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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.

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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.

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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

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