2. Prove:

(a) If A is a subset of S, then |A| ≤ |S|.

(b) For any subset A of S, we have |S| = |A|+|S\A|, where S\A is the set of all elements of S that are not in A.

(c) For any subsets A and B of S, let A ∩ B be their intersection and A U B their union; then

|A| + |B| = |A ∩ B| + |A U B|.

(2.20)