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The Axiom of Dependent Choice (DC) is often considered to be an "intuitive and non-controversial" version of choice used in the proofs of many theorems in Analysis. Similarly, the Axiom of Determinacy (AD) leads to especially nice properties for real numbers. Thus, in a sense, the system ZF + DC + AD is much "nicer" than ZFC, but maybe it restricts the "richness" of the set theoretic universe in some ways. Hence, I am wondering how this system interacts with large cardinals.

In particular:

  1. Since AD is inconsistent with unrestricted Axiom of Choice (AC), are any of the "standard" large cardinal axioms - strongly inaccessible, Mahlo, measurable, rank-into-rank etc - inconsistent with ZF + DC + AD?
  2. Conversely, can any such large cardinals - for example, strong inaccessibles - be proved to exist in this system? (Since the usual objection to proving their existence doesn't exist in this system)
  3. Will the typical "size relations" between large cardinals in ZFC hold? For instance, is a measurable cardinal going to be inaccessible in this system? (The only proof I know uses well ordering) Is the smallest measurable greater than the smallest inaccessible?

I realize the question is quite broad - I am interested in sampling the kinds of results that may have been proved/conjectured.

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    $\begingroup$ Regarding 2, ZF + AD + DC proves $\omega_1$ is measurable (in the sense that it has a $\omega_1$-complete normal measure.) But I do not think it proves there is an inaccessible: If there is such, then cutting off the universe at the level of the inaccessible would be a model of ZF + AD + DC. $\endgroup$
    – Hanul Jeon
    Commented Mar 22, 2023 at 7:08
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    $\begingroup$ Also, it is known that if we have enough large cardinals (like a proper class of Woodin cardinals) then $L(\mathbb{R})$ is a model of ZF + AD + DC. I guess $L(\mathbb{R})$ also possesses moderate large cardinals like inaccessibles. (A cardinal $\kappa$ is inaccessible if $V_\kappa$ is a model of second-order $\mathsf{ZF}$. This definition is equivalent to a usual definition if we have Choice.) $\endgroup$
    – Hanul Jeon
    Commented Mar 22, 2023 at 7:11
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    $\begingroup$ It is consistent that every regular cardinal bellow $\Theta$ is measurable (specifically, $L(\mathbb R)$ will satisfy this). IIRC, there is a similar result about supercomapcts $\endgroup$
    – Holo
    Commented Mar 22, 2023 at 9:49
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    $\begingroup$ Furthermore, in $ZF+AD$ every regular $\kappa<\aleph_{\omega_1}$ is Jonsson $\endgroup$
    – Holo
    Commented Mar 22, 2023 at 9:58
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    $\begingroup$ The known results are much stronger. Kanamori's book on large cardinals is a good introduction. For Jónsson cardinals and the like, see for instance "Determinacy and Jónsson cardinals in $L(\mathbb R)$", MR3343535. Work by Sargsyan and others has pushed significantly what we know about large cardinals below $\Theta$ in models of $\mathsf{AD}$ under various assumptions. $\endgroup$
    – Andrés E. Caicedo
    Commented Mar 22, 2023 at 14:05

1 Answer 1

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An unpublished theorem due to Woodin (which appears without proof as Theorem 7.35 of "In search of Ultimate L") states that if the $\Omega$ conjecture holds and there is a proper class of Woodin cardinals, then there is a partially ordered set $\mathbb P$ such that if $G\subseteq \mathbb P$ is a $V$-generic filter and $\mathbb R^* = (\mathbb R)^{V[G]}$, then $V(\mathbb R^*)$ is a model of AD. The hypotheses of this theorem are known to be consistent: in fact they hold in any model with a proper class of Woodins that satisfies Woodin's weakly homogeneous iteration hypothesis (WHIH), and this includes many of the current fine structure models by Steel's theorem that such models satisfy UBH; see Woodin's The Axiom of Determinacy, Forcing Axioms, and the Nonstationary Ideal, Definition 10.4 and Theorem 10.151. The model $V(\mathbb R^*)$ will in this context satisfy ZF + DC + AD plus a proper class of Woodin cardinals. If there is a proper class of cardinals with some stronger large cardinal property in $V$, say a proper class of extendible cardinals for concreteness, then $V(\mathbb R^*)$ will have a proper class of extendible cardinals by standard lifting arguments; but the $\Omega$ conjecture is not known to be consistent with the existence of a proper class of extendible cardinals (or even a proper class of measurable Woodin cardinals).

In summary, if ZFC plus the $\Omega$ conjecture is consistent with a proper class of $X$ cardinals, then the Axiom of Determinacy is consistent with a proper class of $X$ cardinals, for any $X\in \{\text{Woodin, Extendible, Superduperhuge, etc}\}$. Moreover the $\Omega$ conjecture is consistent with a proper class of Woodin cardinals assuming the existence of a proper class of Woodin cardinals is consistent.

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    $\begingroup$ Superduperhuge. ☺️ $\endgroup$
    – Andrés E. Caicedo
    Commented Mar 24, 2023 at 14:22

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