Isomorphism Classes of Real Closed Subfields of $\Bbb C$ The real line $\Bbb R$ is a maximal real closed subfield of the complex plane $\Bbb C$. How many such maximal real closed subfields exist(up to isomorphism)? Is there a way to see that there must be at least infinitely many such maximal real closed subfields?
 A: There are $2^{2^{\aleph_0}}$-many maximal real-closed subfields of $\mathbb{C}$ up to isomorphism.
Let me first appeal to a "big theorem": Since the theory $\mathrm{RCF}$ of real-closed fields is unstable, it has the maximal number ($2^\kappa$) of models of cardinality $\kappa$ up to isomorphism for every uncountable cardinal $\kappa$.
Taking $\kappa = 2^{\aleph_0}$, there are $2^{2^{\aleph_0}}$-many real closed fields up to isomorphism of cardinality $2^{\aleph_0}$. It remains to show that each such field embeds as a maximal real-closed subfield of $\mathbb{C}$.
Let $R$ be a real-closed field of cardinality $2^{\aleph_0}$, and let $C = R[i]$ be its algebraic closure. Then $|C| = 2^{\aleph_0}$, so there is an isomorphism $\sigma\colon C\cong \mathbb{C}$ (since any two uncountable algebraically closed fields of the same characteristic and cardinality are isomorphic). Since $R$ is a maximal real-closed subfield of $C$, $\sigma(R)$ is a maximal real-closed subfield of $\mathbb{C}$ which is isomorphic to $R$.

Above I appealed to Shelah's "Many Models Theorem" for unstable theories. The proof of this theorem is rather technical, but it's quite a bit easier in the special case of $\mathrm{RCF}$, as explained in this MathOverflow post by Dave Marker. Briefly:

*

*Show that there are $2^\kappa$-many linear orders of cardinality $\kappa$ up to isomorphism.

*For every linear order $L$, order the field $F _L = \mathbb{Q}(\{x_a\mid a\in L\})$ in such a way that each $x_a$ is greater than every rational number, and $x_a^n < x_b$ for all $n\in \mathbb{N}$ whenever $a<b$ in $L$.

*Let $R_L$ be the real closure of $F_L$. Show that $L$ can be recovered from $R_L$ up to isomorphism, so $L\not\cong L'$ implies $R_L\not\cong R_{L'}$.

