I'm going through Rudin for a real analysis class, and the proof of this theorem is unclear:
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 $\alpha = \inf B$. In particular, $\inf B$ exists in $S$.
Proof:
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 $\alpha$
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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?
Best Answer
edit: Well, it seems that your question is "Why $L\subset S $? The definition of the set $L$ is $L = \left\{ x\in S : x\leq y, \forall y\in B\right\} $. Here, the set $S$ may be thought as an abstract ordered set which has the least upper bound property - and, in the hypothesis of you theorem, the set $S$ is the universe.
I guess that you got confused with the definition of "least upper bound property". An ordered set $S$ has this property if the following holds:
"$L\subset S $ and $L$ has upper bound $ \Longrightarrow $ $L$ has a least upper bound $ x\in S $ " (the least upper bound $ x $ is called supremum of $L$).
you may see http://en.wikipedia.org/wiki/Least-upper-bound_property for further clarifications...
Anyway, Rudin showed that you set $ L $, which is, by definition, a subset of $ S $, has upper bound (in $S$). Therefore, since $S$ has least upper bound property, $L$ has a supremum in $S$.