Added the c0 space.

src/topology/functions/uniform.tex

@@ -11,12 +11,41 @@
     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:
     \begin{enumerate}
         \item $C(X; Y) \subset Y^X$ is closed with respect to the uniform topology.
-        \item If $X$ is a uniform space, then $UC(X; Y) \subset Y^X$ is closed with respect to the uniform topology.
+        \item If $X$ is a uniform space, then $UC(X; Y) \subset Y^X$ is closed with respect to the uniform topology. 
     \end{enumerate}
     In particular, if $Y$ is complete, then the above spaces are complete.
 \end{proposition}
@@ -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 +1,30 @@
+\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. 
+        \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}.
+\end{proof}