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Completeness axiom/Achimedean principle and sup set.

  1. Nov 19, 2012 #1
    1. The problem statement, all variables and given/known data
    Let [tex]B=\left\{{\frac{1}{2},\frac{2}{3},\frac{3}{4},...}\right\} [/tex]
    Prove sup B = 1


    2. Relevant equations
    Archimedean principle:
    Let a<b and a>0 [tex]\exists n \in{N}[/tex] such that an > b.


    3. The attempt at a solution
    Its trivial to see that 1 is an upper bound for B and B is nonempty, so B must have a finite supremum by the completeness axiom.

    I want to show that 1 is the smallest upper bound. So I want to try to do a proof by contradiction:
    Let M be an upper bound for B, and M<0.

    -- Find contradiction ????

    I know each element in B is a/b, with natural numbers a and b and a<b. But how to continue further??
     
    Last edited: Nov 19, 2012
  2. jcsd
  3. Nov 19, 2012 #2

    jbunniii

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    Well, surely you know more than that:
    [tex]\frac{1}{2} = 1 - \frac{1}{2}[/tex]
    [tex]\frac{2}{3} = 1 - \frac{1}{3}[/tex]
    [tex]\frac{3}{4} = 1 - \frac{1}{4}[/tex]
    Does this suggest a formula for the elements of B?
     
  4. Nov 20, 2012 #3
    Yes, each element in B is of the form:

    [tex]1-\frac{1}{k} , k>1, k\in{N}[/tex]

    So intuitively its easy to see you can choose a k such that |1-(1-1/k)|<ε, for ε>0.
     
  5. Nov 20, 2012 #4
    I want to prove that there exists a k, such that 1-1/k > M, when M<1.
     
  6. Nov 20, 2012 #5

    Dick

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    Pick ε=1-M>0. Then your Archimedean principle tells you there is a k such that kε>1. So ε>1/k. So?
     
  7. Nov 20, 2012 #6
    Yes thanks, I think this is correct.

    So the proof will look like this.

    To prove, sup B = 1

    Proof:
    (1) Let M be an upper bound for B and M<1.

    (2) 1-M > 0 (follows from 1)

    (3) Exists natural number n s.t. : n(1-M) > 1 (Archimedean principle)

    (4) (1-M) > 1/n (First multiplicative propery)

    (5) -M > 1/n -1 (Follows from 4)

    (6) M < 1 - 1/n (Contradiction, since M is an upper bound for B and 1-1/n is in B)

    (7) Conclusion: 1 must be the least upper bound, therefore sup B = 1
     
  8. Nov 20, 2012 #7

    Dick

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    That looks ok to me. Since you know 1 is an upper bound, it must be the least upper bound.
     
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