Proof. Let $\mathcal {N}_0=\left \langle N_0,{+},{\cdot },0,1, <\right \rangle $ be any model of arithmetic that admits a truth predicate (for example, the standard model ${\mathbb N}^M$ arising in any model of set theory M). Let T be the theory consisting of the elementary diagram $\Delta (\mathcal {N}_0)$ of this model, in the language of arithmetic with constants for every element of $N_0$ , together with the assertion “ $\mathrm {Tr}$ is a truth predicate,” which is expressible as a single assertion about $\mathrm {Tr}$ , namely, the assertion that it satisfies the recursive Tarskian truth requirements. The theory T is consistent, because by assumption, $\mathcal {N}_0$ itself admits a truth predicate. Furthermore, any model of the theory T provides an elementary extension $\mathcal {N}$ of $\mathcal {N}_0$ , when reduced to the language of arithmetic, together with a truth predicate for $\mathcal {N}$ . Suppose toward contradiction that every elementary extension of $\mathcal {N}_0$ that admits a truth predicate has a unique such class. It follows that any two models of T with the same reduction to the language of the diagram of $\mathcal {N}_0$ must have the same interpretation for the predicate $\mathrm { Tr}$ . Thus, the predicate $\mathrm {Tr}$ is implicitly definable in T, in the sense of the Beth implicit definability theorem (see [Reference Chang and Keisler1, theorem 2.2.22]), and so by that theorem, the predicate $\mathrm {Tr}$ must be explicitly definable in any model of T by a formula in the base language, the language of arithmetic with constants for elements of $\mathcal {N}_0$ . But this violates Tarski’s theorem on the non-definability of truth, which implies that no model of arithmetic can have a definable truth predicate. Thus, there must be models of T with identical reductions $\mathcal {N}$ to the language of the diagram of $\mathcal {N}_0$ , but different truth predicates on $\mathcal {N}$ . In other words, $\mathcal {N}$ is an elementary extension of $\mathcal {N}_0$ having at least two different incompatible truth predicates. (And it is not difficult by similar reasoning to see that there must be an $\mathcal {N}$ having infinitely many distinct truth predicates.)

In the case where $\mathcal {N}_0$ admits an inductive full satisfaction class, we simply add $\mathrm {PA}(\mathrm {Tr})$ to the theory T, with the result by the same reasoning that the elementary extension $\mathcal {N}$ will also have multiple inductive full satisfaction classes, as desired.

Proof. ( $1\to 2$ ) Suppose that $\mathcal {N}$ is a countable nonstandard ZFC-standard model of arithmetic, arising as $\mathcal {N}={\mathbb N}^M$ for some model of set theory $M\models \mathrm {ZFC}$ , which we may assume is countable. Clearly, $\mathcal {N}$ satisfies the theory $\mathrm {Th}({\mathbb N})^{\mathrm {ZFC}}$ . Note also that $\mathcal {N}$ admits an inductive satisfaction class, namely, the collection of arithmetic truths $\mathrm {TA}^M$ as they are defined inside M. An easy overspill argument shows that any model of PA with an inductive satisfaction class is computably saturated (e.g., [Reference Smoryński28, lemma 1.1]). Namely, suppose that $p(x,\vec n)$ is a computable type in the language of arithmetic with parameters $\vec n\in \mathcal {N}$ and that p is finitely realized in $\mathcal {N}$ . Since the type p is computable, the model $\mathcal {N}$ computes its own version of the type $p^{\mathcal {N}}$ , and this will agree with p on all standard formulas. Furthermore, the structure $\left \langle \mathcal {N},\mathrm {TA}^M\right \rangle $ can see that all the standard-finite initial segments of $p^{\mathcal {N}}$ are satisfiable according to the truth predicate $\mathrm {TA}^M$ , which by induction in the meta-theory agrees with actual truth on the standard-finite length formulas. Thus, by overspill (since we have induction for this predicate), it follows that some nonstandard length initial segment of $p^{\mathcal {N}}$ is satisfied in $\mathcal {N}$ according to the truth predicate, and since that truth predicate agrees with actual satisfaction on standard-finite formulas, it follows in particular that p itself is satisfied in $\mathcal {N}$ . So $\mathcal {N}$ is a countable computably saturated model of $\mathrm {Th}({\mathbb N})^{\mathrm {ZFC}}$ , as desired.

