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There are different definitions for topological manifolds, sometimes second-countability or paracompactness are added to being locally euclidian Hausdorff. (Sometimes Hausdorff is also left out, but I won't do that here.) The two additional conditions lead to two new definitions not equivalent to the one without, for example the long line is locally euclidian Hausdorff, but neither second-countable nor paracompact. (Six examples for the independence of the conditions when second-countable is added can be found here.) I was wondering if the two new definitions (secound-countable locally euclidian Hausdorff as well as paracompact locally euclidian Hausdorff) are equivalent? We indeed have an implication:

Lemma 1: Second-countable and locally compact spaces are $\sigma$-compact (See here).

Lemma 2: Locally compact and $\sigma$-compact spaces are paracompact (See here).

For a second-countable locally euclidian Hausdorff space, we have: It is locally euclidian and therefore locally compact. According to lemma 1 it is $\sigma$-compact and according to lemma 2 it is paracompact.

Can the backwards direction, that paracompact locally euclidian Hausdorff spaces are second-countable also be proven or can someone give a counterexample?

Since Sassatelli Guilio has provided a counterexample to this, but it is not connected: What about when connectedness is added?

Samuel Adrian Antz
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    The fact is that the disjoint union of any family of paracompact spaces is paracompact, so you can take $\coprod_{i\in \Bbb \aleph_1} X_i$ with $X_i=\Bbb R^n$ and this is a paracompact, locally Euclidean metric space, but not second countable. – Sassatelli Giulio Jul 18 '22 at 13:29
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    I don't know of a connected example, and apparently neither does pi base (though it may be due to the fact that the database does not have its spaces tagged by local Euclideanity). – Sassatelli Giulio Jul 18 '22 at 13:46
  • Thanks for the counterexample, I have included connectedness as a new question. – Samuel Adrian Antz Jul 18 '22 at 15:04
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    See my answer at https://mathoverflow.net/a/237/75. – Eric Wofsey Jul 18 '22 at 15:12

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Eric Wofsey has answered your question in mathoverflow.

Here is a similar approach.

  1. A connected, locally compact, paracompact Hausdorff space is $\sigma$-compact. See here.

  2. Let $X$ be a paracompact and connected locally euclidian Hausdorff space. Cover $X$ by open subsets $B_\alpha$ which are open euclidean balls. Since $X = \bigcup_n K_n$ with compact $K_n$, countably many $B_k = B_{\alpha_k}$ suffice to cover $X$. Thus $X$ is a countable union of second countable open subsets. Therefore $X$ is second countable.

Note that "locally euclidean" can be replaced by "locally second countable".

Paul Frost
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