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Let $F$ be a meromorphic function on $\mathbb{C}$, and assume that the first-order theory of $(\mathbb{C},F)$ defines $\mathbb{Z}$, which means that there exists a formula $\varphi(z)$ (in the language of fields, so we cannot use complex conjugation, but parameters from $\mathbb{C}$ are allowed) such that

$\{z \in \mathbb{C}\mid \varphi(z)\}=\mathbb{Z}$.

(The last statement is not strictly first-order, but it can be made so: more precisely we can express that $\varphi$ defines a cyclic group that happens to be $\mathbb{Z}$ in the model $(\mathbb{C},F)$).

Now I propose the following statement:

  1. $\mathbb{R}$ is not definable in $(\mathbb{C},F)$;

  2. There are $2^{2^{\aleph_0}}$ automorphisms $\sigma$ of $\mathbb{C}$ such that $$\forall z \in \mathbb{C} (\sigma\circ F(z)=F \circ \sigma (z)).$$

    Is the above statement plausible? Would it be of any possible interest?

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  • $\begingroup$ Maybe there is something I don't understand, but the $\sin$ function both defines $\mathbb Z=\{z : \sin (z)=0\}$ and $\mathbb R=\{z : \sin (z)=\overline{\sin (z)}\}$. $\endgroup$ Mar 17, 2015 at 20:30
  • $\begingroup$ You cannot use complex conjugation, since it already implicitly defines $\mathbb{R}$. $\endgroup$
    – user38200
    Mar 17, 2015 at 20:31
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    $\begingroup$ Is there any meromorphic function at all from which $\mathbb{R}$ is definable? It seems plausible to me that any set which is definable from a meromorphic function must be either countable or cocountable, but I haven't thought about it much. $\endgroup$ Mar 17, 2015 at 20:37
  • $\begingroup$ (More directly, conjugation is not part of the signature or language for this structure.) $\endgroup$
    – Todd Trimble
    Mar 17, 2015 at 20:37
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    $\begingroup$ I think it is still open whether $\mathbb R$ is definable in $\mathbb C_{exp}$ so 1. is very strong. $\endgroup$ Mar 18, 2015 at 21:26

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