How to: Cauchy Proof I feel pretty confident on proving if a sequence is Cauchy or not, but I'm not quite sure, I  understand the following proof.
Suppose $X_n \in \mathbb{Z}$ for $n\geq N$. If $\{X_n\}$ is Cauchy, prove $X_n$ is eventually constant (ie. $X_n=a$).
I know a Cauchy sequence converges, so is this asking to prove it converges to $a$?
 A: You need to show that $\exists\ M \in \mathbb{N}$, such that, for some $a \in \mathbb{Z}$, $x_n = a$ for all $n \geq M$.
Use the definition of Cauchy sequence: for every $\epsilon >0, \exists\ K_{\epsilon} \in \mathbb{N}$ such that $|x_i-x_j|<\epsilon$ for all $i,j \geq K_{\epsilon}$. Take $\epsilon = 0.5$ and obtain the result.
A: As mentioned in the comment, it is asking you to prove that for eventually the sequence take the value $a$ rather than just getting arbitrarily closed to $a$.
Hint:
$X_n \in \mathbb{Z}$, what is the smallest positive distance between two distinct integer? Try to get $\epsilon$ that is motivated from that.
Edit:
Choose $\epsilon=\frac12>0$. Since the sequence is cauchy, $\exists N >0$ , $m,n >N \implies |X_m-X_n|<\frac12.$
but $X_m, X_n \in \mathbb{Z}$. Hence $\exists N >0$ , $m,n >N \implies |X_m-X_n|=0 $
That is for all $\exists N>0$, $x_n=X_{N+1}$.
A: Well, ..., yes (because $\mathbb{Z}$ is discrete).  But the language used is actually much, much stronger.  It is asking you to show that there is $N \in \mathbb{N}$ such that $X_N = X_{N+1} = \cdots = a$.  I.e., you are asked to show that the sequence converges (and is subsequently constant) after a finite number of steps.
What happens to your Cauchy sequence when we want $|X_i - X_j| < 1$ for sufficiently large $i,j$?
