Prove that Baire space $\omega^\omega$ is completely metrizable? When I tried to prove that Baire space $\omega^\omega$ is completely metrizable, I defined a metric $d$ on $\omega^\omega$ as: If $g,h \in \omega^\omega$ then let $d(g,h)=1/(n+1)$ where $n$ is the smallest element in $\omega$ so that $g(n) \ne h(n)$ is such $n$ exists, and $d(g,h)=0$ otherwise.  
I am stuck trying to prove this metric is complete. Can you help me please? 
Thanks in Advance.

Let $x_n$ be a Cauchy sequence.  Then $d(x_n, x_{n+1}) \rightarrow 0$.  In particular, for every $\epsilon > 0$, there exists an $N$ such that $n > N$ implies $d(x_n, x_{n+1}) < \epsilon$.
Exercise:  translate that statement into one relating $\omega^n$ and $\omega^{n+1}$ (or equivalently, prefixes in finite trees).
Exercise:  what happens when $n \rightarrow \omega$?
A different metric might make this easier.  In fact, I'd say that the only thing keeping this problem from being "trivial" is picking a metric that makes the argument trivial.
