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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.

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$\{1,\frac 1 2 ,\frac 1 3,..\}$ 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.

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  • $\begingroup$ Thanks. So, these are counterexamples for the statement "every metric space with discrete topology is complete". $\endgroup$
    – ganesh
    Apr 20, 2022 at 6:05
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    $\begingroup$ @ganesh Completeness is a 'metric property'. Topology alone does not give completeness. $\endgroup$ Apr 20, 2022 at 6:06

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