
Yarema B. answered 11/21/15
Tutor
4.9
(135)
Topology, Modern, Real and Complex Analysis.
The answer to part (a) is 2^n. A bijection between the functions and a powerset of D is the following: given a function f biject it with the element of the powerset of D that contains only elements d∈D for such f(d)=1. Here is another bijection: well order both sets and map the k-th element of the first set into the k-th element of the second set. There are (2^n)!-2 more bijections out there ;).