If a uniform ultrafilter U over an uncountable cardinal κ is not outright countably complete, probably the next best thing is that it have a finest partition: a master function f:κ → ω with ƒ −({n}) ∉ U for each n ϵ ω such that for any g: κ → κ, either (a) it is one-to-one on a set in U, or (b) it factors through ƒ (mod U), i.e. for some function h, {α < κ ∣ h(f(α)) = g(α)} ϵ U. In this paper, it is shown that recent contructions of irregular ultrafilters over ω 1 can be amplified to incorporate a finest partition.

Henceforth, let us assume that all ultrafilters are uniform.

There has been an extensive study of substantial hypotheses, which are nonetheless weaker than countable completeness, on ultrafilters over uncountable cardinals. To survey some results and to establish a context, let us first recall the Rudin-Keisler (RK) ordering on ultrafilters: If U i is an ultrafilter over I i i for i = 1, 2, then U 1 ≤RK U 2 iff there is a projecting function Ψ:I 2 → I 1 such that U 1 = Ψ *(U 2) = {X ⊆, I 1∣ Ψ −1(X) ϵ U 2}· U 1, =RK U 2 iff U 1, ≤RK and U 2 and U 2≤RK U 1; and U 1<RK U 2 iff U 1≤RK U 2 yet U 1 ≠RK U 2. In terms of this ordering, if an ultrafilter U has a finest partition ƒ, then ƒ *(U) over ω is maximum amongst all RK predecessors of U: for any g:κ → κ, if g *(U) <RK U, then g is not one-to-one on a set in U, so since g factors through ƒ with some h,g *(U) = h *(ƒ *(U)). Say now that an ultrafilter U over κ > ω is indecomposable iff whenever ω < λ < κ, there is no V ≤RK U such that V is a (uniform) ultrafilter over λ.