Updated proof for C0 space.

.vscode/settings.json

@@ -5,5 +5,7 @@
             "task": "Watch"
         }
     ],
-    "latex.linting.enabled": false
+    "latex.linting.enabled": false,
+    "latex-workshop.latex.autoBuild.run": "never",
+    "latex-workshop.latex.texDirs": ["${workspaceFolder}"]
 }

document.tex

@@ -1,4 +1,4 @@
-% \documentclass{report}
+%\documentclass{report}
 \usepackage{amssymb, amsmath, hyperref}
 \usepackage{preamble}
 
@@ -12,7 +12,7 @@
 %\input{./src/process/index}
 
 \bibliographystyle{alpha} % We choose the "plain" reference style
-\bibliography{refs} % Entries are in the refs.bib file
+\bibliography{refs.bib} % Entries are in the refs.bib file
 
 
 \end{document}

src/fa/lp/definition.tex

@@ -114,3 +114,49 @@
 
     Therefore $\norm{f_n - f}_{L^p(X; E)} \to 0$ as $n \to \infty$.
 \end{proof}
+
+\begin{theorem}[Markov's Inequality]
+\label{theorem:markov-inequality}
+    Let $(X, \cm, \mu)$ be a measure space, $E$ be a normed space, and $f: X \to E$ be a Borel measurable function, then
+    \begin{enumerate}
+        \item For any $\alpha > 0$, 
+        \[
+        \mu\bracs{|f| \ge \alpha} \le \frac{1}{\alpha}\norm{f}_{L^1(X; E)}
+        \]
+        \item For $\alpha > 0$ and strictly increasing function $\phi: [0, \infty) \to [0, \infty)$,
+        \[
+        \mu\bracs{|f| \ge \alpha} \le \frac{1}{\phi(\alpha)}\norm{\phi \circ |f|}_{L^1(X; E)}
+        \]
+        \item For any $\alpha > 0$,
+        \[
+        \mu\bracs{|f| \ge \alpha} \le \frac{1}{\alpha^p}\norm{f}_{L^p(X; E)}^p
+        \]
+    \end{enumerate}
+\end{theorem}
+\begin{proof}
+    (1): For any $\alpha > 0$, 
+    \begin{align*}
+        \mu\bracs{|f| \ge \alpha} &= \int_{\bracs{f \ge \alpha}} d\mu \le \frac{1}{\alpha}\int |f|d\mu = \frac{\norm{f}_{L^1(X; E)}}{\alpha}
+    \end{align*}
+    
+    (2): By \autoref{lemma:monotone-measurable}, $\phi: [0, \infty) \to [0, \infty)$ is Borel measurable. Since $\phi$ is strictly increasing, $\bracs{|f| \ge \alpha} = \bracs{\phi \circ |f| \ge \phi(\alpha)}$, and the result holds by applying (1) to $\phi \circ |f|$. 
+
+    (3): Since $x \mapsto x^p$ is strictly increasing on $[0, \infty)$, applying (2) to $\phi(x) = x^p$ yields the desired result.
+\end{proof}
+
+\begin{proposition}
+\label{proposition:lp-in-measure}
+    Let $(X, \cm, \mu)$ be a measure space, $E$ be a normed space, $p \in [1, \infty]$, $\seq{f_n} \subset L^p(X; E)$, and $f \in L^p(X; E)$ such that $f_n \to f$ in $L^p$, then $f_n \to f$ in measure.
+\end{proposition}
+\begin{proof}
+    Let $\eps > 0$. If $p < \infty$, then by \hyperref[Markov's Inequality]{theorem:markov-inequality},
+    \[
+    \limv{n}\mu\bracs{|f_n - f| \ge \eps} \le \limv{n}\frac{1}{\eps^p}\norm{f_n - f}_{L^p(X; E)}^p = 0
+    \]
+    
+    If $p = \infty$, then there exists $N \in \natp$ such that $\norm{f_n - f}_{L^\infty(X; E)} < \eps$ for all $n \ge N$. In which case, $\mu\bracs{|f_n - f| \ge \eps} = 0$ for all $n \ge N$. 
+\end{proof}
+
+
+
+

