The complete ordered field is unique up to isomorphism.
I am quite sure that Parts 1,2,3 are correct. Could you please help me verify Parts 4,5,6 at the end of the proof. Do Parts 4,5,6 look fine or contain logical flaws/gaps? Any suggestion is greatly appreciated. Thank you for your help!
My attempt:
Let $\mathfrak{A}=\langle A,<',\oplus,\odot,0',1' \rangle$ and $\mathfrak{R}=\langle \Bbb R,<,+,\cdot,0,1 \rangle$ be complete ordered fields where $\Bbb R$ is the set of real numbers. It suffices to prove that $\mathfrak{R}$ is isomorphic to $\mathfrak{A}$. Let $\langle C,<',\oplus,\odot,0',1' \rangle$ 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 $\langle C,<',\oplus,\odot,0',1' \rangle$ and that $C$ is dense in $A$. Let $\sup, \sup'$ be supremums w.r.t $<$ and $<'$ respectively.
We define a mapping $F:\Bbb R \to A$ by $F(x)=\sup' \{f(p) \mid p\in \Bbb Q \text{ and } p<x\}$ for all $x\in\Bbb R$.
- $F$ is well-defined
$\Bbb R$ has no endpoints and $\Bbb Q$ is dense in $\Bbb R$ $\implies$ there exist $p_0,q_0\in \Bbb Q$ such that $q_0<x<p_0$ $\implies \{ p\in \Bbb Q \mid p<x\}$ is nonempty and bounded from above by $p_0\in\Bbb Q$. Then $\{f(p) \mid p\in \Bbb Q \text{ and } p<x\}$ is nonempty and bounded from above by $f(p_0)$. Hence $F(x)=\sup' \{f(p) \mid p\in \Bbb Q \text{ and } p<x\}$ exists.
- $\forall x\in \Bbb Q:F(x)=f(x)$
$\begin{align}\forall x\in \Bbb Q:F(x)&= \sup' \{f(p) \mid p\in \Bbb Q \text{ and } p<x\}\\ &= \sup' \{f(p) \mid f(p)\in f[\Bbb Q] \text{ and } f(p)<'f(x)\}\\ &= \sup' \{c \mid c\in C \text{ and } c<'c'\},\space c:=f(p) \text{ and } c':=f(x)\\ &= c'\\ &=f(x)\end{align}$
- $\forall x,y \in \Bbb R: x<y \iff F(x)<' F(y)$
Since $<$ and $<'$ are linear orders, it suffices to prove that $\forall x,y \in \Bbb R: x<y \implies F(x)<' F(y)$.
$x,y \in \Bbb R$ , $x<y$ , and $\Bbb Q$ is dense in $\Bbb R$ $\implies$ there exist $p_1,p_2\in \Bbb Q$ such that $x<p_1<p_2<y$.
First, $p_1<p_2$ and $p_1,p_2\in \Bbb Q \implies f(p_1) <' f(p_2) \implies F(p_1) <' F(p_2)$. Second, $x<p_1 \implies \{f(p) \mid p\in \Bbb Q \text{ and } p<x\} \subsetneq \{f(p) \mid p\in \Bbb Q \text{ and } p<p_1\} \implies$ $\sup'\{f(p) \mid p\in \Bbb Q \text{ and } p<x\} \le' \sup'\{f(p) \mid p\in \Bbb Q \text{ and } p<p_1\} \implies F(x)\le' F(p_1)$.
Similarly, $F(p_2)\le' F(y)$. Hence $F(x)\le' F(p_1)<'F(p_2)\le' F(y)$ and thus $F(x)<' F(y)$.
- $F$ is surjective
For $a\in A$, let $X'=\{c\in C \mid c<'a\} \subseteq C$. Since $C$ is dense in $A$, $\sup'X'=a$. Let $X=f^{-1}[X']$. Then $X\subseteq \Bbb Q$ and $f[X]=X'$.
$X'$ is nonempty bounded from above and $f$ is an isomorphism between $\Bbb Q$ and $C$ $\implies$ $X'$ is nonempty bounded from above. Let $x=\sup X \in \Bbb R$. Next we prove $F(x)=a$.
