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Let $M$ be a compact differentiable manifold which can be covered by two open subsets $U$ and $V$. Then $H_{\text{dR}}^n(M)$ is finite-dimensional for all $n$. But how about $U$, $V$ and $U\cap V$? Are their de Rham cohomologies necessarily finite-dimensional? The open sets $U$ and $V$ can be chosen to be very strange but I have not found a counterexample. Also, the Mayer-Vietoris sequence and the rank-nullity theorem seem not to provide any information about this.

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Choose as one of the two subsets the disjoint union of infinitely many non-empty open subsets; in this case the $H^0$ is not finitely dimensional. –  Angelo Jun 13 '12 at 16:48
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I was going to give @Angelo's example, but since he beat me to it, I should just note that this can be done in any dimension up to $n-1$ –  Igor Rivin Jun 13 '12 at 16:51
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I think I can construct a counterexample now. Let $M$ be a circle $\mathbb R/\mathbb Z$, $U=M$ and $V=M−\{1/2^i+\mathbb Z:i\in\mathbb N\}$. Then at least $H^0(V)$ is infinite dimensional. –  Alberto Jermaine Jun 13 '12 at 17:08

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