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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Does this answer your question: A question about local compactness and $\sigma$-compactness ? – PinkyWay Jul 14 '22 at 05:28
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@SummerChild I don't think so. Hausdorff is the assumption used there. – Jimmy Kang Jul 14 '22 at 05:34
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Maybe you should also include that in the title. – PinkyWay Jul 14 '22 at 05:44
2 Answers
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.
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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).
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1Your 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