Throughout assume $\mathsf{ZFC}$ + "There is a proper class of inaccessible cardinals." I'm also happy to strengthen the large cardinal hypothesis if that would help.

Say that a model $M\models\mathsf{ZFC}$ is powerful iff every end extension satisfying $\mathsf{ZFC}$ is a top extension. The transitive powerful models are exactly the $V_\alpha$s where $\alpha$ is a worldly cardinal: that every $V_{\mathsf{worldly}}$ is powerful is trivial, and the large cardinal hypothesis above gives the other half of the result.

The ill-founded situation is on the other hand completely unclear to me:

Is there an ill-founded powerful model of $\mathsf{ZFC}$?

I don't really see where to start with this. Certainly there are no countable powerful models of $\mathsf{ZFC}$, ill-founded or otherwise, since we can (genuinely) force over such models; however, I don't see any useful tools for analyzing uncountable ill-founded models. The existence of rigid ill-founded models (e.g. the substructure of any ill-founded model of $\mathsf{ZFC+V=L}$ consisting of the parameter-freely-definable elements) does make the existence of powerful models feel not totally implausible, but that said I don't see how any of the techniques for building examples of the former type are useful for attempting to build an example of the latter.

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    $\begingroup$ @MonroeEskew An end extension adds no new elements to existing sets; a top extension additionally adds no new sets whose rank is an ordinal in the base model. E.g. forcing extensions are always end extensions but never top extensions, and (nontrivial) top extensions can be elementary but (nontrivial) end extensions can never be. $\endgroup$ Jul 19, 2021 at 22:19
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    $\begingroup$ Aren't there models of $\mathsf{ZFC}$ that have no end extensions? $\endgroup$ Jul 19, 2021 at 23:44
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    $\begingroup$ I thought I remembered seeing something, but that might have been elementary end extensions or something like that. $\endgroup$ Jul 19, 2021 at 23:46
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    $\begingroup$ I am still interested in the transitive case, when you don't have those large cardinals above. Could it be that there is a powerful transitive set $M$ that is not a worldly $V_\alpha$? Perhaps it was worldly in an inner model, but the worldliness of its height was destroyed, so that now it has no end-extensions. $\endgroup$ Jul 20, 2021 at 11:46
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    $\begingroup$ @VladimirReshetnikov Merely that $M$-generic filters are guaranteed to actually exist when $M$ is countable - as opposed to the usual subtleties around "forcing over the universe" and so on. $\endgroup$ Jul 20, 2021 at 19:26

1 Answer 1


Back in the mid 1980s I remember convincing myself (alas, in unpublished work) that there is an ill-founded model $M$ of ZFC that has no end extension to another model of ZFC. Such a model $M$ by default is powerful and technically answers the question.

My unpublished work above used techniques employed in the following related results from the same time period:

1. There are ill-founded models $M$ of any given consistent extension of ZFC that have no top extension to another model of ZFC; see Theorem 1.5 of this paper of mine. Note that top extensions are referred to as rank extensions in the paper.

2. There are ill-founded models $M$ of any given consistent extension of ZFC such that there are no $N \supsetneq M$ such that $N$ is a forcing end extension of $M$. This was noted in this other paper of mine (see the remark after Theorem 1.5, which itself relates to Theorem 1.4).

In (1) and (2) above, the cardinality of $M$ can be arranged to be $\aleph_1$. This is optimum since every countable model of ZF (including ill-founded ones) has lots of top extensions and forcing extensions


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