This question is in connection with the question that I've asked at:

[Where do models of false theories exist?](https://mathoverflow.net/questions/294581/where-do-models-of-false-theories-exist)

The answer to that question was that any consistent theory can have its primitives be re-interpreted in such a manner as to come true. So the difference between a false theory and a true theory is one of reference, a true theory is one whose sentences are satisfiable in the part of the Platonic realm that it refers to.

Now according to that answer, I'll pose the following possibility and the question is what is against that possibility:

Now let's assume that there exists a Platonic world $P^{sets}$ of all sets, and
two Platonic worlds $P^{\in\_1}$ , $P^{\in\_2}$ of primitive ordered pairs of sets, these are taken to represent distinct membership relations between sets. So the ordered pairs in realms $P^{\in\_1}$, $P^{\in\_2}$ only have sets as their projections, so $P^{sets}$ is their domain, so they represent two kinds of membership relations between sets $\in\_1$ and $\in\_2$ relations defined on the same domain, as:

$y \in\_1 x \iff \exists p \in^\* P^{\in\_1} (p=\langle y,x \rangle)$

$y \in\_2 x \iff \exists p \in^\* P^{\in\_2} (p=\langle y,x \rangle)$

So for example the sentence $\exists x \forall y (y \not \in x)$ would be:

$\exists x \in^\* P^{sets} \forall y \in^\* P^{sets} (\not \exists p \in^\* P^{\in\_1} (p=\langle y,x \rangle))$

$\exists x \in^\* P^{sets} \forall y \in^\* P^{sets} (\not \exists p \in^\* P^{\in\_2} (p=\langle y,x \rangle))$

Where $\in^\*$ is the membership relation between sets and $P^{sets}$ and between ordered pairs of sets and the realms $P^{\in\_1}, P^{\in\_2}$.

Now since we are having two membership relations on sets defined after two Platonic realms, then we can have two theories each referring to one of these membership relations, so no confusion of reference would raise (as far as each theory is speaking correctly about the part of the Platonic realm that it refers to), and so both theories would be TRUE in that Platonic world. Accordingly we can have both membership relations obeying all rules of $\text{ZF}$ and yet one of them obeying $\text{Choice}$ while the other negating it.

So this would mean that the answer to as whether choice or negation of choice is true about membership in sets, is to say that there are two kinds of membership in sets, one fulfills choice and the other negates it. 

I don't see anything in the definition of a true theory [from a Platonic perspective] that can go against that possibility. Why should there be just one kind of membership in sets? there is no rule to the effect that no two distinct relations in the Platonic world can have the same domain, actually, this is not the case with the standard model of arithmetic, for it does have distinct relations having the same arity defined on the same domain of standard naturals (exp: the binary relations $Successor$ ,$<$; the ternary relations $+ ,\times $) so why not have the same situation with sets? 

Along the same lines of this argument, we may have two membership relations one obeying $\text{CH}$ and the other negating it, on the SAME domain of all sets.

This question is intended to be answered from a Platonistic perspective.