find sup of $\frac m{m+n}$ 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.
 A: 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 $m\ge n(k-1)$.
A: Your guess is right $sup(A) = 1 $.
We need to prove that:
(1) $\forall m,n \in \mathbb{N} \quad   \frac{m}{m+n} \leq 1$ (which is obviously true)
(2)$\forall \varepsilon >0 \quad \exists m,n \in \mathbb{N} $ such that $ \frac{m}{m+n}>1-\varepsilon$
So let $0<\varepsilon < 1$ (otherwise you can choose any $m,n \in \mathbb{N})$.
Set $n = 1$.
$$\frac{m}{m+1}>1-\varepsilon \iff$$
$$   m >\frac{1}{\varepsilon}-1$$ 
Now you use Archimedes prinicple and you get $m^{*}$ such that $m^{*} >\frac{1}{\varepsilon}-1 $.Finally you found member from set A to satisfy (2) $m=m^{*} ,n =1$.
A: 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 $(m_{0}-1)/m_{0} \in A$.
