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.$\newcommand{\ol}[1]{\overline{#1}}\newcommand{\mc}[1]{\mathcal{#1}}$
A Hausdorff topological space $(X,\mathcal T)$ is called minimal Hausdorff if there is no Hausdorff topology on $X$ that is strictly weaker than $\mathcal T$. (I.e., it is minimal element of the set of all Hausdorff topologies on $X$ with partial ordering $\subseteq$.)
How to show that every minimal Hausdorff space is H-closed?
Is there an easy counterexample for the other direction? (I guess that finding a counterexample is not that easy – otherwise Willard would very probably mention an example.)
This is part of Problem 17M from Willard: General Topology, p.127
In the preceding problems from that book I have already shown some results on H-closed and minimal Hausdorff spaces:
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Let $X$ be a Hausdorff space. $X$ is H-closed if and only if every open filter in $X$ has a cluster point.
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Let $X$ be a Hausdorff space. $X$ is H-closed if and only if every open cover $\mc C$ of $X$ contains a finite subsystem $\mc D$ such that $\bigcup \{\ol D; D\in\mc D\}=X$, i.e., the closures of the sets from $\mc D$ cover $X$.
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Let $X$ be a Hausdorff space. $X$ is minimal Hausdorff $\Leftrightarrow$ every open filter with unique cluster point converges.
Hint it Willard's book suggest to show that if $X$ Hausdorff and not H-closed, then it is not minimal. I've tried to use the characterization of H-closed spaces using filters. Which means that there is an open filter in $X$ which has no cluster point. But if I tried to use a construction similar to the one I used in the proof of the characterization of minimal Hausdorff spaces via open ultrafilters, the new topology was not necessarily strictly weaker than the original one.
Best Answer
I found the following proof in the paper Horst Herrlich: $T_v$-Abgeschlossenheit und $T_v$-Minimalität, Mathematische Zeitschrift, Volume 88, Number 3, 285-294, DOI: 10.1007/BF01111687. The proof is given there in a greater generality; for $T_v$-minimal and $T_v$-closed spaces, where $v\in\{2,3,4\}$.
Here is a translation of H. Herrlich's proof:$\newcommand{\mc}[1]{\mathcal{#1}}$
Some minor details:
The topology $\mc T''$ is Hausdorff: If we have $x_0\ne y$, $y\in X$ then there are $\mc T'$-neighborhoods $U_x\ni x$, $V_1\ni y$ which are disjoint. Similarly, we have $U_a\ni a$, $V_2\ni y$, which are disjoint. Hence $U_x\cup (U_a\cap X)$ and $V_1\cap V_2$ are $\mc T''$-neighborhoods separating the points $x$ and $y$. The points different from $x_0$ have the same neighborhoods as in $\mc T$.
The fact that $\mc T''$ is strictly weaker than $\mc T$ follows from the fact, that $\{a\}$ is not isolated in $X'$ (equivalently, $X$ is not closed in $X'$). Since $(X,\mc T')$ is Hausdorff, we have disjoint neighborhoods $U_{x_0}\ni x$ and $U_a\ni a$, which separate $x_0$ and $a$ in this space. The set $U_{x_0}$ is not open in $(X,\mc T'')$.
The following example is given in Willard's book, Problem 17M/4, as an example of an H-closed space which is not compact.
Let $\newcommand{\N}{\mathbb N}\N^*=\{0\}\cup\{\frac1n; n\in\mathbb N\}$ with the topology inherited from real line. (I.e., a convergent sequence.) Then we take a topological product $\N\times\N^*$, where $\N$ has discrete topology. (I.e. this is just the topological sum of countably many integer sequences.) We adjoin a new point $q$ with the neighborhood basis consisting of the sets of the form $U_{n_0}(q)=\{(n,1/m)\in\N\times\N^*; n\ge n_0\}\cup\{q\}$. (I.e., $U_{n_0}(q)$ consists of isolated points in all but finitely many sequences.) Let us call this space $X$.
Note: A similar space is described as example 100 in Counterexamples in Topology p.119-120. If you look only at the left half of the picture given in this book, it depicts a typical basic neighborhood of the point $q$.
A topological space is called semiregular, if regular open sets form a base. A set $U$ is regular open if $U=\operatorname{Int} \overline U$. It is known that a space $X$ is Hausdorff minimal if and only if it is H-closed and semiregular.
The space $X$ described above is an H-closed space, but it is not semiregular, since closure of each set $U_{n_0}(q)$ contains all points $(n,0)$ for $n\ge n_0$ in its interior. Hence no regular open set containing $p$ is contained in the basic set $U_{n_0}(q)$. Since this space is not semiregular it is not Hausdorff minimal. Thus this is an example of a topological space which is H-closed but not Hausdorff minimal.