- 06 Aug '03 08:16Suppose X is a complete ordered field. Prove that it is uncountable using only definitions of 'complete ordered field' and 'uncountable' and the usual axioms, including Choice (I imagine you might need it.) If that's too easy, try to explain what the technical words in the preceding two sentences mean to a non-mathematician!
- 06 Aug '03 16:26

I'll sit down and try the first part...I think I've got a general idea of how to do it. Oh you evil person ! The second part is the difficult one. I think you are the better pedagogue, so by all means....*Originally posted by Acolyte***Suppose X is a complete ordered field. Prove that it is uncountable using only definitions of 'complete ordered field' and 'uncountable' and the usual axioms, including Choice (I imagine you might need it.) If that's too easy, try to explain what the technical words in the preceding two sentences mean to a non-mathematician!** - 07 Aug '03 15:08 / 2 edits

Okay. I worked out several different proofs last night, one of which just showed that a countable field satisfying the order axiom cannot be complete. However, a better way to bring it down to earth would be to:*Originally posted by Acolyte***Suppose X is a complete ordered field. Prove that it is uncountable using only definitions of 'complete ordered field' and 'uncountable' and the usual axioms, including Choice (I imagine you might need it.) If that's too easy, try to ...[text shortened]... words in the preceding two sentences mean to a non-mathematician!**

1. Show that any two complete ordered fields (COF) are isomorphic (This is a bit stronger than what is needed-we really just need a bijection between any two.)

2. Give some example of a COF (this is more "down-to-earth" because I will use (R,+,X) as an example)

3. Show that the example set is uncountable.

4. Conclude that since all COF are order isomorphic, they must all have the same cardinality, and therefore, since the example is uncountable, they all must be.

So:

1. Claim: Every COF is isomorphic to the real numbers, (R, +, X).

Proof: An ordered field must be dense to satisfy the multiplicative inverse property, and thus must contain some set isomorphic to Q. Furthermore, since it is complete, and the reals (R, +, X) form a complete ordered field, there is a subfield (called R) of X1 that is isomorphic to the reals. Now suppose that there exists a positive x in X1\R, and real k such that x<k. Since x>0, the set S of reals less than x is bounded above by k. Thus supremum S = s and infimum S = t are real and x is in (t,s). However, the reals cannot simultaneously be dense and let S have extrema at s and t. Thus such a k cannot exist. Finally, note that 1 < x so 1/x < 1 and is not real (since x isn't), which is a contradiction. Thus any COF X1 is isomorphic to R. - 07 Aug '03 15:19Now we need only prove that R is uncountable to establish that any COF is uncountable. Furthermore, while Cantor's proof is cool, I like a little variety. For some subset K of the naturals, let xK:N-->{0,1} be the characteristic function of K. For each K, let S(K) = {xK(1), xK(2),...}. Each S(K) represents one unique subset of N and is an ordered sequence of 0's an 1's. Thus the power set of N (P(N)) can be put into 1-1 correspondence with the set of all possible sequences of 0's and 1's. Clearly, the set of all sequences of 0's and 1's is the same as the set of all reals in [0,1), since each decimal number has a binary representation. Furthermore, [0,1) has the same cardinality as the reals. Thus the reals have the same cardinality as P(N) and since card P(N) > card N, the reals are uncountable.

Thus every complete ordered field is uncountable.

- 07 Aug '03 22:43

Rotfliailf. And you always impress me with the humour... Thanks for the vote of confidence.*Originally posted by ChessNut***RC,**

You always impress me with your Mad Math Skills. When I was 15 I didn't know what isomorphic was!!

I'm still trying to figure out if Penicilin will help it or not... - 08 Aug '03 07:15

I understand the general method, but some of the assertions don't seem obvious to me:*Originally posted by royalchicken***Okay. I worked out several different proofs last night, one of which just showed that a countable field satisfying the order axiom cannot be complete. However, a better way to bring it down to earth would be to:**

1. Show that any two complete ordered fields (COF) are isomorphic (This is a bit stronger than what is needed-we really just need a biject ...[text shortened]... not real (since x isn't), which is a contradiction. Thus any COF X1 is isomorphic to R.

"An ordered field must be dense to satisfy the multiplicative inverse property, and thus must contain some set isomorphic to Q."

That's true if it's isomorphic to a subfield of R, but seeing as you're trying to prove the field is isomorphic to R you can't assume this. If you aren't assuming this, what does 'dense' mean, and why is the statement still true?

"Furthermore, since it is complete, and the reals (R, +, X) form a complete ordered field, there is a subfield (called R) of X1 that is isomorphic to the reals."

Clearly the reals are an example of a complete ordered field (though completeness is effectively axiomatic), but I don't see why all COFs must be an extension of R.

"infimum S = t"

S doesn't look like it has a real infimum. I don't think this affects the rest of the proof, though. - 08 Aug '03 16:10(All bold material was originally posted by Acolyte.)

"That's true if it's isomorphic to a subfield of R, but seeing as you're trying to prove the field is isomorphic to R you can't assume this. If you aren't assuming this, what does 'dense' mean, and why is the statement still true?"

First, each element of X1 has a multiplicative inverse. Now if X1 were not dnse, then there would be two elements x and y such that there is no element of X1 between x and y. Let x be the smallest positive x for which this is true. Say x>1. Then x>1/x>1/y and there are no elements of X1 between 1/x and 1/y, since X1 is ordered. This is a contradiction. If x<y<1, then for some a in X1 such that a<x, x-a and y-a are in X1 since X1 is closed under addition. Since X1 is ordered, and there are no elements of X1 between x an y, there can be no elements of X1 between x-a and y-a either, another contradiction. Thus between any elements of X1 there is at least one element of X1. This is what I mean by dense, and it implies that there must be a copy of Q in X1.

"Clearly the reals are an example of a complete ordered field (though completeness is effectively axiomatic), but I don't see why all COFs must be an extension of R."

We have a copy of Q in X1. Since X1 has the lub property, we define a subfield R of X1 that is the set of all least upper bounds of subsets of our copy of Q. This is isomorphic to the reals. The rest of the argument just shows that X1\R = { }.

**"S doesn't look like it has a real infimum. I don't think this affects the rest of the proof, though."**

My mistake. I meant that i is the infimum of the set {y in R such that x<y}, which is not S.

Anyway, since the real numbers are formally defined ONLY by the field axioms, the order axiom, and the completeness axiom, they seem to be isomorphic to any field defined by those axioms anyway. But the proof works.

I always enjoy your critiques, and they always seem to improve what I say . If these explanations are still not to your satisfaction, I would be happy to give them another look.