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Let $A_n$ be a sequence of sets having the same cardinality $k$ (possibly uncountable, and may not be equal to the cardinality of the continuum). Also assume they are disjoint. How could I show that $\cup_n A_n$ has cardinality $k$?

Tongou Yang
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  • I assume by the $A_n$ being countable, you specifically mean countably infinite, i.e. $|A_n| = \aleph_0$? Depending on who asks it, sometimes countable can include finite sets, which points to many counterexamples. EDIT: Then again, you might mean there are a countable number of $A_n$, so I might have totally missed the point. – PrincessEev Oct 04 '19 at 23:02
  • @EeveeTrainer Thanks for pointing that out! – Tongou Yang Oct 04 '19 at 23:07
  • See here – Matematleta Oct 05 '19 at 00:11

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