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        riddles >> putnam exam (pure math) >> Complete Ordered Fields
        
(Message started by: Icarus on Oct 8th, 2005, 7:19am)
Title: Complete Ordered Fields
Post by Icarus on Oct 8th, 2005, 7:19am
The most easily stated definition of the real numbers is that it is the minimal topologically complete ordered field.

But is the "minimal" necessary? Are there any other complete ordered fields?
Title: Re: Complete Ordered Fields
Post by Neelesh on Oct 8th, 2005, 9:23am
By "topologically complete" I guess we mean that every subset which has an upper bound also has a least upper bound.

In that case, R is the only Complete Ordered Field, since all Complete Ordered Fields are order-isomorphic (ie. only the names change)

I guess that there a proof which uses Cauchy sequences and some other things. I don't know the complete proof. But this link gives some steps.

http://www.ms.uky.edu/~ken/ma570/lectures/lecture1/html/real.htm
Title: Re: Complete Ordered Fields
Post by Icarus on Oct 8th, 2005, 6:11pm
Completeness is definable for metric spaces and topological semigroups, and for any other topological space in which you can establish a uniformity condition on the open sets - allowing you to pick open sets of the same size around every point in the space. For metric spaces, the sized sets are the open balls of the same radius around each point. For topological semigroups, they are translations of each other.

Completeness means that every Cauchy sequence converges. A sequence {xn} is Cauchy if for any size of open set, there is some set A of that size, and an integer N such that xn is in A for all n > N.

Whenever you have a linearly ordered set, and such a uniformity structure upon its order-topology, the space is complete if and only if it satisfies the supremum property (every set bounded above has a least upper bound, a.k.a. a supremum).


Your link stops short of the what I am really asking for, though (it states the result, but does not give a hint of how to prove the part I am concerned with).

Following the steps it outlines proves that any complete ordered field must contain a copy of the Real numbers. But nothing is given towards showing that such fields cannot be extensions of the Reals.



So my challenge is: Prove that the real numbers has no complete ordered extension fields (without looking it up).
Title: Re: Complete Ordered Fields
Post by Icarus on Oct 9th, 2005, 11:18am

on 10/08/05 at 18:11:11, Icarus wrote:
Whenever you have a linearly ordered set, and such a uniformity structure upon its order-topology, the space is complete if and only if it satisfies the supremum property (every set bounded above has a least upper bound, a.k.a. a supremum).


My apologies. This isn't true. For example, take the real numbers with 0 removed (R*. Define a metric d on R* by

d(x, y) = |1/x - 1/y| if xy > 0,  d(x, y) = 1 if xy < 0.

It is easily checked that d is a metric, and that it provides the standard topology, which is also the order topology, on R*.

But, sequences converging to 0 (were it there) are not Cauchy under this metric. Therefore R* with this metric is complete, even though it does not satisfy the supremum property.

(This also provides a nice example that completeness is not actually a topological property, but rather a property of the metric, or other uniform structure.)
Title: Re: Complete Ordered Fields
Post by Eigenray on Oct 9th, 2005, 2:49pm
A field is called Dedekind complete if every non-empty subset with an upper bound has a least upper bound.  It is topologically complete if there exists a metric (inducing its topology) in which all Cauchy sequences converge.

The reals are the only Dedekind complete ordered field.

Observe that if F is Dedekind complete, it must be Archimedean: for any x in F, there's some positive integer n such that |x| < n.  For otherwise, |x| would be an upper bound for {1,2,3,...}, a set which plainly has no least upper bound (|x|-1 is also an upper bound).

More is true: The reals are the only (topologically) complete Archimedean ordered field.

For an example of a non-Archimedean field, consider R((x)), the field of (formal) Laurent series, where we say f<g if f(x)<g(x) in some interval (0,d).  Thus x < 1/n, and 1/x > n, for any n.  Let ord(f) be the order of f at 0 (the exponent of the first non-zero term), and set |f|=exp(-ord f), d(f,g)=|f-g|.  Note that |fn-fm|->0 iff for each k, the coefficient of xk in fn is eventually constant.  In particular, R((x)) is complete in this metric.
Title: Re: Complete Ordered Fields
Post by Eigenray on Oct 9th, 2005, 5:25pm

on 10/09/05 at 11:18:07, Icarus wrote:
But, sequences converging to 0 (were it there) are not Cauchy under this metric. Therefore R* with this metric is complete, even though it does not satisfy the supremum property.

What about sequences converging to infinity?
But d(x,y)=|x-y|+|1/x-1/y| should work, as well as d(x,y)=|log x - log y| (since it is the pullback of the standard metric under log : R* -> R x {0,[pi]i}).
Title: Re: Complete Ordered Fields
Post by Icarus on Oct 9th, 2005, 8:25pm
True - I was so busy thinking about what is going on at zero that I forgot to think about what happened at infinity.

Of course, the problem is easily solved by changing the definition of the metric to the normal one at some intermediate point. But I have thought of a better way of expressing the same example:

Consider the space {0, 1} x R under the lexigraphic order ( (x,y) < (a,b) iff x < a or (x = a and y < b) ), and give it the standard metric on R2 (of which it is a subspace consisting of two vertical lines). The metric topology and the order topology agree, but the space is metrically complete, but not Dedekind complete.
_________________________________________

So now the question is: Does there exist a topologically complete (it need not be metric - completeness in this case is defined by the topological algebra), non-Archimedean, ordered field?
Title: Re: Complete Ordered Fields
Post by Icarus on Oct 13th, 2005, 3:06pm
Not wanting this thread to died unsolved:

The answer to the question is "yes": The is an ordered extension field of the reals which is topologically complete.

