#### Question

Show that the following four conditions are equivalent:

**(i)** A ⊂ B

**(ii)** A – B = Φ

**(iii)** A ∪ B = B

**(iv)** A ∩ B = A

#### Solution

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 cup B`

Clearly, `B subset A cup B`

Let `x in A cupB`

`=>x in A or x inB`

Case I : `x in A`

`=>x in B` `[because AsubsetB]`

`therefore A cup B subset B`

Case II : `x in B`

Then, `A cup B = B`

Conversely, let `A cup B = B`

Let *x* ∈ A

`=> x in A cup B` `[because A subset A cup B]`

`=> x in B` `[because A cup B = B]`

`therefore A subset B`

Hence, (i) ⇔ (iii)

Now, we have to show that (i) ⇔ (iv).

Let A ⊂ B

Clearly `A nn B subset A`

Let *x *∈ A

We have to show that `x in A nn B`

As A ⊂ B, *x *∈ B

`therefore x in A nn B`

`therefore A subset A nn B`

Hence, A = A ∩ B

Conversely, suppose A ∩ B = A

Let *x* ∈ A

⇒ x ∈ A ∩ B

⇒ *x* ∈ A and *x* ∈ B

⇒ *x *∈ B

∴ A ⊂ B

Hence, (i) ⇔ (iv).