How to find the form of Cauchy sequences in this metric let S be a set consisting of all sequences of positive integers, and if $A = (a_i)$ and $B = (b_i)$ are elements of $S$.
$  d(A,B) =
\begin{cases}
\frac{1}{n},  & \text{if $a_i = b_i$ for $i < n$ and $a_n \neq b_n$} \\
0, & \text{if $a_i = b_i$ for all $i$}
\end{cases}$
What are the cauchy sequences in this metric? Is $S$ a complete spaces in terms of this metric?
 A: The Cauchy sequences $A=(a^i)$ (of sequences, $a^i=(a^i_j)$) are precisely those that satisfy the following condition:
For all $n\in\mathbb{N}$ there exists some $N\in\mathbb{N}$ so that if $i\geq N$ then $a^i$, $a^N$ agree on the first $n$ terms: $a^i_1=a^N_1,\ldots,a^i_n=a^N_n$.
Intuitively, for sufficiently large $i$, if $a^{i-1}$ and $a^i$ agree on the first n terms, then $a^i$ and $a^{i+1}$ will agree on the first $m$ terms for some $m\geq n$.
To see this, let $A=(a^i)$ be a Cauchy sequence (of sequences, $a^i=(a^i_j)$) wrt this metric, and choose $n\in \mathbb{N}$ and $0<\epsilon_n <\frac{1}{n}$. By the definition of a Cauchy sequence, there exists some $N\in\mathbb{N}$ so that if $i,j\geq N$ one has either $d(a^i, a^j) = \frac{1}{m} <\epsilon_n < \frac{1}{n}$ for $m\in\mathbb{N}$, or $d(a^i, a^j) =0$. In the first case $n\leq m - 1$ and by definition of the metric the sequences $a^i$, $a^j$ agree on the first $m-1$ terms. In either case then, the sequences agree on the first $n$ terms.
Conversely, it is easily verified that any such sequence satisfying the above property is Cauchy.
