Matthew S. answered • 04/26/20
Tutor
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Math needn't be a mystery! I can help you make sense of it.
I'm assuming A is a set of finite cardinality, let's call it n. Then cardinality of P(A) is 2n.
Put the elements of A into an ordered list. (The order is arbitrary.) Imagine a subset of A, and write a string of 0s and 1s as follows: if the ith element of A is in the subset, put a 1 in the ith position of the string. Otherwise, put a 0 there. You will end up with a string of length n.
Strings created in this fashion are in 1-to-1 correspondence with subsets of A. They are also in 1-to-1 correspondence with the numbers in the set {0, 1, 2, ... , 2n-1}, and that set has cardinality 2n.
