Theorem 27.6.4 (Runge).label Let $K \subset \complex$ be compact and $P \subset \complex_{\infty} \setminus K$ such that $P$ intersects every connected component of $\complex_{\infty} \setminus K$, then $\complex(x) \cap H(\complex_{\infty} \setminus P; \complex)$ is dense in $H(K; \complex)$ with respect to the uniform topology.
Proof. Let $U \in \cn_{\complex}(K)$, $f \in H(U; \complex)$, and $\eps > 0$. By Proposition 27.6.1, there exists closed rectifiable curves $\seqf{\gamma_j}$ in $U \setminus V$ such that for each $z_{0} \in K$,
Let $T$ be the union of the images of $\seqf{\gamma_j}$, then by Lemma 27.6.2, there exists $\seqf{R_j}\subset \complex(z) \cap H(\complex \setminus T; \complex)$ such that for each $z_{0} \in K$ and $1 \le j \le n$,
so
for all $z_{0} \in K$. By the pole pushing lemma, there exists $S \in \complex(z) \cap H(\complex_{\infty} \setminus P)$ such that $|S(z_{0}) - \sum_{j = 1}^{n} R_{j}(z_{0})| < \eps$ for all $z_{0} \in K$.$\square$
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