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It is well known that any continuous map from a compact space to a Hausdorff space must be a closed map. Does this fact characterize compactness?

That is, if for a space $X$, every continuous map to any Hausdorff space is closed, does it imply that $X$ is compact?

My guess is no (especially since this is NOT a common result), but I can't find a counterexample.

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up vote 6 down vote accepted

I shall try today to ask my friend Oleg Gutik, who considered such spaces, but now for me seems the following.

Let $\mathscr{AC}$ be the class of all spaces $X$ such that every continuous map of $X$ to any Hausdorff space is closed.

The compactness of a space $X$ of the class $\mathscr{AC}$ mostly depends on separation axioms which hold for the space $X$.

Since each continuous map of an antidiscrete space to any Hausdorff space is constant and hence closed, each antidiscrete space $X$ belongs to $\mathscr{AC}$.

From the other side, each Hausdorff space $X\in\mathscr{AC}$ is $H$-closed, that is $X$ is a closed subspace of any Hausdorff space.

In particular, since each Tychonoff space is dense in its compactification, we see that a Tychonoff space $X$ belongs to $\mathscr{AC}$ iff $X$ is compact.

Moreover, there is the following well-known characterization of $H$-closed spaces: a Hausdorff space $X$ is $H$-closed iff each open cover of $X$ has a finite subfamily such that the union of the closures of its members covers the space $X$. This characterization implies that each regular $H$-closed space is compact.

Update. So, I spoke with Oleg and I have to tell the following.

In my Russian version of Ryszard Engelking’s “General topology” there is a section devoted to $H$-closed and $H$-minimal spaces, containing Exercise 3.12.5. It contains four equivalent conditions characterizing $H$-closed Hausdorff spaces. Moreover, each Hausdorff continuous image of a Hausdorff $H$-closed space is $H$-closed again (I suggest that this result can be easily proved from the characterization of Hausdorff $H$-closed spaces, which I write above). So, a Hausdorff space $X$ belongs to $\mathscr{AC}$ iff $X$ is $H$-closed.

There is the following well-known example of a Hausdorff non-regular and non-compact $H$- closed space $X$. Let $X$ be the unit interval $[0,1]$ where each non-zero point has a base from standard topology of the unit interval $[0;1]$ and the zero has a base consisting of its open neighborhoods in the standard topology without the converging to zero sequence $\{1/n: n\in\mathbb N\}$. The covering criterion of $H$-closedness which I wrote above implies that the space $X$ is $H$-closed.

At last, there exists a $T_1$ space $X$ such that any continuous image of $X$ into a Hausdorff space in constant, so $X\in\mathscr{AC}$. Put $X=\mathbb Z$ and a neighborhood base at the point $n\in X$ is $\{\{n\}\cup [m,\infty)\cap\mathbb Z:m\in\mathbb Z\}$. Then every two non-empty open subsets of the space $X$ have a non-empty intersection, hence each map of $X$ into a Hausdorff space should be constant.

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antidiscrete? Is that a space where no two points can be separated by neighborhoods? And isn't that equivalent to the non-existence of disjoint open sets? – Stefan Hamcke Aug 22 '13 at 13:22
@StefanH.: Antidiscrete space is one with trivial topology – only itself and empty set are open. – user87690 Aug 22 '13 at 13:54
@Stefan: Antidiscrete topology is another name for what I call the indiscrete topology and some call the trivial topology: the coarsest possible topology on a set. – Brian M. Scott Aug 22 '13 at 16:36
@BrianM.Scott: Okay, I thought antidiscrete may be no disjoint open sets, since that would suffice for the map to be constant. – Stefan Hamcke Aug 22 '13 at 16:39

Let $\tau=\{\varnothing\}\cup\{U\subseteq\Bbb N:0\in U\}$; then $\langle\Bbb N,\tau\rangle$ is not compact, but every continuous map $f$ from $X$ to a Hausdorff space is constant: $f[\Bbb N]=\{f(0)\}$. (I’m still thinking about the case in which $X$ has more than $T_0$ separation.)

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It seems that I know such $T_1$ space $X$. Now I am writing about it and other related things in an update of my answer. – Alex Ravsky Aug 22 '13 at 19:19

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