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I'm looking for an example of the thing in the title. So far I tried many topologies but none seemed to work.

Thoughts: I obviously have to find some non-Hausdorff space because otherwise the example doesn't exist. I thought of some non-Hausdorff spaces in which the compacts are easy to detect, but then I found that all the open sets in them are in fact locally compact.

Thanks in advance.

Cauchy
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    You may want to see http://math.stackexchange.com/questions/289521/closed-subspaces-of-a-locally-compact-space-are-locally-compact – Ashwin Iyengar Nov 19 '16 at 18:48
  • @AshwinIyengar that's very relieving! I'll change my question to "open" instead of "closed". – Cauchy Nov 19 '16 at 18:56
  • What is your definition of "locally compact"? – Daniel Fischer Nov 19 '16 at 19:00
  • @DanielFischer each point has a local base consisting of compact sets. – Cauchy Nov 19 '16 at 19:01
  • Then every open subspace of a locally compact space is locally compact. For each $x \in O$, there is a compact neighbourhood (in $X$) of $x$ contained in $O$, since the compact neighbourhoods form a local base. – Daniel Fischer Nov 19 '16 at 19:02
  • If your definition were "each point has a (quasi)compact neighbourhood", there'd be examples. – Daniel Fischer Nov 19 '16 at 19:03
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    @DanielFischer I don't understand how your argument proves that $O$ is locally compact. You have to show that for each open subset $U$ of $O$, there exists a compact neighborhood of $x$ in $O$ which is contained in $U$. – Cauchy Nov 19 '16 at 19:07
  • If $O$ is open, then every $O$-neighbourhood $U$ of $x$ is also an $X$-neighbourhood of $x$, so by the local compactness of $X$, there is a compact $X$-neighbourhood $V$ contained in $U$. But $V$ is of course also an $O$-neighbourhood of $x$. – Daniel Fischer Nov 19 '16 at 19:11
  • @DanielFischer got it.. thanks – Cauchy Nov 19 '16 at 19:20

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