( $2\to 1$ ) Conversely, suppose that $\mathcal {N}$ is a countable computably saturated model of $\mathrm {Th}({\mathbb N})^{\mathrm { ZFC}}$ . Consider the theory T consisting of the ZFC axioms together with the assertions $\sigma ^{\mathbb N}$ for every sentence $\sigma \in \mathrm {Th}(\mathcal {N})$ , which is consistent precisely because $\mathcal {N}\models \mathrm { Th}({\mathbb N})^{\mathrm {ZFC}}$ . Indeed, $\mathcal {N}$ agrees on the consistency of any particular finite subtheory of T, because for any finite fragment $\varphi $ of ZFC and any arithmetic sentence $\sigma $ , we have $\mathrm {ZFC}\vdash \sigma ^{\mathbb N}\mathrel {\rightarrow } \mathop {\mathrm {Con}}(\varphi +\sigma ^{\mathbb N})$ by the reflection theorem, and so this statement is true in $\mathcal {N}$ , along with its antecedent $\sigma $ , and so $\varphi +\sigma ^{\mathbb N}$ is consistent in $\mathcal {N}$ .

By computable saturation, the theory T is coded in $\mathcal {N}$ , since we may write down a computable type for this, and so there is a nonstandard finite theory $t\in \mathcal {N}$ whose standard part is exactly T.

Since as we have mentioned $\mathcal {N}$ thinks the finite subtheories of T are consistent, by overspill we may assume by cutting down to a nonstandard initial segment if necessary that t is consistent in $\mathcal {N}$ . Inside $\mathcal {N}$ , we can now build the canonical complete consistent Henkin theory H extending t, and let $M\models H$ be the corresponding Henkin model. In particular, $M\models \mathrm {ZFC}$ , since this is a part of t, and so ${\mathbb N}^M$ is a ZFC-standard model of arithmetic and hence computably saturated. Note also that ${\mathbb N}^M$ has the same theory as $\mathcal {N}$ , because this is part of t and hence H. The structure $\mathcal {N}$ can construct an isomorphism from itself with an initial segment of ${\mathbb N}^M$ , because for every $a\in \mathcal {N}$ it has a Henkin constant $\check a$ witnessing $\check a=\overbrace {1+\cdots +1}^a$ and it must be part of the theory that any $x<\check a$ is some $\check b$ for some $b<a$ in $\mathcal {N}$ , since otherwise $\mathcal {N}$ would think H is inconsistent. It follows that $\mathcal {N}$ and ${\mathbb N}^M$ have the same standard system. But any two countable computably saturated models of arithmetic with the same standard system are isomorphic, by the usual back-and-forth argument, and so $\mathcal {N}\cong {\mathbb N}^M$ , showing that $\mathcal {N}$ is ZFC-standard, as desired.

Proof. The proof of Proposition 3 adapts to accommodate the expanded structure. If a countable nonstandard model $\left \langle \mathcal {N},\mathrm {Tr}\right \rangle $ arises as $\left \langle {\mathbb N},\mathrm {TA}\right \rangle ^M$ for some $M\models \mathrm {ZFC}$ , then it admits an inductive truth predicate in the expanded language, and this implies that it is computably saturated just as above. Conversely, any countably computably saturated model of $\mathrm {Th}({\mathbb N},\mathrm {TA})^{\mathrm {ZFC}}$ can build as before a model of the corresponding Henkin theory extending $\mathrm {ZFC}+\{\,{\sigma ^{\left \langle {\mathbb N},\mathrm {TA}\right \rangle }\mid \left \langle \mathcal {N},\mathrm { Tr}\right \rangle \models \sigma }\,\}$ . The corresponding Henkin model M will have $\left \langle \mathcal {N},\mathrm {Tr}\right \rangle $ as an initial segment of $\left \langle {\mathbb N}^M,\mathrm {TA}\right \rangle $ , and so these two models have the same standard system, and since they also are elementarily equivalent and computably saturated, they are isomorphic by the back-and-forth construction. So $\left \langle \mathcal {N},\mathrm { Tr}\right \rangle $ is ZFC-standard.

