Using The Completeness Axiom To Find Supremum and Saximum.

  • #1

Homework Statement



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

{r [itex]\in \mathbb{Q} [/itex] : 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 [itex]\sqrt{5}[/itex], but since [itex]\sqrt{5} \notin \mathbb{Q}[/itex], then the set has no maximum. Additionally, since [itex]\sqrt{5} \notin \mathbb{Q}[/itex] 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, [itex]\sqrt{5}[/itex], this least upper bound is not in the set of rationals and therefore the set has no supremum?
 

Answers and Replies

  • #2
STEMucator
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The supremum does not have to be inside of your set. ##\sqrt{5}## is indeed the supremum.
 
  • #3
The book gives some other example where the set T = {q [itex] \in \mathbb{Q}[/itex]: 0 ≤ q ≤ [itex]\sqrt{2}[/itex]}.

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

Wouldn't the same exact argument apply for [itex]\sqrt{5}[/itex]?
 
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  • #4
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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.
 
  • #5
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?
 
  • #6
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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.
 
  • #7
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.
 
  • #8
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The field Q is not complete.
R is complete.
You're supposed to deal with the sup in R. The argument [itex]\sqrt{5}[/itex] is not in Q should only be applied to show that the set aforementioned has no sup over Q.
 

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