Prove that number of subsets of a set containing n distinct elements is 2^n, for all n ∈ N.

Let P(n): Number of subset of a set containing n distinct elements is 2″, for all ne N.

For n = 1, consider set A = {1}. So, set of subsets is {{1}, ∅}, which contains 2¹ elements.

So, P(1) is true.

Let us assume that P(n) is true, for some natural number n = k.

P(k): Number of subsets of a set containing k distinct elements is 2^k To prove that P(k + 1) is true,

we have to show that P(k + 1): Number of subsets of a set containing (k + 1) distinct elements is 2^k+1

We know that, with the addition of one element in the set, the number of subsets become double.

Number of subsets of a set containing (k+ 1) distinct elements = 2×2^k = 2^k+1

So, P(k + 1) is true. Hence, P(n) is true.