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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?

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  • $\begingroup$ 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
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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.

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  • $\begingroup$ 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
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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 $M$.

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