Space of functions vanishing at infinity

Remarks

Example: $f(x) = e^{-x^2}$ on $\R$. For $0 < \varepsilon \le 1$, the set $\set{\abs{f} \ge \varepsilon}$ is the closed bounded interval $\abs{x} \le \sqrt{\log(1/\varepsilon)}$, which is compact. For $\varepsilon > 1$ it is empty.

Non-example: the constant function $1$ on $\R$. For $\varepsilon = 1/2$, the set $\set{\abs{f} \ge 1/2}$ is all of $\R$, which is not compact.

Each $f \in C_0(X)$ is bounded: $\abs{f} < 1$ off the compact set $\set{\abs{f} \ge 1}$, and $\abs{f}$ attains a maximum on that set if it is nonempty. So $C_0(X)$ is a subspace of the space of bounded continuous functions, and it is closed there, so it is a Banach space. If $X$ is compact, then $C_0(X) = C(X)$. For $X = \N$ with the discrete topology, $C_0(\N)$ is the sequence space $c_0$. The compactly supported continuous functions are dense in $C_0(X)$.