3.1 The model

We shall construct an $\omega $ -model $M_\infty \subset \mathcal {P}(\mathbb {N})$ which satisfies $\Delta ^1_1$ - $\mathsf {CA}_0$ but not finite choice. The basic setup follows [Reference Montalbán10].

To define $M_\infty $ we will construct a generic object

$$\begin{align*}\langle T^G, \{f_i^G: i \in \mathbb{N}\}, h^G\rangle, \end{align*}$$

where $T^G$ is a tree, each $f_i^G$ is a distinct branch through $T^G$ , and $h^G: T^G \to \omega _1^{CK} \cup \{\infty \}$ is the well-founded rank function on $T^G$ , i.e., for all $\sigma $ in the well-founded part of $T^G$ , we have $h^G(\sigma ) = \sup \{h^G(\tau )+1: \tau \in T^G, \, \tau \supseteq \sigma \}$ , while for all $\sigma $ which is extendible in $T^G$ , we have $h^G(\sigma ) = \infty $ . Our convention is that $\infty = \infty +1$ and $\infty> \infty > \alpha $ , so $h^G(\sigma ) = \sup \{h^G(\tau )+1: \tau \in T^G, \, \tau \supseteq \sigma \}$ for all $\sigma $ in $T^G$ .

Then, for each finite $F \subset \mathbb {N}$ (written $F \subset _{\mathrm {f}} \mathbb {N}$ ), we define

$$\begin{align*}M_F = \{X \subseteq \mathbb{N}: \exists \mu < \omega_1^{CK} (X \leq_T (T^G \oplus \langle f_i^G \rangle_{i \in F})^{(\mu)})\}, \end{align*}$$

where $Y^{(\mu )}$ denotes the transfinite iteration of the Turing jump of Y up to level $\mu $ . (For background on transfinite iterations of the Turing jump, see [Reference Sacks13]. Later, we will express $M_F$ as the sets which are computable in $H_{\mu ,F}$ for some $\mu < \omega _1^{CK}$ , defined below.) Notice that $\mu $ ranges over the computable ordinals, rather than the ordinals which are computable in $\omega _1^{T^G \oplus \langle f_i^G \rangle _{i \in F}}$ . Nonetheless we will show in Corollary 3.14 that $M_F = \mathrm {HYP}(T^G \oplus \langle f_i^G \rangle _{i \in F})$ .

Finally, our desired $\omega $ -model is defined by

$$\begin{align*}M_\infty = \bigcup_{F \subset_{\mathrm{f}} \mathbb{N}} M_F. \end{align*}$$

Notice that $h^G$ does not appear in the definitions of $M_F$ or $M_\infty $ . Nonetheless it will play a crucial role in showing that $M_\infty $ has the properties we desire.

Let us sketch why finite choice fails in $M_\infty $ . We will show (Lemma 3.9) that for each $F \subset _{\mathrm {f}} \mathbb {N}$ , $M_F \cap [T^G] = \{f_i^G: i \in F\}$ . This implies that the branches through $T^G$ in $M_\infty $ are exactly the branches $f_i^G$ , for $i \in \mathbb {N}$ . This also implies that $M_\infty $ does not contain any infinite sequence of distinct branches $f_i^G$ , for such a sequence has to lie in some $M_F$ , contradicting Lemma 3.9.

This allows us to construct a failure of finite choice: For each n, let $T_n$ be the subtree of $T^G$ passing through $\langle n \rangle $ . Our forcing will ensure that each $[T_n]$ contains (nonzero) finitely many $f_i^G$ . Hence $M_\infty $ thinks that $\langle T_n \rangle _n$ is an instance of finite choice. Any finite choice solution to $\langle T_n \rangle _n$ would be an infinite sequence of distinct branches $f_i^G$ (they are distinct because the nth branch has first entry n), and hence would not lie in $M_\infty $ , by the previous paragraph.

In preparation for describing the forcing, we shall give every element of $M_\infty $ a name. Define by recursion on $\mu < \omega _1^{CK}$ :

$$\begin{align*}H_{1,F} = T^G \oplus \langle f^G_i \rangle_{i \in F}, \quad S_{\mu,F,e} = W^{H_{\mu,F}}_e, \quad H_{\mu,F} = \bigoplus_{\nu<\mu,e \in \mathbb{N}} S_{\nu,F,e} \end{align*}$$

for $\mu < \omega _1^{CK}$ , $F \subset _{\mathrm {f}} \mathbb {N}$ , $e \in \mathbb {N}$ . (Here $W^Y_e$ is the eth computably enumerable set relative to the oracle Y.)

3.2 The forcing language

We consider a ramified language $\mathcal {L}_\infty $ which extends the language of second-order arithmetic with constants for each element of $M_\infty $ , and various types of restricted set variables.

