Question: I want to prove $\sup(A)=1$ where $A=\frac{m}{m+n}$ where $m,n$ are Natural numbers.  

1- Every element of $A$ is less than $1$ ( $1$ is min ) so it is bounded above.  

2- let $\epsilon>0$ and consider $ 1-\epsilon$.  

Let $a=\frac{M}{M+N}$ I do now know how to use archemidean pronciple and prove $1-\epsilon$ is not an upper bound.  

I have hints and I have done few steps. I need a clean full proof to learn the whole solution please.

Answer: 
So you want to prove that for any $\epsilon > 0$ that there are natural numbers $m,n$ so that

$1-\epsilon < \frac m{m+n} < 1$

Well.... just tool about....

$1-\epsilon < \frac {m}{m+n}=\frac {m+n}{m+n} -\frac n{m+n}=1-\frac n{m+n} < 1 \iff$

$-\epsilon < -\frac n{m+n} < 0 \iff$

$0 < \frac n{m+n} < \epsilon$. 

......

Need more?

......

Well 

$\iff \frac 1{\epsilon} > \frac {m+n}n= \frac mn + 1$.

Need more?

We can find so $m,n$ if we let $n = 1$ and $m > \frac 1{\epsilon} -1$.

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Different tooling around:

For any $\epsilon > 0$ there is an $k\in \mathbb N$ so that $\frac 1k < \epsilon$. (Just let $k > \frac 1 {\epsilon}$).

So we want:

$1 -\epsilon < 1-\frac 1k=\frac{k-1}{k} < \frac m{m+n} < 1$.

Well, that requires that $(m+n)(k-1) < km$.

or $mk +nk -m -n <km$ or

$nk < m+n$.

Let $n$ be anything and $m\ge n(k-1)$.
Let $\epsilon > 0$ such that $1-\epsilon > 0$. Then there exists a natural number $m\_{0}>1$ such that $1/m\_{0} < \epsilon$, so that $-1/m\_{0} > -\epsilon$. Thus, we have $$1 - \frac{1}{m\_{0}} > 1 - \epsilon \Rightarrow \frac{m\_{0}-1}{m\_{0}} > 1 - \epsilon.$$ Take $m = m\_{0}-1$ and $n=1$, so that $(m\_{0}-1)/m\_{0} \in A$.