Updated proof for C0 space.

src/topology/main/c0.tex

@@ -13,7 +13,7 @@
     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 $C_0(X; E)$ 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}
@@ -26,5 +26,15 @@
     \[
     f(\bracs{g \in V}) \subset (f - g)(X) + g(\bracs{g \in V}) \subset V + V = U
     \]
+
     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}