How do I show that a sequence $(x_n)$ in metric space $(S,\rho)$ is Cauchy if and only if $\lim_{n\rightarrow\infty}\sup_{k\geq n}\rho(x_n,x_k)=0$?
This is exercise 3.2.1 in Economic Dynamics by John Stachurski.
The definition of a Cauchy sequence I'm trying to use is that $(x_n)$ is Cauchy if, for all $\epsilon > 0$, there exists an $N$ such that $\rho(x_j,x_k)<\epsilon$ for all $j,k\geq N$.
There is an attempted answer here but I don't follow the logic of the first half of the proof and it is corrected in a non-transparent way in the comments.
I think this is a case where writing the definitions explicitly can help a lot:
We want to prove that a sequence $(x_n)$ is a Cauchy sequence $\iff$ it is satisfies the condition you stated.
So, a Cauchy sequence means:
$$(1)\;\;\forall \epsilon>0 \exists N. \forall j,k>N .p(x_j,x_k)\leq\epsilon$$
And the condition says that :
$$(2)\;\;\forall \epsilon>o \exists n_0. \forall n>n_0. |\sup_{k\geq{n}}p(x_n,x_k)|\leq\epsilon$$
So if we assume (1), then we can choose for each $\epsilon$ an $N$, then in particular if we set $n_0=N$ then we know that for every $j>N$ then $p(x_j,x_N)\leq\epsilon$, so it is also true that $$|\sup_{k\geq{n}}p(x_n,x_k)|\leq\epsilon.\forall n>N$$
The same way by assuming (2), we know the existence of such $n_0$ (for every $\epsilon$) such that the $sup$ of the distance between each elements after this $n_0$ is small enough. Do you see what this means?
Exactly, this means that in particular this is true for any two elements after this $n_0$ and so we get (1).
And this completes the proof.