In order to define $\mathcal {L}_\infty $ , we first define a language $\mathcal {L}_F$ for each $F \subset _{\mathrm {f}} \mathbb {N}$ . $\mathcal {L}_F$ consists of first-order formulas in the language of second-order arithmetic generated by the following set variables and constants: for each $D \subseteq F$ , there are unranked set variables of the form $X_D$ and ranked set variables of the form $X^\lambda _D$ for each $\lambda < \omega _1^{CK}$ , while the constants are intended to name every element of $M_F$ :

1. – $\mathbf {T}$ , $\mathbf {f}_i$ for $i \in F$ ;

2. – for each $\nu < \omega _1^{CK}$ , $\mathbf {H}_{\nu ,F}$ and $\mathbf {S}_{\nu ,F,e}$ for each $e \in \mathbb {N}$ .

Let $C_F$ denote the above set of constants. If $\textbf {S}$ is of the form $\textbf {S}_{\nu ,F,e}$ , we define $\mathrm {dom}(\textbf {S})$ to be F. We define $C^\mu $ to be the set of all constants of the form $\textbf {H}_{\nu ,F}$ or $\textbf {S}_{\nu ,F,e}$ for some $\nu < \mu $ .

The language $\mathcal {L}_\infty $ consists of $\bigcup _{F \subset _{\mathrm {f}} \mathbb {N}} \mathcal {L}_F$ , unranked set variables of the form X, and ranked set variables of the form $X^\lambda $ for each $\lambda < \omega _1^{CK}$ .

Each formula $\psi $ of $\mathcal {L}_\infty $ can be assigned an ordinal rank as follows:

$$\begin{align*}\mathrm{rk}(\psi) = \omega_1^{CK}\cdot u(\psi) + \omega^2\cdot o(\psi) + \omega \cdot r(\psi) + n(\psi), \end{align*}$$

where:

1. – $u(\psi ) < \omega $ denotes the number of unranked quantifiers in $\psi $ ;

2. – $o(\psi ) < \omega _1^{CK}$ denotes the least upper bound of

   $$ \begin{align*} &\{\nu: \nu \text{ is the superscript of a bound variable in }\psi\} \\ \cup \; &\{\nu+1: \text{ some constant of the form }\textbf{S}_{\nu,F,e} \text{ or } \textbf{H}_{\nu,F} \text{ occurs in }\psi\}; \end{align*} $$

3. – $r(\psi ) < \omega $ denotes the number of ranked set quantifiers in $\psi $ ;

4. – $n(\psi ) < \omega $ denotes the number of connectives.

We say that a formula $\psi $ of $\mathcal {L}_\infty $ is ranked if all of its bound variables are ranked, or equivalently, if $\mathrm {rk}(\psi ) < \omega _1^{CK}$ .

A variable of the form $X^\lambda _D$ or $X_D$ is F-restricted if $D \subseteq F$ . A constant of the form $H_{\mu ,D}$ or $S_{\mu ,D,e}$ is F-restricted if $D \subseteq F$ . A formula of $\mathcal {L}_\infty $ is F-restricted if all of its bound variables and constants are F-restricted.

A formula is $\Sigma $ -over- $\mathcal {L}_F$ if it is built up from ranked F-restricted formulas using $\land $ , $\forall n$ , and $\exists X$ .

For any formula $\psi $ and any $\mu < \omega _1^{CK}$ , we define $\psi ^\mu $ by replacing every unranked set variable in $\psi $ with its ranked counterpart, i.e., X is replaced by $X^\mu $ and $X_F$ is replaced by $X^\mu _F$ .

3.3 The forcing notion

The forcing $\mathbb {P}$ consists of tuples

$$\begin{align*}p = \langle T^p, f^p, h^p \rangle \end{align*}$$

such that:

1. (C1) $T^p \subseteq \mathbb {N}^{<\mathbb {N}}$ is a finite tree;

2. (C2) $f^p$ is a finite partial function from $\mathbb {N}$ to $T^p\backslash \{\emptyset \}$ such that each $\sigma \in T^p$ of length $1$ is an initial segment of some $f^p(i)$

   (we view $f^p$ as a finite collection of finite paths in $T^p$ );

3. (C3) $h^p: T^p \to \omega _1^{CK} \cup \{\infty \}$ satisfies the following:

   1. (a) $h^p$ is a rank function, i.e., if $\tau \subsetneq \sigma $ , then $h^p(\tau )> h^p(\sigma )$

      (by fiat $\infty> \infty > \alpha $ for every $\alpha $ );

   2. (b) for all $i \in \mathrm {dom}(f^p)$ , $h^p(f^p(i)) = \infty $ ;

   3. (c) $h^p(\emptyset ) = \infty $ .

We think of $h^p$ as a labeling or tagging of nodes in $T^p$ . Condition (C2) is novel; ordinary Steel forcing only requires that $f^p$ is a finite partial function from $\mathbb {N}$ to $T^p$ . We impose (C2) in order to ensure that every node of length $1$ is extendible in the generic tree $T^G$ . Note that by (C2) and (C3)(b), for each $\sigma $ in $T^p$ of length $1$ , we must have $h^p(\sigma ) = \infty $ . Therefore, if $T^p$ contains any node of length $1$ , then (C3)(c) follows from (C2) and (C3)(b). There may be nodes $\sigma $ (of length greater than $1$ ) such that $h^p(\sigma ) = \infty $ yet there is no i such that $\sigma \subseteq f^p(i)$ .

