Added path lemma.

src/dg/complex/derivative.tex

@@ -280,5 +280,63 @@
     (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/space.tex

@@ -38,3 +38,15 @@
 
     (C1) $\Leftrightarrow$ (C2): By the \hyperref[Arzelà-Ascoli Theorem]{theorem:arzela-ascoli}.
 \end{proof}
+
+\begin{definition}[Space of Holomorphic Functions Near a Set]
+\label{definition:holomorphic-function-space-near}
+    Let $E$ be a complete separated locally convex space over $\complex$ and $A \subset \complex$. Direct $\cn_{\complex}^o(A)$ under reverse inclusion, then the inductive limit
+    \[
+    H(A; E) = \varinjlim_{U \in \cn_{\complex}^o(A)} H(U; E)
+    \]
+
+    is the \textbf{space of holomorphic functions near} $A$, and is of type (LF). 
+\end{definition}
+
+