Let $X$ be a locally compact Hausdorff space. Let $K \subset X$ be compact and $U \subset X$ be open with $K \subset U$. Then there exists a continuous function $f: X \to [0, 1]$ such that
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\begin{align*}
f \equiv 1 \text{ on } K, \quad \operatorname{supp}(f) \subset U, \quad \text{and} \quad \operatorname{supp}(f) \text{ is compact}.
\end{align*}