\section{Runge's Theorem}
\label{section:runge}

\begin{proposition}
\label{proposition:existence-curves}
    Let $K \subset \complex$ be compact and $U \in \cn_\complex(K)$, then there exists closed rectifiable curves $\seqf{\gamma_j}$ such that for any separated locally convex space $E$, $f \in H(U; E)$, and $z_0 \in K$, 
    \[
    f(z) = \sum_{j = 1}^n \int_{\gamma_j} \frac{f(z)}{z - z_0}dz
    \]
\end{proposition}
\begin{proof}[Proof, {{\cite[Proposition VIII.1.1]{ConwayComplex}}}. ]
    Via \hyperref[fattening]{proposition:distance-compact}, let $V \in \cn_\complex^o(K)$ precompact with $\ol V \subset U$. Identify $\complex = \real^2$, then since $V$ is precompact, there exists $\delta > 0$ and $\seqf{(x_j, y_j)} \subset U$ such that:
    \begin{enumerate}
        \item For each $1 \le j \le n$, $R_j = [x_j, x_j + \delta] \times [y_j, y_j + \delta] \subset U$. 
        \item $\bigcup_{j = 1}^n R_j \supset V$. 
        \item $x_j, y_j = 0 \mod \delta$. 
    \end{enumerate}

    In other words, $V$ is covered with a grid of squares with side-length $\delta$ and corners $\seqf{(x_j, y_j)}$. Now, for each $1 \le j \le n$ and $t \in [0, 1]$, let
    \begin{align*}
        \gamma_{j, \rightarrow}(t) &= (1 - t)(x_j, y_j) + t(x_j + \delta, y_j) \\
        \gamma_{j, \uparrow}(t) &= (1 - t)(x_j + \delta, y_j) + t(x_j + \delta, y_j + \delta) \\
        \gamma_{j, \leftarrow}(t) &= (1 - t)(x_j + \delta, y_j + \delta) + t(x_j, y_j + \delta) \\
        \gamma_{j, \downarrow}(t) &= (1 - t)(x_j, y_j + \delta) + t(x_j, y_j) \\
    \end{align*}

    and $\gamma_j = \gamma_{j, \downarrow} \cdot \gamma_{j, \leftarrow} \cdot \gamma_{j, \uparrow} \cdot \gamma_{j, \rightarrow}$ be their concatenation, then for each $z \in U \setminus \partial R_j$, 
    \[
    \int_{\gamma_j} \frac{f(z)}{z - z_0}dz = \begin{cases}
        f(z) &z \in R_j^o \\
        0 &z \in U \setminus R_j
    \end{cases}
    \]

    by \hyperref[Cauchy's Integral Formula]{theorem:cauchy-formula}. Let $\seqf[N]{\mu_j}$ be an enumeration of
    \[
    \bigcup_{j = 1}^n \bracs{\gamma_{j, \downarrow}, \gamma_{j, \leftarrow}, \gamma_{j, \uparrow}, \gamma_{j, \rightarrow}}
    \]

    then for each $z_0 \in \bigcup_{j = 1}^n R_j^o$ and $f \in H(U; E)$, 
    \[
    f(z) = \sum_{j = 1}^N \int_{\mu_j}\frac{f(z)}{z - z_0}dz
    \]

    From here, it is sufficient to eliminate line segments that intersect $K$ and ensure that the remaining segments form a collection of loops. Let $1 \le j \le k \le N$, then $(\mu_j, \mu_k)$ is \textit{redundant} if for each $t \in [0, 1]$, $\mu_j(t) = \mu_k(1 - t)$. If $(\mu_j, \mu_k)$ are redundant, then
    \begin{enumerate}[start=3]
        \item There exists no $1 \le l \le N$ with $l \ne j, k$ such that $(\mu_j, \mu_l)$ or $(\mu_k, \mu_l)$ is redundant. 
        \item $\int_{\mu_j}\frac{f(z)}{z - z_0}dz + \int_{\mu_k}\frac{f(z)}{z - z_0}dz = 0$.
    \end{enumerate}

    so every line segment either cannot form a redundant pair, or forms a unique one. 

