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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?

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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 $S$.

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  • $\begingroup$ $L$ is defined to be the set of all lower bounds of $B$, which is why we can be sure that the glb of $B$ is in $L$. Restricting $L$ to be the subset of $S$ containing lower bounds of $B$ doesn't allow this. For example, with $S=[1,5]$, $B=[1,2]$, the set of lower bounds of B is $(-\infty,1)$, but this is not a subset of $S$. $\endgroup$ – Ankit Soni Jan 29 '14 at 1:59
  • $\begingroup$ Well, perhaps you got confused with the statement of the theorem. When he says "B is bounded below", he is saying $B$ is a subset of $S$ (which is a ordered set). And, considering the order in $S$, $B\subset S $ is bounded below (by elements of $S$). You aren't in any moment considering that $S$ is a subset of any other bigger subset (not even $\mathbb{R}$). $\endgroup$ – Fernando Jan 29 '14 at 2:20
  • $\begingroup$ You may allow you to think more abstractly. Forget about the real line. You are considering an abstract set $S$ which has the property of least upper bound. And, when you say that $B$ is a subset bounded below, it means that $L$, as I defined above, isn't empty. $\endgroup$ – Fernando Jan 29 '14 at 2:21
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    $\begingroup$ By the way, in your example (in the comment above) the set $L$ wouldn't be $\left( -\infty , 1\right) $. It would be $ L = \left\{ 1 \right\} $. The set $ L $ is, by definition, the set of lower bounds of $B$ (considering the "universe" being the ordered set $S$). As I said, you are not saying that $S$ is a subset of $\mathbb{R} $ or $S$ is a subset of $\mathbb{R}\times \mathbb{R}$. $S$ is your universe. And, in your example, $L$ would be $\left\{ 1\right\} $. And you could apply Rudin' s argument. $\endgroup$ – Fernando Jan 29 '14 at 2:27
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Define $L=\{s\in S: s\text{ is a lower bound for } B \}$.

Now for the last point you can argue as follows. Let $a= \text{lub}(L)$ and suppose for the sake of contradiction that $a$ is not the glb of $B$, then there is some $a'>a$ and for all $b\in B$ we have $a'\le b$. So $a'\in L$ (why?) but $a'>a$ which contradicts that $a$ is an upper bound for $L$.

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