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Since I found out about it, I've always been interested in the Axiom of Determinacy rather than the Axiom of Choice. Along these lines, I've kept flipping back to, and occasionally looking on google, because I keep thinking ZF+AD should be able to prove non-obvious things about it, although I haven't found anything other than (I think) something saying it must be regular. So, I'm finally asking here.

$\Theta := \operatorname{sup}(\{\alpha \in \operatorname{Ord} : (\exists f \in \alpha^\mathbb{R})(\operatorname{Range}(f) = \alpha)\})$

What is known about $\Theta$ in ZF+AD? In particular, how big is it?
For example, is it known to be different from $\omega_2$?

Is anything more known in ZF + AD + V=L($\mathbb{R}$) ?

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up vote 10 down vote accepted


A good reference for this question specifically and determinacy in general is Kanamori's book on large cardinals, "The Higher infinite". The last chapter is devoted to determinacy.

We know a huge deal about $\Theta$. For example, it is much larger than $\omega_2$. It does not need to be regular, but it is regular if in addition we assume $V=L({\mathbb R})$.

The key fact to see that $\Theta$ is large is Moschovakis's coding lemma which, in its simplest version, says that:

(Under AD) if there is a surjection $f:{\mathbb R}\to\alpha$, then there is a surjection $f:{\mathbb R}\to{\mathcal P}(\alpha)$.

Harvey Friedman used this to prove that $\Theta$ is a limit cardinal. This is easy; the point is that there is a definable bijection between ${\mathcal P}(\tau)$ and ${\mathcal P}(\tau\times\tau)$ for any infinite ordinal $\tau$. But if $\tau$ is a cardinal, there is a surjection from ${\mathcal P}(\tau\times\tau)$ onto $\tau^+$: If $A\subseteq\tau\times\tau$ codes a well-ordering, send it to its order type. Else, to 0.

With a bit more effort, you can check that $\Theta=\aleph_\Theta$ and in fact it is limit of cardinals $\kappa$ such that $\kappa=\aleph_\kappa$, and it is a limit of limits of these cardinals, etc.

In $L({\mathbb R})$, $\Theta$ is regular (Solovay was first to prove this). In fact, if $V=L(S,{\mathbb R})$ for $S$ a set of ordinals, or $V=L(A,{\mathbb R})$ for $A\subseteq{\mathbb R}$, then $\Theta$ is regular.

(A technical aside: If AD holds, it holds in $L({\mathcal P}({\mathbb R}))$. Woodin defined a strengthening of AD that is now called $AD^+$. It is open whether $AD^+$ is strictly weaker than AD, since all known models of AD are models of $AD^+$ and any current technique that gives us a model of one gives us a model of the other. If $L({\mathcal P}({\mathbb R}))$ is a model of $AD^+$, then it is either of the form $L(A,{\mathbb R})$ for some $A\subseteq{\mathbb R}$, or else it is a model of $AD_{\mathbb R}$, the strengthening of AD where we allow reals (rather than integers) as moves of the games.)

However, ZF+AD does not suffice to prove that $\Theta$ is regular. If DC holds, the "obvious" diagonalization shows that ${\rm cf}(\Theta)>\omega$. But Solovay proved that $ZF+AD_{\mathbb R}+{\rm cf}(\Theta)>\omega$ implies the consistency of $ZF+AD_{\mathbb R}$, so by the incompleteness theorem, ZF+AD or even the stronger $ZF+AD_{\mathbb R}$ cannot prove that ${\rm cf}(\Theta)>\omega$.

Nowadays we know much more. For example, $\Theta$ is a Woodin cardinal in the HOD of $L({\mathbb R})$, and the computation of the large cardinal strength of $\Theta$ in the HOD of the models of AD is a guiding principle of what is now known as descriptive inner model theory.

You may be interested in the slides of recent talks by Grigor Sargsyan on the core model induction (which should be available somewher online, or from him by email). You will see there that the large cardinal strength of AD assumptions is calibrated by the large cardinal character of $\Theta$ inside HOD, and this is associated with the length of the so called Solovay sequence which keeps track of how difficult it is to define the surjections $$f:{\mathbb R}\to\alpha$$ as $\alpha$ increases. This difficulty is related to the Wadge degree of sets of reals present in the AD model.

(I can be much more detailed, but this will require me to get significantly more technical. Let me know.)

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This is good enough to be accepted even if you ignored this, but there's one more detail that I'm curious about: Asking in both ZF+AD and ZF+AD+V=L(R), is $L_\Theta$ a model of ZF? – Ricky Demer Nov 23 '10 at 1:46
@Ricky : Yes, ZF+AD proves that $\Theta$ is strongly inaccessible in HOD, and inaccessibility relativizes down, so $\Theta$ is strongly inaccessible in $L$, and $L_\Theta$ is therefore a model of ZFC. – Andrés Caicedo Nov 23 '10 at 2:22
A lower-powered alternative to Andres's proof (in the preceding comment) that AD implies $L_\Theta$ is a model of ZF: AD implies (with lots of room to spare) the existence of $0^{\#}$, which in turn implies that $L_\kappa$ is a model of ZF for every uncountable cardinal $\kappa$. – Andreas Blass Nov 24 '10 at 20:48
@Andreas: Hehe. Yes, this is easier. – Andrés Caicedo Nov 25 '10 at 17:49

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