We say that q extends p, written $q \leq p$ , if:

1. (E1) $T^q \supseteq T^p$ ;

2. (E2) for all $i \in \mathrm {dom}(f^p)$ , $f^q(i)$ is defined and $f^p(i) = f^q(i) \cap T^p$

   (old paths cannot be extended in the old tree);

3. (E3) for all $i \in \mathrm {dom}(f^q)\backslash \mathrm {dom}(f^p)$ , $f^q(i) \cap T^p = \emptyset $

   (new paths can only intersect the old tree at the root);

4. (E4) $h^q \supseteq h^p$ .

Conditions (E2) and (E3) are needed for us to control the complexity of the forcing relation. Condition (E2) is present in ordinary Steel forcing [Reference Steel16, p. 68]. One could weaken (E2) by considering only conditions p where each $f^p(i)$ is a leaf in $T^p$ (such conditions are dense in the above forcing): In this case we could replace (E2) by “for all $i \in \mathrm {dom}(f^p)$ , $f^q(i)$ is defined and $f^p(i) \subseteq f^q(i)$ .” Condition (E3) is not in [Reference Steel16] but needs to be added to it, as pointed out in [Reference Montalbán10, Section 2.2].

We will take G to be a sufficiently $\mathbb {P}$ -generic filter (G must decide all $\Sigma $ -over- $\mathcal {L}_F$ formulas for all $F \subset _{\mathrm {f}} \mathbb {N}$ , as well as all formulas of the form $\forall n(\psi (n) \leftrightarrow \neg \varphi (n))$ , where $\varphi $ and $\psi $ are $\Sigma $ -over- $\mathcal {L}_F$ for some $F \subset _{\mathrm {f}} \mathbb {N}$ ). Then we may define $T^G = \bigcup _{p \in G} T^p$ , $f_i^G = \bigcup _{p \in G} f^p(i)$ for $i \in \mathbb {N}$ , and $h^G = \bigcup _{p \in G} h^p$ . By genericity, each $f_i^G$ is an infinite branch in $T^G$ , the branches $f_i^G$ are distinct, and $h^G$ is the well-founded rank function on $T^G$ .

We encode each condition as a number as follows. Fix a computable linear ordering with well-founded part $\omega _1^{CK}$ . (Such orderings are known as pseudo-well-orderings; see [Reference Harrison7].) We identify each $\alpha < \omega _1^{CK}$ with its corresponding element in the well-founded part. When we write $\alpha < \beta $ , we always refer to their order as ordinals rather than the natural number ordering. Using the above identification and standard pairing functions, we may encode $\mathbb {P}$ as a $\Pi ^1_1$ subset of $\mathbb {N}$ . For each $\alpha < \omega _1^{CK}$ , define $\mathbb {P}_\alpha $ to be the set of all conditions p such that the range of $h^p$ is contained in $\alpha \cup \{\infty \}$ . We have $\mathbb {P} = \bigcup _{\alpha < \omega _1^{CK}} \mathbb {P}_\alpha $ . Each $\mathbb {P}_\alpha $ is computable, uniformly in $\alpha $ .

3.4 The forcing relation

The forcing relation for formulas in $\mathcal {L}_\infty $ is defined by recursion as follows:

1. (1) for quantifier-free formulas of arithmetic $\psi $ , $p \Vdash \psi $ if and only if $\psi $ is true;

2. (2) $p \Vdash \boldsymbol {\sigma } \in \mathbf {T}$ if either $|\sigma | < 2$ and $\sigma \in T^p$ , or $\sigma ^- \in T^p$ and $h^p(\sigma ^-) \geq 1$ ;

3. (3) $p \Vdash \langle \mathbf {n},\mathbf {m} \rangle \in \textbf {f}_i$ if $i \in \mathrm {dom}(f^p)$ and $f^p(i)(n) = m$ ;

4. (4) $p \Vdash \langle 0,\sigma \rangle \in \textbf {H}_{1,F}$ if $p \Vdash \boldsymbol {\sigma } \in \textbf {T}$ ;

5. (5) $p \Vdash \langle 1,\langle \textbf {i},\langle \textbf {n},\textbf {m} \rangle \rangle \rangle \in \textbf {H}_{1,F}$ if $i \in F$ and $p \Vdash \langle \textbf {n},\textbf {m} \rangle \in \textbf {f}_i$ ;

6. (6) for $\nu> 1$ , $p \Vdash \mathbf {n} \in \mathbf {S}_{\nu ,F,e}$ if $p \Vdash \exists s R(\mathbf {H}_{\nu ,F}; \mathbf {e},s,\mathbf {n})$ where R codes a universal Turing machine;

7. (7) for $\nu> 1$ , $p \Vdash \langle \mathbf {e},\mathbf {n},\mu \rangle \in \mathbf {H}_{\nu ,F}$ if $\mu < \nu $ and $p \Vdash \mathbf {n} \in \mathbf {S}_{\mu ,F,e}$ ;

8. (8) $p \Vdash \forall x\psi (x)$ if for all $n \in \mathbb {N}$ , $p \Vdash \psi (\mathbf {n})$ ;

9. (9) $p \Vdash \forall X^\lambda _F \psi (X^\lambda _F)$ if for all $\nu < \lambda $ , $e \in \mathbb {N}$ , $p \Vdash \psi (\mathbf {S}_{\nu ,F,e})$ ;

