Completeness of a metric with discrete topology

Question

Let $\mathbb{N}$ be a set of natural numbers. Define a metric $d$ as follows, $$d(m,n)=\left |\frac{m-n}{mn} \right | \quad for \: all \: m,n \in \mathbb{N}.$$ With this metric, every singleton subset of $\mathbb{N}$ is an open set. Hence topology generated by this metric is a discrete topology. Every discrete metric space is a complete metric space. But here $\mathbb{N}$ is not a complete metric space, because $1,2,3, \dots$ is a Cauchy sequence in $(\mathbb{N},d)$, which is not convergent. I don't understand why this happened? The set $\mathbb{N}$ with discrete topology is complete with one metric while it is not complete with another metric generating the same topology.

Answer 1

$\{1,\frac 1 2 ,\frac 1 3,..\}$ has discrete topology w.r.t. the usual metric but this is not complete. A set with the discrete metric is complete but not every metric space with discrete topology is complete.