The question was to prove $\{\frac{n-m}{n+m}|n,m\in \mathbb{N}, m
Proving it's bounded was easy, But I got stuck on proving 1 is it's supremum.
This is what I have:
i. We'll hypothesize that $\sup\left(A\right)=1$
This means that $\forall\epsilon>0,\exists x\in A,x>1-\epsilon$
$\frac{n-m}{n+m}>1-\epsilon\implies n-m<\left(1-\epsilon\right)\left(n+m\right)\implies$ $\implies n-m
But I'm not sure how I continue from here. If epsilon is $\geq1$ then this can never be right, and I'm actually confused as to why (or even if) the steps I did untill now are true. Can I just find an example for when $\epsilon\geq1$ and then expain why a solution exist for when $\epsilon<1$ (Though I'm not sure how to do that either...)?
I'm assuming proving supremum is the same, so I guess this is a double question :p
p.s. I don't know how to tag this, but it's homeworks in infi so I guess calculus it is...