(this are notes for a talk)

We can approach the real numbers axiomatically:

Field axioms:

1. (a+b)+c=a+(b+c)

2. a+0=a

3. For every a, there’s -a, a+(-a)=0

4. a+b=b+a

5. (ab)c=a(bc)

6. a1=a

7. For every a!=0, there’s a^(-1), a(a^(-1))=1

8. ab=ba

9. a(b+c)=ab+ac

10. 1!=0

Examples of fields: Q, Z/p, rational functions over Z

Order axioms:

1. For every a,b, exactly one of a<b, b<a or a=b

2. If a<b<c, a<c

Examples of orders: Alphabet

Ordered field axioms:

2. If a<b, then a+c<b+c

3. If a<b, 0<c, then ac<bc

Examples of ordered fields: Q, rational functions over Z

Example of field which cannot be ordered: Z/p

Proof:

1<0 ==> 0<-1 ==> 0<-1*-1 ==> 0<1

0<1 ==> 1<2 ==> 0<2 and by induction 0<k ==>0<p=0 contradiction

In general, if a field is ordered, it is of 0 characteristic (n!=0)

Complete ordered field axioms:

1. If A is not empty and there is an b, such that for a in A, a<b, then there is a minimal such b. We call this sup(A).

Theorem:

If F,G are COF, F and G are isomorphic with a unique isomorphism.

Proof:

We know it is of 0 characteristic. We can prove that the image of N in F (and G) is unbounded: otherwise there is a minimal

upper limit t. n+1<t for every n, so n<t-1, which contradicts t being a least upper bound. Therefore for every e>0, let’s look at 1/e>0, and at n>1/e, then 1/n<e. Now assume t,s are in F (or G) and t<s. s-t>0, find an n such that 1/n<s-t. Let’s look at k/n. There is some whole k such that k>nt and so k/n>t. Let’s take the least such k, then k/n>t but if k/n>s then k-1/n>s-1/n>s-(s-t)=t in contradiction to k minimality, and so t<k/n<s. Now we define a function from F to G:

E(t)=sup({k/n: k/n<t})

To prove E is well-defined, note that if l>t is an integer, than the set is bounded by l, so has a least upper bound.

It is straightforward to prove that E is order preserving, addition preserving and multiplication preserving and is 1-1.

To prove it is onto, note that we can explicitly construct the inverse.

To prove uniqueness, note that an isomorphism carries 1 to 1, and so k/n into k/n, and sup to sup.

QED

Definition: The “real numbers” are a complete ordered field.

This definition is good enough — it is unique up to a unique isomorphism.

Do they exist?

Lemma:

In the COF axiom, it’s enough to assume A is a set of rationals.

Proof:

Let A be a set of real numbers, and construct A’={q:q rational, q<a for some a in A}. Let’s take sup(A’). Suppose there is some a in A, sup(A’)<a, then there is a rational q, sup(A’)<q<a, which is a contradiction, so sup(A’)>=a for every a in A. Suppose a<b<sup(A’) for every a, then if q in A’, q<a<b for some a, so b would be an upper limit for A’. Contradiction, and thereofre sup(A’) is a least upper bound for A.

Construction:

An ultrafilter on a set M is a set U of subsets of M such that:

1. For every set A contained in M, either A in U or M\A in U

2. If A in U, A contained in B, B in U

3. If A,B in U, A intersection B in U

4. The empty set is not in U.

Theorem (not proven here): If F is a set of subsets of M which fullfills 2,3,4 then there is a ultrafilter containing it.

Let M=N the natural numbers, F be “the complement is finite”, and U an ultrafilter from the theorem.

Let Q be the rationals.

Take {f:f:N->Q}, and divide it by “almost everywhere equal”: f~g if {n:f(n)!=g(n)} is not in the ultrafilter.

The result is an ordered field. Let’s prove for example the existence of inverse. If f!~0, then let’s define g(n)=1/f(n) if f(n)!=0 otherwise g(n)=1. Then fg(n)=1 unless f(n)=0, which is a set not in the ultrafilter.

It does not have COF! Let’s take the “finites”: {f:f<n for some n}. It’s a ring, with a maximal ideal: I={f:f<1/n for all n}.

Let’s divide F/I, it’s again an ordered field. Is it complete?

Let’s use the lemma, and let A be a set of rationals. The rationals are countable so A is. Now define f: f(n)=max(first n elements of A). f defines a finite element because A is bounded. f>=a for every a in A. If b<f, then b(n)<f(n) for at least one element, so b(n)<a for some a in A (since the image of f is in A). So f=sup(A). QED.

Why are they useful?

Here is a simple existence theorem: let P be a polynomial of odd degree. Then P has a root.

Proof:

Assume without loss of generality that the top coefficient is 1.

We can take x big enough so we know P(x)>0 (for example, sum the absolute values of all coefficients). We can take x small enough so we know P(x)<0 (ditto, but invert). between x_0 and x_1, we take the set of all numbers, P(x)<0. This is by definition bounded, so let’s take a supremum s. If P(s)<0, we can find 1/n such that P(s+1/n)<0, in contradiction. Similarily, if P(s)>0, we can find n such that to be <0, x has to be smaller than s-1/n, in contradiction. So P(s)=0. QED

Do they exist?

Your definition of sup needs work. it should be <= not <. As written, the set {0} has no sup as there are plenty of reals larger than 0, but no smallest positive real. 🙂