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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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I think you may be missing some information. If you let $A=B=C$ be some non-empty set, then the intersection is non-empty! Perhaps you want that $B$ and $C$ are disjoint? – Ravi Dec 9 '12 at 4:21
I'm not sure. By information do you mean that there should be more to the question? – blutuu Dec 9 '12 at 4:55
@blutuu yeah Fortuon does mean that. Just to make it concrete, let $A=B=C=\{1\}$. Then $B\cup C=\{1\}$ and so $A=\{1\}\subseteq \{1\}=B\cup C$ and $A\cap B \cap C = \{1\} \ne \emptyset$. So this is a counterexample; the claim which you are meant to prove is false. Is there any other information in the question as it was given to you? Incidentally, you need to start your comment with @[name] to get [name]'s attention. – crf Dec 9 '12 at 5:15
@crf No this was a previous exam's question. My professor had proven it in class, but I don't have my notes with me to look at. I'm sure he's done this one a different way. – blutuu Dec 9 '12 at 5:24
@blutuu well it's clearly false as Fortuon showed. – crf Dec 9 '12 at 5:30
up vote 0 down vote accepted

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.

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