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.