Why is the following collection of sets equals the following? Suppose $(A_n)$ is a sequence of events, For any $I \subset \{1,2,\ldots \} $, set 
$$ C_I = \bigg( \bigcap_{n \in I} A_n \bigg) \cap \bigg( \bigcap_{n \notin I } A_n^c \bigg) $$
I am trying to show that for any $n \geq 1 $ we have 
$$ \bigcup_{|I| < \infty,\ n \in I} C_I = A_n $$
I find kind of hard to understand this identity. For example, if I take $I = \{1,2,3 \} $, then 
$$ C_I = ( A_1 \cap A_2 \cap A_3) \cap ( A_4^c \cap A_5^c \cap \cdots) $$
But, then how can I understand and compute $ \bigcup_{n \in I } C_I $ in this situation? 

Look at how something gets into a given $C_I$.  Let $I$ be a finite index set and $a\in C_I$
then we have $\forall k\in I,a\in A_k$.  We also have $a\in A_j^C$  for $j\notin I$, or in other words, $a\notin A_j$   Therefore, for a given index set, we're collecting all the events which are in every one of those index sets, and in NO other ones. 
Now,   fix $n$ and let $I$ range over every finite index set that includes $n$.  How does something get into one of these sets?   Well,  it has to be in $A_n$,  since we are only taking index sets that include n,  so we have your union is a subset of $A_n$   So,  the only question is,  does it miss anything from $A_n$?  The answer is no,   because if $a\in A_n$,  there will be some index set $I$  that $a\in C_I$. 
In all honestly, the reason for that last step is evading me, but I started typing before I had the final answer!  Hopefully it'll come to you, me, or someone else shortly.
The general proof method though is to try and show both sets are subsets of each other when trying to show a complicated set equality, and thus look at how things get into each one.
