Saved Bookmarks
| 1. |
For any sets A and B, show that: P(A ∩ B), = P(A) ∩ P(B) |
|
Answer» Let X ∈ P (A ∩ B),then A ⊂ (A ∩ B). So, X ⊂ A and X ⊂ B. ∴ X ∈ P(A) and X ∩ P(B) ⇒ X ∈ P(A) and P(B) Thus, P(A ∩ B) ⊂ P(A) ∩ P(B) ..(i) Again, Let Y ∈ P(A) ∩ P(B),then Y ∈ P(A) and Y ∈ P(B) So, Y ⊂ A and Y ⊂ B. ∴ Y ⊂ A ∩ B ⇒ Y ∈ P(A ∩ B) Then P(A) ∩ P(B) ∈ P(A ∩ B) ..(ii) Hence from (i) and (ii), we gets P (A ∩ B) = P(A) ∩ P(B) |
|