Using the definitions from Peter May's [A Concise Course in Algebraic Topology](https://www.math.uchicago.edu/%7Emay/CONCISE/ConciseRevised.pdf), a topological space $X$ is weak Hausdorff if for every compact Hausdorff space $K$ and continuous function $f:K\to X$, $f(K)$ is closed in $X$. A subset $A$ of $X$ is said to be compactly closed if for any compact Hausdorff space $K$ and continuous map $f:K\to X$, $f^{-1}(A)$ is closed in $K$. Finally, $X$ is said to be a $k$-space if every compactly closed subset is closed.


In the lemma at the top of page 40, May gives a characterization of weak Hausdorff spaces which can be paraphrased as such:



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> A $k$-space $X$ is weak Hausdorff if and only if the diagonal $\Delta\_X$ is compactly closed in $X\times X$ under the product topology.
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I was searching for a counterexample to this lemma when the $k$-space assumption is omitted. I came up with a proof of this lemma, and the forward direction didn't require the $k$-space assumption. So in other words, I am asking



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> What is an example of a topological space $X$ such that $\Delta\_X$ is compactly closed in $X\times X$, but $X$ is not weak Hausdorff?
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By the lemma, such an example must not be a $k$-space.



Space with compactly closed diagonal but which is not weak Hausdorff
The space $X$ constructed in Theorem 1.5 of this [preprint](https://arxiv.org/abs/2211.12579) has the required properties. This space contains a non-closed compact metrizable subspace $K$, so is not weakly Hausdorff.


On the other hand, the space $X$ is $k\_2$-metrizable, which means that $X$ is the image of a metrizable space $M$ under a continuous bijective proper map $f:M\to X$ such that for every continuous map $g:H\to X$ from a compact Hausdorff space $H$, the map $f^{-1}\circ g:H\to M$ is continuous. This property implies that the diagonal $\Delta\_X$ is compactly closed in $X\times X$.


