I read that there are $(M,E)$ well-founded model of ZF that in fact not to be. However I don't understand in what sense "not really to be ". I think that $(M,E)$ is well-founded because $(M,E)\vDash (\text{foundation})$ so does "not really to be " mean that any infinite $E$-descending sequence is not an element of M ?


$(M,E)$ satisfies the axiom of regularity if whenever $A\in M$ is a non-empty set, then there is some $B\in M$, such that $M\models B\mathrel{E} A$, and $M\models A\cap B=\varnothing$.

In the presence of choice, this is equivalent to saying that there is no $F\in M$, such that $M\models \operatorname{dom}(F)=\omega^M\text{ and }\forall n, F(n+1)\mathrel{E}F(n)$.

But it could be that such $F$ exists outside of $M$. If you are familiar with ultrapowers (ultraproduct where all the structures are identical), you can show that an ultrapower of $\Bbb N$ with a free ultrafilter over $\Bbb N$ is never well-founded. The same argument will also work for models of $\sf ZF$.

  • $\begingroup$ Thank you for the answer! To understand more about your answer I have some questions: 1. When you write the intersection of A and B is the empty set do you mean that $\lnot \exists z\in M (zEA\land zEB)$? and also do $\emptyset^M $ and $\omega^M$ are defined by the same formulas that usually define them in V but relativized to M,E ? ( for example $\emptyset^M $ is an element $x\in M $ such that $\forall z\in M \lnot zEx $) finally is it correct to say that "assuming axiom of choice" is the same that V satisfies this axiom or M satisfies it ? $\endgroup$ – eagle Jun 26 '15 at 14:32
  • $\begingroup$ Yes to the first question (on all of its sub-parts); $M$ for the second question, since we're interested in something happening inside $M$ (note that the existence of such $F$ always contradicts the axiom of foundation). $\endgroup$ – Asaf Karagila Jun 26 '15 at 14:36

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