Complete regularity of locally compact Hausdorff spaces with functions vanishing at infinity?

Question

Let $X$ be a locally compact Hausdorff space ; then I know that $X$ is completely regular i.e. for every closed set $F$ in $X$ and $y\in X\setminus F$ , there is a continuous function $f: X\to [0,1]$ such that $f(y)=1$ and $f(F)=\{0\}$ . My question is , can we find such a continuous function in $C_0(X)$ i.e for every closed set $F$ in $X$ and $y\in X\setminus F$ , does there exist a continuous function $f: X\to [0,1]$ such that $f(y)=1$ , $f(F)=\{0\}$ and for every $\epsilon >0$ , there exist a compact $K \subseteq X$ such that $|f(x)|<\epsilon , \forall x \in X\setminus K$ ?

Answer 1

Yes. In Aleksandrov compactification $\alpha X=X\cup\{\alpha\}$ of $X$ there exists a continuous function $\bar f: X\to [0,1]$ such that $\bar f(y)=1$ and $\bar f(F\cup\{\alpha\})=\{0\}$. Put $f=\bar f|_X$.

Answer 2

Use the results that a compact subspace of $X$ is a normal space (because a compact Hausdorff space is normal) and that disjoint closed subsets of a normal space are functionally separated.

Notation: The closure bar denotes closure in $X.$

Let $U_1,U_2$ be open subsets of $X$ such that $$y\in U_2\subset \bar U_2 \subset U_1\subset \bar U_1\subset X \setminus F$$ and such that $\bar U_1$ is compact.

Now $C=\bar U_1$ is a normal space, and $\bar U_2,\; (\bar U_1 \setminus U_1)$ are disjoint closed subsets of $C,$ so take a continuous $f:C\to [0,1]$ with $f(\bar U_2)=\{1\}$ and $f(\bar U_1 \setminus U_1)\subset \{0\}.$

Extend the domain of $f$ to $X$ by letting $f(x)=0$ for $x\in X \setminus \bar U_1.$ (Check that $f:X\to [0,1]$ is continuous.)

Let $K=f^{-1}[0,\epsilon].$ Then $y\in K=\bar K\subset \bar U_1.$ Recall that $\bar U_1$ is compact and disjoint from $F$, so $K=\bar K$ is compact and disjoint from $F$. And (of course) $f(y)=1$ while $$f(F)\subset f(X\setminus \bar U_1)\cup f(\bar U_1\setminus U_1)\subset \{0\}.$$ And $f$ is $0$ except on the compact set $\bar U_1$ so $f\in C_0(X).$