End extensions of models which do not preserve axioms

Question

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.

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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.

Answer 1

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 $V_\kappa$.

Answer 2

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.