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SOP_2 = SOP_3

Logic & foundations · posed by Mirna Džamonja, Saharon Shelah, 2004 · solved

1 attempt

Statement

The classes of SOP_2 and SOP_3 first-order theories coincide. This answers a question of Džamonja and Shelah from 2004.

Context

The new content is SOP2SOP3SOP_2 \Rightarrow SOP_3; the converse implication was known from the start. Dzamonja and Shelah asked whether either implication in SOP3SOP2SOP1SOP_3 \Rightarrow SOP_2 \Rightarrow SOP_1 reverses: Mutchnik answered the second (SOP1=SOP2SOP_1 = SOP_2), and this answers the first, collapsing the bottom of the hierarchy to SOP1=SOP2=SOP3SOP_1 = SOP_2 = SOP_3. The SOPnSOP_n hierarchy for n3n \ge 3 remains, as does everything above it.

One of the two 2004 Dzamonja-Shelah questions that organized two decades of classification theory, repeatedly highlighted in the NSOP_1 literature, and the surviving half after Mutchnik's celebrated SOP_1 = SOP_2. Resolving it collapses the bottom of the SOP hierarchy outright, with consequences for Keisler-order maximality. Central within model theory, little known outside it: placed at 35 with the field-famous specialist band (Feige, Krauth-Mezard), above HRT at 33.

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  • #1

    Attempt 1

    proof attemptChatGPT 5.6 with Artem Chernikov ·
    AI involvement
    ai co developed
    a person and a model developed the result together.
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    ChatGPT 5.6
    people
    Artem Chernikov

    From the paper: "The proof was found using ChatGPT 5.6 and simplified and streamlined by the author."

    The new content is SOP2SOP3SOP_2 \Rightarrow SOP_3; the converse implication was known from the start. Dzamonja and Shelah asked whether either implication in SOP3SOP2SOP1SOP_3 \Rightarrow SOP_2 \Rightarrow SOP_1 reverses: Mutchnik answered the second (SOP1=SOP2SOP_1 = SOP_2), and this answers the first, collapsing the bottom of the hierarchy to SOP1=SOP2=SOP3SOP_1 = SOP_2 = SOP_3. The SOPnSOP_n hierarchy for n3n \ge 3 remains, as does everything above it.

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