Question

Show that for any sets A and B, $A = (A ∩ B) ∪ (A – B)$ and $A ∪ (B – A) = (A ∪ B)$

Answer 1

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 $x ∈ A ∪ (B – A)$
$⇒ x ∈ A$ or $x ∈ (B – A)$
$⇒ x ∈ A$ or $(x ∈ B \space and \space x ∉ A)$
$⇒ (x ∈ A \space or x ∈ B)$ and $(x ∈ A \space or x ∉ A)$
$⇒ x ∈ (A ∪ B)$
$∴ A ∪ (B – A) ⊂ (A ∪ B) … (3)$
Next, we show that $(A ∪ B) ⊂ A ∪ (B – A).$
Let $y ∈ A ∪ B$
$⇒ y ∈ A \space or\space y ∈ B$
$⇒ (y ∈ A$ 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 $A ∪ (B – A) = A ∪B.$