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If $A$ and $B$ are subests of the universal set $U$, then show that
(i) $A \subset A \cup B$
(ii) $A \subset B \Leftrightarrow A \cup B=B$
(iii) $(A \cap B) \subset A$
(i) $A \subset A \cup B$
(ii) $A \subset B \Leftrightarrow A \cup B=B$
(iii) $(A \cap B) \subset A$
Solution:
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Verified Answer
(i) Let $x \in A$
$\Rightarrow \mathrm{x} \in A$ or $\mathrm{x} \in B \Rightarrow x \in A \cup B$
Hence, $A \subset A \cup B$
(ii) If $A \subset B$
Let $x \in A \cup B$
$\Rightarrow x \in A$ or $x \in B$
$\Rightarrow x \in B \quad[\therefore A \subset B]$
$\Rightarrow A \cup B \subset B \quad$... (i)
But $A \subset B \cup B$
From Eqs. (i) and (ii), we have
$A \cup B=B$
Next if $A \cup B=B$
Nowlet $y \in A$
$\Rightarrow y \in A \cup B \Rightarrow y \in B[\therefore A \cup B=B]$
$\Rightarrow A \subset B$
Hence, $A \subset B \Leftrightarrow A \cup B=B$
(iii) Let $x \in A \cap B$
$\Rightarrow x \in A$ and $x \in B \Rightarrow x \in A \Rightarrow A \cap B \subset A$
$\Rightarrow \mathrm{x} \in A$ or $\mathrm{x} \in B \Rightarrow x \in A \cup B$
Hence, $A \subset A \cup B$
(ii) If $A \subset B$
Let $x \in A \cup B$
$\Rightarrow x \in A$ or $x \in B$
$\Rightarrow x \in B \quad[\therefore A \subset B]$
$\Rightarrow A \cup B \subset B \quad$... (i)
But $A \subset B \cup B$
From Eqs. (i) and (ii), we have
$A \cup B=B$
Next if $A \cup B=B$
Nowlet $y \in A$
$\Rightarrow y \in A \cup B \Rightarrow y \in B[\therefore A \cup B=B]$
$\Rightarrow A \subset B$
Hence, $A \subset B \Leftrightarrow A \cup B=B$
(iii) Let $x \in A \cap B$
$\Rightarrow x \in A$ and $x \in B \Rightarrow x \in A \Rightarrow A \cap B \subset A$
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