algebras have closure under finite disjoint countable unions

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Lemma

Let $A$ be an algebra and let ${E_{n}}$ be a countable collection of elements in $A$ . Then there exists ${F_{n}}$ a disjoint countable collection of elements in $A$ such that

⋃_{n} E_{n} = ⋃_{n} F_{n}

ie we can get closure under countable unions when we have closure under countable disjoint unions

Proof

Let $G_{n} = ⋃_{k = 1}^{n} E_{k}$ so that $G_{1} \subset G_{2} \subset G_{3} \subset \dots$ and $⋃_{n} G_{n} = ⋃_{n} E_{n}$ . Now, let $F_{1} = G_{1}$ and for all $k$ define

F_{k + 1} = G_{k + 1} ∖ G_{k}

Then we also have that the disjoint $⋃_{n} F_{n} = ⋃_{n} G_{n} = ⋃_{n} E_{n}$ .

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algebras have closure under finite disjoint countable unions

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