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Here is the background material from which I am working:

  1. The Cantor set is an uncountable compact Hausdorff space with empty interior.
  2. In a locally compact Hausdorff space, each countable set has empty interior.
  3. The rational numbers with the subspace topology is a non-locally compact Hausdorff space in which all compact sets have empty interior.

I am trying to find a non-locally compact Hausdorff space in which all infinite compact sets have nonempty interior. I am guessing the example will be an exotic function space.

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    $\begingroup$ Your point 2 doesn't hold, $\mathbb{Z}$ is a locally compact Hausdorff space with the discrete topology. $\endgroup$ – Daniel Fischer Jul 11 '13 at 22:57
  • $\begingroup$ But in a locally compact Hausdorff space each countable union of nowhere dense sets has empty interior, i.e. it is a Baire space. So if a point is nowhere dense, which means it is not open, then 2) is right. $\endgroup$ – Stefan Hamcke Jul 11 '13 at 23:02
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Let $X=\Bbb R\cup\{\infty\}$ with the discrete topology on $\Bbb R$, and a neighborhood of $\infty$ is a set with a countable complement. This $X$ is clearly Hausdorff.
Now, the funny thing about this space is that each compact set is finite. Such a space is called anti-compact. To see this, let $S$ be an infinite subset. If it avoids $\infty$, then it has the discrete topology and is not compact. If $\infty\in S$, then choose a countable subset $Q$ of $S$ and note that $(S-Q)\cup\bigcup_{q\in Q}\{q\}$ is an infinite open cover of $S$ without a finite subcover. This means that no infinite set is compact. In particular, $\infty$ has no compact neighborhood, so $X$ is not locally compact.

Now, the property you wish for is vacuously satisfied since there are no infinite compact subsets :-)

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  • $\begingroup$ I was this close to figuring out something similar. Argh! Now of course the question needs to change to eliminate such.... $\endgroup$ – dfeuer Jul 11 '13 at 23:35
  • $\begingroup$ Well, you could try to find a space where there do exist infinite compact subsets :) $\endgroup$ – Stefan Hamcke Jul 11 '13 at 23:37
  • $\begingroup$ Thank you, Stefan, for this creative solution. I am not sure of the protocol on this site - is it OK for me to change my poorly posed question to "Is there a Hausdorff space which: (1) is not locally compact, (2) contains at least one infinite compact subset, and (3) all infinite compact sets have nonempty interior?" If so, do I edit the original posting or inline in the comments? $\endgroup$ – Wayne Jul 15 '13 at 4:50
  • $\begingroup$ @Wayne: I suggest you ask a new question since it is actually a different question with a more difficult solution. You should mention this question in the new post and point out the differences, so the readers won't accidentally mark it as a duplicate. $\endgroup$ – Stefan Hamcke Jul 15 '13 at 11:10

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