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This question arose from considering for a connected smooth Hausdorff manifold the (possible) equivalence of the following properties:

(1) paracompact,
(2) metrizable,
(3) second countable,
(4) countable at infinity,
(5) $\sigma$−compact,
(6) Lindelöf,
(7) separable.

I know proofs for the equivalence of the first six, and that they imply (7), but it is problematic, whether this implies the others. By countable at infinity I mean existence of a sequence of compact sets $K_i$ whose union covers the space and which satisfies $K_i\subseteq{\rm Int\ }K_{i+1}$ . Of course, locally euclidean means that each point has a neighbourhood homeomorphic to $\mathbb R^k$ with the standard topology and $k\in\mathbb N$ .

flag	asked Feb 22 2011 at 12:44 TaQ 606●2●11

	Take a look at the article I linked to in [my answer](mathoverflow.net/questions/229/…) to the question [Is the long line paracompact](mathoverflow.net/questions/229/…). It has 107 conditions for a connected locally Euclidean Hausdorff space equivalent to that it be metrisable. I'm pretty sure that your 7 are amongst that list. – Andrew Stacey Feb 22 2011 at 12:54
	@ Andrew Stacey. Thanks for the reference. Theorem 2 there has conditions 57,58,59,67,77 of all 108 which are of the form "separable and sth. else". In Example 5 on page 15 there is given a manifold which is claimed to be separable but not metrizable. I have to think it through carefully. – TaQ Feb 22 2011 at 13:40

TaQ · Accepted Answer · 2011-02-22 22:40:41Z UTC

I answer my our question: Separability of a connected locally euclidean Hausdorff topological space does not imply second countability, or any of the equivalent conditions (1), ... (6) given in the question. A counterexample is given in Example 5 on page 15 in David Gauld's preprint. There is constructed a separable Hausdorff topological space, which is not second countable. The space can be equipped with a compatible analytic atlas modelled on $\mathbb R^2$ . One such is $\lbrace{\rm id\ }S\rbrace\cup\lbrace\phi_{\eta,\zeta}:\eta,\zeta\in\mathbb R\rbrace$ , where $\phi_{\eta,\zeta}$ is given by $(0,\eta,z)\mapsto(0,z-\zeta)=(0,v)$ when $|v|<1$ , and $(x,y)\mapsto(x,|x|^{-1}(y-\eta)-\zeta)=(u,v)$ when $0<|x|<1$ and $|v|<1$ . So Gauld's space does not satisfy any of the conditions (1), ... (6).

Henno Brandsma · Answer 2 · 2011-02-22 15:26:28Z UTC

1

Indeed, check the paper by Gauld. Your (4) implies his condition hemicompact. His example at p15, that you saw refutes the just separable condition (7). Note that (1)-(6) imply imply metrisability for just continuous manifolds, so it still might be that the situation vis à vis separability is different for smooth manifolds instead of continuous ones, though I suspect not.

answered Feb 22 2011 at 15:26

Henno Brandsma
2,236●7●19

	I do not understand what essential new information your answer would provide. In Gauld's Example 5 on page 15, there is constructed a topological space which is claimed to be a nonmetrizable separable 2−manifold. However, I do not (yet) clearly see how this topological space would be made into a manifold by providing a homeomorphism between some neighbourhood of each point and $\mathbb R^2$ . – TaQ Feb 22 2011 at 16:59
	Possibly I now begin to understand how Gauld's Example 5 should be interpereted. If I am right, the space is separable since the set $\{(x,y):x\not=0\text{ and }x,y\in\mathbb Q\}$ is dense, and it is not second countable, hence not metrizable, since $(0,\eta_1,\zeta_1)\not\in W_{\eta,\zeta,r}$ holds when $\eta_1\not=\eta$, and so uncountably many sets are needed for a base of the topology. It also seems that an atlas, maybe even a smooth one, modelled on $\mathbb R^2$ can be constructed, but I have not yet checked the details. – TaQ Feb 22 2011 at 19:52

Is a connected separable locally euclidean Hausdorff topological space second countable?

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