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Must a Hausdorff Baire space be $σ-$compact?

A Baire space is a topological space in which the union of every countable collection of closed sets with empty interior has empty interior.

A topological space X is called $σ-$compact if it is the countable union of compact subsets.

Just now, from the http://en.wikipedia.org/wiki/Σ-compact_space, I found that conclusion as follows:

A Hausdorff, Baire space that is also σ-compact, must be locally compact at at least one point.

How to prove it?

Thanks a lot.

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  • $\begingroup$ If $X$ is Hausdorff Baire $\sigma$-compact space then it is countable union of compact sets, and at least such compact set has nonempty interior. $\endgroup$ – Hanul Jeon May 21 '15 at 2:25
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Suppose $X$ is a Hausdorff Baire space and is $\sigma$-compact. Thus $X$ is the union of countably many compact sets, which are also closed sets since $X$ is Hausdorff. Since $X$ is a Baire space, it can't be the union of countably many nowhere dense sets. Therefore, at least one of those compact sets, call it $Y,$ has a nonempty interior. Let $p$ be an interior point of $Y.$ Since $Y$ is a compact neighborhood of $p,$ the space $X$ is locally compact at $p.$

The space of irrational numbers is a Hausdorff Baire space which is not locally compact at any point, and is not $\sigma$-compact.

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Certainly not without additional hypotheses: the discrete topology on $\aleph_1$-many points is Hausdorff and Baire (trivially Baire - the only closed set with no interior is the empty set), but since the only compact subsets are finite it is not $\sigma$-compact.

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