Number of distinct sets in a $\sigma$-algebra 
Let $F = \{E_1, E_2, ..., E_n\}$ be a collection of $n$ subsets of a set $X$, where $n$ is a positive integer. How many distinct sets could there be in $\sigma(F)$? 

Here are my thoughts: the largest that $\sigma(F)$ could be would be the power set of $F$, with $2^n$ elements. 
This doesn't satisfy me (and is probably wrong!) because a) our prof said that the answer is much larger than $2^n$, and b) this doesn't use the hypothesis about the sets $E_i$ in any way.
How can I begin to wrap my head around this?
 A: For each $x\in X$, consider the function $\chi_x\colon \mathcal{P}(X)\to \{0,1\}$ given by the "membership function"; that is, $\chi_x(E) = 1$ if $x\in E$, and $\chi_x(E)=0$ if $x\notin E$.
Define a relation on $X$ by $x\sim y$ if and only if $\chi_x(E_i)=\chi_y(E_i)$ for $i=1,\ldots,n$. It is easy to see that this is an equivalence relation. 
Question. If $x\sim y$, will $\chi_x(E) = \chi_y(E)$ for all $E\in\sigma(F)$?
To see that the answer is "yes", let $M(x,y)\subseteq \mathcal{P}(X)$ be the set of all subsets of $X$ on which $\chi_x$ and $\chi_y$ agree. Clearly, $F\subseteq M(x,y)$. Prove that $M(x,y)$ is a $\sigma$-algebra. 
Then show that we can consider the question in $X/\sim$ instead of $X$ (intuitively: if the sets in $F$ cannot distinguish between $x$ and $y$, then neither can the sets in $\sigma(F)$, so you may as well consider $x$ and $y$ "the same point"). This set has at most $2^n$ elements (the possible values of the membership function on $F$). What does that tell you about $\sigma(F)$? Since it is a subset of $\mathcal{P}(X)$, it has at most $2^{2^n}$ elements.
To verify that this is indeed the best you can say, try to construct a set $X$ with $2^n$ elements, and an $F$ with $n$ such that $\sigma(F)$ is all of $\mathcal{P}(X)$. Hint: Take $X=\{0,1,\ldots,2^{n}-1\}$, and think binary expansion.
