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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?

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    @GerryMyerson No I didn't mess up the question. I have the actual test sheet right here, but here is where everything you all are saying comes together. He stated in the test instructions that at least one of the problems is false and must be proven by counterexample. I guess this is it. Thanks for the help.2012-12-09

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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.