Proof of Theorem 1

(as corollary to Proposition 4). By Beth’s theorem as in the Proof of Theorem 2, we may find two different truth predicates on the same model of arithmetic, with both $\left \langle \mathcal {N},\mathrm {Tr}_1\right \rangle $ and $\left \langle \mathcal {N},\mathrm {Tr}_2\right \rangle $ being computably saturated models of $\mathrm {Th}({\mathbb N},\mathrm {TA})^{\mathrm {ZFC}}$ . It follows by Proposition 4 that they arise as the standard model inside two different models of set theory, with $\mathcal {N}={\mathbb N}^{M_1}={\mathbb N}^{M_2}$ and $\mathrm {Tr}_1=\mathrm {TA}^{M_1}$ and $\mathrm { Tr}_2=\mathrm {TA}^{M_2}$ , establishing theorem 1.

Proof of Theorem 1

(direct argument). Suppose that $M_1$ is any countable $\omega $ -nonstandard model of set theory. It follows that $M_1$ ’s version of the standard model of arithmetic $\left \langle {\mathbb N},{+},{\cdot },0,1, <,\mathrm { TA}\right \rangle ^{M_1}$ , augmented with what $M_1$ thinks is the true theory of arithmetic, is countable and computably saturated. Since $\mathrm { TA}^{M_1}$ is not definable in the reduced structure ${\mathbb N}^{M_1}=\left \langle {\mathbb N},{+},{\cdot },0,1, <\right \rangle ^{M_1}$ , it follows by saturation that there is an arithmetic sentence $\sigma \in \mathrm {TA}^{M_1}$ realizing the same $1$ -type in ${\mathbb N}^{M_1}$ as another sentence $\tau $ in ${\mathbb N}^{M_1}$ with $\tau \notin \mathrm {TA}^{M_1}$ ; that is, ${\mathbb N}^{M_1}\models \varphi (\sigma )\mathrel {\leftrightarrow }\varphi (\tau )$ for every formula $\varphi $ in the language of arithmetic. To see this, consider the type $p(s,t)$ containing all formulas $\varphi (s)\mathrel {\leftrightarrow }\varphi (t)$ for $\varphi $ in the language of arithmetic, plus the assertions $s\in \mathrm {TA}$ and $t\notin \mathrm {TA}$ . This is a computable type, and it is finitely realized in $\left \langle {\mathbb N},{+},{\cdot },0,1, <,\mathrm { TA}\right \rangle ^{M_1}$ , precisely because otherwise we would be able to define $\mathrm {TA}^{M_1}$ in ${\mathbb N}^{M_1}$ . Thus, there are arithmetic sentences $\sigma $ and $\tau $ in $M_1$ , whose Gödel codes realize the same $1$ -type in ${\mathbb N}^{M_1}$ , but such that $M_1$ thinks that $\sigma $ is true and $\tau $ is false in ${\mathbb N}^{M_1}$ . Since these objects have the same $1$ -type in ${\mathbb N}^{M_1}$ , it follows by the back-and-forth construction that there is an automorphism $\pi :{\mathbb N}^{M_1}\to {\mathbb N}^{M_1}$ with $\pi (\tau )=\sigma $ .

Since $M_1$ sits over ${\mathbb N}^{M_1}$ in the way that it does and ${\mathbb N}^{M_1}$ is isomorphic to its image under $\pi $ , we may build a copy $M_2$ of $M_1$ in such a way that extends the isomorphism $\pi $ .

That we may do so is an instance of the trivial general observation about structures that whenever you have made an isomorphic copy of part of a structure, then you may extend this to an isomorphic copy of the whole structure. In our case, we had an isomorphism $\pi $ of ${\mathbb N}^{M_1}$ with its copy (which happened to be ${\mathbb N}^{M_1}$ itself), and so we may extend this to an isomorphism $\pi ^*$ of $M_1$ with a suitable structure $M_2$ extending ${\mathbb N}^{M_1}$ . So we have a model $M_2$ which is a copy of $M_1$ , witnessed by the isomorphism $\pi ^*:M_1\to M_2$ extending $\pi $ , in such a way that every element $m\in {\mathbb N}^{M_1}$ sits inside $M_1$ the same way that $\pi (m)$ sits inside $M_2$ .

Since $\pi $ was an automorphism of ${\mathbb N}^{M_1}$ , the situation is therefore that $M_1$ and $M_2$ have exactly the same natural numbers ${\mathbb N}^{M_1}={\mathbb N}^{M_2}$ . Yet, $M_1$ thinks that the arithmetic sentence $\sigma $ is true in the natural numbers, while $M_2$ thinks $\sigma $ is false, because $M_1$ thinks $\tau $ is false, and so $M_2$ thinks $\pi (\tau )$ is false, but $\pi (\tau )=\sigma $ .

