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A topological space that is a countable union of compact subsets is called $\sigma$-compact.

My intuition says that the follow property should be equivalent to $\sigma$-compactness: $$ \text{Every closed set is a countable union of compact sets} $$

Is this actually the case?

Note: I'm thinking only of Hausdorff spaces here (it seems like this should fail in the non-Hausdorff case).

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Your intuition is correct, also if the space is not Hausdorff.

Let $X$ denote the underlying set of the topological space you are working in.

If every closed set is a countable union of compact sets then $X$ is a countable union of compact sets. This because $X$ is closed.

If conversely $X$ is a countable union of compact subsets and $F\subset X$ is closed then the intersections of these compact sets with $F$ are compact and $F$ equals the union of these intersections. Any intersection of a compact and a closed set is compact. You do not need Hausdorff for this:

Let $C$ be compact, $F$ closed and let $\mathcal{U}$ denote an open cover of $C\cap F$. Then $\mathcal{U}\cup\left\{ F^{c}\right\} $ is an open cover of compact $C$. It has a finite subcover $\mathcal{U}'$ and $\mathcal{U}'\backslash\left\{ F^{c}\right\} \subset\mathcal{U}$ is a finite subcover of $C\cap F$.

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  • $\begingroup$ Thanks for the refinement! $\endgroup$ – pre-kidney Feb 8 '14 at 6:03
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If $X = \bigcup_{i \in \mathbb{N}} K_i$ is $\sigma$-compact and Hausdorff (with the $K_i$ compact, and hence closed), then given any closed $F \subseteq X$ we have $F = \bigcup_{i \in \mathbb{N}} ( F \cap K_i )$ with each $F \cap K_i$ compact.

The other direction is trivial since $X$ is closed in itself.

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