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Prove that a set S of numbers has a maximum iff it is bounded above and sup S ε S

  1. Apr 27, 2012 #1
    1. The problem statement, all variables and given/known data
    Sorry that should read and sup S[itex]\in[/itex]S


    2. Relevant equations
    just the definitions of sup, inf, boundedness, and max



    3. The attempt at a solution
    I have tried a few different things, but this question is posted for advice, not just an answer. I am trying to think of a good way to start and we don't use contrapositive proofs in my class. I was thinking I could try proof by contradiction, but then I will have to do 2 cases, first assuming S is not bounded, then assuming supS is not in S. I understand the definitions of the terms, and I could explain why this is true verbally, but I am stuck trying to do it formally. Again, I am not just fishing for an answer, I really need to get these down or I wont pass anyway.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Apr 27, 2012 #2
    This depends on your precise definition of maximum. If you assume S has a maximum, then what do you know about it?

    Similarly, if S is bounded above, and the sup S, call it s0, is an element inside of S, then what do we know about s0?
     
  4. Apr 27, 2012 #3
     
  5. May 4, 2012 #4
    Case 1: S is not bounded above. So this is a direct contradiction to the definition given for a maximum in the book, "a member c of S is called the max of S provided that it is an upper bound for S". Since S is assumed to be unbounded, no number c exists s.t. c is an upper bound for S.

    Case 2: Sup S is NOT in S. Then we know for every x in S, x<c, but x can never equal c, since we have assumed sup S is NOT in S. Then we take any positive number epsilon= |(c-x_0)/2|, which can never be equal to 0 because c can never equal x. If we assume some number x_0 is the max of S, then we can take (x_0 + epsilon), where x_0 < (x_0 + epsilon) < c. This is a contradiction to the choice of x_0 as the choice of a maximum. Basically, if Sup S is not in S, then no matter what number x_0 in S that you choose, there exists a number x_0 + |(c-x_0)/2|, s.t. x_0 < (x_0 + |(c-x_0)/2| ) < c. This will contradict any choice of x_0 as a maximum of S.
     
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