I know that ZF is not finitely axiomatizable, but what about Z (i.e. ZF without Replacement)?
Is Zermelo set theory finitely axiomatizable?
7
$\begingroup$
logic
set-theory
model-theory
1 Answers
9
No. It is not.
You can find the proof as Theorem 8 in:
Mathias A. R. The Strength of Mac Lane Set Theory, Annals of Pure and Applied Logic, 110 (2001) 107--234.
(The article also appears on Mathias' homepage without the need for a paywall)
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0Thank you very much. In the article that you quote Mathias writes (above Theorem 8) that this result has been known since work of Wang and others in the fifties. I did not find anything relevant in the references. Do you know if there is any previous proof of this fact? – 2012-12-02
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0I have no idea really. I have to admit this is the first time I saw this result explicitly... In the commentary file Mathias writes: "Theorem 8 gives a new, set-theoretical, proof of a result that proof-theorists, but not set-theorists, would regard as standard". It might have been known to proof theorists as a folklore (or a common exercise), but I can't really tell. – 2012-12-02
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0I know. It seems like it is a wll known fact but I have searched through all standard set theory (e.g. Jech, Kunen) and model theory books (e.g. Chang & Keisler, Marker), and I can't find anything that is relevant. It would be interesting if some old logician could show us a way to prove this through model theoretic methods. – 2012-12-02
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0If I were to look for this I would look for proof-theoretic papers dealing with finitely axiomatizable theories, and showing why ZF is not one of them. Just following my nose I would guess that Shoenfield's book about proof theory would be a good place to start. – 2012-12-02
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1@Sumac - see Hao Wang, [A Survey of mathematical logic](https://books.google.it/books), Chapter XVII: Relative Strength and Reducibility, page 438. – 2017-05-10