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Thread: supS = -inf(-S) But how?

  1. #1
    Feb 2010

    supS = -inf(-S) But how?

    Hi, our instructor left this question as an exercise for us. And I first need to prove
    supS ≥ -inf(-S)
    -inf (-S) ≥ supS

    I think I could do st on the first part.

    Say x S

    supS ≥ x

    For -x -S

    -x ≥ inf(-S) If we multiply it with - ;

    -inf(-S) ≥ x
    Then we get
    supS ≥ x
    -inf(-S) ≥ x

    Since the greatest value of x is supS, then supS ≥-inf(-S).

    But I cannot find a way to prove the reverse. Any thoughts? Thanks in advance.
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  2. #2
    Junior Member
    Aug 2010
    yes do the same for the set (-S) and you get the inequality that sup(S) $\displaystyle \le$ -inf(-S); combine two inequalities and you have your equality.
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