# Detecting uncountable cardinals in $(\mathbb{R};+,\times,\mathbb{N})$

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

• Can there be a projective well-ordering but not one of length $\mathfrak c$? Jul 27, 2021 at 5:08
• @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. Jul 27, 2021 at 5:21
• @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 $\omega_1$. Jul 27, 2021 at 6:34
• @NoahSchweber Do you know a consistency proof for $\aleph_2$ being undetectable (with $\aleph_2\leq 2^{\aleph_0}$)? Jul 30, 2021 at 9:12

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 $$\left_{\alpha<\omega_1}$$ 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 $$\left_{\alpha<\omega_1}$$ 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 $$\left_{n<\omega}\subseteq A$$ 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 $$n$$.

• Fantastic, thanks! Jul 30, 2021 at 12:42
• Is it obvious what happens under $\mathsf{ZFC+MA+2^{\aleph_0}=\aleph_{\omega_1+1}}$? Jul 31, 2021 at 8:53
• 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. Aug 1, 2021 at 16:39
• 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. Aug 1, 2021 at 16:41
• 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. 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 $$p\not\Vdash \neg T.$$

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