every ordered field with the least upper bound property is isomorphic to $\mathbb{R}$ , proof that

Let $F$ be an ordered field with the least upper bound property. By the order properties of $F$ , $0<1_{F}$ and by an induction argument $0<n\cdot 1_{F}$ for any positive integer $n$ . Hence the characteristic of the field $F$ is zero, implying that there is an order-preserving embedding $j\colon\mathbb{Q}\to F$ .

We would like to extend this map to an embedding of $\mathbb{R}$ into $F$ . Let $r\in\mathbb{R}$ and let $D_{r}=\{q\in\mathbb{Q}\colon q<r\}$ be the associated Dedekind cut. Since $D_{r}$ is nonempty and bounded above in $\mathbb{Q}$ , it follows that the set $j(D_{r})$ is nonempty and bounded above in $F$ . Applying the least upper bound property of $F$ , define a function $\widetilde{\jmath}\colon\mathbb{R}\to F$ by

\widetilde{\jmath}(r)=\sup\left(j(D_{r})\right).

One can check that $\widetilde{\jmath}$ is an order-preserving field homomorphism. By replacing $F$ with the isomorphic field $F\setminus\widetilde{\jmath}(\mathbb{R})\cup\mathbb{R}$ , we may assume that $\mathbb{R}\subset F$ .

We claim that in fact $\mathbb{R}=F$ . To see this, first recall that since $F$ is a partially ordered group with the least upper bound property, $F$ has the Archimedean property (http://planetmath.org/DistributivityInPoGroups). So for any $f\in F$ , there exists some positive integer $n$ such that $-n<f<n$ . Hence the set $D^{\prime}_{f}=\{r\in\mathbb{R}\colon r<f\}\subset\mathbb{R}$ is nonempty and bounded above, implying that $f^{\prime}=\sup_{\mathbb{R}}D^{\prime}_{f}$ lies in $\mathbb{R}$ . Now observe that applying the least upper bound axiom in $F$ gives us that $f=\sup_{F}D^{\prime}_{f}$ . Since $f^{\prime}$ is an upper bound of $D^{\prime}_{f}$ in $F$ , it follows that $f\leq f^{\prime}$ .

Seeking a contradiction, suppose $f<f^{\prime}$ . By the Archimedean property, there is some positive integer $n$ such that $f<f^{\prime}-n^{-1}<f^{\prime}$ . Because $f^{\prime}=\sup_{\mathbb{R}}D^{\prime}_{f}$ , we obtain $f^{\prime}-n^{-1}<f$ , which implies the contradiction $f<f$ . Therefore $f=f^{\prime}$ , and so $f\in\mathbb{R}$ . This completes the proof.

Title	every ordered field with the least upper bound property is isomorphic to $\mathbb{R}$ , proof that
Canonical name	EveryOrderedFieldWithTheLeastUpperBoundPropertyIsIsomorphicTomathbbRProofThat
Date of creation	2013-03-22 14:10:51
Last modified on	2013-03-22 14:10:51
Owner	mps (409)
Last modified by	mps (409)
Numerical id	9
Author	mps (409)
Entry type	Proof
Classification	msc 12E99
Classification	msc 54C30
Classification	msc 26-00
Classification	msc 12D99

every ordered field with the least upper bound property is isomorphic to ℝ, proof that

every ordered field with the least upper bound property is isomorphic to $\mathbb{R}$ , proof that