# Every Archimedean ordered field is real.
Take $F$ to be an Archimedean ordered field.
We know $F$ has a copy of $\mathbf{Q}$ that is dense in $F$. Identify the rationals in $F$ as the rationals in reals.
Consider the map $f:F\to\mathbf{R}$ as follows $$
f(x) = \{q\in \mathbf{Q}: q < x\}.
$$Here we are attempting to define a Dedekind cut. And indeed it is. $\blacktriangleright$ Note $f(x)$ is not all of $\mathbf{Q}$ since by $F$ being Archimedean, there is some positive integer $N > x$, whence $N \not \in f(x)$. Nor if $f(x)$ empty, otherwise $x$ is smaller than every rational, which makes $-x$ larger than every rational, contradicting $F$ Archimedean. This set $f(x)$ is closed downwards in $\mathbf{Q}$, since if $q\in f(x)$, then $q < x$, and any $q' < q$ must be $q' < x$, namely $q' \in f(x)$. And this set $f(x)$ has no maximal element, if $q \in f(x)$, then $x - q >0$. So by Archimedean, there exists positive integer $N$ such that $N(x-q) > 1$. Which gives $\frac{1}{N} +q < x$. So the rational number $\frac{1}{N} + q \in f(x)$, and $f(x)$ has no maximal element. This shows $f(x)$ is a Dedekind cut. $\blacksquare$
Since we can identify the reals as the collection of all Dedekind cuts of the rationals, $f$ is a well-defined map from $F$ to $\mathbf{R}$.
We can also see that $f$ is an an order preserving injection. Suppose $x < y$ in $F$. Take $t \in f(x)$, then $t < x$ and hence $t For any Archimedean ordered field $F$, there exists an order-preserving one-to-one map $f:F\to \mathbf{R}$.
In fact, this map $f$ is also a ring homomorphism. I give a sketch here.
Take $x,y \in F$. We wish to show $f(x)+f(y)$ is the same as $f(x+y)$. On one hand, the sum of the Dedekind cuts $f(x)+f(y):=\{a+b:a\in f(x),b\in f(y)\} = \{a+b:a < x, b < y\}$, and $f(x+y)=\{q\in \mathbf{Q}:q < x+y\}$. So if we have some element $a+b \in f(x)+f(y)$, with rationals $a,b$ where $a < x$ and $b < y$, then $a+b < x + y$, whence $a+b \in f(x+y)$. So this shows $f(x)+f(y) \subset f(x+y)$. And if $q \in f(x+y)$, then $q\in \mathbf{Q}$ with $q < x+y$.
Note there is a rational number $q'\in(x-\frac{1}{N},x)$ for every positive integer $N$, and note $q-q' < q-x+\frac{1}{N}$. Now $q-x < y$, so there exists positive integer $N$ such that $N(y-q+x) > 1$, or $q -x + \frac{1}{N} < y$. So take this $N$, which gives this $q' \in (x- \frac{1}{N} ,x)$. Then we have $q-q' < y$. So the rational $q' \in f(x)$ and the rational $q-q' \in f(y)$, this shows $q = q' + (q-q') \in f(x) + f(y)$. Hence $f(x+y) \subset f(x) + f(y)$, and so we have $f(x+y) = f(x) + f(y)$.
Note, $f(0) = 0$ and $f(1) =1$ as $f$ preserves the rationals. So $f(-x)+f(x)=f(0)=0$, namely $f(-x) = -f(x)$ for any $x \in F$.
Now, suppose $x, y > 0$, we wish to show $f(xy) = f(x) f(y)$. Since $f$ preserves order, $f(x),f(y) > 0$. So $f(x)f(y) = \{ab: a\in f(x), b\in f(y)\}$. So if we take $ab\in f(x)f(y)$, we have $ab < xy$, hence $ab \in f(ab)$. This shows $f(x)f(y) \subset f(xy)$. Now suppose $q \in f(xy)$, some rational $q < xy$. Note for any positive integer $N$, there exists some rational $a$ such that $x- \frac{1}{N} < a < x$. We need to pick $N$ such that the resulting $a$ satisfies $\frac{q}{a} < y$. Note that for any such $N$ and $a$, we have $$
\frac{q}{x-\frac{1}{N}} > \frac{q}{a} > \frac{q}{x}
$$So suffices to pick $N$ where $\frac{q}{x- \frac{1}{N}} < y$. This requires $q < xy - \frac{y}{N}$ or $\frac{y}{N} < xy - q$. Since $xy - q > 0$, by Archimedean we can pick such positive integer $N$ such that $N(xy-q) > y$. So pick rational $a$ where $x-\frac{1}{N} < a < x$ and this gives rational $\frac{q}{a} < y$. Hence $a \in f(x)$ and $\frac{q}{a} \in f(y)$, so $q = a (\frac{q}{a}) \in f(x)f(y)$ . In other words, $f(x) \subset f(x)f(y)$. This shows equality $f(x)=f(x)f(y)$.
If $x < 0$, then we have $f(x)f(y) = -f(-x)f(y) = -f(-xy) = f(xy)$, and similarly when $y < 0$ or both. And I leave the details of $x=0$ or $y=0$ to the reader.
Once showing all of these, we have
> For any Archimedean ordered field $F$, there exists an injective order-preserving, ring homomorphism $f:F\to \mathbf{R}$. That is, $F$ is isomorphic to a subfield of $\mathbf{R}$.
B / 7 2024