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Lower bound proof/definition

  1. Aug 22, 2007 #1
    in my book this is called the lower bound but it implies that it might be called the greatest lower bound elsewhere.

    lower bound: some quantity m such that no member of a set is less than m but there is always one less than m + [itex]\epsilon[/itex]

    definition using Dedekind section

    there are quantities a such that no member of a set is less than any a. there are quantities b such that some members of a set are less than any b. all a(s) are less than all b(s)[anyone have a problem with this???]. and all quantities of the same dimension are either a(s) or b(s). a(s) define the L class and b(s) define the R class. now make a partition, m, that has to be in L. for if it were in R then there would be some member of the set, k, less than it and and there would be no a(s) between m and k and hence it wouldn't partition correctly. m is then the greatest lower bound. m + [itex]\epsilon[/itex] is in R which by prior definition must have a member less than it.

    the part in brackets i'm not sure about specifically besides not being sure about the whole thing, despite the fact that i basically followed the Dedekind proof/defition for least upper bound. this is the first mathematical proof by argument i have ever done and boy was it more abstract than any physics problem i've ever done.
     
    Last edited: Aug 22, 2007
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  3. Aug 23, 2007 #2

    HallsofIvy

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    You are right. If your book really does define "lower bound" that way, it is very peculiar. The definition of "lower bound" I have always seen is "a number that is less than or equal to any member of the set". If there is always a a value less than [itex]m+\epsilon[/itex] there cannot be any number larger than m which is also an lower bound and so m is the greatest lower bound. A set can have an infinite number of lower bounds but only one greatest lower bounmd.

    What, exactly, are you trying to prove? That if a set has a lower bound, then it must have a greatest lower bound? I would not "copy" the proof of the least upper bound. I would start with a set having a lower bound, multiply everything by -1 and then USE the least upper bound property.
     
  4. Aug 23, 2007 #3
    i was trying to define the greatest lower bound using a dedekind cut. in writing this definition i was guided by the definition of the least upper bound in the book
     
  5. Aug 23, 2007 #4

    HallsofIvy

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    Actually, I find your use of "L" and "Q" a bit peculiar. I learned the definition of Dedekind cut (from Rudin) as a set of rational numbers satisfying:
    1) It is non-empty
    2) It does not contain all rational numbers
    3) If a< b and b is in the cut, then so is a
    4) It has no largest member

    In other words, what I would call a Dedekind cut corresponds only to your set "L" which is guarenteed to be non-empty by 1. Your set "R" is the complement of L (in the rational numbers).
    The proof that a set of real numbers with an upper bound has a least upper bound consists of taking the union of all dedekind cuts (your "L" sets) in the set and showing that that is itself a dedekind cut and is the least upper bound.
    To find the greatest lower bound of a set having a lower bound, you probably want to look at the union of all the "R" sets. Take its complement (in the rational numbers) and call that "L". Does that make a cut in your sense? Is it the greatest lower bound>?
     
  6. Aug 23, 2007 #5
    I learned the definition of Dekekind cut as follows: A pair of subsets [tex] A,B [/tex] of [tex] \mathbb{Q} [/tex] such that (i) [tex] A \cup B = \mathbb{Q}, A \neq \emptyset, B \neq \emptyset, A \cap B = \emptyset [/tex], (ii) if [tex] a \in A [/tex] and [tex] b \in B [/tex] then [tex] a < b [/tex] and (iii) [tex] A [/tex] contains no largest element. Denote the cut by [tex] x = A|B [/tex].
     
  7. Aug 24, 2007 #6

    HallsofIvy

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    Basically, the same, then. Your "A" is, as I said, what I would consider "the cut" and your "B" is its complement in the rational numbers.
     
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