Let A, B and C be sets such that ϕ ≠ A B ⊆ C . Then which of the...

For A = C, A - C = ϕ$<br />$⇒ ϕ ⊆ B$<br />But A ⊈ B$
Hence, If $( A - C ) ⊆ B,$ then A ⊆ B$is not true<br />Let$x \in C, ⇒ x \in $\left(C \cup A\right)$ \cap \left(C \cup B\right)$<br />$⇒ x \in \left(C \cup A\right)$and x \in\left(C \cup B\right)$
$⇒ ( x \in C$ or $x \in A ) and ( x \in C or x \in B )$
$⇒ x \in C$ or $x \in\left(A \cap B\right)<br />$⇒ x \in C or x \in C\left(asA \cup B ⊆ C\right)$<br />$⇒ x \in C$<br />$⇒ \left(C \cup A\right) \cap \left(C \cup B\right) ⊆ C$.... (i)<br />Now$x \in C ⇒ x \in \left(C \cup A\right)$and x \in\left(C \cup B\right)$
$⇒ x \in ( C \cup A ) \cap \left(C \cup B\right)<br />$⇒ C ⊆ ( C \cup A ) \cap \left(C \cup B\right)$..... (ii)<br />$⇒$From (i) and (ii)<br />$C = \left(C \cup A\right) \cap \left(C \cup B\right)
Hence, $( C \cup A ) \cap ( C \cup B ) = C$ is true
Let $x \in A$ and $x \notin B$
$⇒ x \in \left(A - B\right)$
$⇒ x \in C \left(asA - B ⊆ C\right)<br />Let$x \in A$and$x \in B$<br />$⇒ x \in \left(A \cap B\right)$<br />$⇒ x \in C \left(asA \cap B ⊆ C\right)
Hence $x \in A ⇒ x \in C$
$⇒ A ⊆ C$
Hence, If $( A - B ) ⊆ C,$ then A ⊆ C$is true<br />as$C ⊇ \left(A \cap B\right)
$⇒ B \cap C ⊇ \left(A \cap B\right)$
as A \cap B \ne ϕ$<br />$⇒ B \cap C \ne ϕ$<br />Hence,$B \cap C \ne ϕ$is true</p><p>Alternate solution</p><p>Given </p><p>$ϕ \ne A \cap B ⊆ C$</p><p>Let answer that</p><p>$C = \left\{1, 2, 3, 4, 5, 6\right\}$</p><p>A = \left\{1, 2, 3\right\}$

$B = \left\{3, 4, 5\right\}</p><p>A \cap B = \left\{3\right\}</p><p>Option (a) </p><p>$B \cap C = \left\{3, 4, 5\right\} \ne ϕ$</p><p>Option (b)$C \cup A = \left\{1, 2, 3, 4, 5, 6\right\}

$C \cup B = \left\{1, 2, 3, 4, 5, 6\right\}$

\left(C \cup A\right) \cap \left(C \cup B\right) = \left\{1, 2, 3, 4, 5, 6\right\} = C$</p><p>Option (c) A - B = \left\{1, 2, 3\right\}$ ⊆ CA ⊆ C$True</p><p>Option (d) A - C = ϕ ⊆ BA ⊆ C$ Not correct

$∴$ Option (d) is correct