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I'd like to have a reference for the proof of the following fact of complex analysis. I think it follows from Runge's theorem, but I don't know how to prove it.

Fact. Let $U \subseteq V \subseteq \mathbb{C}$ be open sets such that no one connected component of $V \setminus U$ is compact. Then the restriction map $\rho_{VU} \colon \mathcal{O}(V) \to \mathcal{O}(U)$ has dense image.

(If $U$ is an open subset of $\mathbb{C}$, the algebra $\mathcal{O}(U)$ of holomorphic functions on $U$ is endowed with the topology of uniform convergence over compact subsets, i.e. the compact-open topology.)

Thanks to all!

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Indeed, Runge's theorem is just the thing for this problem. Given a compact subset $K\subset V$ and a number $\epsilon>0$, we must show that for every $f\in \mathcal O(U)$ there exists $g\in\mathcal O(V)$ such that $\sup_K|f-g|<\epsilon$. We want $g$ to be a rational function with poles outside of $V$. According to Runge, we must choose a point in each connected component of $\overline{\mathbb C}\setminus K$ where the poles of $g$ may be located. Let us be careful here: we don't want $K$ to have extra holes in it. By enlarging $K$ we can make sure that each connected component of $\overline{\mathbb C}\setminus K$ contains a component of $\overline{\mathbb C}\setminus U$. One way to do it is to let $K$ consist of all points of $U$ whose spherical distance to $\overline{\mathbb C}\setminus U$ is at least $\epsilon$.

It remains to prove that no connected component of $\overline{\mathbb C}\setminus K$ is contained in $V$. Suppose, to the contrary, that $\Omega$ is a connected component of $\overline{\mathbb C}\setminus K$ that is contained in $V$. Let $E$ be a connected component of $\overline{\mathbb C}\setminus U$ that is contained in $\Omega$. Being a connected component, $E$ is closed in $\overline{\mathbb C}\setminus U$. And since $\overline{\mathbb C}\setminus U$ is a compact set, so is $E$. Finally, $E$ is also a component of $V\setminus U$ because $E\subset V$. A contradiction.

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I don't understand the last three lines. Could you be clearer? –  Andrea Jul 8 '12 at 20:59
    
@Andrea I rewrote the proof with more details and fewer errors. –  user31373 Jul 9 '12 at 21:01
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