hich is a subset of
| U |^| V.  To show the converse, let !B! be a family with the
| specified property and let !T! be the family of all unions
| of members of !B!.  A union of members of !T! is itself
| a union of members of !B! and is therefore a member
| of !T!, and it is only necessary to show that the
| intersection of two members U and V of !T! is
| a member of !T!.  If x in U |^| V, then we
| may choose U' and V' in !B! such that:
|
| x  in  W  c  U' |^| V'  c  U |^| V.
|
| Consequently U |^| V is th