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Spivak Exercice, Least Upper Bounds Chapter, Proof

  1. Aug 26, 2012 #1
    Hi there!

    I always think whether I am posting this correctly, or this belongs to the homework section. If so, my apologies.

    I am trying to understand the solutions for a problem in Spicak Calculus, 3[itex]^{rd}[/itex] edition.

    #8-13

    The Problem:

    "Let A and B bt two nonempty sets of numbers which are bounded above, and let A+B denote de set of all numbers x+y, with x in A and y in B. Prove that sup(A+B) = supA+supB. Hint: The inequality sup(A+B) ≤ sup A + sup B is easy. Why? To prove that supA + supB ≤ sup(A+B) it suffices to prove that supA + supB ≤ sup(A+B) + ε for all ε >0; begin by choosing x in A, and y in B with supA - x < [itex]\frac{ε}{2}[/itex] and supB -y <[itex]\frac{ε}{2}[/itex]."

    Well, proving sup(A+B) ≤ supA + supB seems to be straightforward, but it is proving that supA + supB ≤ sup(A+B) that bothers me.

    The explanation in the "Answers" chapter of the book, doesn't explain very clearly anything in this exercice. So, I have tried to prove it on my own:

    1 - First of all, why is it sufficient to prove that supA + supB ≤ sup(A+B) + ε for all ε >0?

    supA + supB - sup(A+B) ≤ ε, for all ε > 0.
    Lets denote X = { ε : ε > 0}.
    Then X is bounded below, and it is nonempty. The greatest lower bound is 0. And, since supA + supB - sup(A+B) is also a lower bound for X, we should have
    supA + supB - sup(A+B) ≤ 0 ⇔ supA + supB ≤ sup(A+B).

    2 - How do you prove that supA + supB ≤ sup(A+B) + ε for all ε >0?

    If supA and supB are the greatest upper bounds for the A and B sets, respectly, then, for every ε > 0, there is an x in A , and an y in B such that:

    →supA - x < [itex]\frac{ε}{2}[/itex];
    →supB - y < [itex]\frac{ε}{2}[/itex];

    →supA < [itex]\frac{ε}{2}[/itex] + x;
    →supB < [itex]\frac{ε}{2}[/itex] + y;

    Which means : supA + supB < ε + (x+y) ⇔ (x+y) > supA + supB - ε.. All this means, that for every ε > 0 , there is an x in A and y in B such that (x+y) > supA + supB - ε.

    And finally, since sup(A+B)≥ x+y, for every x in A and y in B, we have for every ε > 0

    sup(A+B) ≥ supA + supB -ε. Which, according to the first point is suffient.

    Well, I feel pretty confident about this. However I would really appreciate if you could check it. I have spent way to much time around this one, and wanted to move on. ;D

    Regards,

    c.teixeira
     
  2. jcsd
  3. Aug 26, 2012 #2

    Bacle2

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    I'm not sure I understand your first question, but if you show, in general, that:

    x≤ y+1/n , for all n (subbing-in 1/n for ε ),

    then it follows that x≤y

    A way of showing this is that if x-y≤ 1/n for all n , then x≤y , by the

    Archimedean Property. In a metric space,two elements cannot be

    indefinitely-close ( using x-y as d(x,y) in ℝ ) , so we can conclude

    x-y≤ 0 , so x≤y .
     
  4. Aug 26, 2012 #3

    micromass

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    Your solution seems fine.
     
  5. Aug 26, 2012 #4

    Bacle2

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    Sorry, I did not read your question carefully. I agree, it seems O.K.
     
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