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We often assume manifolds to be paracompact Hausdorff. Clearly, this implies normal.

However, there is a manifold (I mean locally Euclidean Hausdorff space) which is not paracompact. Without paracompactness, they are still regular because manifolds are locally compact, but does it imply normal?

The only example of non paracompact manifold which I know is the "long line". However, I hear this is normal. I want to know whether manifold implies normal or not.

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3 Answers 3

up vote 12 down vote accepted

The fiberwise one-point-compactification of the tangent bundle of the long line (pick any smooth structure) is not normal.

The two distinguished sections of this $S^1$ bundle (the zero section & the section at infinity) cannot be separated by open subsets.

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André's answer is great, but this example might be simpler.

Let $\mathbb{L}_+=\omega_1\times[0,1)-\langle 0,0\rangle$ be the longray, and set $\Omega=\{\langle \alpha, 0\rangle\,:\,\alpha\in\omega_1\}\subset\mathbb{L}_+$. Take $M=\mathbb{L}_+\times(-1,1) - \Omega\times\{0\}$. Then $M$ is a $2$-manifold since $\Omega$ is closed in $\mathbb{L}_+$. But the two closed sets $A=\cup_{\alpha\in\omega_1}\{\langle\alpha,0\rangle\}\times(0,1)$ and $B=(\mathbb{L}_+-\Omega)\times\{0\}$ cannot be separated by disjoint open sets.

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There are even non-normal separable complex manifolds. The paper Complex analytic manifolds without countable base by Calabi and Rosenlicht gives examples $M,S$ of a non-normal separable complex manifolds of complex dimension 2 and hence real dimension 4. It was proven by Rado that every Riemann surface is paracompact, so the dimension of these manifolds is optimal.

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Prüfer surface (which is talked about in the article you cite) is also separable and non-normal, and has real dimension 2 (but is not a complex surface). –  Mathieu Baillif Jul 21 at 1:36

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