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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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No. Pick a point x in your base set. For every other point y pick a set A(y) containing x but not y, unless that's impossible. Intersect all the sets A(y) you just found. This is an atom containing x. – François G. Dorais Apr 25 '10 at 5:06
Another way to see it is to notice that chains under inclusion are countable, so Zorn's lemma can be applied to yield a minimal element containing x, which is necessarily an atom. – Jonas Meyer Apr 25 '10 at 5:15
I see (smacks self on head). In the finite case, it is possible to get an atom by intersecting all sets containing x. In countable case, this approach doesn't work, since such a collection could be uncountable. But yes,your example gives a countable collection. Thanks a lot. – Cosmonut Apr 25 '10 at 5:19
Jonas, for a countable set, chains under inclusion are not countable. Which is what was tripping me up. As an example, consider the rationals in (0,1), call it Q(0,1) Now look at the chain of sets {x: x \in Q(0,1), x < r} as runs increases from 0 to 1. – Cosmonut Apr 25 '10 at 5:23
Yes, what a silly error! Thanks for the correction and example. – Jonas Meyer Apr 25 '10 at 5:27
up vote 3 down vote accepted

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.

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Hi Francois, I used the method in your comment this morning to prove the following. Let S be a set of cardinality K. Let F be a collection of subsets such that - S is in F - F is closed under complements - F is closed under unions of cardinality less than or equal to k Then every element of x is contained in some atom of F. In fact, the atoms of F form a partition of S. This is somewhat different from your result, which does not depend on cardinality of the underlying set (which is why you have the K-chain conditions), but I thought you may find it interesting. – Cosmonut Apr 25 '10 at 15:53
could you please provide a reference to this theorem of Tarski? Thanks! Mirna – Mirna Dzamonja Sep 17 '15 at 11:08
Hi Mirna, unfortunately I'm drawing a blank on what I was thinking when I wrote this in 2010. I wish I had included the reference in the post. I'm digging through some old stuff in the hope of jolting my memory. I will let you know if I find anything. – François G. Dorais Sep 18 '15 at 1:21

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