Proposition 1   If $K\subseteq U\subseteq X$ with $K$ compact and $U$ open, then there exists an open subset $V$ of $X$ with compact closure such that $K\subseteq V\subseteq\overline{V}\subseteq U$ .

Proof. Since $K$ is a compact subset of the Hausdorff space $X^*$ , it is closed, and because $X$ is open in $X^*$ , $U$ is as well. Therefore, by normality, there exists an open subset $V$ of $X^*$ such that $K\subseteq V\subseteq\overline{V}\subseteq U$ (note that the closure of $V$ in $X^*$ coincides with that of $V$ in $X$ , since the former set is contained in $X$ and the latter set is equal to the former intersected with $X$ ). As $\overline{V}$ is closed in $X^*$ , it is compact, and because $V$ is open in $X^*$ and $V\subseteq X$ , $V$ is open in $X$ . Thus $V$ possesses the desired properties.

Corollary 1   For each $x\in X$ and each open subset $U$ of $X$ containing $x$ , there exists an open subset $V$ of $X$ with compact closure such that $x\in V$ and $\overline{V}\subseteq U$ .

Proof. Take $K=\set{x}$ in the preceding proposition.

Theorem 1 (Urysohn's Lemma for LCH Spaces)   If $K\subseteq U\subseteq X$ with $K$ compact and $U$ open, then there exists $f\in C_c(X)$ such that $0\leq f\leq 1$ , $f\vert_K\equiv 1$ , and $\supp f\subseteq U$ .

Proof. By the first Proposition, there exists an open subset $V$ of $X$ with compact closure such that $K\subseteq V\subseteq\overline{V}\subseteq U$ ; since $K$ and $X^*-V$ are disjoint closed subsets of the normal space $X^*$ , Urysohn's Lemma furnishes $g\in C(X^*)$ such that $0\leq g\leq 1$ , $g\vert_K\equiv 1$ , and $g\vert_{X^*-V}\equiv 0$ . Put $f=g\vert_X$ . Then $f\in C(X)$ , $0\leq f\leq 1$ , and $f\vert_K\equiv 1$ . Moreover, $f$ vanishes outside $\overline{V}$ because $g$ does, so $\set{x\in X:f(x)\neq 0}\subseteq\overline{V}\subseteq U$ ; since $\overline{V}$ is compact, and consequently closed, the last inclusion gives $\supp f\subseteq\overline{V}\subseteq U$ and $f\in C_c(X)$ .

Theorem 2 (Tietze Extension Theorem for LCH Spaces)   If $K\subseteq X$ is compact and $f\in C(K)$ is real, then there exists a real $g\in C_c(X)$ extending $f$ .

Corollary 2   $C_0(X)$ is the uniform closure of $C_c(X)$ in $C_b(X)$ .

Proof. We first show that $C_0(X)$ is closed in $C_b(X)$ . To this end, assume that $(f_n)_{n=1}^\infty$ is a uniformly convergent sequence of functions in $C_0(X)$ with limit $f$ and let $\epsilon>0$ be given. Select $N\in\mathbb{Z}^+$ such that $\norm{f-f_N}_\infty<\epsilon/2$ , and select a compact subset $K$ of $X$ such that $\abs{f_N}<\epsilon/2$ for $x\in X-K$ . We then have, for all such $x$ , \begin{equation*} \abs{f(x)}=\abs{f(x)-f_N(x)+f_N(x)}\leq\abs{f(x)-f_N(x)}+\abs{f_N(x)}\leq\norm{f-f_N}_\infty+\abs{f_N(x)}< \epsilon\text{.} \end{equation*}Thus $f$ vanishes at infinity; since the uniform limit of continuous functions is continuous, we obtain $f\in C_0(X)$ , whence $C_0(X)$ is closed. It remains to establish the density of $C_c(X)$ in $C_0(X)$ . Given $f\in C_0(X)$ and $\epsilon>0$ , select a compact subset $K$ of $X$ such that $\abs{f(x)}<\epsilon/2$ for $x\in X-K$ . By Theorem 1, there exists $g\in C_c(X)$ with range in $[0,1]$ satisfying $g\vert_K\equiv 1$ . The function $h=fg$ is continuous and supported inside $\supp g$ , hence lies in $C_c(X)$ ; moreover, if $x\in K$ , then we have $\abs{f(x)-h(x)}=\abs{f(x)-f(x)}=0$ , while if $x\notin K$ , then \begin{equation*} \abs{f(x)-h(x)}=\abs{f(x)-f(x)g(x)}=\abs{f(x)}\abs{1-g(x)}\leq\abs{f(x)}<\dfrac{\epsilon}{2}\text{.} \end{equation*}It follows that $\norm{f-h}_\infty<\epsilon$ , hence that $f\in\overline{C_c(X)}$ , completing the proof.