Added Runge's theorem.

src/cat/notation/index.tex

@@ -14,6 +14,8 @@
     $\mathbb{D}_n$, $\mathbb{D}$ & Dyadic rationals of level $n$; all dyadic rationals. & \autoref{definition:dyadic} \\
     $\mathrm{rk}(q)$ & Dyadic rank of $q \in \mathbb{D}$. & \autoref{definition:dyadic-rank} \\
     $M(x)$ & Unique $M(x) \subset \mathbb{N}^+ \cap [1, \mathrm{rk}(x)]$ such that $x = \sum_{n \in M(x)} 2^{-n}$. & \autoref{proposition:dyadic-subset} \\
-    $[n]$ & $\bracs{1, \cdots, n}$ & N/A
+    $[n]$ & $\bracs{1, \cdots, n}$ & N/A \\
+    $R[x]$ & Ring of polynomials over $R$. & N/A \\
+    $F(x)$ & Field of fractions over $F$. & N/A \\ 
 \end{tabular}
 

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/dg/notation.tex

@@ -17,5 +17,8 @@ Differential geometry is the study of things invariant under change of notation.
     $\tilde C_\sigma^n(U; F)$ & $n$-fold continuously $\tilde \sigma$-differentiable functions. & \autoref{definition:continuously-differentiable-space} \\
     $L^{(n)}_\sigma(E; F)$ & Codomain of derivatives. $L^{(0)}_\sigma(E; F) = F$, $L^{(n)}_\sigma(E; F) = L(E; L_\sigma^{(n-1)}(E; F))$, equipped with the $\sigma$-uniform topology. & \autoref{definition:higher-derivatives-codomain} \\
     $x^{(k)}$ & Tuple of $x$ repeated $k$ times. & \autoref{theorem:taylor-peano} \\
-    $D^+f(x)$ & Right derivative of $f$ at $x$. & \autoref{definition:right-differentiable-mvt}
+    $D^+f(x)$ & Right derivative of $f$ at $x$. & \autoref{definition:right-differentiable-mvt} \\
+    $\omega_{z, r}$ & Standard path of winding number 1. & \autoref{definition:winding-number-1} \\
+    $H(U; E)$ & Space of $E$-valued holomorphic functions on $U$. & \autoref{definition:holomorphic-function-space} \\
+    $H(A; E)$ & Space of $E$-valued holomorphic functions near $A$. & \autoref{definition:holomorphic-function-space-near} 
 \end{tabular}

src/topology/main/compactify.tex

@@ -71,5 +71,59 @@
     
 \end{proof}
 
+\begin{lemma}
+\label{lemma:lch-compactification-open}
+    Let $X$ be a LCH space and $(Y, \varphi)$ be a compactification of $X$, then $\varphi(X) \subset Y$ is open. 
+\end{lemma}
+\begin{proof}
+    For each $x \in X$, let $U \in \cn_X(x)$ be a compact neighbourhood. Since $Y$ is a compact Hausdorff space, $\varphi(U)$ is closed by \autoref{proposition:compact-closed}. As $\varphi \in C(X; Y)$ is an embedding, there exists $V \in \cn_Y(\varphi(x))$ such that $\varphi(U) = \varphi(X) \cap V$. Given that $\varphi(X)$ is dense in $Y$, $\varphi(U) = \ol{\varphi(X) \cap V} \supset V$. Therefore $\varphi(U) \in \cn_{Y}(\varphi(x))$, and $\varphi(X)$ is open in $Y$. 
+\end{proof}
+
+
+
+\begin{definition}[One-Point Compactification]
+\label{definition:alexandroff-compactification}
+    Let $(X, \mathcal{T})$ be a LCH space, then there exists a pair $(X^*, \iota)$ such that:
+    \begin{enumerate}
+        \item $(X^*, \iota)$ is a compactification of $X$. 
+        \item[(U)] For any pair $(Y, \varphi)$ satisfying (1), there exists a unique $\varphi^* \in C(Y; X^*)$ such that the following diagram commutes:
+        \[
+        \xymatrix{
+        Y \ar@{->}[rd]^{\varphi^*} &  \\
+        X \ar@{->}[r]_{\iota} \ar@{->}[u]^{\varphi} & X^*
+        }
+        \]
+    \end{enumerate}
+    
+    The pair $(X^*, \iota)$ is the \textbf{one-point compactification} of $X$.
+\end{definition}
+\begin{proof}
+    Let $\infty$ be a point not in $X$, $X^* = X \sqcup \bracs{\infty}$, and $\mathcal{T}^* \subset 2^{X^*}$ such that for each $U \in \mathcal{T}^*$, one of the following holds:
+    \begin{enumerate}[label=(\alph*)]
+        \item $U \in \mathcal{T}$.
+        \item $\infty \in U$ and $U^c \subset X$ is compact. 
+    \end{enumerate}
+
+    Let $\seqi{U} \subset \mathcal{T}^*$ be an open cover of $X$, then there exists $i \in I$ such that $\infty \in U$. In which case, $U_i$ must satisfy (b), so there exists $J \subset I$ finite such that $\bigcup_{j \in J}U_j \supset U_i^c$, and $\bracsn{U_j|j \in J \cup \bracs{i}}$ is a finite subcover. Now, let $x \in X$, then since $X$ is locally compact, there exists a precompact neighbourhood $U \in \cn_X^o(x)$. In which case, $\ol{U}^c \in  \cn_{X^*}(\infty)$ with $U \cap \ol{U}^c = \emptyset$. Therefore $X^*$ is a compact Hausdorff space. 
+    
+    Let $\iota: X \to X^*$ be the inclusion map. For each $U \in \mathcal{T}^*$ satisfying (b), $\iota^{-1}(U) = U \cap X$. Since $U^c \subset X$ is compact, $U \cap X$ is open by \autoref{proposition:compact-closed}, so $\iota \in C(X; X^*)$. By (a), $\iota$ is an embedding. 
+
+    Finally, let
+    \[
+    \varphi^*: Y \to X^* \quad x \mapsto \begin{cases}
+        \varphi^{-1}(x) &x \in \varphi(X) \\
+        \infty &x \not\in \varphi(X)
+    \end{cases}
+    \]
+
+    Let $U \subset X^*$ with $\infty \not\in U$, then $(\varphi^*)^{-1}(U) = \varphi(U)$ is open in $\varphi(X)$ because $\varphi$ is an embedding, and open in $Y$ by \autoref{lemma:lch-compactification-open}. On the other hand, for each $V \in \cn_{X^*}^o(\infty)$, 
+    \[
+    (\varphi^*)^{-1}(V) = V \cup (Y \setminus \varphi(X))
+    \]
+
+    Since $\varphi \in C(X; Y)$ is an embedding, $V$ is relatively open in $\varphi(X)$, so $V \cup (Y \setminus \varphi(X))$ is open in $Y$. 
+\end{proof}
+
+
 
 

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}
+
+