5
$\begingroup$

$ZFC_2$, 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?

$\endgroup$
3
  • 1
    $\begingroup$ If there’s $\mathfrak{c}^+$ inaccessibles, some $V_{\kappa}$ has the same theory as a smaller one. Now treat that as your $V.$ $\endgroup$ Nov 27 '21 at 21:24
  • $\begingroup$ @ElliotGlazer So under those conditions there’s no consistent extension of $ZFC_2$ which has no set models? $\endgroup$ Nov 27 '21 at 21:36
  • 1
    $\begingroup$ A similar question is covered by Hamkins and Solberg. See the following arXiv paper or the following recorded talk. $\endgroup$
    – Hanul Jeon
    Nov 28 '21 at 0:16
6
$\begingroup$

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).

$\endgroup$
5
  • $\begingroup$ Wouldn't that be something similar to indescribable cardinals? $\endgroup$
    – Asaf Karagila
    Nov 27 '21 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. $\endgroup$ Nov 27 '21 at 23:11
  • 2
    $\begingroup$ 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. $\endgroup$ Nov 28 '21 at 0:32
  • $\begingroup$ @ElliotGlazer Isn't that $M$-superscript a problem though? $\endgroup$ Nov 28 '21 at 0:33
  • 1
    $\begingroup$ $\kappa$ has the property $`` \exists \alpha < \kappa$ such that $V_{\alpha} \equiv_{\text{SOL}} V_{\kappa}"$ because $M$ thinks $j(\kappa)$ has that property. $\endgroup$ Nov 28 '21 at 1:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.