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This claim seems intuitively true to me, but I'm having trouble coming up with a rigorous proof. My first idea was to construct two subsets whose union is the entire set, but I'm having trouble proving that two such subsets must exist.

It could be true that this claim is false, but so far I haven't been able to find any counterexamples.

aadithyaa
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  • If X is infinite, it contains a copy of the natural numbers, but this needs the axiom of choice. Are you admitting that? – Cranium Clamp Mar 20 '22 at 01:56
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    I suspect that the axiom of choice will help you show that any infinite set $X$ has a countably infinite subset. Just choose an element of $X$, then another, and so on – Henry Mar 20 '22 at 01:56
  • You need the axiom of choice for this. Without the axiom of choice, there can exist not finite sets all of whose subsets are finite or cofinite. – TomKern Mar 20 '22 at 03:04
  • Btw, a set whose countable subsets are all finite is commonly called a Dedekind-finite set. – Jonathan Schilhan Mar 20 '22 at 10:18
  • Thank you all for your comments! I learned a lot about set theory that I didn't know before. It's surprising to me that something that seems so intuitive can't just be proven in ZF. – aadithyaa Mar 20 '22 at 16:55

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As the comments mention, it depends on which axioms of set theory are being used. Without a sufficiently strong form of Choice Principle, it need not be true. In particular, if we assume that every infinite set has a countably-infinite subset (that is, that every infinite set is Dedekind-infinite), then it is immediately true. However, if our set theory allows for infinite, Dedekind-finite sets (infinite sets without any countably-infinite subsets), then such sets will have the property of being infinite, while all of their countable subsets are finite.

Cameron Buie
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