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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?
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.
