The title says is all.
To motivate the problem, here is a theorem for finite sets.
Theorem: If S is a finite set, then it can be proved that the atoms of any sigma algebra on S form a partition of S.
I am trying to extend this theorem to a countable set.
- It is easy to show that the atoms must be disjoint - this does not need finiteness.
-The part that does use finiteness is to show that every point is S belongs to some atom. The idea is: If x is any element of S, one can create a nested sequence of proper subsets containing x. Since S is finite, there must be a smallest set in the sequence and that's an atom.
-This part breaks down for countably infinite sets
Thinking further, I have found that you can still extend the theorem if the sigma algebra F, has the following property: Every member of F contains an atom of F
Proof: Since atoms are disjoint, there are only countably many of them. Hence, if you consider the complement of all the atoms, you are still left with a set in F. If this set is nonempty, it must contain an atom. Contradiction !
So, the only way the theorem can fail to extend is if you have a member of F (necessarily infinite) which has no atoms. But I'm not sure if that would be consistent with the requirements of a sigma algebra.
Finding such a set would give you a countably infinite set with a sigma algebra on it without any atoms. Hence my question.