Exclude the "closure under complementary" axiom from $\sigma$-algebra

Question

I hope this is not answered already on stackexchange because I cannot quite search it using minimal number of words, but I have heard of the Borel Hierarchy and suspected that the axiom "closure under complementary " is why the construction of minimal $\sigma$-algebra generated from a family of sets is quite complicated and not in finite number of steps.

My question is: if we remove the axiom "closure under complementary", say a $\sigma$-lattice on a set $X$ is a collection of subsets of $X$ closed under countable union and countable intersection, then is it true that the smallest $\sigma$-lattice generated by $\mathcal{A}$ can be constructed by adding countable intersection of sets in $\mathcal{A}$ (say the new family of sets is $\mathcal{A}_1$) and then add countable union of those in $\mathcal{A}_1$?

Answer 1

Let $\mathcal{A}$ be the set of open intervals in $(0,1)$, and let $E$ be the set of irrationals. Then $E$ is in the lattice, since we can write $$E = \bigcap_{q \in \mathbb{Q}} ((0,q) \cup (q,1)).$$ But it can't be written as a countable union of countable intersections. Any intersection of open intervals is either an interval (open, closed, or half-open) or a singleton. $E$ doesn't contain any intervals, and it is not a countable union of singletons.