User cosmonut - MathOverflow

Sigma algebra without atoms ?

2010-04-25T03:07:33Z

I'm looking for an example of a set S, and a sigma algebra on it, which has no atoms.

Motivation: It seems to me that a lot of definitions in probability and stochastic processes - conditional probability, filtrations, adapted processes - become a lot simpler if phrased in terms of a sample space partitioned into atoms.

In the book I'm reading, this is done for finite sample spaces. But the only problem I find with extending the definitions to general spaces seems to be that, in general, you might have a sigma algebra without any atoms. (Note: All this definitions can be made without introducing a measure, so objections on grounds of uncountability don't apply.)

Does anyone have an example ?

2) Related question: What if you replace sigma algebra by algebra - closed under finite, rather than countable unions ?

I suspect that the algebra of subsets of R generated by open intervals has no atoms, but can't prove it rigorously.

Existence of probability measure defined on all subsets

2010-03-26T12:51:35Z

Let S be an uncountable set. Does there exist a probability measure which is defined on all subsets of S, with P({x}) = 0 for any element x of S ?

If I remove the condition P({x}) = 0, then I can trivially get a measure defined on all subsets as follows: Fix some a in S. For any subset U of S, define P(U ) = 1 if a is in U and 0 otherwise.

But what happens if I am not allowed to put nonzero probability on individual points ?

Peculiar examples with Axiom of Countable Choice ?

2010-05-02T10:10:42Z

I've been going over the extremely interesting discussions about Axiom of Choice.

It looks to me like all the "weird" consequences of AC (Banach-Tarski etc) come from using it on uncountable collections of sets.

If, instead, we only believe the Axiom of Countable Choice, do we still get unintuitive consequences in the same sense ?

Apologies in advance if the question is vague.

Is there a sigma algebra without atoms on a countably infinite set ?

2010-04-25T04:52:55Z

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.

Question about exotic spheres

2010-03-12T12:28:38Z

From what I understand, an exotic n-sphere is a manifold which is homeomorphic to the n-sphere but not diffeomorphic to it. Now I have read that there are no exotic 2-spheres. But isn't something like a tetrahedron an example of a manifold which is homeomorphic to the sphere, but not diffeomorphic ? (Because of the corners and edges.) What am I missing ?

A trigonometry problem

2010-02-27T08:48:16Z

Let x = pi/(2k+1), for k>0. Prove that
cosxcos2xcos3x...coskx = (1/2)^k

I've confirmed this numerically for n from 1 to 30. I'm finding it surprisingly difficult using standard trig formula manipulation. Even for the case k = 2, I needed to actually work out cosx by other methods to get the result.

Please let me know if you have a neat proof.

Comment by Cosmonut

2010-05-18T05:04:48Z

Hi Gian, do you know who proved this first ? I came up with precisely this proof. It was published in the American Mathematical Monthly, November 2000 issue.

Comment by Cosmonut

2010-04-25T15:53:35Z

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.

Comment by Cosmonut

2010-04-25T05:48:28Z

This thread is about sigma fields, not sigma notation.

Comment by Cosmonut

2010-04-25T05:23:08Z

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.

Comment by Cosmonut

2010-04-25T05:19:44Z

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.

Comment by Cosmonut

2010-04-25T05:01:44Z

Thanks, Joel. Very helpful.

Comment by Cosmonut

2010-03-26T12:57:03Z

Joel, could you please elaborate a little or give me a reference ? I am not very familiar with foundations of mathematics issues. Thanks a lot.

Comment by Cosmonut

2010-03-14T04:30:41Z

Thanks Henrik, that makes a lot of sense ! So, loosely speaking, once you make the tetrahedron a differentiable manifold, you've smoothened away the corners and edges. I see.

Comment by Cosmonut

2010-02-27T12:03:30Z

Thanks Steve. Very neat !