# 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!!

• Every linearly ordered set satisfies your definition. – azarel Sep 9 '13 at 0:47
• I'm not sure I get what you're saying..My question is why is he putting the definition in the form of an implication? it's in the form A,B-->C But why? I don't know what a linearly ordered set is put thanks either way! – wwbb90 Sep 9 '13 at 0:49
• Because he's suppressing the "For all E" (and doing other things I don't like). $\;$ – user57159 Sep 9 '13 at 0:58

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 $\;\;\;$.

.

• very clearly stated – Betty Mock Sep 9 '13 at 1:09
• I find the blueness of your empty set philosophically curious. – Pete L. Clark Sep 9 '13 at 2:01
• @Pete : $\:$ What color are links? $\;\;\;$ – user57159 Sep 9 '13 at 2:03
• @Ricky: They're blue. I understand the phenomenon and am not objecting to it. It's just that I read your answer and thought, "Hmm, I wasn't expecting the empty set to be blue". Then I realized what a striking statement that was. – Pete L. Clark Sep 9 '13 at 2:16

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 $\mathbb{Q}$.

• Thank you so much that's exactly what I was looking for!!, it's annoying that I can't upvote your post though.. – wwbb90 Sep 9 '13 at 7:14