Re: Proofs about Sets
a.
Since A ~ B, there exists a 1-1 and onto function θ. So consider all the power sets of A. Just take every a in there, and replace it with θ(a). All these elements must be distinct, so as long as the power sets of A are distinct (which they are), then these will be the power sets of B.
b.
Taking a countable number of elements away from an uncountable set, it should be clear this will leave you with an uncountable set.
Let's assume the B - A is countable. Then B - A = {x : x∈B and x is not in A}. Consider B - A + (B intersect A) = B. B intersect A is countable since A is countable. Thus, we have B - A + (B intersect A) which must be countable since both B-A and B intersect A are countable. So B is countable. Contradiction.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."