Lemma 4.1.4.label Let $X$ be a set and $\cali{A}\suf 2^{X}$ be an algebra, then the following are equivalent:
- (1)
For any $\curl{A_n }_{n\in\N}\suf \cali{A},\cups{}{n\in\N}A_{n}\in\cali{A}$.
- (2)
For any $\curl{A_n }_{n\in\N}\suf \cali{A}$ with $A_{n}\suf A_{n+1}$ for all $n\in\N,\cups{}{n\in\N}A_{n}\in\cali{A}$.
- (3)
For any $\curl{A_n }_{n\in\N}\suf \cali{A}$ pairwise disjoint, $\dcups{}{n\in\N}A_{n}\in\cali{A}$.
Proof. (2)$\implies$(1): For each $n\in\N$ let $B_{n}=\cups{}{k\leq n}A_{k}\in\cali{A}$ then $\cups{}{n\in\N}A_{n}=\cups{}{n\in\N}B_{n}\in\cali{A}$.
(3)$\implies$(2): Let $A_{0}=\emp$ and $B_{n}=A_{n}\del A_{n-1}$ for all $n\in\N$ then $\curl{B_n}_{n\in\N}\suf\cali{A}$ is pairwise disjoint and $\cups{}{n\in\N}A_{n}=\dcups{}{n\in\N}B_{n}\in\cali{A}$.$\square$
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