Assertion: filter | /r/AtLocation | ultrafilter
Facets:
Source sentences:
text: One can show that every filter on a Boolean algebra (or more generally, any subset with the finite intersection property) is contained in an ultrafilter (see Ultrafilter lemma) and that free ultrafilters therefore exist, but the proofs involve the axiom of choice (AC) in the form of Zorn's lemma.
text: On the other hand, the statement that every filter is contained in an ultrafilter does not imply AC.