I know that ZF is not finitely axiomatizable, but what about Z (i.e. ZF without Replacement)?
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.
Here's how it works:
- Anybody can ask a question
- Anybody can answer
- The best answers are voted up and rise to the top
No. It is not.
You can find the proof as Theorem 8 in:
(The article also appears on Mathias' homepage without the need for a paywall)