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# 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. 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.

## Mathematics Subject Classification

12E99*no label found*54C30

*no label found*26-00

*no label found*12D99

*no label found*

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