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Please see Joel Hamkins answer: https://mathoverflow.net/q/51786.

A section from his argument:

Suppose that M is a model of ZFC. Thus, in particular, ZFC is consistent. If it happens that M is ω-standard, meaning that it has only the standard natural numbers, then M has all the same proofs and axioms in ZFC that we do in the meta-theory, and so M agrees that ZFC is consistent. In this case, by the Completeness theorem applied in M, it follows that there is a model m which M thinks satisfies ZFC, and so it really does.

I don't understand the following two inferences:

1) "then M has all the same proofs and axioms in ZFC that we do in the meta-theory, and so M agrees that ZFC is consistent"

2) "it follows that there is a model m which M thinks satisfies ZFC, and so it really does"

Mainly, for (1) M agrees that ZFC is conistent because its an arithmetical statement and those are absolute for $\omega$-models, so what does this have to do with having "all the same proofs and axioms in ZFC that we do in the meta-theory". And for (2) why does $M \models (m \models \text{ZFC})$ imply $m \models \text{ZFC}$? After all, we didn't assume $M$ was transitive. I only know that the $\models$-relation is absolute for transitive models of $\text{ZF-P}$ (I think?).

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    $\begingroup$ For (2) I think that, since $M$ is $\omega$-standard, what $M$ believes to be $\mathsf{ZFC}$ and the "true" $\mathsf{ZFC}$ of the meta theory agree, so the structure $m$ that $M$ believes to satisfy $\mathsf{ZFC}$ actually satisfies the true $\mathsf{ZFC}$ $\endgroup$
    – Alessandro Codenotti
    Jan 30, 2020 at 14:47
  • $\begingroup$ @AlessandroCodenotti Why would the fact that M and the universe agree on what ZFC is imply that the universe agrees with M on what $\models$ means? $\endgroup$
    – Jori
    Jan 30, 2020 at 14:51

1 Answer 1

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The question is what do you mean by a "model of ZFC". If you mean the axioms as we enumerate them, in the meta-theory, this may or may not be the same thing as those of the universe of sets in which we are looking. The same can be told about the inference rules of FOL.

But since we can code everything into integers, that means that if we have a model whose integers are standard (read: agree with the meta-theory), then these problems go away.

What does that mean? Well, if the model was a model of ZFC, then it proves the completeness theorem, and since the arithmetic statement "ZFC is consistent" is true, that means that we can find a model of ZFC, and that it is the same ZFC as our meta-theory's one.

You might want to argue that the relation of that model might not be "correct" in some way, but we can make this model internally countable and the relation is then coded by a subset of $\omega$. So it's all good.

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  • $\begingroup$ I'm not sure about your last sentence. Do you mean to say that inside M we can code $m \models \text{ZFC}$ as a statement about reals (subsets of $\omega$) because m can be assumed countable (from M)? But then 1) don't we need m to be hereditarily countable? 2) are these statements absolute for $\omega$-models? I am not familiar with coding, except from what I know of en.wikipedia.org/wiki/Code_(set_theory) . $\endgroup$
    – Jori
    Jan 30, 2020 at 15:58
  • $\begingroup$ Inside $M$ we can assume that the model is $\omega$ with some relation on $\omega$. Now, you since the definition of $\models$ is recursive (using the atomic diagram as a parameter), this relation is really modeling ZFC on $\omega$. $\endgroup$
    – Asaf Karagila ♦
    Jan 30, 2020 at 17:12
  • $\begingroup$ (Generally speaking, when you say "coding" in set theory you mean a bijection into some robust object, e.g. a set of ordinals. This means that all the pairs, tuples, etc, are all coded uniformly into a single set of ordinals.) $\endgroup$
    – Asaf Karagila ♦
    Jan 30, 2020 at 17:12
  • $\begingroup$ I've took some time of to digest, things are clearer now, but: from $M \models (m \models \text{ZFC}^M)$ and $\text{ZFC}^M = \text{ZFC}$, why does it follow that $m \models \text{ZFC}$? Because $M$ is not necessarily transitive, we cannot make use of the result that $\models$ is absolute for transitive models of ZF-P, correct? $\endgroup$
    – Jori
    Apr 26, 2020 at 16:37
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    $\begingroup$ If $M$ and $V$ disagree on $\omega$, then $M$ has non-standard integers, and thus non-standard variable indices (think $x_n$ for non-standard $n$), and thus non-standard axioms in $\sf ZFC$. Since the two agree on the axioms, it means that they agree on $\omega$ as well. But now look at the definition of the satisfaction relation, it's a recursive definition on $\omega$, and on formulas. As $M$ and $V$ agree on all of these things, they agree on the satisfaction relation. Note that $(m,\in)$ might not be a model of $\sf ZFC$, but there is some relation $E$ such that $(m,E)\models\sf ZFC$. $\endgroup$
    – Asaf Karagila ♦
    Apr 26, 2020 at 16:51

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