TheFacemore said:Assume that phi is a surjection of A onto P(A) and consider the set U= {a in A : a not in phi(a)} of A. Since phi is a surjection there must be a u in A satisfying phi(u)=U.
A power set is a set that contains all the subsets of a given set. For example, if the set A = {1, 2}, then the power set of A is P(A) = {{}, {1}, {2}, {1,2}}.
The cardinality of a power set is equal to 2^n, where n is the number of elements in the original set. In other words, if a set has n elements, its power set will have 2^n elements.
Power sets and cardinalities are closely related because the cardinality of a power set is determined by the number of elements in the original set. The power set will always have a larger cardinality than the original set.
One way to prove this is by using mathematical induction. First, show that the formula holds for a set with one element. Then, assume it holds for a set with n elements and use this to show that it also holds for a set with n+1 elements. This will prove that the formula holds for all natural numbers and therefore, for any set.
Yes, the cardinality of a power set can be infinite. For example, the power set of the set of natural numbers (N) has a cardinality of 2^N which is infinite. This means that the power set of N contains an infinite number of subsets.