Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

It is well known that any continuous map between 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.

share|improve this question

2 Answers 2

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 subspase 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 Hausdoff 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 criterium 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 Hasdorff 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.

share|improve this answer
    
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
1  
@StefanH.: Antidiscrete space is one with trivial topology – only itself and empty set are open. –  user87690 Aug 22 '13 at 13:54
2  
@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.)

share|improve this answer
1  
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.