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Complete ordered fields are Archimedean

  1. May 5, 2007 #1

    quasar987

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    1. The problem statement, all variables and given/known data
    I'm troublde by the proof appearing in my book that complete ordered fields are Archimedean. It says, "Suppose F is not Archimedean, i.e. that given an x in F, there are no "integer" N with x < N. And consider the monotone sequence 1,2,3,..." (and then it goes on to show that this sequence, although increasing and bounded by x, does not converge becuz if it did to y, then we would have 1=|n+1-1|<|n+1-y|+|y-n|<2*epsilon ==><== as soon as epsilon < 1/2)

    But if the list of integers ends abruptly as hypothesized, the sequence is ill defined, is it not? When one writes "1,2,3,...", one means that the actual sequence is the map from [itex]\mathbb{N}[/itex] to F that sends 1 to 1, 2 to 2, 3 to 3, etc. But what does the map send N to if N is not in F?
     
    Last edited: May 5, 2007
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  3. May 5, 2007 #2

    Hurkyl

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    Where was that hypothesized?
     
  4. May 5, 2007 #3

    quasar987

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    Well, there is no integer greater than x. So there is a largest integer.
     
  5. May 5, 2007 #4

    Hurkyl

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    How does that follow?
     
  6. May 5, 2007 #5

    quasar987

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    Well, we know that 1 and x are in F. x is greater than 1. There is no integer greater than x. Therefor, either 1 is the greatest integer, or there exists another one btw 1 and x. In any case, there is a greatest integer.

    At least this is how I think about all this. What is wrong?
     
  7. May 5, 2007 #6

    Hurkyl

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    The conclusion does not follow from anything you've said. How do you conclude the existence of a greatest integer? Where do you prove that there exists an integer n such that, for every other integer m, m < n?

    You argued that either 1 is the largest integer, or there exists an integer between 1 and x. Why does that imply there is a largest integer?


    Try rigorously proving your claim -- hopefully you'll see that there's a problem, and why anyone would have ever thought to state the Archimedian axiom.


    If you want an explicit example of a nonarchmedian ordered ring, consider the ring of all polynomials over the reals with the reverse dictionary ordering: if you write down the coefficients (ending with the constant one), the one that would come first in the dictionary is the smaller one. e.g.

    1 + 3 x^2 < 2 x + x^3, because if we write the coefficients:

    ... 0 0 3 0 1 | for 1 + 3 x^2
    ... 0 1 0 2 0 | for 2x + x^3

    and 0 3 0 1 comes before 1 0 2 0 in the dictionary order.

    Observe that x > n for all integers n. This ordering can be extended to the field of rational functions over the reals.

    I suppose a more analytic way to describe this ordering is:

    f < g iff there exists a C such that for all x > C we have f(x) < g(x).
     
    Last edited: May 5, 2007
  8. May 5, 2007 #7

    quasar987

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    I understand.
     
  9. Oct 10, 2009 #8
    hi there,

    in this proof i didnt understand why |n+1-y| < epsilon. I think that If |n-y|< epsilon then |n+1-y|<epsilon+1.

    Where am I wrong?

    Also, I didnt understand why the sequence is bounded above by x, isnt it unbounded? How can we bound 1,2,3 ... ?
     
    Last edited: Oct 10, 2009
  10. Oct 10, 2009 #9
    That's what happens if the field is non-Archimedean.
     
  11. Oct 11, 2009 #10
    I got it, thanks a lot.

    Can someone explain why |n+1-y| < epsilon when |n-y| < epsilon ?
     
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