# 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. Nov 29, 2013 at 7:33
• @Andrej: Is the resulting model well-founded? Can you at least ensure it has only standard integers? Nov 29, 2013 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. Nov 29, 2013 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. Nov 29, 2013 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. Nov 29, 2013 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? Nov 29, 2013 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. Nov 29, 2013 at 16:30
• I cannot help but to observe that according to this terminology set theory is topless. Nov 29, 2013 at 19:57
• Joel, yes that was my guess. Thank you. Nov 29, 2013 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. Nov 29, 2013 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. Nov 29, 2013 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 $T$. Nov 29, 2013 at 15:18