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.
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$$\forall l \in L$, $l \leq x$, thus L is bounded above. How does this satisfy Definition 1.7 since he is saying all x in B and not saying there exists an x in B. $\forall x \in B$$\forall l \in L$, $l \leq x.$ doesn't match definition 1.7.
A few other questions I have pertain to when B is empty.
If B is empty and L is not empty, $\forall x \in B$$\forall l \in L$, $l \leq x.$ is vacuously true but can we say that B is bounded above? In Rudin's proof of Theorem 1.11 he says all x's in B are upper bounds of L. In this situation where B is empty can we say that?
If B is empty and L is empty, would $\forall x \in B$$\forall l \in L$, $l \leq x.$ be vacuously true? Would B be bounded above and could we say all x's in B are upper bounds of L.
With the questions when B is empty how is $\forall x \in B$$\forall l \in L$, $l \leq x.$ related to Definition 1.7?