If is a transitive set, then by definition: whenever , and , then . That is it. Now, if is nonempty, as is the case by your assumption, then there is some element . If is the empty set, then the case satisfies our proof. On the other hand, let us suppose is nonempty. Then there is some . By the definition of transitivity, . The same reasoning applied to can be applied to now--i.e., either y is empty or it contains something that is also in . It might be fruitful to index these elements as a sequence. Thus, what we end up having to prove is that this member sequence will "bottom out" at some point. Now, how to go about this proof, I am unsure, and maybe I've approached it in an inefficient or less fruitful manner.