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If we solve it by taking examples, let $A=\{1,2\}, B=\{2,3\}, C=\{3,4\}$.

Then

$$A\cap B= \{2\}, \quad B\cap C= \{3\}, \quad C\cap A= \emptyset.$$

So the intersection with $\emptyset$ will always be empty. So shouldn't the answer be $\emptyset$?

The answer given is $A\cap B\cap C$.

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    $\begingroup$ As an alternative example, what do you get when $A = B = C$? $\endgroup$ – Eric Towers Jul 31 at 16:06
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    $\begingroup$ It may be easy to see that set intersection is commutative and associative, and obviously $A\cap A=A$. $\endgroup$ – Dave Jul 31 at 16:16
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    $\begingroup$ Try drawing a Venn diagram. $\endgroup$ – gandalf61 Jul 31 at 16:35
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Remember those properties:

  • $A\cap A = A$
  • $(A \cap B )\cap C = A \cap B \cap C = A\cap (B \cap C)$

Then: $(A\cap B)\cap (B\cap C)\cap (C\cap A) = A \cap B \cap B \cap C \cap C \cap A$

so: $$=A\cap B\cap C$$

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Just because it holds true for one example that you obtain the \emptyset, does not mean, that it holds true, for every choice of $A, B, C$.

Just imagine $A=B=C=\{1\}$. Then the answer, would be $\{1\}$. You might want to try to proof

$(A∩B)∩(B∩C)∩(C∩A)=A\cap B\cap C$

It is a basic proof.

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  • $\begingroup$ Sure I get that, but shouldn’t it’s hold true for every example I take? $\endgroup$ – Aditya Jul 31 at 16:17
  • $\begingroup$ @Aditya It does, since in your example $A\cap B\cap C=\emptyset$. But that does not mean, that $A\cap B\cap C$ always has to be the emptyset. $\endgroup$ – Cornman Jul 31 at 16:33
  • $\begingroup$ So technically my answer is right? That’s what I need to know. It might be a special example, but it’s an example nonetheless. So it has to be valid. $\endgroup$ – Aditya Jul 31 at 16:35
  • $\begingroup$ Yes, your example is correct, but it seems that you think now, that because of one examples resulting into the emptyset EVERY example has the emptyset as result, which is not true. You are asked to proof this statement. Do you know how to proof equality of sets? $\endgroup$ – Cornman Jul 31 at 16:41
  • $\begingroup$ It wasn’t asked to prove, it was an MCQ. And there was an option for $\phi$. I would have circled that, and that’s basically why I am concerned about the correctness of my example. And yes, I do know how to prove equality of sets. $\endgroup$ – Aditya Jul 31 at 17:11
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Try evaluating either $A \cap B$ or $B \cap C$ first, then take the intersection of that set with either $C$ or $A$, respectively. You should come to the same conclusion.

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Once you have the inuition that the whole thing equals $A \cap B \cap C$, an "element by element" proof may be simpler:

Let $x \in A \cap B \cap C$, then $x \in A \cap B$, $x \in A \cap C$ and $x \in B \cap C$

Conversely, let $x \in (A \cap B) \cap (A \cap C) \cap (B \cap C)$. Then, $x \in A$, $x \in B$ and $x \in C$

Of course, you may have been explicitely required to not proceed like this

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\begin{align*} (A \cap B)\cap(B\cap C)\cap(C\cap A) &= A \cap B \cap B \cap C \cap C \cap A &&\text{drop brackets, because of the associativity of the operation } \cap \\ &= A \cap B \cap C \cap A &&B \cap B = B \text{ and } C \cap C = C \\ &= A \cap A \cap B \cap C &&\text{commutativity and associativity: } (A \cap B \cap C) \cap A = A \cap (A \cap B \cap C) \\ &=A \cap B\cap C && A \cap A = A \end{align*}

Consider $A,B,C$ as numbers and $\cap$ as a multiplication sign. https://en.m.wikipedia.org/wiki/Algebra_of_sets

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  • $\begingroup$ you need to fix your latex $\endgroup$ – Luis Felipe Jul 31 at 16:17
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We won't have

$(A \cap B) \cap (B \cap C) \cap (C \cap A) = \emptyset \tag 1$

in general, since it is possible that

$A = B = C \ne \emptyset; \tag2$

then

$(A \cap B) \cap (B \cap C) \cap (C \cap A) = (A \cap A) \cap (A \cap A) \cap (A \cap A) = A \cap A \cap A = A \ne \emptyset; \tag 3$

however, we do have

$(A \cap B) \cap (B \cap C) \cap (C \cap A)$ $= (A \cap B) \cap (B \cap C)) \cap (C \cap A) = (A \cap (B \cap B) \cap C) \cap (C \cap A)$ $= (A \cap B \cap C) \cap (C \cap A) = (A \cap A) \cap B \cap (C \cap C) = A \cap B \cap C, \tag 4$

where we have used the usual properties of the "$\cap$" operation, such as associativity, commutativity, and so forth. Note especially that we always have

$D \cap D = D \tag 5$

for any set $D$.

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Note that $$(A\cap B)\cap (B\cap C)\cap (C\cap A)=A\cap (B\cap B)\cap( C\cap C)\cap A=$$

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

Thus the answer in general is $A\cap B\cap C$ and in your special example we have $$ A\cap B\cap C = \phi $$

There is no problem here.

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  • $\begingroup$ $A,B,C$ are non empty sets, not disjoints $\endgroup$ – Luis Felipe Jul 31 at 16:18
  • $\begingroup$ I did not make that assumption. $\endgroup$ – Mohammad Riazi-Kermani Aug 1 at 21:26

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