Tony J. answered • 12/10/17

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By trying a few small sets, we see the following:

|A| = 0 ---> |P(A)| = 1

|A| = 1 ---> |P(A)| = 2

|A| = 2 ---> |P(A)| = 4

|A| = 3 ---> |P(A)| = 8

This evidence suggests that maybe, for a set with n elements, the power set will have 2

^{n}elements. We prove this by induction.The base case is established by the observations above. Now, consider a set A with n elements, and add a single new element: call it z. Let's denote A ∪ {z} as set B, which has n+1 elements. We can enumerate the subsets of B, in pairs, as follows: any subset of A is a subset of B, and any subset of A, with z added, is a different subset of B. For every subset of A, we obtain precisely two subsets of B in this way, and every subset of B is accounted for in this enumeration. Thus, the number of subsets of B is precisely twice the number of subsets of A. So:

|P(B)| = (2)|P(A)| = (2)(2

^{n}) = 2^{n+1},as desired.