Question 4:
Show that the following four conditions are equivalent:
(i) A ⊂ B (ii) A – B = Φ
(iii) A ∪ B = B (iv) A ∩ B = A
Answer
First, we have to show that (i) ⇔ (ii).
Let A ⊂ B
To show: A−B ≠ Φ
If possible, suppose A − B ≠ Φ
This means that there exists x ∈ A, x ≠ B, which is not possible as A ⊂ B. ∴ A −B = Φ
∴ A ⊂ B ⇒ A − B = Φ
Let A− B = Φ
To show: A ⊂ B
Let x ∈ A
Clearly, x ∈ B because if x ∉ B, then A − B ≠ Φ ∴ A − B = Φ ⇒ A ⊂ B
∴ (i) ⇔ (ii)
Let A ⊂ B
To show: A ∪ B= B
Clearly, B ⊂ A ∪ B
Let x ∈ A ∪ B
⇒ x ∈ A or x ∈ B
Case I : x ∈ A
⇒ x ∈ B [∵ A ⊂ B]
∴ A ∪ B ⊂ B
Case II: x ∈ B
Then, A ∪ B = B
Conversely, let A ∪ B = B
Let x ∈ A
⇒ x ∈ A ∪ B [∵ A ⊂ A ∪ B]
⇒ x ∈ B [∵ A ∪ B = B]
∴ A ⊂ B
Hence, (i) ⇔ (iii)
Now, we have to show that (i) ⇔ (iv).
Let A ⊂ B
Clearly A ∩ B ⊂ A
Let x ∈ A
We have to show that x ∈ A ∩ B
As A ⊂ B, x ∈ B
∴ x ∈ A ∩ B
∴ A ⊂ A ∩ B
Hence, A = A ∩ B
Conversely, suppose A ∩ B = A
Let x ∈ A
⇒ x ∈ A ∩ B
⇒ x ∈ A and x ∈ B
∴ A ⊂ B
Hence, (i) ⇔ (iv).
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