# Is Zermelo set theory finitely axiomatizable?

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

## 1 Answer

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)

• 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? Dec 2, 2012 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. Dec 2, 2012 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. Dec 2, 2012 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. Dec 2, 2012 at 17:02
• @Sumac - see Hao Wang, A Survey of mathematical logic, Chapter XVII: Relative Strength and Reducibility, page 438. May 10, 2017 at 10:20