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 $X$.


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 $Y$.

  • $\begingroup$ 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)$? $\endgroup$ May 21 '20 at 23:29
  • 1
    $\begingroup$ @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 $\overline Z\subset \overline {Y\setminus V}=Y\setminus V.$ $\endgroup$ May 22 '20 at 13:21

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