Does ZF have an initial model?

Question

Definition 1: A class $\mathcal{K}$ of countable transitive models of $\text{ZF}$ has an "initial member" $M$ if each member of $\mathcal{K}$ is a forcing extension of $M$ for some partial order $\mathbb{P}\in M$ and some $\mathbb{P}$-generic $G$ over $M$.

Definition 2: An extension $T$ of $\text{ZF}$ has an "initial c.t.m" if the collection of all countable transitive models of $T$ has an "initial member".

Question 1: Assuming some consistency assumptions is it consistent that $\text{ZF}$ has an initial c.t.m?

Question 2: Assuming some consistency assumptions is it consistent that $\text{ZF}$ has a consistent extension like $T$ with an initial c.t.m?

Answer 1

Your question has a certain affinity with the concept of a solid bedrock model, which arises in the theory of set-theoretic geology. Namely, $W$ is bedrock for $V$ if $V$ is a forcing extension of $W$ and $W$ satisfies the ground axiom, meaning that it is not a forcing extension of any deeper ground, or in other words, that it is minimal among all the grounds of $V$. The model $W$ is a solid bedrock if $W$ is least among all the grounds of $V$.

Although assertions about whether there is a bedrock or whether there are grounds of a certain nature seem at first to be second-order assertions about $V$, because they quantify over the inner models $W$ that might be grounds, in fact these are all first- order expressible in the language of set theory. The reason is that the collection of grounds of $V$ is uniformly definable, in that there is a definable family of classes $W_r$ such that every $W_r$ is a ground of $V$; every ground of $V$ is $W_r$ for some $r$; and the relation $x\in W_r$ is definable in $x$ and $r$. Thus, one may quantify over the collection of grounds by quantifying over the parameter $r$ used in this definition.

In his dissertation, Jonas Reitz proved that there are bottomless models $V$, which have no bedrock models; that is, a bottomless model $V$ can be realized as a set-forcing extension $V=W[G]$ of a ground $W$, but one can always go deeper, and realize $W=W_0[G_0]$ as a forcing extension of a still deeper ground, with no bottom.

Answer 2

Let $T$ be the statement that says that $V=L$ and no $L_\alpha$ is a model of $\mathsf{ZF}+V=L$. This extension $T$ is consistent if $\mathsf{ZF}$ is, and has an initial member if there are any transitive set models of $\mathsf{ZF}$, namely, $L_\alpha$ for the smallest height $\alpha$ of a transitive model of $\mathsf{ZF}$. The point is that $L_\alpha$ is the only transitive set model of $T$, as any model is some $L_\beta$, and if $\beta>\alpha$, then $L_\beta$ sees that there is a transitive set model of $V=L$.

On the other hand, $\mathsf{ZF}$ itself has no initial member, since (provably in $\mathsf{ZF}$) there are proper class forcing extensions that are no set forcing extensions so, if $\mathsf{ZF}$ has any transitive set models at all, none of them can be an initial member.

An interesting (and harder) question is whether we can have a (recursively enumerable) $T$ as in the first example, with a unique transitive set model, but such that $T$ implies $V\ne L$. Such a $T$ must imply that there are no significant large cardinals, and cannot be proved consistent by forcing.