# Sequence with integral values ultimately constant

Dear Stackexchange Community,

I would like to ask how to show that a Cauchy sequence that takes on integral values is ultimately constant.

Being unfamiliar with this, do 'integral' values refer to integer values?

My solution currently is:

Let the sequence $(x_n: n \in \mathbb N)$ be Cauchy.

Then by definition, $\forall \varepsilon > 0, \exists K \in \mathbb N$ such that $\forall n, m > K, |x_n-x_m| < \varepsilon$.

I would assume to prove the statement by contradiction. Assume otherwise, that the sequence takes on more than one value (and thus not constant). However I am finding it tough to establish a contradiction.

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Yes, integral values means integer values. –  copper.hat Mar 6 '13 at 3:19
If $x_n,x_m$ are two integer values, and $|x_n-x_m| < 1$, what does that say about $x_n,x_m$? –  copper.hat Mar 6 '13 at 3:20
What I think: If $x_n$ and $x_m$ are two integer values, then for $|x_n-x_m| < 1$, would it imply that one of those values is not an integer, because difference of two integer values must be an integer? If $|x_n-x_m| = 0$, that would imply that $x_n = x_m$? –  Julian Park Mar 6 '13 at 3:35
The latter. The point is that if the sequence is Cauchy, then for any $\epsilon>0$, there is some $N$ such that $|x_n-x_m| < \epsilon$ for all $n,m \ge N$. So if you choose $\epsilon = 1$, then if $n,m \ge N$, we have $|x_n-x_m|<1$, which means that (since we are dealing with integers) $x_n = x_N$ for all $n \geq N$. –  copper.hat Mar 6 '13 at 3:45

Let $\epsilon=1/2$, can you take it from here?

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Since $x_n$ and $x_m$ are two integer values, then would we have that $|x_n-x_m|$ must also be an integer: for if $|x_n-x_m| < 1/2$, then $x_n$ and $x_m$ would not be integer values (if $|x_n-x_m| = 0$, then $x_n = x_m$). –  Julian Park Mar 6 '13 at 3:39
yes, $x_n-x_m$ is an integer. If its norm is smaller than $1/2$, then there is only one option for what it can be. –  Jonathan Mar 6 '13 at 3:41
That option would be if $x_n = x_m$, and thus is constant? –  Julian Park Mar 6 '13 at 3:43
yes, that's right –  Jonathan Mar 6 '13 at 3:43
I should also mention that it doesn't have to be $1/2$. Any $\epsilon\le 1$ works just as well. –  Jonathan Mar 6 '13 at 3:44