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

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.

share|improve this question
add comment

2 Answers

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.

share|improve this answer
add comment

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.

share|improve this answer
add comment

Your Answer

 
discard

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.