How to represent the member of generated sigma algebra in terms of unions and complements? Let $\mathscr{A}$ be an algebra over a set $\Omega$. For any $E\in\sigma(\mathscr{A})$, prove $E$ can be represented as countable union and complements which belong to $\mathscr{A}$. Can someone give a hint?

The difficulty I have with this problem is that the definition of $\sigma(\mathscr{A})$ is not constructive:
$$\sigma(\mathscr{A})=\cap\{\mathscr{F}:\mathscr{A}\subseteq\mathscr{F},\mathscr{‌​F}\text{ is a sigma algebra over }\Omega\}$$
 A: Since $\sigma(\scr A)$ is the smallest $\sigma$-algebra containing $\scr A$, the usual way to prove that every $E\in\sigma(\scr A)$ has a property is to show that the set of all sets with the property is a $\sigma$-algebra containing $\scr A$. So let $${\scr P}=\{\bigcup_{n\in\Bbb N}\bigcap_{k\in\Bbb N}A_{nk}:A_{nk}\in A\}.$$
Obviously $\scr A\subseteq P$; we need to show that $\scr P$ is closed under complement and countable union. A countable union of sets in $\scr P$ is $\bigcup_{i,n\in\Bbb N\times\Bbb N}\bigcap_{k\in\Bbb N}A_{ink}$, which is in $\scr P$ because $\Bbb N\times\Bbb N$ is countable. Showing that $\scr P$ is closed under complement is more interesting - can you prove it? (It may actually be false; if so then you will need to extend the characterization of $\scr P$ to sets of the form $\bigcup_{n_1\in\Bbb N}\bigcap_{k_1\in\Bbb N}\bigcup_{n_2\in\Bbb N}\bigcap_{k_2\in\Bbb N}\dots A_{n_1n_2\dots k_1k_2\dots}$ for either countably many or $\omega_1$ many alternations of intersection and union of elements of $\scr A$. See also the proof that Borel sets are explicitly generated by iterating countable union and complement on the open sets $\omega_1$ times.)
