Archimedean ordered fields are real

In this entry, we shall show that every Archimedean ordered field is isomorphic toa subfield of the field of real numbers. To accomplish this, we shall construct anisomorphism using Dedekind cuts.

As a preliminary, we agree to some conventions. Let $𝔽$ denote an orderedfield with ordering relation $>$. We will identify the integers with multiples ofthe multiplicative identity of fields. We further assume that $𝔽$ satisfiesthe Archimedean property: for every element $x$ of $𝔽$, there exists aninteger $n$ such that .

Since $𝔽$ is an ordered field, it must have characteristic zero. Hence,the subfield generated by the multiplicative identity is isomorphic to the fieldof rational numbers. Following the convention proposed above, we will identifythis subfield with $\mathbb{Q}$. We use this subfield to construct a map $\rho$from $𝔽$ to $ℝ$:

Definition 1.

For every $x\mathrm{\in }\mathrm{F}$, we define

$$\rho (x)=(\{y\in \mathbb{Q}\mid y>x\},\{y\in \mathbb{Q}\mid y\le x\})$$

Theorem 1.

For every $x\mathrm{\in }\mathrm{F}$, we find that $\rho \mathit{}\mathrm{(}x\mathrm{)}$ is a Dedekind cut.

Proof.

Because, for all $y\in \mathbb{Q}$, we have either $y>x$ or $y\le x$, the twosets of $\rho (x)$ form a partition of $\mathbb{Q}$. Furthermore, every elementof the latter set is less than every element of the former set. By the Archimedeanproperty, there exists an integer $n$ such that ; hence the former set isnot empty. Likewise, there exists and integer $m$ such that , or $x>-m$,so the latter set is also not empty.∎

Having seen that $\rho$ is a bona fide map into the real numbers, we now show thatit is not just any old map, but a monomorphism of fields.

Theorem 2.

The map $\rho \mathrm{:}\mathrm{F}\mathrm{\to }\mathrm{R}$ is a monomorphism.

Proof.

Let $p$ and $q$ be elements of $𝔽$; set $(A,B)=\rho (p)$, set $(C,D)=\rho (q)$, and set $(E,F)=\rho (p+q)$. Since $a>p$ and $b>q$ implies $a+b>p+q$for all rational numbers $a$ and $b$, it follows that $a\in A$ and $b\in C$ implies that$a+b\in E$. Likewise, since $a\le p$ and $b\le q$ implies $a+b\le p+q$for all rational numbers $a$ and $b$, it follows that $a\in B$ and $b\in D$ implies$a+b\in F$. Hence, $\rho (p)+\rho (q)=\rho (p+q)$.

Since a rational number is positive if and only if it is greater than $0$, itfollows that $\rho (0)=0$. Together with the fact proven in the last paragraph,this implies that $\rho (-x)=-\rho (x)$ for all $x\in 𝔽$.

Suppose that $p$ and $q$ are positive elements of $𝔽$. As before, set$(A,B)=\rho (p)$, set $(C,D)=\rho (q)$, and set $(E,F)=\rho (p\cdot q)$. Since$a>p$ and $b>q$ implies $a\cdot b>p\cdot q$ for all rational numbers $a$and $b$, it follows that $a\in A$ and $b\in C$ implies that $a\cdot b\in E$.Likewise, since $a\le p$ and $b\le q$ implies $a\cdot b\le p\cdot q$for all rational numbers $a$ and $b$, it follows that $a\in B$ and $b\in D$ implies$a\cdot b\in F$. Hence, $\rho (p)\cdot \rho (q)=\rho (p\cdot q)$.

By using the fact demonstrated previously that $\rho (-x)=-\rho (x)$, we may extendwhat was shown above to the statement that $\rho (p\cdot q)=\rho (p)\cdot \rho (q)$for all $p,q\in 𝔽$. Thus, $\rho$ is a morphism of fields. Since $𝔽$is Archmiedean, if $p\ne q$, there must exist a rational number $r$ between $p$ and $q$,hence $\rho (p)\ne \rho (q)$, so $\rho$ is a monomorphism.∎

Since $\rho$ is a morphism of fields, its image is a subring of $ℝ$. Since $\rho$is a monomorphism, its restriction to this image is an isomorphism, hence $𝔽$ isisomorphic to a subfield of $ℝ$.