Urysohn lemma for locally compact spaces

Question

How can I prove the following theorem?

Let $X$ locally compact, $K \subseteq X$ compact, $O \subseteq X$ open with $K \subseteq O$. Then there exists $f \in C_c(X)$ with $0 \leq f \leq 1$, $f|_K \equiv 1$ and $\operatorname{supp}(f) \subseteq O$.

(Where $C_c(X)=\{f ∈ C(X, \mathbb C) : \operatorname{supp}(f) \text{ compact}\}$).

I think that I would need the topological lemma for locally compact spaces, but I couldn't come up with a suitable proof. Thanks in advance!

Answer 1

Here is one more proof (which is similar to Paul Frost's proof, but not using the one-point compactification).

It is well-known that for a locally compact (Hausdorff) $X$ there exists an open $U \subset X$ such that $C = \overline U$ is compact and $K \subset U \subset C \subset O$.

The sets $K$ and $C \setminus U$ are disjoint closed subsets of $C$. Because $C$ is normal, there exists a continous $g : C \to [0,1]$ such that $g \mid_K \equiv 1$ and $g \mid_{C \setminus U} \equiv 0$. Define $$f : X \to [0,1], f(x) = \begin{cases} g(x) & x \in C \\ 0 & x \in X \setminus U\end{cases}$$

1. This is well-defined because $C \cap (X \setminus U) = C \setminus U $ and $g \mid_{C \setminus U} \equiv 0$.

2. $f$ is continous by the pasting lemma because the sets $C$ and $X \setminus U$ are closed in $X$ and $f \mid_C = g$ and $f \mid_{X \setminus U} = 0$ are continuous.

3. By construction $\operatorname{supp}(f) = \overline{\{ x \in X \mid f(x) \ne 0 \}} \subset \overline U = C \subset O$.

Answer 2

You can see the proof of Urysohn's Lemma (a weaker version) for locally compact spaces on Walter Rudin's Real and Complex Analysis

The problem is that the Urysohn Lemma is valid for Normal Spaces and we are trying to prove something similar, in fact weaker, but for locally compact spaces. Recall that a locally compact space is completely regular, but there are locally compact spaces that are not normal (I am aware of the existence of such spaces, although a counterexample does not come to mind) and hence the full force of the Urysohn's Lemma does not apply.

Still, for what you are looking for, Rudin has it all covered up excellently and I couldn't do it in a more concise way here other than copy and pasting it, so you'd better look it upon pages 35 40 of the aforementioned book.

Answer 3

It is not clear whether you assume that "locally compact" includes "Hausdorff". Let us make this assumption here.

You can reduce the Urysohn lemma for locally compact spaces to the Urysohn Lemma for normal spaces which is a well-known theorem in general topology.

Let $X^* \supset X$ denote the one-point compactification of $X$. This a compact Hausdorff space, hence it is normal. The point added to $X$ will be deoeted by $\infty$.

$K$ and $L = X^* \setminus O$ are disjoint closed subsets of $X^*$. Note that $\infty \in L$. Choose disjoint open neigborhoods $U$ of $K$ and $V$ of $L$ in $X^*$. Then $\infty \in V$, thus $U \subset X$. Since $X^* \setminus V$ is a compact set contained in $X$ and $U \subset X^* \setminus V$, we get $\overline U^X \subset \overline{X^* \setminus V}^X = X^* \setminus V$, thus $\overline U^X$ is compact.

$K$ and $M = X^* \setminus U$ are disjoint closed subsets of $X^*$, thus there exists a continuous $F : X^* \to \mathbb R$ such that $F \mid_K \equiv 1$ and $F \mid_M\equiv 0$.

The restriction $f = F \mid_X : X \to \mathbb R$ has the desired properties:

• Obviously $f \mid_K \equiv 1$.

• The set $s(f) = \{x \in X \mid f(x) \ne 0\}$ is contained in $X^* \setminus M = U$. Thus $\operatorname{supp}(f) \subset \overline U^X$. Hence $\operatorname{supp}(f)$ is compact.

• $\operatorname{supp}(f) \subset \overline U^X \subset X^* \setminus V \subset X^* \setminus L = O$.