# $\mathbb{N}$- a complete metric space with $d(x,y)=|x-y|$

$\mathbb{N}$- a complete metric space with $d(x,y)=|x-y|.$

This seems quite intuitively correct, but I do not know how to prove this formally, does anyone know how they would go about this?

• Can you define completeness? Can you describe what a Cauchy sequence looks like in this space? We can't help you if you don't help us understand your difficulty. – Milo Brandt Aug 1 '15 at 19:23
• The metric $\delta(x,y)=1$ if $x\ne y$ and $=0$ when $x=y$ generates the same topology. It has the name discrete where every point is an open set. – A.Γ. Aug 1 '15 at 19:35
• @A.G.: Being completely is not topologically invariant. $(0,1)$ can be given a complete metric which generates the same topology as generated by the usual metric (which is incomplete). More severely, the irrational numbers can be given a complete metric which generates the same topology as the usual metric generates, and there it is very far from being complete. – Asaf Karagila Aug 1 '15 at 19:37
• @AsafKaragila Thanks for the comment. I feel that I have to dig more in this topic. – A.Γ. Aug 1 '15 at 19:57

If $x_n$ is a Cauchy sequence, pick $\varepsilon=\frac12$, then for some $N$, it holds that every $n,m>N$ satisfy $|x_n-x_m|<\frac12$.
Assume we know that $(\mathbb{R},d(x,y))$ is a complete metric space, then the set of natural numbers $\mathbb{N}$ is a closed subset of $\mathbb{R}$, so it must hold that $(\mathbb{N},d(x,y))$ is also a complete metric space with respect to the same metric since closed subsets of complete spaces are complete too. See also here.
If $\{x_n\}_n$ is Cauchy in this space and given $1>\epsilon>0$, if we have $|x_m-x_n|<\epsilon$ then $x_n=x_m$, in particular $x_n$ converge and your limit is a natural number.