Show that the following four conditions are equivalent:

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: $\mathrm{A} \cup \mathrm{B}=\mathrm{B}$

Clearly, $\mathrm{B} \subset \mathrm{A} \cup \mathrm{B}$

Let $x \in A \cup B$

$\Rightarrow x \in \mathrm{A}$ or $x \in \mathrm{B}$

Case I: $x \in \mathrm{A}$

$\Rightarrow x \in \mathrm{B} \quad[\because \mathrm{A} \subset \mathrm{B}]$

$\therefore \mathrm{A} \cup \mathrm{B} \subset \mathrm{B}$

Case II: $x \in B$

Then, $\mathrm{A} \cup \mathrm{B}=\mathrm{B}$

Conversely, let $\mathrm{A} \cup \mathrm{B}=\mathrm{B}$

Let $x \in A$

$\Rightarrow x \in \mathrm{A} \cup \mathrm{B} \quad[\because \mathrm{A} \subset \mathrm{A} \cup \mathrm{B}]$

$\Rightarrow x \in \mathrm{B} \quad[\because \mathrm{A} \cup \mathrm{B}=\mathrm{B}]$

$\therefore A \subset B$

Hence, (i) $\Leftrightarrow$ (iii)

Now, we have to show that (i) $\Leftrightarrow$ (iv).

Let $A \subset B$

Clearly $\mathrm{A} \cap \mathrm{B} \subset \mathrm{A}$

Let $x \in \mathrm{A}$

We have to show that $x \in A \cap B$

As $A \subset B, x \in B$

$\therefore x \in A \cap B$

$\therefore \mathrm{A} \subset \mathrm{A} \cap \mathrm{B}$

Hence, $A=A \cap B$

Conversely, suppose $A \cap B=A$

Let $x \in A$

$\Rightarrow x \in A \cap B$

$\Rightarrow x \in A$ and $x \in B$

$\Rightarrow x \in B$

$\therefore A \subset B$

Hence, (i) $\Leftrightarrow$ (iv).