As a general comment on this kind of construction, I might suggest reading this answer.
Generally speaking, what you describe is the "top-down" approach to the construction. It is not, generally, practical to actually know the description of the elements in the generated object. For that, you want the "bottoms-up" construction. In the question linked-to above, Asaf Kargila provides the description of the bottoms-up construction.
In your particular case, of course, since $E$ is small (3 elements), so $P(E)$ is small (8 elements), the number of possible $\sigma$-algebras is somewhat manageable (though still large).
What are all $\sigma$-algebras on $E=\{1,2,3\}$ that contain $\{1\}$ as an element? They must contain $\emptyset$, $\{1\}$, $\{2,3\}$, and $E$. There are 4 other elements in $P(E)$ which may or may not be in a $\sigma$-algebra.
- One $\sigma$-algebra is just $\{\emptyset, \{1\}, \{2,3\}, E\}$.
- If the $\sigma$-algebra contains either $\{2\}$ or $\{3\}$ in addition to those, then it must also contain the other (symmetric difference with $\{2,3\}$), and hence be all of $P(E)$.
- If the $\sigma$-algebra contains any other $2$-element subset, then it must contain another singleton (symmetric difference with $\{2,3\}$), hence must be all of $P(E)$.
So in fact the only $\sigma$-algebras on $E$ that contain $\{1\}$ are $\{\emptyset, \{1\},\{2,3\}, E\}$ and $P(E)$. The intersection is just $\{\emptyset,\{1\},\{2,3\},E\}$, so the $\sigma$-algebra generated by $\{1\}$ is $\{\emptyset, \{1\}, \{2,3\}, E\}$.
Recall the definition of a $\sigma$-algebra: it is a collection of subsets of some particular $X$, such that:
- It is closed under countable unions;
- It is closed under taking complement (relative to $X$).
From this, countable intersections follow by DeMorgan's Laws.
Now suppose that $X_i\in\mathcal F$ for $i\in\mathbb N$. So for all $\alpha\in A$ we have $X_i\in\mathcal F_\alpha$. Since $\mathcal F_\alpha$ is a $\sigma$-algebra we have that $\bigcap X_i\in\mathcal F_\alpha$ for all $\alpha\in A$ and therefore $\bigcap X_i\in\bigcap\mathcal F_\alpha=\mathcal F$.
For complements, the principle is the same.
Furthermore, the same idea works to show that $\mathcal F$ contains $\mathcal C$. Lastly we need to show that this is indeed the smallest:
Suppose that $\mathcal S$ is a $\sigma$-algebra which contains $\mathcal C$, then for some $\alpha\in A$ we have $\mathcal S=\mathcal F_\alpha$, so it took part in the intersection which generated $\mathcal F$, therefore $\mathcal F\subseteq\mathcal S$.
Further Reading:
- The $\sigma$-algebra of subsets of $X$ generated by a set $\mathcal{A}$ is the smallest sigma algebra including $\mathcal{A}$
Best Answer