10. (10) $p \Vdash \forall X^\lambda \psi (X^\lambda )$ if for all $\nu <\lambda $ , $e \in \mathbb {N}$ , $F \subset _{\mathrm {f}} \mathbb {N}$ , $p \Vdash \psi (\mathbf {S}_{\nu ,F,e})$ ;

11. (11) $p \Vdash \forall X_F \psi (X_F)$ if for all $\nu < \omega _1^{CK}$ , $e \in \mathbb {N}$ , $p \Vdash \psi (\mathbf {S}_{\nu ,F,e})$ ;

12. (12) $p \Vdash \forall X \psi (X)$ if for all $\nu < \omega _1^{CK}$ , $e \in \mathbb {N}$ , $F \subset _{\mathrm {f}} \mathbb {N}$ , $p \Vdash \psi (\mathbf {S}_{\nu ,F,e})$ ;

13. (13) $p \Vdash \varphi \land \psi $ if $p \Vdash \varphi $ and $p \Vdash \psi $ ;

14. (14) $p \Vdash \neg \psi $ if for every $q \leq p$ , $q \not \Vdash \psi $ .

Just as we encoded each forcing condition as a number, we may also encode each formula in $\mathcal {L}_\infty $ as a number. We will show that the forcing relation for certain classes of formulas (encoded as a relation on numbers) can be computed within $M_\infty $ (Lemma 3.12).

3.5 Analyzing the forcing relation for ranked formulas

In order to control the complexity of the forcing relation, we will show that conditions that are sufficiently similar to each other force the same formulas of sufficiently low complexity. The basic notion of similarity in Steel forcing is known as retagging:

We make some observations:

1. – $\mu $ -F-retagging is an equivalence relation on conditions.

2. – If p and $p^\ast $ are $\mu $ -F-retaggings, then for any $\mu ' \leq \mu $ and any $F' \subseteq F$ , p and $p^\ast $ are $\mu '$ - $F'$ -retaggings as well.

Central to our analysis will be the ability to replace a quantifier over $\mathbb {P}$ with a quantifier over some subset $\mathbb {P}_\beta $ (recall that $\mathbb {P}$ is $\Pi ^1_1$ while $\mathbb {P}_\beta $ is computable). Once we have established connections between retagging and the forcing relation, the following lemma will be useful for achieving the above:

Proof Define $T^{q^\ast } = T^q$ , $f^{q^\ast } = f^q$ , and

$$\begin{align*}h^{q^\ast}(\tau) = \begin{cases} \infty, & h^q(\tau) \geq \beta, \\ h^q(\tau), & h^q(\tau) < \beta. \end{cases} \end{align*}$$

It is easy to see that $q^\ast $ satisfies the desired properties.

A cornerstone of the method of Steel forcing is the following basic retagging lemma.

The first part of the lemma is identical in form to [Reference Montalbán10, Lemma 2.5]. The second part has no analog in [Reference Montalbán10], but its analog would follow immediately from the proof of [Reference Montalbán10, Lemma 2.5]. For our forcing we will construct $q^\ast $ (specifically $f^{q^\ast }$ ) slightly differently in order to prove the second part. The second part will be used in the proof of Lemma 3.11.

The following consequences of the basic retagging lemma are routine (see [Reference Montalbán10, Lemma 2.4 and Corollary 2.6]). We include their proofs for completeness.

We may now analyze which branches on $T^G$ lie in each $M_F$ . The proof below follows [Reference Montalbán10, Lemma 2.7], with modifications in order to account for (C2).

Proof Suppose towards a contradiction that $S = S_{\nu ,F,e} \in [T^G]$ is not $f^G_i$ for any $i \in F$ . Then there is some prefix $\sigma \subset S$ such that $\sigma \not \subset f^G_i$ for any $i \in F$ . For later purposes, fix such $\sigma $ with length greater than $1$ . By Lemma 3.2, fix $p \in G$ such that $p \Vdash \varphi (\textbf {S})$ , where $\varphi (\textbf {S})$ is

$$\begin{align*}\textbf{S} \in [\textbf{T}] \; \land \; \boldsymbol{\sigma} \subset \textbf{S} \; \land \; (\forall i \in F)(\boldsymbol{\sigma} \not\subset \textbf{f}_i). \end{align*}$$

Note $\varphi (\textbf {S})$ is a ranked formula in $\mathcal {L}_F$ .

By genericity, we may assume that $\sigma \in T^p$ . Next, fix $\beta < \omega _1^{CK}$ large enough so that $\beta> \omega \cdot \mathrm {rk}(\varphi (\textbf {S}))$ and $p \in \mathbb {P}_\beta $ .

