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$ZFC_2$, 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?

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    $\begingroup$ If there’s $\mathfrak{c}^+$ inaccessibles, some $V_{\kappa}$ has the same theory as a smaller one. Now treat that as your $V.$ $\endgroup$
    – Elliot Glazer
    Commented Nov 27, 2021 at 21:24
  • $\begingroup$ @ElliotGlazer So under those conditions there’s no consistent extension of $ZFC_2$ which has no set models? $\endgroup$
    – Keshav Srinivasan
    Commented Nov 27, 2021 at 21:36
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    $\begingroup$ A similar question is covered by Hamkins and Solberg. See the following arXiv paper or the following recorded talk. $\endgroup$
    – Hanul Jeon
    Commented Nov 28, 2021 at 0:16

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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).

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  • $\begingroup$ Wouldn't that be something similar to indescribable cardinals? $\endgroup$
    – Asaf Karagila ♦
    Commented Nov 27, 2021 at 22:42
  • $\begingroup$ @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. $\endgroup$
    – Noah Schweber
    Commented Nov 27, 2021 at 23:11
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    $\begingroup$ 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. $\endgroup$
    – Elliot Glazer
    Commented Nov 28, 2021 at 0:32
  • $\begingroup$ @ElliotGlazer Isn't that $M$-superscript a problem though? $\endgroup$
    – Noah Schweber
    Commented Nov 28, 2021 at 0:33
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    $\begingroup$ $\kappa$ has the property $`` \exists \alpha < \kappa$ such that $V_{\alpha} \equiv_{\text{SOL}} V_{\kappa}"$ because $M$ thinks $j(\kappa)$ has that property. $\endgroup$
    – Elliot Glazer
    Commented Nov 28, 2021 at 1:43

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