Let $(x_n)$ be an integer Cauchy sequence. Prove that $x_n$ is eventually constant. I'm working on the following problem.

Let $(x_n)$ be an integer Cauchy sequence, i.e. $x_n \in \mathbb{Z}$. Prove that $x_n$ is eventually constant, i.e. there exists $N$ so that $x_n = c$ whenever $n \ge N$.

My attempt:
Let $(x_n)$ be an integer Cauchy sequence. By the Cauchy convergence theorem, $x_n \to x$, for some $x$. Given $\epsilon >0$, there exists $N \in \mathbb{N}$ so that
$$
|x_n - x | < \epsilon 
$$
whenever $n \ge N$.
Take $\epsilon < 1$ in the definition so that 
$$
|x_n - x| < 1
$$
But, if the distance between two integers is less than 1, they must be equal. Thus, $x_n = x$ whenever $n \ge N$.
Now, I think my logic is sound except for the huge assumption that $x_n$ converges to an integer.
Would anyone be able to tell me if i'm on the right track? Is there any way to show that $x_n$ converges to an integer?
 A: Let $(x_n)$ be a sequence. If there for any $\epsilon > 0$ exists an $N \in \mathbb{N}$ such that $|x_n - x_m| < \epsilon$ whenever $n, m \geq N$, then $(x_n)$ is a Cauchy sequence.
The assumption that $x_n$ converges is thus unnecessary. In fact, the main benefit of Cauchy sequences is that one does not have to construct a limit point. Convergence is simply an attribute in complete metric spaces. 
So, let $(x_n)$ be a Cauchy sequence in $\mathbb{Z}$. For any $\epsilon > 0$, there exists an $N \in \mathbb{N}$ such that $|x_n - x_m| < \epsilon$ whenever $n, m \geq N$. Especially, let $\epsilon = 1$. Then $|x_n - x_m| < \epsilon = 1$ implies $x_n = x_m$ for all $n, m \geq N$. Hence $(x_n)$ is eventually constant.
A: You are on the right path.
You should not at first assume that $x$ is an integer: it's enough to initially infer that $x_n\to x$ for some real $x$. Then, following the same argument as you have shown, we see that there is some $N$ such that for all $n\geq N$, $|x_n-x|<\frac{1}{3}$. But then, for all $m,n\geq N$:
$$
|x_n-x_m|\leq|x_n-x|+|x-x_m|<\frac{2}{3}\implies|x_n-x_m|=0.
$$
This means that $x_n$ is constant for all $n\geq N$. This constant must in fact be $x$, which in turn implies that $x$ is an integer.
A: I think it's easier just to use the definition of Cauchy:
There must exist an $N$ such that for all $m,n \geq N$ we have
$$ |x_m - x_n | <1$$
Since $x_m$ and $x_n$ are integers, this implies as you have remarked, that $x_m = x_n$ for alle $m,n \geq N$. But this means that sequence is eventually constant, since $x_n = x_N$ for all $n \geq N$.
