Finding a PA cut in a nonstandard model of PA 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 $\neg\text{Con}(\PA_k)$.
 A: 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 


*

*$M|_a=K|_a$,

*$M|_{2^a}\subseteq K$,

*$K\models\exists d<2^{a^c}Pr_{{\bf I}\Sigma_r}(d,\ulcorner\bot\urcorner)$,

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


Proof. See this paper.

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.

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:


*

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

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

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

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:


*

*$Y(k,(c)_0)=(c)_1$,

*$(c)_1$ is nonstandard,

*$k < (c)_0$,


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


*

*$I\models PA$,

*$k\in I$,

*$(c)_0\not \in I$, and hence $c\not\in I$.