    By relabeling, let $\bracs{\mu_j|1 \le j \le m}$ such that for each $1 \le j \le m$, there exists no $1 \le k \le n$ such that $(\mu_j, \mu_k)$ is redundant. By (5), for each $z_0 \in \bigcup_{j = 1}^n R_j^o$ and $f \in H(U; E)$, 
    \[
    f(z) = \sum_{j = 1}^m \int_{\mu_j}\frac{f(z)}{z - z_0}dz
    \]

    Let $1 \le j \le N$ such that $\mu_j([0, 1]) \cap K \ne \emptyset$. Since $V \in \cn_\complex(K)$, there exists $1 \le k \le N$ such that $(\mu_j, \mu_k)$ are redundant by (2) and (3). Therefore $\bigcup_{j = 1}^m \mu_j([0, 1]) \cap K = \emptyset$. Since $\bigcup_{j = 1}^n R_j^o$ is dense in $K$, the above also holds for all $z \in K$. 

    Finally, let $1 \le j \le m$. Since $\mu_j$ does not form a redundant pair, $\mu_j(1)$ intersects at most two distinct squares by (3). In which case, there must exist $1 \le k \le m$ such that $\mu_k(0) = \mu_j(1)$. Therefore $\seqf[m]{\mu_j}$ forms a collection of closed rectifiable curves. 
\end{proof}

\begin{lemma}
\label{lemma:rational-curve-approximation}
    Let $\gamma \in C([a, b]; \complex)$ be a rectifiable curve, $K \subset \complex$ such that $K \cap \gamma([a, b]) = \emptyset$, $f \in C(\gamma([a, b]); \complex)$, and $\eps > 0$, then there exists $R \in \complex(z)$ such that:
    \begin{enumerate}
        \item $R \in H(\complex \setminus \gamma([a, b]); \complex)$. 
        \item For each $z_0 \in K$, 
        \[
        \abs{\int_{\gamma} \frac{f(z)}{z - z_0}dz - R(z)} < \eps
        \]
    \end{enumerate}
\end{lemma}
\begin{proof}[Proof, {{\cite[Lemma VIII.1.5]{ConwayComplex}}}. ]
    Since the mapping
    \[
    \varphi: K \times [a, b] \to \complex \quad (z_0, t) \mapsto \frac{f \circ \gamma(t)}{\gamma(t) - z}
    \]

    is continuous, it is uniformly continuous by \autoref{proposition:uniform-continuous-compact}. Hence the mappings $\bracs{\varphi(z_0, \cdot)|t \in [a, b]}$ are uniformly equicontinuous. Thus there exists $\delta > 0$ such that for each $s, t \in [a, b]$ with $|s - t| \le \delta$, 
    \[
    \abs{\frac{f \circ \gamma(s)}{\gamma(s) - z_0} - \frac{f \circ \gamma(t)}{\gamma(t) - z_0}} < \eps
    \]

    for all $z_0 \in K$. 

    Let $(P = \seqfz{t_j}) \in \scp([a, b])$ with $\sigma(t) < \delta$, and
    \[
    R(z) = \sum_{j = 1}^n f \circ \gamma(t_j)\frac{\gamma(t_j) - \gamma(t_{j-1})}{\gamma(t_j) - z}
    \]

    then $R \in \complex(z) \cap H(\complex \setminus \gamma([a, b]); \complex)$, and for each $z_0 \in K$, 
    \begin{align*}
        \abs{\int_\gamma \frac{f(z)}{z - z_0}dz - R(z)} &\le 
        \sum_{j = 1}^n \int_{t_{j-1}}^{t_j}\abs{\frac{f \circ \gamma(t)}{\gamma(t) - z_0} - \frac{f \circ \gamma(t_j)}{\gamma(t_j) - z}}\gamma(dt) \\
        &\le \eps \norm{\gamma}_{\text{var}}
    \end{align*}
\end{proof}

\begin{lemma}[Pole Pushing]
\label{lemma:pole-pushing}
    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 for every $a \in \complex \setminus K$ and $\eps > 0$, there exists $R \in \complex(z)$ such that:
    \begin{enumerate}
        \item $R \in H(\complex_\infty \setminus P; \complex)$. 
        \item For every $z \in K$, $|R(z) - 1/(z - a)| < \eps$. 
    \end{enumerate}
\end{lemma}
\begin{proof}[Proof, {{\cite[Lemma III.1.10]{ConwayComplex}}}. ]
    First suppose that $\infty \not\in P$. Let $A \subset \complex_\infty \setminus K$ be the collection of elements such that the lemma holds\footnote{Under the identification that $1/(z - \infty) = 0$.}. 

