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What are two definitions of an object or class of objects, which are equivalent assuming the axiom of choice, but are inequivalent assuming only ZF set theory? I am not looking for theorems that are inequivalent, more like interesting examples of definitions.

user107952
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  • yes for example dedekind finite sets ( doesn't have proper subset of same cardinality) correspond to sets that have finite cardinality (are in bijection with a finite cardinal) in ZFC but not in ZF – MIO Apr 07 '21 at 22:32
  • "vector space" and "free module over a field" are equivalent with choice, but aren't without – Hagen von Eitzen Apr 07 '21 at 22:34
  • @Arthur my bad I meant the opposite – MIO Apr 07 '21 at 22:34

1 Answers1

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Theorems tend to translate into definitions.

As people mentioned in the comments, in $\sf ZFC$ we can prove the theorem that a set is infinite if and only if it contains a countably infinite subset. In $\sf ZF$ this isn't true, so it becomes a type of definition.

In $\sf ZFC$ a ring is Noetherian if one of the two equivalent definitions hold:

  1. Every collection of ideals contains a maximal element;
  2. every increasing chain of ideals is finite.

In $\sf ZF$ the equivalent fails, and they turn into two separate definitions.

This extends to every theorem of $\sf ZFC$ that fails in $\sf ZF$.

Asaf Karagila
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  • Wait, can you give me examples of rings in ZF which differ among those properties? Also, in ZF does 1. imply 2., or vice versa? – user107952 Apr 08 '21 at 01:54
  • Yes, 1 implies 2. As for examples, Hodges has a nice paper "Six Impossible Rings" where he has an example (as far as "has an example" to a counterexample to AC can go). – Asaf Karagila Apr 08 '21 at 08:39
  • If you want something that is inequivalent: (I) $X$ can be linearly ordered without a maximum element; (II) $X$ can be partitioned into infinitely many infinite sets. In ZFC those are the same, but in ZF neither one implies the other.. – Asaf Karagila Apr 08 '21 at 08:43