## A trichotomy theorem in natural models of AD+

June 30, 2009

[Updated: July 24, 2009]

Richard Ketchersid and I have submitted the paper A trichotomy theorem in natural models of ${\sf AD}^+$ to the Proceedings of BEST. The preprint is available at my papers page. In the paper we provide references and background for the results we discuss, so here I will only mention briefly what the paper is about.

${\sf AD}^+$ is a strengthening, due to Woodin, of the more familiar axiom of determinacy. In all known models of determinacy, it is the case that in fact ${\sf AD}^+$ holds. Since ${\sf AD}^+$ is an axiom about sets of reals, its natural models are those of the form $L({\mathcal P}({\mathbb R})),$ although there are models of ${\sf AD}^+$ not of this form.

In this paper, we prove the following result:

Theorem. Assume that $V=L({\mathcal P}({\mathbb R}))$ and that ${\sf AD}^+$ holds. Let $(X,\le)$ be any partially ordered set. Then either there is an injection of the full binary tree $2^{\mathbb N}$ into $X$ such that no two points in its image are $\le$-comparable, or else $X$ can be written as a well-ordered union of $\le$-chains.

This statement should be reminiscent of the Harrington-Marker-Shelah theorem on Borel orderings, and in a sense our argument is a generalization of this result.

Two corollaries are worth pointing out: Suppose first that $\le$ is simply the diagonal on $X.$ Then the theorem gives us:

Corollary. Assume that $V=L({\mathcal P}({\mathbb R}))$ and that ${\sf AD}^+$ holds. Let $X$ be a set. Then either ${\mathbb R}$ injects into $X,$ or else $X$ is well-orderable.

This can be seen as a generalization of Silver’s theorem on co-analytic equivalence relation. In particular, we have the following basis result:

Corollary. Assume that $V=L({\mathcal P}({\mathbb R}))$ and that ${\sf AD}^+$ holds. Then $\aleph_0$ injects into every infinite set, and if $X$ is uncountable, then either $\aleph_1$ or ${\mathbb R}$ injects into $X.$

Our arguments make use of technology developed by Woodin. First, any model of ${\sf AD}^+$ of the form $L({\mathcal P}({\mathbb R}))$ either satisfies ${\sf AD}_{\mathbb R}$ or else it has the form $L(S,{\mathbb R})$ for some set $S$ of ordinals.

In the second case, one argues via an analysis of the $\infty$-Borel sets. Essentially, one uses what is sometimes called code compression to obtain, given an $\infty$-Borel code for a set $A,$ local versions of this code, that are sufficiently absolute in that they compute traces of $A$ correctly both in small inner models of choice, and their forcing extensions. Once this is obtained, the result essentially follows from soft forcing arguments as if the original sets under consideration were Borel.

In the first case, one uses the argument above to see that a set $X$ is expressible as a well-ordered union of smaller sets, for which the result applies. This uses that, under ${\sf AD}_{\mathbb R},$ models of the form $L({\mathcal P}({\mathbb R}))$ are of the form ${\sf OD}((<\Theta)^\omega),$ where $(<\Theta)^\omega$ denotes the family of countable bounded subsets of $\Theta.$ One then uses the uniqueness of the supercompactness measures on ${\mathcal P}_{\omega_1}(\gamma)$ for $\gamma<\Theta$ to “paste together” the smaller pieces that make up $X$ in a coherent way. The idea in this case was suggested by Woodin, and it is surprisingly flexible.

As an application, we consider the countable-finite game due to Scheepers. In this game, one fixes a set $S,$ and two players, I and II, alternate for $\omega$-many moves, with I moving first, so that each move of I is a countable subset of $S,$ and each move of II is a finite subset of $S.$ Player II wins if and only if the union of the finite sets covers the union of the countable sets. If choice holds, it is obvious that player II has a winning strategy. The same argument shows that, without choice, player II has a winning strategy when $S$ is countable. In contrast, we prove:

Corollary. Assume that $V=L({\mathcal P}({\mathbb R}))$ and that ${\sf AD}^+$ holds. Then the countable-finite game on $S$ is undetermined for all uncountable sets $S.$

For a brief presentation of these results, see the talks that Richard and I gave at BEST 18, available here or at my talks page.

(The “trichotomy” in the title refers to an additional clause in the main theorem, related to the Glimm-Effros dichotomy. I expect to post about an extension of this part of the result soon.)