LAUSR

Abstract We show in the Zermelo-Fraenkel set theory ZF without the axiom of choice: Given an infinite set X, the Stone space S(X) is ultrafilter compact. For every infinite set X, every countable filterbase of X extends to an ultra-filter iff for every infinite set X, S(X) is countably compact. ω has a free ultrafilter iff every countable, ultrafilter compact space is countably compact. We also show the following: There are a permutation model ???? and a set X ∈ ???? such that X has no free ultrafilters and S(X) is not compact but S(X) is countably compact and every countable filterbase of X extends to an ultrafilter. It is relatively consistent with ZF that every countable filterbase of ω extends to an ultrafilter but there exists a countable filterbase of ℝ which does not extend to an ultrafilter. Hence, it is relatively consistent with ZF that ℝ has free ultrafilters but there exists a countable filterbase of ℝ which does not extend to an ultrafilter.