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On the wikipedia page of the axiom of infinity it says that we can’t prove the axiom of infinity from the other axioms because ZFC implies cons(ZFC-infinity) and then use Gödel’s second incompleteness theorem.

Why doesn’t this strategy work for the axiom of choice?

Gödel showed you can’t disprove AC using ZF, so (assuming ZF is consistent) ZFC is consistent. However ZFC implies cons(ZF), so if you could prove AC from ZF we’d have ZFC implies cons(ZFC), which contradicts the second incompleteness theorem.

I know this isn’t a proper proof of the independence of AC, but I don’t see where the error is.

H. de Gracht
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    Are you sure ZFC proves Con(ZF)? – Patrick Stevens Apr 22 '22 at 15:06
  • The axiom of choice is independent of ZF. That means : If ZF is consistent, then it remains consistent no matter if you add the axiom that the axiom of choice is true or the axiom that the axiom of choice is false. – Peter Apr 22 '22 at 15:08
  • @PatrickStevens I think so, removing an axiom from a consistent system can never make it inconsistent. – H. de Gracht Apr 22 '22 at 15:11
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    There is a difference between $Consistency(ZFC)\implies Consistency(ZF)$ and $ZFC\implies Consistency(ZF)$. – Michael Apr 22 '22 at 15:12
  • @Michael Wikipedia (not the best source, I know) says that ZFC implies cons(ZFC-infinity), I don’t understand why choice would be different than infinity. – H. de Gracht Apr 22 '22 at 15:15
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    "we can’t prove the axiom of infinity from the other axioms because ZFC implies cons(ZFC-infinity) and then use Gödel’s second incompleteness theorem." . This is incorrect. Even if we assume that ZFC is consistent and it follows that ZFC-infinity is consistent, this does not contradict Goedel's second incompleteness theorem because ZFC-infinity is not the same as ZFC , it is a weaker system and ZFC might be able to prove its consistency. – Peter Apr 22 '22 at 15:24
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    There's an easy model of ZFC without Infinity: the hereditarily finite sets. ZFC can express this model and can show that it satisfies all the axioms of ZFC without Infinity. By contrast, there's no model of ZFC without Choice which is "sufficiently nice" to do this. (None of the things I just said are obviously true; they all require proof.) See e.g. https://math.stackexchange.com/a/351580/259262. – Patrick Stevens Apr 22 '22 at 15:24
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    @Peter it’s a proof that infinity can’t be proven using ZFC-infinity, so ZFC implies cons(ZFC-infinity), but if infinity was a theorem in ZFC-infinity, this would mean ZFC implies cons(ZFC) which contradicts Gödel’s second incompleteness theorem. – H. de Gracht Apr 22 '22 at 15:45
  • @PatrickStevens oh, that makes sense. Thx for the answer. – H. de Gracht Apr 22 '22 at 15:46
  • @PatrickStevens : In rethinking my original interpretation that $ZF$ is distinct from $Consistency(ZF)$, I wonder if it is better to assume ZF (or any other axiom scheme) asserts its own consistency. For example it seems reasonable that $ZF\implies Consistency(ZF)$ by the following proof that uses the logical statements (i) $q \implies (p \implies q)$; (ii) $(p \mbox{AND} Not(p) )\implies q$. Define $q=Cons(ZF)$. If $q$ is true then by (i) we are done. Now suppose $Cons(ZF)$ is false. Then ZF contains a contradiction of the type $(p \mbox{ AND} Not(p))$ and so by (ii) we are done. $\Box$ – Michael Apr 22 '22 at 22:45
  • @Peter : I think the answer to my question above is that indeed any axiom scheme asserts its own consistency, but asserting your own consistency is not the same as a proof of consistency, which requires demonstrating a model for which consistency holds. – Michael Apr 22 '22 at 22:55

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