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The title kind of says it all. I've been working through Axiomatic Set Theory, Suppes and Mathematical Logic, Kleene. And I haven't thoroughly studied ordinals and incompleteness yet. But, skimming through it I started thinking about this question.

In suppes, the set of natural numbers, $\omega$ is constructed using the ZFC axioms. However, this seems like it goes against Gödel because ZFC set theory is a consistent theory.

Where is the discrepancy? Is there a limitation in the definition of $\omega$? Suppes later constructs the rational numbers out of the ordinals. Do these rational numbers to satisfy the field axioms?

BENG
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    The incompleteness theorem doesn't say that no consistent theory can prove the consistency of PA; it says that no consistent theory $T$ at least as strong as PA can prove the consistency of $T$. – Malice Vidrine Feb 11 '19 at 00:46
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    How do you know ZFC is a consistent theory? – Derek Elkins left SE Feb 11 '19 at 00:48
  • Oh, I guess it was a false assumption. I remember reading under the section of completeness of predicate logic, that if the set of axioms are simultaneously satisfiable then they are consistent (Kleene 320). I was assuming that that could have been done for ZFC – BENG Feb 11 '19 at 00:52
  • Can ZFC be shown to be consistent computationally through model theory by building a model that satisfies ZFC? – BENG Feb 11 '19 at 00:56
  • @BENG ZFC may well be consistent, but Godel's theorem implies if it is, we cannot prove this in ZFC. This includes the construction of a model of ZFC. Stronger set theories can construct a model of ZFC and thus prove consistency e.g. ZFC plus axiom of inaccessibles. – spaceisdarkgreen Feb 11 '19 at 00:57
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    ZFC proves that PA is consistent. There's no problem with that, the Incompleteness theorem lets us infer that PA does not prove that ZFC is consistent (unless PA was inconsistent to begin with). – Asaf Karagila Feb 11 '19 at 01:09
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    Your last paragraph doesn't make sense to me. Yes they satisfy the field axioms. But I don't see the connection that would imply we shouldn't care about incompleteness. – spaceisdarkgreen Feb 11 '19 at 01:25
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    You haven't explained why you think the construction of the ordinals in ZFC "violates Gödel's incompleteness theorem". – bof Feb 11 '19 at 01:58

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I suspect what you're thinking is something like:

"Godel's incompleteness theorem says that we can't prove that PA is consistent. Yet ZFC lets us produce a model of PA, and so proves that PA is consistent. Isn't this a contradiction?"

The issue here is that when we say

we can't prove that PA is consistent

we have to be very careful what we mean. Here are a couple of true statements:

  • No consistent recursively axiomatizable theory extending (in any of a few different senses) PA can prove its own consistency.

  • If PA proves "PA is consistent," then PA is inconsistent.

  • If ZFC proves that PA is consistent and PA proves that ZFC is consistent, then each is inconsistent.

The takeaway, essentially, should be "any theory proving the consistency of PA must be significantly stronger$^*$ than PA" - but that doesn't mean that no nice theory can prove the consistency of PA! ZFC proves the consistency of PA, is recursively axiomatizable and stronger (in an appropriate sense) than PA, and - as far as we know currently - is consistent (in particular, ZFC doesn't prove its own consistency).

Put another way, here's the right conclusion: since ZFC proves that PA is consistent, ZFC must be significantly more powerful than PA (and indeed, we know lots of ways in which it's immensely more powerful). In particular, when you write

However, this seems like it goes against Gödel because ZFC set theory is a consistent theory

you're taking for granted the consistency of ZFC itself, which is even less trivial than the consistency of PA.

$^*$Making this fully precise is a bit subtle - ignore it for now.

Noah Schweber
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  • Thanks that helps a lot. I see know how incompleteness (ii) works with this question. But I no have some confusion about the first part of incompleteness. (To the best of my knowledge) it says that if a first order Theory $T$ that is (a) algorithmically denumerable (b) allows some of PA and (c) is consistent, then $T$ is not complete. But completeness says any first order theory is complete. Could you make the conclusion through contradiction that a theory $T$ that is both (a) and (b) is inconsistent? – BENG Feb 11 '19 at 06:53
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    @BENG: The completeness theorem states that first order logic is complete, as a logic, not that every theory is complete. That is demonstrably false. – Asaf Karagila Feb 11 '19 at 08:08
  • @AsafKaragila. So there is a theory $T$ that is incomplete such that $T {\vDash} {\phi}$ and not $T {\vdash} {\phi}$? – BENG Feb 11 '19 at 08:55
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    @BENG: That is the completeness of the logic. A theory is complete if for every $\varphi$ either $T\vdash\varphi$ or $T\vdash\lnot\varphi$. It might be a good idea to review all the relevant definitions again, and not work from memory. – Asaf Karagila Feb 11 '19 at 08:58
  • @AsafKaragila: Hm, that definition of completeness of a theory wasn't given in the text I was working out of, (Mathematical Logic, Kleene). I see that the same word is used to speak about two different things. – BENG Feb 11 '19 at 09:02
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    @BENG: Yes, "completeness" is quite ubiquitous in mathematics. Logic can be complete, a theory can be complete, a metric space can be complete... – Asaf Karagila Feb 11 '19 at 09:04
  • @AsafKaragila: Thanks! Could I ask you for a text recommendation? I've worked through pretty much all of Kleene, but there were a lot of things that it left out. What would a good second pass of logic be? – BENG Feb 11 '19 at 09:07