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Consider $ZFC_2$, the second-order version of $ZFC$ set theory. Are there extra first-order theorems of $ZFC_2$ that are not in $ZFC$?

user107952
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1 Answers1

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Since $\sf ZFC_2$ is equiconsistent with the existence of inaccessible cardinals, it follows that it proves that $\sf ZFC$ is consistent, which is a first-order statement about arithmetic; and using the completeness theorem, we can show there is a model of $\sf ZFC$, and in fact even much more than that, there is a transitive model of $\sf ZFC$.

If we also fix a meta-theory, then all the models of $\sf ZFC_2$ must contain all the real numbers, set of real numbers, and much more, and therefore agree about the continuum function below the first inaccessible. It does not mean that there is a proof of $\sf CH$, as second-order logic does not have a completeness theorem, but this sort of phenomenon is worth mentioning.

Asaf Karagila
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  • Yes, but I always find it curious that one does not specify what one means by theorem in this context. – Andrés E. Caicedo Aug 21 '17 at 20:29
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    Yes, I guess because for most people "natural deduction" is a one-fix-for-all solution, and they are unaware of the differences in deductive strengths and enumerability of these when you change the logic. – Asaf Karagila Aug 21 '17 at 20:33
  • Asaf, this may be outside the topic, but could you be more specific about 'agree about the continuum function'? – Trixie Wolf Aug 21 '17 at 20:52
  • @Trixie: That means that any model of $\sf ZFC_2$ must agree on the value of $2^{\aleph_0}$ on the $\aleph$-numbers, as well as any other cardinal below the first inaccessible. – Asaf Karagila Aug 21 '17 at 20:53
  • So, if $2^{\aleph_0}$ = $\aleph_2$ in one model, it does in all of them, in some hoof-wavy sense? I suspect I lack the understanding of model theory (or maybe just second-order set theory, which I have zero exp with) to grasp why you can't construct some models of ZFC$_2$ that satisfy CH and construct others which do not. (Feel free to ignore if I'm being super-ignorant.) – Trixie Wolf Aug 21 '17 at 21:12
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    @Trixie: Yes. The point is that $2^{\aleph_0}$ is a very small cardinal compared to the least inaccessible cardinal, and since all models of $\sf ZFC_2$ are of the form $V_\kappa$ for an inaccessible cardinal $\kappa$, they all must agree with the meta-universe and with one another about the value of $2^{\aleph_0}$. – Asaf Karagila Aug 21 '17 at 21:17
  • At the meta level that makes sense. But that doesn't mean that the models themselves need to agree, does it? If CH holds in the meta-universe, you can still construct models of ZFC where it doesn't hold in the model. I think I'm still missing something subtle about the distinction between models of ZFC and ZFC_2 here. I'll think it over some on my own, though. Thanks for the help! – Trixie Wolf Aug 21 '17 at 21:28
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    @Trixie: Models live in the same meta-theory. Sure, we can switch meta-theory, but then you are talking about meta-meta-theories. The point is that you can build models of *first-order* $\sf ZFC$ where CH holds and fails, but with second-order it is impossible. See the thread I linked in the comments to the question itself, there I explain why second-order logic is not a good fit for foundations for $\sf ZFC$. – Asaf Karagila Aug 21 '17 at 21:30