Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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 $k\in\mathbb N$ .

share|improve this question
Take a look at the article I linked to in my answer to the question Is the long line paracompact‌​. 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. –  Loop Space Feb 22 '11 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 '11 at 13:40

2 Answers 2

up vote 3 down vote accepted

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).

share|improve this answer

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.

share|improve this answer
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 '11 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 '11 at 19:52

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.