# Non-equivalence of $\mathbb R$ in different set theories

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?

• No, rather it means that the surrounding set theory determines the content of $Z_2$. Sep 5, 2017 at 13:02
• 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. Sep 5, 2017 at 13:21
• @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? Sep 5, 2017 at 13:21
• Yes, essentially. Sep 5, 2017 at 13:23

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)."
• 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). Sep 5, 2017 at 16:09
• Go ahead. ${ }$ Sep 6, 2017 at 11:23
• 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? Sep 20, 2017 at 8:56