**$mathbb{N}$ complete metric space**without wasting too much if your time.

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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 $Nin mathbb{N}$ 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 $lin mathbb{N}$

we want to show that

$left |a_{n}-lright|<epsilon$ $(*)$

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

$left |a_{n}-lright|<left|a_{n}-a_{n_{k}}right|+left|a_{n_{k}}-lright|<epsilon$

My approach is correct ?? because I’m confused and I’m not 100% sure.

## Answer

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 $linmathbb{N}$ such that $a_nto l$.

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?

**$mathbb{N}$ complete metric space**- If you find the proper solution, please don't forgot to share this with your team members.