src/measure/measurable-maps/in-measure.tex

@@ -0,0 +1,83 @@
+\section{Convergence in Measure}
+\label{section:convergence-in-measure}
+
+\begin{definition}[Convergence in Measure]
+\label{definition:convergence-in-measure}
+    Let $(X, \cm, \mu)$ be a measure space, $(Y, d)$ be a metric space, and $\seq{f_n}$ and $f$ be Borel measurable functions from $X$ to $Y$, then $f_n \to f$ \textbf{in measure} if for every $\eps > 0$,
+    \[
+    \lim_{n \to \infty}\mu(\bracs{d(f_n, f) > \eps}) = 0
+    \]
+\end{definition}
+
+\begin{definition}[Cauchy in Measure]
+\label{definition:cauchy-in-measure}
+    Let $(X, \cm, \mu)$ be a measure space, $(Y, d)$ be a metric space, and $\seq{f_n}$ be Borel measurable functions from $X$ to $Y$, then $\seq{f_n}$ is \textbf{Cauchy in measure} if for every $\eps > 0$,
+    \[
+    \mu(\bracs{d(f_m, f_n) > \eps}) \to 0 \quad \text{as}\ m, n \to \infty
+    \]
+\end{definition}
+
+\begin{lemma}
+\label{lemma:ae-in-measure}
+    Let $(X, \cm, \mu)$ be a finite measure space, $(Y, d)$ be a metric space, and $\seq{f_n}$ and $f$ be Borel measurable functions from $X$ to $Y$ such that $f_n \to f$ almost everywhere, then $f_n \to f$ in measure.
+\end{lemma}
+\begin{proof}
+    Let $\eps > 0$, then for almost every $x \in X$, there exists $N \in \natp$ such that $d(f_n(x), f(x)) < \eps$ for all $n \ge N$, so
+    \[
+    \mu\paren{\bigcup_{N \in \natp}\bigcap_{n \ge N}\bracs{d(f_n, f) < \eps}} = \mu(X)
+    \]
+
+    By continuity from above (\autoref{proposition:measure-properties}), 
+    \[
+    \limv{N}\mu{\bracs{d(f_N, f) \ge \eps}} \le \limv{N}\mu\paren{\bigcup_{n \ge N}\bracs{d(f_n, f) \ge \eps}} = 0
+    \]
+\end{proof}
+
+
+\begin{proposition}[{{\cite[Theorem 2.30]{Folland}}}]
+\label{proposition:cauchy-in-measure-limit}
+    Let $(X, \cm, \mu)$ be a measure space, $(Y, d)$ be a complete metric space, and $\seq{f_n}$ be Borel measurable functions from $X \to Y$, then:
+    \begin{enumerate}
+        \item There exists a Borel measurable function $f: X \to Y$ such that $f_n \to f$ in measure.
+        \item There exists a subsequence $\seq{n_k}$ such that $f_{n_k} \to f$ almost everywhere.
+        \item For any Borel measurable function $g: X \to Y$ such that $f_n \to g$ in measure, $f = g$ almost everywhere.
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (2): Since $\seq{f_n}$ is Cauchy in measure, there exists a subsequence $\seq{n_k}$ such that for each $k \in \natp$, $\mu(\bracsn{d(f_{n_k}, f_{n_{k+1}}) > 2^{-k}}) \le 2^{-k}$.
+    
+    In this case, for any $K \in \natp$ and $j \ge k \ge K$, 
+    \[
+    \bracsn{d(f_{n_j}, f_{n_k}) > 2^{-K+1}} \subset \bigcup_{\ell = k}^{j - 1}\bracsn{d(f_{n_\ell}, f_{n_{\ell + 1}}) > 2^{-\ell}}
+    \subset \bigcup_{\ell = K}^{\infty}\bracsn{d(f_{n_\ell}, f_{n_{\ell + 1}}) > 2^{-\ell}}
+    \]
+
+    By monotonicity and subadditivity (\autoref{proposition:measure-properties}), 
+    \[
+    \mu\paren{\bigcup_{j, k \ge K}\bracsn{d(f_{n_j}, f_{n_k}) > 2^{-K+1}}} 
+    \le \sum_{\ell \ge K}\mu\bracsn{d(f_{n_\ell}, f_{n_{\ell + 1}}) > 2^{-\ell}}
+    \le 2^{-K+1}
+    \]
+
+    so
+    \[
+    \mu\braks{\paren{\bigcap_{j, k \ge K}\bracsn{d(f_{n_j}, f_{n_k}) \le 2^{-K+1}}}^c} \le 2^{-K+1}
+    \]
+
+    By the \hyperref[First Borel-Cantelli Lemma]{lemma:borel-cantelli-1},
+    \[
+    \mu\braks{\limsup_{K \to \infty}\paren{\bigcap_{j, k \ge K}\bracsn{d(f_{n_j}, f_{n_k}) \le 2^{-K+1}}}^c} = 0
+    \]
+
+    Thus, for almost every $x \in X$, there exists $K \in \natp$ such that $d(f_{n_j}(x), f_{n_k}(x)) < 2^{-K+1}$ for all $j, k \ge K$. Therefore $\seq{f_n(x)}$ is Cauchy for almost every $x$, and converges to a Borel measurable function $f: X \to Y$.
+
+    (1): Let $\eps, \delta > 0$, then there exists $N \in \natp$ such that $\mu(\bracs{d(f_m, f_n) > \delta/2}) < \eps/2$ for all $m, n \ge N$. Let $k \in \natp$ such that $n_k \ge N$ and $\mu(\bracs{d(f, f_{n_k}) > \delta/2}) < \eps/2$, then for any $m \ge N$,
+    \[
+    \mu\bracs{d(f_m, f) > \delta} \le \mu\bracs{d(f_m, f_{n_k}) > \delta/2} + \mu\bracs{d(f, f_{n_k}) > \delta/2} < \eps
+    \]
+
+    (3): By (2), there exists a subsequence $\seq{n_k}$ such that $f_{n_k} \to f$ and $f_{n_k} \to g$ almost everywhere, so $f = g$ almost everywhere. 
+\end{proof}
+
+
+

