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.


3 Answers 3


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.

  • $\begingroup$ Perhaps a dumb question: why not? (I realize this is an old MO question, but I just found it!) $\endgroup$ Aug 9, 2016 at 4:34
  • $\begingroup$ An everywhere non-zero vector field would integrate to a faithful action of the reals on the long line. But that's impossible, as the limit as $t \to \infty$ of the flow applied to some given point $p$ cannot be long line's infinity, and must therefore be a fixed point of the action. $\endgroup$ Sep 2, 2016 at 23:11
  • $\begingroup$ Thanks. I agree there's no everywhere nonzero vector field (this would also imply that the tangent bundle is trivial, which is known to be false: ams.org/journals/proc/1969-023-02/S0002-9939-1969-0246318-X/… ). And I agree that if there were such a vector field it would give open sets that separate the two sections mentioned. What I'm still not seeing is how not having such a vector field guarantees there's no way to put disjoint open sets around those sections. Or in other words, why would such open sets necessarily get you a nonvanishing vector field? $\endgroup$ Sep 4, 2016 at 2:47
  • $\begingroup$ Ok, here's a complete argument. Assume two disjoint opens. The boundary of an open defies a multi-valued vector field (for each point, a closed subset of the tangent line, bounded away from $0$ and from $\infty$). Take the max. That's a discontinuous everywhere non-zero vector field. Integrate to get an action of $\mathbb R$ on the long line. This action is faithful because the vector field is everywhere non-zero. Now apply the same argument as above. Contradiction. $\endgroup$ Sep 4, 2016 at 10:04

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.


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.

  • $\begingroup$ 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). $\endgroup$ Jul 21, 2015 at 1:36

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