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I believe I miss something simple as I'm not quite good in foundations. Please tell me if I use some notion incorrectly.

As you know $\mathbb R$ is defined uniquely up to isomorphism. The theory of $\mathbb R$ is given by the second-order arithmetic, which is categorical and denoted by $Z_2$. The reals also can be defined in set theory. But taking into account the axiom of determinacy (AD) in fact we have two incompatible set theories: ZFC and ZF+AD. The question of existence of non-measurable sets is answered differently in these theories. And so the theories of reals in two set theories are not equivalent. Does it mean $Z_2$ is incompatible with one of them? If yes, then with which of them, and does it compatible with the other? Does it mean the unique (standard) model of $Z_2$ is also model in only one set theory?

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    $\begingroup$ No, rather it means that the surrounding set theory determines the content of $Z_2$. $\endgroup$
    – Andrés E. Caicedo
    Sep 5, 2017 at 13:02
  • $\begingroup$ Well, this is under assumption that the uniqueness of $\mathbb{R}$ also preserves things like the existence of a non-measurable sets. Does it? Analogously $\mathbb{N}$ and $\mathbb{Z}$ are equinumerous (isomorphic in Set) but when you introduce additional structure (e.g. semigroup structure) then they suddenly become non-isomorphic. $\endgroup$
    – freakish
    Sep 5, 2017 at 13:21
  • $\begingroup$ @AndrésE.Caicedo Ahh, so that's the point. The definition of $Z_2$ assumes we have set theory, and so we have different $Z_2$'s for different set theories, right? $\endgroup$
    – Seriy
    Sep 5, 2017 at 13:21
  • $\begingroup$ Yes, essentially. $\endgroup$
    – Andrés E. Caicedo
    Sep 5, 2017 at 13:23

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The categoricity of $\mathbb R$ is always in reference to a fixed model of set theory. You can have the same set theory but different models thereof. In this sense I disagree with the conclusion of the exchange of comments following the question and agree with Andrés's adjective "essentially" which however is rather significant when spelled out. For example, there are models of ZF+ACC that feature nonmeasurable sets and other models that don't.

Remark by Andrés: "The issue manifests itself much earlier. For instance, (conceivably) there are models of ZF where Con(ZF) holds and models where it fails, and this is a Π$_1^0$ statement, much simpler complexity-wise than a question about measurable sets, which actually is not even within the scope of $Z_2$ (unless the nonmeasurability manifests itself fairly early)."

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  • $\begingroup$ The issue manifests itself much earlier. For instance, (conceivably) there are models of $\mathsf{ZF}$ where $\mathrm{Con}(\mathsf{ZF})$ holds and models where it fails, and this is a $\Pi^0_1$ statement, much simpler complexity-wise than a question about measurable sets, which actually is not even within the scope of $Z_2$ (unless the nonmeasurability manifests itself fairly early). $\endgroup$
    – Andrés E. Caicedo
    Sep 5, 2017 at 16:09
  • $\begingroup$ @AndrésE.Caicedo, that's a very good point. While the kind of issue I mentioned may be more familiar to a traditionally trained mathematician, the point you make deserves being elaborated more fully. What I was mainly pointing out that one needs to be careful about the model/theory distinction. Would you like to post your comment as a separate answer? Otherwise with your permission I could add your comment at the end of my answer since the point deserves to be made more prominently. $\endgroup$
    – Mikhail Katz
    Sep 6, 2017 at 7:06
  • $\begingroup$ Go ahead. ${ }$ $\endgroup$
    – Andrés E. Caicedo
    Sep 6, 2017 at 11:23
  • $\begingroup$ Hi, thanks for your answer. I've investigated the subject a little and didn't get why do we need a fixed set theory model for the categoricity. This answer helped me to see how the theories and the results of my question are related to each other. Namely, we have $Z_2$, set theory and model theory formulated in that set theory, and that model theory tells us $Z_2$ is categorical. So we need a particular set theory, not a model. What I've missed? $\endgroup$
    – Seriy
    Sep 20, 2017 at 8:56
  • $\begingroup$ True, categoricity refers to a theory but your question express puzzlement over "non-equivalent R's". What I was pointing out is that the uniqueness of R can only make sense in the context of a fixed model of set theory. People losely speak of "categoricity" as characterizing R uniquely (in fact my using the term "categoricity" at the beginning of my answer is a case of such loose usage) but the uniqueness in question is not absolute in any sense but rather relative to a background model of set theory. @Seriy $\endgroup$
    – Mikhail Katz
    Sep 20, 2017 at 9:05

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