Let X be a set and S a sigma-algebra on X. Let us name borelian sigma-algebra on X a sigma-algebra that is generated by a topology T on X. Given that it is possible for a set X that some sigma-algebras on X are not borelian, do we know a specific property of S separating the borelian and non-borelian sigma-algebras on X ? Gérard Lang
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$\begingroup$ If the $\sigma$-algebra $\mathcal S$ is countably-generated then it is generated by some topology: just take a countable generating subfamily $\mathcal B$ and consider the topology $\tau$ having the family $\{U,X\setminus U:U\in \mathcal B\}$ as a subbase. This topology is zero-dimensional and second-countable and hence hereditarily Linedelof. Consequently all open sets will be in the $\sigma$-algebra $\mathcal S$ and $\mathcal S$ will be the $\sigma$-algebra of Borel sets of the zero-dimensional second-countable topological space $(X,\tau)$. $\endgroup$– Taras BanakhCommented Oct 9, 2019 at 5:06
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$\begingroup$ If the countably-generated $\sigma$-algebra $\mathcal S$ separated points of $X$, then the zero-dimensional second-countable topological space $(X,\tau)$ from the above comment is metrizable and separable. $\endgroup$– Taras BanakhCommented Oct 9, 2019 at 5:08
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$\begingroup$ The product $\sigma$-algebra on $2^\mathbb{R}$ where $2$ has its discrete $\sigma$-algebra is a $\sigma$-algebra that cannot be obtained as the Borel $\sigma$-algebra of any topology on the underyling set: mathoverflow.net/q/87838/61785 $\endgroup$– Robert FurberCommented Oct 9, 2019 at 22:12
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