दिखाइए कि निम्नलिखित चार प्रतिबंध तुल्य हैं

$(i)$ $A \subset B$

$(ii)$ $A-B=\phi$

$(iii)$ $A \cup B=B$

$(iv)$ $A \cap B=A$

First, we have to show that $(i) \Leftrightarrow(i i)$

Let $A \subset B$

To show: $A-B \neq \varnothing$

If possible, suppose $A-B \neq \varnothing$

This means that there exists $x \in A, x \neq B,$ which is not possible as $A \subset B$

$\therefore A-B=\varnothing$

$\therefore A \subset B \Rightarrow A-B=\varnothing$

Let $A-B=\varnothing$

To show: $A \subset B$

Let $x \in A$

Clearly, $x \in B$ because if $x \notin B$, then $A-B \neq \varnothing$

$\therefore A-B=\varnothing \Rightarrow A \subset B$

$\therefore(i) \Leftrightarrow(i i)$

Let $A \subset B$

To show: $A \cup B=B$

Let $x \in A$

Clearly, $x \in B$ because if $x \notin B$, then $A-B \neq \varnothing$

$\therefore A-B=\varnothing \Rightarrow A \subset B$

$\therefore(i) \Leftrightarrow(i i)$

Let $A \subset B$

To show: $A \cup B=B$

Clearly, $B \subset A \cup B$

Let $x \in A \cup B$

$\Rightarrow x \in A$ or $x \in B$

Case $I:$ $x \in A$

$\Rightarrow x \in B$           $[\because A \subset B]$

$\therefore A \cup B \subset B$

Case $II:$ $x \in B$

Then, $A \cup B=B$

Conversely, let $A \cup B=B$

Let $x \in A$

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

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

$\therefore A \subset B$

Hence, $(i) \Leftrightarrow(\text {iii})$

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

Let $A \subset B$

Clearly $A \cap B \subset A$

Let $x \in A$

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

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

$\therefore x \in A \cap B$

$\therefore A \subset A \cap B$

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, $(i) \Leftrightarrow(i v)$