TITLE: Detecting uncountable cardinals in $(\mathbb{R};+,\times,\mathbb{N})$
QUESTION [24 upvotes]: For a structure $\mathcal{X}=(X;...)$, say that a cardinal $\kappa$ is $\mathcal{X}$-detectable iff there is some sentence $\varphi$ in the language of $\mathcal{X}$ together with a fresh unary predicate symbol $U$ such that for all $A\subseteq X$, the expansion of $\mathcal{X}$ gotten by interpreting $U$ as $A\subseteq X$ satisfies $\varphi$ iff $\vert A\vert\ge\kappa$.
For example, $\omega_1$ is $(\omega_1;<)$-detectable since the uncountable subsets of $\omega_1$ are exactly the unbounded ones. By contrast, Alex Kruckman observed that by a result of Robinson no uncountable cardinal is $\mathcal{R}=(\mathbb{R};+,\times)$-detectable.
I'm interested in the expansion $\mathcal{R}_\mathbb{N}:=(\mathbb{R};+,\times,\mathbb{N})$ of $\mathcal{R}$ gotten by adding a predicate naming the natural numbers (equivalently, adding all projective functions and relations). Since we can talk about one real enumerating a list of other reals, $\omega_1$ is $\mathcal{R}_\mathbb{N}$-detectable ("there is no real enumerating all elements of $U$"). More pathologically, if $\mathfrak{c}=2^\omega$ is regular and there is a projective well-ordering of the continuum of length $\mathfrak{c}$ then $\mathfrak{c}$ is $\mathcal{R}_\mathbb{N}$-detectable. So for example it is consistent with $\mathsf{ZFC}$ that $\omega_2$ is $\mathcal{R}_\mathbb{N}$-detectable.
I'm curious whether this type of situation is the only way to get $\mathcal{R}_\mathbb{N}$-detectability past $\omega_1$. There are multiple ways to make this precise, of course. At present the following two seem most natural to me:

Is it consistent with $\mathsf{ZFC}$ that there are at least two distinct regular cardinals $>\omega_1$ which are $\mathcal{R}_\mathbb{N}$-detectable?

Is it consistent with $\mathsf{ZFC}$ that there is a singular cardinal which is $\mathcal{R}_\mathbb{N}$-detectable?


Note that an affirmative answer to either question requires a large continuum, namely $\ge\omega_3$ and $\ge\omega_{\omega+1}$ respectively. Although my primary interest is in first-order definability, I'd also be interested in answers for other logics which aren't too powerful (e.g. $\mathcal{L}_{\omega_1,\omega}$).

REPLY [6 votes]: Farmer S asked in the comments about the consistency of $\aleph_2$ being undetectable. I hope to dispel any doubt: adding enough Cohen reals does work. Specifically, adding $\kappa$ Cohen reals will ensure that no cardinal $\aleph_1<\lambda\leq \kappa$ is $\mathcal R_{\mathbb N}$-detectable, at least.
Let $\dot{G}$ be the canonical name for a function $\kappa\to\mathbb R$ enumerating the new generic reals. A name $\dot{r},$ thought of as a name for a real, is "supported by" a set $I\subseteq\kappa$ if the Boolean value of each statement "$\check{i}\in \dot{r}$" ($i\in\omega$) is invariant under permutations that fix every element of $I.$ Every name has a countable support. Countability comes from the fact that these Boolean values can be represented by a regular open set in $2^{\kappa\times\omega},$ which is the union of some countable maximal antichain of basic opens, and the indices used in the basic opens constitute a support.
For the purposes of induction I'll use a relative kind of indetectability.
Fix disjoint sets $F,U,V\subseteq \kappa$ with $U,V$ uncountable, and names $\dot{r_1},\dots,\dot{r_k}$ of reals supported by $\kappa\setminus(U\cup V).$ Fix also a formula $\phi$ in prenex normal form in the language of $\mathcal R_{\mathbb N}$ with a unary predicate, translated to a statement of set theory by restricting quantifiers to the reals. The unary predicate is written as the first parameter. I will argue that
$$\Vdash S\iff T\tag{*}$$
with
$$S=\phi(\dot{G}[F\cup U],\dot{r_1},\dots,\dot{r_k}),$$
$$T=\phi(\dot{G}[F\cup V],\dot{r_1},\dots,\dot{r_k}).$$
To get indetectability of $\lambda$ plug in $F=\emptyset$ and $k=0,$ with $U$ a subset of $\kappa$ of cardinality $\lambda,$ and $V$ a disjoint subset of $\kappa$ of cardinality $\aleph_1.$ I think it might not be too hard to extend this to any formula $\phi$ in $L(\mathbb R)^{M[G]},$ and analogous models using HOD or symmetric extensions.
Suppose (*) does not hold. Negate $\phi$ if necessary so that the first quantifier, if any, is existential. Swapping $(S,U)$ and $(T,V)$ if necessary we can assume there is a condition $p’$ such that $p’\Vdash S\wedge \neg T.$ Let $p$ be the condition obtained by ignoring indices in $U\cup V,$ i.e. $p=p’\setminus((U\cup V)\times\omega\times 2).$
The following standard argument shows that $p\Vdash  S\wedge \neg T.$ Permutations of $\kappa$ act on forcing conditions and on names by their action on the Boolean algebra of regular opens of $2^{\kappa\times \omega}.$ For any $q\leq p$ we can find a permutation $\pi$ of $\kappa$ fixing $\kappa\setminus(U\cup V)$ elementwise and fixing $U$ and $V$ setwise, so that no index $i\in U\cup V$ is used by both $q$ and by $\pi p’.$ This means that the conditions $q$ and $\pi p’$ are compatible. Applying symmetry we have $\pi p’\Vdash S\wedge \neg T$ and hence $q \cup \pi p’\Vdash S\wedge \neg T,$ and since $q$ was arbitrary we have $p\Vdash  S\wedge \neg T.$
If $\phi$ is quantifier-free then (*) is easy to see. Otherwise we arranged that $\phi$ is of the form $(\exists x\in\mathbb R)\psi.$ Then there exists a name $\dot{x}$ for a real, and a condition $q\leq p,$ such that
$$q\Vdash \psi(\dot{G}[F\cup U],\dot{r_1},\dots,\dot{r_k},\dot{x}).$$
Partition $U$ as $U_1\cup U’$ with $U_1$ countable, and similarly $V=V_1\cup V’,$ such that  $q$ and $\dot{x}$ are supported by $\kappa\setminus(U’\cup V’)$ (extending the definition of "supported by" to conditions). By induction on quantifiers we can apply (*) to $\psi$ to get
$$q\Vdash \psi(\dot{G}[F\cup U_1\cup V’],\dot{r_1},\dots,\dot{r_k}, \dot{x}).$$
Pick a permutation $\pi$ fixing each element of $\kappa\setminus(U\cup V)$ but with $\pi[U_1\cup V’]=V.$ Then
$$\pi q\Vdash \psi(\dot{G}[F\cup V],\dot{r_1},\dots,\dot{r_k}, \pi\dot{x}).$$
The condition $\pi q$ and name $\pi \dot{x}$ therefore witnesses $p\not\Vdash \neg T.$