Two Ordered Fields Satisfying the Completeness Axiom

Question

I can't seem to figure out the following problem.

Suppose that for $i=1,2$, $\mathbb{F}_{i}$ is an ordered field satisfying the Completeness Axiom. Show that there is a unique isomorphism $\alpha$ from $\mathbb{F}_{1}$ onto $\mathbb{F}_{2}$. That is, $\alpha$ is a field isomorphism that preserves the the order ($x\leq_{1} y\Rightarrow \alpha(x)\leq_{2}\alpha(y)$, where $\leq_{i}$ is the order on $\mathbb{F}_{i}$). Thus, there is at most one field of "real numbers".

Would the mapping be the function which takes the supremum of sets in $\mathbb{F}_{1}$ to the supremum of sets in $\mathbb{F}_{2}$? I'm completely lost and I have no idea where to start.

Answer 1

Here's how I would approach it: Every field is guaranteed by definition to contain at least two elements: the additive identity ($0$), and the multiplicative identity ($1$). Also, if the field is ordered, then $0<1<1+1<1+1+1<\ldots$, so now you've effectively constructed $\mathbb{N}$. Use the fact that every number in a field has an additive inverse, and now you have $\mathbb{Z}$. Next, give every non-zero integer a multiplicative inverse, and you have $\mathbb{Q}$. Since applying the completeness axiom gives you $\mathbb{R}$, you have now shown that every complete ordered field must contain a copy of $\mathbb{R}$. All you need to prove now is that you can't put any new number in your complete ordered field without causing problems.

Answer 2

Prove that every ordered field contains an unique copy of $\Bbb Q$.

Prove that every complete, ordered field satisfies the Archimedean property. It will be very handy in the final steps.

Let call $\Bbb Q_i$ the copy of $\Bbb Q$ in $\Bbb F_i$. Let $\beta$ be the isomorphism between these copies.

Now define for $x\in\Bbb F_1$ $$\alpha(x)=\sup\{\beta(y):y<x\}$$

To finish the proof of existence, you must show that the sets $\{\beta(y):y<x\}$ are bounded in $\Bbb F_2$ and that $\alpha$ is an isomorphism of ordered fields.

For uniqueness, consider other isomorphism $\alpha'$ and assume that $\alpha(x)<\alpha'(x)$. Can you derive a contradiction?

Answer 3

Others have pointed out that within every ordered field (whether complete or not) there is an isomorphic copy of $\mathbb Q.$ The field $\mathbb Q$ is the unique smallest ordered field.

Now bring in completeness. In some ordered fields that are not complete, $\mathbb Q$ has upper bounds. But in a complete ordered field, the existence of an upper bound of $\mathbb Q$ implies the existence of a least upper bound, which let us call $q.$ So $q-1$ cannot be an upper bound of $\mathbb Q$; thus some rational number $r$ is at least as big as $q-1,$ and so $r+ \frac 3 2$ is a rational number bigger than $q.$ But then $q$ is not an upper bound of $\mathbb Q.$ So $\mathbb Q$ can have no upper bound within a complete ordered field. That implies the set of positive members of $\mathbb Q$ cannot have any positive lower bound $\varepsilon,$ since then $1/\varepsilon$ would be an upper bound of $\mathbb Q.$ That means the length $b-a$ of an open interval $(a,b)$ cannot be a lower bound of $\mathbb Q,$ so find a rational number $1/n,$ $n\in\mathbb Z,$ that is smaller than $b-a,$ and conclude that some member of $\{k/n : k\in\mathbb Z\}$ must be in the interval $(a,b).$ Therefore: between any two members $a,b$ of a complete ordered field there is a member of $\mathbb Q.$ In some ordered fields that are not complete, that doesn't happen; i.e. there are some open intervals that do not intersect $\mathbb Q.$

Now look at your two complete ordered fields $\mathbb F_1$ and $\mathbb F_2.$

Map the rationals within $\mathbb F_1$ to the rationals within $\mathbb F_2$ in the obvious way. Suppose $a\in\mathbb F_1$ is not rational. All rational numbers less than $a$ correspond to rationals in $\mathbb F_2$ and all rational numbers greater than $a$ similarly correspond to rationals in $\mathbb F_2.$ Since an isomorphism preserves order, the image in $\mathbb F_2$ of $a\in\mathbb F_1$ must be greater than some rationals and less than others. There can be only one member of $\mathbb F_2$ that can serve as that image, because if there were more than one, then we would have members $e,f\in\mathbb F_2$ with no rationals between them, and we've already seen why that cannot happen.