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## Nonstandard Reals in the Complex Plane

[Edited so as to reflect answers and comments given so far]

Let $F$ be a real closed field. Then by 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) [ANSWERED] Is it possible for an uncountable such $F$ to be at least Lebesgue measurable ?*

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 have a subfield isomorphic to $\Bbb{R}$.

-
Can we rule out a) if we also ask that ${\mathbb C}=F[i]$? – Dave Marker May 28 2011 at 3:01
Dave: good question; I will add it to the list of questions. – Ali Enayat May 28 2011 at 18:22
A Borel subfield $F$ of $\mathbb C$ which is equal neither to $\mathbb C$ nor to $\mathbb R$ must have Hausdorff dimension $0$. And then, of course, $F+iF$ also has Hausdorff dimension $0$. ams.org/journal-getitem?pii=S0002-9939-02-06653-4 – Gerald Edgar May 28 2011 at 20:29
@Gerald--Thanks I had forgotten your result with Chris – Dave Marker May 28 2011 at 21:59
If $F$ is a Borel subfield of $\mathbb C$ (or more generally any Borel field") that is isomorphic to the reals. Then, as Ali points out, the ordering is Borel. There is a real coding the embedding $q\mapsto a_q$ of the rationals into $F$. The isomorphism between the reals and $F$ is given by $f(x)= y$ if $x<q \Leftrightarrow y<a_q$ for all rationals, which has a Borel graph. – Dave Marker Jun 2 2011 at 12:52
show 4 more comments

Is there a Cantor set $K$ in $\mathbb R$ of algebraically independent elements?
Thinning it out, if necessary, we can assume $K^n$ has Hausdorff dimension $0$ for all $n$, and there are still other things algebraically independent of it. Then $F_1 = \mathbb Q(K)$ is at least an analytic subfield, still of dimension zero, and its algebraic completion $F_2$ is again analytic, now isomorphic to $\mathbb C$, but not all of $\mathbb C$ and not equal to $\mathbb R$. So $F_2$ is still of Hausdorff dimension $0$. Of course it has a subfield $F_3$ isomorphic to $\mathbb R$. (Probably many different ones?) So if we now add something algebraically independent of that, specifying that it is infinitely large, say, compared to $F_3$, then can't we now do a real-completion to get something still analytic and nonarchimedean real closed?
 Edgar: there certainly is a Cantor set &K$in$\BBb{R}$of algebraically independent elements (so far, so good). But let me ponder the rest of your argument. – Ali Enayat May 28 2011 at 21:58 @Gerald-- There are perfect sets of algebraically independent reals. (For a logical argument--a perfect set of mutually generic Cohen reals has this property and the existence of a perfect tree of algebraically independent reals is absolute. This can probably be translated into a reasonably easy Baire Category proof). One difficulty I see with your sketch is that it is not clear you can find$F_3$as Borel subfield of$F_2$. – Dave Marker May 28 2011 at 22:02 ... and that last real-completion maybe doesn't make sense, either. In any case, though, as noted in the original question, if$F$is any real closed field of power c, then$F[i]$is isomorphic to$\mathbb C$, and my argument here says there is a Lebesgue null subfield of$\mathbb C$isomorphic to$\mathbb C$, and thus inside it a Lebesgue null copy of$F$. Answering (b) with no need for axioms beyond ZFC. But probably not Borel or even analytic. – Gerald Edgar May 29 2011 at 0:16 Edgar and Dave: it appears to me that the null copy of$F$can be arranged to be analytic since there is an infinite binary tree all of whose branches are algebraically independent (by a fine-tuning of the argument sketched by Dave). I will edit the question to reflect this development. – Ali Enayat May 29 2011 at 0:51 I would like to see more details of the$F$analytic claim – Gerald Edgar May 29 2011 at 12:55 show 1 more comment A partial answer to (b): Consistently, yes. LĂ¶wenheim-Skolem implies that there are non-Archimedean real closed fields of cardinality$\aleph_1$, and it is consistent that all sets of size$\aleph_1$are of measure zero. (For example, this follows from MA plus non-CH.) Alternatively: Take any model$V$of set theory, let$K\in V$be an uncountable non-Archimedean real closed field, and add a Cohen real$c$to$V$. In$V[c]$, the set of old reals has now measure zero, and$K$is still a non-Archimedean real closed field. I have a feeling that an absoluteness argument should now help to get the existence of$K$in$V$, but I cannot get it to work. I do not even know if I can get a definable (say: analytic)$K$in$V[c]$(though it seems to me that I can get a$K$containing a perfect set). - Why bother with$\aleph_1$? There are non-archimedean real closed fields of cardinality$\aleph_0$, and certainly when (non-constructively?) embedded into$\mathbb C$they are Borel sets. Of course the interesting question would be the case where the real closed field extends$\mathbb R$and thus has cardinal of the continuum. – Gerald Edgar May 27 2011 at 13:44 Gerald: precisely because of the point you made about countable non-archimedean real closed fields, my questions all are formulted for uncountable real closed fields. The most interesting case is when such a field$F$contains$Bbb{R}$as a subfield [part (c) of the question], but it seems to me that even arranging for$F$to be uncountable and measurable is nontrivial in$ZFC$. – Ali Enayat May 27 2011 at 14:17 Martin: thanks for your observation about the consistency of (b). Here is a thought: in the presence of sufficiently large cardinals, Woodin's$\Sigma^2_{1}$-absoluteness theorem, when coupled with your answer, assures us (it seems) that the answer to (b) is positive in forcing extensions satisfying$CH$. So perhaps already in$ZFC+CH\$ the answer to (b) is positive? – Ali Enayat May 27 2011 at 14:25