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?

Homework Helper
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 $\sqrt{5}$?

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Homework Helper
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?

Homework Helper
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.