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I want to show that subspace of $\mathbb R$ is locally compact.

Let $W$ be subspace of $\mathbb R$. Let $x\in W$ and $y\in \mathbb R \setminus W$. Since $\mathbb R$ is Hausdorff space, there exists disjoint open sets $U$ and $V$ such that $x\in U$ and $y\in V$ for $x\in W$, $y\in \mathbb R \setminus W$. Then, $U\subseteq X \setminus V$ and $X\setminus V$ is closed in $\mathbb R$, so the closure of $U$ is contained in $X \setminus V $. Since the closure of $U$ is bounded and closed in $\mathbb R$, the closure is compact. Hence, there exists open set containing $x$ such that compact subset is contained in the open set. That means $W$ is locally compact.

Please let me know whether it's right or not.

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  • $\begingroup$ You have to produce a neighborhood of $x$ in the subspace topology of $W$ whose closure in $W$ is compact. Your argument does not involve the relative topology on $W$ so it is not correct. $\endgroup$ – Kabo Murphy Dec 3 '18 at 8:13
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It is well known (and not hard to prove) that $\mathbb Q$ is not locally compact. See my comment about the mistakes in your argument.

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