Proof. Fix such a structure $\mathcal {N}$ and non-definable class S inside a countable model of set theory M. Let $M_1$ be any countable computably saturated elementary extension of M, in the language with constants for every element of M. In particular, the structure $\left \langle \mathcal {N},S,m\right \rangle _{m\in \mathcal {N}^M}^{M_1}$ is countable and computably saturated (and this is all we actually require). Since S is not definable from parameters, there are objects s and t in $\mathcal {N}^{M_1}$ with the same $1$ -type in $\left \langle \mathcal {N},m\right \rangle _{m\in \mathcal {N}^M}^{M_1}$ , yet $s\in S^{M_1}$ and $t\notin S^{M_1}$ . Since this latter structure is countable and computably saturated, it follows that there is an automorphism $\pi :\mathcal {N}\to \mathcal {N}$ with $\pi (t)=s$ and $\pi (m)=m$ for every $m\in \mathcal {N}^M$ . That is, $\pi $ is an automorphism of $\mathcal {N}^{M_1}$ mapping t to s, and respecting the copy of $\mathcal {N}^M$ inside $\mathcal {N}^{M_1}$ . Let $M_2$ be a copy of $M_1$ containing M, witnessed by an isomorphism $\pi ^*:M_1\to M_2$ extending $\pi $ and fixing the elements of M. It follows that $\mathcal {N}^{M_1}=\mathcal {N}^{M_2}$ and $M_1$ thinks $s\in S$ , but since $M_1$ thinks $t\notin S$ , it follows that $M_2$ thinks $\pi (t)\notin S$ and so $M_2$ thinks $s\notin S$ , as desired.

Proof. By ‘sufficiently robust’ we mean that $\mathcal {N}=\left \langle N,\ldots \right \rangle $ interprets the standard model of arithmetic ${\mathbb N}$ , so that it can handle the Gödel coding of formulas, and also that it has a definable pairing function, so that it contains (Gödel codes for) all finite tuples of its elements. We are assuming that $\mathcal {N}$ is sufficiently robust from the perspective of the model M. It follows by Tarski’s theorem on the non-definability of truth that the satisfaction relation of M for $\mathcal {N}$ —that is, the relation $ \mathrm {Sat}(\varphi ,\vec a)$ , which holds just in case $\mathcal {N}\models \varphi [\vec a]$ from the perspective of M—is not definable in $\mathcal {N}$ . And so the current theorem is a consequence of Theorem 5, using that relation.

Proof. What we mean by “respectively” is that each case may be exhibited separately, using different pairs of extensions $M_1$ and $M_2$ . Each statement in the theorem is an immediate consequence of Corollary 6 or of Theorem 5. Thanks to Roman Kossak for pointing out statement (2), which follows from Theorem 5 because multiplication is not definable in Presburger arithmetic $\left \langle {\mathbb N},+, <\right \rangle $ , as the latter is a decidable theory. Statement (1) is proved similarly, since the set of primes and the powers of two are not definable in $\left \langle {\mathbb N},S, <\right \rangle $ , since any nonstandard model of this structure consists of infinitely many ${\mathbb Z}$ chains above the standard part, and these admit numerous automorphisms by translation. By picking a suitable translation, we may move a power of two to an odd prime, and apply the argument in the Proof of Theorem 1. For statement (3), we apply Corollary 6 to the structure $\mathcal {N}=\left \langle {\mathbb N},{+},{\cdot },0,1, <\right \rangle ^M$ . For statement (4), one should clarify exactly what is meant by ‘projective truth,’ since it can be viewed variously as the full second-order theory of the standard model of arithmetic, using $P({\mathbb N})^M$ , or as the theory of the ordered real field $\left \langle {\mathbb R},{\mathbb Z},{+},{\cdot },0,1, <\right \rangle $ with a predicate for the integers, or as the theory of the set-theoretic structure $\left \langle V_{\omega +1},{\in }\right \rangle $ , or of the structure of hereditarily countable sets $\mathop {\mathrm {HC}}=\left \langle \mathop {\mathrm {HC}},{\in }\right \rangle $ . Nevertheless, these models are all bi-interpretable with each other, and one can view projective truth as the satisfaction relation for any of them; in each case, we can make the conclusion of statement (4) by using that structure in Corollary 6. Notice that one can similarly view arithmetic truth as residing in the structure of hereditarily finite sets $\left \langle \mathrm {HF},{\in }\right \rangle $ , which is mutually interpretable with the standard model of arithmetic via the Ackermann encoding, and in this case, statements (3), (4) and (5) can be seen as a progression, concerning the structures $H_\omega $ , $H_{\omega _1}$ and $H_{\omega _2}$ , a progression which continues, of course, to higher orders in set theory. Statement (6) is similarly an immediate consequence of Corollary 6, using the structure $\mathcal {N}=\left \langle V_\delta ,{\in }\right \rangle ^M$ .

