If $A,B \in \mathcal{S}$ implies that $A \cup B \in \mathcal{S}$, is $\mathcal{S}$ closed over countable, or merely finite union?

general-topology elementary-set-theory

Not at all. Consider, for example, the set of finite subsets of $\Bbb{N}$. Clearly, the union of two finite sets is finite, but the countable union $$\bigcup_{n \in \Bbb{N}} \{n\}$$ is infinite.

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