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eigenguy

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Registered: 2014-03-18

Posts: 78

The only complete ordered field

Even though apparently no one else found the recursion formula as nice as I did, I am going to put another favorite discovery, this one more recent - I only realized this last year. (Both results were of course known well before. They were just discoveries for me.)

I had long heard that the Real numbers were the only complete ordered field, but knew only half of why this was so. Last year while studying the Surreals, I finally noticed the missing half. First, the easy part (okay - it takes more work to dot the i's and cross the t's, but it is a well-known construction.)

Theorem: Every ordered field contains the rational numbers.

(Some would say that should be "contains a copy of the rational numbers". But the rationals (and the reals) are defined only by their properties, so in some sense any copy of the rationals is the rationals.)
I'll only give an outline of the proof:

Let F be an ordered field. Then F contains 0 and 1. Define inductively a map f from the set Q of rationals into F by f(0) = 0[sub]F[/sub] and for all non-negative integers n,  f(n + 1) =  f(n)  + 1[sub]F[/sub]. Note that since F is ordered, for all x in F, x + 1 > x, which allows us to prove inductively that if n < m, then f(n) < f(m), so f is 1-1. That f also obeys f(n +m) = f(n) + f(m), and  f(nm) =  f(n) f(m) are also straight-forward inductive proofs.

f can be expanded to negatives by f(-n) = -f(n), and to all rationals by f(n/m) = f(n) / f(m). It is easy to check that these are well-defined, and that the homomorphic and injective properties still hold. Just f injects the rationals into F.
QED

For ordered sets, the concept of "completeness" is defined in terms of the order. An ordered set is complete if and only if it possesses the supremum property: every non-empty subset that is bounded above has a least upper bound. That is, if A is a non-empty subset, and if there is some M such that for all x ∈ A, x < M, then there is some S such that:
      A ≤ S  (that is, for all x ∈ A, x ≤ S).
      If for some y, A ≤ y, then S ≤ y.
define sup(A) = S.

Theorem: Any complete ordered field contains the real numbers.

One again, an outline proof: Every r in the set R of real numbers is the supremum of the rational numbers below it: r = sup{q ∈ Q : q < r}. So extend the injection f of Q into F to all of R by f(r) = sup{f(q) : q is rational and less than r}.
f is well-defined and still 1-1 and a homomorphism. Thus it injects R into  F.
QED

That is one side: any complete ordered field must contain the real numbers. That is, it is an extension field of the real numbers. The other side of this theorem is:

Theorem: Any proper ordered extension field of the real numbers is not complete.

First we prove a lemma:
Lemma: there is at least one infinite element in a proper ordered extension field of the real numbers.

By "infinite", I mean greater than all real numbers. The field must have at least one element, x, which is not a real number (otherwise it wouldn't be proper). Since -x also cannot be real, we can assume x > 0. If x is greater than all real numbers, then we are done. So assume there is at least one real number greater than x. Define the set A = { r ∈ R : r < x }. Then 0 ∈ A and A is bounded above in the reals, so, since the reals are complete, there is a real number s = sup(A). Since s is real, s ≠ x. s could be either above or below x. But either way, there cannot be any other real number between s and x. For by definition, s ≥ A, which include all real numbers less than x, and since all real numbers above x are upper bounds for A, s ≤ all real numbers above x.

Define y = | x - s |. y > 0 since s ≠ x. And if 0 < r < y, then either s + r < x, or x < s - r. In either case, r cannot be real, as one of s + r or s - r is closer to x than s. Therefore there is no real number between y and 0. If t is any positive real number, y < 1/t. Which means also that t < 1/y. Hence 1/y is an infinite element of the field.
QED

Now assume that the proper ordered extension field is complete. Then by the lemma, R is bounded above. So R must have a supremum w. But w - 1 < w, which means that it cannot be an upper bound for R. So there exists some real number r such that w - 1 < r. But then w < r + 1, which is also real. This contradicts the definition of w. Therefore there cannot a complete proper ordered extension field of R.
QED

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"Having thus refreshed ourselves in the oasis of a proof, we now turn again into the desert of definitions." - Bröcker & Jänich