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Homework Help: Question About Spivak Proof Of Bounds

  1. Aug 31, 2010 #1
    1. The problem statement, all variables and given/known data

    Let A and B be two nonempty sets of numbers which are bounded above, and let A+B denote the set of all numbers x+y with x in A and y in B. Prove that sup(A+B) = sup(A) + sup(B).

    Hint: The inequality sup(A+B) <= sup(A) + sup(B) is easy. To prove that sup(A) + sup(B)<= sup(A+B) it suffices to prove that sup(A) + sup(B) <= sup(A+B) + E for all E > 0; begin by choosing x in A and y in B with sup(A) - x < E/2 and sup(B) - y < E/2.

    2. Relevant equations


    3. The attempt at a solution

    My attempt was not this complicated, so it's pretty surely wrong, yet I'm having trouble seeing where.

    Since A is a nonempty set that is bounded above, then it must have a least upper bound. Call this least upper bound x0. Then sup(A) = x0, and x0 is the greatest possible value for x. Similarly, since B is a nonempty set that is bounded above, then it must have a least upper bound. Call this least upper bound y0. Then sup(B) = y0 and y0 is the greatest possible value for y. Thus sup(A) + sup(B) = x0 + y0.

    sup(A+B) is the least upper bound for A+B, and therefore occurs where A+B is its highest. Then the least upper bound for A+B is wherever x+y is highest. Since we already determined that x0 is the highest x and y0 is the highest y, then x0 + y0 is the highest x+y and thus sup(A+B) = x0 + y0.

    Therefore, sup(A+B) = x0 + y0 = sup(A) + sup(B).

    Also, I don't see how in Spivak's hint it's possible for him to simply pick an x in A such that sup(A) - x < E/2 if E>0. What if A is just the set {...-3, -2, -1, 0}? Then sup(A) - x >= 1 unless x is sup(A). And if E is, say, 1, then the inequality seems to fail. What's up, here?

    Thank you.
  2. jcsd
  3. Aug 31, 2010 #2
  4. Aug 31, 2010 #3
    I think I understand your proof, and I think that's helped me see what I may be misunderstanding in Spivak's proof.

    I see your statement (in your proof) that [tex]\alpha = inf(A) + inf(B) + \epsilon[/tex] and understand that, since [tex]\alpha[/tex] > inf(A) + inf(B) there must be some [tex]\epsilon[/tex] for which this is true. What throws me off about Spivak's proof is that he seems to assert that it is possible to prove that sup(A) - x < E/2 can be proved for any E, which seems wrong to me, as with my previous example.
  5. Aug 31, 2010 #4


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    x is allowed to be sup(A) if that happens to be in the set.

    By definition you can find elements of A arbitrarily close to sup(A). Instead of saying pick one within epsilon of sup(A), he just picked one within epsilon/2
  6. Aug 31, 2010 #5
    I understand how one could do that if the set was, say, all real numbers from x to x + 5. But what if the set is {1, 2, 3}? Then how is it possible to find and element that is arbitrarily close to A? Isn't the closest one can get 1?

    Or are you saying that one could at that point allow x to be 3, so that sup(A) = x and therefore sup(A) - x = 0 < E/2 for any E?
  7. Aug 31, 2010 #6


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    That's exactly right
  8. Aug 31, 2010 #7
    Ah, then I think I understand Spivak's proof now.

    But can someone point out the flaw in mine?
  9. Aug 31, 2010 #8
    The problem with your proof is that you assumed sup(A) and sup(B) are in the set A and B respectively. You concluded that sup(A) is the highest x from A but this is not always true. The supremum of a set need not belong to the set.
  10. Aug 31, 2010 #9
    What you actually showed is that [tex](A+B)\leq sup(A)+sup(B)[/tex]. Basically, you establised an upper bound for (A+B) and you are yet to show that [tex]sup(A+B)= sup(A)+sup(B)[/tex]. So there is nothing "wrong" with your approach apart from that little mistake and the fact that it is not yet complete.

    I will suggest you finish it by arguing that there is no upper bound smaller than sup(A) + sup(B).
    Last edited: Aug 31, 2010
  11. Sep 1, 2010 #10
    Just to make sure, you're referring to sets like {-(1/n) : n E N}, which has a least upper bound of 0, but 0 is not in the set?
  12. Sep 1, 2010 #11
    Yes, sets like that one and many others.
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