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. $\mathcal{L}_{\omega_1,\omega}$).

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    $\begingroup$ Can there be a projective well-ordering but not one of length $\mathfrak c$? $\endgroup$ Jul 27, 2021 at 5:08
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    $\begingroup$ Sure, I am just wondering about this side remark. $\endgroup$ Jul 27, 2021 at 5:14
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    $\begingroup$ @AndrésE.Caicedo Sorry, I misread your comment (which led to a couple off-topic now-deleted comments of my own). That's a neat question - I've played with similar questions on the computability-theoretic side of things, but that's a natural one I haven't thought of at all. My instinct is that this should be possible - that is, that it is consistent with $\mathsf{ZFC}$ that there is a (genuine) well-ordering $W$ of $\mathbb{R}$ which is projective but no projective well-ordering of ordertype exactly $\mathfrak{c}$ (so the ordertype of $W$ is $>\mathfrak{c}$) - but this is really pure guesswork. $\endgroup$ Jul 27, 2021 at 5:21
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    $\begingroup$ @AndrésE.Caicedo If you force CH by $\text{Col}(\omega_1,\mathfrak{c})$, the projective wellorders of $\mathbb R$ in the extension are the same as those in ground model (since you haven't changed $\mathbb R$). So if CH failed in the ground model but the reals had a projective wellorder (consistent by Harrington's "Long Projective Wellorders"), then in the extension, there is a projective wellorder of $\mathbb R$ but none of length $\omega_1$. $\endgroup$ Jul 27, 2021 at 6:34
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    $\begingroup$ @NoahSchweber Do you know a consistency proof for $\aleph_2$ being undetectable (with $\aleph_2\leq 2^{\aleph_0}$)? $\endgroup$
    – Farmer S
    Jul 30, 2021 at 9:12

2 Answers 2


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.

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 $n$.

  • $\begingroup$ Fantastic, thanks! $\endgroup$ Jul 30, 2021 at 12:42
  • $\begingroup$ Is it obvious what happens under $\mathsf{ZFC+MA+2^{\aleph_0}=\aleph_{\omega_1+1}}$? $\endgroup$ Jul 31, 2021 at 8:53
  • $\begingroup$ Well "$\leq\aleph_{\omega_1}$" is detectable much like "$\leq\aleph_{\omega}$ was in the second argument above, but also using a coded subset of size $\aleph_1$ in place of the $\omega$-sequence there. We then also get that "$\geq\aleph_{\omega_1+1}$" and "$\geq\aleph_{\omega_1}$" are detectable much as there. But since there are only countably many formulas available, only countably many cardinals are detectable. The least undetectable is a limit cardinal. I don't see whether the set of detectables $<\aleph_{\omega_1}$ is transitive. $\endgroup$
    – Farmer S
    Aug 1, 2021 at 16:39
  • $\begingroup$ Yeah, I was thinking specifically about what happens below $\aleph_{\omega_1}$ in that situation - I should have specified. The transitivity question is a good one, that hadn't occurred to me. $\endgroup$ Aug 1, 2021 at 16:41
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    $\begingroup$ How crucial is it that we start with $L$? For example, is it consistent with $\mathsf{ZFC+MA_{\aleph_1}}$ that $\aleph_2$ is not detectable? My instinct is that first adding $\aleph_2$-many Cohens (per Harry West's answer) and then forcing $\mathsf{MA}_{\aleph_1}$ appropriately over that will result in $\aleph_2$ being undetectable, but I don't immediately see the details. $\endgroup$ Aug 2, 2021 at 6:04

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 $p\not\Vdash \neg T.$

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    $\begingroup$ Great, doubts dispelled! In the definition of $p$ (from $p'$), should it be restricting to $\kappa\backslash(U\cup V)$ instead of $F$? $\endgroup$
    – Farmer S
    Aug 2, 2021 at 19:39
  • $\begingroup$ Thanks, this helps round things out! I wonder what happens with other forcing notions ... $\endgroup$ Aug 2, 2021 at 19:40
  • $\begingroup$ @FarmerS: yes, thank you $\endgroup$
    – Harry West
    Aug 3, 2021 at 5:01

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