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Using The Completeness Axiom To Find Supremum and Saximum.

  1. Apr 28, 2013 #1
    1. The problem statement, all variables and given/known data

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

    {r [itex]\in \mathbb{Q} [/itex] : r2 ≤ 5}


    2. Relevant 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.


    3. 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?
     
  2. jcsd
  3. Apr 28, 2013 #2

    Zondrina

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    The supremum does not have to be inside of your set. ##\sqrt{5}## is indeed the supremum.
     
  4. Apr 28, 2013 #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]?
     
    Last edited: Apr 28, 2013
  5. Apr 28, 2013 #4

    Zondrina

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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.
     
  6. Apr 28, 2013 #5
    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?
     
  7. Apr 28, 2013 #6

    Zondrina

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    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.
     
  8. Apr 28, 2013 #7
    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.
     
  9. Apr 28, 2013 #8
    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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