3
$\begingroup$

On page 2 in "The Hyperuniverse Project and Maximality" is written:

The models $\mathcal{M}$ of MK are of the form $\langle M, \in, \mathcal{C} \rangle$, where $M$ is transitive model of ZFC, $\mathcal{C}$ the family of classes of $\mathcal{M}$ (i.e. every element of $\mathcal{C}$ is a subset of $M$) and $\in$ is the standard $\in$ relation.

Why can we assume that $M$ is transitive? If we knew, that every model of MK is well-founded, we could use the Mostowski collapse lemma. But is this the case?

Remark: In the book MK is formulated in a two-sorted version (have a look at the link).

$\endgroup$
  • 2
    $\begingroup$ By "model" read instead "standard model", these are the ones we typically want to consider. (But no, arbitrary models need not be well-founded.) $\endgroup$ – Andrés E. Caicedo Nov 24 '18 at 13:31
  • $\begingroup$ If there are set models at all, you can use compactness to get illfounded ones. $\endgroup$ – Andrés E. Caicedo Nov 24 '18 at 17:38
  • $\begingroup$ Or you can start with an illfounded model of ZFC+"there is an inaccessible", and show that it contains a model of MK, and that this model is illfounded (from the outside). :-) $\endgroup$ – Andrés E. Caicedo Nov 24 '18 at 17:41
  • $\begingroup$ Should I delete this question? $\endgroup$ – Popov Florino Nov 24 '18 at 17:51
  • 3
    $\begingroup$ I would suggest instead that you post an answer yourself explaining the situation, so that people who may have a similar question in the future may find this post. $\endgroup$ – Andrés E. Caicedo Nov 24 '18 at 20:08
2
$\begingroup$

As Caicedo pointed out in the comments by "models" only "standard models" are meant. The argument that there are illfounded models of MK is exactly the same as for ZFC (e.g. here). If we assume the axiom of foundation at the background level such an illfounded model is necessarily not isomorphic to a transitive model. If we don't have foundation in the metatheory this is not always the case (see this question).

$\endgroup$

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.