$C_0(X)$ is isometrically isomorphic to a subspace of $C(X_\infty)$ Let $X$ be a locally compact space and let $X_{\infty}$ be the one-point compactification of $X$. How can I prove that $C_0(X)$ is isometrically isomorphic to $\{f:X_{\infty}\to \mathbb K: f \text{ is continuous and }f(\infty)=0\}$, where $C_0(X)=\{f:X\to \mathbb K: f \text{ is continuous and for each } \epsilon>0, \{x\in X:|f(x)|\geq \epsilon\} \text{ is compact}\}\}$. I am not finding any clue. Please suggest something.
 A: By definition, the open sets in $X_{\infty}$ are sets of the form $U$ or $U \cup \{\infty\}$ for $U$ open in $X$, where in the second case we require that $X - U$ be compact.
If $f \in C_0(X)$, extend $f$ to a function $\bar{f}$ on $X_{\infty}$ by setting $f(\infty) = 0$.  Let's check that this extension is still continuous.  Let $V = \{z \in \mathbb{C} : |z-a| < r \}$ be an open ball in $\mathbb{C}$.  We need to show that $\bar{f}^{-1}V$ is open in $X_{\infty}$.  Since $f$ is continuous, $$f^{-1}V = \{ x \in X : |f(x) - a| < r \}$$
is open in $X$.  If $0$ is not in  $V$, then $f^{-1}V = \bar{f}^{-1}V$.  On the other hand, if $0$ is in $V$, then $\bar{f}^{-1}V = f^{-1}V \cup \{\infty\}$, which is open in $X_{\infty}$, because the complement of $f^{-1}V$ in $X$ is
$$\{x \in X : |f(x) - a| \geq r \}$$
which is compact, because it is a closed subset of the compact set $\{x \in X : |f(x)| \geq r - |a|\}$.  This shows that $\bar{f}$ is a continuous function.
Now, the assignment $f \mapsto \bar{f}$ obviously defines a linear map $C_0(X) \rightarrow C(X)$.  It is an isometry, because the supremum (maximium) of the set of values $|f(x)| : x \in X$ is not changed by adding the point $|f(\infty)| = 0$.  
The last thing to check is surjectivity.  If $F \in C(X_{\infty})$, and $F(\infty) = 0$, we are done if we can show that the restriction $f$ of $F$ to $X$ lies in $C_0(X)$.  Since $F$ is continuous, so is $f$.  And for any $\epsilon > 0$, setting $V = \{z \in \mathbb{C} : |z| < \epsilon\}$, we have
$$F^{-1}V = f^{-1}V \cup \{\infty \}$$
Since $X_{\infty}$ is compact, the complement of $F^{-1}V$ in $X_{\infty}$ is a  compact subset of $X_{\infty}$.  Moreover, this complement is contained in $X$, so it is a compact subset of $X$.  This complement is exactly
$$X - f^{-1}V = \{ x \in X : |f(x)| \geq \epsilon \}$$
