A problem in my Real Analysis class asks for me to show that this is true, but I am a bit confused by it. If I take the set $S = (\sqrt{2}, \sqrt{3}) \cap \mathbf{Q}$, then inf $S$ = $\sqrt{2}$ but it is not within $S$, correct?
2026-03-30 00:23:02.1774830182
Does any non-empty set of the reals contain its own greatest lower bound if is bounded from below?
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Correct, $\sqrt{2}$ is the greatest lower bound (glb) of your set $S$ (even though it isn't an element of $S$), since the rationals are dense in $\mathbb{R}$. If you had defined the set $S$ as $[2, 3] \cap \mathbb{Q}$, then the glb would have been $2$, and $2$ is in the set.