Assuming the axiom of choice there is a neat way defining inaccessible cardinals as uncountable, regular, strong limit cardinals.

Without the axiom of choice we have several notions of inaccessibility (cf. this paper by Blass, Dimitriou, and Lowe). For the purpose of the question, when we say that $\kappa$ is inaccessible in $ZF$ we mean that it is an uncountable, regular limit cardinal, and for $\alpha<\kappa$ we have no surjection from $V_\alpha$ onto $\kappa$.

Consider the following example:

Suppose now that $V\models ZF+DC_\kappa+\lnot DC_{\kappa^+}$ and $\kappa\in V$ is inaccessible. In particular it is clear that $V_\kappa\models ZFC$. The last is quickly proved since $|V_\alpha|<\kappa$ (since the cardinals are comparable, and $\kappa\nleq |V_\alpha|$ from its inaccessibility).

If we now consider $V$, it is an end extension of $V_\kappa$, however $V$ violates the axiom of choice.

Suppose that there are two inaccessibles, $\lambda<\kappa$, then $V_\lambda$ has two end extensions, one is $V_\kappa$ which is a model of $ZFC$, and the other is $V$ which is not a model of choice.

Question: Can we violate other axioms a well? Could we somehow arrange a model that for an inaccessible $\kappa$ we have $V_\kappa\models V=L$, while above it we may have something wider such as a measurable cardinal? Or the negation of the axiom of choice?

Is it possible to "gradually" break some axioms? That is a model with a class of large cardinals such that at each point we get slightly more complicated than the previous step?

The motivation comes from several points at which it seems that we "measure how far $V$ is from $L$" by using forcing axioms, or large cardinals, and so on.


You can certainly have $V=L$ up to an inaccessible cardinal $\kappa$ and violations of choice higher up. Just choose a larger regular cardinal $\lambda$, adjoin a lot of generic subsets of $\lambda$ with conditions of size smaller than $\lambda$, and then form a symmetric submodel. In other words, do at $\lambda$ what Cohen did at $\omega$.

But you can't have a measurable cardinal in this situation, because that would give the existence of $0^\#$, violating $V=L$ in $V_\kappa$.

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  • $\begingroup$ Ah, of course. How about other axioms, though, for example $V=HOD$ or something else which I cannot think of right now..? $\endgroup$ – Asaf Karagila Nov 1 '11 at 20:15
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    $\begingroup$ The construction in the first paragraph of my answer also works for HOD --- below $\kappa$, you have $V=HOD$ because $V=L$, and higher up you have $V\neq HOD$ because choice fails. If you want a model where choice holds, just skip the "form a symmetric submodel" part. The Cohen subsets of $\lambda$ won't be OD. As for "something else which I cannot think of right now", this comment box is too small to contain the things I can't think of right now (or ever). $\endgroup$ – Andreas Blass Nov 1 '11 at 22:13

The following theorem of McAloon provides a striking analogue of your phenomenon in the realm of models of arithmetic. The idea here is that every nonstandard model of the weak theory $I\Delta_0$, which includes induction only for $\Delta_0$ formulas, has a nonstandard initial segment that is a model of the comparatively stronger theory PA. Thus, this is a situation where a model of a strong theory, PA, was end-extended to a model of a much weaker theory higher up, and the theorem is that this is unavoidable in nonstandard models of the weak theory.

Theorem. (McAloon, Kenneth, On the complexity of models of arithmetic. J. Symbolic Logic 47 (1982), no. 2, 403–415. Every nonstandard model of $I\Delta_0$ has an initial segment that is a nonstandard model of PA.

There has been further work with these models. For example, in Strong initial segments of models of $I\Delta_0$, Paola D'Aquino, Julia F. Knight, Fund. Math. 195 (2007), 155-176, the authors investigate the possibility of stronger features in the initial segments.

I am not sure to the extent to which there is a ZFC analogue of McAloon theorem. One natural analogue of it would be the question of whether every nonstandard model of KP has an initial segment that is a model of ZFC. I recall hearing once about this, but I don't recall now whether it was as a theorem or as a question.

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  • $\begingroup$ The actual McAloon theorem is much stronger. It says that for every $\Sigma_1$-sound recursively axiomatized theory $T\supseteq I\Sigma_1$, you can find (arbitrarily short) nonstandard initial segments that are models of $T$. For example, one can take for $T$ the set of all arithmetical consequences of ZFC (plus whatever large cardinals you please). Moreover, the theorem also holds for nonstandard models of the weaker theory $IE_1$ in place of $I\Delta_0$, as shown by Wilmers. $\endgroup$ – Emil Jeřábek Nov 2 '11 at 11:38
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    $\begingroup$ @Emil: The combination of "$\Sigma_1$-sound" near the beginning of your message and "whatever large cardinals you please" near the end suggests considerable confidence in my taste in large cardinal axioms. Thank you. $\endgroup$ – Andreas Blass Nov 2 '11 at 12:59
  • $\begingroup$ (In case I was unclear, I’m referring to the arithmetical hierarchy, not Levy hierarchy.) Under normal circumstances, the $\Sigma^0_1$-soundness of a particular large cardinal is implied by the existence of a larger large cardinal (or even of two specimens of the same cardinal). If the large cardinals in whose consistency you believe can be put in an increasing chain, you are covered. $\endgroup$ – Emil Jeřábek Nov 2 '11 at 14:13
  • $\begingroup$ Emil, I took Andreas to mean that your comment suggests that all the large cardinals that please us are actually also consistent... $\endgroup$ – Joel David Hamkins Nov 2 '11 at 15:43
  • $\begingroup$ I see. Well, I find inconsistent large cardinals rather unpleasant, but whatever. Large cardinals are not the point, the point is that you can take as strong a theory as you want, as long as it does not prove things that are too obviously false. $\endgroup$ – Emil Jeřábek Nov 2 '11 at 15:59

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