# Pointwise definable models of determinacy

Suppose that the Axiom of Determinacy (AD) holds in $$L(ℝ)$$, and for some statement φ, $$α$$ is minimal such that $$L_α(ℝ)⊨φ$$. Do definable (in $$L_α(ℝ)$$) elements of $$L_α(ℝ)$$ form an elementary substructure $$L_α(ℝ)$$?

Extension: Assume ZF+AD (or if needed $$\text{AD}^+$$), and let $$W_α$$ consist of all sets of reals of Wadge rank $$<α$$. Suppose that for some statement $$φ$$, $$(α,β)$$ is lexicographically minimal such that $$L_β(W_α)⊨φ$$. Do definable (in $$L_β(W_α)$$) elements of $$L_β(W_α)$$ form an elementary substructure $$L_β(W_α)$$?

Notes:
* To handle $$α=0$$, we assume that all hereditarily finite sets are included. The nonextended version holds for $$α=0$$ because of lightface projective uniformization.
* A positive answer would likely extend to existence of lightface definable scales (or if we added a real number $$r$$ that could be referenced by $$φ$$, to scales to definable from $$r$$).
* A weakening of AD that might suffice is determinacy for $$L_{β+1}(W_α)$$.
* Under large cardinal axioms, the extension extends far beyond $$L(ℝ)$$ — and beyond Wadge ranks and definability in the minimal inner model of $$\text{AD}_ℝ$$ + "$$Θ$$ is Mahlo" containing all the reals.
* The extension is reducible to the use of $$W_α$$ (with different $$α$$) in place of $$L_β(W_α)$$ if we can show that under the conditions, $$L_{β+1}(W_α)$$ has a set of real numbers that codes $$L_β(W_α)$$, and such that quantification over the corresponding $$W_{α'}$$ allows effective use of that set.

Motivation

The axiom of choice, while very natural and very useful, leads to 'paradoxical' sets that are apparently not definable. To a mathematician objecting to such sets, we can try to reply that if you insist that all sets be definable, you will end up with a model of the axiom of choice. And indeed, a theory extending ZF (without adding new predicates) has a pointwise definable model iff the theory is consistent with V=HOD.

However, ZF is not finitely axiomatizable. And a positive answer to the question would imply that if we lower our requirement to a single statement (which can still have $$Σ_{100}$$ replacement), then canonical pointwise definable models abound. Or if we insist on the full replacement schema (not in this question), we can (perhaps canonically) add infinitely many predicates $$R_1,R_2,...$$ to the language and replacement in the extended language, and still have pointwise definable models of ZF (or ZFC) with $$R_{i+1}$$ permitting definition of some sets not in $$\text{HOD}_{R_1,...,R_i}$$.

• Your first paragraph reminds me of Lemma 2.17 and 2.25 of the Koellner-Woodin handbook article; however, they are using parameters from $\mathbb{R}$ and their language includes a predicate for the set of reals. With the use of these parameters, Lemma 2.25 says your set is equal to $L_\alpha(\mathbb{R})$. – William Mar 10 at 0:27

The least ordinal $$\kappa$$ such that $$L_\kappa(\mathbb R)$$ satisfies KP is a counterexample. This essentially reduces to Theorem 1.3 of Tony Martin's "The Largest Countable This, That, and the Other." Of course, we are assuming some determinacy; $$\text{AD}^{L(\mathbb R)}$$ is more than enough, and Martin used the optimal hypothesis.
Here goes. Since KP is finitely axiomatizable, $$L_\kappa(\mathbb R)$$ is the least level of $$L(\mathbb R)$$ satisfying some sentence, as you have required. The sets that are (lightface) $$\Sigma_1$$-definable over $$L_\kappa(\mathbb R)$$ are precisely the inductive sets. There is a largest countable inductive set, which is just the set of reals $$x$$ such that for some $$\alpha < \kappa$$, $$x$$ is definable in $$L_\alpha(\mathbb R)$$. But Martin's theorem states that this set in fact contains every real definable in $$L_\kappa(\mathbb R)$$. (We are translating Martin's theorem from the notation of his paper, where the collection of reals definable in $$L_\kappa(\mathbb R)$$ is denoted by $$\bigcup \Sigma^*_n$$.) The complement $$A$$ of the largest countable inductive set is of course coinductive, or in other words $$\Pi_1$$-definable over $$L_\kappa(\mathbb R)$$. Yet $$A$$ contains no reals that are definable in $$L_\kappa(\mathbb R)$$: by Martin's theorem it is equal to the set of all reals that are not definable in $$L_\kappa(\mathbb R)$$. It follows that the definable elements of $$L_\kappa(\mathbb R)$$ do not form an elementary substructure of $$L_\kappa(\mathbb R)$$, since any elementary substructure of $$L_\kappa(\mathbb R)$$ would contain an element of the nonempty definable set $$A$$.
• Thank you. I will accept the answer. Do you happen to know though of conditions that would give a positive answer? One possibility is for $α$ (or $α$ and $β$ for the extension) to have definable countable cofinality. – Dmytro Taranovsky Mar 10 at 3:26
• When $\varphi$ is a $\Sigma_1$ formula, I think you'll get an elementary substructure if and only if $\alpha$ is not admissible by the analysis from Steel's "Scales in $L(\mathbb R)$," Corollary 2.8 and Theorem 2.9. I don't know about the case when $\varphi$ is more complex, but there is probably a generalization. My guess would be it fails for all ordinals inside a $\Sigma_1$-gap, and holds for ordinals $\beta$ at the end of a gap if and only if $\beta$ fails to be strongly $\Pi_n$-reflecting for some $n<\omega$. (See Definition 3.1 of Steel.) But I never looked closely at the ends of gaps. – Gabe Goldberg Mar 10 at 20:33