Proposition 4.21.2. Let $X$ be a topological space and $E$ be a TVS over $K \in \RC$, then:
$C_{0}(X; E) \subset BC(X; E)$.
$C_{0}(X; E)$ is a closed subspace of $BC(X; E)$ with respect to the uniform topology. In particular, if $E$ is complete, then so is $C_{0}(X; E)$.
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.
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
Thus $\bracs{f \not\in U}$ is a closed subset of $\bracs{g \not\in V}$, which is compact by Proposition 4.16.2.
If $E$ is complete, then $BC(X; E)$ is complete by Definition 8.11.3. Since $C_{0}(X; E)$ is a closed subspace, it is complete by Proposition 5.5.3.
(3): Let $f \in C_{0}(X; E)$ and $U \in \cn_{E}^{o}(0)$ be balanced. By Urysohn’s lemma, 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
so $f \in \ol{C_c(X; E)}$.$\square$