Every finite $\sigma$-algebra is of the form...? Let $\mathcal{F}$ be a finite $\sigma$-algebra.
The problem asks to show there exists a partition $G = \{ G_1,\dots,G_n \}$ of $\Omega$ such that for all $A \in \mathcal{F}$, $A$ is the union of all or some $G_i$:
$$A = \bigcup_{i \in I} G_i$$
The existence of a partition is immediate from the definition of a $\sigma$-algebra, but I'm not sure how to use the fact $\mathcal{F}$ is finite to construct the generating set $G$. Specifically, to show that every member of $\mathcal{F}$ can be generated from a single partition using only union.
Can someone give me a hint? 
Not a homework problem, I am working through a textbook for self-study.
 A: Let $\mathcal F =\{B_1,B_2,\cdots ,B_n\}$. Consider all sets of the type $A_1\cap A_2\cap \cdots \cap A_n$ where $A_i$ is either $B_i$ or $B_i^{c}$ for each $i$. You get $2^{n}$ such intersection but many of them be just empty. It is now routine to check that the nonempty sets in this collection form a partition of $\Omega$ and that each $B_i$ is a union of some sets in this partition. For example, you can write $B_1$ as  a union of such sets by fixing $A_1$ to be $B_1$ and varying $A_2,A_3,\cdots,A_n$.
A: Here is another way:
Note that $\mathcal F$ is a partially ordered set with respect to set inclusion. Let's call a set $A\in\mathcal F$ minimal if $B\subseteq A$ implies $B\in\{\emptyset,A\}$ for all $B\in\mathcal F$. In plain English: A set $A\in\mathcal F$ is minimal if the only sets smaller than or equal to $A$ are the empty set and $A$ itself.
We can now show that the set of all minimal sets in $\mathcal F$ are a partition of $\Omega$, and every element of $\mathcal F$ is a union of minimal sets.
Minimal sets form a partition:
Here we need to show that the minimal sets are pairwise disjoint and that every element of $\Omega$ is contained in at least one minimal subset.
The minimal sets are disjoint because if we take two elements $A,B\in\mathbb F$, then $A\cap B\in\mathcal F$, since $\mathcal F$ is a $\sigma$-algebra. So $A\cap B\subseteq A,B$, meaning that $A$ and $B$ can only be minimal if $A\cap B=\emptyset$ or $A\cap B=A$ and $A\cap B=B$. So either $A=B$ or $A\cap B=\emptyset$.
And every element $x\in\Omega$ is contained in at least one minimal set because the intersection
$$A_x:=\bigcap_{A\in\mathcal F\\x\in A}A$$
is in $\mathcal F$, contains $x$, and is minimal. It is in $\mathcal F$ because the intersection is finite and $\sigma$-algebras are stable even under countable intersections. It contains $x$ for obvious reasons. And it is minimal because if $B\subseteq A_x$, then either $x\in B$, in which case $A_x\subseteq B$, making $B=A_x$. Or $x\not\in B$, in which case $x\in A_x\backslash B$, and since $\sigma$-algebras are stable under set differences, this makes $A_x\backslash B$ an element of $\mathcal F$ containing $x$, so $A_x\subseteq A_x\backslash B$. And since $B\subseteq A$, this makes $B=\emptyset$.
Every element of $\mathcal F$ is a union of minimal sets:
Let $B\in\mathcal F$. We have that
$$B=\bigcup_{x\in B}A_x.$$
The union is finite since there are only finitely many elements of $\mathcal F$, so the minimal sets are also finitely many. The inclusion $\subseteq$ is obvious. And the inclusion $\supseteq$ is true because every element $x\in B$ is contained in the set $B\cap A_x,$ which is not empty and contained in $A_x$, and thus equal to $A_x$, since $A_x$ is minimal. So this union is equal to
$$\bigcup_{x\in B}B\cap A_x=B\cap\bigcup_{x\in B}A_x,$$
which is obviously a subset of $B$.
As a side note, I don't see where it is necessary to use finiteness of $\mathcal F$. The critical parts are where unions and intersections of possibly many sets are concerned, but it is sufficient if these are countable unions and intersections, no finiteness needed.
