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I can sort of see this intuitively, seeing as it's a similar argument to the Baire Category theorem, but does anyone have a proof I could look at?

Would it suffice to say that every locally compact Hausdorff space is also a Baire space $\implies$ not the countable union of nowhere dense sets?

If so, how about this for a proof?

Since X is locally compact Hausdorff, it is homeomorphic to an open subspace of a compact Hausdorff space, say Y. By the Baire Category Theorem, Y being a compact Hausdorff space implies that Y is a Baire space. Since open subspaces of Baire spaces are Baire spaces Then X is a Baire Space.

This is under the assumption that a Baire space $\implies$ not the countable union of nowhere dense sets? Unless I misinterpreted the Baire Category Theorem.

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  • $\begingroup$ Your conclusion is correct. Baire's Theorem for locally compact Hausdorf spaces is Theorem 10.1 in Dugundji's 'Topology'. $\endgroup$ – Kavi Rama Murthy Dec 13 '18 at 8:30
  • $\begingroup$ (contd) [Chapter 11. See also Theorem 10.5] $\endgroup$ – Kavi Rama Murthy Dec 13 '18 at 8:36
  • $\begingroup$ More generally, Baire's theorem holds for: locally compact regular space, even if not Hausdorff. $\endgroup$ – GEdgar Dec 13 '18 at 11:30