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Supremum is the least upper bound

  1. Nov 9, 2008 #1
    1. The problem statement, all variables and given/known data

    Prove that the supremum is the least upper bound

    2. Relevant equations

    3. The attempt at a solution

    Proof: let x be an upper bound of a set S then x>=supS (by definition). If there exists an upper bound y and y<=SupS then y is not an upper bound (contradiction) therefore every upper bound is greater than SupS so SupS is the least upper bound.

    Is that proof correct?

    Thank you.
  2. jcsd
  3. Nov 9, 2008 #2


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    Science Advisor

    Please tell what your definitions of "supremum" and "least upper bound" are! As far as I know "supremum" is just another name for "least upper bound" and there is nothing to prove.
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