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Dedekind cuts

  1. Sep 19, 2004 #1
    Hi I'm having trouble understanding dedekind cut.

    Suppose ScQ is a Dedekind cut.

    i) One might try to define the negative of S as a set

    -S={-s in Q : s in S}

    Explain why this is not a Dedekind cut.

    I have no idea how Dedekind cuts work even after reading over the defn of it 3 or 4 times. Any tips would be great
     
  2. jcsd
  3. Sep 19, 2004 #2

    Hurkyl

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    Well, to begin, can you say the definition of a Dedekind cut?
     
  4. Sep 20, 2004 #3

    HallsofIvy

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    The definition of "Dedekind cut" is usually given in 3 parts. Which of those parts is violated by {-s | s is in S}? (By the way, it is not necssary to say "-s in Q". Since s is in S, s and -s must be in Q.)
     
  5. Sep 21, 2004 #4
    Definition of dedekind cut:

    A set ScQ is a Dedekind set if

    1)S is not 0, S is not Q
    2)r<s in S -> r in S
    3) S doesn't contain it's upper bound.
    ie: if s in S, then there exists r in S with r>s


    I think I just cannot see which one it disobeys, I might confused with the upper bound part. Anyways any thing will help
     
  6. Sep 21, 2004 #5

    Hurkyl

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    Next step: can you give a couple examples of Dedekind cuts?
     
  7. Sep 21, 2004 #6

    mathwonk

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    cat: so your definition of a cut basically says it is all rational numbers equal to or to the left of some real number, does that seem right?
     
  8. Sep 21, 2004 #7
    Nvm I understand the question now, thanks for all your help guys
     
  9. Sep 21, 2004 #8

    HallsofIvy

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    You "understand" the question. Does that mean you can answer it? ;)

    One very simple example of a Dedekind cut is the set of all negative rational numbers. The set of all {-s} where s is in the set of all negative rational numbers is simply the set of all positive integers. That set is not a Dedeking cut because it does not satisfy "if x is in the set and y< x then y is in the set". 1/2 is in the set of all positive rational numbers, 0< 1/2 but 0 is not in the set.
     
  10. Sep 23, 2004 #9

    mathwonk

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    I haven't taught this topic in a long time and just now I keep thinking " there must be a better way!"

    But anyway: suppose we take your suggestion and look at all negatives of elements in our dedekind cut. then that seems to give us the opposite of what we want. so we could then take the complement in the set of rationals of those numbers. now unfortunately that could give us a set containing its lub, so if so we must throw that out. kind of clumsy.


    lets try another way. presumably addition is easier, i.e. to add two dedekind cuts probably you just add all their pairs of elements. so we know that the negative of a dedekind cut X should be the solution of the equation X+Y = 0. so maybe we should just take for -X, where X is a dedekind cut, the set of all rationals y such that for all x in X, x+y is negative.

    What does that give? Shoot, that also gives a set with a lub in it. Ok, last try:

    given a cut X, take the set of all those y in the rationals, such that for each y there exists a positive K, such that for all x in X, x+y is less than -K.

    gee this is awful. it would probably be bettter to just notice that you only need to use this construction to construct the positive reals. after that just construct the negative reals formally as a copy of the positive reals with a minus sign attached.

    i.e. it is unnecessary to carry the cumbersome construction around for the whole process.
     
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