Various typo fixes.
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Bokuan Li
2026-05-30 20:35:32 -04:00
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@@ -135,12 +135,12 @@
where the convergence is uniform on $K$, so there exists a neighbourhood of $\infty$ that is contained in $A$.
Since $P \cup \bracs{\infty} \subset A$, and $A$ is an open and closed subset of $C_\infty \setminus K$, $A$ is a union of connected components of $C_\infty \setminus K$ by \autoref{lemma:union-connected-components}. Given that $P$ intersects every connected component of $C_\infty \setminus K$, the lemma holds for all $a \in \complex \setminus K$.
Since $P \cup \bracs{\infty} \subset A$, and $A$ is an open and closed subset of $C_\infty \setminus K$, $A$ is a union of connected components of $\complex_\infty \setminus K$ by \autoref{lemma:union-connected-components}. Given that $P$ intersects every connected component of $\complex_\infty \setminus K$, the lemma holds for all $a \in \complex \setminus K$.
\end{proof}
\begin{theorem}[Runge]
\label{theorem:runge}
Let $K \subset \complex$ be compact and $P \subset C_\infty \setminus K$ such that $P$ intersects every connected component of $C_\infty \setminus K$, then $\complex(x) \cap H(C_\infty \setminus P; \complex)$ is dense in $H(K; \complex)$ with respect to the uniform topology.
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.
\end{theorem}
\begin{proof}
Let $U \in \cn_\complex(K)$, $f \in H(U; \complex)$, and $\eps > 0$. By \autoref{proposition:existence-curves}, there exists closed rectifiable curves $\seqf{\gamma_j}$ in $U \setminus V$ such that for each $z_0 \in K$,