MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In the presence of Godel's Completeness Theorem, the 2nd Incompleteness Theorem has the following strictly model theoretic interpretation: if there exists any model at all of (say) ZFC, there also exists a model M of ZFC such that M contains no (internal) model of ZFC. That is the direct model-theoretic translation of: if ZFC is consistent, then ZFC + (ZFC is inconsistent) is consistent.

My question(s): Either 1) Does this model theoretic statement possibly enjoy a "purely" model-theoretic proof? or 2) Is there a good reason not to expect a purely model-theoretic proof of this statement?

Since category theory affords a rich context for studying internalization, perhaps I should substitute "category-theoretic" for "model-theoretic."

[As a graduate student years ago I wondered whether "every model of ZFC contains an internal model of ZFC" would simply run directly afoul of the Axiom of Foundation until someone refined my understanding of "internal model" and convinced me that the story couldn't be that simple.]

share|cite|improve this question
up vote 7 down vote accepted

Hi David. The following is ("probably", he says) not what you want, but I am leaving it here, as it may explain some of the context. It is an argument of Woodin, similar to one of Jech, which though mathematically alike, is more "proof theoretic" in nature. I wrote some notes on it; they are here.

The key fact you refer to is however not proved model theoretically, but by an appeal to the arithmetic fixed point lemma, also a key fact in Gödel’s approach.

The idea is the following (the note provide the missing details and makes this precise): A property $P$ of models of set theory is hereditary iff whenever $P(M)$ and $N\in M$ is a model of set theory such that $M$ thinks that $P(N)$, then in fact $P(N)$.

Then the following holds:

For any hereditary $P$, either $P(N)$ fails for all $N$, or else there is an $M$ such that $P(M)$ but $P(N)$ fails for all $N\in M$.

This is proved by an application of the fixed point lemma: Take $\phi$ such that (ZFC proves tha) $\phi$ holds iff for all $N$, $P(N)$ implies $N\models\lnot\phi$.

It is immediate that if $P(M)$ holds for some $M$, then it holds for some $M$ such that $M\models\phi$. But then $M$ thinks that there is no $N$ satisfying $P$.

Ok. If $P(N)$ is "$N$ is a model of set theory", then $P$ is hereditary, and we have the second incompleteness theorem. As shown in the note, quite a few similar results follow all at once by considering appropriate properties $P$.

share|cite|improve this answer
Thanks Andres. Read your blog, and now I'm sorry I didn't stick with the program back in 1988. I let experts talk me out of it (but I was certainly missing some critical ideas). – David Feldman Jan 9 '11 at 7:50

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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