



Maths Crash Course - Live lectures + all videos + Real time Doubt solving!
Examples
Example 2 Important
Example 3
Example 4
Example 5 Important
Example 6 (i)
Example 6 (ii) Important
Example 6 (iii)
Example 6 (iv)
Example 6 (v)
Example 7 Important
Example 8 (i)
Example 8 (ii)
Example 9 Important
Example 10
Example 11 Important
Example 12
Example 13 Important
Example 14
Example 15
Example 16
Example 17
Example 18 Important
Example 19
Example 20
Example 21
Example 22 Important
Example 23
Example 24 Important
Example 25 Important
Example 26
Example 27 Important
Example 28
Example 29 Important
Example 30
Example 31 Deleted for CBSE Board 2023 Exams You are here
Example 32 Important
Example 33 Important
Example 34 Important
Last updated at Sept. 3, 2021 by Teachoo
Maths Crash Course - Live lectures + all videos + Real time Doubt solving!
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 ⊂ 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 ).