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Dedekind Cuts and Sequences

  1. Dec 25, 2009 #1


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    1. The problem statement, all variables and given/known data

    Prove that every non-decreasing, bounded sequence of rational numbers converges to some real number using Dedekind cuts.

    2. Relevant equations

    A real number is a set [itex]\alpha[/itex], of rational numbers, with the following four properties:
    • If [itex]x \in \alpha[/itex] and [itex]y[/itex] is a rational number with [itex]y < x[/itex], then [itex]y \in \alpha[/itex].
    • [itex]\alpha \neq \emptyset[/itex]
    • [itex]\alpha \neq \mathbb{Q}[/itex]
    • There is no greatest element in [itex]\alpha[/itex]; in other words, if [itex]x \in \alpha[/itex], then there is some [itex]y \in \alpha[/itex] with [itex]y > x[/itex]

    3. The attempt at a solution

    This seems to be a straight forward proof, but I can't seem to complete it satisfactorily; any advice and/or suggestions are appreciated. Well, here goes nothing . . .

    Let [itex]\{a_n\}[/itex] be a non-decreasing, bounded sequence of rational numbers and define the set [itex]\alpha \subset \mathbb{Q}[/itex] such that [itex]\alpha = \{x \in \mathbb{Q}: x < a_n \;\mathrm{where}\; n \in \mathbb{N}\}[/itex]. Now, to prove that [itex]\alpha[/itex] is a real number, we need only verify that our four conditions hold:
    • Suppose [itex]x\in\alpha[/itex]. Then [itex]x < a_n[/itex] for some [itex]n \in \mathbb{N}[/itex]. Now, let [itex]y < x[/itex], then [itex]y < a_n[/itex] and consequently [itex]y\in\alpha[/itex]
    • Clearly [itex]\alpha \neq \emptyset[/itex]
    • Since [itex]\{a_n\}[/itex] is bounded above, this means that there must be some rational number [itex]x > a_n[/itex] for all [itex]n\in\mathbb{N}[/itex]. Thus [itex]x \notin \alpha[/itex] and consequently [itex]\alpha \neq \mathbb{Q}[/itex]
    • Suppose [itex]x\in\alpha[/itex]. Then, for some [itex]n\in\mathbb{N}[/itex] we have that [itex]x < a_n[/itex]. Since [itex]x,a_n\in\mathbb{Q}[/itex] we know that [itex]\varepsilon = \frac{a_n - x}{2} \in \mathbb{Q}[/itex]. Consequently, [itex]x < x + \varepsilon < a_n[/itex]. Therefore, [itex]\alpha[/itex] contains no greatest element.
    Now, I just need to prove that [itex]\alpha[/itex] is the real number that this sequence converges to. Intuitively, it seems like it should be, but how might I articulate an appropriate argument? Thanks!
    Last edited: Dec 26, 2009
  2. jcsd
  3. Dec 26, 2009 #2


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    First show that every n, [itex]\alpha_n< \alpha[/itex]. Then show that for any [itex]epsilon[/itex], there exist n such that [itex]\alpha- \alpha_n< \epsilon[/itex].
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