Prove that and .
Proof. First, it is clear that (see Exercise 12 of Section I.2.5, or simply note that implies since is in ).
For the reverse inclusion, if is any element of then or . But implies (and ). So, in either case ; hence, .
Proof. From Exericse 12 (Section I.2.5) we know that for any set ; hence, .
For the reverse inclusion, if is any element in , then and ; hence, . Therefore, .