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Least Upper bounds proof

  1. Sep 20, 2009 #1
    1. The problem statement, all variables and given/known data
    Let A contained in R be a set of real numbers. For c in R define set cA as
    cA: {x in R|x=ca for some a in A}

    a. prove that if c is greater than or equal to zero, then cA is bounded above and sup(cA)=cSup(A).

    b. prove that if c is less than zero, then cA is bounded below and that the inf(cA)=csup(A)

    2. Relevant equations



    3. The attempt at a solution

    a. Assume A contained in R is bounded above and let c in R be greater than or equal to zero. Since A is bounded above, that means for all a in A, there exists a b in R so that:
    a is less then or equal to b for all a in A.
    Multiplying by c yields:
    ca is less than or equal to cb, where ca is in cA and cb is in R.
    Therefore there exists an element r in R so that ca is less than or equal r for all ca in CA.
    Meaning, that cA is bounded above.

    Since cA is bounded above, that means by the completeness axiom, that cA has a least upper bound, call it Sup(cA). Where for any epsilon greater than zero,
    sup(cA) - epsilon is less than ca for all ca in cA.

    Similarly, since A is bounded above, that means that A has a least upper bound, call it Sup(A). Where for any epsilon greater than zero,
    sup(A)-epsilon is less than a
    multiplying by c yields:
    cSup(A) -epsilon is less than ca.

    This is where I get stuck, I don't know how to show that cSup(A) is equal to Sup(A). Any help/hints would be appreciated!!
     
  2. jcsd
  3. Sep 20, 2009 #2

    Office_Shredder

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    I'm assuming the question statement includes that A is bounded in it?

    I noticed an error in your proof attempt:

    This is a false statement.
    Fortunately for you, you don't really need anything similar to this for your proof... you demonstrated cA is bounded above quite nicely, which implies the existence of sup(cA).

    We also know sup(A) exists by assumption. So the question is to prove sup(cA) = csup(A).

    I would do this by showing sup(cA) <= csup(A) and csup(A) <= sup(cA). Often a contradiction argument is best.

    Suppose cSup(A) < Sup(cA). Then it must be there exists an element of a such that ca > cSup(A) (otherwise cSup(A) is an upperbound for cA, and hence is greater than or equal to Sup(cA)). But Sup(A) >= a by definition of Sup(A), so cSup(A) >= ca. That's a contradiction, so it must be cSup(A) <= Sup(cA).

    Try proving the other direction
     
  4. Sep 20, 2009 #3
    So for the other direction of this proof would I just start off by assuming
    sup(cA) < cSup(A)?
     
  5. Sep 20, 2009 #4

    Office_Shredder

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    That's right
     
  6. Sep 20, 2009 #5
    How does this look?

    Assume sup(cA) is less than csup(A). This means there must exist an element of a so that:
    c(a) is greater than sup(cA) and
    ca is greater than sup(cA).

    However, sup(cA) bounds cA above, and by definition that means, for all ca in cA
    ca is less than or equal to sup(cA). Therefore the statement that ca is greater than sup(cA) is a contradiction. Therefore,
    sup(cA) is greater than or equal to csup(A)

    **In the previous part it was proven that cSup(A) is greater than or equal to sup(CA), and now we just proved that sup(cA) is greater than or equal to csup(A), which means:

    csup(A) is greater than or equal to sup(cA) is greater than or equal to csup(A)
    which implies that csup(A) is equal to sup(cA)
     
  7. Sep 20, 2009 #6

    Office_Shredder

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    Why?
     
  8. Sep 20, 2009 #7
    because if no such element a existed where c(a) is greater than sup(cA), then the statement
    sup(cA) is less than or equal to csup(A) would be incorrect and either sup(cA) would be equal to or greater than or equal to csup(A)
     
  9. Sep 20, 2009 #8

    Office_Shredder

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    You still haven't shown WHY though. You just said that it's true.

    cSup(A) is (maybe) strictly larger than any number of the form ca, so why can't it be cSup(A)>sup(cA)> ca for every element a in A?
     
  10. Sep 20, 2009 #9
    Is it because if there was no element c(a) greater than sup(cA) then sup(cA) would be an upper bound for cSup(A) and therfore the inequality would be incorrect?
     
  11. Sep 20, 2009 #10

    Office_Shredder

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    I picked to do the direction which had an easier argument intentionally. Trying to directly copy the argument I posted will not lead you to a correct answer.

    If cSup(A)>Sup(cA), how can you PROVE there is a number of the form ca such that ca>Sup(cA)? Hint: You have to use the definition of Sup(A) to demonstrate that elements of the form ca get arbitrarily close to cSup(A)
     
  12. Sep 20, 2009 #11
    well, the definition of Sup(A) is that for every epsilon greater than zero:
    sup(A)-epsilon<a for some a in A.
    I said in my original post that
    csup(A)-epsilon<ca for some ca in cA
    perhaps i'd have to define what epsilon is equal to in order to prove this, but I don't know what to pick.
     
  13. Sep 20, 2009 #12
    Ok i think I got this now:

    Proof by contradiction:
    assume a is greater than supA
    By definition of an upper bound, a is less than or equal to supA for all a in A. Therefore the statement, a is greater than supA is a contradiction. Which means a is less than or equal to supA

    now assume a is less than supA
    Let n in N where there exists an a in A so that
    a+(1/n)>supA.
    Now, define n as (1/(supA-a)). using substitution we get:
    a+(1/1/supA-a)>supA
    which is equivalent to:
    a+supA-a>supA
    which implies that supA>supA, which is a contradiction, because something cannot be greater than itself. Therefore
    a is greater than or equal to supA

    Since we proved that a is less than or equal to supA AND a is greater than or equal to supA. that means that a equals supA and furthermore, ca=csupA

    **so if i'm assuming csup(A)>sup(cA), then i can now assume that there exists a a in A so that ca-csupA and therefore ca>sup(cA)
     
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