applications of Urysohn’s Lemma to locally compact Hausdorff spaces

Let $X$ be a locally compact Hausdorff space (LCH space) and $X^{*}$ its one-point compactification. We employ the following notation:

• $C(X)$ denotes the set of continuous complex functions on $X$;

• $C_b(X)$ denotes the set of continuous and bounded complex functions on $X$;

• $C_0(X)$ denotes the set of continuous complex functions on $X$ which vanish at infinity;

• $C_c(X)$ denotes the set of continuous complex functions on $X$ with compact support

Note that we have $C_c(X)\subseteq C_0(X)\subseteq C_b(X)\subseteq C(X)$, and that when we replace $X$ with $X^{*}$ (in general, when $X$ is compact), these four classes of functions coincide.

Now, while Urysohn’s Lemma does not directly apply to $X$ (since $X$ need not in general be normal), it does apply to $X^{*}$, for being compact Hausdorff, $X^{*}$ is necessarily normal. One may therefore indirectly apply Urysohn’s Lemma to $X$ by way of $X^{*}$ to obtain various results asserting the existence of certain continuous functions on $X$ with prescribed properties. The following results and their proofs illustrate this technique and are frequently useful in analysis.

Proposition 1.

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

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

Proof.

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

Theorem 1.

(Urysohn’s Lemma for LCH Spaces)If $K\mathrm{\subseteq }U\mathrm{\subseteq }X$ with $K$ compact and $U$ open, then there exists $f\mathrm{\in }{C}_c\mathit{}\mathrm{(}X\mathrm{)}$ such that $\mathrm{0}\mathrm{\le }f\mathrm{\le }\mathrm{1}$, ${f\mathrm{|}}_K\mathrm{\equiv }\mathrm{1}$, and $\mathrm{supp}\mathit{}f\mathrm{\subseteq }U$.

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 $0\le g\le 1$, ${g|}_K\equiv 1$, and ${g|}_{X^{*}-V}\equiv 0$. Put $f={g|}_X$. Then $f\in C(X)$, $0\le f\le 1$, and ${f|}_K\equiv 1$. Moreover, $f$ vanishes outside $\overline{V}$ because $g$ does, so $\{x\in X:f(x)\ne 0\}\subseteq \overline{V}\subseteq U$; since $\overline{V}$ is compact, and consequently closed, the last inclusion gives $\mathrm{supp}f\subseteq \overline{V}\subseteq U$ and $f\in C_c(X)$.∎

Theorem 2.

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

Corollary 2.

$C_0(X)$ is the uniform closure of $C_c\mathit{}\mathrm{(}X\mathrm{)}$ in $C_b\mathit{}\mathrm{(}X\mathrm{)}$.

Proof.

We first show that $C_0(X)$ is closed in $C_b(X)$. To this end, assume that ${(f_n)}_{n=1}^{\mathrm{\infty }}$ is a uniformly convergent sequence of functions in $C_0(X)$ with limit $f$ and let $ϵ>0$ be given. Select $N\in {\mathbb{Z}}^{+}$ such that , and select a compact subset $K$ of $X$ such that for $x\in X-K$. We then have, for all such $x$,

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 $ϵ>0$, select a compact subset $K$ of $X$ such that for $x\in X-K$. By Theorem 1, there exists $g\in C_c(X)$ with range in $[0,1]$ satisfying ${g|}_K\equiv 1$. The function $h=fg$ is continuous and supported inside $\mathrm{supp}g$, hence lies in $C_c(X)$; moreover, if $x\in K$, then we have $|f(x)-h(x)|=|f(x)-f(x)|=0$, while if $x\notin K$, then

It follows that , hence that $f\in \overline{C_c(X)}$, completing the proof.∎