Greatest Lower Bounded Irrational?

Question

If all elements of $S$ are irrational and bounded from below by $\sqrt 2$ then $\inf S$ must be irrational .

I would say this statement is true since $S=\{ \sqrt 2, \sqrt 3, \sqrt 5,\ldots\}$ the greatest lower bound is $\sqrt 2$ which is irrational and bounded from below the sequence.

Is this correct?

Answer 1

Consider $$S = \left\{2 + \frac{\sqrt{2}}{n} : n = 1, 2, 3, \dots\right\}$$

Then every element of $S$ is larger than $\sqrt{2}$, $S$ contains no rational entries, and $\inf S = 2$ is rational.

Answer 2

HINT: All elements of $S=[2,3]\setminus\Bbb Q$ are irrational and bounded below by $\sqrt2$, but $\inf S>\sqrt2$; what is $\inf S$?