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Homework Help: Prove inf(S)=-Sup(-S)?

  1. Sep 20, 2012 #1
    Prove inf(S)=-Sup(-S)??

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

    Let S,T be subsets of ℝ, where neither T nor S are empty and both Sup(S) and Sup(T) exist.

    Prove inf(S)=-sup(-S).

    Starting with =>

    I let x=inf(S). Then by definition, for all other lower bounds y of S, x≥y.

    I'm stuck at this point...

    Any help please?

  2. jcsd
  3. Sep 20, 2012 #2
    Re: Prove inf(S)=-Sup(-S)??

    You have to show two things:

    • -x is an upper bound of -S
    • If y is another upper bound of -S, then [itex]-x\leq y[/itex]

    So, in order to prove that -x is an upper bound of -S. Take an element -s from -S and prove [itex]-s\leq -x[/itex]. Why is that true?
  4. Sep 20, 2012 #3
    Re: Prove inf(S)=-Sup(-S)??

    I'm not following along, why would -s≤-x? Unless you mean if I multiply both side by (-1), then it would be x≤s? Or am I totally off on a tangent?
  5. Sep 20, 2012 #4


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    Re: Prove inf(S)=-Sup(-S)??

    If [itex]-s \in -S[/itex], then [itex]s \in S[/itex]. And by definition, [itex]x[/itex] is a lower bound of [itex]S[/itex]...
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