The existence of free ultrafilters on ω does not imply the extension of filters on ω to ultrafilters. Issue 4 (5th August 2013) - Hall, Eric J.
Abstract Let X be an infinite set and let BPI(X) and UF(X) denote the propositions "every filter onXcan be extended to an ultrafilter" and "Xhas a free ultrafilter", respectively. We denote by S(X) the Stone space of the Boolean algebra of all subsets of X. We show: For every well‐ordered cardinal number ℵ, UF(ℵ) iff UF(2ℵ). UF(ω) iff "2ωis a continuous image ofS(ω)" iff "S(ω)has a free open ultrafilter " iff "every countably infinite subset ofS(ω)has a limit point". BPI(ω) implies "every open filter on2ωextends to an open ultrafilter" implies "2ωhas an open ultrafilter" implies UF(ω). It is relatively consistent with ZF that UF(ω) holds, whereas BPI(ω) fails. In particular, none of the statements given in (2) implies BPI(ω).