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Let $(X,T)$ be a topological space. The claim is that $E\in\mathscr{B}(X):=\sigma(T)$ if and only if $E$ can be obtained by countable union/intersection/complement of open sets in $X$.

For the reverse direction, if $E$ is a countable union/intersection/complement of open sets in $X$, then $E\in S_\alpha$ for each $\sigma$-algebra $S_\alpha$ such that $T\subseteq S_\alpha$, so $E\in\mathscr{B}(X)$.

But I'm not sure how to prove the other direction. If $E\in\mathscr{B}(X)$, then $E\in S_\alpha$ for every $\sigma$-algebra $S_\alpha$ such that $T\subseteq S_\alpha$. Why should $E$ be a countable union/intersection/complement of open sets?

Sid Caroline
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  • It isn't. Not every Borel set is the countable union, etc. of open sets. It's defined as the smallest collection of sets containing the open sets which is closed under these operations. – Chris May 02 '18 at 22:45
  • @Chris Can you give an example of a Borel set that is not a countable union/intersection/complement of open sets, say in $\mathbb{R}$? – Sid Caroline May 02 '18 at 23:29
  • See https://math.stackexchange.com/questions/73296/example-of-a-borel-set-that-is-neither-f-sigma-nor-g-delta – Chris May 02 '18 at 23:34

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The set of all rational numbers is not a countable union/intersection of complement of open sets. The fact that it is not a countable intersection of open sets is not elementary, but it is a standard application of Baire Category Theorem. Of course this set is a Borel set because it is a countable union of singleton sets.

Kavi Rama Murthy
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