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Assume ZF. A more general version of this question is the following. Given the three statements

  1. Axiom of countable choice
  2. Every infinite set has a countably infinite subset
  3. Infinite implies Dedekind-infinite

we know that 1 implies 2 and 2 implies 3. However can anyone prove either reverse implication (in ZF), or else has anyone heard of a model where 3 holds but but not 2, or where 2 holds but not 1?

Noah Schweber
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Alex Eustis
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    Statements 2 and 3 are the same because Dedekind-infinite = has a countably infinite subset. – bof Oct 25 '21 at 06:06
  • What is your definition of an infinite set? – Erik D Oct 25 '21 at 06:15
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    @ErikD The correct definition of infinite set in set theory is "a set not in bijection with any finite ordinal." When working under the axiom of choice we can fudge this, but in $\mathsf{ZF}$ alone this is the definition of "infinite" that is used. – Noah Schweber Oct 25 '21 at 06:17
  • The wikipedia entry for Dedekind-infinite set claims that "the equivalence of the two definitions [infinite respectively Dedekind-infinite] is strictly weaker than the axiom of countable choice (CC)", but no specific reference is given. – Erik D Oct 25 '21 at 06:28
  • That infinite sets are Dedekind infinite does not imply countable AC. This is problem 8.3.4 in Jech's AC book. He attributes this result to Pincus, who proved it in his dissertation. – Jason Zesheng Chen Oct 25 '21 at 07:04

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