Proof. In other words, we will have $({\mathbb N}_n)^{M_1}=({\mathbb N}_n)^{M_2}$ , yet $(\mathrm {Tr}_n)^{M_1}\neq (\mathrm { Tr}_n)^{M_2}$ . This is an immediate consequence of Corollary 6, using the model $\mathcal {N}=({\mathbb N}_n)^M$ .

Proof. These are each consequences of Theorem 5. For statement (1), consider the structure $\mathcal {N}={\mathbb N}^M$ , with the predicate $S= {\mathrm {WO}}^M$ , the set of indices of computably enumerable well-orderings on ${\mathbb N}$ in M. This is a $\Pi ^1_1$ -complete set of natural numbers, and hence not first-order definable in the structure $\mathcal {N}$ from the perspective of M. Thus, by Theorem 5, we get models of set theory $M_1$ and $M_2$ , elementarily extending M, which agree on ${\mathbb N}^{M_1}={\mathbb N}^{M_2}$ , yet disagree on $ {\mathrm { WO}}^{M_1}\neq {\mathrm {WO}}^{M_2}$ . So there is some c.e. relation $\vartriangleleft $ in common, yet $M_1$ (we may assume) thinks it is a well-order and $M_2$ does not. Since every c.e. order is isomorphic to a computable order, we may furthermore assume that $\vartriangleleft $ is computable, and the models will compute it with the same program, and both see that it is a linear order, as desired. Statement (2) follows as a consequence, since using the program computing $\vartriangleleft $ , we may construct deviations between $\mathcal {O}^{M_1}$ and $\mathcal {O}^{M_2}$ ; or alternatively, statement (2) follows immediately from Theorem 5, since $\mathcal {O}$ is not definable in ${\mathbb N}$ . Statement (3) also follows from Theorem 5, since the class of Silver indiscernibles below $\kappa $ is not definable in $\left \langle L_\kappa ,{\in }\right \rangle $ in M, as from it we could define a truth predicate for that structure. So there must be extensions which agree on $\left \langle L_\kappa ,{\in }\right \rangle ^{M_1}=\left \langle L_\kappa ,{\in }\right \rangle ^{M_2}$ and on the fact that $\kappa $ is an uncountable cardinal and that $0^\sharp $ exists, yet disagree on the Silver indiscernibles below $\kappa $ . Similarly, s $0^\sharp $ also is not definable in $\left \langle L_\kappa ,\in \right \rangle $ , we may make them disagree on $0^\sharp $ .

Proof. We identify ${\mathbb N}\times {\mathbb N}$ with ${\mathbb N}$ via Gödel pairing. We use $X'$ and $X^{(n)}$ to denote the Turing jump of X and the $n^{\mathrm {th}}$ Turing jump of X, respectively. Recall that a set $X\subseteq {\mathbb N}$ is n-generic if for every $\Sigma ^0_n$ subset $S\subseteq 2^{ <\omega }$ , there is an initial segment of X that either is in S or has no extension in S. A set is arithmetically generic if it is n-generic for every n. It is a standard fact that if X is $1$ -generic, then $X' \equiv _T 0' \oplus X$ , and if X is n-generic and Y is $1$ -generic relative to $X \oplus 0^{(n-1)}$ , then $X \oplus Y$ is n-generic. Let $A = 0^{(\omega )} = \oplus _{n} 0^{(n)}$ , which is Turing equivalent to the set of true sentences of first order arithmetic. We shall construct a set B with the following features:

1. (1) $\{\,{n \in \mathbb {N} \mid B_n\text { is arithmetic}}\,\}=A$ . More specifically, if $n \in A$ , then $B_n$ is $(n+1)$ -generic and computable from $0^{(n+1)}$ , and if $n \notin A$ , then $B_n$ is arithmetically generic.

