How do you prove that Z (the set of integers) equipped with the Euclidean metric (induced from real numbers) is complete? How do you prove that $\mathbb{Z}$ (the set of integers) equipped with the Euclidean metric (induced from real numbers) is complete? 
I am having trouble with this question, I don't really know where to start!
 A: Directly: try and show a Cauchy sequence in $\mathbb{Z}$ is eventually constant. 
A little more theoretical: If you know $\mathbb{R}$ is complete, and $\mathbb{Z}$ is closed in $\mathbb{R}$, try and show that closed subsets of complete spaces are complete.
A: Complete means if we start with a cauchy sequence, it actually converges.
So, let $x_n$ be a cauchy sequence in $\mathbb Z$.   Then,  for for any $\epsilon >0$, $\exists N\in \mathbb N$,$\forall n,m\ge N,|x_n -x_m |<\epsilon$. 
So,  lets pick an $\epsilon$.  Say.  $\frac 1 {10}$.   Then after a certain point,  $|x_n-x_m|<\frac 1 {10}$.   So we have $-\frac 1 {10}<x_n - x_m <\frac 1 {10}$,  hence
$x_m - \frac 1 {10}<x_n<x_m+\frac 1 {10}$.   But integers are at least 1 apart,   so the only way for this to be true is for $x_n=x_m$.   Thus,  $\forall n\ge N$,$x_n =x_N$
So your sequence is eventually constant, and thus converges to that constant.
A: Let $(x_n)$ be a Cauchy sequence. Then there exists $k$ such that, for all $m,n\ge k$,
$$
|x_m-x_n|<\frac{1}{2}
$$
which means $x_m=x_n$. Then…
