space of continuous function on real line, completeness and separability Let $C[0,\infty)$ be the space of all continuous real valued functions on $[0,\infty)$ with metric:
$\rho (\omega_{1},\omega_{2})=\sum_{n=1}^{\infty}(1/2^{n})\max_{0\leq t\leq n}(|\omega_{1}(t)-\omega_{2}(t)|\wedge 1)$.
Show that under $\rho$, the space is complete and separable.
Any reference is okay.
 A: Suppose $w_k$ is Cauchy. Note that
$\max_{t \in [0,l]} (1,|w_n(t) -w_m(t)|) \le 2^l \rho(w_n,w_m)$, hence since
$C[0,l]$ is complete (and separable, but we're not there yet), we see that $w_n \to w \in C[0,l]$. Since $l$ is arbitrary, this defines a continuous $w$ on all of $[0,\infty)$. Furthermore, the convergence is uniform on any compact interval.
Let $\epsilon>0$ and choose $N$ such that $\sum_{l> N} {1 \over 2^l} < {\epsilon \over 2}$. Now choose $N' \ge N$ such that if $n \ge N'$, we have $\max_{[0,N]}|w(t)-w_n(t)| < {\epsilon \over N}$. Then if $n \ge N'$, we have $d(w,w_n) < N \max_{[0,N]}|w(t)-w_n(t)| + {\epsilon \over 2} = \epsilon$. Hence the space is complete.
The same idea shows that the space is separable. By choosing polynomials with rational coefficients, we see that $C[0,l]$ is separable. Let $w$ be an element of $C[0,\infty)$ and let $\epsilon>0$. Choose $N$ as above, and choose a polynomial $q$ with rational coefficients so that $\max_{[0,N]}|w(t)-q(t)| < {\epsilon \over N}$, then we have $d(w,q) < \epsilon$, and so the space is separable.