src/measure/measurable-maps/index.tex

@@ -6,3 +6,4 @@
 \input{./real-valued.tex}
 \input{./simple.tex}
 \input{./metric.tex}
+\input{./in-measure.tex}

src/measure/measurable-maps/measurable-maps.tex

@@ -6,11 +6,17 @@
     Let $(X, \cm)$ and $(Y, \cn)$ be measurable spaces and $f: X \to Y$ be a mapping, then $f$ is \textbf{$(\cm, \cn)$-measurable} if $f^{-1}(E) \in \cm$ for all $E \in \cn$.
 \end{definition}
 
+
 \begin{definition}[Borel Measurable]
 \label{definition:borel-measurable-function}
     Let $(X, \cm)$ be a measurable space, $Y$ be a topological space, and $f: X \to Y$ be a mapping, then $f$ is \textbf{Borel measurable} if $f$ is $(\cm, \cb_Y)$-measurable.
 \end{definition}
 
+\begin{definition}[Convergence Almost Everywhere]
+\label{definition:convergence-ae}
+    Let $(X, \cm, \mu)$ be a measure space, $Y$ be a topological space, $\seq{f_n}$ and $f$ be Borel measurable functions from $X$ to $Y$, then $f_n \to f$ \textbf{almost everywhere/a.e.} if there exists a $\mu$-null set $N \in \cm$ such that $f_n \to f$ pointwise on $X \setminus N$.
+\end{definition}
+
 \begin{lemma}
 \label{lemma:continuous-borel-measurable}
     Let $X, Y$ be topological spaces and $f: X \to Y$ be continuous, then $f$ is Borel measurable.
@@ -34,3 +40,40 @@
 
