Define the set $$T=\lbrace \frac{m-n}{m+n} : m,n\in\mathbb{N} \rbrace$$
Where $\mathbb{N}$ is the set of natural numbers, starting from 1.
Use the Archimedean Property to prove that the infimum of this set $T$ is $-1$ (and similarly the supremum is 1).
I find this problem to be really tricky to approach since there are two variables. Although I can get an intuitive feel as to why the infimum and supremum are what they are, I'm really having trouble formalizing it. How would you prove this?
Since $ \frac{m-n}{m+n} >-1$ for any m, n $\in \mathbb{ N} $, the infimum of T is greater than or equal to -1. Choose $m, n \in \mathbb{N}$ such that $ \frac{m}{n} < \frac{\epsilon}{1-\epsilon}$ using Archimedean property( where $ \epsilon$ is an arbitrary positive number less than 1)
$\Rightarrow 0 <2m < (\epsilon +1)m + \epsilon n$
$\Rightarrow -m -n <m -n < -m - n + \epsilon (m +n) $
$ \Rightarrow -1 < \frac{m-n}{m+n} < -1+ \epsilon $
So, by the definition of infimum, the infimum of T is 1
Now, for the supremum of T -
Since $ \frac{m-n}{m+n} <1 $ $\forall m, n \in \mathbb{N}$, hence the supremum of T is less than or equal to 1.
Also, for any $ 0 <\epsilon <1 $, by Archimedean property, choose $m, n \in \mathbb{N} $ such that
$\frac {n}{m} < \frac{\epsilon}{2 - \epsilon} $ $\Rightarrow (1-\epsilon)(m+n) < m -n$ $\Rightarrow (1-\epsilon)(m+n) < m-n <m+n$ $\Rightarrow 1-\epsilon < \frac{m-n}{m+n} <1$
And hence by the definition of supremum, the supremum of T is 1