Let $A\subset(0,\infty)$ non empty. Show that $u=\sup A$ iff:
i) $\forall x \in A \quad x \leq u$
ii) $\forall d>1 \quad \exists a \in A: \frac{u}{d}<a$
One implication:
If $u=\sup A$, then we have i) from the definition. Let $d>1$, since $u>0$ we have $\frac{u}{d}<u$ and from another characterization of the supremum there exist $a\in A$ such that: $$u-u+\frac{u}{d}=\frac{u}{d}<a$$ since $u-\frac{u}{d}$ is positive.
For the other side, we have that $u$ is an upper bound from i), but I have not been able to show that is the least upper bound, not even using the previous characterization.