Least-upper-bound property Rudin book In baby Rudin:
1.10 Definition:
An ordered set S is said to have the least-upper-bound property if the following is true:
If E $\subset$ S, E is not empty, and E is bounded above, then sup E exists in S.
I find the way it's written to be weird, shouldn't he instead have written:
If E $\subset$ S, E is not empty, and E is bounded above, and sup E exists in S.
I mean it's a definition, he can't come to conclusions in a definition.. Please someone explain I'm really confused thanks a lot!!
 A: It should actually be as follows:

An ordered set S is said to have the least-upper-bound property if and only if the following is true:
For all E, $\;\;\;$ if $\;$ $\:$$\{\hspace{-0.02 in}\}$ $\neq$ E $\subset$ S$\:$ and E is bounded above $\:\:$ then $\:\:$ sup E exists in S $\;\;\;$.

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A: There's nothing wrong with putting implications in a definition, and in this case, it's quite necessary.  If the implication is true for a given ordered set $S$, then the set has the least upper bound property, and otherwise, it does not have the least upper bound property.
For example, if $S=\mathbb{Q}$ is the ordered set of rational numbers, then the implication fails: The set $E=\{x \in \mathbb{Q}: x^2<2\}$ is nonempty and bounded above, but $\sup E$ does not exist.  Therefore, $\mathbb{Q}$ does not have the least upper bound property.
On the other hand, consider the ordered set $S=\mathbb{Z}$, the integers.  Any subset of integers $E \subset \mathbb{Z}$ which is nonempty and bounded above must have a maximum element, and this maximum element is $\sup E$.  Therefore , $\mathbb{Z}$ does have the least upper bound property.
Later, you will prove that the set of real numbers $\mathbb{R}$ has the least upper bound property, which is the characteristic property that distinguishes it from sets like $\mathbb{Q}$.
