Proving that the metric space of all sequences of positive integers is complete Consider the set of all sequences of positive integers with the following metric:given $x=(n_j)$, $y=(m_j)$
$$d(x,y)= 1/\inf\{j: n_j \ne m_j\}$$ if $x\ne y$ and $0$ otherwise. 
I want to show that it is complete, i.e. that all cauchy sequences in this space converge to a point in the space. 
Let $x_n$ cauchy in this space, then $x_n$ is a sequence of sequences. I want to find the sequence of positive integers to which it converges. I'm not too sure how to approach this problem. I tried the following sequence: $x = \{x_{ii}\}$ where for each i, $x_{ii}$ is the ith element of the sequence $x_i$. However I was unable to show that the cauchy sequence converges to this sequence. Am I on the right track, if not to what sequence should I be proving the convergence of the cauchy sequence to?
 A: It helps to just think about the behavior of a Cauchy sequence for a little while. 
Consider a Cauchy sequence $x_n$. Let's fix our notation, $x_n = (x_{n,i})_{i=1}^\infty$. 
For any two $m,n \ge 1$, the distance $d(x_m,x_n)$ takes values in the set $\{1,\frac{1}{2},\frac{1}{3},\ldots\}$. 
Thinking about the Cauchy criterion, it seems interesting to pick an integer $K \ge 1$ and consider $\epsilon < \frac{1}{K}$. The Cauchy criterion guarantees the existence an integer $M$ so that if $m,n \ge M$ then $d(x_m,x_n) < \epsilon$, and from that inequality it follows that the two sequences $x_m$ and $x_n$ have the same entries for $i=1,...,K$: $x_{m,1}=x_{n,1}$; $x_{m,2}=x_{n,2}$, and so on to $x_{m,K}=x_{n,K}$.
From this one can conclude: 
Step 1. The first entry $x_{m,1}$ becomes constant for sufficiently large $m$, say $m \ge M_1$. Let that entry be denoted $a_1$.
Step 2. The second entry $x_{m,2}$ becomes constant for sufficiently large $m$, say $m \ge M_2 \ge M_1$. Let that entry be denoted $a_2$.
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Step $k$. The $k^{\text{th}}$ entry $x_{m,k}$ becomes constant for sufficiently large $m$, say $m \ge M_k \ge M_{k-1}$. Let that entry be denoted $a_k$.
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Continuing by induction, we obtain a sequence $a = (a_1,a_2,\ldots)$, which looks to me like a very good guess for the limit of the sequence $x_n$.
