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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.

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  • $\begingroup$ 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 $T$? $\endgroup$ Commented 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. $\endgroup$ Commented 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 $\in$?) $\endgroup$ Commented 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). $\endgroup$ Commented Jan 10, 2023 at 2:51
  • $\begingroup$ Indeed, that's an excellent argument! $\endgroup$ Commented Jan 10, 2023 at 2:55

3 Answers 3

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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 "$V=\mathrm{HOD}$".

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  • $\begingroup$ 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. $\endgroup$ Commented 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. $\endgroup$ Commented 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
    Commented 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
    Commented Jan 10, 2023 at 15:31
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    $\begingroup$ 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). $\endgroup$ Commented Jan 11, 2023 at 3:54
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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.

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    $\begingroup$ 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. $\endgroup$ Commented Jan 8, 2023 at 18:47
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    $\begingroup$ 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. $\endgroup$ Commented Jan 8, 2023 at 20:15
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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 $\mathrm{V = HOD}$.

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  • $\begingroup$ A very nice observation! $\endgroup$ Commented 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. $\endgroup$ Commented Jan 9, 2023 at 16:57
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    $\begingroup$ @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
    Commented Jan 9, 2023 at 21:28

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