Q:

Understanding how to rigorously prove the supremum of a set is a given number

Im thinking of the following set of real numbers 
$$ S = \left\{ \dfrac{m}{n} : m,n \in \mathbb{N} \; \text{and} \; \; m < 2n \right\}  $$
Certainly, since $n > 0$, one has that $\dfrac{m}{n} < 2 $ $\; \; \;$ ${\bf \forall} m,n$.
I want to claim that $\sup S = 2 $. If $u$ is an upper bound of $S$ that is if $ \dfrac{m}{n} \leq u $ for all $m,n$, ${\bf then}$ if we somehow can manage to prove that $2 \leq u$, then we would have proved our statement.
I ${\bf Dont}$ see a way to prove this directly (Is it possible??) so I would argue by contradiction. What if there is a $u_0$ such that $u_0 < 2 $, then $2 - u_0 > 0$ and by the archimedean principle there is a natural number $N$ such that $\dfrac{1}{N} < 2 -u_0 $
Next, $N(2-u_0) > 0$ so again can find natural $M$ so that $M > N(2-u_0)$ and thus 
$$ \dfrac{M}{N} > 2-u_0 > \dfrac{1}{N} $$
But, I dont see how to derive a contradiction from here. Am I on the right track?
As for the infimum, I can see that $\inf S \geq 0$, But I get stuck here again trying to show that this lower bound is actually the greatest. Any help would be greatly appreciated.

A:

It is not useful to have $\frac MN >2-u_0$. Instead of this choose $N$ such that $\frac 1 N <2-u_0$ and take $M=2N-1$. Then $\frac M N >u_0$. Thus, for every $u_0 <2$ there is an  element $\frac M N$ of our set which is greater than $u_0$. This proves that the supremum is $2$. 
For the infimum use the fact that for any $\epsilon >0$ we can choose $n$ such that $\frac  1 n <\epsilon$  and $\frac  1 n$ is an element of the given set. Hence the infimum is $0$.  

