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I am having trouble showing one direction of the proof that a locally compact Hausdorff space is compactly generated.

Specifically, my question is how do I show that: if X is a locally compact Hausdorff space and A is a subset of X with the property that $$A\cap K$$ is closed for every compact K, then A is closed.

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HINT: If $A$ is not closed, there is a point $x\in(\operatorname{cl}A)\setminus A$. Let $U$ be an open nbhd of $x$ with compact closure, and let $K=\operatorname{cl}U$. Now consider $K\cap A$ to get a contradiction.

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  • $\begingroup$ $K \cap A$ must be closed by hypothesis, Since $x \in cl(A) \setminus A$, there is a sequence $(a_n)$ in $A$ converging to x. This sequence is eventually in $U$ as well, since $U$ is an open set containing $x$. Hence $(a_n)$ is eventually in $A \cap U$ and hence in $A \cap K$. But limit of $(a_n)$ is $x$ which is not in $K \cap A$, contradicting that $K \cap A$ is closed. Is this right? $\endgroup$ – eatfood Mar 27 at 2:37
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Let $x$ be an element in the complementary of $A$, since $X$ is locally compact, there exists a neighborhood $U$ of $x$ which has a compact adherence $U'$, $A\cap U'$ is compact thus closed. It implies that $U-A\cap U'$ is open, thus the complementary of $A$ is open. done.

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  • $\begingroup$ How do we know that $U-A \cap U' $ is contained in the complement of $A$ ? $\endgroup$ – fn2197 Dec 22 '15 at 22:07

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