# Bounded above implies there exists a $\sup B$?

This is Rudin's mathematical analysis book's theorem about sup and least-upper-bound

### 1.11 Theorem

Suppose $$S$$ is an ordered set with the least-upper-bound property, $$B\subset S$$, $$B$$ is not empty, and $$B$$ is bounded below. Let $$L$$ be the set of all lower bounds of $$B$$. Then

\begin{align*}\alpha =\sup L\end{align*}

exists in $$S$$, and $$\alpha =\inf B$$. In particular, $$\inf B$$ exists in $$S$$.

Does this theorem imply this proposition(Is this one theorem or definition?)

### proposition

Every subset $$B$$ of $$R$$(or ordered sets with least-upper-bound?) which is bounded above must have a $$\sup B$$?

If not, where is this proposition referred or implied in Rudin's book?

wiki's edition explicitly say: The least-upper-bound property states that any non-empty set of real numbers that has an upper bound must have a least upper bound in real numbers.

But in Rudin's book it is constructing Real numbers from Rational numbers. And is on ordered sets firstly?

### 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 $$S$$.

If the proposition above is just the Least-Upper-Bound property, then it seems to be an $$\color{green}{definition}$$, and that is definition 1.10 in Rudin's book. And I found it's an axiom in Royden's Real Analysis.

So, proof: Since $$R$$ is an ordered set with least upper bound property, then proposition is true. Is this not necessary?

• Look at the construction of real number system in chapter 1, listed as an appendix. Most people look at this as an axiom, but it is provable. It is definitely not a definition Jul 25, 2013 at 14:50
• @Vishesh wiki's edition explicitly say: The least-upper-bound property states that any non-empty set of real numbers that has an upper bound must have a least upper bound in real numbers. Jul 25, 2013 at 15:06
• @Vishesh If this is just the Least-Upper-Bound property, then it's an definition, and that is definition 1.10 in Rudin's book, And I found it's an axiom in Royden's Real Analysis. Jul 26, 2013 at 15:12
• Well the least upper bound property is itself a definition. But the fact that the reals have it is an axiom or theorem in different books. Your theorem 1.11 is of course a theorem. Thats what I meant. Jul 26, 2013 at 15:56

The property says

If $S$ is nonempty and bounded above, $\sup S$ exists.

Now Rudin is proving

If $S$ is nonempty and bounded below, $\inf S$ exists.

He does this by noticing that the set $L$ of lower bounds of $S$ is nonempty and bounded above, so $\sup L$ exists. Can you see why $\sup L=\inf S$?

• Your answer is like my intuition, I hope it's right. Now I still have some confusions. I'll digest it after sleep. good night. :) Jul 25, 2013 at 15:37
• So, an ordered set has the greatest-lower-bound property implies it also has the least-upper-bound property? Jul 26, 2013 at 15:04
• @HyperGroups Yes, and conversely.
– Pedro
Jul 26, 2013 at 15:18

The general result applies to any poset (what is called above ordered set) $P$. If $P$ is such that every nonempty bounded above subset has a supremum, then every nonempty bounded below subset has an infimum. It is also the case that if $P$ is such that every nonempty bounded below subset has an infimum, then every nonempty bounded above subset has a supremum. The proof above is the proof of the former. A very similar proof gives the latter.

In the special case of the real numbers, one either assumed axiomatically that $\mathbb R$ satisfies that every nonempty bounded above subset has a supremum. Or one develops a model of the real numbers from a presupposed model of the rational numbers (typically using Dedekind cuts or Cauchy sequences). One then proceeds to prove that every nonempty bounded above subset has a supremum. In any case, it thus follows that every nonempty bounded below subset has an infimum.

Remark: The general results stated for posets is an example of the principle of duality, which is really what's going on here.

• So, an ordered set has the greatest-lower-bound property implies it also has the least-upper-bound property? Is these obvious? And can be used without proof? Jul 26, 2013 at 15:06
• If the proposition above is just the Least-Upper-Bound property, then it seems to be an definition in Rudin's book? Jul 26, 2013 at 15:17

This is the least upper bound property.

• Least upper bound property discuss whether the bound is in the set or not. Jul 25, 2013 at 14:48
• wiki's edition explicitly say: The least-upper-bound property states that any non-empty set of real numbers that has an upper bound must have a least upper bound in real numbers. But in Rudin's book it is constructing Real numbers from Rational numbers. Are you familiar with that book? Jul 25, 2013 at 15:07
• @HyperGroups: Yes I think he uses Dedekind cuts? Jul 25, 2013 at 15:09
• @HyperGroups Rudin constructs the reals and then proves the least-upper-bound property for them. This property is also known as the completeness of the reals. Jul 25, 2013 at 15:10
• @IttayWeiss oh, as your sayings, then seems this should be implied in the Appendix. Jul 25, 2013 at 15:21

The fact that if $B\subset S$ is bounded above it has a supremum in $S$ is precisely the least upper bound property for $S$ which is being assumed.

• An ordered set has the greatest-lower-bound property implies it also has the least-upper-bound property? Is these obvious? And can be used without proof? Jul 26, 2013 at 15:10
• If the proposition above is just the Least-Upper-Bound property, then it seems to be an definition in Rudin's book? Jul 26, 2013 at 15:17
• @HyperGroups In the situation you have described, the least upper bound property is being assumed - and is used to prove the greatest lower bound property. One can similarly assume the greatest lower bound property and prove the least upper bound property from it. But you have to begin somewhere. There are ordered fields which have neither property. Jul 26, 2013 at 15:24