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. ∎

Title	applications of Urysohn’s Lemma to locally compact Hausdorff spaces
Canonical name	ApplicationsOfUrysohnsLemmaToLocallyCompactHausdorffSpaces
Date of creation	2013-03-22 18:33:31
Last modified on	2013-03-22 18:33:31
Owner	azdbacks4234 (14155)
Last modified by	azdbacks4234 (14155)
Numerical id	23
Author	azdbacks4234 (14155)
Entry type	Topic
Classification	msc 54D15
Related topic	UrysohnsLemma
Related topic	TietzeExtensionTheorem
Related topic	VanishAtInfinity
Related topic	SupportOfFunction
Related topic	LocallyCompact
Related topic	T2Space
Related topic	NormalTopologicalSpace