How to prove the supremum of this set is 1 Let $B = \{n/(n+m) : m, n \in \mathbb{N}\}$, find the supremum and prove it is indeed the supremum.
I claim that $\sup(B) = 1$ and I prove it using the following definition:

So to prove $(i)$, note that $m/n > 0 \implies m/n+1 > 1 \implies 1/(m/n+1) = n/(n+m) < 1$, so $1$ is an upper bound for the set $B$.
I am not exactly sure how to prove $(ii)$, any help would be appreciated.
EDIT: with regards to @Daniel Fischer's comment, I think the following should complete the proof.
So we need to show that if $b<1$, then $b$ is not an upper bound for $B$. That is, there is some element $n/(n+m) > b$.
So fix $m$ and choose $n$ (which exists according to the Archimedean Property) such that $1/n < 1/m\left(1/b-1 \right) \implies m/n < 1/b-1 \implies 1/(m/n+1) >b \implies n/(n+m) > b$ as required.
 A: 
Proposition Let $A\subseteq \Bbb R$. Then $s=\sup A$ if and only if $s$ is an upperbound of $A$ and there exists a sequence of elements in $A$, $a_1<a_2<\cdots$ with $a_n\to s$.

Proof 
$(\Rightarrow)$ Suppose $s=\sup A$. Then for every $\varepsilon >0$ there exists $a\in A$ such that $s-a<\varepsilon$.  For $\varepsilon_1 =1$ there is $a_1$ with $s-a_1<\varepsilon$. Let $\varepsilon_2=\min(s-a_1,1/2)$. Then there is $a_2$ with $s-a_2<\varepsilon_2$, hence $s-a_2<1/2$ and $s-a_2<s-a_1$, so $a_2>a_1$. Suppose we have constructed $a_1<a_2<\cdots <a_{k-1}$, with $s-a_i<1/ i$. Pick $\varepsilon_{k}=\min (s-a_{k-1},1/k)$. Then there exists $a_k\in A$ with $s-a_k<\varepsilon_k$, so $s-a_k<1/k$ and $s-a_k<s-a_{k-1}$ whence $a_k>a_{k-1}$. It is clear $a_k\to s$. Hence our sequence is constructed.
$(\Leftarrow)$ Suppose such a sequence exists. If $s'<s$ we can find $N$ large enough so that $s-a_N<s-s'=\varepsilon >0$, so that $A\ni a_N>s'$ and $s'$ is not an upper bound.
Consider your set and the increasing (!) sequence $x_n=\dfrac{n}{n+1}\to 1$.