$\begin{align} F(x)&=\sup' \{f(p) \mid p\in \Bbb Q \text{ and } p<x\} \\ &= \sup' \{f(p) \mid p\in \Bbb Q \text{ and } p<\sup X\}\\ &= \sup' \{f(p) \mid p\in \Bbb Q \text{ and } \exists p'\in X:p<p'\}\\ &= \sup' \{f(p) \mid f(p)\in f[\Bbb Q] \text{ and } \exists f(p')\in f[X]:f(p)<'f(p')\}\\ &= \sup' \{f(p) \mid f(p)\in C \text{ and } \exists f(p')\in X':f(p)<'f(p')\}\\ &= \sup' \{c \mid c\in C \text{ and } \exists c'\in X':c<'c'\},\space c:=f(p) \text{ and }c':=f(p')\\ &= \sup' \{c \mid c\in C \text{ and } c<'\sup' X'\}\\ &= \sup' \{c \mid c\in C \text{ and } c <' a\},\space \sup' X'=a\\ &= a\end{align}$
- $\forall x,y \in \Bbb R: F(x+y) = F(x) \oplus F(y)$
By definition, $x+y = \sup \{p+q \mid p,q\in\Bbb Q \text{ and } p<x \text{ and } q<y\}$. Furthermore, $r\in \Bbb Q$ and $r < x+y \iff$ $r\in \Bbb Q$ and $r<p+q$ for some $p,q\in\Bbb Q$ such that $p<x$ and $q<y \iff$ $r=p+q$ for some $p,q\in\Bbb Q$ such that $p<x$ and $q<y$.
$F(x)=\sup' \{f(p) \mid p\in \Bbb Q \text{ and } p<x\}=\sup' A$
$F(y)=\sup' \{f(q) \mid q\in \Bbb Q \text{ and } q<y\}=\sup' B$
$\begin{align}F(x+y)&=\sup' \{f(r) \mid r\in \Bbb Q \text{ and } r<x+y\} \\&= \sup' \{f(p+q) \mid p,q\in \Bbb Q \text{ and } p<x \text{ and } q<y\}\\&=\sup' \{f(p) \oplus f(q) \mid p,q\in \Bbb Q \text{ and } p<x \text{ and } q<y\}\end{align}$
It follows that $F(x+y)=\sup ' (A+B)$ and thus $F(x+y)=F(x) \oplus F(y)$.
- $\forall x,y \in \Bbb R: F(x \cdot y) = F(x) \odot F(y)$
WLOG, we assume $x>0$ and $y>0$.
By definition, $x \cdot y = \sup \{p \cdot q \mid p,q\in\Bbb Q \text{ and } 0<p<x \text{ and } 0<q<y\}$. Furthermore, $r\in \Bbb Q$ and $0 < r < x \cdot y \iff$ $r\in \Bbb Q$ and $0 < r <p \cdot q$ for some $p,q\in\Bbb Q$ such that $0<p<x$ and $0<q<y \iff$ $r=p \cdot q$ for some $p,q\in\Bbb Q$ such that $0<p<x$ and $0<q<y$.
$F(x)=\sup' \{f(p) \mid p\in \Bbb Q \text{ and } p<x\}=\sup' \{f(p) \mid p\in \Bbb Q \text{ and } 0<p<x\}=\sup' A$
$F(y)=\sup' \{f(q) \mid q\in \Bbb Q \text{ and } q<y\}=\sup' \{f(q) \mid q\in \Bbb Q \text{ and } 0<q<y\}=\sup' B$
$\begin{align}F(x \cdot y)&=\sup' \{f(r) \mid r\in \Bbb Q \text{ and } r<x \cdot y\} \\&=\sup' \{f(r) \mid r\in \Bbb Q \text{ and } 0<r<x \cdot y\} \\&= \sup' \{f(p \cdot q) \mid p,q\in \Bbb Q \text{ and } 0<p<x \text{ and } 0<q<y\}\\&=\sup' \{f(p) \odot f(q) \mid p,q\in \Bbb Q \text{ and } 0<p<x \text{ and } 0<q<y\}\end{align}$
It follows that $F(x \cdot y)=\sup ' (A \cdot B)$ and thus $F(x \cdot y)=F(x) \odot F(y)$.