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.
Thanks for your answers.
$\begingroup$
$\endgroup$
3
asked Nov 8, 2013 at 16:34
-
$\begingroup$ 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? $\endgroup$– mdpCommented Nov 8, 2013 at 16:38
-
$\begingroup$ yes I did to prove that $\endgroup$– Madrit ZhakuCommented Nov 8, 2013 at 16:43
-
$\begingroup$ Every discrete space is metrizable and complete under any metric giving the discrete topology. $\endgroup$– egregCommented Nov 8, 2013 at 16:46
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
8
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 $\Bbb R$.
answered Nov 8, 2013 at 16:39
-
$\begingroup$ Thank you for your reply but your answer if you can give me the solution and deteli be complete, thank you $\endgroup$ Commented Nov 8, 2013 at 16:45
-
$\begingroup$ @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 $1$? $\endgroup$ Commented Nov 8, 2013 at 16:51
-
$\begingroup$ @StefanH doesn't this imply that Z is not complete? $\endgroup$ Commented Nov 8, 2013 at 16:58
-
1
-
$\begingroup$ @StefanH because elements are integers and so the min distance is 1. You say choose $\epsilon < 1$, but no two integers satisfy that. $\endgroup$ Commented Nov 9, 2013 at 3:49
$\begingroup$
$\endgroup$
2
Big hint: Show that every Cauchy sequence in $\mathbb{Z}$ is eventually constant (take $\epsilon <1$).
-
$\begingroup$ Thank you for your reply but your answer if you can give me the solution and deteli be complete, thank you $\endgroup$ Commented Nov 8, 2013 at 16:46
-
2$\begingroup$ I think the hint I've given is sufficiently complete for you to solve the problem on your own. $\endgroup$– Dan RustCommented Nov 8, 2013 at 16:49