Proof. Let be any element in . Then, since . Further, there is some such that (since means that is a proper subset of ). Then, since we have both . Hence, (since every element of is in , and there is at least one element of not in , so ).∎
Proof. Let be any element of . Then, since . Further, implies since . Thus, .∎
Proof. This was established in part (a) since we didn’t need in that proof. (Since a proper subset of guaranteed there was some in that wasn’t in , and since is a subset, proper or otherwise, of , this is in ; hence, is a proper subset of .)∎
Proof. Since we know every element of is in . Hence, if , then .∎
If and , then we cannot conclude that . For example, let and . Then, (since is the set and contains this set). However, , but (since contains the set which contains 1, but does not contain the element 1).