Assertion: proper filter | /r/HasProperty | free
Facets:
Source sentences:
text: Since the intersection of any finite number of members of a filter is also a member, no proper filter on a finite set is free, and indeed is the principal filter generated by the common intersection of all of its members.
text: Since the intersection of any finite number of members of a filter is also a member, no proper filter on a finite set is free, and indeed is the principal filter generated by the common intersection of all of its members.