Closed under pairwise unions ⇒ closed under arbitrary unions? – math.stackexchange.com

Closed under pairwise unions ⇒ closed under arbitrary unions? – math.stackexchange.com

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Let $X$ be a set and $S$ be a collection of subsets of $X$, such that given any $U,V\in S$, $U\cup V\in S$. Intuitively it seems like this should imply that arbitrary unions are also in $S$. That is, ...

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