Given Any Measurable Space, Is There Always a Topological Space Generating it?

[linulysses]

As well known, for any topological space (X,T), there is a smallest measurable space (X,M) such that T\subset M. We say that (X,M) is generated by (X,T). Right now, I was wondering whether the "reverse" is true: for any measurable space (X,M), there exists a finest topological space (X,T) such that the smallest measurable space generated by (X,T) is exactly (X,M).

 

[Hurkyl]

Does T=M not work?

 

[micromass]

Not really, since M is not always a topology. For example, the Borel subsets of $$\mathbb{R}$$is not a topology, since it contains the singletons, thus the topology would have to be discrete.

I have no counterexample, but I guess the Lebesgue sigma-algebra could be one. I don't immediately see a topology that generates it...

 

[linulysses]

Not really, since M is not always a topology. For example, the Borel subsets of $$\mathbb{R}$$is not a topology, since it contains the singletons, thus the topology would have to be discrete.

I have no counterexample, but I guess the Lebesgue sigma-algebra could be one. I don't immediately see a topology that generates it...

I would guess it is true and have a candidate proof. Let S be the collection of all topological spaces (X,T') such that T' \subset M. Then S is nonempty and partially ordered. If we could prove that every chain in S has an upper bound in S, then we can apply Zorn's lemma to assert the existence of the maximal (X,T). Next, we can show that (X,T) generates (X,M). Unfortunately, I have no idea about proving the essential statement that "every chain in S has an upper bound in S".

Any idea? or that statement is actually false?

 

[micromass]

Well, we all know what the upper bound should be: the topology generated by

$$\bigcup_{i\in I}{\mathcal{T}_i}$$

But I don't think it's necessarily true that this is a subset of M. Furthermore, it's not even obvious to me that the maximal element should generate the sigma-algebra, but I think that would be true: if T does not generate M, then there is a set A which is not generated by T, then we should look at $\mathcal{T}\cup\{A\}$...

I'm checking out books of descriptive set theory and boolean algebras, perhaps I'll find something useful in there...

 

[linulysses]

Well, we all know what the upper bound should be: the topology generated by

$$\bigcup_{i\in I}{\mathcal{T}_i}$$

But I don't think it's necessarily true that this is a subset of M. Furthermore, it's not even obvious to me that the maximal element should generate the sigma-algebra, but I think that would be true: if T does not generate M, then there is a set A which is not generated by T, then we should look at $\mathcal{T}\cup\{A\}$...

I'm checking out books of descriptive set theory and boolean algebras, perhaps I'll find something useful in there...

Yes, exactly, to prove that $$\bigcup_{i\in I}{\mathcal{T}_i}$$ is a subset of M or to find out a counterexample puzzles me...

 

[Hurkyl]

Not really, since M is not always a topology. For example, the Borel subsets of $$\mathbb{R}$$is not a topology, since it contains the singletons, thus the topology would have to be discrete.

Yah, you're right. That'll teach me to make a quick reply!

 

[micromass]

Hmm, an interesting result is Maharam's theorem: http://en.wikipedia.org/wiki/Maharam's_theorem

I think this implies that we only need to show that the Lebesgue algebra on [0,1] is generated by a topology...

 

[micromass]

Here's another interesting thing I've found

Let X be a topological space satisfying Baire's theorem and let M be the set of meager Borel sets (which is a sigma-complete ideal of Bor(X)), then Bor(X)/M=RO(X) (=the regular open subsets).

So, this would give an indication of how to find the topology that generates the Lebesgue algebra: take all the meager Lebesgue sets, and find representatives of Leb(X)/M. The topology generates by these representatives should generate the Lebesgue algebra.

So, I kind of think your question is true. And using this and my previous post could provide a method to prove this.

More information can be found in "Handbook of Boolean algebra's" by Monk & Bonnet in p182...

 

[linulysses]

Here's another interesting thing I've found

Let X be a topological space satisfying Baire's theorem and let M be the set of meager Borel sets (which is a sigma-complete ideal of Bor(X)), then Bor(X)/M=RO(X) (=the regular open subsets).

So, this would give an indication of how to find the topology that generates the Lebesgue algebra: take all the meager Lebesgue sets, and find representatives of Leb(X)/M. The topology generates by these representatives should generate the Lebesgue algebra.

So, I kind of think your question is true. And using this and my previous post could provide a method to prove this.

More information can be found in "Handbook of Boolean algebra's" by Monk & Bonnet in p182...

Well, definitely I need some time to consume those you provided. I will get back to you if I make breakthrough on this direction. Thanks a lot~