State and prove the inclusion-exclusion principle by induction.

Proof:

We set P(n) P ( n ) to be the number of element for the union of n n exact sets.We make the sets to be A1,A2,A3…… A 1 , A 2 , A 3 … … , distinguished by different symbols.

BASIS STEP:

P(1)=|A| P ( 1 ) = | A |
P(2)=|A∪B|=|A|+|B|−|A∩B| P ( 2 ) = | A ∪ B | = | A | + | B | − | A ∩ B |

INDUCTIVE STEP:

For n>2 n > 2
We assume that n−1 n − 1 is true, which means

P(n−1)=|⋃i=1n−1Ai|=∑i=1n−1(−1)i−1∑1≤b1<...<bi≤n−1|⋂j=1iAbj| P ( n − 1 ) = | ⋃ i = 1 n − 1 A i | = ∑ i = 1 n − 1 ( − 1 ) i − 1 ∑ 1 ≤ b 1 < . . . < b i ≤ n − 1 | ⋂ j = 1 i A b j |

For n n , we have

|⋃i=1nAi|======|(⋃i=1n−1Ai)⋃An||⋃i=1n−1Ai|+|An|−|(⋃i=1n−1Ai)⋂An||⋃i=1n−1Ai|+|An|−|⋃i=1n−1(Ai⋂An)|∑i=1n|Ai|+∑i=2n−1(−1)i−1∑1≤b1<...<bi≤n−1|⋂j=1iAbj|+∑i=2n−1(−1)i−1∑1≤b1<...<bi≤n−1|⋂j=1iAbj⋂An|+(−1)n−1|⋂i=1nAi|∑i=1n|Ai|+∑i=2n−1(−1)i−1∑1≤b1<...<bi≤n|⋂j=1iAbj|+(−1)n−1|⋂i=1nAi|∑i=1n(−1)i−1∑1≤b1<...<bi≤n|⋂j=1iAbj| | ⋃ i = 1 n A i | = | ( ⋃ i = 1 n − 1 A i ) ⋃ A n | = | ⋃ i = 1 n − 1 A i | + | A n | − | ( ⋃ i = 1 n − 1 A i ) ⋂ A n | = | ⋃ i = 1 n − 1 A i | + | A n | − | ⋃ i = 1 n − 1 ( A i ⋂ A n ) | = ∑ i = 1 n | A i | + ∑ i = 2 n − 1 ( − 1 ) i − 1 ∑ 1 ≤ b 1 < . . . < b i ≤ n − 1 | ⋂ j = 1 i A b j | + ∑ i = 2 n − 1 ( − 1 ) i − 1 ∑ 1 ≤ b 1 < . . . < b i ≤ n − 1 | ⋂ j = 1 i A b j ⋂ A n | + ( − 1 ) n − 1 | ⋂ i = 1 n A i | = ∑ i = 1 n | A i | + ∑ i = 2 n − 1 ( − 1 ) i − 1 ∑ 1 ≤ b 1 < . . . < b i ≤ n | ⋂ j = 1 i A b j | + ( − 1 ) n − 1 | ⋂ i = 1 n A i | = ∑ i = 1 n ( − 1 ) i − 1 ∑ 1 ≤ b 1 < . . . < b i ≤ n | ⋂ j = 1 i A b j |

So it’s proofed.

P(n)=|⋃i=1nAi|=∑i=1n(−1)i−1∑1≤b1<...<bi≤n|⋂j=1iAbj| P ( n ) = | ⋃ i = 1 n A i | = ∑ i = 1 n ( − 1 ) i − 1 ∑ 1 ≤ b 1 < . . . < b i ≤ n | ⋂ j = 1 i A b j |