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Hello :) i have a question on topology:

Let X be locally compact and Hausdorff. I want to prove the following: If $Y\subset X$ is open then Y is locally compact.

How to prove this? I have prove this for closed subsets. Also i want to conculude with this two lemmas that the following holds:

If $Y\subset X$ is locally closed then Y is locally compact.

Can somesone help me?! Thank you

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1 Answer 1

Hint: It suffices to show that given any open $U \subseteq X$ and any $x \in U$ there is an open $V \subseteq X$ such that $x \in V \subseteq \overline{V} \subseteq U$ and $\overline{V}$ is compact. To show that this holds:

  • By local compactness there is an open neighbourhood $W$ of $x$ such that $Y = \overline{W}$ is compact.
  • As $Y$ is regular and $U \cap Y$ is an open neighbourhood of $x$ in $Y$ there is an open neighbourhood $V_0$ of $x$ in $Y$ such that $\mathrm{cl}_Y (V_0 ) \subseteq U \cap Y$.
  • If $V \subseteq X$ is open such that $V \cap Y = V_0$, then $V \cap W$ is required.
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