Verify that |P(A)| = 2^n , if |A| = n for n = 0, 1, 2, 3.

Answer:We have to prove that,|P(A)| = [tex]2^n[/tex] , if |A| = n for n = 0, 1, 2, 3.For n = 0,A = {}P(A) = { {} } = [tex]2^0[/tex] = 1For n = 1, A = { a }     ( suppose )P(A) = { {}, a } = [tex]2^{1}[/tex] = 2,For n = 2,A = { a, b }P(A) = { {}, {a}, {b}, {a, b} = [tex]2^2[/tex] = 4,For n = 3,A = { a, b, c },P(A) = { {}, {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, {a, b, c} } = [tex]2^3[/tex] = 8Thus, it is verified for n = 0, 1, 2, 3.Now, suppose it is valid for a set B having k elements,That is, |P(B)| = [tex]2^k[/tex]Also, there is a set A,Such that, A = B ∪ {x}Since, after including the element x in set B,The element x will be come with every element of set B in the power set of B,i.e. P(A) = [tex]2^k+2^k[/tex] = [tex]2^k(1+1)[/tex] = [tex]2^{k}.2[/tex] = [tex]2^{k+1}[/tex]Hence, by the induction it has been proved,|P(A)| = [tex]2^n[/tex] , if |A| = n, Where, n∈ N ( set of natural numbers )