Final answer:
The intersection of a nonempty family of nonempty complete sets is complete because any Cauchy sequence in the intersection will converge within each set, and thus also in the intersection. The union of complete sets is not necessarily complete because the limit of a Cauchy sequence is not guaranteed to be unique within the union.
Step-by-step explanation:
In mathematics, specifically in the context of metric spaces or more generally complete metric spaces, a set is said to be complete if every Cauchy sequence of points in the set has a limit that is also within the set. To show that the intersection of a nonempty family of nonempty complete sets is also complete, let us consider a family of complete sets {Ai} where i is an index running over some index set I. Suppose xn is a Cauchy sequence in the intersection ∩Ai. Since xn is in the intersection, it means xn is in each of the Ais. Since each Ai is complete, the sequence xn must converge to a limit x in each Ai, implying that x is also in their intersection, thus proving the intersection is complete.
However, the union of a family of complete sets is not necessarily complete. To understand why, consider two disjoint complete sets A and B in a metric space. Take a sequence that converges to a point in A and another sequence that converges to a different point in B. In the union A ∪ B, you could create a sequence which alternates between elements of these two sequences. This new sequence may not converge within A ∪ B because the limit is not unique, thus the union is not necessarily complete.