# 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 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 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. Hence the set $D^{\prime}_{f}=\{r\in\mathbb{R}\colon 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. By the Archimedean property, there is some positive integer $n$ such that $f. Because $f^{\prime}=\sup_{\mathbb{R}}D^{\prime}_{f}$, we obtain $f^{\prime}-n^{-1}, which implies the contradiction $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 EveryOrderedFieldWithTheLeastUpperBoundPropertyIsIsomorphicTomathbbRProofThat 2013-03-22 14:10:51 2013-03-22 14:10:51 mps (409) mps (409) 9 mps (409) Proof msc 12E99 msc 54C30 msc 26-00 msc 12D99