     is the smallest $\sigma$-algebra on $X$ such that each $\seqi{f}$ is measurable.
 \end{definition}
+
+
+\begin{theorem}[Egoroff]
+\label{theorem:egoroff}
+    Let $(X, \cm, \mu)$ be a finite measure space, $(Y, d)$ be a metric space, and $\seq{f_n}$ and $f$ be Borel measurable functions from $X$ to $Y$ such that $f_n \to f$ almost everywhere, then for any $\eps > 0$, there exists $E \in \cm$ such that:
+    \begin{enumerate}
+        \item $f_n \to f$ uniformly on $E$.
+        \item $\mu(X \setminus E) < \eps$
+    \end{enumerate}
+\end{theorem}
+\begin{proof}
+    Since $f_n \to f$ almost everywhere, for any $\eps > 0$, 
+    \[
+    \mu\paren{\limsup_{n \to \infty}\bracs{d(f_n, f) > \eps}} = 0
+    \]
+    
+    By continuity from above (\autoref{proposition:measure-properties}), 
+    \[
+    \lim_{N \to \infty}\paren{\bigcup_{n \ge N}\bracs{d(f_n, f) > \eps}} = 0
+    \]
+
+    For each $k \in \natp$, let $N_k \in \natp$ such that
+    \[
+    \paren{\bigcup_{n \ge N_k}\bracs{d(f_n, f) > 1/k}} < 2^{-k}\eps
+    \]
+    then by subadditivity, 
+    \[
+    \mu\paren{\bigcup_{k \in \natp}\bigcup_{n \ge N_k}\bracs{d(f_n, f) > 1/k}} \le \eps
+    \]
+    and $f_n \to f$ uniformly on
+    \[
+    \braks{\bigcup_{k \in \natp}\bigcup_{n \ge N_k}\bracs{d(f_n, f) > 1/k}}^c
+    \]
+\end{proof}
+
+
+

src/measure/measurable-maps/real-valued.tex

@@ -50,3 +50,12 @@
 
     Finally, if the above functions are $\real$-valued, then they are $(\cm, \cb_\real)$-measurable by \autoref{lemma:extended-real-measurable}.
 \end{proof}
+
+\begin{lemma}
+\label{lemma:monotone-measurable}
+    Let $f: \real \to \real$ be a non-decreasing or non-increasing function, then $f$ is Borel measurable.
+\end{lemma}
+\begin{proof}
+    By taking $-f$, assume without loss of generality that $f$ is non-decreasing. In which case, for any $a \in \real$, $x \in f^{-1}((-\infty, a])$, and $y \le x$, $f(y) \le f(x) \le a$, so $y \in f^{-1}((-\infty, a])$. Thus $f^{-1}((-\infty, a))$ is an interval and hence measurable. Since the open rays generate $\cb_\real$ (\autoref{proposition:borel-sigma-real-generators}), $f$ is Borel measurable.
+\end{proof}
+

src/measure/measure/measure.tex

@@ -64,15 +64,19 @@
 
 
     (3): Denote $E_0 = \emptyset$. For each $n \in \natp$, let $F_n = E_n \setminus E_{n - 1}$, then $\seq{F_n}$ is pairwise disjoint with $\bigsqcup_{n \in \natp}F_n = \bigcup_{n \in \natp}E_n$. In which case,
-    \[
+    \begin{align*}
-    \mu\paren{\bigcup_{n \in \natp}E_n} = \mu\paren{\bigsqcup_{n \in \natp}F_n} = \limv{n}\sum_{k = 1}^n \mu(F_n) = \limv{n}\mu(E_n)
+    \mu\paren{\bigcup_{n \in \natp}E_n} &= \mu\paren{\bigsqcup_{n \in \natp}F_n} \\
-    \]
+    &= \limv{n}\sum_{k = 1}^n \mu(F_n) = \limv{n}\mu(E_n)
+    \end{align*}
+    
 
 
     (4): Since $\mu(E_1) < \infty$,
-    \[
+    \begin{align*}
-    \limv{n}\mu(E_n) = \mu(E_1) - \limv{n}\mu(E_1 \setminus E_n) = \mu(E_1) - \mu\paren{E_1 \setminus \bigcap_{n \in \natp}E_n} = \mu\paren{\bigcap_{n \in \natp}E_n}
+    \limv{n}\mu(E_n) &= \mu(E_1) - \limv{n}\mu(E_1 \setminus E_n) \\
-    \]
+    &= \mu(E_1) - \mu\paren{E_1 \setminus \bigcap_{n \in \natp}E_n} \\
+    &= \mu\paren{\bigcap_{n \in \natp}E_n}
+    \end{align*}
 
     by (3).
 
