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](https://math.stackexchange.com/a/4203688/28111) 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. $\mathcal{L}\_{\omega\_1,\omega}$).



Detecting uncountable cardinals in $(\mathbb{R};+,\times,\mathbb{N})$
For the first question (distinct regular cardinals $>\aleph\_1$): Force ZFC + MA + $2^{\aleph\_0}=\aleph\_3$ over $L$ in the usual way (see Jech, Theorem 16.13; note the forcing is ccc and it forces MA + $2^{\aleph\_0}=\aleph\_3$, which is all we need here). Then in $L[G]$, $\aleph\_2$ and $\aleph\_3$ are both $\mathcal{R}\_{\mathbb{N}}$-detectable.


$\aleph\_2$: By MA$\_{\aleph\_1}$, every $\omega\_1$-sequence of reals is coded via almost disjoint forcing with respect to the canonical almost disjoint sequence $\left<A\_\alpha\right>\_{\alpha<\omega\_1}$ in $L$. This a.d. sequence is lightface projective (in the standard codes for countable ordinals), so the relation "$y$ is a real enumerated in the $\omega\_1$-sequence of reals $\vec{z}\_x$ coded by $x$" is lightface definable (over $\mathcal{R}\_{\mathbb{N}}$). So just let the statement $\varphi$ be "$A$ is uncountable and there is no real $x$ such that every element of $A$ is enumerated in $\vec{z}\_x$" (the "uncountable" part is dealt with as in the original post). Then $\varphi$ is true exactly when $A\subseteq\mathbb{R}$ has cardinality $\geq\aleph\_2$.


$\aleph\_3$: (It doesn't seem obvious to me that there is a lightface projective wellorder of $\mathbb{R}$ in $L[G]$, so we seem to need another argument than that in the original post.) Let $A\subseteq\mathbb{R}$ with $A\in L[G]\models$"$A$ has cardinality $\leq\aleph\_2$".
Then we can definably talk about ordered pairs of reals and $A^2$ over $(\mathcal{R}\_{\mathbb{N}},A)$, and we can talk about subsets of $A^2$ coded by reals $x$, again via disjoint forcing, but this time with respect to the set $(A^2)'$,
where the prime ' means that we convert the family $A^2$ into a disjoint family $(A^2)'$ in the usual manner. I.e., although we had a wellordered family $\left<A\_\alpha\right>\_{\alpha<\omega\_1}$ in the previous case, this is not relevant. The almost disjoint forcing for coding a subset of $A^2$ is ccc (in fact $\sigma$-centered), and there is an $\aleph\_2$-sized family of dense sets which ensures that the generic real codes a given set $\subseteq A^2$, so by MA$\_{\aleph\_2}$ we will have a real coding any given $X\subseteq A^2$). Note that there is a wellorder of $A$ in ordertype $\leq\omega\_2$, and this is a subset of $A^2$, so we have a code for it, and moreover, every proper segment of this wellorder has cardinality $\leq\aleph\_1$. Since "$\geq\aleph\_2$" is already known to be detectable, hence so is "$\leq\aleph\_1$", so we can detect whether there is such a wellorder of a given $A$. I.e. let $\psi$ be the statement (in the augmented language with symbol $\dot{A}$) saying "there is a real $x$ which codes a subset $X\subseteq\dot{A}^2$ with respect to the family $(\dot{A}^2)'$,
$X$ is a wellorder of $\dot{A}$, every proper segment of $X$ has cardinality $\leq\aleph\_1$". Note that given any $A\subseteq\mathbb{R}$ in $L[G]$, we have $(\mathcal{R}\_{\mathbb{N}},A)\models\psi$ iff $A$ has cardinality $\leq\aleph\_2$ in $L[G]$. Therefore $\aleph\_3$ is also $\mathcal{R}\_{\mathbb{N}}$-delectable.




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Edit: For the second question: Proceed as above but forcing MA + $2^{\aleph\_0}=\aleph\_{\omega+1}$. Then all cardinals $\kappa\leq\aleph\_{\omega+1}$ are $\mathcal{R}\_{\mathbb{N}}$-detectable in $L[G]$. For $\aleph\_n$ where $n<\omega$ this is basically as above. However, the complexity of the formulas used for the $\aleph\_n$'s seems to increase with $n$, when done just as above, so this doesn't seem to immediately yield $\aleph\_{\omega}$. Instead we can use a slight variant. We first observe that "$\leq\aleph\_\omega$" is detectable: Note that $A\subset\mathbb{R}$ has cardinality $\leq\aleph\_{\omega}$ iff there is a wellorder of $A$ in order type $\leq\omega\_{\omega}$, and any such wellorder will be coded by a real (via a.d. forcing as before). We can assert that the wellorder $<^\*$ has ordertype $\leq\omega\_{\omega}$ by saying that there is a sequence $\left<x\_n\right>\_{n<\omega}\subseteq A$ which is cofinal in $<^\*$ and such that $x\_0$ has only countably many predecessors and for each $n<\omega$ and each $y\in A$ with $x\_n<^\*y<^\*x\_{n+1}$, the set of predecessors of $y$ and the set of predecessors of $x\_n$ have the same cardinality, as witnessed by a bijection coded by some real.


It follows that "$\geq\aleph\_{\omega+1}$" is detectable. To get "$\geq\aleph\_\omega$", note that $A$ has card $\geq\aleph\_\omega$ iff $A$ has card $\geq\aleph\_{\omega+1}$ or there is a wellorder of $A$ exactly in ordertype $\omega\_\omega$, and the latter condition can be expressed as above, together with the extra requirement that there is no real coding a bijection between the predecessors of $x\_n$ and those of $x\_{n+1}$, for each $n$.


