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

  • Thread starter jmac85
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  • #1
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Homework Statement



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)

Homework 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


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)
 

Answers and Replies

  • #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.
 
  • #3
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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.
 
  • #4
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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.
 
  • #5
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Thank you all! I'll be working on the problem again tonight, this time with some new things to try!
 

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