Question

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

Answer 1

Answer:

We have to prove that,

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

For n = 0,

A = {}

P(A) = { {} } = $2^0$ = 1

For n = 1,

A = { a }     ( suppose )

P(A) = { {}, a } = $2^{1}$ = 2,

For n = 2,

A = { a, b }

P(A) = { {}, {a}, {b}, {a, b} = $2^2$ = 4,

For n = 3,

A = { a, b, c },

P(A) = { {}, {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, {a, b, c} } = $2^3$ = 8

Thus, 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)| = $2^k$

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) = $2^k+2^k$ = $2^k(1+1)$ = $2^{k}.2$ = $2^{k+1}$

Hence, by the induction it has been proved,

|P(A)| = $2^n$ , if |A| = n, Where, n∈ N ( set of natural numbers )