Set theory (very basic question) I have a very basic question about set theory (I think naive set theory because it's a intro to probability class). I can't seem to be able to prove that 
$(A \cap B) \cap (A \cap C) = A \cap B \cap C$ without the Venn Diagram...
Maybe my manuel is missing a property or something.
Could anyone enlightnen me on this cause I've been studying for too long and can't concentrate anymore and this simple question is making me crazy :D
Thank you everyone
 A: A direct approach would be to prove that $x \in (A \cap B) \cap (A \cap C)$ if and only if $x \in A \cap B \cap C$ using the definition of intersection at each step.
Another approach would be to use basic algebraic properties of intersection:
$$\begin{align*}
(A \cap B) \cap (A \cap C) &= ((A \cap B) \cap A) \cap C && \text{by associativity} \\
&= (A \cap (A \cap B)) \cap C && \text{by commutativity} \\
&= ((A \cap A) \cap B) \cap C && \text{by associativity} \\
&= (A \cap B) \cap C && \text{by idempotence}
\end{align*}$$
This requires you to have already proved that intersection satisfies associativity, commutativity and idempotence, so I recommend the direct approach.
A: Intersection of sets satisfy the associative, commutative, and idempotent laws (see https://en.wikipedia.org/wiki/Algebra_of_sets), and you can use them to prove this identity, for example, like this: 
$(A \cap B) \cap (A \cap C) = ((A \cap B) \cap A) \cap C$ (by associativity)
$((A \cap B) \cap A) \cap C = (A \cap (B \cap A)) \cap C $  (by associativity)
$(A \cap (B \cap A)) \cap C = (A \cap (A \cap B)) \cap C $  (by commutativity)
$(A \cap (A \cap B)) \cap C = ((A \cap A) \cap B) \cap C $  (by associativity)
$((A \cap A) \cap B) \cap C = (A \cap B) \cap C $  (by idempotency)
Since intersection of sets is an associative operation, you can drop the parentheses:
$(A \cap B) \cap C = A \cap B \cap C $
Thus:
$(A \cap B) \cap (A \cap C) = A \cap B \cap C$
A: As I wrote in my comment, a good way to solve set theory problems like this is to take an element from the set on the LHS of the equation and prove it belongs in the other set, and vice versa.
Here, let $x\in((A\cap B)\cap(A\cap C))$. What does this mean? Well, $x$ is in (A and B) and it is also in (A and C). Therefore $x$ is in A and $x$ is in B and $x$ is in C. So $x\in (A\cap B\cap C)$.
Then take $y$ in $(A\cap B\cap C)$, and prove $y\in((A\cap B)\cap(A\cap C))$.
A: When it comes to elementary set theory, don't think in terms of properties. Think in terms of what things mean.
We say two sets are equal if they have the same elements - that is, $A = B$ if whenever $x \in A$, $x$ is also $\in B$, and vice versa. That means that a good strategy for proving two sets are equal is to take an element of one and prove that it's in the other.
The intersection of two sets, written $A \cap B$, is the set of all things that are in both of them; so $x \in A \cap B$ if and only if $x \in A$ and $x \in B$. So, in particular, $x \in (A \cap B) \cap (A \cap C)$ if and only if $x$ is in both $A \cap B$ and in $A \cap C$. $x \in A \cap B \cap C$ if and only if $x$ is in $A$, $B$, and $C$.
So, what you want to show is that if $x$ is in both $A \cap B$ and $A \cap C$, then $x$ is in $A$, $B$, and $C$, and vice versa. Here's part of it: if $x$ is in both $A \cap B$ and $A \cap C$, then $x \in A$, because everything in $A \cap B$ is in $A$. Can you see how to show that $x \in B$? How about $x \in C$?
Finally, you need to do it the other way around: if $x$ is in $A$, $B$, and $C$, show that $x$ is in $A \cap B$ and in $A \cap C$. Think about what it means for $x$ to be in $A \cap B$.
