Showing that $\{A_1, \ldots, A_n\}$ and $\{ \bigcap_{j=1}^{n} B_j : B_j \in \{ A_j, A_j^{\mathsf{c}} \} \}$ generate the same $\sigma$-algebra Let $\Omega$ be an arbitrary set and $\mathcal{G}_1 := \{A_1, \ldots, A_n\} \subseteq \mathcal{P}(\Omega)$ a collection of subsets of $\Omega$. Now set
$$
\mathcal{G}_2 : = \bigg\{ \bigcap_{j=1}^{n} B_j : B_j \in \{ A_j, A_j^{\mathsf{c}} \} \bigg\}.
$$
I want to prove that $\sigma(\mathcal{G}_1)=\sigma(\mathcal{G}_2)$, where the $\sigma$-operator means "$\sigma$-algebra generated by".
Idea of proof: For $B_j \in \{ A_j, A_j^{\mathsf{c}}\}$ we have that $B_j \in \sigma(\mathcal{G}_1)$ and $\bigcap_{j=1}^{n} B_j \in \sigma(\mathcal{G}_1)$. This shows that $\mathcal{G}_2 \subseteq \sigma(\mathcal{G}_1)$. Since $\sigma$ preserves inclusions it follows that $\sigma(\mathcal{G}_2) \subseteq \sigma(\mathcal{G}_1)$. Now it remains to show that $\sigma(\mathcal{G}_1) \subseteq \sigma(\mathcal{G}_2)$. Therefore it suffices to show that each $A_i$ can be written as countable union of sets of the form $\bigcap_{j=1}^{n} B_j$. For example in the case $n=3$ we have
$$
A_1 = (A_1 \cap A_2 \cap A_3) \uplus (A_1 \cap A_2^{\mathsf{c}} \cap A_3) \uplus (A_1 \cap A_2 \cap A_3^{\mathsf{c}}) \uplus (A_1 \cap A_2^{\mathsf{c}} \cap A_3^{\mathsf{c}}).
$$
I think I have an idea how to continue this schema for $n>3$ but I have problems to write it down formally correct. Maybe there is a more obviously way to write each $A_i$ in the desired form.
 A: We define the sets
$$
M_1^{[1]}, \ldots ,M_{\binom{n-1}{1}}^{[1]}, \\
M_1^{[2]}, \ldots ,M_{\binom{n-1}{2}}^{[2]}, \\
\vdots \\
M_{\binom{n-1}{n-1}}^{[n-1]} = M_{1}^{[n-1]}
$$
by 
$$
M_{b}^{[a]} := A_j \cap \bigcap_{k=1, k \neq j}^{n} B_k,
$$
where $b$ denotes the number of complements and $a$ the position of the complements.
In total we have $\binom{n-1}{1} + \binom{n-1}{2} + \ldots + \binom{n-1}{n-1}=2^{n-1}$ such sets.
Now we have to show that
$$A_j = \bigcup_{a,b} M_{b}^{[a]}.$$
But this equation is clear, since the $B_k$'s consist of all possible combinations of complements and non-complements.

Possibly simpler than the above: 
To prove that $\mathcal G_1\subseteq\sigma(\mathcal G_2)$, let $N=\{1,2,\ldots,n\}$ and, for every $I\subseteq N$, $$B_I=\bigcap_{i\in I}A_i\cap\bigcap_{i\in N\setminus I}A_i^c.$$
Then, each $B_I$ is in $\mathcal G_2$ and, for each $j$ in $N$, $A_j$ is the union of the sets $B_I$ over every $I\subseteq N$ containing $j$. As such, $A_j$ is in $\sigma(\mathcal G_2)$.
