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Let $X$ be a Hausdorff topological space. A subspace $A\subseteq X$ is called $k$-closed if and only if for every continuous map $h \colon K \to X$ from a compact Hausdorff space the preimage $h^{-1}(A)$ is closed in $K$. Let $\bigcup K_i = X$ be a covering of $X$ where each $K_i$ is a compact Hausdorff subspace. Further assume that for each compact Hausdorff subspace $K \subseteq X$ there exists an index $i$ such that $K$ is a (closed) subset of $K_i$.

My question is now whether it is sufficient for $A$ to be k-closed if $A \cap K_i$ is closed in $K_i$ for all $i$? I am assuming that this is the case but don't know how to prove it... Thanks in advance!

benjaminroos
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  • Just curoous, can you name a k-closed set which is not closed? – Arctic Char Oct 17 '20 at 13:12
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    @ArcticChar See this example from Henno https://math.stackexchange.com/questions/411434/an-example-of-a-space-which-fails-to-be-compactly-generated?rq=1 or my example in this question https://math.stackexchange.com/questions/3673020/example-of-a-non-compactly-generated-complete-locally-convex-topological-vector. – benjaminroos Oct 17 '20 at 13:20
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    Clearly this fails when we cover $X$ with singletons $K_x={x}$. – freakish Oct 17 '20 at 13:30
  • @freakish Of course you're right about this. I added another assumption with regards to the covering. Thank you! – benjaminroos Oct 17 '20 at 13:54

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Well, the added assumption on the covering is already giving the answer to the question:

Let $h \colon K \to X$ be a continuous map from a compact Hausdorff space $K$. Since $X$ is Hausdorff it is an easy exercise to see that $h$ is a quotient map to its image. Thus checking whether $h^{-1}(A)$ is closed in $K$ is equivalent to find out if $A \cap h(K)$ is closed in $h(K)$. By the assumption there exists $i$ with $h(K) \subseteq K_i$ closed and further $A \cap K_i$ is closed in $K_i$. Hence $A \cap h(K) = (A \cap K_i) \cap h(K)$ is closed in $h(K)$ by the definition of the subspace topology.

For the interested reader: This is a step of simplification in the proof of the fact that $\mathbb{R}^{\mathbb{R}}$ is not compactly generated in http://groupoids.org.uk/pdffiles/tentopologies.pdf, section 6.

benjaminroos
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    Observe that your proof works equally well if you replace your assumption with the more general statement "each compact $K\subseteq X$ is covered by finitely many of the $K_i$". This is the statement that the covering ${K_i}$ is compact-finite. Note that each locally-finite covering is compact-finite. – Tyrone Oct 18 '20 at 16:58
  • @Tyrone This is even better! I think one can expand the result to weakly Hausdorff spaces without any problems. Then the only thing that is left to be shown is that $h(K)$ is also Hausdorff, but this is not so difficult and can be looked up in lemma 1.4 in Neil Strickland's famous paper https://neil-strickland.staff.shef.ac.uk/courses/homotopy/cgwh.pdf. – benjaminroos Oct 19 '20 at 11:28