Since p forces that $\sigma $ is extendible in $T^G$ , we have $h^p(\sigma ) = \infty $ (by Lemma 3.2). We now define $p^\ast $ which is a $\beta $ -F-retagging of p, such that $h^{p^\ast }(\sigma ) \in [\beta ,\omega _1^{CK})$ . Define $T^{p^\ast } = T^p$ . Define $f^{p^\ast }$ to be the union of $f^p\restriction F$ and a function g such that $\mathrm {dom}(g)$ is disjoint from F and $\mathrm {range}(g) = \{\sigma \in T^p: |\sigma |=1\}$ . Finally, define

$$\begin{align*}h^{p^\ast}(\tau) = \begin{cases} \beta+|\tau|_Q, & \text{if }\tau \supseteq \sigma \land h^p(\tau) = \infty, \\ h^p(\tau), & \text{otherwise}, \end{cases} \end{align*}$$

where $Q = \{\tau : \tau \subseteq \sigma \lor (\tau \supseteq \sigma \land h^p(\tau ) = \infty )\}$ .

Since $h^p(\sigma ) = \infty $ and $|\sigma |> 1$ , it is straightforward to check that $h^{p^\ast }$ is a rank function and $h^{p^\ast }(\emptyset ) = \infty $ . In order to show that $p^\ast $ is a condition, it suffices to check that $h^{p^\ast }(f^{p^\ast }(i)) = \infty $ for all $i \in \mathrm {dom}(f^{p^\ast })$ . If $i \in F$ , we have $\sigma \not \subseteq f^p(i) = f^{p^\ast }(i)$ , so $h^{p^\ast }(f^{p^\ast }(i)) = h^p(f^{p^\ast }(i)) = h^p(f^p(i)) = \infty $ . If $i \in \mathrm {dom}(f^{p^\ast })\backslash F$ (that is, $\mathrm {dom}(g)$ ), then $f^{p^\ast }(i)$ has length $1$ , so $h^{p^\ast }(f^{p^\ast }(i)) = h^p(f^{p^\ast }(i)) = \infty $ .

Finally, it is clear that $p^\ast $ is a $\beta $ -F-retagging of p, and hence an $(\omega \cdot \mathrm {rk}(\varphi (\textbf {S})))$ -F-retagging of p. By Lemma 3.6 we have

$$\begin{align*}p^\ast \Vdash \textbf{S} \in [\textbf{T}] \land \boldsymbol{\sigma} \subset \textbf{S}, \end{align*}$$

which is impossible because $h^{p^\ast }(\sigma ) \neq \infty $ .

The following lemma is new; (C2) was designed in order to prove it.

Proof Consider the following arithmetical predicate $\varphi (n,X)$ (with parameter $T^G$ ):

$$\begin{align*}X \in [T^G] \; \land \; X(0) = n. \end{align*}$$

First, we claim that for each n, there is some $X \in M_\infty $ such that $\varphi (n,X)$ holds. By genericity, fix some $p \in G$ such that $\langle n \rangle \in T^p$ . By (C2), there is some i such that $\langle n \rangle \subseteq f^p(i)$ . Then $\varphi (n,f^G_i)$ holds.

Second, we claim that for each n, there are at most finitely many $X \in M_\infty $ such that $\varphi (n,X)$ holds. By Lemma 3.9, it suffices to show that there are only finitely many i such that $\varphi (n,f^G_i)$ holds. As above, fix $p \in G$ such that $\langle n \rangle \in T^p$ . For each i, we claim that if $i \notin \mathrm {dom}(f^p)$ , then $\langle n \rangle \not \subseteq f^G_i$ . Suppose to the contrary that there is some $i \notin \mathrm {dom}(f^p)$ such that $\langle n \rangle \subseteq f^G_i$ . Fix some $q \in G$ such that $q \Vdash \langle \textbf {n} \rangle \subseteq \textbf {f}_i$ . Let $r \in G$ be a common refinement of p and q. Then $\langle n \rangle \subseteq f^r(i)$ . But then $f^r(i) \cap T^p \neq \emptyset $ , violating (E3) for $r \leq p$ .

This shows that $\varphi (n,X)$ is an instance of finite choice in $M_\infty $ . By Lemma 3.9, $M_\infty $ does not contain any infinite sequence of distinct branches through $T^G$ , so $M_\infty $ does not contain any finite choice solution to $\varphi (n,X)$ .

3.6 Analyzing the forcing relation for $\Sigma $ -over- $\mathcal {L}_F$ formulas

In order to show that $M_\infty \models \Delta ^1_1$ - $\mathsf {CA}_0$ (Section 3.7), we need to analyze the forcing relation for $\Sigma $ -over- $\mathcal {L}_F$ formulas $\psi $ . We begin by considering formulas $\psi ^\rho $ where $\rho < \omega _1^{CK}$ . Note that such formulas may not lie in any $\mathcal {L}_{F'}$ because they may contain (existential) quantifiers over set variables of the form $X^\rho $ .

