If $A, B$ and $C$ are non-empty sets, simplify $(A\cap B)\cap (B\cap C)\cap (C\cap A)$. 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 $A\cap B\cap C$.
 A: 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: 
$$=A\cap B\cap C$$
A: 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. 
A: \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
A: 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.
A: 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
A: 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.
A: 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 $D$.
