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Are $\sigma$ -compact Hausdorff spaces normal?

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2 Answers

Yes. Every $\sigma$-compact space is Lindelöf, every $\sigma$-compact Hausdorff space is $T_3$ (regular and $T_1$), and every $T_3$ Lindelöf space is $T_4$ (normal and $T_1$).

Correction: I was apparently not yet awake when I wrote that. There are countable Hausdorff spaces that are not regular, so the result is actually false in general. However, if you have any property that ensures regularity, you do get normality for free.

For an example of a countable Hausdorff space that isn’t regular, see this answer to an earlier, related question.

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There may be some confusion based on definitions. Engelking's text, in particular, defines $\sigma$-compact spaces to be regular (T$_3$) spaces which are the countable union of compact subspaces. With this Brian's (original) answer works. (I cannot stress enough that his current answer works perfectly well.)

However if you remove regularity from the definition of $\sigma$-compactness, you can get counterexamples, as Brian mentions above. Another example is the following: Let $A = \{ \frac{1}{n} : n \in \mathbb{N} \}$, and give $\mathbb{R}$ the topology by declaring the open sets to be of the form $U \setminus B$ where $U \subseteq \mathbb{R}$ is open in the usual metric topology and $B \subseteq A$.

  • As this topology is finer than the usual metric topology, it follows that this space is Hausdorff.
  • It is $\sigma$-compact since you can cover $\mathbb{R}$ by countably many closed (bounded) intervals which each contain only finitely many elements of $A$.
  • However this space is not regular because $A$ is closed in the new topology, but there is no pair of disjoint open sets around $0$ and $A$, respectively.
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Now that I know how you sound like, it's impossible to read your posts without hearing you in my head. –  Asaf Karagila Sep 11 '12 at 20:15
    
@Asaf: You're not the only one. –  Arthur Fischer Sep 11 '12 at 20:18
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