Example 31 - Chapter 1 Class 11 Sets
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
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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.