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Supremum, Infimum (Is my proof correct?)

  1. Apr 11, 2016 #1
    1. The problem statement, all variables and given/known data
    Let ##A## be a nonempty set of real numbers which is bounded below. Let ##-A## be the set of all numbers ##-x##, where ##x \in A##. Prove that
    ##\inf A = -\sup(-A)##.

    2. Relevant equations
    Definition:
    Suppose ##S## is an ordered set, ##E\subset S##, and ##E## is bounded above. Suppose there exists an ##\alpha \in S## with the following properties:
    (i) ##\alpha## is an upper bound of ##E##.
    (ii) If ##\gamma < \alpha## then ##\gamma## is not an upper bound of ##E##.
    Then ##\alpha## is called the supremum of ##E## and we write ##\alpha = \sup E##.
    (Equivalently for infimum)

    3. The attempt at a solution
    From the definition of supremum ##\exists \alpha > y, \forall y \in -A## or equivalently ##\exists \alpha > -x, \forall x \in A##.
    Then ##-\alpha < x, \forall x \in A##, hence ##-\alpha = -\sup(-A)## is a lower bound of ##A##.

    It's left to show that if ##\gamma > -\alpha## then ##\gamma## is not an lower bound of ##A##.
    Suppose ##\gamma## is a lower bound of ##A## with ##\gamma > -\alpha##. Then ##\gamma < x \forall x\in A## or equivalently ##-\gamma > -x \forall x \in A##. But this means that ##\gamma## is an upper bound of ##-A## and since ##\sup(-A) = \alpha## we have that ##-\gamma \ge \alpha## or equivalently ##\gamma \le -\alpha## a contradiction! Hence ##\inf A = -\sup(-A)##.

    Is the above correct? Anything I could do to improve it? I'm quite new to proofs so I'm not sure if I'm doing this right.
     
  2. jcsd
  3. Apr 11, 2016 #2

    Samy_A

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    Maybe I'm nitpicking, but it seems that you set ##\alpha=-\sup(-A)##, and then prove that ##-\alpha=\inf A##.
    Shouldn't it be the other way around? You are given that ##A## is bounded below, meaning ##\inf A## exists. Set ##\alpha=\inf A## and prove that ##\alpha = -\sup(-A)##.
    Or, you could do it as you did, but then first prove that if ##A## has a lower bound, ##-A## has an upper bound.
     
  4. Apr 11, 2016 #3
    Good point! I'm going to go with trying to prove that ##-A## have an upper bound. It seemed easier to start on the more complicated side.
    ##A## is bounded below, that is ##\exists \xi \in \mathbf R## such that ##\xi < x, \forall x\in A##. But this means that ##-\xi > -x, \forall x\in A## that is ##-\xi > y, \forall y \in -A##, hence ##-A## is bounded above since ##-\xi \in \mathbf R##.

    If I put that in before the start of my earlier proof, would that do it?

    Edit: I'm also using the proposition that
    If ##x<0## and ##y<z## then ##xy>xz##
    And others, but I guess I don't have to write thing like this out for every step.
     
  5. Apr 11, 2016 #4

    Samy_A

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    Looks fine.
     
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