# Minimum transitive models and V=L

Is there a c.e. theory $$T⊢\text{ZFC}$$ in the language of set theory such that the minimum transitive model of $$T$$ exists but does not satisfy $$V=L$$?

You may assume that ZFC has transitive models. Note that $$M$$ is minimal iff $$∀M' \, M'⊈M$$ and minimum iff $$∀M' \, M'⊇M$$.

It may be tempting to consider ZFC + $$0^\#$$ (assuming large cardinal axioms), but while this theory has a minimum inner model (i.e. $$L[0^\#]$$), it has incomparable minimal transitive models. Model comparability uses iterability, but transitiveness does not suffice for iterability. Moreover, for every c.e. $$T⊢\text{ZFC\P}+0^\#$$ having a model $$M$$ with $$On^M = α < ω_1$$, the intersection of all such $$M$$ equals $$L_α$$, and furthermore a subset of $$L_α$$ is definable (with parameters) in all such $$M$$ iff it is in $$L_{α^{+,\mathrm{CK}}}$$. To see this (briefly), $$0^\#$$ allows $$M$$ to 'continue' $$L$$ beyond $$α$$, and $$L_{α^{+,\mathrm{CK}}}⊆(L_{α^{+,\mathrm{CK}}})^M$$ (the well-founded part of any model of KP being admissible), and so $$L_{α^{+,\mathrm{CK}}}∩V_α=L_α$$. Also, existence of $$M$$ is $$Σ^1_1(α)$$, so the intersection of all $$M$$ is at most $$L_{α^{+,\mathrm{CK}}}$$.

A minimum transitive model of ZFC + $$A$$ for a statement $$A$$ cannot be produced through set forcing either. Also, I think that minimal models need not satisfy $$V=HOD$$; not sure about minimum models.

Thus, minimum transitive models usually either satisfy $$V=L$$ or do not exist. However, I suspect that exceptions exist, including in the form $$L_α[r]$$ with $$r∈ℝ$$, but require an interesting coding argument. Perhaps there is such an $$r$$ computable from the theory of the least transitive model of ZFC $$L_α$$, with $$r$$ self-verifying (in $$L_α[r]$$) and yet weak enough for $$L_α[r]⊨\text{ZFC}$$.

Update: I accepted Farmer Schlutzenberg's positive answer (see also the partial answers by Hamkins and Enayat). A remaining open problem is whether such a $$T$$ can be obtained by extending ZFC with a single statement.

• A related question: if $T,S$ are c.e. theories extending $\mathsf{ZFC}$ with transitive models and some transitive model of $T$ does not contain any transitive model of $S$, must every transitive model of $S$ contain a transitive model of $T$? Jan 8, 2023 at 21:54
• @NoahSchweber No, for example consider ZFC + $V=L[G]$ for $G$ generic for $\operatorname{Add}(ω,1)$ for $S$ and $\operatorname{Add}(ω_1,1)$ for $T$. Then, no transitive model of $S$ of the minimum height is a superset of a transitive model of $T$, and vice versa. Jan 8, 2023 at 23:02
• Sorry, I mistyped my question: $T,S$ are supposed to each have a minimum transitive model. So an affirmative answer to my question would a fortiori imply an affirmative answer to yours, since levels of $L$ are comparable. (Maybe the following phrasing is better: say that a transitive set $A$ is basic iff there is some c.e. theory $T$ such that $A$ is the minimum transitive model of $T$. Are the basic sets linearly orderd by $\in$?) Jan 8, 2023 at 23:05
• @NoahSchweber With Farmer Schlutzenberg's answer below, there are such $S$ and $T$ (by using different generics for the forcing in the answer). Jan 10, 2023 at 2:51
• Indeed, that's an excellent argument! Jan 10, 2023 at 2:55

Yes, I claim you can in fact get one whose minimum model is a set forcing extension of a segment of $$L$$. Let $$L_\alpha$$ be least modelling ZFC. Let $$\mathbb{P}=\mathbb{P}^{L_\alpha}$$ be Jensen's forcing for adding a $$\Pi^1_2$$-singleton, as defined over $$L_\alpha$$; ZFC proves that forcing with $$\mathbb{P}^L$$ over $$L$$ adds a unique $$L$$-generic filter. Working in $$V$$, define the sequence $$\left_{n<\omega}$$ through $$\mathbb{P}$$, which will be $$L_\alpha$$-generic, as follows. Fix a recursive enumeration $$\left<\varphi_k\right>_{k<\omega}$$ of all formulas in the language of set theory in one free variable. Now let $$p_0=\emptyset$$. Suppose we have defined $$p_n$$. Let $$\alpha_n$$ be the least $$\beta$$ such that $$L_\beta\preccurlyeq_{\Sigma_n}L_\alpha$$. Let $$p_{n+1}$$ be the least $$p\in\mathbb{P}$$ such that $$p\leq p_n$$ and $$p$$ is in all open dense subsets $$D$$ of $$\mathbb{P}$$ which are defined over some $$L_{\alpha_i}$$ by $$\varphi_j$$, for some $$(i,j)$$ with $$i,j\leq n$$. (That is, for some such $$(i,j)$$, $$D$$ is the unique $$D'\in L_{\alpha_i}$$ such that $$L_{\alpha_i}\models \varphi_j(D')$$.) This $$p$$ exists since there are only finitely many dense sets we consider here. This defines the sequence.

