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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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    There is a exercise in Rudin related to this, although I forget what chapter.2012-06-06

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No, your example does not meet the hypotheses of the theorem, because $B$ is not bounded below in $S$. Note that $L=\{a\in S:\text{for all }b\in B, a\leq b\}=\emptyset$.

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    Thank you Jonas for your explanation and DonAntonio for your question. This was very helpful.2012-06-07