# Complete Metric Spaces and discrete spaces

$$\textbf{Definition:}$$ 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!

• Isn't $\{ \frac{1}{n}: n\geq 1\}\subset \mathbb R$ a discrete metric space that is not complete? – Zircht Apr 20 '19 at 23:25

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 $$M$$.
• Yes you are right, the proposition is true if $(M,d)$ where $d$ is the discrete metric. – Orested Apr 20 '19 at 23:47
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.