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Scott Spectral Gaps are Bounded for Linear Orderings

David Gonzalez, Matthew Harrison-Trainor

TL;DR

This work establishes that for linear orders, the gap between the complexity of a theory and the complexity of its simplest model is uniformly bounded: any satisfiable $\Pi_\alpha$ sentence extending the theory of linear orders has a model with a $\Pi_{\alpha+4}$ Scott sentence, giving Scott rank at most $\alpha+3$. The authors develop new syntactic hierarchies $\mathfrak{E}_\alpha$/$\mathfrak{A}_\alpha$ tailored to back-and-forth analysis, and a forcing construction to produce generic linear orders with controlled ranks. They prove the main bound via a two-stage scheme: (i) a verification that generic structures with a Property $(*)$ have low Scott rank, and (ii) a construction that attains such a generic model satisfying the target $\mathfrak{E}_\alpha$ sentence. The paper also exhibits nontrivial Scott spectral gaps through examples at limit and successor ordinals, introduces almost-bi-interpretations to transfer complexity up the hyperarithmetic hierarchy, and connects these results to faithful Borel reductions, concluding that linear orders are not faithfully Borel complete. These results illuminate why linear orders resist universal interpretations and have implications for the structure of Scott spectra across logical and descriptive-set-theoretic contexts.

Abstract

We demonstrate that any $Π_α$ sentence of the infinitary logic $L_{ω_1 ω}$ extending the theory of linear orderings has a model with a $Π_{α+4}$ Scott sentence and hence of Scott rank at most $α+3$. In other words, the gap between the complexity of the theory and the complexity of the simplest model is always bounded by $4$. This contrasts the situation with general structures where for any $α$ there is a $Π_2$ sentence all of whose models have Scott rank $α$. We also give new lower bounds, though there remains a small gap between our lower and upper bounds: For most (but not all) $α$, we construct a $Π_α$ sentence extending the theory of linear orderings such that no models have a $Σ_{α+2}$ Scott sentence and hence no models have Scott rank less than or equal to $α$.

Scott Spectral Gaps are Bounded for Linear Orderings

TL;DR

This work establishes that for linear orders, the gap between the complexity of a theory and the complexity of its simplest model is uniformly bounded: any satisfiable sentence extending the theory of linear orders has a model with a Scott sentence, giving Scott rank at most . The authors develop new syntactic hierarchies / tailored to back-and-forth analysis, and a forcing construction to produce generic linear orders with controlled ranks. They prove the main bound via a two-stage scheme: (i) a verification that generic structures with a Property have low Scott rank, and (ii) a construction that attains such a generic model satisfying the target sentence. The paper also exhibits nontrivial Scott spectral gaps through examples at limit and successor ordinals, introduces almost-bi-interpretations to transfer complexity up the hyperarithmetic hierarchy, and connects these results to faithful Borel reductions, concluding that linear orders are not faithfully Borel complete. These results illuminate why linear orders resist universal interpretations and have implications for the structure of Scott spectra across logical and descriptive-set-theoretic contexts.

Abstract

We demonstrate that any sentence of the infinitary logic extending the theory of linear orderings has a model with a Scott sentence and hence of Scott rank at most . In other words, the gap between the complexity of the theory and the complexity of the simplest model is always bounded by . This contrasts the situation with general structures where for any there is a sentence all of whose models have Scott rank . We also give new lower bounds, though there remains a small gap between our lower and upper bounds: For most (but not all) , we construct a sentence extending the theory of linear orderings such that no models have a Scott sentence and hence no models have Scott rank less than or equal to .

Paper Structure

This paper contains 14 sections, 41 theorems, 54 equations.

Key Result

theorem 1.3

For any countable ordinal $\alpha$, there is a satisfiable $\Pi_2$ sentence $T$ such that all of the models of $T$ have Scott rank $\geq \alpha$.

Theorems & Definitions (81)

  • definition 1.1
  • theorem 1.3: Harrison-Trainor
  • theorem 1.5
  • theorem 1.8
  • corollary 1.11
  • definition 2.1
  • Remark 2.1
  • definition 2.2
  • theorem 2.3: Karp Karp
  • lemma 2.4: Chen, Gonzalez, and Harrison-Trainor CGHT
  • ...and 71 more