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I previously asked (see Compact refinement of a covering) about whether there is a compact set covered by 2 open sets which cannot be decomposed as the union of 2 (not necessarily disjoint) compact subsets, each of which is contained in one of the open sets. That compactification of $\mathbb{Q}$ solved that question. Now I'm curious whether there is such a compact set if you add the condition that the space must be locally compact. (This is asking for another counterexample to the previous question which is additionally locally compact.) By locally compact I mean every point has a neighborhood base of compact sets.

davik
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This is not an answer, but it is too long for a comment and shows some restrictions if we search for a counterexample (which, as I believe, exists).

Known facts:

(1) For compact $T_1$-spaces, Hausdorff is equivalent to normal.

(2) For $KC$-spaces, the "compact refinement property" implies normal. The converse holds for compact $KC$-spaces.

Thus, any non-Hausdorff compact $KC$-space is a counterexample to your first question. But now we have

(3) For locally compact spaces, $KC$ is equivalent to Hausdorff.

The implications Hausdorff $\Rightarrow$ $KC$ $\Rightarrow$ $T_1$ are true without any further assumptions. So let $X$ be a locally compact (which means that each point has a neighborhood base consisting of compact sets) $KC$-space. Consider two distinct points $x,y \in X$. There exist an open neighborhhod $U$ of $x$ and a compact $C$ such that $U \subset C \subset X \setminus \{ y \}$. But $C$ is closed, hence $V = X \setminus C$ is an open neighborhood of $y$ such that $U \cap V = \emptyset$.

This means that we cannot find a counterexample among compact locally compact $KC$-spaces.

Here is some related material:

Is a locally compact space a KC-space if and only if it is Hausdorff?

Künzi, Hans-Peter A., and Dominic van der Zypen. "The Construction of Finer Compact Topologies." Dagstuhl Seminar Proceedings. Schloss Dagstuhl-Leibniz-Zentrum für Informatik, 2005. https://www.researchgate.net/publication/30814746_The_Construction_of_Finer_Compact_Topologies

Paul Frost
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  • Hi, this isn't related to this question but since you mentioned the one point compactification of Q before, I'm curious whether you know if it's a k-space? I suspect it is since someone else gave its square as an example of a non k-space, but I'm sort of lost for a proof. – davik Oct 22 '18 at 16:51
  • @davik Each compact space is a $k$-space! – Paul Frost Oct 22 '18 at 17:03
  • I think we aren't using the same definition. My k-space is any space such that a subset is closed iff the inverse image under a map from a compact Hausdorff space is closed. And of course my compact spaces don't have to be Hausdorff – davik Oct 22 '18 at 17:14
  • Also possible I'm just missing something obvious – davik Oct 22 '18 at 17:14
  • Sorry should say closed iff inverse image under any such map is closed – davik Oct 22 '18 at 19:40
  • @davik Okay, my understanding was as in https://en.wikipedia.org/wiki/Compactly_generated_space. I also do not assume that compact spaces are Hausdorff. So your definition is: $A \subset X$ is closed iff for all continuous $f : K \to X$, where $K$ is compact Hausdorff, $f^{-1}(A)$ is closed? – Paul Frost Oct 22 '18 at 22:18
  • Yup. I can't see how to show it is a k space – davik Oct 22 '18 at 22:47
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    Oh I see how to show it's a k-space. For future reference the sticking point for me was if $A$ doesn't contain infinity but $\bar{A}$ does, how to find a sequence to $\infty$ but we can do this because $A$ is not precompact in $\mathbb{Q}$ so either it's unbounded or it has a nonrational limit, in either case we explicit construct a sequence to $\infty$. – davik Oct 24 '18 at 22:34