# Does there always exist a categorical extension of $ZFC_2$ with no set models?

$$ZFC_2$$, i.e. second-order Zermelo-Fraenkel set theory with Choice, has only one proper class model upto isomorphism, namely $$V$$. But it may or may not also have set models. If $$V$$ has no inaccessible cardinals, then $$ZFC_2$$ has no set models, making $$V$$ its only class model upto isomorphism. But if $$V$$ has inaccessible cardinals then $$ZFC_2$$ can have some set models of the form $$V_\kappa$$ for some inaccessible cardinal $$\kappa$$.

My question, no matter what the truth is about the nature, existence, and number of inaccessible cardinals in $$V$$, does there always exist an extension of $$ZFC_2$$ which has no set models, making $$V$$ its only class model upto isomorphism? Or are there some conditions under which no consistent extension of $$ZFC_2$$ has no set models?

• If there’s $\mathfrak{c}^+$ inaccessibles, some $V_{\kappa}$ has the same theory as a smaller one. Now treat that as your $V.$ Nov 27 '21 at 21:24
• @ElliotGlazer So under those conditions there’s no consistent extension of $ZFC_2$ which has no set models? Nov 27 '21 at 21:36
• A similar question is covered by Hamkins and Solberg. See the following arXiv paper or the following recorded talk. Nov 28 '21 at 0:16

Since "consistent" is a weird notion in the context of second-order set-theories and moreover we can't even directly talk about a second-order theory being true of $$V$$ within $$V$$, I think it's usefully demystifying to rephrase the question in a "set-ish" way as follows:

Is it consistent with $$\mathsf{ZFC}$$ that there is some inaccessible cardinal $$\kappa$$ such that for every second-order theory $$T$$ in the language of set theory, if $$V_\kappa\models T$$ then $$V_\alpha\models T$$ for some $$\alpha<\kappa$$?

Note that whether or not $$V_\gamma\models S$$ for $$\gamma\le\kappa$$ and $$S$$ a second-order set theory is detected by the first-order diagram of $$V_{\kappa+1}$$, so the above question does make sense.

As Elliot Glazer observes, there is in this case a simple counting argument we can employ: if there are more than continuum-many inaccessibles, then some pair of inaccessibles $$\alpha<\kappa$$ have $$V_\alpha\equiv_{\mathsf{SOL}}V_\kappa$$.

Of course this is somewhat unsatisfying. What we have is a set-theoretic assumption which guarantees the existence of some appropriate $$\kappa$$, but we don't have a concrete property which would identify such a $$\kappa$$. So at this point it's natural to ask:

Is there a natural set-theoretic property - e.g. an already-studied large cardinal property - which guarantees that any $$\kappa$$ with that property has $$V_\alpha\equiv_{\mathsf{SOL}}V_\kappa$$ for some $$\alpha<\kappa$$?

Note that, trivially, such a property would have to be non-second-order-definable. And this pushes us into the realm of very strong properties indeed (see e.g. the discussion here).

• Wouldn't that be something similar to indescribable cardinals? Nov 27 '21 at 22:42
• @AsafKaragila Yes, basically $\Sigma^1_\omega$-indescribability (for theories instead of individual sentences, but I don't think that changes the strength too much). I guess measurables, being $\Pi^2_1$-indescribable, would probably do the trick. Nov 27 '21 at 23:11
• Yes measurables work, because for $j: V \rightarrow M$ with critical point $\kappa,$ $V_{\kappa+1}^M$ and $V_{j(\kappa)+1}^M$ have the same theory. Nov 28 '21 at 0:32
• @ElliotGlazer Isn't that $M$-superscript a problem though? Nov 28 '21 at 0:33
• $\kappa$ has the property $ \exists \alpha < \kappa$ such that $V_{\alpha} \equiv_{\text{SOL}} V_{\kappa}"$ because $M$ thinks $j(\kappa)$ has that property. Nov 28 '21 at 1:43