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I know $\sigma$-compact locally compact Hausdorff spaces are hemicompact. What if we remove the Hausdorff condition or replace Hausdorff with $T_1$?

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As you probably know, without the Hausdorff property there are various inequivalent notions of locally compact space. The weakest of them being weakly locally compact, that is every point has a compact nbhd (= condition (1) in Wikipedia).

Now the planetmath proof of the result mentioned in @SamuelAdrianAntz's answer seems to make some assumptions (open sets with compact closure, ...) which do not always hold. But the following more general result is true.

Theorem: Every weakly locally compact $\sigma$-compact space is hemicompact.

Proof: As shown in this answer, being weakly locally compact and $\sigma$-compact is equivalent to being exhaustible by compact sets. So we can find a sequence of compact sets $K_n$ in the space $X$, each contained in the interior of the next one, and whose union is the whole set. Note that the interiors of all the $K_n$ also cover the whole space. Now given any compact set $K\subseteq X$, it is contained in the union of the interiors of the $K_n$. Thus, by compactness it is contained in the interior of one of the $K_n$, and also in $K_n$. In other words, $X$ is hemicompact.

PatrickR
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Yes, you can drop the Hausdorff condition completely. In analogy to the theorem, that locally compact and $\sigma$-compact spaces are paracompact (See here), we have:

Theorem: Locally compact and $\sigma$-compact spaces are hemicompact (Proof see here).

There is an example of a hemicompact, but not locally compact space found in Example 10, Chapter 4, Section 2 in Introduction to General Topology by K. D. Joshi. Nonetheless, a first countable and hemicompact space is locally compact (Proof see here). The inverse direction holds for the other condition though: Hemicompact spaces are $\sigma$-compact (See here). On the other hand, $\mathbb{Q}$ is a $\sigma$-compact, but not hemicompact space (See here).

Samuel Adrian Antz
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    Your answer saves my day! Thanks a lot. I once thought Hausdorff is needed. I concluded that LCH and sigma-compact imply hemicompact from Folland's Real Analysis Proposition 4.39 and 4.40's proof. Now I know, it actually does not need Hausdorff. – Jimmy Kang Jul 15 '22 at 15:11