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.



Is there a sigma algebra without atoms on a countably infinite set ?
There is an old result of Tarski which says that any algebra A of sets which is 


1. κ-complete (i.e. A is closed under unions and intersections of fewer than κ sets from A), and
2. satisfies the κ-chain condition (i.e. there is no family of κ many pairwise disjoint nonempty sets in A),


then A is necessarily atomic. 


The proof of atomicity is along the lines of what Jonas suggested, but with a wellordered descending chain. Given a0 ∈ A and x ∈ a0. Starting from a0, recursively construct a strictly descending wellordered chain (aζ: ζ < δ) of sets in A, each containing x, for as long as possible. Since the differences aζ \ aζ+1 are pairwise disjoint and nonempty, this construction must terminate with δ < κ. Therefore, the intersection a = ∩ζ<δ aζ is an element of A by κ-completeness and x ∈ a. This intersection must in fact be an atom in A, otherwise we could extend the chain further. It follows that a0 is the union of all atoms below a0 and hence that A is atomic.


In your case, you have an ω1-complete algebra of sets. If the underlying set of the algebra is countable, then there certainly cannot be a family of ω1 many pairwise disjoint nonempty sets in A. Note also that the case κ = ω shows that finite algebras of sets are atomic.


