If $M=\sup \left( A \right)$ and $A\subseteq \mathbb{Q}$ (rational numbers)
$\forall\ \varepsilon>0$ of $\mathbb{Q}, \exists\ x\in A$ s.t. $x > M–\varepsilon$.
Is this property true for rationals ? because in Wikipedia $A$ always a subset of $\mathbb{R}$ (real numbers)
That really confuse me.
The definition of supremum is valid in all kinds of underlying sets with an ordering relation $\leq$, so it also works in the rational numbers $\mathbb{Q}$. The property you state there is also true there.
The only decisive difference is that in $\mathbb{R}$, every (bounded) set has a supremum. In $\mathbb{Q}$, you can construct many sets that do not have a supremum because that would be a number in $\mathbb{R} \setminus \mathbb{Q}$.