The isomorphism between two complete ordered fields is unique.
Does my attempt look fine or contain logical flaws/gaps? Any suggestion is greatly appreciated. Thank you for your help!
My attempt:
Let $\mathfrak{R}=\langle \Bbb R,<,+,\cdot,0,1 \rangle,\mathfrak{A}=\langle A,\prec,\oplus,\odot,0',1' \rangle$ be complete ordered fields where $\Bbb R$ is the set of real numbers. Let $\mathfrak{B}=\langle B,\prec,\oplus,\odot,0',1' \rangle$ be the smallest subfield of $\mathfrak{A}$ and $\mathfrak{Q}=\langle \Bbb Q,<,+,\cdot,0,1 \rangle$.
Lemma 1: $\mathfrak{R}$ is isomorphic to $\mathfrak{A}$.
Lemma 2: $\mathfrak{Q}$ is uniquely isomorphic to $\mathfrak{B}$.
By Lemma 1, let $\Phi:\Bbb R \to A,\Psi:\Bbb R \to A$ be isomorphisms between $\mathfrak{R}$ and $\mathfrak{A}$. By Lemma 2, let $f:\Bbb Q \to B$ be the unique isomorphism between $\mathfrak{Q}$ and $\mathfrak{B}$.
Let $X \subseteq \Bbb Q$ be bounded from above and $\sup,\sup'$ supremums w.r.t $<,\prec$ respectively. We next prove that $\Phi(\sup X) = \sup' f[X]$.
$\forall x\in X: x \le \sup X \implies \forall x\in X: \Phi(x) \preccurlyeq \Phi(\sup X) \implies \forall x\in X: f(x) \preccurlyeq \Phi(\sup X) \implies \sup' f[X] \preccurlyeq \Phi(\sup X).$
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Assume the contrary that $\sup' f[X] \prec \Phi(\sup X)$. Since $B$ is dense in $A$, there exists $b\in B$ such that $\sup' f[X] \prec b \prec \Phi(\sup X)$. Then there exists $p\in \Bbb Q$ such that $f(p)=b$. Thus $\sup' f[X] \prec f(p)=\Phi(p) \prec \Phi(\sup X).$
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We have $\Phi(p) \prec \Phi(\sup X) \implies p<\sup X \implies p<p'$ for some $p'\in X \implies$ $f(p) \prec f(p')$ for some $p'\in X$ $\implies f(p) \prec \sup' f[X]$. This is a contradiction.
Hence $\Phi(\sup X)=\sup' f[X]$. Similarly, $\Psi(\sup X)=\sup' f[X]$.
Let $X_x=\{p\in\Bbb Q \mid p<x\} \subseteq \Bbb Q$. Since $\Bbb Q$ is dense in $\Bbb R$, $x=\sup X_x$ for all $x\in\Bbb R$. Then $\Phi(x)=\Phi(\sup X_x)=\sup' f[X_x]=\Psi(\sup X_x)=\Psi(x)$ for all $x\in\Bbb R$. It follows that $\Phi=\Psi$.
Best Answer
This is an extended comment on the uniqueness of the isomorphism.
(1a). Let $F, G$ be sub-fields of $\Bbb R$ such that $\psi:F\to G$ is a field-isomorphism. ($\psi$ is not assumed to preserve order.) If $\forall x\in F\,(0\le x\implies \sqrt x\in F) $ then $\psi=id_F$ and $G=F.$
Proof: $\Bbb Q\subset F$ and $\psi|_{\Bbb Q}=id_{\Bbb Q}$ so for any $x\in F$ and any $q\in \Bbb Q$ we have $$x\ge q\iff \psi(x)-q=\psi(x)-\psi(q)=(\psi(\sqrt {x-q}))^2\ge 0\iff$$ $$\iff \psi(x)\ge q$$ so $\Bbb Q \cap (-\infty,x]=\Bbb Q\cap (-\infty,\psi(x)],$ so $x=\psi(x).$
(1b).In particular, letting $F=\Bbb R$ in (1a), the only sub-field of $\Bbb R$ that is field-isomorphic to $\Bbb R$ is $\Bbb R$ itself, and the only field-isomorphism of $\Bbb R$ to $\Bbb R$ is $id_{\Bbb R}.$
(2). If $ B $ is a field and $\psi_1, \psi_2$ are field-isomorphisms from $\Bbb R$ to $B$ then by (1b), $\psi_2^{-1}\psi_1=id_{\Bbb R},$ so $\psi_1=\psi_2.$