2. (2) For each natural number k, the set $C_k = \oplus _{i \in \omega } B_{m_i}$ is $(k+1)$ -generic, where $m_i$ is the ith element of the set $\{\,{m \in {\mathbb N} \mid m \notin A \text { or } k\leq m}\,\}$ .

Any set B with these features, we claim, fulfills the lemma. To see this, we argue first by argue by induction that $B^{(n)} \equiv _T 0^{(n)} \oplus C_{n}$ for any natural number n. This is immediate for $n=0$ , since $C_0=\oplus _m B_m$ . If $B^{(n)} \equiv _T 0^{(n)} \oplus C_{n}$ , then $B^{(n+1)} \equiv _T (0^{(n)} \oplus C_n)' \equiv _T 0^{(n+1)} \oplus C_n$ , since $C_n$ is $(n+1)$ -generic and hence $1$ -generic relative to $0^{(n)}$ , but $0^{(n+1)} \oplus C_n \equiv _T 0^{(n+1)} \oplus C_{n+1}$ because either $n \notin A$ and so $C_n = C_{n+1}$ or $n \in A$ so $C_{n+1} \equiv _T B_n \oplus C_n$ , since $B_n \leq _T 0^{(n+1)}$ . So we have established $B^{(n)}\equiv _T 0^{(n)}\oplus C_n$ . Since $C_n$ is $(n+1)$ -generic, it follows that $0^{(n)}\oplus C_n$ does not compute $0^{(n+1)}$ , and so also $B^{(n)}$ does not compute $0^{(n+1)}$ . In particular, $B^{(n)}$ does not compute A for any n, and so A is not arithmetically definable from B, as desired.

It remains to construct the set B with features (1) and (2). We do so in stages, where after stage n we will have completely specified $B_0, B_1, \ldots , B_n$ and finitely much additional information about B on larger coordinates. To begin, let $B_0$ be any set satisfying the requirement of condition (1). We will ensure inductively that after each stage n, the set $C_{k,n} = B_{m_0} \oplus \ldots \oplus B_{m_j}$ is $(k+1)$ -generic, where $k\leq n$ and $m_0, \ldots , m_j$ are the elements of $\{\,{m \in {\mathbb N} \mid m \notin A \text { or }k\leq m}\,\} \cap \{0, \ldots , n\}$ .

At stage $n> 0$ , for each of the finitely many pairs $(i,k)$ with $i,k < n$ , we let $S_{i,k}$ be the ith $\Sigma ^0_{k+1}$ subset of $2^{<\omega }$ , and if possible, we make a finite extension to our current approximation to B so that the resulting approximation to $C_k$ extends an element of $S_{i,k}$ , thereby ensuring this instance of (2). If there is no such extension, then since inductively $C_{k,n-1}$ is $(k+1)$ -generic, there is already a finite part of our current approximation to B that cannot be extended to extend an element of $S_{i,k}$ , and this also ensures this instance of (2).

We complete stage n by specifying $B_n$ . If $n \notin A$ , then we simply extend our current approximation to B by ensuring that $B_n$ is arithmetically generic relative to $B_0 \oplus \ldots \oplus B_{n-1}$ . This ensures this instance of (1) while maintaining our induction assumption that $C_{k,n}$ is $(k+1)$ -generic for each $k < n$ , since $C_{k,n-1}$ is $(k+1)$ -generic and $B_n$ is $(k+1)$ -generic relative to it; and similarly, $C_{n,n}$ is now $(n+1)$ -generic. If $n \in A$ , then we let $B_n$ be any $0^{(n+1)}$ -computable $(n+1)$ -generic set extending the finitely many bits of $B_n$ specified in the current approximation. For each $k < n$ , let $j_0, \ldots , j_t$ be the elements of A in the interval $[k,n)$ . Since our new $B_n$ is $1$ -generic relative to $0^{(n)}$ , which can compute $B_{j_0} \oplus \ldots \oplus B_{j_t}$ , it follows that that $B_{j_0} \oplus \ldots \oplus B_{j_t} \oplus B_n$ is $(k+1)$ -generic, and so $C_{k,n}$ is $(k+1)$ -generic, as the remaining elements in the finite join defining $C_{k,n}$ are mutually arithmetically generic with this; and since $C_{n,n}$ is $n+1$ -generic, we maintain our induction assumption.

