Defining a sequence of models of ZFC satisfying a certain countability requirement Assuming whatever large cardinal axioms we please, is there a way to define a sequence of sets $M^\omega$ such that for all $n < \omega$, the following hold?


*

*$M_n$ is a transitive model of ZFC

*$M_n \in M_{n+1}$

*$M_{n+1}$ believes that every element of $M_n$ is countable


I do not especially mind if $M_{n+1}$ believes that $M_n$ is uncountable, but what I would really like is an actual explicit description of such a sequence, as opposed to a mere existence or consistency statement.
A related question.
If $M$ is a countable transitive model of ZFC, is there a canonical way of finding another countable transitive model $M'$ such that $M \in M'$ and every $x \in M$ is believed countable by $M'$?
 A: Just from the assumption that there is a transitive set model of set theory we cannot conclude that there is another one of different height. Similarly, just knowing that there are ordinals $\alpha_1<\dots<\alpha_n$ such that each $\alpha_i$ is the height of a transitive model of set theory, is not enough to conclude that there is a further height $\alpha_{n+1}$. To see this, simply assume that there is such an ordinal $\alpha_{n+1}$. By working in $L$ if needed, we may also assume that $V=L$. We may also assume that $\alpha_{n+1}$ is least possible. Now, consider the model $L_{\alpha_{n+1}}$. In this model, the statement that there are $n$ distinct ordinals that are heights of transitive models is true, but the statement that there are $n+1$ such ordinals is false, by absoluteness considerations. This indicates that there is no "canonical" way of finding models as in your last question.
To ensure that there are $\omega$ such heights, assuming that there is an inaccessible cardinal is an overkill. Assuming that there is a worldly cardinal is also much more than it is needed: If $V_\alpha\models\mathsf{ZFC}$, consider a countable elementary substructure of $V_\alpha$, and its transitive collapse. That gives us a transitive model $M_1$ of countable height. Now consider a countable elementary substructure of $V_\alpha$ with $M_1$ as an element. Its collapse is a transitive model $M_2$, again of countable height, and now $M_1\in M_2$ (since the collapse of $M_1$ is $M_1$), and $M_1$ is countable in $M_2$, by elementarity. Exactly the same argument gives us countable chains of such models, and beyond.
If you want the sequence a bit more explicit than obtained through the procedure just outline, note again that $L_\alpha$ is a transitive model of set theory if there is a transitive model of height $\alpha$. The argument using elementary substructures (and condensation) shows that the least such $\alpha$ is countable, that the second least such ordinal $\beta$ (if it exists) is also countable, and $L_\alpha$ is countable in $L_\beta$, that the third such ordinal $\gamma$ (if it exists) is also countable, and $L_\beta$ is countable in $L_\gamma$, etc.
A: Lets assume there is an inaccessible cardinal $\kappa$. Now $V_{\kappa}$ is a model of 
ZFC form a generic extension $V[G]$ where we collapse $\kappa$ to be countable then 
$V_{\kappa} \in V[G]$ is a model of ZFC and $V[G]$ thinks all elements of $V_{\kappa}$ are countable. Of course $V[G]$ is not a set model just a class model, so we can assume there is another inaccessible cardinal and repeat the same argument.
Thus briefly, we assume there is an infinite sequence of inaccessible cardinals and successively collapse them to be countable. The $V_{\kappa}$ are then your sequence.  
