Prove that Baire space $\omega^\omega$ is completely metrizable?

Question

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.

Answer 1

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.

Answer 2

The point here is that two functions are close iff they agree on an initial segment, that is, $d(f,g)\le 1/(n+1)$ iff $f(0)=g(0),f(1)=g(1),\dots,f(n-1)=g(n-1)$. Now, if $(f_n)_n$ is a Cauchy sequence, then, for each $n$, there is $N_n$ such that for all $m,k>N_n$ we have $d(f_m,f_k)\le1/(n+1)$. That is, all functions $f_m$ with $m>N_n$ agree on their first $n$ values.

This suggests naturally what the limit of the sequence $(f_n)_n$ should be: Define $f:\omega\to\omega$ simply by setting $f(k)$ to be the common value $f_m(k)$ for all $m$ large enough (say, for all $m>N_{k+1}$). To verify that $f$ is indeed the limit of the $f_n$, note that, by construction, for any $k$, $d(f_m,f)<1/(k+1)$ as long as $m>N_{k+1}$. But this is precisely what $\lim_m f_m=f$ means.

One of the many things that make $\omega^\omega$ with this metric interesting is that, as a topological space, this is just the irrationals. (Of course, the metric is not the Euclidean metric restricted to the irrationals, since the irrationals are clearly not a complete metric space under the standard metric.) A nice proof of this is at the very beginning of Arnie Miller's monograph on descriptive set theory and forcing.