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Greatest upper bound analysis proof

  1. Sep 14, 2009 #1
    1. The problem statement, all variables and given/known data
    Use the completeness axiom to prove that to prove that every non-empty subset of real numbers, which is bounded below, has a greatest lower bound.


    2. Relevant equations
    N/A


    3. The attempt at a solution
    Assume A is a nonempty subset of real numbers which is bounded below. Define B as the set of of lower bounds for A, meaning for all b in B, b is less than or equal to all a in A. Since every element in A is greater than every element in B, A is the set of upper bounds for B. According to the completeness axiom, every non-empty set of real numbers that is bounded above has a least upper bound, which would mean that the B is bounded above and SupB is an element in A.

    **This is where i get stuck. I need to show that the SupB=InfA, which would mean that the element is in both sets**

    Any help/tips would be greatly appreciated :)
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 14, 2009 #2
    I think that B being the set of lower bounds for A implies that any [tex]b\in B\le a\in A[/tex] and so [tex]sup(B)\le inf(A)[/tex]. If sup(B) is an element in A as you said, then [tex]inf(A)\le sup(B) [/tex] hence they are equal from:

    [tex]inf(A) \le sup(B)[/tex]

    [tex]sup(B) \le inf(A)[/tex]
     
    Last edited: Sep 14, 2009
  4. Sep 14, 2009 #3
    I don't understand how you got that sup(B) is less than or equal to inf(A)? Just because something is the least upper bound, does not necessarily mean that it is in the set that it is bounding.
     
  5. Sep 14, 2009 #4

    Office_Shredder

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    First, supB is not necessarily an element of A, for example if A=(0,1) then supB=0

    We want to show SupB is the greatest lower bound of A. What does that mean. Ok, by that we mean given any a in A, supB is less than or equal to a. Furthermore, for any e>0, there exists a in A such that |supB-a|<e. The first condition shows that supB is a lower bound, and the second condition that you can have no greater lower bound (since if K is a proposed lower bound and K-supB>0, pick e=(K-supB)/2

    Suppose there exists a in A less than supB. Then by definition of least upper bound: let d=|supB-a|/2. There exists some element in the set of lower bounds, call this b, such that |supB-b|<d. But then b>a. b is supposed to be a lower bound of A, so this is a contradiction. supB is in fact a lower bound of A.

    Now try the second part on your own
     
  6. Sep 14, 2009 #5
    I'm confused, wouldn't you say for any e>0, there exists a b in B such that |supB-b| <e?
     
  7. Sep 14, 2009 #6

    Office_Shredder

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    That's what you need to prove supB is in fact the least upper bound of B. But we're assuming it is (since we have the completeness axiom, and that's the definition of sup). To show supB is the greatest lower bound of A, you have to show that elements of A are arbitrarily close to supB.
     
  8. Sep 14, 2009 #7
    I think I may be getting confused by your wording, but I'm not exactly sure what you are saying. Are you trying to say we are proving that SupB is the least upper bound of A by assuming that it is NOT the least upper bound? I thought by the axiom of completeness, it is assumed that supB is the least upper bound of A.

    I understand what you are saying about needing to show it is arbitrarily close, but I just don't understand what you are using for your proof by contradiction...
     
  9. Sep 14, 2009 #8

    Office_Shredder

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    We have supB is the least upper bound of B. Now we want to prove it is also the greatest lower bound of A. You do that in two steps:
    1) Show it's a lower bound of A.
    2) Show there is no larger lower bound of A.

    The way you do part 1 I demonstrated in my post, using the property that supB is the least upper bound of B. Now you have to do part 2.... I gave one way to start that part, but of course you can use any method you think of
     
  10. Sep 14, 2009 #9
    Ok, this makes a lot more sense now (sorry I'm really tired and a bit frustrated by this proof :/ thanks for being patient with me). This is what i got so far and where i get lost:
    Claim: SupB is the greatest lower bound of A
    *proof by contradiction*
    Assume supB is not a lower bound of A. Meaning, there exists an a in A so that a is less than supB. Since supB is the least upper bound of B, it follows that for every e>0, there exists a b in B so that supB-b>e

    *now I don't understand the whole d=|supB-a|/2 part... where did you get that from??
     
  11. Sep 14, 2009 #10

    Office_Shredder

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    This is just a statement regarding properties of the least upper bound.

    I realized I might be complicating things... what's your classes definition for greatest lower bound?
     
  12. Sep 14, 2009 #11
    We never had a formal definition for it, however the book says it is similar to the definition for the least upper bound, so i'm guessing this would be it:
    A real number s is the greatest lower bound for a set A contained in R if it meets the following two criteria:
    1) s is an lower bound for A
    2) if b is any lower bound for A then s is greater than or equal to b.

    It really isn't saying much though.
     
  13. Sep 14, 2009 #12
    Ok I think I understand the first part without using all the e's and d's and such. Let me know if this is a reasonable way to prove it.
    Claim: Sup(B) is the greatest lower bound of A
    *proof by contradiction*
    1) assume B is not a lower bound of A, meaning there exists an a in A so that a is less than the Sup(B).
    Since the Sup(B) is the least upper bound of B, that means that if a in A is an upper bound of B, then:
    sup(B) is less than or equal to a. Which is a contradiction, because we previously stated that in order for SupB not to be a lower bound for A, that there must be an a in A so that a is less than SupB.
    Therefore, SupB is a lower bound of A.
     
  14. Sep 15, 2009 #13

    Office_Shredder

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    It looks good, except you probably want to justify why a is an upper bound of B.
     
  15. Sep 15, 2009 #14
    Alright, now that I understand the first part, how would I go about doing the second part. I'm guessing this is the part where I need to start using the error estimates. So If SupB is a lower bound of A that means for all e>0 there exists an a in A so that |SupB-a|<e. But wouldn't this also mean for all e>0, there exists a b in B so that |supB-b|<e as well?
     
  16. Sep 15, 2009 #15
    Nevermind, I found another way to prove that it is the greatest lower bound by making a similar argument to the one I used for proving that SupB is a lower bound. Thank you so much for your help, I understand it a lot better now!!!
     
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