A bijection f : A -> B can be "lifted" to a bijection between P(A) and P(B). Namely, consider where , i.e., g applies f to every element of X and collects the results. Prove that g : P(A) -> P(B) and that g is a bijection.
I have another question
Q: Prove that if A ~ B then P(A) ~ P(B).
first I started with stating that since A and B are equinumerous
so there exists f:A ->B
and suppose given a subset X of A
then find b in B and a exists in A such f(a) = b
and then stuck....