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Let $F$ be a countable family of subsets of $X$. Let $\sigma(F)$ be the $\sigma$-algebra generated by $F$. Let $\tau(F)$ be the topology generated by $F$.

The collection of finite intersections of sets from $F$ is countable, so:

Question 1. If I'm not mistaken, the topology $\tau(F)$ is second-countable.

For a second-countable space, arbitrary unions of open sets can be replaced by countable unions of open sets, so:

Question 2. If I'm not mistaken, $\sigma(F)$ is the Borel $\sigma$-algebra of $\tau(F)$, i.e. $\sigma(F)=\sigma(\tau(F))$.

In conclusion, a $\sigma$-algebra is countably generated if and only if it is the Borel $\sigma$-algebra of a second-countable space.

For a $\sigma$-algebra $\Sigma$, let's say that $\mathcal{T}$ is the finest topology that generates $\Sigma$, if $\Sigma=\sigma(\mathcal{T})$ and if $\Sigma=\sigma(\mathcal{T'})$ implies $\mathcal{T'}\subset \mathcal{T}$ for any topology $\mathcal{T'}$.

Question 3. Let $\Sigma$ be a countably generated $\sigma$-algebra. Is there always a finest topology that generates $\Sigma$?

Due to the answer to a related question asking for the coarsest topology, I fear that the answer will be no. I guess a construction where one adds an arbitrary "single point" as open set to the finest topology will show that such a finest topology does not exist in general.

In the comment to an answer about topological groups, I wondered how to "prevent" the use of the axiom of choice. I want things to be countable, without loosing too much generality.

Question 4. Let $\Sigma$ be a countably generated $\sigma$-algebra. Will any topology $\mathcal{T}$ with $\mathcal{T}\subset \Sigma$ be second-countable?

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Thomas Klimpel
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1 Answers1

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For Q1 and Q2: Yes, you are right. Actually, every countably generated $\sigma$-algebra is induced by a real valued function, so it comes from a separable pseudometrizable space.

For Q3: Yes, the Borel $\sigma$-algebra on $\mathbb{R}$ is a counterexample. You can add a countable number of points to any topology generating the usual Borel sets without changing the generated $\sigma$-algebra.

For Q4: The $\sigma$-algbera consisting of countable sets and sets with countable complements on $\mathbb{R}$ is a sub-$\sigma$-algebra of the Borel $\sigma$-algebra and is generated by the cofinite topology, which is not second countable. Since the countable-cocountable $\sigma$-algebra is also not countably generated, it doesn't come from any second countable topological space.

Michael Greinecker
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  • The cofinite topology is a good counter example. I wonder: If I take the cofinite topology on a Cantor set and add these sets as "open sets" to the usual topology on $\mathbb R$, will the (generated) topology no longer be second-countable, but still generate the "normal" Borel $\sigma$-algebra on $\mathbb R$? – Thomas Klimpel Feb 18 '12 at 21:33
  • The topology will still be second countable. Every finite set is closed in the usual topology, so every cofinite set is open in the usual topology. So you don't get a larger topology by cofinite sets. The point is that aa topology that is coarser than a second countable topology may fail to be second countable. Similarly, a sub $\sigma$-algebra of a countably generated $\sigma$-algebra may not be countably generated. – Michael Greinecker Feb 18 '12 at 21:51
  • You're right, the topology will still be second countable. (But it is larger, because it now contains a Cantor set as an open set.) My point was to have a non-second countable topology generate a countably generated $\sigma$-algebra. – Thomas Klimpel Feb 19 '12 at 09:49
  • You are right, the new topology amounts to adding the Cantor set as an open set. – Michael Greinecker Feb 19 '12 at 10:58