Added Runge's theorem.

src/dg/complex/derivative.tex

@@ -280,63 +280,4 @@
     (5) $\Rightarrow$ (3): By the equivalence of the prior points, for any $\phi \in E^*$, $\phi \circ f$ satisfies (3). By the \hyperref[Hahn-Banach Theorem]{proposition:hahn-banach-utility}, $f$ also satisfies (3). 
 \end{proof}
 
-\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}
-
 

src/dg/complex/index.tex

@@ -3,8 +3,9 @@
 
 
 \input{./derivative.tex}
+\input{./sphere.tex}
 \input{./space.tex}
 \input{./log.tex}
 \input{./entire.tex}
-
+\input{./runge.tex}
 

src/dg/complex/runge.tex

@@ -0,0 +1,170 @@
+\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 $C_\infty \setminus K$ by \autoref{lemma:union-connected-components}. Given that $P$ intersects every connected component of $C_\infty \setminus K$, the lemma holds for all $a \in \complex \setminus K$. 
+\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. 
+\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 $1 \le j \le n$, 
+    \[
+    \abs{\int_{\gamma_j}\frac{f(z)}{z - z_0}dz - R_j(z)} < \frac{\eps}{n}
+    \]
+
+    so
+    \[
+    \abs{f(z_0) - \sum_{j = 1}^n R_j(z)} < \eps
+    \]
+    
+    for all $z_0 \in K$. By the \hyperref[pole pushing lemma]{lemma:pole-pushing}, there exists $S \in \complex(z) \cap H(\complex_\infty \setminus P)$ such that $|S(z_0) - \sum_{j = 1}^n R_j(z_0)| < \eps$ 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}
+
+
+

src/dg/complex/sphere.tex

@@ -0,0 +1,16 @@
+\section{The Riemann Sphere}
+\label{section:riemann-sphere}
+
+
+\begin{definition}[Extended Complex Plane]
+\label{definition:extended-complex-plane}
+    Let $\complex$ be the complex plane, then its one-point compactification $\complex_\infty = \complex \sqcup \bracs{\infty}$ is the \textbf{extended complex plane}. 
+\end{definition}
+
+\begin{definition}[Holomorphic on $\complex_\infty$]
+\label{definition:holomorphic-on-sphere}
+    Let $E$ be a separated locally convex space and $f \in C(\complex_\infty; E)$, then $f$ is \textbf{holomorphic at $\infty$} if $z \mapsto f(z^{-1})$ (under the identification that $1/0 = \infty$) is holomorphic at $0$. 
+\end{definition}
+
+
+

src/topology/main/connected.tex

@@ -45,3 +45,13 @@
 \begin{proof}
     Let $C$ be the union of all connected sets that contain $A$, then $C$ is connected by \autoref{proposition:connected-union}, and is the maximum connected set containing $A$ by definition.
 \end{proof}
+
+\begin{lemma}
+\label{lemma:union-connected-components}
+    Let $X$ be a topological space and $A \subset X$ be both open and closed, then $A$ is a union of connected components of $X$. 
+\end{lemma}
+\begin{proof}
+    Let $C \subset X$ be a connected component, then $C \cap A$ and $C \setminus A$ are both open. Since $C$ is connected, either $C \cap A = \emptyset$ and $C \subset A^c$, or $C \setminus A = \emptyset$ and $C \subset A$. 
+\end{proof}
+
+