Infimum and supremum of the set of all numbers whose square is greater than 2

Question

Suppose, $S= \{ x \in \mathbb R\mid x^2 > 2\}$. Then $\sup S = +\infty$. What is $\inf S$?

I'm guessing that $\inf S \in (-\sqrt 2, \sqrt 2)$. Is that true?

Answer 1

No. What if $x = -2?\quad -4? \quad -333337899999 ?$

Can you see that $\inf S = -\infty$

The interval you suggest, actually the interval $[-\sqrt 2, \sqrt 2]$ is excluded from $S$, so that $S = \mathbb R \setminus [-\sqrt 2, \sqrt 2]$.