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Thanks!

BiP

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- Thread starter Bipolarity
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- #1

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Thanks!

BiP

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- #3

jgens

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- For each element x in F let A
_{x}be the collection of rationals in F that are less than x. - Define f(x) = sup A
_{x}where the supremum is taken in**R**. We can do this because of the identification I mentioned before.

- #4

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Q. So we have a way of identifying the rationals in F with the rationals inR. Then you can map this guy into the reals as follows:

So now we have a map f:F→

- For each element x in F let A
_{x}be the collection of rationals in F that are less than x.- Define f(x) = sup A
_{x}where the supremum is taken inR. We can do this because of the identification I mentioned before.Rand 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.

I see! Thanks!

What definition of completeness are you using?

BiP

- #5

jgens

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Order-completeness. Metric-complete ordered fields are actually not unique.

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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]

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

http://math.caltech.edu/~ma108a/defreals.pdf [Broken]

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