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Question on Theorem 1.11 in Baby Rudin

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Definition 1.7 in Baby Rudin says: Suppose S is an ordered set, and E $\subset$ S. If there exists a $\beta \in$ S such that $x \leq \beta$ for every $x \in E$, we say E is bounded above and call $\beta$ an upper bound of E.

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In the proof of Theorem 1.11 in Baby Rudin, Rudin says that all x in B are upper bounds of L, thus L is bounded above. So Rudin is saying $\forall x \in B$$\forall l \in L$, $l \leq x$, thus L is bounded above. How can we say all x in B are upper bounds of L if the definition uses there exists? $\forall x \in B$$\forall l \in L$, $l \leq x.$ doesn't match definition 1.7. If B were the empty set $\forall x \in B$$\forall l \in L$, $l \leq x.$ would be vacously true but saying all x in the empty set are upper bounds of L doesn't make sense to me because there are no x in the empty set. Also how would those x's in the empty set be elements of S? Can someone explain this please? Thank you!


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