Consider the metric space $\{$ $a_0a_1a_2....$ $:$ $a_i=0,1$ $\}$
with the metric $d(x,y)=min(I: a_i\neq b_i)$ and 0 otherwise.
Edit: I meant to say $d(x,y)=\frac{1}{2^{min(I:a_I\neq b_I)}}$ when $x\neq y$.
Show that the space is complete.
I'm not really sure how to construct a limit, perhaps there's a theorem I must consider? May I have some hints?
On
Let $(x_n)_n$ be a Cauchy sequence in $2^w$. So there exists $N$ such that whenever $m,n\geq N$, we have $d(x_n,x_m)<\frac{1}{2^K}$. This is equivalent to saying that $x_N|i=x_m|i$ are equal for all $i\leq K$ for all $m\geq N$. Define $x\in 2^w$ to be such that $x|i=x_N|i$ for all $i\leq K$. So choosing $K$ to be such that $\frac{1}{2^K}<\epsilon$. whenever $n\geq N$, we have $x_n|i=x|i$ for all $i\leq K$. Thus, $d(x,x_n)\leq\frac{1}{2^K}<\epsilon$.
Note, the following lemma has been used:
Lemma 1: For $x,y \in 2^w$, $d(x,y)<\frac{1}{2^N}$ if and only if $x_i=y_i$ for all $i\leq N$.
Proof: Suppose there exists $i\leq N$ such that $x_i\neq y_i$. Put $K=min(j : x_j\neq y_j)$. Therefore, $d(x,y)=\frac{1}{2^K} \geq \frac{1}{2^i}$ $\geq$ $\frac{1}{2^N}$. Contradiction. The converse holds similarly.
Hint: $d(x,y) <\frac 1 N$ implies $x_i=y_i$ for $1 \leq i \leq N$. In any Cauchy sequence any particular coordinate, say the $k-$th coordinate becomes constant after some stage so the limit of the $k-$th coordinates exists. This gives the limit of the Cauchy sequence.