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Math Help - Proof of Inf S= -sup(-S)

  1. #1
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    Proof of Inf S= -sup(-S)

    I'm trying to prove that inf S = -sup(-S) where -S={-s : s is in S} and S is a nonempty subset of the Reals.

    I was able to do the bounded above and below case, and the bounded below case and not above, but I having trouble with the not bounded below and bounded above and not bounded at all cases.

    Help Appreciated.
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  2. #2
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    Re: Proof of Inf S= -sup(-S)

    If S is not bounded below, then -S is not bounded above. What do we say inf S and sup(-S) are in this case?

    - Hollywood
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  3. #3
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    Re: Proof of Inf S= -sup(-S)

    Quote Originally Posted by hollywood View Post
    If S is not bounded below, then -S is not bounded above. What do we say inf S and sup(-S) are in this case?

    - Hollywood
    Yes, I understand that if S is not below then inf S=-infinity and if -S is not above, then sup -S = infinity. But is it really just a matter of saying then since inf S = -infinity and sup -S= infinity then -sup -S is -infinity also?

    Or am I trying to make it harder than it has to be?
    Last edited by JSB1917; September 30th 2012 at 06:11 PM.
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  4. #4
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    Re: Proof of Inf S= -sup(-S)

    Yes, that's all it is.

    To be 100% technically correct, the problem statement would have to be "Inf S = -Sup(-S) when they are finite and Inf S = -\infty if and only if Sup(-S)= +\infty" since the expression -(+\infty) is undefined. But that's a technicality - no one thinks -(+\infty) is anything other than -\infty.

    The final case is easy now, right?

    - Hollywood
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