$C_0(X)$ is isometrically isomorphic to a subspace of $C(X_\infty)$

Question

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.

Answer 1

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

$$X - f^{-1}V = \{ x \in X : |f(x)| \geq \epsilon \}$$