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Open and closed intervals and real numbers

  1. Oct 16, 2010 #1
    1. The problem statement, all variables and given/known data

    Show that:

    Let S be a subset of the real numbers such that S is bounded above and below and
    if some x and y are in S with x not equal to y, then all numbers between x and y are in S.


    then there exist unique numbers a and b in R with a<b such that S is one of the intervals (a,b), [a,b), (a,b], or [a,b].




    2. Relevant equations



    3. The attempt at a solution

    Assume if x and y are elements of S with x not equal to y, then all numbers between x and y are in S and S is bounded above and below.

    Thus there exists a M, N such that M is greater than or equal to the maximal element of S and N is smaller than the minimal element of S. Also all elements between x and y are inside (M,N).
     
    Last edited: Oct 16, 2010
  2. jcsd
  3. Oct 16, 2010 #2

    tiny-tim

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    hi reb659! :smile:

    i think i'd start by proving that there must be a greatest lower bound and a least upper bound, and then call them a and b, and carry on from there. :wink:
     
  4. Oct 16, 2010 #3
    Good idea.

    Isn't it an axiom that if a nonempty subset of R has an upper bound, then it has a least upper bound/sup(S)?
     
  5. Oct 16, 2010 #4

    tiny-tim

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    I don't think it's an axiom, I think it's the Dedekind-completeness theorem:

    A bounded real-valued function has a least upper bound and a greatest lower bound.

    (See Rolle's theorem in the PF Library :wink:)
     
  6. Oct 17, 2010 #5
    So far:

    Since S is bounded above and below, by Dedekind completeness there exists a supremum of S. Call it b. Again by dedekind completeness we can say there exists an infimum of S. Call it b. By definition of sup and inf, a<b. We are left to show a,b are unique and that S is exactly one of the intervals in the OP.

    To show this, can't we consider four simple different cases in which a,b are either in S or outside of it?
     
  7. Oct 18, 2010 #6

    tiny-tim

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    Yup, that's the proof! :smile:
     
  8. Oct 19, 2010 #7
    Yay!

    How exactly does uniqueness follow though? It seems like its trivial to prove.
     
  9. Oct 19, 2010 #8

    tiny-tim

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    as you said, "by Dedekind completeness there exists a supremum of S" …

    there can't be two supremums, can there? :wink:
     
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