@@ -125,5 +129,23 @@
     \[
     \mu(F) = \limv{n}\mu(E_n \cap F) = \limv{n}\nu(E_n \cap F) = \nu(F)
     \]
-
 \end{proof}
+
+\begin{lemma}[First Borel-Cantelli Lemma]
+\label{lemma:borel-cantelli-1}
+    Let $(X, \cm, \mu)$ be a measure space and $\seq{E_n} \subset \cm$. If $\sum_{n \in \natp}\mu(E_n) < \infty$, then
+    \[
+    \mu\paren{\limsup_{n \to \infty}E_n} = 0
+    \]
+\end{lemma}
+\begin{proof}
+    For any $n \in \natp$, by monotonicity and subadditivity (\autoref{proposition:measure-properties}), 
+    \[
+    \mu\paren{\limsup_{k \to \infty}E_k} \le \mu\paren{\bigcup_{k \ge n}E_k} \le \sum_{k \ge n}\mu(E_k)
+    \]
+    As $\sum_{k \in \natp}\mu(E_k) < \infty$,
+    \[
+    \mu\paren{\limsup_{k \to \infty}E_k} \le \inf_{n \in \natp}\sum_{k \ge n}\mu(E_k) = 0
+    \]
+\end{proof}
+

src/measure/measure/radon.tex

@@ -159,5 +159,47 @@
     \[
     \norm{x \cdot \one_A - x \cdot f}_{L^p(X; E)} \le \norm{x}_E \mu(\bracs{f \ne \one_A})^{1/p} \le \norm{x}_E\mu(U \setminus K)^{1/p} < \eps^{1/p}\norm{x}_E
     \]
-
 \end{proof}
+
+\begin{theorem}[Lusin, {{\cite[Theorem 7.10]{Folland}}}]
+\label{theorem:lusin}
+    Let $X$ be a LCH space, $\mu$ be a Radon measure on $X$, $E$ be a normed vector space, and $f: X \to E$ be a measurable function with $\mu\bracs{f \ne 0} < \infty$, then for any $\eps > 0$,
+    \begin{enumerate}
+        \item There exists $A \subset \bracs{f \ne 0}$ such that $f|_A$ is continuous and $\mu(\bracs{f \ne 0} \setminus A) < \eps$
+        \item If $E = \complex$, then there exists $\phi \in C_c(X; E)$ such that $\mu\bracs{f \ne \phi} < \eps$.
+        \item If $E = \complex$ and $f$ is bounded, then $\phi$ can be taken such that $\norm{\phi}_u \le \norm{f}_u$.
+    \end{enumerate}
+\end{theorem}
+\begin{proof}
+    First assume that $f$ is bounded.
+
+    (1, bounded): If $f$ is bounded, then $f \in L^1(X; E)$. By \autoref{proposition:radon-cc-dense}, there exists $\seq{\phi_n} \subset C_c(X)$ such that $\phi_n \to f$ in $L^1(\mu)$. Since $\phi_n \to f$ in $L^1(\mu)$, $\phi_n \to f$ in measure by \autoref{proposition:lp-in-measure}. By taking a subsequence using \ref{proposition:cauchy-in-measure-limit}, assume without loss of generality that $\phi_n \to f$ almost everywhere. 
+
+    By \hyperref[Egoroff's Theorem]{theorem:egoroff}, there exists $A \subset \bracs{f \ne 0}$ such that $\phi_n \to f$ uniformly and $\mu(\bracs{f \ne 0} \setminus A) < \eps/3$. By \autoref{proposition:radon-regular-sigma-finite}, there exists $K \subset A$ compact such that $\mu(A \setminus K) < \eps/3$. 
+
+    Let $\phi_0 = \lim_{n \to \infty}\phi_n$ on $K$, then $\phi_0 \in C(K; E)$ by \autoref{proposition:uniform-limit-continuous}, $\phi_0 = f$ almost everywhere on $K$, and $\mu(\bracs{f \ne 0} \setminus K) < \eps$.
+
+    (2, bounded): By outer regularity, there exists $U \in \cn^o(\bracs{f \ne 0})$ such that $\mu(U \setminus \bracs{f \ne 0}) < \eps/3$. 
+
+    By the \hyperref[Tietze Extension Theorem]{theorem:lch-tietze}, there exists $\phi \in C_c(X; \complex)$ such that $\phi|_K = \phi_0$ and $\supp{\phi} \subset U$. In which case, 
+    \[
+    \mu\bracs{\phi \ne f} \le \underbrace{\mu(\bracs{f \ne 0} \setminus K)}_{< 2\eps/3} + \underbrace{\mu(U \setminus \bracs{f \ne 0})}_{< \eps/3} < \eps
+    \]
+
+    (3): Let
+    \[
+    \psi: \complex \to \complex \quad z \mapsto \begin{cases}
+        z &|z| \le \norm{f}_u \\
+        z\norm{f}_u/|z| &|z| > \norm{f}_u
+    \end{cases}
+    \]
+    then $\psi$ is continuous with $\bracs{\phi = f} = \bracs{\psi \circ \phi = f}$ and $\norm{\psi \circ \phi}_u \le \norm{f}_u$. 
+
+    Now assume that $f$ is arbitrary.
+
+    (1, unbounded): Since $\mu\bracs{f \ne 0} < \infty$, there exists $\alpha > 0$ such that $\mu\bracs{|f| > \alpha} < \eps/2$. Let $g \in L^\infty(X; E)$ such that $g = f$ on $\bracs{|f| \le \alpha}$. By (1) applied to $g$, there exists $A \subset \bracs{|f| \le \alpha}$ such that $f|_A$ is continuous and $\mu(\bracs{|f| \le \alpha} \setminus A) < \eps/2$. In which case, $\mu(\bracs{f \ne 0} \setminus A) < \eps$, as desired. 
+
+    (2, unbounded): By (1) applied to $g$, there exists $\phi \in C_c(X; \complex)$ such that $\mu\bracs{\phi \ne g} < \eps/2$, and $\mu(\bracs{\phi \ne f}) < \eps$.
+\end{proof}
+
+

