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So I'm going through Otto Forster's "Lectures on Riemann Surfaces", and I need another hint (shame). This is in the "Cohomology Groups" sections, as part of a problem to show that for $X$ a compact Riemann Surface, $H^1(X,\mathbb{C})$ is a finite dimensional $\mathbb{C}$-vector space.

Question

Suppose $X$ is a manifold, $U \subseteq X$ is open, and $V \subseteq U$ is relatively compact in $U$. Show that $V$ meets at most finitely many connected components of $U$.

My Attempt (so far)

Let $\{U_i\}_{i \in I}$ be the connected components of $U$, with the $U_i$ disjoint, and $I$ perhaps infinite. Since all the $U_i$ are open (in $U$), and $V \subseteq U$, $\{V \cap U_i\}_{i \in I}$ is an open cover of $V$ (relative to $V$). Since, by assumption, the closure of $V$, which I'll denote $\overline{V}$, is compact in $U$, and $\{\overline{V} \cap U_i\}_{i \in I}$ is an open cover of $\overline{V}$, there exists a finite subcover $\{\overline{V} \cap U_i\}_{i=1}^n$ (sorry for the run-on sentence...).

I then (want to) claim that $V$ meets only the finite set of components $\{U_i\}_{i=1}^n$ given by the finite subcover of $\overline{V}$. This is where I'm getting stuck. Obviously, I need to work in the assumption that $X$ is a manifold someone in here, but I'm not seeing it yet.

Am I on the right track, or should I scrap it?

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1 Answer 1

up vote 1 down vote accepted

Finitely many $\bar{V} \cap U_i$ covers $\bar{V}$, and remember that $U_i$ are disjoint to start with. What can you say about all the $\bar{V} \cap U_j$ that does not show up in your finite subcover?

P.S. You already used that $X$ is a manifold when you say that $U_i$ are open. Connected components in general may not be open.

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Ahhh, thank you. So, suppose there is some $j$ such that $\overline{V} \cap U_j \neq \emptyset$, with $j \neq 1,\cdots, n$. Then $x \in \overline{V}$, and since the $\{\overline{V} \cap U_i\}_{i=1}^n$ cover $\overline{V}$, there is some $i=1,\cdots, n$ such that $x \in \overline{V} \cap U_i \cap U_j$. But the $U_i$ are disjoint, so this is a contradiction. Gotcha. And since $V \subseteq \overline{V}$, this means $V$ meets only finitely many of the $U_i$. –  Brian H Feb 10 '13 at 20:38

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