Assuming that $\sf ZF$ is consistent, is it the case that for every consistent recursively enumerable theory $\sf T $ that extends the theory $\sf ZF+\neg C $, in the language of set theory, there is a model $(M, \in)$ of $\sf ZFC$ and a set $ M^- \subseteq M $ such that $(M^-,\in^{M^-})$ is a model of $\sf T$?

Where: $\in ^ {M^-}$ is the restriction of $\in^M$ to $M^-$

Also, is the opposite true? That is if we swap $\sf ZF+\neg C$ and $\sf ZFC$ in the above statement, would the result be true?

  • $\begingroup$ Do you require that $M^-$ be an element, not just subset, of $M$? $\endgroup$ Nov 27 '21 at 19:53
  • $\begingroup$ @NoahSchweber, No $M^-$ is just a subset of $M$. $\endgroup$
    – Zuhair
    Nov 27 '21 at 21:04

Any countable model of $\sf ZF$ has a top extension satisfying $V=L$, although it may or may not be well-founded. That is known as the Barwise extension theorem.

So if you at least allow us to focus on countable models, the answer is positive.

In the other direction, well, forcing and symmetric extensions provide you an easy solution.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.