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Defining Cuts in the Rationals

  1. Dec 2, 2013 #1
    I'm having some trouble understanding why cuts are defined with property 3 below:

    "A cut in Q is a pair of subsets A, and B of Q such that

    (a) [itex]A\cup B =Q[/itex], [itex]A\ne ∅[/itex], [itex]B \ne ∅ [/itex], [itex]A\cap B = ∅ [/itex]

    (b) If [itex] a \in A [/itex] and [itex] b \in B [/itex], then [itex] a < b [/itex]

    (c) A contains no largest element." (Pugh, 2001)

    I'm not quite sure why property three must be in place. Why does A have "no largest element" or I guess, what does it mean by "no largest element" in this context?
     
  2. jcsd
  3. Dec 2, 2013 #2
    It means that there must not be an ##a\in A## such that ##a\geq x## for all ##x\in A##. Equivalent to this is demanding that for each ##a\in A##, there is a ##b\in A## such that ##a<b##.

    Why do we demand this? Well, the intuition is the following. With any cut ##(A,B)## we will define a real number. Intuitively, the number associated with the cut ##(A,B)## will be ##\textrm{sup}(A)##. For example, if you take

    [tex]A = \{x\in \mathbb{Q}~\vert~x^2<2\}~\text{and}~B=\mathbb{Q}\setminus A[/tex]

    then this will correspond to the real number ##\sqrt{2}##.

    But this leaves us with some technical problem if (3) were not there. Indeed, if we only demand (1) and (2) then both

    [tex]A = \{x\in \mathbb{Q}~\vert~x<0\}~\text{and}~B=\mathbb{Q}\setminus A[/tex]
    and
    [tex]A^\prime = \{x\in \mathbb{q}~\vert~x\leq 0\}~\text{and}~B^\prime=\mathbb{Q}\setminus A[/tex]

    are valid cuts. But both of these correspond to ##\textrm{sup}(A) = \textrm{sup}(A^\prime) =0##. So both cuts would define the same number. This is an unwanted situation. The goal of (3) is to eliminate ##A^\prime## from being a valid cut.
     
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