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? 


But, then how can I understand and compute $\bigcup_{n\in I} C_I$ in this situation? 

No, $n$ is a free variable in the required union sequence; it is not bound as a 'dummy variable'.   The interval $I$ is the dummy variable in this case.
You want the union of all $C_I$, for all finite intervals $I$ that contain a particular $n$.  
$$\bigcup_{I\ni n: |I|<\infty} C_I \quad=\quad C_{\{n\}} \cup C_{\{1, n\}} \cup C_{\{2,n\}} \cup\cdots\cup C_{\{1, 2, n\}}\cup\cdots$$
Show that for that $n$ this is equal to $A_n$.
