Number of elements in a finite $\sigma$-algebra

Let $\Sigma$ be the $\sigma$-algebra. Choose $x \in X$, and define $M_x = \cap_{M \in \Sigma, x \in M} M$. Clearly $M_x \neq \emptyset$, and $M_x \in \Sigma$.

Furthermore, the collection $F = \{M_x \} \subset \Sigma$ is a partition of $X$ (and finite, of course). To see this, suppose $M_x \cap M_y \neq \emptyset$. Then we must have $M_x = M_y$, or else either $M_x \setminus M_y $ or $M_y \setminus M_x $ would be strictly smaller sets contradicting the definition of either $M_x$ or $M_y$.

Furthermore, it is clear that if $M \in \Sigma$, then $M = \cup_{x \in M} M_x$, hence every element of $\Sigma$ is the (disjoint) union of members of $F$ (the empty set taken as the union of no members of $F$), hence $|\Sigma| = 2^{|F|}$.

Here's a different argument. Suppose $\mathcal{B}$ is a $\sigma$-algebra of subsets of some set $X$, or even just an algebra. Define an addition operation on $\mathcal{B}$ by $A+B=(A\setminus B)\cup(B\setminus A)$ (this operation is also known as "symmetric difference"). Straightforward computations show that this operation is commutative and associative, has the empty set as an identity, and satisfies $A+A=\emptyset$ for all $A\in\mathcal{B}$. This operation thus makes $\mathcal{B}$ an abelian group, and it is in fact a vector space over the field $\mathbb{Z}/2$ since $A+A=\emptyset$ for all $A$.

Now we just use linear algebra. Every vector space over $\mathbb{Z}/2$ has a basis. If $\mathcal{B}$ is finite, the basis is finite, so $\mathcal{B}$ is isomorphic to the vector space $(\mathbb{Z}/2)^n$ for some $n$. In particular, $\mathcal{B}$ has $2^n$ elements.

Note that a finite $\sigma$-algebra $\mathcal{A}$ has minimal nonempty elements. Show that every element of $\mathcal{A}$ is a union of these minimal elements.