Least-upper-bound property Rudin book

Question

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!!

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 09 '13 at 00:49
Because he's suppressing the "For all E" (and doing other things I don't like). $;$ — , Sep 09 '13 at 00:58

score 6 · Answer 1 · 2013-09-09T01:08:54.640

6

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 $\;\;\;$.

.

edited Sep 09 '13 at 01:08

answered Sep 09 '13 at 00:49

very clearly stated – Betty Mock Sep 09 '13 at 01:09
1

I find the blueness of your empty set philosophically curious. – Pete L. Clark Sep 09 '13 at 02:01
@Pete : $:$ What color are links? $;;;$ – Sep 09 '13 at 02:03
1

@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 09 '13 at 02:16

score 3 · Answer 2 · answered Sep 09 '13 at 01:59

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 09 '13 at 07:14

Least-upper-bound property Rudin book

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