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Real Analysis - Prove the following: inf A = -sup(-A)

  1. Sep 21, 2009 #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 is in A. Prove that:
    inf A = -Sup(-A)

    2. Relevant equations

    What should I use as a starting statement? I understand that its true. It makes sense in my head but I can't get it out on paper. l believe this to be right, although I'm not sure if my assumptions are correct or if what I did is actually legal to do. This is from Rudin's Principlese of Mathematical Analysis - Page 22 - Problem 6


    3. The attempt at a solution

    By definition --
    inf A --> x <= y where y is in A
    sup(-A) --> -x >= -y where y is in A

    Thus --
    -sup(-A) --> -(-x >= -y) --> x <= y

    So we see --
    -sup(-A) --> x <= y where y is in A

    Therefore --
    inf A = -sup(-A)
     
  2. jcsd
  3. Sep 21, 2009 #2

    Office_Shredder

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    What you've done is show that the negative of a lower bound of A is an upper bound of -A. But you haven't proven that the negative of the greatest lower bound is the least upper bound of -A.
     
  4. Sep 22, 2009 #3
    You may want to approach the problem by proving these 2 inequalities :

    1) inf(A) <= -sup(-A) , or equivalently, sup(-A) <= -inf(A)

    2) inf(A) >= -sup(-A)

    Combining these 2 inequalities will give you the desired result.
     
  5. Sep 22, 2009 #4
    Start with definition of infimum and supremum.
    Then take sup A where A is a set of real numbers.
    Define a set B = -A. Take inf of B.
     
  6. Sep 22, 2009 #5
    Thank you all! I'll be working on the problem again tonight, this time with some new things to try!
     
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