6 Updated the status of the questions, and fixed a typo

Let $F$ be a real closed field. Then by Artin-Schrier Artin-Schreier theory, $F[i]$ is algebraically closed. If $F$ is further assumed to have cardinality at most continnum, then by a classical theorem of Steinitz [stating the isomorphism of any two uncountable algebraically closed fields of the same cardinality and characteristic], we can conclude that $F[i]$ is isomorphic to a subfield of $\Bbb{C}$ of complex numbers.

In particular, if $F$ is a non-archimedean real closed field of cardinality continuum, then $F[i]$ is isomorphic to $\Bbb{C}$ [The proof uses the axiom of choice in a serious way, by the way].

We can therefore conclude:

Theorem. Every nonarchimedean real closed field of power at most continuum is isomorphic to a subfield of $\Bbb{C}$.

As a special case, we may conclude that there is a subfield $F$ of $\Bbb{C}$ such that $F$ is a non-archimdean real closed field that, furthermore, has a subfield isomorphic to the field $\Bbb{R}$ of real numbers.

The above considerations allow me to state my questions.

Questions.

(a) [UNANSWERED] Is there an uncountable Borel non-Archimedean real closed field $F$ of $\Bbb{C}$?

NOTE: In his comment below Dave Marker asks whether this question has a negative answer if we further assume that $F[i]=\Bbb{C}$, then Gerald Edgar pointed out in his comment that this is indeed the case; based on a result that appears in a joint paper of his with Chris Miller.

(b) If the answer to (a) is negative, is [ANSWERED] Is it possible for an uncountable such $F$ to be at least Lebesgue measurable ? [if so, the measure has to be 0 by Steinhaus' theorem]. *

NOTE. (b) has been answered. First Martin Goldstern pointed out that (b) follows from $MA + \lnot CH$; and that (b) is also true in any universe of set theory obtained by adding a Cohen real. Then Gerald Edgar pointed out that (b) is provable outright in $ZFC$ [see the answers below].

(c) [UNANSWERED] If the answer to (a) is positive, does the answer change if we insist for $F$ to satisfy have a subfield isomorphic to $\Bbb{R}$.

5 added 227 characters in body

Let $F$ be a real closed field. Then by Artin-Schrier theory, $F[i]$ is algebraically closed. If $F$ is further assumed to have cardinality at most continnum, then by a classical theorem of Steinitz [stating the isomorphism of any two uncountable algebraically closed fields of the same cardinality and characteristic], we can conclude that $F[i]$ is isomorphic to a subfield of $\Bbb{C}$ of complex numbers.

In particular, if $F$ is a non-archimedean real closed field of cardinality continuum, then $F[i]$ is isomorphic to $\Bbb{C}$ [The proof uses the axiom of choice in a serious way, by the way].

We can therefore conclude:

Theorem. Every nonarchimedean real closed field of power at most continuum is isomorphic to a subfield of $\Bbb{C}$.

As a special case, we may conclude that there is a subfield $F$ of $\Bbb{C}$ such that $F$ is a non-archimdean real closed field that, furthermore, has a subfield isomorphic to the field $\Bbb{R}$ of real numbers.

The above considerations allow me to state my questions.

Questions.

(a) Is there an uncountable Borel non-Archimedean real closed field $F$ of $\Bbb{C}$? [in

NOTE: In his comment below Dave Marker asks whether this question has a negative answer if we further assume that $F[i]=\Bbb{C}$]. F[i]=\Bbb{C}$, then Gerald Edgar pointed out in his comment that this is indeed the case; based on a result that appears in a joint paper of his with Chris Miller. (b) If the answer to (a) is negative, is it possible for an uncountable such$F$to be at least Lebesgue measurable ? [if so, the measure has to be 0 by Steinhaus' theorem]. NOTE. (b) has been answered. First Martin Goldstern has kindly pointed out in his answer below that (b) follows from$MA + \lnot CH$; and that (b) is also true in any universe of set theory obtained by adding a Cohen real. The question remains whether Then Gerald Edgar pointed out that (b) is provable in$ZFC$, or at least outright in$ZFC+CH$ZFC$ [see the answers below].

(c) If either the answer to (a) or (b) have a is positiveanswer, does the answer change if we insist for $F$ to satisfy have a subfield isomorphic to $\Bbb{R}$\Bbb{R}$. 4 added 384 characters in body; added 2 characters in body; added 2 characters in body Let$F$be a real closed field. Then by Artin-Schrier theory,$F[i]$is algebraically closed. If$F$is further assumed to have cardinality at most continnum, then by a classical theorem of Steinitz [stating the isomorphism of any two uncountable algebraically closed fields of the same cardinality and characteristic], we can conclude that$F[i]$is isomorphic to a subfield of$\Bbb{C}$of complex numbers. In particular, if$F$is a non-archimedean real closed field of cardinality continuum, then$F[i]$is isomorphic to$\Bbb{C}$[The proof uses the axiom of choice in a serious way, by the way]. We can therefore conclude: Theorem. Every nonarchimedean real closed field of power at most continuum is isomorphic to a subfield of$\Bbb{C}$. As a special case, we may conclude that there is a subfield$F$of$\Bbb{C}$such that$F$is a non-archimdean real closed field that, furthermore, has a subfield isomorphic to the field$\Bbb{R}$of real numbers. The above considerations allow me to state my questions. Questions. (a) Is there an uncountable Borel non-Archimedean real closed field$F$of$\Bbb{C}$? [in his comment below Dave Marker asks whether this question has a negative answer if we further assume that$F[i]=\Bbb{C}$]. (b) If the answer to (a) is negative, is it possible for an uncountable such$F$to be at least Lebesgue measurable ? [if so, the measure has to be 0 by Steinhaus' theorem]. NOTE. Martin Goldstern has kindly pointed out in his answer below that (b) follows from$MA + \lnot CH$; and that (b) is also true in any universe of set theory obtained by adding a Cohen real. The question remains whether (b) is provable in$ZFC$, or at least in$ZFC+CH$. (c) If either (a) or (b) have a positive answer, does the answer change if we insist for$F$to satisfy have the extra property of having a subfield isomorphic to$\Bbb{R}\$?

3 Fixed typos; added 5 characters in body; deleted 1 characters in body; deleted 1 characters in body
2 Fixed Latex code, and some wording/spacing; edited body; added 2 characters in body
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