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In Memoirs of a Proof Theorist, Gaisi Takeuti relates how Gödel taught him about nonstandard models in an "interesting" way:

It went as follows. Let T be a theory with a nonstandard model. By virtue of his Incompleteness Theorem, the consistency proof of T cannot be carried out within T. Consequently, T and the proposition "T is inconsistent" is consistent. There is, therefore, a natural number N which is the Gödel number of a proof leading to a contradiction from T. Such a number is obviously an infinite natural number.

Can someone elaborate on the final sentence of this quote?

(Also, in what sense is this teaching nonstandard models if it assumes them from the start?)

Asaf Karagila
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logically challenged
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    We need the Completeness Theorem (there is a model). In such a model, there is an $N$. This $N$ cannot be identified with an ordinary natural number, else we could ungodelize. – André Nicolas Jul 01 '14 at 16:30

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About the final sentence: If the Goedel number, in some model, of a proof of a contradiction were a standard natural number, then it would encode an actual proof of a contradiction. But there is no actual proof of a contradiction in $T$ because, by hypothesis, $T$ has a model.

About assuming that $T$ has a nonstandard model: That assumption could be replaced with "$T$ has a model" and the argument would still work. So I conjecture that "nonstandard" in the first sentence was just a mistake.

This argument applies only to theories $T$ that are recursively (or at least definably-in-$T$) axiomatizable, because one needs to apply the incompleteness theorem.

Finally, in nonstandard analysis, one usually uses "nonstandard model" to mean an elementary extension of the standard model. That isn't what this argument produces, though. It produces a model satisfying "$T$ is inconsistent" whereas the standard model satisfies "$T$ is consistent."

Andreas Blass
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