I am doing problems in my instructor's notes, which is the following
1) If $X$ is a compact metric space, then $C(X)$ is separable.
2) If $X$ is a locally compact $\sigma$-compact metric space,
show that $C_0(X)$, the space of functions vanishing at infinity, is separable. Hint: One approach is to use
the fact that the 1-point compactification of $X$ is metrizable.
I did the first part by constructing a dense countable sequence of functions. I am stuck at the second one. So my guess is this: Let $Y$ be the 1-point compactification of $X$. Then $C(Y)$ is separable.
I believe there is some relations between $C_0(X)$ and $C(Y)$. I looked up that a subspace of separable metric space is separable, but is it true that $C_0(X) \subseteq C(Y)$ and if so, how do we prove it?
One more question, where do we use the locally compact $\sigma-$compact hypothesis? Which is required to prove the fact that $1-$point compactification of $X$ is metrizable?
Best Answer
$X$ being locally compact and $\sigma$-compact metric implies that the one-point compactification $Y$ of $X$ is metrizable. Note that local compactness is needed to form the one-point compactification and that for $Y$ being metrizabale we need $\sigma$-compactness (since $\infty$ must have a countable basis of open neigborhoods; their complements are compact).
Note that all functions in $C_0(X)$ are bounded, thus $C_0(X)$ can be endowed with the $\sup$-norm.
Now let $i : X \to Y$ denote the inclusion map. Then $$i^* : C(Y) \to C(X), i^*(f) = f \circ i = f \mid_ X$$ is a well-defined (linear) function.
If $C'(Y) = \{ f \in C(Y) \mid f(\infty) = 0 \}$, then $i^*(C'(Y)) = C_0(X)$. To see this, note that if $f \in C'(Y)$, then for each $\varepsilon > 0$ there exist an open neighborhood $U$ of $\infty$ in $Y$ such that $\lvert f(y) \rvert < \varepsilon$ for all $y \in U$. But $U = Y \setminus C$ for some compact $C \subset X$ which implies that $i^*(f)$ vanishes at infinity. Conversely, if $g \in C_0(X)$, define $e(g) : Y \to \mathbb R$ by $e(g)(x) = g(x)$ for $x \in X$ and $e(g)(\infty) = 0$. Continuity of $e(g)$ in all $x \in X$ is obvious because $X$ is open in $Y$. To check continuity in $\infty$, note that for each $\varepsilon > 0$ there exist a comapct $C \subset X$ such that $\lvert g(x) \rvert < \varepsilon$ for all $x \in X \setminus C$. Hence $\lvert e(g)(y) \rvert < \varepsilon$ for all $y \in Y \setminus C$.
Clearly, if $f, f' \in C'(Y)$ such that $i^*(f) = i^*(f')$, then $f = f'$. Hence we obtain a (linear) bijection $$i^* : C'(Y) \to C_0(X) .$$ By construction it is an isometry. Since $C'(Y)$ is separable (as a subset of separable space), also $C_0(X)$ is separable.