Let each of $A, B$, and $C$ be a set and suppose $A \subseteq B \cup C$. Prove that $A \cap B \cap C = \varnothing$.
I start this problem by letting $x$ be an element of $A \subseteq B \cup C$ and stating that $x$ is an element of $A$ and also $B \cup C$. After that I get confused. Could someone provide some hints?
Answer from comments:
If you let A=B=C be some non-empty set, then the intersection is non-empty.
et A=B=C={1}. Then B∪C={1} and so A={1}⊆{1}=B∪C and A∩B∩C={1}≠∅. So this is a counterexample.
The question is found to be false.