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Show that for any sets A and B, A = (A ∩ B) ∪ (A – B) and A ∪ (B – A) = (A ∪ B) |
Answer» To show: A = (A ∩ B) ∪ (A – B) Let x ∈ A We have to show that x ∈ (A ∩ B) ∪ (A – B) Case I x ∈ A ∩ B Then, x ∈ (A ∩ B) ⊂ (A ∪ B) ∪ (A – B) Case II x ∉ A ∩ B ⇒ x ∉ A or x ∉ B ∴ x ∉ B [x ∉ A] ∴ x ∉ A – B ⊂ (A ∪ B) ∪ (A – B) ∴ A ⊂ (A ∩ B) ∪ (A – B) … (1) It is clear that A ∩ B ⊂ A and (A – B) ⊂ A ∴ (A ∩ B) ∪ (A – B) ⊂ A … (2) From (1) and (2), we obtain A = (A ∩ B) ∪ (A – B) To prove: A ∪ (B – A) ⊂ A ∪ B Let x ∈ A ∪ (B – A) 8 ⇒ x ∈ A or x ∈ (B – A) ⇒ x ∈ A or (x ∈ B and x ∉ A) ⇒ (x ∈ A or x ∈ B) and (x ∈ A or x ∉ A) ⇒ x ∈ (A ∪ B) ∴ A ∪ (B – A) ⊂ (A ∪ B) … (3) Next, we show that (A ∪ B) ⊂ A ∪ (B – A). Let y ∈ A ∪ B ⇒ y ∈ A or y ∈ B ⇒ (y ∈ A or y ∈ B) and (y ∈ A or y ∉ A) ⇒ y ∈ A or (y ∈ B and y ∉ A) ⇒ y ∈ A ∪ (B – A) ∴ A ∪ B ⊂ A ∪ (B – A) … (4) Hence, from (3) and (4), we obtain A ∪ (B – A) = A ∪B. |
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