Closure of set of continuous functions with extension to 1-point compactification is $\mathbb{R}^X$ with compact open topology.

Question

Let $X$ be a locally compact Hausdorff space. Let $C_0(X) \subset \mathbb{R}^X$ be the set of continuous functions $f: X \to \mathbb{R}$ for which there exists an extension to the one-point compactification of $X$. What is the easiest way to show that $\overline{C_0(X)} = \mathbb{R}^X$ where $\mathbb{R}^X$ is given the compact open topology?

Answer 1

Let $K_1, \dots, K_n \subset X$ be compact subsets, let $V_1, \dots, V_n \subset \mathbb{R}$ be open subsets. Let$$U = U_{K_1, V_1} \cap \dots \cap U_{K_n, V_n}$$ be the corresponding basis element for the compact-open topology. The one-point compactification of $X$ is compact Hausdorff space, and hence T4. By Urysohn's lemma, for each $i$, there exists a function$$f_i: X \cup \{\infty\} \to [0, 1]$$ satisfying$$f(x) = 1 \text{ for all }x \in K_i,\text{ }f(x) = 0 \text{ for all }x \in K_1 \cup \dots \cup K_{i-1} \cup K_{i+1} \cup \dots \cup K_n \cup \{\infty\}.$$ For each $V_i$, choose a point $p_i \in V_i$. The linear combination, $$f = \sum_{i=1}^n p_i f_i \in C_0(X) \cap U,$$is a function which extends to the one-point compactification and is an element of $U$. Therefore, $C_0(X)$ is dense in $\mathbb{R}^X$.