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

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

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

$$\tilde{ȷ}(r)=sup\left(j(D_r)\right).$$

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

We claim that in fact $ℝ=F$. To see this, first recall thatsince $F$ is a partially ordered group with the least upperbound property, $F$ has the Archimedean property (http://planetmath.org/DistributivityInPoGroups).So for any $f\in F$, there exists some positive integer $n$ such that .Hence the set is nonempty andbounded above, implying that $f^{\prime }={sup}_{ℝ}{D}_f^{\prime }$ lies in $ℝ$.Now observe that applying the least upper bound axiom in $F$ gives us that$f={sup}_F{D}_f^{\prime }$. Since $f^{\prime }$ is an upper bound of $D_f^{\prime }$ in $F$, it followsthat $f\le f^{\prime }$.

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