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.

  • $\begingroup$ 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 $T$? $\endgroup$ Jan 8, 2023 at 21:54
  • $\begingroup$ @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. $\endgroup$ Jan 8, 2023 at 23:02
  • $\begingroup$ 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 $\in$?) $\endgroup$ Jan 8, 2023 at 23:05
  • $\begingroup$ @NoahSchweber With Farmer Schlutzenberg's answer below, there are such $S$ and $T$ (by using different generics for the forcing in the answer). $\endgroup$ Jan 10, 2023 at 2:51
  • $\begingroup$ Indeed, that's an excellent argument! $\endgroup$ Jan 10, 2023 at 2:55

3 Answers 3


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 $\left<p_n\right>_{n<\omega}$ 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 $\left<p_n\right>_{n<\omega}$. 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 $\left<T_n\right>_{n<\omega}$ 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 "$V=\mathrm{HOD}$".

  • $\begingroup$ 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. $\endgroup$ Jan 10, 2023 at 2:16
  • $\begingroup$ 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. $\endgroup$ Jan 10, 2023 at 9:52
  • $\begingroup$ 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/…) $\endgroup$
    – Ali Enayat
    Jan 10, 2023 at 10:14
  • $\begingroup$ 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. $\endgroup$
    – Ali Enayat
    Jan 10, 2023 at 15:31
  • 1
    $\begingroup$ 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). $\endgroup$ 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.

  • 1
    $\begingroup$ 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. $\endgroup$ Jan 8, 2023 at 18:47
  • 1
    $\begingroup$ 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. $\endgroup$ 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 $\mathrm{V = HOD}$.

  • $\begingroup$ A very nice observation! $\endgroup$ Jan 9, 2023 at 15:23
  • $\begingroup$ 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. $\endgroup$ Jan 9, 2023 at 16:57
  • 2
    $\begingroup$ @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. $\endgroup$
    – Ali Enayat
    Jan 9, 2023 at 21:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.