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Second-order ZFC offers partial categoricity in the sense that, given any two models, one of them must be isomorphic to an initial segment of the other [1]. However, this leaves questions regarding the height of the cumulative hierarchy unspecified.

So my question is if second-order ZFC can at least be said to be categorical with regard to its proper class models (i.e. having a unique proper class model up to isomorphism)? And in that sense is it also correct to say that ZFC2 has a unique 'largest' model (up to isomorphism), since the proper class models are 'larger' than the set models.

I am assuming standard semantics for second order logic (e.g. not Henkin semantics).

[1] http://web.mit.edu/arayo/www/puzzle.pdf

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    $\begingroup$ I don't think "proper class model with standard semantics" makes sense. If we are using 1st order ZFC to talk about models of 2nd order ZFC, then there is no such thing as the collection of all classes. Furthermore, 2nd order ZFC proves the consistency of 1st order ZFC in a straightforward way, so there is no hope of giving a natural proper-class interpretation of 2nd order ZFC. $\endgroup$ – Monroe Eskew May 10 '15 at 5:38
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    $\begingroup$ The real challenge is to decide how you want to formalize the question. As Joel David Hamkins' answer explains, if we choose an arbitrary model of $V$ of ZFC as out metatheory to give a definition of full second-order semantics, then that model $V$ will think that $V$ is itself the only class model of full second-order ZFC. Does this really show what you wanted to show? $\endgroup$ – Carl Mummert May 10 '15 at 12:37
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The only second-order part that is needed for this is the second-order power set axiom.

That is, if $M$ is any proper class model of (a weak fragment of) first-order ZFC and satisfies the second-order power set axiom, then $M$ contains all sets. The reason is that if we iterate the power set operation inside $M$, then we will construct the actual $V_\alpha$'s, since $M$ is correct about power sets. And so every $V_\alpha$ is contained in $M$ and so $M=V$.

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  • $\begingroup$ Thanks for the response! Do you agree with Carl Mummert's interpretation of your answer (above)?: "As Joel David Hamkins' answer shows, if we choose an arbitrary model of V of ZFC as the definition of our second-order semantics, then that model V will think that V is the only model of second-order ZFC." $\endgroup$ – Andy May 10 '15 at 12:49
  • $\begingroup$ Yes, I think that Carl is exactly right. $\endgroup$ – Joel David Hamkins May 10 '15 at 12:53
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    $\begingroup$ In particular, for people who have some favored set theoretic background (such as a "platonisitic $V$" consisting of "all sets"), the argument goes through in their background like any other. So rather than being more restrictive, the viewpoint that second-order logic is only defined relative to a set-theoretic background is more general. The source of categoricity is not only the use of second-order semantics; it is also in the choice of a particular set-theoretic background in which to interpret those semantics. @Andy $\endgroup$ – Carl Mummert May 10 '15 at 14:54
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    $\begingroup$ Andy, my view is that second-order logic must take place with respect to some set-theoretic meta-theoretic background, since it makes use of the concept of subsets of the domain. Which sets are there? I find all talk of second-order logic to be essentially set-theoretic in nature. Whether a given second-order theory is categorical or satisfiable or whatever will depend on the set-theoretic ontological commitments of one's meta-theory. Thus, one might as well have asked a set theory question about the meta-theory instead. $\endgroup$ – Joel David Hamkins May 11 '15 at 0:05
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    $\begingroup$ The concept of "standard" power set semantics is dependent on one's set-theoretic background. Every set-theoretic background thinks its own power set concept is "standard", and they can disagree. $\endgroup$ – Joel David Hamkins May 12 '15 at 11:14

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