If a manifold $M$ is $\sigma$-compact, does $M$ possess any countable dense subset?
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It means 'If $M$ is $\sigma$-compact then $M$ has a countable dense subset'? – Hanul Jeon Oct 03 '13 at 03:05
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Manifolds are usually required to be second-countable by definition, and any second-countable space is separable (i.e., has a countable dense subset). Does your definition of "manifold" require second-countability? – Zev Chonoles Oct 03 '13 at 03:06
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@tetori Yes, I wanted to describe as that is. – lotz84 Oct 03 '13 at 03:10
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@ZevChonoles Thank you, so much!! I understand it is a property of just manifold. – lotz84 Oct 03 '13 at 03:13
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But if you wanted to use $\sigma$-compactness, and not manifold-ness, you could do that too.
Every compact set has a countable dense subset, namely the union of finite $(1/n)$-nets.
And for a countable union of compact sets, you take the union of those countable dense subsets.
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1The ordinal $:$$\omega_1$$+1:$ with the order topology is compact but does not have a countable dense subset. $\hspace{.84 in}$ – Oct 03 '13 at 23:57