applications of Urysohn’s Lemma to locally compact Hausdorff spaces
Let $X$ be a locally compact Hausdorff space^{} (LCH space) and ${X}^{*}$ its onepoint compactification. We employ the following notation:

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$C(X)$ denotes the set of continuous^{} complex functions on $X$;

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${C}_{b}(X)$ denotes the set of continuous and bounded^{} complex functions on $X$;

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${C}_{0}(X)$ denotes the set of continuous complex functions on $X$ which vanish at infinity;

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${C}_{c}(X)$ 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 $\u03f5>0$ be given. Select $N\in {\mathbb{Z}}^{+}$ such that $$, and select a compact subset $K$ of $X$ such that $$ for $x\in XK$. We then have, for all such $x$,
$$ 
Thus $f$ vanishes at infinity; since the uniform limit of continuous functions is continuous, we obtain $f\in {C}_{0}(X)$, whence ${C}_{0}(X)$ is closed. It remains to establish the density of ${C}_{c}(X)$ in ${C}_{0}(X)$. Given $f\in {C}_{0}(X)$ and $\u03f5>0$, select a compact subset $K$ of $X$ such that $$ for $x\in XK$. 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  20130322 18:33:31 
Last modified on  20130322 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 