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Proof of rational density using Dedekind cuts

  1. Aug 7, 2014 #1
    The Problem

    Let [itex]x[/itex] and [itex]y[/itex] be real numbers such that [itex]y<x[/itex], using the Dedekind cut construction of reals prove that there is always a rational [itex]q[/itex] such that [itex]y<q<x [/itex]

    What I've done

    Since I can associate a cut to every real number, let [itex]x^∗[/itex] be the cut associated to [itex]x[/itex] and [itex]y^∗[/itex] the one associated with [itex]y[/itex].
    Since [itex]y<x \implies y^* \subsetneq x^*[/itex] then [itex]\exists q \in \Bbb Q[/itex] such that [itex]q \in x^*[/itex] and [itex]q\not\in y^*[/itex]. Next I associate a cut [itex]q^∗[/itex] to [itex]q[/itex]. Now how can I deduce from there that [itex]q^* \subsetneq x^*[/itex] and [itex]y^* \subsetneq q^*[/itex], thus proving [itex]y<q<x[/itex]

    Any help will be appreciated,
    M.
     
    Last edited: Aug 7, 2014
  2. jcsd
  3. Aug 7, 2014 #2

    HallsofIvy

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    If you are simply asking if that is a valid proof, yes, it looks good to me.
     
  4. Aug 7, 2014 #3
    Thank HallsofIvy for replying.
    The thing is that my professor told me that its not immediate from what I previously wrote that [itex]q^* \subsetneq x^*[/itex] and [itex]y^* \subsetneq q^*[/itex] so I can't conclude solely on my argument [itex] y<q<x[/itex]. But I can't see what I'm missing, I was hoping you guys would guide me towards finding that subtlety to complete the proof.
     
  5. Aug 7, 2014 #4

    jbunniii

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    I'm assuming that in your definition of Dedekind cut, if ##q## is rational then ##q## is NOT an element of ##q^*## but rather is the smallest element of ##\mathbb{Q}\setminus q^*##. If you use the opposite convention, then adjust my argument below appropriately.

    If ##x## and ##y## are irrational, then your proof is fine.

    If ##x## and ##y## are rational, then you can simply choose ##q = (x+y)/2##.

    If ##y## is irrational and ##x## is rational, then your proof is fine: ##q## cannot be ##x## because you chose ##q \in x^*## whereas ##x \not\in x^*##.

    Finally, if ##y## is rational and ##x## is irrational, then your proof does not exclude the possibility that ##q=y##. In other words, ##q## might be the smallest element of ##x^* \setminus y^*##. But if this is the case, note that ##x^*## does not have a largest element, so you can simply replace ##q## with a larger rational in ##x^*##.
     
  6. Aug 8, 2014 #5

    disregardthat

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    Let the real numbers x and y be represented by cuts [itex](X_1,X_2)[/itex] and [itex](Y_1,Y_2)[/itex], where a cut [itex](A,B)[/itex] is a partition of [itex]\mathbb{Q}[/itex] such that any element of A is less than any element of B, and A has no greatest element. To say that [itex]y < x[/itex], is to say that [itex]Y_1 \subset X_1[/itex] by the ordering of cuts. Since this is a proper subset, we may find a rational [itex]p \in X_1 - Y_1[/itex]. But since [itex]X_1[/itex] has no greatest element, we may find another rational [itex]q \in X_1-Y_1[/itex] such that [itex]q>p[/itex]. It is easily seen that its cut [itex](Q_1,Q_2)[/itex] is such that [itex]Y_1 \subset Q_1 \subset X_1[/itex], because if we let the cut [itex](P_1,P_2)[/itex] represent p, we have the following: [itex]Y_1 \subseteq P_1 \subset Q_1 \subset X_1[/itex], where the last inclusion is true because q cannot be the greatest element of [itex]X_1[/itex].
     
  7. Aug 9, 2014 #6
    Oh, now I get it. Thanks for the replies and the help guys, it was very helpful.
     
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