Let x be any set and let lambda Solution Proof. Obviously and X are contained in T . For the second criterion, look at a subfamily {U : A} of T . Without loss of generality, assume that all U have a countable complement. Then X \\ S A U = T A X \\ U is the intersection of countable sets and therefore countable. Also, the finite union of countable sets is countable which yields the third criterion..

1. Let x be any set and let lambda
Solution
Proof. Obviously and X are contained in T . For the second criterion, look at a subfamily {U :
A} of T . Without loss of generality, assume that all U have a countable complement. Then X
S A U = T A X U is the intersection of countable sets and therefore countable. Also, the finite
union of countable sets is countable which yields the third criterion.