Let \(A_{1}, A_{2}, \ldots, A_{N}\) be non-empty subsets of \(X\).

(i) If the \(A_{n}\) are disjoint and \(\biguplus A_{n}=X\), then \(\# \sigma\left(A_{1}, A_{2}, \ldots, A_{N}ight)=2^{N}\).

Remark. A set \(A\) in a \(\sigma\)-algebra \(\mathscr{A}\) is called an atom, if there is no proper subset \(\emptyset eq B \varsubsetneqq A\) such that \(B \in \mathscr{A}\). In this sense all \(A_{n}\) are atoms.

(ii) Show that \(\sigma\left(A_{1}, A_{2}, \ldots, A_{N}ight)\) consists of finitely many sets.

[show that \(\sigma\left(A_{1}, A_{2}, \ldots, A_{N}ight)\) has only finitely many atoms.]