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$\omega_1$-anchored labels in minimal counterexamples to Vaught's conjecture: a per-witness trichotomy and an unconditional stationary dichotomy

Aug 2026 · 0 citations · 33 references
Mathematics

Abstract

Let $\varphi$ be a minimal counterexample to Vaught's conjecture in the sense of Montalban; such a $\varphi$ exists if Vaught's conjecture fails, by Steel and Harnik-Makkai. We bring into contact, at the level of statements, two bodies of work on the models of such a $\varphi$: the analysis of Gonzalez-Rossegger-Turetsky, in which at every countable level $\beta$ exactly one back-and-forth class $C_\beta$ is uncountable and, at fixed points of an associated function, has a distinguished member of least Scott rank (its label); and the supply of models with prescribed $\omega_1^A$ from higher recursion theory, namely Montalban's Gandy-basis lemma and Sacks'$\Sigma_1$-hull club. All results are theorems of ZFC under standing hypotheses (H0)-(H3). We prove: (i) a per-witness trichotomy -- for every limit $\lambda$ in the fixed-point club above $qr(\varphi)$, every model $A$ of $\varphi$ with $\omega_1^A=\lambda$ and Scott rank at least $\lambda$ either is the label $K_\lambda$, or lies outside $C_{\lambda+1}$ and so forces two non-isomorphic models of Scott rank $\lambda+1$, or is the label $K_{\lambda+1}$ and attains the Nadel bound; (ii) coordinate identities: the first and third branches are equivalent, level by level, to computations of $\omega_1$ of the labels; (iii) a seeding theorem: on a club, Sacks'construction supplies at every level a model of top rank with prescribed $\omega_1$, and his atomic chain consists of the labels; (iv) an unconditional stationary dichotomy: on that club, either stationarily many successor levels carry two non-isomorphic models of Scott rank $\lambda+1$, or stationarily many labels attain the Nadel bound. On the fiber of models with $\omega_1=\lambda$ we further prove uniqueness of the node-saturated model and give a complete isomorphism invariant with countable spectrum. We prove nothing bearing on Vaught's conjecture itself.

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