Extending a Model of $ T + \operatorname {Con} ( T ) $ to a model of $ T + \neg \operatorname {Con} ( T ) $

Question

Let $ T $ be a recursively axiomatizable extension of $ \mathsf {PA} $ and $ \mathfrak M $ be a model of $ T + \operatorname {Con} ( T ) $. Is it true that there must exist a model $ \mathfrak N $ such that $ \mathfrak M \subset \mathfrak N $ and $ \mathfrak N \models T + \neg \operatorname {Con} ( T ) $?

My intuition is that the answer should be affirmitive. I think since $ \neg \operatorname {Con} ( T ) $ is $ \Sigma _ 1 $, we may be able add a new element to $ | \mathfrak M | $ coding a (nonstandard) proof of contradiction in $ T $. I suppose some sort of argument similar to that of Gödel's incompleteness theorems might suffice for proving the existence of such extension; but I don't know if that's true.

What I tried to do was taking the positive diagram $ \Delta $ of $ \mathfrak M $, and trying to prove that $ T \cup \Delta + \neg \operatorname {Con} ( T ) $ is consistent. As my first attempt, I tried to find theorems from model theory, like Robinson's joint consistency theorem, that would do the job. After failing to find such theorems, I tried to prove the consistency of the mentioned theory via an argument by contradiction: if it's not consistent, finitely many members of $ \Delta $, say $ \delta _ 0 , \dots , \delta _ { n - 1 } $, would suffice to derive a contradiction. Letting $ \delta = \bigwedge _ { i < n } \delta _ i $, we would then have $ T + \delta \vdash \operatorname {Con} ( T ) $. As the parameters from $ \mathfrak M $ do not appear in $ T $ or $ \operatorname {Con} ( T ) $, we can replace the parameters appearing in $ \delta $ with variables and take the existential closure of the resulting formula to get a $ \Sigma _ 1 $ sentence $ \gamma $ in the language of arithmetic such that $ T + \gamma \vdash \operatorname {Con} ( T ) $. Now, since $ \operatorname {Con} ( T ) $ is $ \Pi _ 1 $, I doubt that the existence of such $ \gamma $ would be possible, but I couldn't go further and prove this fact.

Answer 1

The answer is yes.

Suppose $T$ is as above, $\mathfrak{M}\models T$, and $\mathfrak{M}$ has no extension to a model of $T+\neg Con(T)$. Then by compactness, we can extract an existential sentence $\varphi$ such that $T\vdash\varphi\rightarrow Con(T)$ - basically, $\varphi$ corresponds to a finite subset of the atomic diagram of $\mathfrak{M}$ which together with $T$ prevents $\mathfrak{M}$ from being expanded to a model of $\neg Con(T)$.

However, we now can use the internal $\Sigma_1$ completeness of $\mathsf{PA}$ (see here). Consider the theory $S=T+\varphi$. Since $T$, and hence $S$, contains $\mathsf{PA}$, we have that $S$ proves "$T$ proves every true $\Sigma_1$ sentence." In particular, $S$ proves "If $\varphi$ is true then $T$ proves $\varphi$," and so a fortiori "If $T$ is consistent then $T+\varphi$ is consistent." By Godel's second incompleteness theorem, $S$ cannot prove its own consistency, so $S$ cannot prove $T$'s consistency either. But this is exactly saying that $T+\varphi+\neg Con(T)$ has a model, contradicting our assumption on $\varphi$ above.

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Note that this leaves open the situation for extremely weak theories such as $\mathsf{EA}$, in the context of which extensions by true $\Sigma_1$ sentences can be quite odd; see Visser's paper Oracle bites theory.