The smallest subfield of a complete ordered field is dense in itself

Question

The smallest subfield of a complete ordered field is dense in itself.

This theorem is part of my attempt to prove that A complete ordered field is unique up to isomorphism.

My questions:

1. Could you please verify my attempt? Does it look fine or contain logical flaws/gaps? Any suggestion is greatly appreciated.

2. My proof may be unnecessarily long. I would like to ask for a shorter or more elegant proof.

Many thanks for your help!

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My attempt:

Let $\mathfrak{A}=\langle A,<,+,\cdot,0',1' \rangle$ be a complete ordered field and $\langle C,<,+,\cdot,0',1' \rangle$ be the smallest subfield of $\mathfrak{A}$. It is well-known that there exists a unique isomorphism $f:\Bbb Q \to C$ between $\langle \Bbb Q,<,+,\cdot,0,1 \rangle$ and $\langle C,<,+,\cdot,0',1' \rangle$ where $\Bbb Q$ is the set of rationals.

1. For all $a\in A$, there exists $c\in C$ such that $a<c$

If not, the set $X=\{ a\in A \mid \forall c\in C: c \le a\}$ is nonempty and bounded from below by some $c\in C$, and thus has an infimum. We have $a\in X \implies a-1'\in X$. If not, $a-1' \notin X$ and thus $\exists c\in C:a-1'<c$. Then $a<c+1'\in C$ and thus $a \notin X$, which is a contradiction. To sum up, we have both $X$ has an infimum and $(a\in X \implies a-1'\in X)$. This again leads to a contradiction.

2. For all $a\in A$, there exists $c\in C$ such that $c<a$

The proof is similar to above reasoning.

3. $C$ is dense in $A$

Assume $a_1,a_2\in A$ such that $a_1<a_2$. Then there exist $c_1,c_2\in C$ such that $c_1<a_1<a_2<c_2$. Let $\Delta=c_2-c_1,\Delta_1=a_1-c_1,\Delta_2=a_2-a_1$. There exist $c_0\in C$ such that $c_0>\dfrac{\Delta}{\Delta_2}$ and $n_0\in\Bbb N$ such that $f(n_0)>c_0$. Let $I=\left\{n\in\Bbb N \mid f\left(\dfrac{n}{n_0}\right)\le\dfrac{\Delta_1}{\Delta}\right\}$.

• For $n=0$, $f\left(\dfrac{n}{n_0}\right)=f(0)=0'<\dfrac{\Delta_1}{\Delta}$. Then $0\in I$ and thus $I \neq \emptyset$.

• Furthermore, $\forall n\ge n_0:\dfrac{n}{n_0} \ge 1 \implies \forall n\ge n_0:f\left(\dfrac{n}{n_0}\right) \ge f(1)=1'>\dfrac{\Delta_1}{\Delta}$ $\implies \forall n\ge n_0:n\notin I$. It follows that $I$ is bounded from above by $n_0$.

Hence $i:=\max I$ exists.

It is clear that $f\left(\dfrac{i+1}{n_0}\right) > \dfrac{\Delta_1}{\Delta}$ or $\Delta\cdot f\left(\dfrac{i+1}{n_0}\right)>\Delta_1$ or $\Delta\cdot f\left(\dfrac{i+1}{n_0}\right)>a_1-c_1$ or $c_1+\Delta\cdot f\left(\dfrac{i+1}{n_0}\right)>a_1$. Next we prove $c_1+\Delta\cdot f\left(\dfrac{i+1}{n_0}\right)<a_2$.

• If not, $c_1+\Delta\cdot f\left(\dfrac{i+1}{n_0}\right) \ge a_2 \iff \Delta\cdot f\left(\dfrac{i+1}{n_0}\right) \ge a_2-c_1 \iff$ $\Delta\cdot \left( f\left( \dfrac{i}{n_0} \right)+\dfrac{1'}{f(n_0)} \right) \ge \Delta_1 + \Delta_2 \iff f\left( \dfrac{i}{n_0} \right)+\dfrac{1'}{f(n_0)} \ge \dfrac{\Delta_1}{\Delta} + \dfrac{\Delta_2}{\Delta}$.

• Moreover, $\dfrac{\Delta_1}{\Delta} \ge f\left(\dfrac{i}{n_0}\right)$. It follows that $\dfrac{\Delta_1}{\Delta}+\dfrac{1'}{f(n_0)} \ge f\left( \dfrac{i}{n_0} \right)+\dfrac{1'}{f(n_0)} \ge \dfrac{\Delta_1}{\Delta} + \dfrac{\Delta_2}{\Delta}$. Hence $\dfrac{1'}{f(n_0)} \ge \dfrac{\Delta_2}{\Delta}$ or equivalently $f(n_0) \le \dfrac{\Delta}{\Delta_2}$, which is a contradiction.

To sum up, $a_1<c_1+\Delta\cdot f\left(\dfrac{i+1}{n_0}\right)<a_2$ where $c_1+\Delta\cdot f\left(\dfrac{i+1}{n_0}\right) \in C$.