Axiom of Determinacy

Question

It is quite easy to see that $ZF + AD$ (the Axiom of Determinacy) implies the countable axiom of choice ($AC_\omega$), yet $AC$ is inconsistent with $AD$. The dependent choice principle $DC$ is however stronger than $AC_\omega$. In the chapter of Jech's book concerning $AD$, the author writes "we shall be working in $ZF+DC$". But how do we know a priori that $DC$ cannot turn out to be inconsistent with $AD$?

Answer 1

Although Asaf's answer is correct, a few clarifications are in order:

• The consistency strength of $\mathsf{AD}$ is $\omega$ Woodin cardinals (we do not need the extra measurable). That is, if there are $\omega$ Woodin cardinals, then in a forcing extension there is an inner model where $\mathsf{ZF}+\mathsf{AD}$ holds. And, conversely, if there is a model of $\mathsf{ZF}+\mathsf{AD}$, then there is a forcing extension of an inner model that satisfies $\mathsf{ZFC}+$ there are $\omega$ Woodin cardinals. This equiconsistency is due to Hugh Woodin.
• If $\mathsf{AD}$ holds, then it holds in $L(\mathbb R)$. This is straightforward, as winning strategies can be coded by reals.
• It is a theorem of Kechris that, provably in $\mathsf{ZF}$, if $L(\mathbb R)\models\mathsf{AD}$, then also $L(\mathbb R)\models\mathsf{DC}$. This is a delicate result; it anticipates much of the modern theory of determinacy by arguing through a kind of induction on the levels of $L(\mathbb R)$, using the theory of scales due to Steel.

It also happens that if $\mathsf{ZF}+\mathsf{DC}$ holds in the universe, then it holds in $L(\mathbb R)$, but the point of Kechris's result is that the compatibility of $\mathsf{AD}$ with $\mathsf{DC}$ is a deeper issue. The modern way of proving this result uses the derived model theorem, due to Woodin.

In $\mathsf{ZFC}$, if we have $\omega$ Woodins and a measurable above (and less than this is needed), then in fact $L(\mathbb R)$ is a model of $\mathsf{AD}$. We do not need to pass to a forcing extension anymore.

Also, to recover the $\omega$ Woodins from $\mathsf{AD}$, passing to a forcing extension is needed in general, that is, for every $n$, we can find inner models of $\mathsf{ZFC}$ with $n$ Woodin cardinals, but it may not be possible to find a way of "weaving" them together into a model with $\omega$ Woodins. The forcing extension (a version of Prikry forcing) is needed for that. Conversely, to see an inner model of determinacy, $\omega$ Woodins do not suffice in general (though they are more than enough for projective determinacy and somewhat more), and we need to go an inner model of a forcing extension (this is the "derived model" construction).

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The best references for these issues are two papers in the Handbook of Set Theory:

• Large cardinals from determinacy, by Woodin and Koellner, and
• Determinacy in $L(\mathbb R)$, by Neeman.

Answer 2

We also know that it is consistent relative to infinitely many Woodin cardinals and a measurable above them that $L(\mathbb R)\models ZF+DC+AD$.