not H-closed then $$X$$ has an open cover $$C$$ such that $$\overline {\cup D} \ne X$$ for every finite $$D\subset C.$$

Now we suppose that $$X$$ is Hausdorff and that $$C$$ is an open cover of $$X$$ such that $$\overline {\cup D}\ne X$$ for every finite $$D\subset C,$$ and construct a Hausdorff space $$Y$$ such that $$X$$ is a non-closed subspace of $$Y.$$

Take $$y\not \in X$$ and let $$Y=X\cup \{y\}.$$

Let $$T_X$$ be the topology on $$X.$$ Let $$[C]^{<\omega}$$ denote the set of all finite subsets of $$C.$$ (Notation borrowed from Set Theory).

Let $$F=\{Y\setminus Cl_X (\cup D): D\in [C]^{<\omega}\}.$$

I will leave it to the reader to confirm that

(i). $$T_X\cup F$$ is a base for a Hausdorff topology $$T_Y$$ on $$Y$$.

(ii). $$\{t\cap X: t\in T_Y\}=T_X.$$ That is, $$(X,T_X)$$ is a subspace of $$(Y,T_Y).$$

(iii). $$y\in Cl_Y(X).$$ So $$X$$ is not closed in $$Y$$.

• In PART ONE, $V$ is disjoint from $\bigcup_{x\in E}U_x$, but why is it disjoint from $\text{Cl}_Y(\bigcup_{x\in E}U_x)$? May 21 '20 at 23:29
• @VictorGustavoMay . $V$ is open in $Y$. An open set $V$ in a space $Y$ that's disgoint from some $Z\subset Y$ is always disjoint from $\overline Z$. Because $\overline Z\subset \overline {Y\setminus V}=Y\setminus V.$ May 22 '20 at 13:21