Question

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.

Answer 1

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 (a_ζ: ζ < δ) of sets in A, each containing x, for as long as possible. Since the differences a_ζ \ a_ζ+1 are pairwise disjoint and nonempty, this construction must terminate with δ < κ. Therefore, the intersection a = ∩_ζ<δ a_ζ 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.