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.

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