How to prove the infimum of this subset

Question

consider $A\:=\:\left\{\frac{n-m}{n+m}\:;\:m<n\:\:n,m\:\in \mathbb{N}\right\}$
How to prove that $1$ is the supremum and $0$ is the infimum?

I go to the definition of infimum, and see that by given any $\epsilon >0$, i need to show that there is $a\in A$, $a\:=\: \frac{n-m}{n+m}$ for some n and m, such that $a\:<\:0\:+\:\epsilon $

why i just can say that " ok, lets pick some $n_0\:$,$m_0\:$ such $\frac{n_{0\:}-m_0}{n_0+m_0}\:\:<\:\epsilon $ and its equal to $\frac{n_{0\:}-m_0}{n_0+m_0}\:\:<\: 0+ \epsilon $ and we finish?"

need help here, tnx!

Answer 1

"Some" can (and must) be more concrete: $m_0=n_0-1$, for example, and

$$\frac{n_{0}-m_0}{n_0+m_0}=\frac{1}{2n_0-1}<\epsilon$$ i.e., $n_0>\cdots$

Answer 2

Hint: Pick $m_0$ such that $\dfrac{1}{2m_0+1}<\epsilon$, then pick $n_0=m_0+1$.