This is what I have to prove: If $A_1, A_2, \ldots , A_N$ partition the set $X$ then cardinality of $\sigma$-algebra generated by $A_1, A_2, \ldots , A_N$ is $2^N$.
I claim that $\sigma \left( \{ A_1, A_2 , \ldots A_N \} \right)$ is a boolean algebra with least element $\emptyset$ and greatest element $X$.
That is, I need to $(\sigma \left( \{ A_1, A_2 , \ldots A_N \} \right), \cup, \cap, \emptyset, X)$ is a complemented distributive lattice.
If $A, B$ are members of sigma algebra then $A\cup B$ and $A\cap B$ are in the sigma algebra by definition. It is clear that $A \lor B=\sup \{ A,B \} = A \cup B$ because $A\subset A\cup B$ and $B\subset A\cup B$ and so $A\cup B$ is an upperbound; and now if $A\subset C$ and $B \subset C$ then $A\cup B\subset C$. Hence, $A\cup B$ is the least upper bound of $\{ A , B \}$. Similarly, we can argue that $A\land B=A\cap B$. So it is indeed an lattice.
Distributivity is clear as $\cup$ distributes over $\cap$ and $\cap$ distributes over $\cup$.
Complemented is clear because if $A$ is in the sigma algebra then so is $A^{c}$.
I claim that all the atoms of this boolean algebra are $A_1, A_2, \ldots , A_N$. Let's first show that $A_i$ is an atom. If we show that there are no element "smaller" than $A_i$, we will be done. (I'm not sure on how to do this but intuitively, the sigma algebra contains all the countable union of all possible countable intersection of the sets of the form $A_i$ or $A_i^{c}=\cup_{j\ne i} A_j$, so it seems to me that there cannot be anything smaller than $A_i$'s).
Now if $A$ is an atom then since $A\cap A_i \ne \emptyset$ for some $i$ because $A_i$'s partition $X$. Since $A$ and $A_i$ are both atoms, we must have $A\cap A_i =A$ and $A\cap A_i =A_i$ and so $A=A_i$.
And now since the boolean algebra is finite because the sigma algebra contains all the countable union of all possible countable intersection of the sets of the form $A_i$ or $A_i^{c}=\cup_{j\ne i} A_j$. We have that any element can be written uniquely as a join of atoms, so, there must be $2^{N}$ elements in the sigma algebra.
Is this proof correct? I would appreciate it if someone could suggest me on how to fill the gaps in my proof.