# Rudin's Principles of Analysis Theorem 1.11

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 $\alpha = \inf B$.

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It would be better to state the theorem in this space. Many viewers, including some who could give very good answers, will not have the book at hand. –  Dylan Moreland Jun 6 '12 at 2:21
–  Jonas Meyer Jun 6 '12 at 2:23
... and I do have the book, but apparently a different edition (mine is Second Edition), because Theorem 1.11 in mine is about the $<$ relation on Dedekind cuts. –  Robert Israel Jun 6 '12 at 2:27
There is a exercise in Rudin related to this, although I forget what chapter. –  rckrd Jun 6 '12 at 14:52

No, your example does not meet the hypotheses of the theorem, because $B$ is not bounded below in $S$. Note that $L=\{a\in S:\text{for all }b\in B, a\leq b\}=\emptyset$.
@DonAntonio: The ordered set in which bounds are considered is $S$ (not $L$, and in the context of the OP's example, not $\mathbb R$). While the statement of the theorem would perhaps be clearer if "in $S$" were explicitly stated for emphasis, in this context $S$ is the "universe". The most common application is where $S=\mathbb R$, but Rudin's statement abstracts the essential order properties, and the OP's example does not satisfy those properties. –  Jonas Meyer Jun 6 '12 at 2:34