Prove that if is transitive and , then .
I can only say that if is transitive, then . How would I have to do in order to continue from here?
Thanks in advance.
Prove that if is transitive and , then .
I can only say that if is transitive, then . How would I have to do in order to continue from here?
Thanks in advance.
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.
Hi
we can use the axiom of regularity , which is one of the axioms of
Zermelo–Fraenkel set theory. Look up here
Axiom of regularity - Wikipedia, the free encyclopedia
axiom says
"Every non-empty set A contains an element B which is disjoint from A"
Since , using this axiom,
and A is transitive , so
and
means
Now lets assume that
since
and
we have
which is contradiction. so the assumption that
is wrong.
hope this helps....