# Show that for any sets A and B,

Question:

Show that for any sets A and B,

$A=(A \cap B) \cup(A-B)$ and $A \cup(B-A)=(A \cup B)$

Solution:

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, ∈ (A ∩ B) ⊂ (A ∪ B) ∪ (A – B)

Case II

x ∉ A ∩ B

⇒ ∉ A or ∉ B

∴ ∉ B [∉ A]

∴ ∉ 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 ∈ A ∪ (B – A)

⇒ ∈ A or ∈ (B – A)

⇒ ∈ A or (∈ B and ∉ A)

⇒ (∈ A or ∈ B) and (∈ A or ∉ A)

⇒ ∈ (A ∪ B)

∴ A ∪ (B – A) ⊂ (A ∪ B) … (3)

Next, we show that (A ∪ B) ⊂ A ∪ (B – A).

Let ∈ A ∪ B

⇒ ∈ A or ∈ B

⇒ (∈ A or ∈ B) and (∈ A or ∉ A)

⇒ ∈ A or (∈ B and ∉ A)

⇒ ∈ A ∪ (B – A)

∴ A ∪ B ⊂ A ∪ (B – A) … (4)

Hence, from (3) and (4), we obtain A ∪ (B – A) = A ∪B.