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Homework Help: Supremum Proof

  1. Sep 7, 2009 #1


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    1. The problem statement, all variables and given/known data

    Consider the open interval [itex](a,b)[/itex]. Prove that [itex]\mathrm{sup}{(a,b)} = b[/itex].

    2. Relevant equations


    3. The attempt at a solution

    I'm terrible at these proofs so I would appreciate it if someone could verify (or correct) my solution.

    Proof: Clearly [itex]b[/itex] is an upper bound since [itex]\forall{x} \in (a,b)[/itex] we have the strict inequality [itex]a < x < b[/itex]. Now, suppose that [itex]b[/itex] is not the least upper bound. Letting [itex]c = \mathrm{sup}{(a,b)}[/itex] there must be some real [itex]\varepsilon > 0[/itex] such that [itex]b - \varepsilon = c[/itex]. However, since [itex]\varepsilon > 0[/itex] this implies that [itex]\frac{\varepsilon}{2} > 0[/itex] and similarly that [itex]b > b - \frac{\varepsilon}{2} > b - \varepsilon = c[/itex] which contradicts the fact that [itex]c = \mathrm{sup}{(a,b)}[/itex]. This completes the proof.

    I know that my proof is definately wordy, but is it correct? Thanks for any suggestions!
  2. jcsd
  3. Sep 7, 2009 #2


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    You may have the basics of a proof there, but it is definitely unclear. Why dose [itex]b > b - \frac{\varepsilon}{2} > b - \varepsilon = c[/itex] contradict the assumption that c is the supremum? If you clear that up, then the proof should be fine. Remember, there are 2 properties c should satisfy to be the supremum: It should be an upper bound, and every other upper bound should be greater than it. Which, if any, of these properties have you shown c not to satisfy?
  4. Sep 7, 2009 #3
    The proof seems fine to me.
  5. Sep 7, 2009 #4
    Yeah it looks clear to me. I mean the OP demonstrates that b - e/2 is in the open interval and (a,b) is greater than the supposed sup (a,b), so there's really not much else to say...
  6. Sep 7, 2009 #5


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    But he doesn't. He doesn't explicitly say it and he doesn't restrict e to eliminate the possibility that b-e/2 < a so that it is not in the interval.
  7. Sep 7, 2009 #6
    He did say it. He said "which contradicts the fact that [tex] c = sup{(a,b)} [/tex] . This completes the proof. ".

    Also, it's pretty obvious that b - e is not less than a (because it's the LUB for the set), and so b - e/2 cannot be less than a.

    I see what you mean, but in my opinion, being that wordy would just make the proof longer than it has to be, without really contributing anything.
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