I'm going through Rudin for a real analysis class, and the proof of this theorem is unclear:
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 $\alpha = \inf B$. In particular, $\inf B$ exists in $S$.
Since $B$ is bounded below, $L$ is not empty. Since $L$ consists of exactly those $y \in S$ which satisfy the inequality $y \leq x$ for every $x \in B$, we see that every $x \in B$ is an upper bound of $L$. Thus $L$ is bounded above. Our hypothesis about $S$ implies therefore that $L$ has a supremum in $S$; call it $\alpha$
My question is on the last line above, the proof applies the least-upper-bound property to $L$ treating it as a subset of $S$. But is it true that $L \subset S$? I don't think it's usually true, so we can't apply the least-upper-bound property to $L$. Where am I going wrong here?