src/measure/sets/index.tex

@@ -4,3 +4,4 @@
 \input{./algebra.tex}
 \input{./lambda.tex}
 \input{./elementary.tex}
+\input{./limits.tex}

src/measure/sets/limits.tex

@@ -0,0 +1,35 @@
+\section{Limits of Sets}
+\label{section:set-limits}
+
+\begin{definition}[Limit of Sets]
+\label{definition:set-limits}
+    Let $X$ be a set and $\seq{E_n} \subset 2^X$, then the \textbf{limit superior} of $\seq{E_n}$ is
+    \[
+    \limsup_{n \to \infty} E_n = \bigcap_{n \in \natp}\bigcup_{k \ge n}E_k
+    \]
+    and the \textbf{limit inferior} of $\seq{E_n}$ is
+    \[
+    \liminf_{n \to \infty}E_n = \bigcup_{n \in \natp}\bigcap_{k \ge n}E_k
+    \]
+\end{definition}
+
+
+\begin{lemma}
+\label{lemma:set-limits-characterisation}
+    Let $X$ be a set and $\seq{X_n} \subset 2^X$, then
+    \[
+    \limsup_{n \to \infty} E_n = \bracs{x \in X|x \in E_n \text{for infinitely many }n}
+    \]
+    and
+    \[
+    \liminf_{n \to \infty}E_n = \bracs{x \in X|x \in E_n \text{for all but finitely many }n}
+    \]
+\end{lemma}
+\begin{proof}
+    For any $x \in X$, $x \in \limsup_{n \to \infty}E_n$ if and only if for any $n \in \natp$, there exists $k \ge n$ such that $x \in E_k$, if and only if $x \in E_n$ for infinitely many $n \in \natp$.
+
+    For any $x \in X$, $x \in \liminf_{n \to \infty}$ if and only if there exists $n \in \natp$ such that $x \in E_k$ for all $k \ge n$, if and only if $x \in E_n$ for all but finitely many $n \in \natp$.
+\end{proof}
+
+
+

src/topology/functions/uniform.tex

@@ -11,6 +11,35 @@
     then the \textbf{uniform topology} on $X^T$ is the topology induced by the uniformity generated by $\bracs{E(U)| U \in \fU}$.
 \end{definition}
 
