I have to find the supremum and infimum of $A=\{x\in \mathbb{Q}: 4<x^2\leq7\}$
My proof:
$4<x^2\leq7\implies 2<x\leq\sqrt{7}$
I will prove that $\sup A=\sqrt7$ and $\inf A=2$
Since $\sqrt7\geq x$, $\forall x\in A$ then it's an upper bound for $A$
For $\sqrt7$ to be the supremum $\forall ε>0$ there should be a $x_0\in A$ such that
$\sqrt7-ε<x_0$
Which is true due to the density of rationals in $\mathbb{R}$
So $\sup A=\sqrt7$
Since $2<x, \forall x\in A$ then it's a lower bound
For $2$ to be the infimum $∀ε>0$ there should be a $x_0\in A$ such that
$2+ε>x_0$ again from the density of rationals the statement is true
Is this proof correct/sufficient? Thanks in advance