What is the proof that the total number of subsets of a set is $2^n . WİTH İNDUCTİON.

Question

We know for any $n\in \mathbb{N}$ ,we can find find n+1. Because we wants to prove it with ınduction. For example$ A=\{a_1,a_2,...a_n\}$ and I proof with Induction $B=\{a_1,a_2,...a_{n+1}\}$ . How should I begin?

Answer 1

The total number of subsets of a set A with $n$ elements is the sum over all sets that have exactly zero elements () plus all sets with one elements $n\choose 1$ plus the sets that have exactly two elements $n\choose 2$ and so on.

Thus, the total number of subsets is $$\sum_{i=0}^n {n\choose i}$$ and that is known to be $2^n$.

Answer 2

With each subset of n-element set we can choose to have or not to have the (n+1)th element. so total number of subset of n+1 elements = $2^n\cdot2=2^{n+1}$

Answer 3

A direct answer is what @Nurator said.

A inductive answer is what @SurendraJain said.

For another one notice in an arbitrary subset, each element has two options, be in, or not. So by multiplication axiom we have $2^n$ subsets.