# Is there a 'largest' second-order categorical axiomatization of set theory, extended from ZFC2

While it's possible to obtain categorical second-order axiomatizations of set theory by extending ZFC2 with additional axioms (see ), these axioms seem 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 there is a 'largest' axiomatization, in the sense that no categorical extension of ZFC2 has a larger model?

In particular, my current understanding is that the largest model for a categorical 2nd-order theory corresponds to the 2nd-order lowenheim number (which is apparently larger than the first measurable cardinal, if such a cardinal exists ). So it would seem that a natural candidate for an additional axiom would be one that fixes the size of the unique model to be equal to the lowenheim number, e.g. perhaps specifying that the largest inaccessible cardinal corresponds to the one immediately prior to the second-order lowenheim number (possibly with the lowenheim number specified by an encoded description, since I believe the actual cardinal is not currently known?).

Unlike the axiomatizations from the link, the intention is that this would have the advantage that no 'larger' categorical axiomatization could be constructed to replace it. I suppose one potential issue is the possibility that an even higher order logic than 2nd-order could be used to axiomatize an even larger categorical ZFC2 extension; however, my understanding is that given the equi-satisfiability results for 2nd-order logic versus nth-order logic, the lowenheim number shouldn't be larger than the 2nd order value, in which case even the higher order ZFC extensions would not be larger than the second order one (though even if they were it would be interesting to know the maximal second order axiomatization)

• This really depends on the meta-theory. If we assume the existence of a unique inaccessible cardinal, then there is only one model of $\sf ZFC_2$ to begin with, so it is already categorical. If we assume that there is a proper class of inaccessible cardinals, then of course $\sf ZFC_2$ has arbitrarily large models, so it is not categorical to begin with. May 3, 2015 at 21:50
• @AsafKaragila I'm not sure I understand. You say that by specifying a unique inaccessible cardinal, the system is 'already categorical', but the concern raised by the question is that there are larger categorical axiomatizations than that one (e.g. an axiom specifying two inaccessible cardinals)? So the question is if there is a maximal categorical axiomatization? And the suggestion was that potentially the axiom could be chosen so that the size of the categorically specified model would be the 2nd-order lowenheim number (and so the largest possible)?
– Andy
May 3, 2015 at 22:04
• @AsafKaragila Also, do you agree with the claim that categorical 2nd order axiomatizations cannot have arbitrarily large models; rather the maximum possible model size of any categorical second order theory is bounded by the 2nd-order lowenheim number?
– Andy
May 3, 2015 at 22:10
• I meant that in the sense that if the universe of set theory has only one inaccessible cardinal, then $\sf ZFC_2$ has a unique model, so it is already categorical and has no larger model. If there are two inaccessible cardinals in the universe, there are two models, and two categorical completions; etc. etc. and under some conditions every second-order property will necessarily reflect down, so it becomes very unclear if there will be a largest categorical completion. May 3, 2015 at 22:11
• No. You forget that $\omega$ has no largest element, but it's still bounded. I agree that there can only be countably many categorical axiomatizations; or if you want, $2^{\aleph_0}$ completions to begin with. And internally to the universe we can of course bound them. But there is no guarantee that there is a largest. May 3, 2015 at 22:12