Examples

Chapter 1 Class 11 Sets
Serial order wise

### Transcript

Question 6 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 ⊂ Subset A ⊂ B (all elements of set A in set B) 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 ).

#### Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.