Let the given statement be defined as

P(n): The number of subsets of a set containing n distinct

elements=2ⁿ, for all n ϵ N.

Step1: For n=1,

L.H.S=As, the subsets of the set containing only 1 element are:

Φ and the set itself

i.e. the number of subsets of a set containing only element=2

R.H.S=2¹=2

As, LHS=RHS, so, it is true for n=1.

Step2: Let the given statement be true for n=k.

P(k): The number of subsets of a set containing k distinct

elements=2^k

Now, we need to show P(k+1) is true whenever P(k) is true.

P(k+1):

Let A={a₁, a₂, a₃, a₄,…, a_k, b} so that A has (k+1) elements.

So the subset t of A can be divided into two collections:

first contains subsets of A which don t have b in them and

the second contains subsets of A which do have b in them.

First collection: { }, {a₁},{a₁, a₂},{a₁, a₂, a₃},…,{a₁, a₂, a₃, a₄,…, a_k} and

Second collection: {b}, {a₁,b},{a₁,a₂,b },{a₁,a₂,a₃,b},…,{a₁,a₂,a₃,a₄,…,a_k, b}

It can be clearly seen that:

The number of subsets of A in first collection

= The number of subsets of set with k elements i.e. { a₁, a₂, a₃, a₄,…, a_k}=2^k

Also it follows that the second collection must have

the same number of the subsets as that of the first = 2^k

So the total number of subsets of A=2^k+2^k=2^k+1

Thus, by the principle of mathematical induction P(n) is true.