Let $$G$$ be the filter generated by $$\left_{n<\omega}$$. Since $$L_\alpha$$ is pointwise definable, $$G$$ is $$L_\alpha$$-generic. So $$L_\alpha[G]\models$$ ZFC.

Now let $$T$$ be the following recursive (not just r.e.) theory extending ZFC, which will correspond to the construction above: for $$n<\omega$$ let $$T_n$$ be ZFC + "$$V=L[g]$$ for some $$(L,\mathbb{Q})$$-generic filter $$g$$, where $$\mathbb{Q}$$ is Jensen's forcing in $$L$$, and defining $$q_n$$ with respect to $$L$$ as in the construction above, we have $$q_n\in g$$". (That is, note that there is a recursive sequence $$\left<\psi_n\right>_{n<\omega}$$ of formulas such that for each $$n$$, we have that $$p_n$$ is the unique $$p\in L_\alpha$$ such that $$L_\alpha\models\psi_n(p)$$. Have $$T_n$$ assert that there is $$q\in g$$ (where $$g$$ is as above) such that $$L\models\psi_n(q)$$. Thus, the sequence $$\left_{n<\omega}$$ is recursive. Of course, the complexity of $$\psi_n$$ increases with $$n$$. Note that the "$$g$$" is a bound variable, not some new constant, so I am working with only the language of set theory.) Now set $$T=\bigcup_{n<\omega}T_n$$. So $$T$$ is recursive.

Claim: $$L_\alpha[G]$$ is the minimum transitive model of $$T$$.

Proof: Let $$P$$ be any transitive model of $$T$$. Then certainly $$L_\alpha\subseteq P$$, and easily if $$\alpha<\mathrm{OR}^P$$ then $$L_\alpha[G]\subseteq P$$. So we may assume $$\alpha=\mathrm{OR}^P$$. So $$L^P=L_\alpha$$. But then letting $$G'$$ be the (unique) $$(L_\alpha,\mathbb{P})$$-generic filter such that $$P=L_\alpha[G']$$, we get $$p_n\in G'$$ for each $$n<\omega$$, since $$P\models T_n$$ for each $$n$$. But therefore $$G'=G$$, so $$P=L_\alpha[G]$$, which suffices.

Remark: This construction is somewhat related to that for Proposition 34 of "On a Conjecture Regarding the Mouse Order for Weasels", arXiv:2207.06136, joint with Jan Kruschewski; (that proposition is stated rather generally, but in its simplest instantiation it gives an example of $$G$$ which is (only just) Cohen generic over $$L_{\omega_1^{\mathrm{ck}}}$$ but with KP failing in $$L_{\omega_1^{\mathrm{ck}}}[G]$$).

Remark 2: The question is rather related to a question of Harvey Friedman's, on which Woodin and Koellner made recent (boldface) progress. The question was (if I recall it precisely) whether there can be an ordinal $$\alpha$$ and a single sentence $$\varphi$$ such that there is a unique transitive model $$M$$ such that $$\mathrm{OR}^M=\alpha$$ and $$M\models$$ ZFC + "$$V\neq L$$" + $$\varphi$$. It was already known that any such model must satisfy "$$0^\sharp$$ does not exist", and I think also that it must satisfy "$$V=\mathrm{HOD}$$".

• Thank you. I mistakenly thought that set forcing cannot produce such a model but that limitation only holds if we are extending ZFC with a single statement. Also, technically, the axiomatization is recursive, while the theory (as in the set of provable statements) is r.e. Jan 10, 2023 at 2:16
• Ah, Great! The underlying idea here is also similar to the main idea of arxiv.org/abs/1105.4597, achieving pointwise definability in a class forcing extension, which it seems will also work with the same method. Jan 10, 2023 at 9:52
• Beautiful. By the way, to my recollection, the construction of the sequence of conditions $p_n$ is reminiscent of a similar construction by Stephen Simpson, which was used to prove that every countable model M of PA or ZFC carries a class C such that (M,C) is pointwise definable (ams.org/journals/proc/1974-043-01/S0002-9939-1974-0434801-5/…) Jan 10, 2023 at 10:14
• Your construction also provides a foil an old theorem of Harvey Friedman that mentioned in my answer to an MO question ( mathoverflow.net/questions/190902/…). Moreover, it answers the open question posed in my second remark there. Jan 10, 2023 at 15:31
• Remarkably, your argument can be extended to get "ordinal-categorical" c.e. theories extending ZFC + V≠L and having arbitrarily-large transitive models; see my answer to the question linked by @AliEnayat (here). Jan 11, 2023 at 3:54

This is not a full answer, but I found it interesting to notice that if we relax the c.e. requirement somewhat, then there is a sweeping positive answer.

