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 $$ \widetilde{\jmath}(r) = \sup D_r. $$ 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 $D'_f = \{ r\in\mathbb{R} \colon r < f \}\subset\mathbb{R}$ is nonempty and bounded above, implying that $f' = \sup_{\mathbb{R}} D'_f$ lies in $\mathbb{R}$ . Now observe that applying the least upper bound axiom in $F$ gives us that $f = \sup_F D'_f$ . Since $f'$ is an upper bound of $D'_f$ in $F$ , it follows that $f\le f'$ .

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