Talk:Algebra of sets

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"A simple inductive procedure allows to "construct" $\mathcal{A}$ as follows. $\mathcal{A}_0$ consists of all elements of $\mathcal{B}$ and their complements. For any $n\in\mathbb N\setminus \{0\}$ we define $\mathcal{A}_n$ as the collection of those sets which are finite unions or finite intersections of elements of $\mathcal{A}_{n-1}$. Then $\mathcal{A}=\bigcup_{n\in\mathbb N} \mathcal{A}_n$."

— Really, you do not need infinitely many steps; rather, on the first step take finite intersections, on the second step – finite unions, and you are done. --Boris Tsirelson 17:14, 31 July 2012 (CEST)

Absolutely right, I'll fix it Camillo 18:58, 31 July 2012 (CEST)
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