I'm having a problem with this theorem. What if set $B$ is all $x$ such that $\sqrt{2} < x \le 2$, and $S$ is the set of all $y$ such that $\sqrt{2} < y \le 3$. $\sup L = \sqrt{2}$, which does not exist in $S$. Does this prove the theorem wrong by contradiction?
1.11 Theorem Suppose $S$ is an ordered set with the least-upper-bound property, $B \subset S$, $B$ is not empty, and $B$ is bounded below. Let $L$ be the set of all lower bounds of $B$. Then $\alpha = \sup L$ exists in $S$, and $\alpha = \inf B$.