Let $A = \{x \in \Bbb Q \; | x^2 < 2\}$.
Prove that $\sup_{x \in A}\, (x) = \sqrt{2}\;$ in real line $\Bbb R.$
I could prove the case in the rational line where no supremum exists. In $\Bbb R, $ it is easy to see that the supremum exists due to least-upper bound property (we know that 2 is an upper bound of $A$). But I don't know how to prove that supremum is equal to $\sqrt{2}\,$ in $\Bbb R$.
$\sqrt{2}$ is an upper bound on $A$, since $ x\in A$ implies $x < \sqrt{2}$. To show it is the least upper bound, suppose there were another, say $A< x' < \sqrt{2}$. There are infinitely many rationals in $(x',\sqrt{2})$, so that there would be an element of $A$ less than $\sqrt{2}$ but greater than $x'$, which contradicts $x'$ as an upper bound of $A$.