$\mathbb{N}$ complete metric space I'm trying to prove that $(\mathbb{N},d)$ is a complete space where $d=\left | m-n \right |$.
So I define $a_{n}:=n$ , if it's cauchy we know that $\forall \epsilon$>0 there exist $N\in \mathbb{N}$ s.t for any $m,n>N$
$\left |a_{n}-a_{m}\right|<\epsilon $
In order to show that $\mathbb{N}$ is complete we have to show that $a_{n}$ is convergent in $\mathbb{N}$.
Suppose that $l\in \mathbb{N}$
we want to show that 
$\left |a_{n}-l\right|<\epsilon$ $(*)$
We know that $a_{n}$ is cauchy so it's bounded and it has a convergent subsequence $a_{n_{k}}$ which is convergent and lets say that $a_{n_{k}}\rightarrow l$
From $(*)$ we have 
$\left |a_{n}-l\right|<\left|a_{n}-a_{n_{k}}\right|+\left|a_{n_{k}}-l\right|<\epsilon$
My approach is correct ?? because I'm confused and I'm not 100% sure.
 A: To show a metric space is complete, you have to show that every Cauchy sequence converges to some limit point. So I'm going to take issue with "I define $a_n := n$." Instead you should let $a_n$ be an arbitrary Cauchy sequence.
I'm also going to take issue with "[$a_n$] has a convergent subsequence $a_{n_k}$." We don't know that $a_n$ has such a convergent subsequence without doing a bit of extra legwork --in fact, any Cauchy sequence with a convergent subsequence must converge! (Exercise: Prove this.)

So how to fix the issues? You need to start with a Cauchy sequence $a_n$, which you know has for every $\epsilon > 0$ there exists some $N$ so for $i,j > N$ we have $|a_i - a_j|<\epsilon$. Given this sequence, you need to find some $l\in\mathbb{N}$ such that $a_n\to l$.
So we need to understand $\mathbb{N}$ well enough that we can guess what $l$ should be. Say we have such a sequence of positive integers. What do you know about the distance between integers? If two integers are $\epsilon$ away from each other, what can we say about them?
