TITLE: Prove that $\sigma(F)=\Omega$
QUESTION [5 upvotes]: Let $F=\{A_1,...,A_n\}\subset P(X)$; $F_a=A_1^{a_1}\cap A_2^{a_2}\cap\cdots \cap A_n^{a_n}$ $ a=(a_1,...,a_n)\in \{0,1\}^n$  
$$A^{a_i} =
\begin{cases}
A, & \text{if  } a_i=0 \\
A^c, & \text{if  } a_i=1
\end{cases}$$
Define $\Omega=\{\bigcup_{a\in D} F_a : D\subset \{0,1\}^n\}$ (by convention $\bigcup_{a\in \varnothing} F_a =\varnothing$)
I need to prove that the sigma algebra generated by $F$, $\sigma(F)=\Omega$
what I have done so far:
1)$\Omega \subset \sigma(F)$: We know that $F\subset \sigma(F)$ then for all $a=(a_1,...,a_n)\in\{0,1\}^n$ $F_a\in \sigma(F)$, hence $\bigcup_{a\in D} F_a \in \sigma(F)$
2)$\sigma(F)\subset \Omega$: for this part I wanted to prove that $\Omega$ is a sigma-algebra that contains $F$: 
$a) \varnothing \in \Omega$
$b)$ Since $\Omega$ is finite ($|\Omega|\le 2^{2^n}$) then we consider just finite unions: Let $A,B\in \Omega$: $$A\cup B= (\bigcup_{a\in D_1}F_a)\cup (\bigcup_{a\in D_2}F_a)=\bigcup_{a\in D_1\cup D_2}F_a\in \Omega$$
$c)$ I´m having trouble checking the complements: Let $A=\bigcup_{a\in D}F_a\in \Omega$; $A^c=(\bigcup_{a\in D}F_a)^c=\bigcap_{a\in D}F_a^c$ but from here how can I check that $A^c\in \Omega$?
I would really appreciate if you can help me with this problem (I hope this question won't be marked as a duplicate)

REPLY [3 votes]: If $A=\bigcup_{a\in D}F_a$, then $A^c=\bigcup_{a\in \{0,1\}^n\setminus D}F_a$.