To this end we exploit automorphisms of $\mathbb {P}$ , as was done in [Reference Steel16, Lemma 10]. Each permutation $\pi : \mathbb {N} \to \mathbb {N}$ induces an automorphism $\hat {\pi }$ of $\mathbb {P}$ by permuting the finite paths in each condition: $T^{\hat {\pi }(p)} = T^p$ , $h^{\hat {\pi }(p)} = h^p$ , and $f^{\hat {\pi }(p)}(\pi (i)) = f^p(i)$ . Each permutation $\pi $ also induces an automorphism of $\mathcal {L}_\infty $ : For each formula $\psi $ , let $\pi \psi $ denote the formula obtained from $\psi $ by replacing each $\textbf {f}_i$ by $\textbf {f}_{\pi (i)}$ and each $\textbf {S}_{\nu ,F,e}$ by $\textbf {S}_{\nu ,\pi "F,e}$ . (In the proof below, we will denote $\textbf {S}_{\nu ,\pi "F,e}$ by $\pi \textbf {S}_{\nu ,F,e}$ .) By induction on $\mathrm {rk}(\psi )$ , one can show that $p \Vdash \psi $ if and only if $\hat {\pi }(p) \Vdash \pi \psi $ .

Proof We prove by induction on k that if $\psi $ is built from ranked, F-restricted formulas using k steps and p is a condition such that $p \Vdash \psi ^\mu $ , $\mathrm {dom}(f^p) \subseteq F$ , and p and $p^\ast $ are $(\omega \cdot \mathrm {rk}(\psi ^\mu )+\omega \cdot 2k)$ -F-retaggings, then $p^\ast \Vdash \psi ^\mu $ . The base case holds by Lemma 3.6.

The only nontrivial inductive step is if $\psi $ has the form $\exists X\varphi (X)$ . We want to prove that for all $q^\ast \leq p^\ast $ , there is some $r^\ast \leq q^\ast $ and some $\textbf {S} \in C^\mu $ such that $r^\ast \Vdash \varphi ^\mu (\textbf {S})$ . Our plan for doing so is illustrated in Figure 1.

Figure 1 Illustration of the proof of Lemma 3.11. Arrows correspond to extension in the forcing. Dotted lines correspond to retaggings.

Given $q^\ast \leq p^\ast $ , there is some $q \leq p$ such that q and $q^\ast $ are $(\omega \cdot \mathrm {rk}(\psi ^\mu ) + \omega \cdot (2k+1))$ -F-retaggings. Furthermore, since $\mathrm {dom}(f^p) \subseteq F$ , we can choose such q with $f^q \subseteq f^{q^\ast }$ (by the second part of Lemma 3.5).

Since $p \Vdash \psi ^\mu $ and $q \leq p$ , there is some $r \leq q$ and some $\textbf {S} \in C^\mu $ such that $r \Vdash \varphi ^\mu (\textbf {S})$ . By extending r, we may assume that $F \cup \mathrm {dom}(\textbf {S}) \subseteq \mathrm {dom}(f^r)$ . To complete the inductive step, we want to define $r^\ast \leq q^\ast $ such that r and $r^\ast $ are $(\omega \cdot \mathrm {rk}(\psi ^\mu )+\omega \cdot 2k)$ - $\mathrm {dom}(f^r)$ -retaggings. By inductive hypothesis, it would follow that $r^\ast \Vdash \varphi ^\mu (\textbf {S})$ as desired.

The naive definition of $r^\ast $ fails because there may be conflict between $f^r$ and $f^{q^\ast }$ . In order to avoid this conflict, we use an automorphism of $\mathbb {P}$ to permute r and $\textbf {S}$ as follows. Consider a permutation $\pi $ which fixes $F \cup \mathrm {dom}(f^q)$ and moves $\mathrm {dom}(f^r)\backslash (F \cup \mathrm {dom}(f^q))$ to some set which is disjoint with $\mathrm {dom}(f^{q^\ast })$ . Since $\pi $ fixes $\mathrm {dom}(f^q)$ , we have $\hat {\pi } r \leq q$ . Since $\pi $ fixes F and $\varphi $ is $\Sigma $ -over- $\mathcal {L}_F$ , we have $\pi \varphi ^\mu = \varphi ^\mu $ . So $\hat {\pi } r \Vdash \varphi ^\mu (\pi \textbf {S})$ . Therefore, by replacing r and $\textbf {S}$ with $\hat {\pi } r$ and $\pi \textbf {S}$ , we may assume that $\mathrm {dom}(f^r)\backslash (F \cup \mathrm {dom}(f^q))$ and $\mathrm {dom}(f^{q^\ast })$ are disjoint.

We claim that q and $q^\ast $ are $(\omega \cdot \mathrm {rk}(\psi ^\mu ) + \omega \cdot (2k+1))$ - $\mathrm {dom}(f^r)$ -retaggings. Since $f^q \subseteq f^{q^\ast }$ , this reduces to showing that $\mathrm {dom}(f^{q^\ast }) \cap \mathrm {dom}(f^r) \subseteq \mathrm {dom}(f^q)$ . Suppose $i \in \mathrm {dom}(f^{q^\ast }) \cap \mathrm {dom}(f^r)$ . By assumption on r, it follows that $i \in F \cup \mathrm {dom}(f^q)$ . If $i \in \mathrm {dom}(f^q)$ , then we are done. Otherwise $i \in F$ , but since $i \in \mathrm {dom}(f^{q^\ast })$ and q and $q^\ast $ are $(\omega \cdot \mathrm {rk}(\psi ^\mu ) + \omega \cdot (2k+1))$ -F-retaggings, it follows that $i \in \mathrm {dom}(f^q)$ as well. This proves the claim.