+\begin{remark}
+\label{remark:uniform-uniformity}
+    Properties such as completeness require the presence of a uniform structure. However, referring to it as the uniform uniformity is a little ridiculous. As such, in this section, completeness with respect to the uniform topology corresponds to completeness with respect to the uniform uniformity. 
+\end{remark}
+
+
+\begin{definition}[Space of Bounded Functions]
+\label{definition:bounded-function-space}
+    Let $T$ be a set, $E$ be a TVS over $K \in \RC$, and $f: T \to E$ be a function, then $f$ is \textbf{bounded} if $f(T)$ is a bounded subset of $E$. The set $B(T; E) \subset E^T$ is the space of all bounded functions from $T$ to $E$, and:
+    \begin{enumerate}
+        \item $B(T; E)$ equipped with the uniform topology and pointwise operations is a TVS over $K \in \RC$. 
+        \item $B(T; E)$ is a closed subset of $E^T$ with respect to the uniform topology. In particular, if $E$ is complete, then so is $B(T; E)$.
+    \end{enumerate}
+\end{definition}
+\begin{proof}
+    (1): For any $f, g \in B(T; E)$ and $\lambda \in K$, $(\lambda f + g)(T) \subset (\lambda f)(T) + g(T)$, which is bounded by \autoref{proposition:bounded-operations}, so $B(T; E)$ is closed under addition and scalar multiplication. 
+    
+    Since $f(E)$ is bounded for all $f \in B(T; E)$, $B(T; E)$ forms a TVS over $K$ by \autoref{proposition:tvs-set-uniformity}.
+
+    (2): Let $f \in \ol{B(T; E)}$ and $U \in \cn_E(0)$ be circled, then there exists $g \in B(T; E)$ such that $(f - g)(E) \subset U$. Since $g \in B(T; E)$, there exists $\lambda \ge 0$ such that $g(E) \subset \lambda U$. In which case,
+    \[
+    f(E) \subset g(E) + (f - g)(E) \subset (\lambda + 1)U
+    \]
+    so $f \in B(T; E)$.
+
+    If $E$ is complete, then $E^T$ with the uniform topology is complete by \autoref{proposition:set-uniform-complete}. Thus $B(T; E)$ is also complete by \autoref{proposition:complete-closed}.
+\end{proof}
+
+
 \begin{proposition}
 \label{proposition:uniform-limit-continuous}
     Let $X$ be a topological space and $Y$ be a uniform space, then:
@@ -39,4 +68,24 @@
     \]
 
     so $(f(x), f(y)) \in W \circ W \circ W \subset V$.
+
+    If $Y$ is complete, then $Y^T$ with the uniform topology is complete by \autoref{proposition:set-uniform-complete}. Thus $C(T; X)$ and $UC(T; X)$ are both complete subspaces by \autoref{proposition:complete-closed}.
 \end{proof}
+
+
+\begin{definition}[Space of Bounded Continuous Functions]
+\label{definition:bounded-continuous-function-space}
+    Let $X$ be a topological space and $E$ be a TVS over $K \in \RC$, then $BC(X; E) = B(X; E) \cap C(X; E)$ is the space of bounded and continuous functions from $X$ to $E$, and
+    \begin{enumerate}
+        \item $BC(X; E)$ equipped with the uniform topology and pointwise operations is a TVS over $K \in \RC$. 
+        \item $BC(X; E)$ is a closed subspace of $C(X; E)$ and $B(X; E)$ with respect to the uniform topology. In particular, if $E$ is complete, then so is $BC(X; E)$.
+    \end{enumerate}
+\end{definition}
+\begin{proof}
+    (1): Since addition and scalar multiplication are continuous, $BC(X; E)$ is a subspace of $B(X; E)$, and hence a TVS over $K \in \RC$ by \autoref{definition:bounded-function-space}.
+
+    (2): Since $B(X; E)$ and $C(X; E)$ are both closed subspaces of $E^X$ by \autoref{definition:bounded-function-space}, $BC(X; E)$ is a closed subspace. 
+    
+    If $E$ is complete, then $B(X; E)$ and $C(X; E)$ are both complete under the uniform topology by \autoref{definition:bounded-function-space}. Therefore $BC(X; E)$ is also complete. 
+\end{proof}
+

