Finding a PA cut in a nonstandard model of PA

Question

For a certain project I am currently working on, I need to be able to find PA cuts in nonstandard models of PA, in desirable intervals. For example, I wonder if the following is true, where $\newcommand\PA{\text{PA}}\PA_k$ refers to the theory with only $\Sigma_k$ induction.

Question. If $M$ is a model of $\PA$ in which $\PA_{k-1}$ is consistent, but $\PA_k$ is not (so $k$ is nonstandard), then is there a $\PA$ cut in $M$ above $k$ in which $\PA_k$ is consistent?

That is, I want to cut $M$ below the first proof of a contradiction in $\PA_k$, but above $k$, and have $\PA+\text{Con}(\PA_k)$.

Alternatively, is there some other $\Sigma_1$ property of $k$, other than $\neg\text{Con}(\PA_k)$, such that I can always find a $\PA$ cut in $M$ between $k$ and the witness of that property? Kameryn Williams suggested that the Paris-Harrington result may provide this, since it is designed to ensure $\PA$ cuts below the corresponding PH-Ramsey number. But I would need, however, that one can always end-extend the model so as to make the $\Sigma_1$ property true. Does the PH construction have both these features?

With the consistency statements, for example, for any nonstandard $k$ in any model $M$ of $\PA$, there is always an end-extension of $M$ to a model of $\PA$ with $\neg\text{Con}(\PA_k)$.

Answer 1

Q1

Theorem1. Let $M$ be a countable nonstandard model of $PA$, let $r,c,a\in M\setminus\mathbb{N}$, $M\models r,c\leq a$, and suppose $M\models Con_{{\bf I}\Sigma_r}$. Then there exists a model $K$ of $PA$ such that $a\in K$ and

1. $M|_a=K|_a$,
2. $M|_{2^a}\subseteq K$,
3. $K\models\exists d<2^{a^c}Pr_{{\bf I}\Sigma_r}(d,\ulcorner\bot\urcorner)$,
4. $K\models Con_{{\bf I}\Sigma_{r-1}}$.

Proof. See this paper.

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The proof of theorem 1 uses finite Godel's theorem and some other provability results:

Theorem2. Let $T\supseteq {\bf I}\Delta_0+\mathsf{EXP}$ be a consistent theory with provability predicate $Pr_T(x,y)$ such that $Pr_T(x,y)\in\mathsf{P}$, then there exists a $\epsilon >0$ such that the length of the shortest $T$-proof of $Con_T(\bar{n})$ defined by $\forall x(|x|\leq \bar{n}\to \neg Pr_T(x,\ulcorner \bot \urcorner))$ is at least $n^\epsilon$.

Proof. See this paper.

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Let $M$ be a countable nonstandard model of $PA+Con_{{\bf I}\Sigma_{r}}$ for some nonstandard $r\in M$, then by theorem 1, there exists a countable nonstandard model $K\models PA$ such that:

1. $M|_{2^{r+1}}\subseteq K$,

2. $K\models\exists d<2^{(r+1)^r}Pr_{{\bf I}\Sigma_r}(d,\ulcorner\bot\urcorner)$,

3. $K\models Con_{{\bf I}\Sigma_{r-1}}$.

This implies that for any $PA$-cut $K'$ in $K$ above $r$, $K|_{2^{(r+1)^r}}\subseteq K'$, hence $K'\models \exists d<2^{(r+1)^r}Pr_{{\bf I}\Sigma_r}(d,\ulcorner\bot\urcorner)$, therfore the answer of the question is no.

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Q2

Let $Y(x,y)$ be the $PA$ indicator defined in theorem 3.23 of metamathematics of first-order arithmetic. Suppose $(Y(x,y)=z) \equiv \exists w \psi(x,y,z,w)$ for some $\psi\in\Delta_0$. Define $\phi(x)$ by $\exists y \left(\psi(x,(y)_0,(y)_1,(y)_2)\land x< (y)_0\land x< (y)_1\right)$. Let $M$ be a nonstandard model of $PA$. Suppose for some nonstandard element $k\in M$, $M\models \phi(k)$. This implies that there exists a least element $c\in M$ such that $M\models \psi(k,(c)_0,(c)_1,(c)_2)\land k< (c)_0\land k< (c)_1$. This implies:

1. $Y(k,(c)_0)=(c)_1$,
2. $(c)_1$ is nonstandard,
3. $k < (c)_0$,

therefore by definition of indicator there exists a cut $I$ in $M$ such that:

1. $I\models PA$,
2. $k\in I$,
3. $(c)_0\not \in I$, and hence $c\not\in I$.