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.
Best Answer
0) $2$ is an upper bound of $S$, since $m/n < 2n/n =2$
2) Consider the sequence $a_n=\frac{2n-1}{n}=2-1/n \in S;$
Assume $b <2$ is a smaller upper bound of $S$.
$|a_n-2|=|-1/n|=1/n$.
Let $\epsilon >0$ given.
Archimedean principle:
There is a $n_0$ s.t. $n_0 >1/\epsilon$.
For $n\ge n_0$ we have
$|a_n-2| =1/n < 1/n_0 <\epsilon;$
$2 -\epsilon <a_n < 2+\epsilon;$
Choose $\epsilon =2-b(>0)$:
Then $b <a_n$, a contradiction .