5
$\begingroup$

Although self-contained, this question is a follow-up to this earlier one. Also, thanks to Fedor Pakhomov for fixing a trivial early version of this question!

Work in $\mathsf{ZFC}$ + "There is a measurable cardinal," and let $\Omega$ be the least measurable. (Measurability is massive overkill here, I just want to make sure I have plenty of "room," so to speak.) Say that a sentence $\sigma$ is an inaccessibility safety net iff $V_\Omega$ satisfies each of the following statements:

  • $\mathsf{ZFC+\sigma}$ has arbitrarily large transitive models, and

  • whenever $M$ is a transitive model of $\mathsf{ZFC+\sigma}$ and $\alpha$ is $M$-inaccessible there is a transitive end extension $N\supseteq M$ such that $N\models\mathsf{ZFC}$, $\alpha$ is $N$-inaccessible, and $M\cap\mathsf{Ord}<N\cap\mathsf{Ord}$.

As far as I can tell, Farmer S's argument at the above-linked question - which relied on pointwise-definable models - does not prove that inaccessibility safety nets cannot exist (consider e.g. $\sigma\equiv\forall x[\mathsf{V\not=HOD}(x)]$).

Question: Is there an inaccessibility safety net?

In some sense an inaccessibility safety net has to axiomatize (a superset of) the consequences which the existence of a "large" inaccessible would have for a "small" inaccessible. Of course this is rather vague, but hopefully it helps motivate this question.

$\endgroup$
5
  • $\begingroup$ Obviously there is such a $\sigma$: "inaccessible cardinals form a proper class". For it we simply could take $N$ to be $M$. $\endgroup$ Commented Aug 23, 2022 at 13:15
  • $\begingroup$ I guess that there are two natural variants of the question that are more interesting: 1. to demand $M \subsetneq N$ and 2. to demand $o(M)<o(N)$. For 1. the answer is still positive, since we could take as $\sigma$: "there is a proper class of inaccessibles, and there are no transitive models with a Mahlo cardinal". Clearly all models of this $\sigma$ are countable and hence have generic extensions where we don't collapse inaccessibles. I don't immediately see how to solve 2., although my guess is that there are no $\sigma$ like this. $\endgroup$ Commented Aug 23, 2022 at 13:31
  • $\begingroup$ @FedorPakhomov Undeleted and fixed! $\endgroup$ Commented Aug 23, 2022 at 16:20
  • $\begingroup$ What theory are you asking for in $N$? You haven't said anything except that $N$ is a transitive end-extension of $M$, with strictly more ordinals, and still thinks $\alpha$ is inaccessible. But that is trivial, if we just add one more ordinal. I guess you want $N\models\text{ZFC}+\sigma$? $\endgroup$ Commented Aug 23, 2022 at 16:40
  • $\begingroup$ @JoelDavidHamkins Whoops, thanks - fixed! (In fact I don't mind if $N\models\neg\sigma$.) $\endgroup$ Commented Aug 23, 2022 at 16:43

1 Answer 1

7
$\begingroup$

There are no sentences $\sigma$ with this property.

For any $\sigma$ having transitive models we could consider the $<_L$-least well-founded model $E_\sigma$ of $\mathsf{ZFC}+\sigma$. Let $M_\sigma$ be the transitive collapse of $E_\sigma$. To answer the question it is sufficient to prove the following claim.

Claim. Suppose $\mathsf{N}\models \mathsf{ZFC}$ and $\mathsf{Ord}\cap N>\mathsf{Ord}\cap M_\sigma$. Then all $M_\sigma$-ordinals are countable from the perspective of $N$.

Note that in particular the claim implies that no $M_\sigma$-ordinal could be inaccessible from the point of view of $N$.

First observe that provably in $\mathsf{ZFC}$, if $E_\sigma$ exists, then it is countable. Indeed, we reason in $\mathsf{ZFC}$ and assume for a contradiction that $E_\sigma$ is uncountable. Consider $E'$ that is an isomorphic copy of a countable elementary submodel of $E_\sigma$ such that the domain of $E'$ is $\omega$. By condensation lemma $E'\in L_{\omega_1}$ and hence $E'$ is $<_L$ below any uncountable set, contradicting the minimality of $E_\sigma$.

Next observe that the property $\varphi(\gamma)$ asserting "there exists a well-founded models of $\mathsf{ZFC}+\sigma$, whose height is $\le \gamma$" is absolute for transitive models of $\mathsf{ZFC}$. For this we use the machinery of $\beta$-proofs, see [1] for definitions. Namely we consider the canonical $\beta$ pre-proof $P$ for the following sequent in two sorted language containing the special sort for ordinals $o$, the second sort which we treat as the sort of sets, predicate $<$ for $o$, predicate $\in$ for sets, and a unary function symbol $f$ from $o$ to sets: the sequent contains the negation of all axioms of $\mathsf{ZFC}$, the negation of $\sigma$ (both formulated in the second sort), and the negation of the axiom asserting that $f$ is an order-preserving embedding of $o$ into $\mathsf{Ord}$. Now if for a given ordinal $\gamma$, the proof-tree $P_\gamma$ is well-founded, then there are no well-founded models of $\mathsf{ZFC}+\sigma$ of the height $\le \gamma$. And from any infinite path through $P_\gamma$ we could construct a well-founded model of $\mathsf{ZFC}+\sigma$ of the height $\le \gamma$. The trees $P_\gamma$ have very simple definitions, in particular uniformly, for limit $\gamma$, they are $\Delta_1$-definable in $L_\gamma$. Thus we could reduce this model existence question to the question about ill-foundedness of the mentioned trees and the latter property is clearly absolute.

Now we are ready to prove the claim. If $\mathsf{Ord}\cap N>\mathsf{Ord}\cap M_\sigma$, then, by absoluteness proved above, in $N$ there is some transitive model of $\mathsf{ZFC}+\sigma$. Hence $N$ thinks that there is a countable transitive model of $\mathsf{ZFC}+\sigma$ and thus by Shoenfield's absoluteness $L^N$ thinks that there is a countable transitive model of $\mathsf{ZFC}+\sigma$. Hence $E_\sigma\in N$. Inside $N$ we define transitive collapse of $E_\sigma$ and observe that it gives us $M_\sigma$. Thus $M_\sigma\in N$ and $M_\sigma$ is a countable transitive model from the perspective of $N$. This immediatelly implies the claim.

[1] J.-Y. Girard. $\Pi^1_2$-logic, part 1: Dilators. Ann. Math. Logic, 21(2):75 – 219, 1981.

$\endgroup$
1
  • 1
    $\begingroup$ +1, thanks as always! I really need to learn some $\Pi^1_2$ logic eventually ... $\endgroup$ Commented Aug 23, 2022 at 17:58

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .