On the density of Hausdorff ultrafilters
Tomek Bartoszyński, Saharon Shelah · Cambridge University Press eBooks · 2007
. An ultrafilter U is Hausdorff if for any two functions f, g ∈ ω ω , f ( U ) = g ( U ) iff f ↾ X = g ↾ X for some X ∈ U . We will show that the statement that Hausdorff ultrafilters are dense in the Rudin-Keisler order is independent of ZFC. Introduction. For f ∈ ω ω and an ultrafilter U on ω define f ( U ) = {X ⊆ ω : f −1 ( X ) ∈ U} , and for f, g ∈ ω ω we say that f = g mod U if there is X ∈ U such that f ( n ) = g ( n ) for n ∈ X . We say that U is Hausdorff if for any two functions f, g ∈ ω ω , if f ( U ) = g ( U ) then f = g mod U . Let FtO be the collection of all finite-to-one functions f ∈ ω ω . Recall that an ultrafilter U is a p-point if for every function f ∈ ω ω either there is n such that f −1 ( {n} ) ∈ U or there exists g ∈ FtO such that f = g mod U . Similarly, U is Ramsey if for every function f ∈ ω ω either there is n such that f −1 ( {n} ) ∈ U or there exists a one-to-one function g ∈ ω ω such that f = g mod U . In this paper we will assume that all ultrafilters U , and their images f ( U ) are non-principal. It is worth mentioning that the following appears as an exercise in [7]. If f ( U ) = U then f = id mod U . Therefore, if U is not Hausdorff, then this is witnessed by two functions, both not one-to-one mod U . It follows from it that Ramsey ultrafilters are Hausdorff.