The answer is Yes. Furthermore, when A is infinite, then such a family can be found of the same cardinality as A (rather than the possibly much larger S).

(I had thought about the case of cuts in the rationals, thinking it might lead to a counterexample. Instead, the solution there led to a solution for all sets S linearly ordered by inclusion, and then I realized the solution is fully general by moving to partial orders.)

Suppose that S is such a collection of subsets of A. Such a family induces a natural partial pre-order on A, namely, a <= b if whenver B in S and b in B, then also a in B. It is easy to see that this relation is transitive and reflexive.

I claim, first, that S consists of exactly the subsets of A that are downward closed in this order. It is clear that every set in S is downward closed in this order. Conversely, suppose that X is downward closed with respect to <=. For any b in X, consider the set X_b, which the intersection of all sets in S containing b as an element. This is in S. Also, X_b consists of precisely of the predecessors of b with respect to <=. So X_b subset X. Thus, X is the union of the X_b for b in X. So X is in S.

Next, define f_a(b) = a if a <= b, and otherwise f_a(b) = b. Let F be the family of all such functions f_a for a in A.

Clearly, every B in S is closed under every f_a, by the definition of <=. Conversely, suppose that X is closed under all f_a. Thus, whenever b is in X and a <= b, then a is in X also. So X is downward closed, and hence by the claim above, X is in S.

Incidently, the sets S are exactly the open sets in the topology on A induced by the lower cones of <=.

The answer is Yes. Furthermore, such a family can be found of size at most the cardinality of A, even when S is much larger.

The key to the solution is to realize that every such family S arises as the collection of downward-closed sets for a certain partial pre-order on A, which I shall define. (Conversely, every such order also leads to such a family.)

An interesting special case occurs when the family S is linearly ordered by inclusion. For example, one might consider the family of cuts in the rational line, that is, downward-closed subsets of Q. (I had thought briefly at first that this might be a counterexample, but after solving it, I realized a general solution was possible by moving to partial orders.)

Suppose that S is such a collection of subsets of A. Define the induced partial pre-order on A by

• a <= b if whenever B in S and b in B, then also a in B.

It is easy to see that this relation is transitive and reflexive.

I claim, first, that S consists of exactly the subsets of A that are downward closed in this order. It is clear that every set in S is downward closed in this order. Conversely, suppose that X is downward closed with respect to <=. For any b in X, consider the set X_b, which the intersection of all sets in S containing b as an element. This is in S. Also, X_b consists of precisely of the predecessors of b with respect to <=. So X_b subset X. Thus, X is the union of the X_b for b in X. So X is in S.

Next, define f_a(b) = a if a <= b, and otherwise f_a(b) = b. Let F be the family of all such functions f_a for a in A.

Clearly, every B in S is closed under every f_a, by the definition of <=. Conversely, suppose that X is closed under all f_a. Thus, whenever b is in X and a <= b, then a is in X also. So X is downward closed, and hence by the claim above, X is in S.

Incidently, the sets S are exactly the open sets in the topology on A induced by the lower cones of <=.

The answer is Yes. Furthermore, when A is infinite, then such a family can be found of the same cardinality as A (rather than the possibly much larger S).

I had thought first about the case of cuts in the rationals, and the solution their led me to think about S that are linearly ordered by inclusion. Then, I realized that the solution is fully general by moving to partial pre-orders.

Suppose that S is such a collection of subsets of A. This induces a natural partial pre-order on A, where a<= b iff whenver B in S and b in B, then also a in B. It is easy to see that this relation is transitive and reflexive.

I claim, first, that S consists of exactly the subsets of A that are downward closed in this order. It is clear that every set in S is downward closed in this order. Conversely, suppose that X is downward closed with respect to <=. For any b in X, consider the set X_b, which the intersection of all sets in S containing b as an element. This is in S. Also, X_b consists of precisely of the predecessors of b with respect to <=. So X_b subset X. Thus, X is the union of the X_b for b in X. So X is in S.

Next, define f_a(b)= a if a<=b, and otherwise f_a(b)=b. Let F be the family of all such functions f_a for a in A.

Clearly, every B in S is closed under every f_a, by the definition of <=. Conversely, suppose that X is closed under all f_a. Thus, whenever b is in X and a<=b, then a is in X also. So X is downward closed, and hence by the claim above, X is in S.

Incidently, the sets S are exactly the open sets in the topology on A induced by the lower cones of <=.

The answer is Yes. Furthermore, when A is infinite, then such a family can be found of the same cardinality as A (rather than the possibly much larger S).

(I had thought about the case of cuts in the rationals, thinking it might lead to a counterexample. Instead, the solution there led to a solution for all sets S linearly ordered by inclusion, and then I realized the solution is fully general by moving to partial orders.)

Suppose that S is such a collection of subsets of A. Such a family induces a natural partial pre-order on A, namely, a <= b if whenver B in S and b in B, then also a in B. It is easy to see that this relation is transitive and reflexive.

I claim, first, that S consists of exactly the subsets of A that are downward closed in this order. It is clear that every set in S is downward closed in this order. Conversely, suppose that X is downward closed with respect to <=. For any b in X, consider the set X_b, which the intersection of all sets in S containing b as an element. This is in S. Also, X_b consists of precisely of the predecessors of b with respect to <=. So X_b subset X. Thus, X is the union of the X_b for b in X. So X is in S.

Next, define f_a(b) = a if a <= b, and otherwise f_a(b) = b. Let F be the family of all such functions f_a for a in A.

Clearly, every B in S is closed under every f_a, by the definition of <=. Conversely, suppose that X is closed under all f_a. Thus, whenever b is in X and a <= b, then a is in X also. So X is downward closed, and hence by the claim above, X is in S.

Incidently, the sets S are exactly the open sets in the topology on A induced by the lower cones of <=.