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A space $X$ is said locally compact if for any $x\in X$ and for any neighbourhood $U$ of $x$ there is a compact neighbourhood $V$ such that $V\subseteq U$.

Does closed subset of locally compact is locally compact?

My friend said that $\{1/n : n \in\mathbb{N}\} \cup \{0\}$ subset of $\mathbb{R}$ is not locally compact. Is it true?

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  • $\begingroup$ What are your thoughts so far? What have you tried? The more you can tell us about your thoughts and efforts, the easier it will be for us to tailor our answers to your needs. We don't want to tell you a bunch of things you already know, nor do we want to try to explain using a bunch of concepts with which you're unfamiliar. $\endgroup$ Apr 5, 2015 at 15:13
  • $\begingroup$ I think that $\{1/n: n \in \mathbb{N}\} \cup \{0\}$ is not locally compact because there is no a compact neighbourhood of $0$. But it contradicts to this property [math.stackexchange.com/questions/698029/… $\endgroup$
    – Hidup
    Apr 5, 2015 at 15:25
  • $\begingroup$ Who says there's no compact neighborhood of $0$? Doesn't $[-1,1]$ qualify? $\endgroup$ Apr 5, 2015 at 15:33
  • $\begingroup$ In fact, $\{1/n:n\in\Bbb N\}\cup\{0\}$ is compact, and so locally compact. You can also see that it's closed and contained in $[-1,1],$ which is compact, so by the result in the link you posted, $\{1/n:n\in\Bbb N\}\cup\{0\}$ is compact $\endgroup$ Apr 5, 2015 at 15:34

1 Answer 1

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If $F$ is a closed subset of $X$, with relative topology, then $\forall x\in F$, $\forall U$ is a neighborhood of $x$ in $X$, such that $\exists V\subset U$ is a neighborhood of $x$ in $X$, and $\overline V$ is compact. Hence, for every neighborhood $U_0$ in $F$, since $U_0=U\cap F$ for some $U$, we know $V\cap F$ satisfies the condition. The last assertion follows from the fact that $V\cap F$ is a closed subset of a compact set.

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