Assume $\sf ZF+AD$. Is there some inner model $M$ containing all the ordinals such that $M\models\sf ZF+AD$ as well?

What if we require $\omega_1$ and/or $\omega_2$ to be computed correctly?

Can we say anything about these models (e.g. $M\models V=L(\Bbb R)$)?

Is it at all consistent?

There's no real reason to expect a minimal model, of course, since Woodin cardinals are involved. But $\sf AD$ is also a very strong axiom and might have unexpected consquences.


1 Answer 1


Assuming AD, Woodin showed that no inner model that is missing a real correctly computes $\omega_1$. (This almost follows from Theorem 9 of Velickovic-Woodin's "Complexity of the set of reals of inner models of set theory.") Therefore if you take an inner model $M$ of AD + $V = L(\mathbb R)$ whose $\omega_1$ is as small as possible, this model has no proper inner model of AD. (This argument shows that inclusion is a wellfounded partial order of the inner models of $\text{AD}+ V = L(\mathbb R)$...)

If one generically wellorders the reals of $M$ without adding reals, one obtains a model of ZFC in which there is a minimum inner model of $\text{AD}$. Update: this might be true, but as Dmytro Taranovsky points out in the comments, it is far from clear.

In general, however, there need not be a minimum inner model of $\text{AD}$. Assume there is a model of $\text{AD}$ containing only countably many reals. (This hypothesis is justified in the following paragraph.) Let $M$ be a minimal model of $\text{AD}$ with this property. Then there is a real $g$ that is $M$-generic for Cohen forcing. Let $N = L(\mathbb R)^{V[g]}$. By another theorem of Woodin, there is an elementary embedding $j : M\to N$. (There are more details in Kechris-Woodin's "Generic codes for uncountable ordinals.") Therefore $N$ is a minimal model of $\text{AD}$, since this is first order expressible, and $N$ is obviously distinct from $M$.

Of course, after forcing with $\text{Col}(\omega,\mathbb R)$ over a model of $\text{AD}^{L(\mathbb R)}$, there is an inner model of $\text{AD}$ that contains countably many reals, so the hypothesis of the above paragraph is consistent. In fact, the hypothesis is true: using $\mathbb R^\#$, one can build an inner model containing only countably many reals that is elementarily embeddable into $L(\mathbb R)$. (Take a countable elementary substructure of the mouse $\mathbb R^\#$ and then iterate away the top measure.) Therefore large cardinals imply that there is no minimum inner model of $\text{AD}$.

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    $\begingroup$ I, for one, would love to see this proof. I hope it is published somewhere... $\endgroup$ Jul 26, 2020 at 15:53
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    $\begingroup$ Yes there will be. Start with a minimal model $M$ with countably many reals. Then there is a Cohen real $g$ over $M$. Now $N= L(\mathbb R)^{M[g]}$ is elementarily equivalent to $M$ (another theorem of Woodin, but this is one is published in Woodin-Kechris “Generic codes for uncountable ordinals”), so $N$ is a minimal model distinct from $M$, but $g\in M[g]\setminus M$. (It may be counterintuitive that $M$ is not contained in $N$.) $\endgroup$ Jul 26, 2020 at 16:11
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    $\begingroup$ @FrançoisG.Dorais I found a reference for Woodin's theorem and updated the answer. $\endgroup$ Jul 26, 2020 at 22:55
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    $\begingroup$ For a variation on the question, we have a strong antiminimality statement: If an inner model $M⊨\text{AD}$ and $|ℝ^M|=ω$, then there is an inner model $L(R')⊨\text{AD}$ with $ℝ^M=ℝ^{L(R')[g]}$ and $g∈ℝ$ Cohen generic over $L(R')$. Assuming the axiom of choice, does this also hold for uncountable $ℝ^M$ (which would contradict the second paragraph of the answer)? On the other hand, the minimal inner model with the reals closed under $M_ω^\#$ is in some ways the minimal canonical inner model of AD (even though it is not a minimal inner model of AD). $\endgroup$ Aug 1, 2020 at 20:27
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    $\begingroup$ Using the converse of the Derived Model Theorem and $M⊨\text{AD}^+$, we get $M$ as a submodel of a generic extension. We can then 'factor out' one of the generic reals to get a generic $M'⊨\text{AD}$ (with $ℝ^{M'}⊂ℝ^M$) missing that real; and using $|ℝ^M|=ω$, $M'$ already exists in $V$. (P.S. Note that for a comment without names, only the author of the answer gets notified.) $\endgroup$ Aug 1, 2020 at 23:55

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