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Let $X$ be a such space. I know that for a Hausdorff space, being locally compact is equivalent to having a basis of pre-compact open sets. But how do I prove that $X$ can be covered by countably many such sets, by using second-countability of X? Thank you.

Boar
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    If $(U_n)$ is countable basis then the subcollection of those $U_n$'s which are pre-com pact is a countable basis of pre-compact open sets. – Kavi Rama Murthy Feb 18 '21 at 08:30
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    Here is one approach: suppose $(U_n)_n$ is a countable base for $X$ and let $x \in X$. By local compactness, $x$ has a compact neighborhood - ie, there is a compact $K$ and an open $U$ such that $x \in U \subseteq K$. From the base, select $U_i$ such that $x \in U_i \subseteq U$. This set is precompact, since its closure is contained in $K$ (a compact set), and a closed set contained in a compact set is necessarily also compact. – Rob Feb 18 '21 at 08:30
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    Yet another approach is to use the fact that second countable spaces are Lindelöf to extract a countable subcover from the one you already have. – Alessandro Codenotti Feb 18 '21 at 08:40
  • @AlessandroCodenotti Thank you. That's all I need. – Boar Feb 19 '21 at 02:23
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    related question: https://math.stackexchange.com/questions/3282738/every-locally-compact-hausdorff-second-countable-space-has-a-countable-basis-c?rq=1 – Chill2Macht Feb 06 '22 at 04:01

2 Answers2

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If we have a countable base $\mathcal{B}$ for $X$ then any other base $\mathcal{C}$ of $X$ has a countable subfamily that is also a base. See here e.g.

We can apply this to the base of pre-compact sets that exists in a locally compact Hausdorff space.

Finally note that a base is in particular a cover of $X$ (for every $x$ there is at least one base element containing it, as $X$ itself is open and so a union of base sets).

Final remark: the closures of the cover elements of course still cover $X$ and show that $X$ is then $\sigma$-compact.

Henno Brandsma
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Every locally compact Hausdorff space is (completely) regular. Every second-countable regular Hausdorff space is metrisable, and hence separable. Any locally compact metrisable space admits a compatible metric, each of whose (proper) open balls is precompact.

Thus if $X$ is locally compact, Hausdorff and second-countable, then it is Polish. It admits a compatible, but possibly noncomplete, metric $d$ for which there is a countable subset $\{x_n\}_{n\in\mathbb{N}}$ such that $\{B_{\frac{1}{m}}(x_n)\mid m,n\in\mathbb{N}\}$ is a countable base of precompact sets for $X$.

Tyrone
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