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?
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One quick caveat: the sets themselves need to be countable. – roundsquare Mar 03 '20 at 16:47
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@roundsquare Last time I checked, finite sets are countable. – Brian61354270 Mar 03 '20 at 16:48
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If all the finite sets lie in a linearly ordered set then it's probably doable without choice. – Henno Brandsma Mar 03 '20 at 16:49
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@Brian My mistake. I misread. :( – roundsquare Mar 03 '20 at 16:49