This statement seems probably true, and for particular examples (e.g. 'all strings of length N in a given alphabet', union over all N) it holds, but I can't find any proof that doesn't rely on the axiom of countable choice, which seems like overkill. Is there a simpler proof?
2026-03-31 23:12:54.1774998774
Countable union of disjoint finite sets is countable
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