I would like to get some assistance with verifying my proof
The point was to prove that set A has a supremum and find this supremum.
is it a legitimate proof ? especially the part when I reach "true statement reached" , is it ok to do it like that ?
And the part that I give counter example to my assumption in proof by contradiction is it ok ?
I am sorry bud I don't know how to type mathematics here in this website so I am attaching my Lyx screenshot here :)
Please please help, it will be much appreciated.
The idea of the first part is correct but I personally don't like how the work is arranged. Starting from a desired conclusion and working towards a true statement is called begging the question and is a logical fallacy. In your case, since all the steps are reversible, it isn't a big deal and the proof still goes through, but you should really start from facts and work towards desired conclusions; not the other way around. The way I think the work should be arranged is as follows. We know that $m \ge 1$ and that $n + 1 \ge n$ for all $m, n \in \mathbb N$. Multiplying this inequalities, we see $$m(n+1) \ge n \,\,\,\,\, \implies \,\,\,\,\, mn + m \ge n \,\,\,\,\, \implies \,\,\,\,\, mn \ge n - m \,\,\,\,\, \implies \,\,\,\,\, 1 \ge \frac{m-n}{mn}$$ for all $m,n \in \mathbb N$. This proves that $1$ is an upper bound for the set.
The second part of your proof is incorrect. You let $\varepsilon$ be arbitrary, but then specified that $\varepsilon > 1$ later, so you actually only proved the claim for $\varepsilon > 1$
Instead, assume that for some $\varepsilon > 0$, $1-\varepsilon$ is an upper bound for the set. Then $$\frac{n-m}{nm} \le 1- \varepsilon$$ for all $m,n \in \mathbb N$. Let us construct an $m, n$ that contradict this. Indeed, setting $m = 1$, we see this implies that $$\frac{n-1}{n} \le 1- \varepsilon \,\,\,\,\, \Longleftrightarrow \,\,\,\,\, 1 - \frac 1 n \le 1 - \varepsilon \,\,\,\,\, \Longleftrightarrow \,\,\,\,\, \frac 1 n \ge \varepsilon \,\,\,\,\, \Longleftrightarrow n \le \frac 1 \varepsilon$$ for all $n \in \mathbb N$. This of course contradicts the unboundedness of the rationals. Thus $1- \varepsilon$ is not an upper bound for the set for any $\varepsilon > 0$.