# A subset , of the space of all continuous real valued functions on a locally compact Hausdorff space ,“ vanishing at infinity” , with compact closure

Let $X$ be a locally compact Hausdorff space , let $C_0(X,\mathbb R)$ be the space of all real valued continuous functions "vanishing at infinity" ( $f:X \to \mathbb R$ is said to be vanishing at infinity if for every $\epsilon >0$ , $\exists K \subseteq X$ , $K$ compact such that $|f(x)|<\epsilon , \forall x \in X \setminus K$ ) . Equip $C_0(X , \mathbb R)$ with the sup norm and let $\mathcal F \subseteq C_0(X,\mathbb R)$ be set with compact closure ; then how to show that $\mathcal F$ " vanishes uniformly at infinity " i.e. that for every $\epsilon >0$ , there is a compact subset $K$ of $X$ such that $|f(x)|<\epsilon , \forall x \in X \setminus K , \forall f \in \mathcal F$ ?

If $\bar X$ denotes the one-point compactification of $X$ then , I can show that $C_0(X , \mathbb R)$ is isometric to a closed subspace of $C(\bar X , \mathbb R)$ namely $M:=\{f \in C(\bar X , \mathbb R) | f(\infty)=0 \}$ where $\{\infty\}:=\bar X \setminus X$ , hence we can consider $\mathcal F$ as a subset of $M$ itself with $cl_M \mathcal F= cl_{C(\bar X , \mathbb R)} \mathcal F$ since $M$ is closed in $C(\bar X , \mathbb R)$ . Hence I can conclude by Arzela-Ascoli that $\mathcal F$ is pointwise bounded and equicontinuous ; but I don't know how to conclude " uniform vanishing at infinity " . Please help . Thanks in advance

Let $\varepsilon>0$ be given.
Since $\mathcal F$ is a relatively compact subset of the space $\left(C_0(X, \mathbb{C}), \,|| \cdot ||_u \right)$, there exists a finite subset $M \subseteq \text{Cl}(\mathcal F)$ such that
$$\delta(\,f, M) =\underset{1\leq i\leq M}{\inf} \, ||\,f-g_i||_u < \frac{\varepsilon}{2}$$ whenever $f \in \mathcal F$ ( here $M$ doubles as a positive integer, meaning $g_i \in M$ ). Now, for each function $g_i$ $(i=1, \ldots, M)$ from the finite collection $M$ there is a compact set, say $K_i$ such that $|g_i(x)|<\frac{\varepsilon}{2}$ whenever $x \in X \setminus K_i$. To complete the proof from here, I advise you to consider the compact set $$K = \bigcup_{i=1}^M \, K_i \; .$$