The definition of the compact-open topology differs slightly depending on whether one is working in the context of compactly generated topological spaces or arbitrary topological spaces.

If $X$ and $Y$ are topological spaces, then the compact-open topology on the set of continuous functions $C(X,Y)$ has, as a sub-basis, subsets of the form $V(K,U)$ where $K$ is a compact subset of $X$, $U$ is an open subset of $Y$, and $V(K,U):= \{f\in C(X,Y)\ \vert \ f(K)\subseteq U\}$.

When working with compactly generated topological spaces, this definition is modified to only allow compact sets $K$ which are the image of a compact Hausdorff space (see https://en.wikipedia.org/wiki/Compact-open_topology).

This suggests that not every compact space is the continuous image of a compact Hausdorff space. What is an example of such a space?

  • $\begingroup$ This is not possible in the French Mathematical doctrine for which a compact is by definition separated, seeking the uniqueness of the limit when it exists. I understand that for Americans a compact can be not separated (i.e. the definitions are different). $\endgroup$ – Piquito May 23 '18 at 22:32

This extended abtract by Künzi and van der Zypen seems of interest. It mentions in passing (remark 3, page 3) a reference

Stone, A.H.: Compact and compact Hausdorff, in: Aspects of Topology, pp. 315–324, London Math. Soc., Lecture Note Ser. 93, Cambridge Univ. Press, Cambridge, 1985.

where it is supposedly shown that a compact space need not be the continuous image of a compact $T_2$ space, based on a theorem

If $Y$ is KC and compact, $f: X \to Y$ is onto and continuous with $X$ compact Hausdorff, then $Y$ is Hausdorff.

I assume, but I have no access to the reference, that this theorem is shown in the Stone paper. I did find the (not so hard proof) in this paper (lemma 1)

Then $\alpha(\mathbb{Q})$ the Alexandroff extension of $\mathbb{Q}$, being a well-known example of a KC but not Hausdorff compact space (see Counterexamples in Topology), must be an example, based on this theorem.

Also the van Douwen example mentioned in this paper of a countable anti-Hausdorff (all non-empty open sets intersect) compact KC space (also sequential and US) is such an example.

  • $\begingroup$ I have access to the Stone paper and can send it to anybody interested. $\endgroup$ – Michael Lee Jun 23 '18 at 14:26
  • $\begingroup$ @MichaelLee I'd like to see it, please: use the address which is my user name, all lowercase, no space at gmail. thx in advance. $\endgroup$ – Henno Brandsma Jun 24 '18 at 5:53
  • $\begingroup$ Sent! Let me know if you don't receive the email. $\endgroup$ – Michael Lee Jun 27 '18 at 0:24
  • $\begingroup$ @MichaelLee received, with gratitude. $\endgroup$ – Henno Brandsma Jun 27 '18 at 3:57

Henno Brandsma has already anwered the question. I shall give an elementary proof that $Y = \alpha(\mathbb{Q})$ is not the continuous image of a compact Hausdorff space. Concerning the Alexandroff compactifaction see Alexandroff compactification: continuous function extension. It is obtained from $\mathbb{Q}$ by adjoining a "point at infinity" $\infty$ and defining the open neighborhoods of $\infty$ as complements of compact subsets of $\mathbb{Q}$. All other open sets in $Y$ are just the open subsets of $\mathbb{Q}$. This makes $\mathbb{Q}$ (with its original topology) an open subspace of $Y$.

$Y$ is a non-Hausdorff $T_1$-space (i.e. all points are closed). Assume that there exists a continuous surjection $f : X \to Y$ defined on a compact Hausdorff space $X$. The closed sets $f^{-1}(0)$ and $f^{-1}(\infty)$ have disjoint open neighborhods $U$ and $V$ in $X$ (compact spaces are normal). Define $A = X \backslash U$, $B = X \backslash V$. These are compact subsets of $X$ so that $f(A)$ and $f(B)$ are compact subsets of $Y$. Since $Y$ is a KC space (i.e. all compact subsets are closed), $f(A)$ is closed in $Y$ so that $A' = f(A) \cap \mathbb{Q}$ is closed in $\mathbb{Q}$. $B' = f(B)$ is a compact subset of $\mathbb{Q}$. We have $A' \cup B' = \mathbb{Q}$ and $0 \notin A'$. Therefore $\mathbb{Q} \backslash A'$ is an open neighborhood of $0$ in $\mathbb{Q}$ which is contained in the compact $B'$. This is a contradiction since $0$ does not have compact neighorhoods.

For the sake of completeness let us show that $Y$ is a KC space. Let $Z \subset Y$ be compact. If $\infty \notin Z$, then $Z$ is a compact subset of $\mathbb{Q}$, hence its complement in $Y$ is open. Consider the case $\infty \in Z$. Assume $Z$ is not closed in $Y$. Then $Z' = Z \cap \mathbb{Q}$ is not closed in $\mathbb{Q}$. Choose $x \in \overline{Z'} \backslash Z'$ and a sequence $(x_n)$ in $Z'$ converging to $x$. The set $K = \lbrace x \rbrace \cup \lbrace x_1, x_2, ... \rbrace$ is compact. We may assume that each $x_n$ has an open neighborhood $U_n$ such that $x_m \notin U_n$ for $m > n$ (construct a subsequence if necessary). Then the $U_n$ and $Y \backslash K$ form an open cover of $Z$. By construction it cannot have a finite subcover which is contradiction.

The above arguments remain valid if $\mathbb{Q}$ is replaced by any non-locally compact metrizable space $M$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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