Added the c0 space.

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}