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 $L_β(W_α)$?

* 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.


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 $\text{HOD}_{R_1,...,R_i}$.

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    $\begingroup$ 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 $L_\alpha(\mathbb{R})$. $\endgroup$ – 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 $A$.

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  • $\begingroup$ 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. $\endgroup$ – Dmytro Taranovsky Mar 10 at 3:26
  • $\begingroup$ 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. $\endgroup$ – Gabe Goldberg Mar 10 at 20:33

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