Correspondences between Borel algebras and topological spaces

Question

Though tangentially related to another post on MathOverflow (here), the questions below are mainly out of curiosity. They may be very-well known ones with very well-known answers, but...

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Suppose $\Sigma$ is a sigma-algebra over a set, $X$. For any given topology, $\tau$, on $X$ denote by $\mathfrak{B}_X(\tau)$ the Borel algebra over $X$ generated by $\tau$.

Question 1. Does there exist a topology, $\tau$, on $X$ such that $\Sigma = \mathfrak{B}_X(\tau)$?

If the answer to the previous question is affirmative, it makes sense to ask for the following too:

Question 2. Denote by ${\frak{T}}_X(\Sigma)$ the family of all topologies $\tau$ on $X$ such that $\Sigma = \mathfrak{B}_X(\tau)$ and let $\tau_X(\Sigma) := \bigcap_{\tau \in {\frak{T}}_X(\Sigma)} \tau$. Is $\Sigma = \mathfrak{B}_X({\frak{T}}_X(\Sigma))$?

Updates. Q2 was answered in the negative by Mike (here).

Answer 1

I think that I can answer the second question. For each point $p \in \mathbb{R}$, let $\tau_p$ be the topology on $\mathbb{R}$ consisting of $\varnothing$ together with all the standard open neighbourhoods of $p$. Unless I've made some mistake, the Borel sigma-algebra generated by $\tau_p$ is the standard one. However, $\bigcap_{p \in \mathbb{R}} \tau_p$ is the indiscrete topology on $\mathbb{R}$.

Answer 2

This is a great question. I am posting not an answer to the question, but an answer to the comment posted by Mike about a proposed counterexample, the elephant in the room.

Theorem. The elephant in the room is not a counterexample. That is, the $\sigma$-algebra of Lebesgue measurable sets is the Borel algebra of the topology consisting of all sets of the form $O-N$, where $O$ is open in the usual topology and $N$ has measure $0$.

Proof. Note that the empty set and the whole of $\mathbb{R}$ have the desired form. Next, note that sets of that form are closed under finite intersection, since $(O-N)\cap(U-M)=(O\cap U)-(N\cup M)$. Next, I claim that they are closed under countable unions. This is because $\bigcup_i (O_i-N_i) = (\bigcup_i O_i)-N$, where $N$ is a certain subset of $\bigcup_iN_i$, which is still measure $0$ since this is a countable union. Next, note that it is closed under arbitrary unions in the case where the open set is the same, $\bigcup_i (O-N_i)$, since this is just the same as $O-(\bigcap_i N_i)$, which is $O$ minus a smaller null set. Finally, since we have a countable basis for the usual topology, using rational intervals, say, it follows now that my sets are closed under arbitrary unions, since for any sized union, we may rewrite it using basic open sets minus null sets, and then group together all the terms arising for each basic open set as a single term, thereby reducing the entire union to a countable union, which we've argued still has the desired form.

Thus, the sets of the form $O-N$ do indeed form a topology, and this topology is clearly contained within the Lebesgue algebra. Furthermore, the Borel algebra generated by my collection of sets includes all measure zero sets, as well as all open sets, and so it is the same as the algebra of all Lebesgue measurable sets. QED

Answer 3

For Q1. How about this. $X = \{0,1\}^A$ for uncountable $A$, and $\Sigma$ is the product $\sigma$-algebra. So each element of $\Sigma$ depends on only countably many coordinates.

Now we just need a proof that this cannot be the Borel algebra of any topology.

Answer 4

EDIT: It's best just to ignore this nonsensical post.

The product space mentioned in the answer by Gerald Edgar does not provide a counterexample. The product $\sigma$-algebra is determined by countably many coordinate, the product topology by finitely many coordinates. A set determined by countably many coordiantes is a countable intersection of sets determined by finitely many coordinates.

Remark: This answer contained a flawed "proof" that the answer to the first question is yes. I've edited it.