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?
Asked
Active
Viewed 311 times
3 Answers
2
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.
Ali Ashja'
- 2,800
- 11
- 13
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 $2^n$.
Nurator
- 1,410
-
2You forgot the empty set. – drhab Oct 24 '19 at 11:38
-
Thanks, I forgot about that! – Nurator Oct 24 '19 at 11:41
1
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 = $2^n\cdot2=2^{n+1}$
Nurator
- 1,410
