Let x be any set and let lambda SolutionProof Obviously and

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.