By Lemma 3.5, there is some $r^\ast \leq q^\ast $ such that r and $r^\ast $ are $(\omega \cdot \mathrm {rk}(\psi ^\mu )+\omega \cdot 2k)$ - $\mathrm {dom}(f^r)$ -retaggings. Since $F \cup \mathrm {dom}(\textbf {S}) \subseteq \mathrm {dom}(f^r)$ , we may apply the inductive hypothesis and the fact that $r \Vdash \varphi ^\mu (\textbf {S})$ to conclude that $r^\ast \Vdash \varphi ^\mu (\textbf {S})$ as desired.

The following lemma is not stated in [Reference Montalbán10] but some form of it is used in the proof of [Reference Montalbán10, Lemma 2.9(1)].

Proof We proceed by induction on the number of steps it takes to construct $\psi $ from ranked F-restricted formulas. The base case holds by Corollary 3.8.

For the inductive step, the only nontrivial case is when $\psi $ has the form $\exists X\varphi (X)$ . Fix any $\mu \geq \omega \cdot \mathrm {rk}(\varphi ^\rho (\textbf {S})) + \omega ^2 + \omega $ such that $p \in \mathbb {P}_\mu $ . We claim that $p \Vdash \exists X^\rho \psi ^\rho (X^\rho )$ if and only if for every $q \leq p$ in $\mathbb {P}_\mu $ , there is some $r \leq q$ in $\mathbb {P}_\mu $ and some $\textbf {S} \in C^\rho $ such that $r \Vdash \varphi ^\rho (\textbf {S})$ . This claim suffices for proving this case of the inductive step.

To prove the forward direction, suppose we are given $q \leq p$ in $\mathbb {P}_\mu $ . By assumption, there is some $r \leq q$ and some $\textbf {S} \in C^\rho $ such that $r \Vdash \varphi ^\rho (\textbf {S})$ . We can retag r to some $r^\ast \leq q$ in $\mathbb {P}_\mu $ such that r and $r^\ast $ are $\mu $ - $(F \cup \mathrm {dom}(f^r))$ -retaggings (Lemma 3.4). Since $\mu \geq \omega \cdot \mathrm {rk}(\varphi ^\rho (\textbf {S})) + \omega ^2$ , it follows from Lemma 3.11 that $r^\ast \Vdash \varphi ^\rho (\textbf {S})$ as desired.

To prove the backward direction, suppose we are given $q \leq p$ . In order to construct some $r \leq q$ and some $\textbf {S} \in C^\rho $ such that $r \Vdash \varphi ^\rho (\textbf {S})$ , we follow Figure 2. We begin by retagging q to some $q^\ast \leq p$ in $\mathbb {P}_\mu $ such that q and $q^\ast $ are $\mu $ - $F'$ -retaggings for every $F' \subset _f \mathbb {N}$ (Lemma 3.4). By assumption, there is some $r^\ast \leq q^\ast $ in $\mathbb {P}_\mu $ and some $\textbf {S} \in C^\rho $ such that $r^\ast \Vdash \varphi ^\rho (\textbf {S})$ .

Figure 2 Illustration of the proofs of Lemmas 3.12 and 3.13. Arrows correspond to extension in the forcing. Dotted lines correspond to retaggings.

To complete the proof of the backward direction, we want to produce some $r \leq q$ which is a retagging of $r^\ast $ . Define $F'$ to be the union of F, $\mathrm {dom}(f^{r^\ast })$ and $\mathrm {dom}(\textbf {S})$ (recall that $\mathrm {dom}(\textbf {S}_{\nu ,D,e})$ is defined to be D.) Then $\varphi ^\rho (\textbf {S})$ is $\Sigma $ -over- $\mathcal {L}_{F'}$ . Since $\mu \geq \omega \cdot \mathrm {rk}(\varphi ^\rho (\textbf {S})) + \omega ^2 + \omega $ , there is some $r \leq q$ such that r and $r^\ast $ are $(\omega \cdot \mathrm {rk}(\varphi ^\rho (\textbf {S})) + \omega ^2)$ - $F'$ -retaggings (Lemma 3.5). It follows from Lemma 3.11 that $r \Vdash \varphi ^\rho (\textbf {S})$ as desired.

Next, we shall show that in order to understand the forcing relation for $\Sigma $ -over- $\mathcal {L}_F$ formulas $\varphi $ , it suffices to consider formulas $\varphi ^\rho $ where $\rho < \omega _1^{CK}$ . This will be useful for showing that $M_\infty \models \Delta ^1_1$ - $\mathsf {CA}_0$ .

Following [Reference Montalbán10, Lemma 2.9(2)], we conclude the following corollary.

We are ready to prove that $M_\infty $ satisfies $\Delta ^1_1$ - $\mathsf {CA}_0$ . The overall strategy follows [Reference Steel16, Lemma 12], with modifications in order to account for (C2).

