Number of elements in a finite $\sigma$-algebra I have been asked to prove that the number of elements in a finite sigma algebra over a set $X$ is $2^n$ for some integer $n$. How do I go about this problem? I have no idea where to start. Thanks in advance for any ideas. 

Do I need to prove that given a set $F$, $\sigma(F)$ is actually a power set of some set say $S$?
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 $|\Sigma| = 2^{|F|}$.
I think all the above answers are wrong, since, in adittion to all the unions and intersections, you have to count all the possible differences and symmetric differences as well as their complements. For instance, for n=2, you will have 16 sets instead of 4. I think it should be 2^{2^n} instead of 2^n.