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.
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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 Nov 08 '13 at 16:38
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yes I did to prove that – Madrit Zhaku Nov 08 '13 at 16:43
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Every discrete space is metrizable and complete under any metric giving the discrete topology. – egreg Nov 08 '13 at 16:46
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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 $\Bbb R$.
Stefan Hamcke
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Thank you for your reply but your answer if you can give me the solution and deteli be complete, thank you – Madrit Zhaku Nov 08 '13 at 16:45
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@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 $1$? – Stefan Hamcke Nov 08 '13 at 16:51
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@StefanH because elements are integers and so the min distance is 1. You say choose $\epsilon < 1$, but no two integers satisfy that. – Tyler Hilton Nov 09 '13 at 03:49
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@TylerHilton: The question is: What are the Cauchy sequences? And once we know what the Cauchy sequences look like, we can check if they are convergent. We know in a Cauchy sequence there must be an $m$ such that $|x_k-x_l|<1$ for $k,l>m$. We can say something about $x_k$ and $x_l$, and with that I don't mean "They don't exist." – Stefan Hamcke Nov 09 '13 at 15:34
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Big hint: Show that every Cauchy sequence in $\mathbb{Z}$ is eventually constant (take $\epsilon <1$).
Dan Rust
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Thank you for your reply but your answer if you can give me the solution and deteli be complete, thank you – Madrit Zhaku Nov 08 '13 at 16:46
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2I think the hint I've given is sufficiently complete for you to solve the problem on your own. – Dan Rust Nov 08 '13 at 16:49