$\sigma(\mathcal{A}) = $ the set of countable unions of countable intersections of elements or complements of elements of $\mathcal{A}$ Let $\mathcal{A} \subseteq \mathcal{P}(\Omega)$, $\Omega$ a set.  Then isn't the set of call countable unions of countable intersections of elements or complements of elements of $\mathcal{A}$ equal to $\sigma(\mathcal{A})$?  More precisely, $\sigma(\mathcal{A}) = \{ \cup_{i=1}^{\infty} \cap_{j=1}^{\infty} A_j^i : A = A_j^i$ or $A = A_j^i$ for some $A \in \mathcal{A}$, for all $i,j\}$.  If that's not true, then I'm pretty sure that if you also contain the countable $\cap$ of countable $\cup$ 's, then it is true.
I found it difficult trying to prove that $\cap_{i=1}^{\infty} \cup_{j=1}^{\infty} A_j^i = \cap_{j=1}^{\infty} \cup_{i=1}^{\infty} A_j^{k(i,j)}$, where $k :\mathbb{N}\times\mathbb{N}\rightarrow\mathbb{N}$ is some onto function.  So that's why I've also said it might have to include the $\cap\cup$'s.
What do you think, is it true?
 A: The question automatically came back (since it has no answer).
As Andrés noted, the answer is "no" in general.  To generate a sigma-algebra "from within" is much more complex than this.
Let $\Omega$ be a set, let $\mathcal{A} \subseteq \mathcal{P}(\Omega)$.  Recursively define sets $\mathcal A_\alpha$ as $\alpha$ ranges over all ordinals:
\begin{align}
\mathcal A_0 &:= \mathcal A\\  
\mathcal A_{\alpha+1}&:= \text{countable unions of sets and complements of sets from } \mathcal A_{\alpha}\\
\mathcal A_\lambda &:= \bigcup_{\alpha < \lambda} \mathcal A_\alpha \text{ for limit ordinal }\lambda
\end{align}
Then:
$$
\sigma(\mathcal A) = \mathcal A_{\omega_1}
$$
where $\omega_1$ is the least uncountable ordinal.  And, in general, all the  $\mathcal A_\alpha$ for $\alpha < \omega_1$ are different, and (therefore) are not sigma-algebras.  
In particular, if $\Omega = \mathbb R$ and $$\mathcal A = \{(a,b) \subseteq \mathbb R : -\infty < a < b < +\infty\},$$ then all $\mathcal A_\alpha$ for $\alpha < \omega_1$ are different.
A: Let $\Sigma$ be the described set.  $\Omega$ is in $\Sigma$ since you can construct a trivial $\cup\cap$ consisting of any $A\in\mathcal{A}$ and its complement.  If $A\in\Sigma$, then $A = \cup\cap A_j^i$, then Demorgan's law gives $A = \cap\cup (A_j^i)^c$, which equals a $\cup\cap$ possibly, or if it doesn't then say we're working with the second described set.  Let $(A_n)_{n\geq 1}$ be a sequence in $\Sigma$.  Then each $A_n = \cup\cap A_{n,j}^i$.  A countable union of a countable union is a countable union, so $\cup_{n=1}^{\infty}A_n = $  Oops!
I see now where my mistake is, we can't let the $\cap\cup$'s in or my last statement gets ruined.  But I think $\cup\cap =$ some $\cap\cup$.  If that's true, then this proof will work.