    Let $a_0 \in \complex \setminus K$, then there exists $V \in \cn_\complex(a)$ compact such that $V \cap K = \emptyset$. In which case, the function
    \[
    K \times V \to \complex \quad (z, a) \mapsto \frac{1}{z - a}
    \]

    is continuous, and hence uniformly continuous by \autoref{proposition:uniform-continuous-compact}. Thus $1/(z - a) \to 1/(z - a_0)$ uniformly on compact sets as $a \to a_0$, and if $a_0 \in \ol{A}$, then $a_0 \in A$ as well. Therefore $A$ is a closed subset of $\complex_\infty \setminus K$. 

    Now, let $a_0 \in A \cap \complex$, then by \autoref{proposition:distance-compact}, $r = d(a_0, K) > 0$. Let $a \in B_\complex(a_0, r)$, then for each $z \in K$, 
    \begin{align*}
    \frac{1}{z - a} &= \frac{1}{z - a_0}\frac{z - a_0}{z - a} = \frac{1}{z - a_0}\braks{\frac{z - a}{z - a_0}}^{-1} \\
    &= \frac{1}{z - a_0}\braks{1 - \frac{a - a_0}{z - a_0}}^{-1}
    \end{align*}
    
    Since $\sup_{z \in K}|a - a_0|/|z - a_0| < 1$, 
    \[
    \frac{1}{z - a} = \frac{1}{z - a_0}\sum_{n = 0}^\infty \braks{\frac{a - a_0}{z - a_0}}^n
    \]

    where the convergence is uniform on $K$. Therefore $B_\complex(a_0, r) \subset A$ as well. 

    Finally, for each $a \in \complex \setminus K$ with $|a| > \sup_{z \in K}|z|$, 
    \[
        \frac{1}{z - a} = -\frac{1}{a\paren{1 - \frac{z}{a}}} = -\frac{1}{a}\sum_{n = 0}^\infty \braks{\frac{z}{a}}^n
    \]

    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 $\complex_\infty \setminus K$ 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 $P \subset \complex_\infty \setminus K$ 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 $z_0 \in K$, 
    \[
    f(z_0) = \sum_{j = 1}^n \int_{\gamma_j}\frac{f(z)}{z - z_0}dz
    \]

    Let $T$ be the union of the images of $\seqf{\gamma_j}$, then by \autoref{lemma:rational-curve-approximation}, there exists $\seqf{R_j} \subset \complex(z) \cap H(\complex \setminus T; \complex)$ such that for each $z_0 \in K$ and $1 \le j \le n$, 
    \[
    \abs{\int_{\gamma_j}\frac{f(z)}{z - z_0}dz - R_j(z)} < \frac{\eps}{n}
    \]

    Thus
    \[
    \abs{f(z_0) - \sum_{j = 1}^n R_j(z)} < \eps
    \]
    
    for all $z_0 \in K$. By the \hyperref[pole pushing lemma]{lemma:pole-pushing}, there exists $S \in \complex(z) \cap H(\complex_\infty \setminus P; \complex)$ such that $|S(z_0) - \sum_{j = 1}^n R_j(z_0)| < \eps$ for all $z_0 \in K$. Therefore $|S(z_0) - f(z_0)| < 2\eps$ for all $z_0 \in K$.
\end{proof}

\begin{corollary}
\label{corollary:runge-rational-approximation}
    Let $K \subset \complex$ be compact, then $\complex(z) \cap H(K; \complex)$ is dense in $H(K; \complex)$. 
\end{corollary}