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$\textbf{Definition:}$ Let $(M,d)$ a metric space. A point $x \in M$ is a isolated point of $M$ if exists $r>0$ such that : $$ B(a,r)=\{a\} $$ A metric space $M$ is called discrete if every point of $M$ is a isolated point.

With these definitions we have the following proposition :

$\textbf{Proposition:}$ A discrete metric space $M$ is complete.

If we consider $M$ with the discrete metric : $d(x,y)=0,$ if $x=y$ and $d(x,y)=1,$ if $x\neq y$. Is obviously that every Cauchy sequence becomes constant from a certain index, so is convergente.

In many books they define a discrete metric space M as the set M provided with the discrete metric.

How could I prove the result, with the definition I give above?. Thanks!

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    $\begingroup$ Isn't $\{ \frac{1}{n}: n\geq 1\}\subset \mathbb R$ a discrete metric space that is not complete? $\endgroup$ – Zircht Apr 20 '19 at 23:25
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The proposition is not true: the set $M=\{ \frac{1}{n}: n\geq 1\}$ as a subspace of the real numbers is a discrete metric space, but it is not complete since the sequence $(\frac{1}{n})_n$ is Cauchy but it doesn't converge to a point in $M$.

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  • $\begingroup$ Yes you are right, the proposition is true if $(M,d)$ where $d$ is the discrete metric. $\endgroup$ – Orested Apr 20 '19 at 23:47
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The proposition (and the argument you gave for it) become true once you assume that there is some uniform $r>0$ such that $d(x,y)>r$ for all distinct points $x$ and $y$ in your space. Without this assumption, the proposition is not true, as pointed out.

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