TITLE: Is there a counterpart of a base of a topology for a $\sigma$-algebra?
QUESTION [7 upvotes]: Update post on Jan 9, 2012:
Given a sigma algebra $\mathcal{F}$ on a set $X$, and a partition $\mathcal{C}$ of $X$. If I am correct, then: 

$\mathcal{C}$ is a generator of $\mathcal{F}$, if and only if any measurable subset is a union of some
  members of $\mathcal{C}$.

Such class of subsets (partition plus the part after "if and only if" characterizes it) to the sigma algebra is like a base to a topology. Allow me to call it the "base" of the sigma algebra.
I wonder if any sigma algebra always has a "base"?
If a sigma algebra has finitely many measurable subsets, then there exists a "base".
If there is a "base", must the sigma algebra has finitely many measurable subsets?
Thanks and regards!

Original post:
A base of a topology is defined as a collection of open sets such that every open set is a union of some of them.
I was wondering if there is a similar concept for a $\sigma$-algebra? My question arose from a notice that a class of subsets that form a partition of the universe seems like a "base" for the $\sigma$-algebra it generates.
Actually I am curious if there is a general concept for a class of subsets closed under some set operation(s).
Thanks and regards!

REPLY [5 votes]: Here are some simple examples which are enough to answer your Jan 9, 2012 questions. Denote by $\mathcal S(X)=\{\{x\};x\in X\}$ the set of singletons of a set $X$.
The power set $2^\mathbb Z=\{A;A\subseteq\mathbb Z\}$ of $\mathbb Z$ is a sigma-algebra on $\mathbb Z$ with $\mathcal S(\mathbb Z)$ as "base". But $2^\mathbb Z$ is neither finite nor countable. In fact, there is no such thing as an infinite countable sigma-algebra.
The Borel sigma-algebra $\mathcal B(\mathbb R)$ has no "base" since any of its bases should contain every singleton, hence the base could only be $\mathcal S(\mathbb R)$, but the subset $\mathbb R_+$ is in $\mathcal B(\mathbb R)$ and is neither countable nor co-countable.