Does visible nonstandardness imply visible ill-foundedness?

Question

For $X\subseteq\mathfrak{M}\models \mathsf{TA}$, say that $X$ is $\mathfrak{M}$-disruptive iff there is some formula $\varphi$ in the language of arithmetic + a new unary predicate symbol $U$ such that

• the expansion of $\mathfrak{M}$ by interpreting $U$ as $X$ satisfies $\varphi$, but

• there is no expansion of the standard model $\mathbb{N}$ which satisfies $\varphi$.

Basically, disruptive sets witness $\mathbb{N}\not\preccurlyeq_{\Pi^1_1}\mathfrak{M}$ (meanwhile we're assuming $\mathbb{N}\preccurlyeq\mathfrak{M}$). I'm curious whether proper cuts are (essentially) the only way for a set to be disruptive:

Suppose $X$ is $\mathfrak{M}$-disruptive. Must there be a proper definable cut in the expansion $(\mathfrak{M};X)$?

Answer 1

Assume for contradiction that the answer is positive. Let $\def\ind{\mathrm{IND}_{L(U)}}\ind\def\N{\mathbb N}\DeclareMathOperator\th{Th}\def\fM{\mathfrak M}$ denote full induction in the language of arithmetic with a new predicate $U$. I claim that for any $L(U)$ formula $\phi(U)$, we have $$\N\models\forall X\,\phi(X)\iff\th(\N)+\ind\vdash\phi(U).\tag1$$ The right-to-left implication is obvious, as the theory $\th(\N)+\ind$ is sound. Conversely, if $\th(\N)+\ind\nvdash\phi(U)$, there exists a model $(\fM,X)\models\th(\N)+\ind+\neg\phi(U)$. Since $(\fM,X)$, being a model of $\ind$, has no definable proper cut, the assumption ensures $X$ is not $\fM$-disruptive. Thus, $\N$ expands to a model of $\neg\phi(U)$, i.e., $\forall X\,\phi(X)$ is false.

But $(1)$ is impossible, as it reduces $\Pi^1_1$ truth to $\varnothing^{(\omega+1)}$.