Showing that $\sup \{ x \in \mathbb{R} : x^2 < 2\} = \sqrt{2}$.

Question

Let $A:=\{x \in \mathbb{R} \mid x^{2} < 2\}$. How to prove $\sup(A)=2^{1/2}$?

I can prove $2^{1/2}$ is an upper bound for $A$.

But I can't prove the next condition in the definition of the supremum.

Answer 1

Establish that any positive real number strictly less than $2^{1/2}$ will be in the set $\{x:x^2<2\}$ and not an upper bound of the set, because $\forall y~((0<y<2^{1/2})\to\exists z~(y^2<z^2<2))$, due to some property of the real numbers.   You can then be sure that there can not exist any other upper bounds between $2^{1/2}$ and $A$.

( I am so dense.   What is that property called again?)