2013/01/03 by Mildenberger, Heike
#Combinatorics (math.CO) #FOS: Mathematics #Logic (math.LO)
paper · doi:10.48550/arxiv.1301.0396
We answer Blass' question from 1989 of whether the inequality \gu < \gro is strictly stronger than the filter dichotomy principle affirmatively. We show that there is a forcing extension in which every non-meagre filter on ω is ultra by finite-to-one and the semifilter trichotomy does not hold. This trichotomy says: every semifilter is either meagre or comeagre or ultra by finite-to-one. The trichotomy is equivalent to the inequality \gu