This question was previously asked here: For all sets A,B, and C, If B ∩ C ⊆ A, then (A-B) ∩ (A-C) ≠∅
However, it didn't help me much so I wanted if someone could verify my working and correct me where I am wrong.
My initial question is:
For all sets $A$, $B$ and $C$, if $(B ∩ C) ⊆ A$, then $(A-B) ∩ (A-C) ≠ ∅$
I know this is true by drawing a Venn Diagram.
Proof:
$(B ∩ C) ⊆ A$
$x∈(B∩C)⇒x∈A$
$x∈(B∩C)^c ∩ A$
$x∈A∩(B^c∩C^c)$
$x∈(A∩B^c)∩(A∩C^c)$
Which is $(A-B)∩(A−C)≠∅$
Thank you for any help.
Best Answer
This statement is not true. For example, take $A = B=C$ then $B \cap C = A$, however $A - B$ and $A-C$ are both empty (Thank you to gt6989b above!).
In your proof, the step $x \in (B \cap C) \implies x \in A$ is correct, but the next,namely the third step is unclear i.e. how was $x$ in $B \cap C$ in statement $2$, and how did it land up in the complement in statement $3$. Also, in the fourth step, $(B \cap C)^c \neq (B^c \color{red}{\cap} C^c)$, but instead equals $B^c \color{blue}{\cup} C^c$, so the fifth step also goes wrong.
The result is not true even after assuming that $B \cap C$ is strictly contained in $A$. Indeed, let $A = C = \{1,2\} $ and $B = \{1\}$, then $A-C$ is empty so the intersection with $A-B$ is empty, although $B \cap C = B$ is strictly contained in $A$. Modifications have to be made to the given statement to make it true.