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 <=.