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Given a set $X$, and $S \subset \mathcal{P}(X)$, we can define $\sigma(S)$ as the intersection of all sigma algebra containing $S$.

I want to have a bottom up approach to construct the generated sigma algebra, and thought of the following: $$ \sigma(S) =\left.\left\{ \bigcup_{i\in I} \bigcap_{j\in J_{i}} T_{i,j}\>\right|\> \text{ countable }, T_{i,j} \in S \text{ or } T_{i,j}^{c} \in S \right\} $$

Intuitively, in this case, I think of $\sigma(S)$ as the union of countably long "words", and each word is the countable intersection of sets that is in $S$ or sets where its complement in $S$.

I planned to ask this a long time ago and read many posts about it, but many bottom up approach seems to be quite complicated. I think I can verify the two axioms in this case? However, maybe I made a mistake as no one proposed anything similar. Also if we can do this, I think there will be a definition of basis and sub-basis as in topology.

Arturo Magidin
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patchouli
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    @geetha290krm I wouldn’t say there is no way to construct $\sigma(S)$ explicitly from $S$ - you could just use a transfinite induction construction up to $\omega_1$, which is what Borel hierarchy is for. – David Gao Jan 22 '24 at 00:08
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    However, while there is a way to describe $\sigma(S)$ in a bottom up way, the OP’s description is incorrect, for example, the collection of sets on the RHS doesn’t have to be closed under taking complements, or taking countable intersections, for that matter. – David Gao Jan 22 '24 at 00:09
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    Let me provide an explicit example showing why the RHS is not closed under taking complements. For example, let $S$ consists of all infinite intervals in $\mathbb{R}$. Note that $S$ is closed under taking complements. Then the RHS contains the set of rational numbers - it is a countable union of singletons, and singletons are intersections of two infinite intervals. But it does not contain the set of irrational numbers... – David Gao Jan 22 '24 at 01:42
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    This answer gives the "bottoms up" construction... but it is not trivial. It requires a transfinite inductive construction. – Arturo Magidin Jan 22 '24 at 01:46
  • This is because any countable intersection of intervals is still an interval, and the only intervals contained in the set of irrational numbers are singletons, but the set of irrational numbers is uncountable, so it is not a countable union of singletons. – David Gao Jan 22 '24 at 01:46
  • @DavidGao Thank you for the example. I made a mistake $\prod_{i\in I}J_{i}$ is not necessarily countable so the complement is not necessarily in the set. – patchouli Jan 22 '24 at 02:02

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