Is the Real Field the Only Complete Ordered Field?

  • Thread starter Thread starter Bipolarity
  • Start date Start date
  • Tags Tags
    Field
Click For Summary
The discussion centers on the uniqueness of the real field as the only complete ordered field under order-preserving isomorphism. A proof sketch is provided, noting that any complete ordered field contains a copy of the rationals, allowing for a mapping to the reals. The mapping is defined using the supremum of rational collections in the field, demonstrating an order-preserving isomorphism. Additionally, the distinction between order-completeness and metric-completeness is highlighted, with the latter not guaranteeing uniqueness. The conversation concludes with a reference to a resource discussing both existence and uniqueness of complete ordered fields.
Bipolarity
Messages
773
Reaction score
2
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
 
Physics news on Phys.org
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.
 
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.
 
jgens said:
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.

I see! Thanks!
What definition of completeness are you using?

BiP
 
Order-completeness. Metric-complete ordered fields are actually not unique.
 
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
 
Last edited by a moderator:
We all know the definition of n-dimensional topological manifold uses open sets and homeomorphisms onto the image as open set in ##\mathbb R^n##. It should be possible to reformulate the definition of n-dimensional topological manifold using closed sets on the manifold's topology and on ##\mathbb R^n## ? I'm positive for this. Perhaps the definition of smooth manifold would be problematic, though.

Similar threads

  • · Replies 30 ·
2
Replies
30
Views
3K
Replies
2
Views
2K
Replies
2
Views
1K
Replies
3
Views
2K
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 27 ·
Replies
27
Views
2K