Let A be a set, and let P(A) denote the power set of A. Prove that|A|<|P(A)|.
Hint: Proceed in two steps.
1. First show that|A| <= |P(A)|. Try defining the function g: A-> P(A) by g(a) ={a}, and verify that g is one-to-one.
2. Then show that we can't have |A|=|P(A)|. Assume not, i.e., suppose that in fact |A|=|P(A)|. Then there exists a bijection f: A->P(A). Let B={aϵA|a Ɇf(a)} ϵ P(A)
Since f is onto, there exists an a0ϵA such that f(a0) =B. How does this lead to a contradiction?