# Every minimal Hausdorff space is H-closed

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.$\newcommand{\ol}{\overline{#1}}\newcommand{\mc}{\mathcal{#1}}$

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:

• Let $X$ be a Hausdorff space. $X$ is H-closed if and only if every open filter in $X$ has a cluster point.

• 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 $X$.

• 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.

• There is some info and many references in the exercises to Engelking's topology book. According to exercise 3.12.5 e) a Hausdorff space is minimal Hausdorff (Engelking calls it H-minimal) if and only if it is H-closed and semiregular (the regular open sets are a basis of the topology). See also exercise 1.7.8. – t.b. Apr 23 '12 at 7:56
• Thanks a lot @t.b. (Engelking is my favorite reference for general topology; I don't know why, but I did not think about checking that book until you reminded me.) – Martin Sleziak Apr 25 '12 at 16:58

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:$$\newcommand{\mc}{\mathcal{#1}}$$

Let $$(X,\mc T)$$ be a space which is not H-closed. Then there exists a $$T_2$$-space $$(X',\mc T')$$ such that $$X'=X\cup\{a\}$$ and $$X$$ is not closed in $$(X',\mc T')$$. If we choose an arbitrary element $$x_0\in X$$ then $$\mc T''=\{M; M\in\mc T; x_0\in M \Rightarrow M\cup\{a\}\in\mc T'\}$$ is a $$T_2$$-topology on $$X$$ which is strictly weaker than $$\mc T$$. Hence $$(X,\mc T)$$ is not $$T_2$$-minimal.

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.

• I find it helpful to view $\langle X,\mathcal{T}''\rangle$ as the quotient of $X'$ obtained by identifying $a$ and $x_0$. – Brian M. Scott Apr 29 '12 at 4:12