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I know that ZF is not finitely axiomatizable, but what about Z (i.e. ZF without Replacement)?

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1 Answer 1

up vote 9 down vote accepted

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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Thank 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? –  Sumac Dec 2 '12 at 16:41
    
I 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. –  Asaf Karagila Dec 2 '12 at 16:46
    
I 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. –  Sumac Dec 2 '12 at 16:57
    
If 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. –  Asaf Karagila Dec 2 '12 at 17:02

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