Theorem. Every complete theory extending ZFC + V=HOD has a minimum transitive model, if it has any transitive models.

Proof. Suppose $$T$$ is a complete theory extending ZFC + V=HOD. Since there is a definable global well order in this theory, we have definable Skolem functions. Therefore, in any model of $$T$$ the parameter-free definable objects will form an elementary substructure. This model will be pointwise definable, and a copy of it will be contained in all other models of the theory. So it will be a minimum model of $$T$$. $$\Box$$

The essence of the idea is that the pointwise definable models of ZFC are exactly the prime models of the theory ZFC + V=HOD.

Corollary. Every transitive model of ZFC + V$$\neq$$L + V=HOD sits above a minimum model of its theory.

So this provides instances of your requested phenomenon using $$T=\text{Th}(M)$$, where $$M$$ is any transitive model of ZFC + V$$\neq$$L + V=HOD.

Meanwhile, similar thinking leads to a negative answer for arithmetically definable complete theories.

Corollary. If a complete theory $$T$$ extends ZFC +V=HOD and is arithmetically definable, then it has no transitive models.

Proof. If it had a transitive model, then it would have a minimum transitive model $$M$$, which must be pointwise definable. But the theory $$T$$ is arithmetically definable, and would therefore be an element of $$M$$. But then $$M$$ would be able to define a copy of itself inside itself by consulting the theory — the model $$M$$ is uniquely isomorphic to the set of definable elements inside any model of $$T$$. But this is impossible since $$M$$ thinks that uncountable sets exist. $$\Box$$

One can relax arithmetically definable to hyperarithmetic — one just wants to know that the theory must be inside any transitive model of the theory.

• Note that completing a c.e. theory is generally an arithmetic process, and indeed, every c.e. theory has a arithmetically low completion. But to find a completion realized in a transitive model will amount to an extra $\Pi^1_1$ requirement, since one wants to insist that the definable objects in the theory admit no infinite descent in the theory. It cannot be hyperarithmetic, for the reasons I mention in the post. Jan 8, 2023 at 18:47
• Let me also notice that the assertion that a given c.e. theory $T$ has a transitive model is a $\Sigma^1_2$ assertion, which is absolute to $L$. I'm not sure how useful this will be for answering the question, however, since in a minimal model of $T$, this is a false assertion by minimality. Jan 8, 2023 at 20:15

Here is another partial result; it complements Joel Hamkins' answer. Note that in the following theorem, $$T$$ is not necessarily a c.e. theory.

Theorem. Suppose $$T$$ is an extension of $$\mathrm{ZF} + \exists a~\mathrm{V}=\mathrm{L}[a]$$ that has a minimum transitive model $$M$$. Then $$M$$ satisfies $$\mathrm{V = HOD}$$.

Proof outline. By a classical result of Vopěnka there is a partial order $$\mathbb{P}$$ in $$\mathrm{HOD}^{M}$$ such that the model $$M$$ is a $$\mathbb{P}$$-generic extension of $$\mathrm{HOD}^{M}$$. Moreover, as shown by Grigorieff (see Theorem 1 of Sec. 5 of this paper ) $$% \mathbb{P}$$ can be arranged to be weakly homogeneous.

Now let $$G_1$$ and $$G_2$$ be mutually generic $$\mathbb{P}$$-filters over $$\mathrm{HOD}^{M}$$. For $$i=1,2$$ let $$N_i$$ denote $$\mathrm{HOD}^{M}[G_i]$$. By weak homogeneity of $$\mathbb{P}$$ we have:

$$(1)~~\mathrm{Th}(M) = \mathrm{Th}(N_1)= \mathrm{Th}(N_2).$$

On the other hand, by an old argument of Solovay, the mutual genericity of $$G_1$$ and $$G_2$$ over $$\mathrm{HOD}^{M}$$ implies:

$$(2)~~N_1 \cap N_2 = \mathrm{HOD}^{M}.$$

(1) and (2) together contradict the assumption that $$M$$ is a minimum model of $$T$$, thus $$M$$ satisfies $$\mathrm{V = HOD}$$.

• A very nice observation! Jan 9, 2023 at 15:23
• Do you know what happens in ZFC without $V=L[a]$? Every set is still generic over HOD (using weakly homogeneously forcing per the reference) but I do not know about complications for $V$ as a whole. Jan 9, 2023 at 16:57
• @DmytroTaranovsky By a result of Hamkins and Reitz, V need not be a class generic extension of HOD , see their paper arxiv.org/pdf/1709.06062.pdf. The same paper mentions an earlier result of Sy Friedman that is also relevant to your question. Jan 9, 2023 at 21:28