Using The Completeness Axiom To Find Supremum and Saximum.

[mliuzzolino]

Homework Statement

For each subset of ℝ, give its supremum and its maximum. Justify the answer.

{r $\in \mathbb{Q}$ : r² ≤ 5}

Homework Equations

Maximum: If an upper bound m for S is a member of S, then m is called the maximum.

Supremum: Let S be a nonempty set of ℝ. If S is bounded above, then the least upper bound of S is called its supremum.

The Attempt at a Solution

Supremum: none, Maximum: none.

We can see that [any positive real number x such that x² ≤ 5 is an upper bound. The smallest of these upper bounds is $\sqrt{5}$, but since $\sqrt{5} \notin \mathbb{Q}$, then the set has no maximum. Additionally, since $\sqrt{5} \notin \mathbb{Q}$ the set does not have a supremum.

I think this is correct, but I'm not exactly sure. Is there no supremum because even though the least upper bound exists, $\sqrt{5}$, this least upper bound is not in the set of rationals and therefore the set has no supremum?

 

[STEMucator]

The supremum does not have to be inside of your set. $\sqrt{5}$ is indeed the supremum.

 

[mliuzzolino]

The book gives some other example where the set T = {q $\in \mathbb{Q}$: 0 ≤ q ≤ $\sqrt{2}$}.

The book then claims that it does not have a supremum when considered as a subset of $\mathbb{Q}$. The problem is that sup T = $\sqrt{2}$, and $\sqrt{2}$ is one of the "holes" in $\mathbb{Q}$ - that is, $\sqrt{2}$ does not exist in $\mathbb{Q}$ and therefore cannot be a Supremum.

Wouldn't the same exact argument apply for $\sqrt{5}$?

 

[STEMucator]

Lets call your set S.

S is nonempty ( clearly, take r = 1 ), S is also bounded above ( Plenty of elements, 5 is an upper bound if you so please ), and you're viewing S as a subset of $\mathbb{R}$ not $\mathbb{Q}$.

So S has to have a supremum by the axiom.

 

[mliuzzolino]

Lets call your set S.
you're viewing S as a subset of $\mathbb{R}$ not $\mathbb{Q}$.

So S has to have a supremum by the axiom.

What exactly does the book mean when it says that when considered as a subset of the rationals, the set does not have a supremum?

The book then goes on to say that "Every nonempty subset of S of the reals that is bounded above has a least upper bound. That is, sup S exists and is a real number."

This seems to completely contradict what it is saying that the set T (as defined by my previous reply) does not have a supremum when considered as a subset of the rationals. It seems that my problem is contingent upon that last part "subset of the rationals," but isn't a subset of the rationals also a subset of the reals, and by the definition, every nonempty subset S of the reals has a supremum?

 

[STEMucator]

What exactly does the book mean when it says that when considered as a subset of the rationals, the set does not have a supremum?

The book then goes on to say that "Every nonempty subset of S of the reals that is bounded above has a least upper bound. That is, sup S exists and is a real number."

This seems to completely contradict what it is saying that the set T (as defined by my previous reply) does not have a supremum when considered as a subset of the rationals. It seems that my problem is contingent upon that last part "subset of the rationals," but isn't a subset of the rationals also a subset of the reals, and by the definition, every nonempty subset S of the reals has a supremum?

When you regard S as a subset of the reals, you have access to the completeness axioms for your arguments. Since $\sqrt{5} \in \mathbb{R}$, and $\sqrt{5} ≥ r$ we know $sup(S) = \sqrt{5}$ in this case.

The problem when you view S as a subset of the rationals is $\sqrt{5} \notin \mathbb{Q}$ so S can't have a supremum. This is part of the reason that $\mathbb{Q}$ is incomplete.

That is, while $\mathbb{Q}$ obeys the ordering axioms, subsets of $\mathbb{Q}$ which are bounded above may fail to have a supremum which we can show using S.

 

[mliuzzolino]

When you regard S as a subset of the reals, you have access to the completeness axioms for your arguments. Since $\sqrt{5} \in \mathbb{R}$, and $\sqrt{5} ≥ r$ we know $sup(S) = \sqrt{5}$ in this case.

The problem when you view S as a subset of the rationals is $\sqrt{5} \notin \mathbb{Q}$ so S can't have a supremum. This is part of the reason that $\mathbb{Q}$ is incomplete.

That is, while $\mathbb{Q}$ obeys the ordering axioms, subsets of $\mathbb{Q}$ which are bounded above may fail to have a supremum which we can show using S.

Wow. You're awesome. Thanks so much for alleviating at least a small part of my vast incomprehension of this material. About 95% of me wants to burn this textbook then ingest like 5000mg of cyanide.

 

[funcalys]

The field Q is not complete.
R is complete.
You're supposed to deal with the sup in R. The argument $\sqrt{5}$ is not in Q should only be applied to show that the set aforementioned has no sup over Q.