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I'd like somebody to specify flaws in my outline of the proof of the above statement. I'm following the definition of topological manifold used in Lee's Introduction to Smooth Manifolds. (it is 2nd countable)

Let $X$ to be a Locally Euclidean Hausdorff topological space.

$\Rightarrow$: If $X$ is a topological manifold, then it has a countable basis of precompact coordinate balls by a lemma. Therefore, there is a countable set of precompact coordinate balls which cover $X$, and the set of closure of these balls is the set of countably many compact subspaces which cover $X$. Thus, $X$ is $\sigma$-compact.

$\Leftarrow$: If $X$ is $\sigma$-compact, there is a set of countably many compact sets which cover $X$. Let $C$ be an arbitrary compact set of this set. Since $X$ (and $C$) is locally Euclidean, $C$ is locally metrizable. Since $C$ is locally metrizable compact Hausdorff space, it is metrizable and therefore 2nd countable. Thus, being the countable union of these compact spaces with countable basis, $X$ is 2nd countable and therefore a topological manifold.

Should I add a proof that locally metrizable compact Hausdorff space is metrizable? I found this statement in Munkre's Topology, but is it really a widely known fact?

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The manifold $\Rightarrow \sigma-$compact looks fine. The other direction probably also works, but you can prove it much more directly/easier.

It suffices to show $X$ the space is second countable. Let $K_n$ be the covering of $X$ by compact sets, and assume they're all nonempty. For each $x\in K_n$, there is an open neighborhood $U_{x,n}$ which is homeomorphic to an open ball in $\mathbb{R}^m$. Then we have $K_n\subset \bigcup_{x\in K_n} U_{x,n}$. By compactness, there are finitely many $U_{x,n}$ which cover each $K_n$. By unioning these all together for each $n$ gives us a countable covering for $X$ by open sets which are homeomorphic to open balls in $\mathbb{R}^m$. Each of these sets has a countable basis. The countable union of these sets is also a countable set, so it remains to show that this set is a basis, which is easy.

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    $\begingroup$ Actually I made the exactly same proof as yours, except for the last part (I forgot to show that the set is a basis). But I wasn't sure whether it worked fine, so I relied on an indirect method, which I realized wasn't practical at all, since who always can recall that locally metrizable compact Hausdorff space is metrizable? I appreciate your response a lot! $\endgroup$ – Math.StackExchange Mar 22 '15 at 17:53
  • $\begingroup$ I wonder where we use the condition hausdorff? $\endgroup$ – Idele Dec 28 '16 at 3:39
  • $\begingroup$ @hctb A manifold has to be Hausdorff. Otherwise you get the line with two origins, which is locally Euclidean but not Hausdorff. $\endgroup$ – Moya Dec 30 '16 at 18:36
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    $\begingroup$ @hctb if you do not have Hausdorffness, you fail to get enough partitions of unity (which follows from Hausdorff+paracompact). Without enough partitions of unity you can't do much at all! $\endgroup$ – David Roberts Oct 8 '17 at 13:15

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