A finitely generated $\sigma$-algebra is finite. I am trying to solve Problem 3.5 (ii) of Schilling book on measure theory for which it must be proven that a finitely generated $\sigma$-algebra is finite.
I can show it, given the following two statements, which I have problems proving.

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*Let $\{B_1,...,B_N\}$ be all the atoms of $\sigma$-algebra $\mathcal A$ of a set $X$ (where atom means non-empty set that contains no other non-empty set in $\mathcal A$). Then any set in $\mathcal A$ can be written as a (countable or finite) union of the $B_j$. In particular $X = \bigcup B_j$.

Idea: It is easy to show that the $B_j$ are disjoint. Otherwise $\mathcal A \ni B_j \cap B_k \subset B_j$ proper for some $k \neq j$. However I can not show $X = \bigcup B_j$. Clearly this is equivalent to $C := \bigcap B_j^c = \emptyset$. My idea was to assume $C$ non-empty. This implies that $C$ must contain a proper non-empty subset $\mathcal A \ni C_1 \subset C$, since otherwise $C$ would be another atom, contrary to hypothesis. The same argument can be applied to $C_1$ and so on, giving an infinite sequence of non-empty sets $(C_j)_{j \in \mathbb N} \subset \mathcal A$ such that $C \supset C_1 \supset C_2 \supset ...$ with all inclusions proper. Then $\bigcap C_j \in  \mathcal A$ must be an atom? But we may have $\bigcap C_j = \emptyset$ (right?) so this does not appear to give the desired contradiction.


*$\sigma(\{ A_1,...,A_N \})$ has at most $2^N$ atoms.

Idea: By induction on $N$. $N=1$ is obvious. Assume the statement is true $N$ and let $\{ B_1, ..., B_M\}$, $M \leq 2^N$, be the atoms of $\sigma(\{ A_1,...,A_N \})$. Then I presume that $C_j := B_j \cap A_{N+1}$ and $D_j := B_j/C_j$ are all the atoms of $\sigma(\{ A_1,...,A_{N+1} \})$ (eventually dropping some $\emptyset$'s) and since $\# (\{ C_j \} \cup \{D_j \}) \leq 2^{N+1}$ this proves the theorem. However I do not see how to show that $\{ C_j \} \cup \{D_j \}$ are indeed all the atoms of $\sigma(\{ A_1,...,A_{N+1} \})$.
This is probably very easy, but I am completely new to the subject. Also, I know that there exists solutions to this, but they do not explain the above steps.
 A: Let $\Sigma$ be the $\sigma$-algebra generated by sets of $A_1,...,A_n$. Then every element $X\in\Sigma$ is obtained by countably many complements, intersections and unions. The de Morgan laws imply that we can take all complements, add them to our collection: $A_1,...,A_n, A_1',...,A_n'$ and then take all unions and then all intersections of them. In every such countable union, only $2n$ terms are different, so there are only finitely many unions. Add them to the collection: $A_1,...,A_n, A_1',...A_n', U_1,...,U_k$. Now the number of countable intersections of sets from this collection is also finite. Hence $\Sigma$ is finite.
A: Let $A_1,\dots, A_n$ be subsets of universal set $X$ and let $\mathcal A$ denote the algebra generated by these sets.
Now let $\mathcal B$ denote the collection of sets of form $B_1\cap\cdots\cap B_n$ where $B_i\in\{ A_i, A_i^c\}$ for $i=1,\dots,n$.
Evidently $\mathcal B\subseteq\mathcal A$ and the elements of $\mathcal B$ are mutually disjoint and cover $X$.
The collection contains at most $2^n$ elements (some might be empty).
Now let $\mathcal C$ be the collection characterized by: $$C\in\mathcal C\iff C\text{ is a union of elements of }\mathcal B$$
Here also the empty union is accepted so that $\varnothing\in\mathcal C$.
Then $\mathcal C$ is a finite collection (having at most $2^{2^n}$ elements) and again $\mathcal C\subseteq \mathcal A$. But also $\mathcal C$ is evidently an algebra.
So we are allowed to conclude that $\mathcal A=\mathcal C$ which means that $\mathcal A$ is a finite collection.
Finally every finite algebra is automatically a $\sigma$-algebra so that $\mathcal A$ can also be classified as the $\sigma$-algebra generated by $A_1,\dots,A_n$.
