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!
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 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.
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.