While it's possible to obtain categorical second-order axiomatizations of set theory by extending ZFC2 with additional axioms (see [1]), these axioms tend to be somewhat arbitrary (e.g. adding an axiom that there are no inaccessable cardinals, or that there is exactly one inaccessible cardinal). In particular, the examples from the referenced question/answer place an arbitrary limit on the height of the cumulative hierarchy, where any one of the suggested 'new' axioms could just be replaced by another axiom specifying an even larger inaccessable cardinal.

So I'm wondering if it can be shown that there is a 'largest' axiomatization, in the sense that no categorical extension of ZFC2 has a larger model (or if the opposite can be shown i.e. that there is no largest)?

Of course, if we place arbitrary restrictions on large cardinals in our meta-theory, we could trivially make this question true (e.g. by assuming there is exactly one inaccessible in the meta-theory). A reasonable set of metatheory assumptions for the purpose of this question would be ZFC + large cardinal axioms i.e. assuming that relevant large cardinals do exist (if necessary). I'm assuming standard (e.g. non-Henkin) semantics for 2nd order logic.

Also, my understanding is that the 2nd order Lowenheim number does at least provide an upper bound on the size of the models associated with categorical extensions of ZFC2 [2], and per this question [3], it sounds like the Lowenheim bound would not be any larger for nth order logic compared to 2nd order logic.

[1] https://math.stackexchange.com/questions/317729/what-axioms-need-to-be-added-to-second-order-zfc-before-it-has-a-unique-model-u

[2] http://www.logic.math.helsinki.fi/people/jouko.vaananen/JV96.pdf

[3] Is the lowenheim-skolem number of nth order logic larger than the corresponding number for 2nd order logic

  • $\begingroup$ By the way, I previously posted this question to math stackexchange but did not get an answer (aside from a couple of comments), so now I'm re-posting here. $\endgroup$
    – Andy
    May 9, 2015 at 21:19
  • $\begingroup$ Could you make the question more precise? As you say, by making suitable hypotheses in the meta-theory, we can ensure that there is a largest such theory, and also that there isn't. But you seem to object to that kind of answer, and so I don't find it clear enough what would count as an answer for you. $\endgroup$ May 9, 2015 at 21:46
  • $\begingroup$ Thanks for considering the question! I tried to clarify it, but please let me know if it still needs improvement. In particular, I added: "A reasonable set of metatheory assumptions for the purpose of this question would be ZFC + large cardinal axioms i.e. assuming that relevant large cardinals do exist (if necessary)." $\endgroup$
    – Andy
    May 9, 2015 at 22:02
  • $\begingroup$ @JoelDavidHamkins By the way, while I did update the question per your recommendation with some suggested constraints on the meta-theory, I would also be happy if whoever answers the question makes their own judgement about what would be 'reasonable' meta-theoretic' assumptions (since they will likely have more expertise in this than I do). $\endgroup$
    – Andy
    May 9, 2015 at 22:35

1 Answer 1


$\newcommand\ZFC{\text{ZFC}}$Assume that $T$ is a categorical extension of second-order set theory $\ZFC_2$. So it has a unique model, which by Zermelo's theorem must be of the form $V_\kappa$ for some inaccessible cardinal $\kappa$.

Let $T^+$ be the theory asserting $\ZFC_2+$ there is a unique model of $T$, and no inaccessible cardinals above the size of that model. It follows (in the theory $\ZFC+$ there is a model of $T$ and a larger inaccessible cardinal) that the unique model of $T^+$ is $V_\delta$, where $\delta$ is the smallest inaccessible cardinal above $\kappa$. So $T$ was not largest in the sense you described. So there can be no such largest theory.

For example, if $T$ is the theory asserting $\ZFC_2+$ there is no inaccessible cardinal, then $T^+$ asserts that there is exactly one inaccessible cardinal, and $T^{++}$ asserts that there are exactly two inaccessible cardinals, and so on. So these theories are the theories that you seem already to have known about.

Notice that I am not placing any anti-large cardinal impositions in the meta-theory, only in the theory $T^+$, which I am considering but not asserting. Any much stronger large cardinal theory (asserting the existence of much larger large cardinals than a $T$-cardinal and one more inaccessible) will agree that $T^+$ is categorical, with a larger model than the unique model of $T$. So this argument seems to agree with your intentions for the large cardinal meta-theory.

  • $\begingroup$ Thanks for the response! One minor clarification: You say that $T^+$ asserts that there are no inaccessibles? Is the idea that $T^+$ asserts that there are no additional innaccessibles beyond those needed for the model of T, or am I misunderstanding? $\endgroup$
    – Andy
    May 9, 2015 at 23:05
  • $\begingroup$ Is there any justification for believing that if "$\ZFC+$ there is a model of $T$" is consistent then so is "$\ZFC+$ there is a model of $T$ and a larger inaccessible cardinal", other than the fact that no one has ever found a $T$ for which this seems to be false? $\endgroup$ May 9, 2015 at 23:39
  • $\begingroup$ Andy, Yes, that is what I meant, and I have edited. @Eric, one cannot justify these axioms, except by proving them in a stronger large cardinal theory. For all the usual large cardinal axioms, the assertion that there is an inaccessible cardinal above is a very mildly stronger theory. But of course, it is relatively consistent with ZFC that there is a model of $T$ and no model of $T^+$, but this is the kind of anti-large cardinal meta-theory to which the OP seemed to be objecting. $\endgroup$ May 9, 2015 at 23:45

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