Question

18. Use induction to prove the following: For any set A, if IA] = n for some finite number n E N, then IP(A) 2"

Answer 1

Proof:

For all n∈N, let P(n) be the proposition:

|A|=n⟹|P(A)|=2ⁿ

where P(A) is the powerset of A

Basis for the Induction:

1. It holds A=∅⟺|A|=0
2. Then P(A)={∅}, such that |P(A)|=1=2⁰
3. from (1.) and (2.) it follows that proposition P(0)holds.

Induction Hypothesis:

We need to show that, if P(k) is true, where k≥2, then it logically follows that P(k+1) is true. So if this is our Induction Hypothesis:

|A|=k⟹|P(A)|=2^k

We need to show that:

|A|=k+1⟹|P(A)|=2^k+1

Induction Step:

1. Let |A|=k+1 and let x∈A
2. Consider the set A′=A∖{x} , where x is some element of A. We see that |A′|=k
3. We now adjoin x to all the subsets of A' . Counting the subsets of A , we have: all the subsets of A' with x adjoined to them.
4. From the Induction Hypothesis, there are 2^k subsets of A'. Adding x to each of these, does not change their number, so there are another 2^k subsets of A , consisting of all the subsets A′ with x adjoined to them.
5. In total that makes 2^k+2^k=2⋅2^k=2^k+1 subsets of A. So P(k)⟹P(k+1) and the result follows by the principle of mathematical induction.
6. Therefore: ∀n∈N:|A|=n⟹|P(A)|=2ⁿ