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I'd like to find a locally compact space $X$ with a subspace $A$ that is NOT locally compact.

As from here, I know that if $A$ is closed and $X$ is Hausdorff, then $A$ is locally compact.

Anyone have an example that satisfies the condition?

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  • $\begingroup$ Try taking $X=\mathbb{R}^2$ $\endgroup$ – Pol van Hoften Mar 3 '14 at 21:10
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$$A=\mathbb Q\subset X=\mathbb R$$

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  • $\begingroup$ this one is perfect, since I know how to prove that $\mathbb{Q}$ is not locally compact. Thanks a lot! $\endgroup$ – PandaMan Mar 3 '14 at 21:40
  • $\begingroup$ Dear PandaMan, I'm glad the answer helped you. $\endgroup$ – Georges Elencwajg Mar 3 '14 at 22:19
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Take Cantor space $\{0,1\}^\omega$ as $\mathbf{X}$ (which is compact), and then note that the subspace $$\mathbf{Y} := \{p \in \{0,1\}^\omega \mid |\{i \in \mathbb{N} \mid p(i)=0\}| = \infty\}$$ is homeomorphic to $\mathbb{N}^\mathbb{N}$, which is not locally compact.

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The subset $$A=(\Bbb R_{>0}\times\Bbb R)\cup\{(0,0)\}$$ of $\Bbb R^2$ is not locally compact, since $\Bbb R^2$ is Hausdorff but $A$ is not open in its closure $\overline A=\Bbb R_{\ge0}×\Bbb R$

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