Is $C([0, \infty))$ a metric space? Is $C([0, \infty))$ = space of real-valued continuous function on $[0, \infty)$   metrizable ? 
 A: It's not a metric space without a metric defined on it. One such example of a metric is 
$$d(f,g) = \sup_{t \ge 0} \frac{|f(t)-g(t)|}{1 + |f(t) - g(t)|}$$
Indeed, let $\rho(a):= a/(1 + a)$ for all $a \ge 0$. Then $d(f,g) = \sup_{t \ge 0} \rho(|f(t)-g(t)|)$. Since $0 \le \rho(a) \le 1$, it follows that $0 \le d(f,g) \le 1$ for all $f$ and $g$. If $d(f,g) = 0$, then $\rho(|f(t)-g(t)|) = 0$ for all $t$, which implies $|f(t) - g(t)| = 0$ for all $t$. Thus $f = g$. The only nontrivial part to prove is the triangle inequality. Note $\rho$ is increasing on $[0,\infty)$ and $\rho(a + b) \le \rho(a) + \rho(b)$ for all $a,b \ge 0$:
$$\rho(a + b) = \frac{a + b}{1 + a + b} = \frac{a}{1 + a + b} + \frac{b}{1 + a + b} \le \frac{a}{1 + a} + \frac{b}{1 + b} = \rho(a) + \rho(b).$$
Therefore, given $f, g, h\in C[0,\infty)$ and $t \ge 0$, then 
\begin{align}\rho(|f(t) - h(t)|) &\le \rho(|f(t) - g(t)| + |g(t) - h(t)|)\\
& \le \rho(|f(t) - g(t)|) + \rho(|g(t) - h(t)|)\\
& \le d(f,g) + d(g,h).
\end{align}
Consequently,
$$d(f,h) \le d(f,g) + d(g,h).$$
A: On any set one can define a metric and make it a metric space. But when it is said that the space $X$ is metrizable, it is assumed that $X$ is a topological space and its metrizability means that there exists a metric on X that induces the same topology as the original topology. So the question makes no sense if $X$ is not equipped with a topology on it.