This completes the construction. We have fulfilled (1) explicitly by the choice of $B_n$ , and we fulfilled (2) by systematically deciding all the required sets $S_{i,k}$ .

Proof of Theorem 10

Fix any countable model of set theory M. Apply Lemma 10.1 inside M, to find a predicate $B\subseteq {\mathbb N}\times {\mathbb N}$ in M, such that the set $S=\{\,{n\in {\mathbb N}\mid B_n\text { is arithmetic}}\,\}$ is not definable in the structure $\mathcal {N}=\left \langle {\mathbb N},{+},{\cdot },0,1, <,B\right \rangle $ from the perspective of M. It follows by Theorem 5 applied to this structure that there are elementary extensions $M_1$ and $M_2$ of M, which agree on $\mathcal {N}^{M_1}=\mathcal {N}^{M_2}$ and in particular on ${\mathbb N}^{M_1}={\mathbb N}^{M_2}$ and $B^{M_1}=B^{M_2}$ , but not on $S^{M_1}\neq S^{M_2}$ . In particular, there is some section $A=B_n$ that is arithmetic in $M_1$ (we may assume), but not in $M_2$ . But since the models agree on the predicate B, they agree on all the sections of B and in particular have the set A extensionally in common.

Proof. Suppose that M is a countable model of ZFC, and consider the theory:

$$ \begin{align*} T_1=\Delta(M) + \ & {V_\delta \prec V}\ + \ \{\,{m\in V_\delta \mid m\in M}\,\}\\ & + \ \text{the least } n \text{ such that } V_\delta\nvDash\Sigma_n\text{-collection is even,} \end{align*} $$

where $\Delta (M)$ is the elementary diagram of M, in the language of set theory having constants for every element of M. Note that the first three components of $T_1$ as it is described above are each infinite schemes, whereas the final assertion “the least n…,” which we take also to assert that there is such an n, is expressible as a single sentence about $\delta $ in the language of set theory, using the ZFC-definable satisfaction relation for $\left \langle V_\delta ,{\in }\right \rangle $ . We claim that this theory is consistent. Consider any finite subtheory $t\subseteq T_1$ . We shall find a $\delta $ in M such that M with this $\delta $ will satisfy every assertion in t. Let k be a sufficiently large odd number, so that every formula appearing in any part of t has complexity at most $\Sigma _k$ , and let $\delta $ be the next $\Sigma _k$ -correct cardinal in M above the largest rank of an element of M whose constant appears in t. We claim now that $\left \langle M,{\in ^M},\delta ,m\right \rangle _{m\in M}\models t$ . First, it clearly satisfies all of $\Delta (M)$ ; and since $\delta $ is $\Sigma _k$ -correct in M, we have $V_\delta ^M\prec _{\Sigma _k} M$ , and since also $\delta $ was large enough to be above any of the constants of M appearing in t, we attain any instances from the first three components of the theory that are in the finite sub-theory t; since $\delta $ is $\Sigma _k$ -correct in M, it follows that $V_\delta ^M$ satisfies every instance of $\Sigma _k$ -collection; but since $\delta $ is not a limit of $\Sigma _k$ -correct cardinals (since it is the “next” one after a certain ordinal), it follows that $V_\delta ^M$ does not satisfy $\Sigma _{k+1}$ -collection, and so the least n such that $V_\delta ^M\not \models \Sigma _n$ -collection is precisely $n=k+1$ , which is even. So $T_1$ is finitely consistent and thus consistent. Similarly, the theory

$$ \begin{align*} T_2=\Delta(M)+ \ & {V_\delta \prec V} \ + \ \{\,{m\in V_\delta\mid m\in M}\,\}\\ & +\text{the least } n \text{ such that } V_\delta\nvDash\Sigma_n\text{-collection is odd} \end{align*} $$

is also consistent.

Let $\left \langle M_1,M_2\right \rangle $ be a computably saturated model pair, such that $M_1\models T_1$ and ${M_2\models T_2}$ . It follows that $\left \langle V_\delta ^{M_1},V_\delta ^{M_2}\right \rangle $ is a computably saturated model pair of elementary extensions of $\left \langle M,{\in ^M}\right \rangle $ , which are therefore elementarily equivalent in the language of set theory with constants for elements of M, and hence isomorphic by an isomorphism respecting those constants. So we may assume without loss of generality that $\left \langle M,{\in ^M}\right \rangle \prec \left \langle V_\delta ,{\in }\right \rangle ^{M_1}=\left \langle V_\delta ,{\in }\right \rangle ^{M_2}$ . Meanwhile, $M_1$ thinks that this $V_\delta $ violates $\Sigma _n$ -collection first at an even n and $M_2$ thinks it does so first for an odd n, since these assertions are part of the theories $T_1$ and $T_2$ , respectively.

