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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?2012-12-09
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    I'm not sure. By information do you mean that there should be more to the question?2012-12-09
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    @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.2012-12-09
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    @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.2012-12-09
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    @blutuu well it's clearly false as Fortuon showed.2012-12-09
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    blutuu, if your professor proved this in class, then she has proved that Mathematics is inconsistent, $0=1$, and we can all go home. Forget what you think your professor said --- look at the examples you have been given here! What you are trying to prove is false! My guess: you copied it in a hurry & messed it up.2012-12-09
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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.