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Prove or find a counterexample to the statement: there are sets $A,B,C$ such that $A\cap B = B \cap C = A \cap C = \emptyset$ but $A \cap B \cap C \neq \emptyset$.

I know that the statement is true and to prove it, you have to let x be an element of some set.

I am stuck at this point. I may also be wrong. I have tried to prove it using the set rules but I had no luck so far. Please help me.

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  • $\begingroup$ Think of three persons of various nationalities; If no two of them have a common language to communicate with each other is there a common language for all three of them to understand each other? $\endgroup$ – P Vanchinathan Apr 6 '14 at 1:23
  • $\begingroup$ What level are you working at? Axiomatic set theory or naive set theory? And if axiomatic set theory, what kind of theorems have you proved? Perhaps $\forall a,b \colon a \cap b \subseteq a$? $\endgroup$ – kahen Apr 6 '14 at 1:50
  • $\begingroup$ I am working on naive set theory. $\endgroup$ – Adarsh Apr 6 '14 at 2:08
  • $\begingroup$ There is no "negation of $A\cap C$". You negate statements, not sets. What is the negation of an apple? Or of a bag of apples? $\endgroup$ – Andrés E. Caicedo Apr 9 '14 at 0:19
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    $\begingroup$ You really need to be more careful. What you need to prove or find a counterexample for is not what you wrote, which makes no sense. Perhaps it is "There are sets $A,B,C$ such that $A\cap B=B\cap C=A\cap C=\emptyset$ but $A\cap B\cap C\ne\emptyset$", or perhaps it is "For any $A,B,C$, if $A\cap B=B\cap C=A\cap C=\emptyset$, then $A\cap B\cap C\ne\emptyset$". Mixing these two options, as you did, and leaving out the quantifiers, as you did, results in nonsense. $\endgroup$ – Andrés E. Caicedo Apr 9 '14 at 0:31
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Here's some intuition: whenever you take the intersection of one set with another set, the result is smaller than both sets you began with. By "smaller," I mean that the intersection is a subset of both of the original sets. In more formal words, if $A$ and $B$ are sets then $A\cap B\subseteq A$ and $A\cap B\subseteq B$. (Can you prove it?)

Do you see where to go from here?

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  • $\begingroup$ Can you give me any more hints please. $\endgroup$ – Adarsh Apr 6 '14 at 10:22
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    $\begingroup$ $A\cap B\cap C\subseteq A\cap B$. $\endgroup$ – user134824 Apr 6 '14 at 14:56
  • $\begingroup$ can you tell me if I am on the right track. I have come up with let x be an element of negation of A intersection B where x is not an element of A and B thus x is an element of negation of A and B hence, x is an element of negation of A intersection B if and only if x is an element of negation of (A intersection B) union (negation of B). Sorry for not using symbols. $\endgroup$ – Adarsh Apr 7 '14 at 2:32
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In general, if $A$ and $B$ are sets then $A\cap B \subset A$ and $A\cap B \subset B$, so in particular if $A\cap B=\emptyset$ then $(A\cap B)\cap C \subset A\cap B = \emptyset$, but the only subset of the empty set is itself.

(So you must have misunderstood or wrote the problem down incorrectly.... possibly its $A\cup B \cup C \cup \neq \emptyset ? )$

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Read the wikipedia page- http://en.wikipedia.org/wiki/Disjoint_sets. Keyword: pairwise disjoint. I also think that you have misunderstood the problem but the converse can be true.

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(copy-paste of my answer to your re-post; please do not ask 2 times the same question)

First of all, as others pointed out, you could state the question as such:

Statement: $\exists A,B,C. \left(A \cap B = B \cap C = A \cap C = \emptyset \right) \wedge A \cap B \cap C \neq \emptyset$

You can now either prove that this statement is false or find an example (sets A, B and C) that would satisfy it. I can't think of any such sets so I carried on with a proof:

By the definition of intersection, you know that

$$(A \cap B) \subset A \wedge (A \cap B) \subset B$$

Now if $A \cap B = \emptyset$, then

$$A \cap B \cap C = \emptyset \cap C$$

Thus

$\emptyset \cap C \subset \emptyset \Rightarrow \emptyset \cap C = \emptyset \Rightarrow A \cap B \cap C = \emptyset$

so the original statement is false.

Note that it makes intuitive sense; if the three sets do not intersect each-others in pairs, how could they possibly all intersect?

PS: drawing a Venn diagram comes in handy in this kind of question ;)

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  • $\begingroup$ Could you please provide an example of a set relating to what you have proven. $\endgroup$ – Adarsh Apr 16 '14 at 5:37
  • $\begingroup$ @Adarsh I'm not sure what you want here; the original statement is false (aka there are no such sets A, B and C that satisfy it). $\endgroup$ – Aegis Apr 16 '14 at 20:20

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