Let A, B and C be sets such that $\phi \neq A \cap B \subseteq C$. Then which of the following statements is not true ?
(A)
If (A – B) $\subseteq$ C, then A $\subseteq$ C
(B)
B $\cap$ C $\neq$ $\phi$
(C)
(C $\cup$ A) $\cap$ (C $\cup$ B) = C
(D)
If (A – C) $\subseteq$ B, then A $\subseteq$ B
MEDIUM
Correct Answer:D
Explanation
The question provides that $A \cap B \subseteq C$ and $A \cap B \neq \phi$. Analyzing each statement:
Statement A: If $(A - B) \subseteq C$, then $A \subseteq C$. This statement can be verified using Venn diagrams. If the part of A that is not in B is contained in C, then A is a subset of C.
Statement B: $B \cap C \neq \phi$. This means that the intersection of B and C is not empty. Given that $A\cap B \subseteq C$, this is true.
Statement C: $(C \cup A) \cap (C \cup B) = C$. Using the distributive property of sets, $(C \cup A) \cap (C \cup B) = C \cup (A \cap B)$. Since $A \cap B \subseteq C$, then $C \cup (A \cap B) = C$. So, this statement is true.
Statement D: If $(A - C) \subseteq B$, then $A \subseteq B$. This statement is NOT necessarily true. It is possible to have $A - C \subseteq B$, $A \cap B \subseteq C$ and $A \cap B \neq \phi$, but it does NOT require that $A \subseteq B$. This is because A can contain elements outside of B that are also outside of C.
Therefore, the statement that is not true is D.