Does ZF have an initial model?

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?

• I know this is not what you are looking for, but I would just like to mention the possibility because it works well in categorical logic. If you widen your notion of model then you can get a sort of initial, actually called universal, model of any theory by a syntactic construction. Essentially you generate the Lindenbaum algebra, but of course this can get technically complicated. I've always wondered why set theory is "against" such methods. – Andrej Bauer Nov 29 '13 at 7:33
• @Andrej: Is the resulting model well-founded? Can you at least ensure it has only standard integers? – Asaf Karagila Nov 29 '13 at 12:33
• @Asaf: It's not a model in the usual sense, but since it embeds in any other model it cannot have nonstandard integers and it cannot have a descending sequence of ordinals. – François G. Dorais Nov 29 '13 at 14:55
• @Andrej: I don't think anybody is against that idea. In fact, set theorists regularly extend the notion of model in not unrelated ways. Do you have a good application or other interesting use for that model? If so, I'm sure set theorists will be interested. – François G. Dorais Nov 29 '13 at 14:59
• ZF(C) is a first-order theory, so presumably one would build a syntactic first-order hyperdoctrine (ncatlab.org/nlab/show/first-order+hyperdoctrine). So yes, we have a category of contexts (the only morphisms are proejctions because ZF has no function symbols), and then above each context the Lindenbaum algebra of formulas in that context. – Andrej Bauer Nov 29 '13 at 20:52

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.

• Joel, is the bottomless model is a model obtained by adding a Cohen to every regular cardinal using an Easton product? – Asaf Karagila Nov 29 '13 at 14:31
• Yes, that's right. If you do an iteration, you get the ground axiom, but with a product, you get a bottomless model. – Joel David Hamkins Nov 29 '13 at 16:30
• I cannot help but to observe that according to this terminology set theory is topless. – Andrej Bauer Nov 29 '13 at 19:57
• Joel, yes that was my guess. Thank you. – Asaf Karagila Nov 29 '13 at 22:00

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.

• Andres, one can modify your theory $T$ to say merely that $V$ is a forcing extension of $L$, and there are no $L_\alpha$ satisfying ZFC. This will still have your initial model, since all transitive models of $T$ will be forcing extensions $L_\alpha[G]$ of the minimal model; but meanwhile, $T$ now has many models. – Joel David Hamkins Nov 29 '13 at 13:30
• For your question at the end, I suppose that you want $Ta$ to be computably axiomatizable, since otherwise one can make examples by asserting that $V$ is a forcing extension of $L$ by a Cohen real whose first digit is $0$, second digit $1$, etc. (listing the digits of a particular generic real, plus no $L_\alpha$ is a model of ZFC. This has a unique model, but we've made the theory complicated. – Joel David Hamkins Nov 29 '13 at 13:49
• @JoelDavidHamkins Hi Joel. Yes, $T$ can be modified as you say in your first comment, at the cost of uniqueness. And yes, as you say in the second comment, I want a recursively enumerable $T$. – Andrés E. Caicedo Nov 29 '13 at 15:18