Generated Sigma Algebras

Consider, for example, the real line and the collection of all one-point sets $\{\{x\}: x \in \mathbb R\}$. One sigma algebra containing these sets is the power set. Another one is the family of all countable sets and their complements. There are many others too. Now take the sets that are common to all such sigma algebras. That gives the sigma algebra generated by our family. In this case this turns out to be exactly countable sets and their complements. Reason: Any sigma algebra that contains singletons must contain all countable sets (since they are countable unions of singletons). It must also contain their complements. Hence the smallest sigma algebra containig all singeltons is exactly the family of all countable sets and their complements.

The point is not to take a $\sigma$-algebra $B$ containing $A$ and intersect it with the powerset of $\Omega$. (Note that if $B$ is any collection of subsets of $\Omega$, and $P(\Omega)$ is the powerset of $\Omega$, then $B\cap P(\Omega)=B$. So intersecting with the powerset of $\Omega$ doesn't do much.) The point is that if we have two $\sigma$-algebras $B_1$ and $B_2$ that contain $A$, then $B_1\cap B_2$ is also a $\sigma$-algebra containing $A$ (exercise). Moreover, $B_1\cap B_2$ is going to be smaller than $B_1$ and $B_2$ (unless one of $B_1$ or $B_2$ contains the other).

So $\sigma(A)$ takes this idea to its extreme: we intersect all $\sigma$-algebras containing $A$. In symbols, let $\mathscr{B}$ be the set of $\sigma$-algebras on $\Omega$ that contain $A$. Then $\sigma(A)=\bigcap_{B\in\mathscr{B}}B$. Then $\sigma(A)$ is a $\sigma$-algebra containing $A$ (exercise), and if $B$ is a $\sigma$-algebra containing $A$ then $\sigma(A)\subseteq B$ by definition. So it makes sense to call $\sigma(A)$ the smallest $\sigma$-algebra containing $A$, or the $\sigma$-algebra generated by $A$.

Now you ask where these bigger $\sigma$-algebras come from, and that very much depends on the particular example. In general, the collection $\mathscr{B}$ above could be quite complicated. The most we can say in general is that there is always at least one $\sigma$-algebra in $\mathscr{B}$, namely, the powerset of $\Omega$.

The construction of $\sigma(A)$ described above is good for a definition, but rather difficult to put into practice because it might be difficult or at least very time consuming to calculate $\mathscr{B}$. Given a particular $A$, if one wants to get a more explicit description of $\sigma(A)$ then this usually involves calculating families of sets that must be in any $\sigma$-algebra containing $A$ until you come up with a family that is itself a $\sigma$-algebra. Indeed, if you can come up with a collection $B$ which is a $\sigma$-algebra containing $A$ and must be contained in any $\sigma$-algebra that contains $A$, then it follows that $\sigma(A)=B$.

When $\Omega$ is finite the brute force idea is a little more reasonable because you can just start closing $A$ under intersections and complements until you get an algebra.