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MathWorld says:

Roughly speaking, Borel sets are the sets that can be constructed from open or closed sets by repeatedly taking countable unions and intersections. Formally, the class of Borel sets in Euclidean is the smallest collection of sets that includes the open and closed sets such that if , , , ... are in , then so are , , and , where is a set difference (Croft et al. 1991).

I was wondering, to get a Borel set, can you take countable unions/intersections countably many times? Or does it have to be finitely many times?

Thank you!

badatmath
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1 Answers1

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This follows from the definition of $\sigma$-algebra, and the fact that the Borel sets are the smallest $\sigma$-algebra containing the open sets.

$\sigma$-algebras are closed under countable unions (and intersections). As a consequence, you can take the union/intersection of a countable number of sets, each of which is itself either a countable union or intersection. Note, that since a countable union of countable sets is still countable, you could eliminate the 'intermediate steps' by combining the original sets itself.

Aryabhata
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