If sets A and B have the same cardinality, then their power sets also have the same cardinality. This is proven by defining a bijection f between A and B. A bijection G is then constructed between the power sets P(A) and P(B) such that for any subset C of A, G maps C to its image under f, which is a subset of B. This G is a bijection because f is a bijection, so each subset of A maps to a unique subset of B and vice versa.

1. Show that if sets A and B have the same cardinality, then the power set of A and the power set
of B have the same cardinality.
Solution
Assume A and B are equipotent (same cardinality), and then assume abiyection f :
A to B. Define G: P(A) to P(B) as follows, Given a subset of A, take the elements, and map
them with f, andtake that set. Meaning, given C a subset of A, take f(C) the image, asubset of B,
be the image of C under G. In other words, G(C) = f(C), for each subset C of A.Note that in one
side of equality we have an element of our domain,and in the other, the set of images of elements
of C under f. This is a biyection between P(A) and P(B) because each image f(C) is uniquely
determined by the hypothesis that f is abiyection, so if two coincide, they are the same subsetof
A. Also every subset C' of B is an image of somesubset C of A because f is a biyection (so
none is leftalone), so there is a biyection between the power sets.