Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm very stuck on the following exercise in the book "A Comprehensive Introduction to Differential Geometry V.1" by Michael Spivak: Let $M^m$ be a smooth connected non-compact manifold. Show that $M$ is the union of a sequence of open sets $U_n$ with the following properties:

  1. $U_n \cap U_{n+1}$ is non-empty for all $n$
  2. For every compact set $C \subset M$ there is $N$ such that $U_n \cap C$ is empty for all $n>N$
  3. $U_n$ is diffeomorphic to $\mathbb{R}^m$ for all $n$.

Any help would be greatly appreciated. Thank you.

share|cite|improve this question
Seems difficult... one can easily get a countable covering by precompact balls diffeomorphic to $\mathbb{R}^m$, but I don't see how to satisfy property 1. I'm even having trouble seeing how to do it on the cylinder $\mathbb{R} \times S^1$. – Anthony Carapetis Aug 19 '13 at 3:14
@AnthonyCarapetis Indeed, since $\{0\}\times S^1$ is a compact subset, the sets $U_n$ will be disjoint from it for large $n $; this implies either $U_n\subset (0,\infty)\times S^1$ for all $n>N$, or $U_n\subset (-\infty,0)\times S^1$ for all $n>N$. I begin to wonder if the exercise statement is misquoted. – user Aug 20 '13 at 4:15
@user89499: Right, so that starting point is useless. Your observation means that all but one of the ends of the manifold must be covered by non-compact $U_n$... still seems feasible but I have no idea how to attack it. – Anthony Carapetis Aug 20 '13 at 4:24

Proof idea: First choose an increasing sequence of connected compact subsets ${C_n}$ so that $\cup C_n = M.$ Then choose a cover $U=${$U_\alpha$} of M by open sets diffeomorphic to $\mathbf{R}^m$ which has no finite subcover. First find a finite cover for $C_1$ by sets in U to start your sequence of sets. Then find finite covers of $C_2 \setminus C_1, C_3\setminus C_2,$ etc to continue your sequence.

share|cite|improve this answer
That is what I initially thought of, but how do I know that $C_n \setminus C_{n-1}$ is connected? – Dan Nov 19 '12 at 19:36
Clarification: if $C_n \setminus C_{n-1}$ is not connected, it is not clear how to continue the sequence in such a way that property 1 is still satisfied – Dan Nov 19 '12 at 19:52
Since $M$ is a manifold, it is metrizable. We can just choose the $C_n$ to be compact balls around a single distinguished point. Then the sets should be connected (they are essentially annuli). – mck Nov 19 '12 at 21:38
Ah actually this still doesn't help. Even in that case the sets can become disconnected. – mck Nov 20 '12 at 1:22

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.