src/topology/main/c0.tex

@@ -0,0 +1,40 @@
+\section{Continuous Functions Vanishing at Infinity}
+\label{section:vanish-at-infinity}
+
+\begin{definition}[Vanish at Infinity]
+\label{definition:vanish-at-infinity}
+    Let $X$ be a topological space, $E$ be a TVS over $K \in \RC$, and $f \in C(X; E)$, then $f$ \textbf{vanishes at infinity} if for every $U \in \cn_E^o(0)$, $\bracs{f \not\in U}$ is compact.
+
+    The set $C_0(X; \complex)$ is the space of all functions that vanish at infinity. 
+\end{definition}
+
+\begin{proposition}
+\label{proposition:c0-properties}
+    Let $X$ be a topological space and $E$ be a TVS over $K \in \RC$, then:
+    \begin{enumerate}
+        \item $C_0(X; E) \subset BC(X; E)$.
+        \item $C_0(X; E)$ is a closed subspace of $BC(X; E)$ with respect to the uniform topology. In particular, if $E$ is complete, then so is $C_0(X; E)$.
+        \item If $X$ is a LCH space, then $C_c(X; E)$ is a dense subspace of $C_0(X; E)$ with respect to the uniform topology.
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (1): Let $U \in \cn_E^o(0)$ be balanced, then $f(X) = \bracs{f \in U} \sqcup \bracs{f \not\in U}$. Since $f(\bracs{f \not\in U})$ is compact, there exists $\lambda > 0$ such that $f(\bracs{f \not\in U}) \subset \lambda U$. In which case, $f(X) \subset (1 \vee \lambda)(U)$.
+
+    (2): Let $f \in \ol{C_0(X; E)}$ and $U \in \cn_E^o(0)$, then there exists $V \in \cn_E^o(0)$ balanced such that $V + V \subset U$. 
+    
+    Let $g \in C_0(X; E)$ such that $(f - g)(X) \subset V$. Since $g \in C_0(X; E)$, $\bracs{g \not\in V}$ is compact, so
+    \[
+    f(\bracs{g \in V}) \subset (f - g)(X) + g(\bracs{g \in V}) \subset V + V = U
+    \]
+
+    Thus $\bracs{f \not\in U}$ is a closed subset of $\bracs{g \not\in V}$, which is compact by \autoref{proposition:compact-extensions}.
+
+    If $E$ is complete, then $BC(X; E)$ is complete by \autoref{definition:bounded-continuous-function-space}. Since $C_0(X; E)$ is a closed subspace, it is complete by \autoref{proposition:complete-closed}.
+
+    (3): Let $f \in C_0(X; E)$ and $U \in \cn_E^o(0)$ be balanced. By \hyperref[Urysohn's Lemma]{lemma:lch-urysohn}, there exists $\phi \in C_c(X; [0, 1])$ such that $\phi|_{\bracs{f \not\in U}} = 1$. In which case, $\phi f \in C_c(X; E)$ with
+    \[
+    (\phi f - f)(X) = \underbrace{(\phi f - f)(\bracs{f \not\in U})}_{0} + \underbrace{(\phi f - f)(\bracs{f \in U})}_{\in U} \in U
+    \]
+    
+    so $f \in \ol{C_c(X; E)}$. 
+\end{proof}

src/topology/main/index.tex

@@ -21,4 +21,5 @@
 \input{./para.tex}
 \input{./support.tex}
 \input{./lch.tex}
+\input{./c0.tex}
 \input{./baire.tex}