# How to prove that $(\mathbb{Z}, d)$ with $d(m,n)=\vert m -n \vert$ is complete

I know that the metric space $(X,d)$ is called complete if each Cauchy sequence is convergent, but I don't now how to prove the following:

$(\mathbb{Z}, d)$ with $d(m,n)=\vert m -n \vert$ is complete.

• I'm a little confused. You have a hypothesis with no conclusion; do you just want to prove that $(\mathbb{Z},d)$ with $d(m,n)=|m-n|$ is complete?
– mdp
Commented Nov 8, 2013 at 16:38
• yes I did to prove that Commented Nov 8, 2013 at 16:43
• Every discrete space is metrizable and complete under any metric giving the discrete topology. Commented Nov 8, 2013 at 16:46

If $(x_n)_n$ is a Cauchy sequences in $(\Bbb Z,d)$, then take $\epsilon<1$. There is an $m\in\Bbb N$ such that $|x_k-x_l|<ϵ$ for $k,l>m$. So how can $x_k$ and $x_l$ be related?

On the other hand, if you know that closed subspaces of complete spaces are complete, and that $(\Bbb R,d)$ is complete, then it suffices to show the closedness of $\Bbb Z$ in $\Bbb R$.

• Thank you for your reply but your answer if you can give me the solution and deteli be complete, thank you Commented Nov 8, 2013 at 16:45
• @MadritZhaku: No, I'm pretty sure you can finish the argument from here. Remember that $x_k$ and $x_l$ are integers. So what can you say if their distance is less than $1$? Commented Nov 8, 2013 at 16:51
• @StefanH doesn't this imply that Z is not complete? Commented Nov 8, 2013 at 16:58
• @TylerHilton: Why? Commented Nov 8, 2013 at 16:59
• @StefanH because elements are integers and so the min distance is 1. You say choose $\epsilon < 1$, but no two integers satisfy that. Commented Nov 9, 2013 at 3:49

Big hint: Show that every Cauchy sequence in $\mathbb{Z}$ is eventually constant (take $\epsilon <1$).

• Thank you for your reply but your answer if you can give me the solution and deteli be complete, thank you Commented Nov 8, 2013 at 16:46
• I think the hint I've given is sufficiently complete for you to solve the problem on your own. Commented Nov 8, 2013 at 16:49