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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...


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).

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    A comment on Question 1: If $\kappa_1<\kappa_2$ are uncountable cardinals, $X$ is a set with cardinality $\kappa_2$, and $\Sigma$ is the set of subsets $S$ of $X$ such that $S$ or $X\setminus S$ has cardinality at most $\kappa_1$, then can $\Sigma$ be the $\sigma$-algebra generated by a topology?2011-12-07
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    @Jonas: Can’t you simply take $\tau$ to be $\{\varnothing\}\cup\{V\subseteq X:|X\setminus V|\le\kappa_1\}$?2011-12-07
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    @Brian: Yes, thanks.2011-12-07
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    OK, elephant in the room: assuming choice, the Lebesgue sigma-algebra on $\mathbb{R}$ is probably not a Borel sigma-algebra. Of course this makes problem 1 into one that depends on your set theory since word on the street is there exist models of ZF in which all subsets of $\mathbb{R}$ are measureable.2011-12-07
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    @Mike. I would agree with you, but I don't see how to prove, assuming the AC, that the Lebesgue sigma-algebra over $\mathbb{R}$ is not generated by a topology. Have you any reference for this?2011-12-07
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    @Mike, I posted an answer arguing that the Lebesgue algebra is not a counterexample.2011-12-07
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    A counterexample to question 1 is $\prod_{i\in I}X_i$ where $I$ is an uncountable set and $X_i$ are non-trivial measurable spaces. This set has size at least $2^{\vert I\vert}$. Under ZFC, it is consistent that this has cardinality the size of the continuum, so there can exist such a sigma-algebra on the reals. In fact, I think you might always be able to restrict this to give an example of such a sigma-algebra on the reals.2011-12-07
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    George, can you post a proof that your counterexample is a counterexample? I don't quite see it yet...2011-12-08
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    Since I am fascinated by question 1, but it has not found an answer here after some time, I have taken the liberty of posting a question about it on mathoverflow: http://mathoverflow.net/questions/87838/is-every-sigma-algebra-the-borel-algebra-of-a-topology2012-02-07

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