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I am trying to find an open subset of a locally compact non Hausdorff space which itself is not locally compact.

Thank you in advance.

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    $\begingroup$ Which definition of local compactness are you using? $\endgroup$ – Eclipse Sun May 20 '18 at 21:13
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    $\begingroup$ A space X is locally compact at a point x if there exists a space U (which contains x) such that the closure of U is compact. X is then locally compact provided that it is locally compact at all its points. $\endgroup$ – P.H. Me May 20 '18 at 21:18
  • $\begingroup$ See here. $\endgroup$ – Eclipse Sun May 20 '18 at 21:32
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The linked question already gave you the answer: Consider $X=\alpha(\mathbb{Q})$ be the Alexandroff compactification of $\mathbb{Q}$. Then the space is compact so certainly locally compact in your definition, which you gave in the comments as $$\forall x \in X: \exists O \text{ open }: x \in O \text{ and } \overline{O} \text{ compact}$$ because we can take $O= X$ for all $x$, and $\mathbb{Q}$ is open in $X$ but not locally compact, as is well-known.

As a reminder: $\alpha X$ for a space $(X,\mathcal{T}_X)$ is defined as $\alpha(X) = X \cup \{\infty\}$ where $\infty$ is some point not in $X$ and as its topology is defined as $$\mathcal{T}_X \cup \{O \subseteq \alpha(X) : \infty \in O \land X \setminus O \text{ closed and compact in } (X,\mathcal{T}_X) \}$$

One can show that $X$ as a subspace of $\alpha(X)$ is just the original space and $\alpha(X)$ is compact.

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