Proof. Fix any countable model $M\models \mathrm {ZFC}$ , such that the worldly cardinals form a stationary proper class in M. That is, every definable proper class club $C\subseteq {\mathrm {Ord}}$ in M contains some $\delta $ that is worldly in M. Let $T_1$ be the theory consisting of the elementary diagram $\Delta (M)$ plus the scheme of assertions $V_\delta \prec V$ , in the language with a new constant symbol for $\delta $ , plus the assertions $a\in V_\delta $ for each constant symbol a for an element $a\in M$ , plus the assertion “ $\delta $ is worldly,” which is to say, the assertion that $V_\delta \models \mathrm {ZFC}$ . Suppose that t is a finite subtheory of $T_1$ , which therefore involves only finitely many instances from the $V_\delta \prec V$ scheme. Let n be large enough so that all the formulas $\varphi $ in these instances have complexity at most $\Sigma _n$ . Since the $\Sigma _n$ -correct cardinals form a closed unbounded class and the worldly cardinals are stationary, there is a $\Sigma _n$ -correct worldly cardinal $\delta $ in M. It follows that M satisfies all the formulas in t using this $\delta $ , and so the theory $T_1$ is finitely consistent and hence consistent.

Let $T_2$ be the theory consisting of the elementary diagram $\Delta (M)$ , the scheme $V_\delta \prec V$ plus the assertion “ $\delta $ is not worldly.” This theory also is finitely consistent, since if $t\subseteq T_2$ is finite, then let n be beyond the complexity of any formula appearing as an instance of $V_\delta \prec V$ in t, and let $\delta $ be a $\Sigma _n$ correct cardinal in M that is not worldly (for example, we could let $\delta $ be the next $\Sigma _n$ -correct cardinal after some ordinal; this can never be worldly since $V_\delta $ will not satisfy $\Sigma _n$ -reflection). It follows that M with this $\delta $ satisfies every assertion in t, showing that $T_2$ is finitely consistent and hence consistent.

Let $\left \langle M_1,M_2\right \rangle $ be a computably saturated model pair, where $M_1\models T_1$ and $M_2\models T_2$ . It follows that $\left \langle V_\delta ^{M_1},V_\delta ^{M_2}\right \rangle $ is also a computably saturated model pair of models of set theory, and these both satisfy the elementary diagram of M. Consequently, they are isomorphic by an isomorphism that respects the interpretation of M in them, and so by replacing with an isomorphic copy, we may assume that $\left \langle V_\delta ,{\in }\right \rangle ^{M_1}=\left \langle V_\delta ,{\in }\right \rangle ^{M_2}$ . The theories $T_1$ and $T_2$ ensure that $V_\delta \prec V$ in both $M_1$ and $M_2$ , that $M\prec V_\delta $ , and furthermore, that $M_1\models \delta $ is worldly and $M_2\models \delta $ is not worldly, or in other words, $M_1\models (V_\delta \models \mathrm {ZFC})$ , but $M_2\models (V_\delta \not \models \mathrm {ZFC})$ , as desired.

Proof. Consider first the case that $\mathcal {M}$ is $\omega $ -nonstandard. By the reflection theorem, any particular finite fragment of ZFC is true in some $\left \langle V_\delta ^M,{\in }\right \rangle ^{\mathcal {M}}$ , and so by overspill there must be some $\delta $ in $\mathcal {M}$ for which $\mathcal {M}$ believes $\left \langle V_\delta ,{\in }\right \rangle ^{\mathcal {M}}$ satisfies a nonstandard fragment of $\mathrm {ZFC}$ , and so in particular it will satisfy the actual ZFC, as desired. Meanwhile, if $\mathcal {M}$ is $\omega $ -standard, then it must satisfy $\mathop {\mathrm { Con}}(\mathrm {ZFC})$ and so it can build the Henkin model of ZFC. So in any case, there is a model of ZFC inside $\mathcal {M}$ .