Solution:

(i) As per the question,

$A$ and $B$ are two subsets

We need to prove: $A \subset A \cup B$

Proof:

Let’s say $x \in A$

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

$\Rightarrow \mathrm{x} \in \mathrm{A} \cup \mathrm{B}$ $\Rightarrow \mathrm{A} \subset \mathrm{A} \cup \mathrm{B}$

As a result, hence proved.

(ii) As per the question,

Two subsets are A and B

We need to prove: $A \subset B \Leftrightarrow A \cup B=B$

Proof:

Let’s say $x \in A \cup B$

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

As, $A \subset B$, we get,

$\begin{array}{l}
\Rightarrow \mathrm{x} \in \mathrm{B} \\
\Rightarrow \mathrm{A} \cup \mathrm{B} \subset \mathrm{B} \ldots \dots \dots(\mathrm{i})
\end{array}$

It is known that,

$B~\subset ~A~\cup ~B~ \ldots \dots \dots(ii)$

From eq.(i) and eq.(ii),

We have,

$A \cup B=B$

So now,

Let’s say $y \in A$

$\Rightarrow \mathrm{y} \in \mathrm{A} \cup \mathrm{B}$

As, $A \cup B=B$, we have,

$\begin{array}{l}
\Rightarrow \mathrm{y} \in \mathrm{B}\} \\
\Rightarrow \mathrm{A} \subset \mathrm{B}
\end{array}$

Therefore,

$A \subset B \Leftrightarrow A \cup B=B$

As a result, hence proved.