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prove that if $a_n$ is a Cauchy sequence, and the set {${ a_n | n\in \mathbb{N} }$} (which means the set of all values the seuqnece $a_n$ can have) is finite, then there is $N_0$ s.t for each $n>N_0$ the sequence $a_n$ is constant.

can someone give me a hint?

Firas Abd El Gani
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    HINT: the finite values the sequence can take have a minimum distance between any two of them. – Thomas Dec 05 '14 at 19:38
  • I still don't know how to formalize this. – Firas Abd El Gani Dec 05 '14 at 19:43
  • Between my comment and your reply a period of 5 minutes elapsed. If you are asking for a hint (and that's what you did) and get one, I'd expect you to spent more (actually much more) time on this befor asking for the next portion of help. (Sorry, but I ould not keep this to myself). – Thomas Dec 05 '14 at 19:47
  • Its just something that Hit my mind before asking the question, but didn't know how to get from there. but now im sure this is the right direction, I will think more about it. thanks for trying to help. – Firas Abd El Gani Dec 05 '14 at 19:52

1 Answers1

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Since the given sequence is a Cauchy sequence, you have that for every positive number $\epsilon$, there is a positive integer $N$ s.t. for all natural numbers $m, \, n > N$: $$ |a_m - a_n| < \epsilon $$

Take $\epsilon$ to be the minimum distance between two distinct points, which exists because the set of pairs of distinct points is also finite. Now, take $N$ such that when $m, \, n > N$, it is true that $|a_m - a_n| < \frac{\epsilon}{2}$. But this means that $a_m = a_n$ by the way $\epsilon$ is defined.

d125q
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  • @Thomas: Thanks, I misread. I edited my answer now. – d125q Dec 05 '14 at 19:47
  • I didn't get why your last line is true. – Firas Abd El Gani Dec 05 '14 at 19:58
  • @FirasAliAbdelGhani: $\epsilon$ is defined as the minimum distance between two distinct points, right? Well, we take that $a_m$ and $a_n$ are at distance $\frac{\epsilon}{2}$ (we are allowed to do this because the definition says "[...] for every positive number $\epsilon$ [...]"), which is not possible, hence they can't be distinct. – d125q Dec 05 '14 at 20:03
  • But in the question I need to Prove that they are not distinct. how did u prove that in here? – Firas Abd El Gani Dec 05 '14 at 20:24
  • @FirasAliAbdelGhani: It follows simply from the definition of $\epsilon$. Again, $\epsilon$ is the minimum distance between two distinct points. We take points $a_m$ and $a_n$ to be at distance smaller than the minimum distance between two distinct points. This can't be true, so $a_m$ and $a_n$ can't be distinct. – d125q Dec 05 '14 at 20:32
  • sorry for prolonging this conversation more than usual. but I missing the part where all this is connected to the set being finite. can u connect the dots? – Firas Abd El Gani Dec 05 '14 at 20:44
  • @FirasAliAbdelGhani: The set being finite ensures the existence of $\epsilon$, a minimum distance between two distinct points. – d125q Dec 05 '14 at 20:46
  • Let us continue this discussion in chat. – Firas Abd El Gani Dec 06 '14 at 10:29