In fact (BIG hint), [hide]the simplest such field is a finite dimensional extension.[/hide]
Title: Re: Complete Ordered Fields
Post by Eigenray on Oct 13th, 2005, 7:46pm

on 10/13/05 at 15:06:53, Icarus wrote:
Not wanting this thread to died unsolved:

Doesn't R((x)) do the trick?


Quote:
In fact (BIG hint), [hide]the simplest such field is a finite dimensional extension.[/hide]

There's only one finite extension of R, and it isn't orderable.
Title: Re: Complete Ordered Fields
Post by Icarus on Oct 14th, 2005, 7:36pm
Are you sure of that?
Title: Re: Complete Ordered Fields
Post by Eigenray on Oct 14th, 2005, 8:58pm

on 10/14/05 at 19:36:34, Icarus wrote:
Are you sure of that?

Suppose F is a finite (non-trivial) extension of R.  Pick x in F \ R, with minimal polynomial f over R.  Since C is algebraically closed, f has a root z in C (and z is not in R since f is irreducible).  Then R(x) is isomorphic to R(z)=C, hence algebraically closed.  Since F is finite over R(x), we therefore have F=R(x) isomorphic to C.

(Note: In fact, R is also the only subfield of finite index in C, but the proof of this is much harder.)

But C is not orderable (formally real).  For, if i>0, then -1=ii > 0; similarly if i<0, then -i>0, and -1=(-i)(-i)>0.  But since -1 > 0, adding 1 gives 0>1, and multiplying by -1 gives 1>0, a contradiction.

Thus the reals are a real closed field, i.e., a maximal (wrt algebraic extensions) formally real field.
Title: Re: Complete Ordered Fields
Post by Icarus on Oct 15th, 2005, 2:34pm
You're right. :-[ I can't believe I did it, but I was thinking of the dictionary order on C = R2, but when I checked to see if it was a field order, I used a coordinate by coordinate product! (a + bi)(c+di) "=" (ab +cdi) :-[ :-[ :-[
Title: Re: Complete Ordered Fields
Post by Icarus on Oct 26th, 2005, 6:35pm

on 10/09/05 at 14:49:00, Eigenray wrote:
For an example of a non-Archimedean field, consider R((x)), the field of (formal) Laurent series, where we say f<g if f(x)<g(x) in some interval (0,d).  Thus x < 1/n, and 1/x > n, for any n.  Let ord(f) be the order of f at 0 (the exponent of the first non-zero term), and set |f|=exp(-ord f), d(f,g)=|f-g|.  Note that |fn-fm|->0 iff for each k, the coefficient of xk in fn is eventually constant.  In particular, R((x)) is complete in this metric.


When you posted this, I somehow failed to notice it. And when you later refered to R((x)), I thought you were refering to the field I will describe below (which still might be true). Sorry about that. Added on top of my stupid mistake about C and the mistaken comment about cauchy completeness implying dedekind completeness, I don't have an impressive track record for this thread!

I am not entirely sure of how to interpret your example. I've always considered Laurent series to be bi-infinite, but your definition of ord(f) requires f to have a term of finite maximal degree. And if f and g are only formal Laurent series (about 0, I presume), then f(x) and g(x) may not be defined on any interval (0, d), as they may not converge. And if we restrict to only convergent Laurent series, I don't believe the space is complete.

Still, I believe the intended field is actually the same as the one I came up with.

Start with R(x), the set of rational functions over R. For f in R(x), define f > 0 if its leading coefficient > 0, and f > g if f - g > 0. It is easily seen that this makes R(x) an ordered field (in particular, if f,g > 0, then f+g > 0, fg > 0, and -f < 0). R(x) is not complete under the induced topology (for example, the sequence {[sum]i=0n x-i/i!}n is cauchy, but not convergent). However, its completion is also an ordered field, which I believe is equivalent to your R((x)).
Title: Re: Complete Ordered Fields
Post by Eigenray on Oct 27th, 2005, 4:24pm

on 10/26/05 at 18:35:55, Icarus wrote:
I am not entirely sure of how to interpret your example. I've always considered Laurent series to be bi-infinite, but your definition of ord(f) requires f to have a term of finite maximal degree.

As I understand it, "formal Laurent series" over some ring R refers to R((x))=R[x][x-1].  If R is a domain, the field of fractions of R[x] is R(0)((x)), which is just R((x)) if R is a field.

On the other hand, bi-infinite series over an arbitrary ring cannot be multiplied, so are not very useful.  But "Laurent series" usually refers to convergent bi-infinite power series over C, i.e., functions analytic in some annulus.


Quote:
And if f and g are only formal Laurent series (about 0, I presume), then f(x) and g(x) may not be defined on any interval (0, d), as they may not converge.  And if we restrict to only convergent Laurent series, I don't believe the space is complete.

Quite right; I don't know why I said that.  For example, the sequence
fn(x) = 1 + x + 2! x2 + ... + n! xn
is Cauchy, but converges only formally.


Quote:
Still, I believe the intended field is actually the same as the one I came up with.

I don't recall whether localization commutes with completion in general, but in this case they do:
R[x]  ->  R[x]
 v                                        v
R(x)  ->  R((x))

(Remark: The completeness of R((x)) does not require the completeness of R.  Q((x)) is just as complete.)


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