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Completeness axiom

  1. Oct 28, 2010 #1
    1. The problem statement, all variables and given/known data

    let A be a set of all positive rational number such that [itex]p^2<2[/itex]

    B be a set of all positive rational number such that [itex]p^2>2[/itex]

    2. Relevant equations

    n/a

    3. The attempt at a solution

    Set A is clearly non empty, and is a subset of real number, anyway i can choose 3 is upperbound, therefore upperbound exist, so by completeness axiom, supremum exist.

    But the book here said

    "Set A is bounded above, in fact every element in B a the upperbound of A. Since B has no smallest element, A has no least upper bound/ supremum in Q."

    i'm really sure i'm not wrong. But am i wrong?

    p/s; i just realised that this book define least-upper-bound property(more general case from completeness axiom), and also above example are the counterexample that proves Q does not have least-upper-bound property(follows from what the book have shown, not mine).

    But aren't this contradicting the completeness axiom?

    since Q is a subset of R, and any non-empty subset of Q that bounded from above has supremum(from completeness axiom), therefore Q has the least-upper-bound property.

    help, where i gone wrong T_T
     
    Last edited: Oct 28, 2010
  2. jcsd
  3. Oct 28, 2010 #2

    fzero

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    The completeness axiom applies to the reals, there is no contradiction. As you say, the set A demonstrates that Q does not have the least-upper-bound property, so Q is not complete. The completeness axiom guarantees that a subset of Q has a least-upper-bound in R. However that least-upper-bound is not in Q.
     
  4. Oct 28, 2010 #3
    AAAAAAAAAAAAAAAAAAA in R, not Q

    Thank YOUUUUUUU, i'm so stupid
     
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