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Real field is unique

  1. Oct 14, 2013 #1
    I was looking for a proof of the fact that the real field is the only complete field up to order preserving field isomorphism under field addition and multiplication and the standard linear ordering defined on ℝ. I haven't been able to find a link online. Could someone perhaps provide me with one?

    Thanks!

    BiP
     
  2. jcsd
  3. Oct 14, 2013 #2

    jbunniii

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    I don't know about online proofs, but Spivak's Calculus contains a sketch of a proof of this result in one of the appendices.
     
  4. Oct 14, 2013 #3

    jgens

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    The argument is pretty simple. Let F be such a field and notice that any such field necessarily contains a copy of Q. So we have a way of identifying the rationals in F with the rationals in R. Then you can map this guy into the reals as follows:
    1. For each element x in F let Ax be the collection of rationals in F that are less than x.
    2. Define f(x) = sup Ax where the supremum is taken in R. We can do this because of the identification I mentioned before.
    So now we have a map f:F→R and it is pretty easy to show that it is an order-preserving isomorphism. If this all seems horribly informal to you, then you can make the identifications I made explicit and the argument goes through just the same, I am just way too lazy to do that.
     
  5. Oct 14, 2013 #4
    I see! Thanks!
    What definition of completeness are you using?

    BiP
     
  6. Oct 14, 2013 #5

    jgens

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    Order-completeness. Metric-complete ordered fields are actually not unique.
     
  7. Oct 17, 2013 #6

    mathwonk

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    the uniqueness proof uses the fact the field is archimedean. least upper bound complete fields are automatically archimedean. otherwise you can prove any complete archimedean ordered field is unique.


    here is a link to a discussion of both existence and uniqueness.

    http://math.caltech.edu/~ma108a/defreals.pdf [Broken]
     
    Last edited by a moderator: May 6, 2017
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