The existence of free ultrafilters on ω does not imply the extension of filters on ω to ultrafilters

摘要

Let X be an infinite set and let BPI(X) and UF(X) denote the propositions "every filter on X can be extended to an ultrafilter" and "X has 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 ?, UF(?) iff UF(2?). UF(ω) iff "2ω is a continuous image of S(ω)" iff "S(ω) has a free open ultrafilter " iff "every countably infinite subset of S(ω) has a limit point". BPI(ω) implies "every open filter on 2ω extends 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).