The method of contradiction is fine for existence, but probably not best for uniqueness. Suppose towards a contradiction that . Then and hence , a contradiction. So there is . Let . Then and hence . Since it follows that .
Here's the theorem I am trying prove
Suppose is a set, and for every family of sets
we have
so prove that has exactly one element. Now there are two parts here. One is the existence part and second is uniqueness part. I have proved the
existence part using method of contradiction. I chose as a particular F to show the contradiction.
Now coming to the uniqueness part, the given is
Given ---->
and the Goal is
I will assume the negation of the goal and try to find the contradiction. Hence the
givens are
and the goal now is contradiction. Now here I am having some difficulty,
how do I choose particular ? Since
and we know that
But there could be more elements in A. Is there any way I can consider different
cases which will exhaust the possibilities ?
please comment.
The method of contradiction is fine for existence, but probably not best for uniqueness. Suppose towards a contradiction that . Then and hence , a contradiction. So there is . Let . Then and hence . Since it follows that .