Bounded above implies there exists a $\sup B$?
Solution 1
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 $\sup L=\inf S$?
Solution 2
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.
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Updated on August 01, 2022Comments

HyperGroups 5 months
This is Rudin's mathematical analysis book's theorem about sup and leastupperbound
1.11 Theorem
Suppose $S$ is an ordered set with the leastupperbound 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 leastupperbound?) 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 leastupperbound property states that any nonempty 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 leastupperbound 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 LeastUpperBound 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?

Vishesh over 9 yearsLook 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

HyperGroups over 9 years@Vishesh wiki's edition explicitly say: The leastupperbound property states that any nonempty set of real numbers that has an upper bound must have a least upper bound in real numbers.

HyperGroups over 9 years@Vishesh If this is just the LeastUpperBound 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.

Vishesh over 9 yearsWell 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.


HyperGroups over 9 yearsLeast upper bound property discuss whether the bound is in the set or not.

HyperGroups over 9 yearswiki's edition explicitly say: The leastupperbound property states that any nonempty 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?

NebulousReveal over 9 years@HyperGroups: Yes I think he uses Dedekind cuts?

Ittay Weiss over 9 years@HyperGroups Rudin constructs the reals and then proves the leastupperbound property for them. This property is also known as the completeness of the reals.

HyperGroups over 9 years@IttayWeiss oh, as your sayings, then seems this should be implied in the Appendix.

HyperGroups over 9 years@FastingGuy yes, cuts is stated with detail in the Appendix, personally I don't like his states...

HyperGroups over 9 yearsYour answer is like my intuition, I hope it's right. Now I still have some confusions. I'll digest it after sleep. good night. :)

HyperGroups over 9 yearsSo, an ordered set has the greatestlowerbound property implies it also has the leastupperbound property?

HyperGroups over 9 yearsSo, an ordered set has the greatestlowerbound property implies it also has the leastupperbound property? Is these obvious? And can be used without proof?

HyperGroups over 9 yearsAn ordered set has the greatestlowerbound property implies it also has the leastupperbound property? Is these obvious? And can be used without proof?

HyperGroups over 9 yearsIf the proposition above is just the LeastUpperBound property, then it seems to be an definition in Rudin's book?

HyperGroups over 9 yearsIf the proposition above is just the LeastUpperBound property, then it seems to be an definition in Rudin's book?

HyperGroups over 9 yearsIf the proposition above is just the LeastUpperBound property, then it seems to be an definition in Rudin's book?

Pedro over 9 years@HyperGroups Yes, and conversely.

Mark Bennet over 9 years@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.