Last updated at May 29, 2018 by Teachoo

Transcript

Example 31 For any sets A and B, show that P(A ∩ B) = P(A) ∩ P(B). To prove two sets equal, we need to prove that they are subset of each other i.e.. we have to prove P (A ∩ B) ⊂ P (A) ∩ P (B) & P(A) ∩ P(B) ⊂ P ( A ∩ B) Let a set X belong to Power set P(A∩ B) i.e. X ∈ P ( A ∩ B ). As set X is in the power set of A ∩ B, X is a subset of A∩ B because power set is the set of all subsets Thus, X is a subset of A ∩ B i.e. X ⊂ A ∩ B. So, X ⊂ A and X ⊂ B. Therefore, Since X is a subset of A & B, X is in power set of A and X is in power set of B i.e. X ∈ P(A) and X ∈ P(B) i.e. X ∈ P(A) and X ∈ P(B) ⇒ X ∈ P(A) ∩ P(B). So, if X ∈ P (A ∩ B), then X ∈ P(A) ∩ P(B) i.e. all elements of set P (A ∩ B) are in set P(A) ∩ P(B) Thus, gives P (A ∩ B) ⊂ P (A) ∩ P (B). Similarly, Let a set Y belong to Power set P(A) ∩ P(B) i.e. Y ∈ P (A) ∩ P(B). Then Y ∈ P (A) and Y ∈ P ( B ). As set Y is in the power set of A & B, Y is a subset of A & Y is a subset of B because power set is the set of all subsets Thus, Y ⊂ A and Y ⊂ B. ∴ Y ⊂ A ∩ B, Therefore, Since Y is a subset of A ∩ B, Y is in power set of A ∩ B ⇒ Y ∈ P ( A ∩ B ). So, if Y ∈ P (A) ∩ P(B) , then Y ∈ P ( A ∩ B ). This gives P (A) ∩ P (B) ⊂ P (A ∩ B) Now since P (A ∩ B) ⊂ P (A) ∩ P (B) & P(A) ∩ P(B) ⊂ P ( A ∩ B) Hence P ( A ∩ B ) = P ( A ) ∩ P ( B ).

Examples

Example 1

Example 2

Example 3

Example 4

Example 5

Example 6

Example 7

Example 8

Example 9

Example 10

Example 11

Example 12

Example 13

Example 14

Example 15

Example 16

Example 17

Example 18

Example 19

Example 20

Example 21

Example 22

Example 23

Example 24

Example 25

Example 26

Example 27

Example 28

Example 29

Example 30

Example 31 You are here

Example 32

Example 33

Example 34

Chapter 1 Class 11 Sets

Serial order wise

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.