3.7 Proof that $M_\infty $ satisfies $\Delta ^1_1$ -comprehension

Suppose $\varphi (n)$ and $\psi (n)$ are $\Sigma $ -over- $\mathcal {L}_F$ with only n free and $M_\infty \models \forall n(\psi (n) \leftrightarrow \neg \varphi (n))$ . We want to define $D \in M_\infty $ such that

$$\begin{align*}M_\infty \models \forall n(\psi(n) \leftrightarrow n \in D). \end{align*}$$

A naive attempt is to consider

$$\begin{align*}\{n \in \mathbb{N}: \exists q \in G(q \Vdash \psi(\mathbf{n}))\}, \end{align*}$$

but there are two obstacles preventing us from showing that the above set lies in $M_\infty $ :

1. – $M_\infty $ does not contain G so we cannot search over $q \in G$ ;

2. – deciding whether $q \Vdash \psi (\mathbf {n})$ is not hyperarithmetic in general.

The second obstacle is overcome using Lemma 3.13. To overcome the first obstacle, we shall use retagging to change the scope of our search to a class of conditions which look like they might lie in G, based on information from $T^G$ and finitely many branches $f^G_i$ (all of which lie in $M_\infty $ , unlike G).

Recall that (by genericity) $h^G$ is the well-founded rank function of $T^G$ . We say that a rank function $h: T \to \omega _1^{CK} \cup \{\infty \}$ is $\nu $ -good if $T \subset T^G$ and for all $\sigma \in T$ :

$$ \begin{align*} h^G(\sigma) < \nu \quad &\Rightarrow \quad h(\sigma) = h^G(\sigma), \\ h^G(\sigma) \geq \nu \quad &\Leftrightarrow \quad h(\sigma) \geq \nu. \end{align*} $$

We will use the notion of $\nu $ -goodness in our definition of D.

We begin the proof by fixing $p \in G$ such that $p \Vdash \forall n(\psi (n) \leftrightarrow \neg \varphi (n))$ . By genericity and by expanding F if necessary, we may assume that $F = \mathrm {dom}(f^p)$ . Since $p \Vdash \forall n(\psi (n) \lor \varphi (n))$ (which is $\Sigma $ -over- $\mathcal {L}_F$ ), by Lemma 3.13, we may fix $\mu < \omega _1^{CK}$ large enough such that $p \Vdash \forall n(\psi ^\mu (n) \lor \varphi ^\mu (n))$ and $\mu $ is greater than the rank of any constant in $\varphi $ and $\psi $ . Since $p \Vdash \forall n(\neg \psi (n) \lor \neg \varphi (n))$ and $\mu $ is large enough, we also have $p \Vdash \forall n(\neg \psi ^\mu (n) \lor \neg \varphi ^\mu (n))$ and so $p \Vdash \forall n(\psi ^\mu (n) \leftrightarrow \varphi ^\mu (n))$ .

Next, fix $\nu < \omega _1^{CK}$ large enough such that $p \in \mathbb {P}_\nu $ and $\mathrm {rk}(\varphi ^\mu (\mathbf {d}) \lor \psi ^\mu (\mathbf {d})) < \nu $ for all $d \in \mathbb {N}$ . Now we define $D \subseteq \mathbb {N}$ as follows: $d \in D$ if and only if there is some $q \in \mathbb {P}_{\omega \nu +\omega ^2+\omega }$ extending p such that:

1. (1) $q \Vdash \psi ^\mu (\mathbf {d})$ ;

2. (2) $T^q \subset T^G$ (this is implied by (3) but we include it for emphasis);

3. (3) $h^q$ is $(\omega \nu +\omega ^2+\omega )$ -good;

4. (4) $\forall i \in F(f^q(i) = f^G_i \cap T^q)$ .

Note that if $q \in G$ , then q satisfies (2) (by (E1)) and (4) (by (E2)). (3) must hold as well, but even more is true: we must have $h^q \subseteq h^G$ . Since $M_\infty $ does not have access to $h^G$ , we make do with sufficiently good approximations. Observe that D is hyperarithmetic in $T^G \oplus \langle f^G_i \rangle _{i \in F}$ , so $D \in M_F \subseteq M_\infty $ . It remains to show that $M_\infty \models \psi (\mathbf {d})$ if and only if $d \in D$ .

The forward direction is straightforward: If $M_\infty \models \psi (\mathbf {d})$ , then $M_\infty \models \psi ^\mu (\mathbf {d})$ , so we may fix $q' \in G$ extending p which forces $\psi ^\mu (\mathbf {d})$ . Using $q'$ , we define q as we did in the proof of Lemma 3.4: $T^q = T^{q'}$ , $f^q = f^{q'}$ , and

$$\begin{align*}h^q(\sigma) = \begin{cases} \infty, & \text{if }h^{q'}(\sigma) \geq \omega\nu+\omega^2+\omega, \\ h^{q'}(\sigma), & \text{otherwise}. \end{cases} \end{align*}$$

It is straightforward to check that q witnesses $d \in D$ (using Lemma 3.11 to show that $q \Vdash \psi ^\mu (\textbf {d})$ ).

To prove the backward direction, suppose $M_\infty \models \varphi (\mathbf {d})$ . Then $M_\infty \models \varphi ^\mu (\mathbf {d})$ , so we may fix $r \in G$ extending p which forces $\varphi ^\mu (\mathbf {d})$ . Suppose towards a contradiction that $d \in D$ , as witnessed by some q. Our plan is to construct conditions $q^\ast $ extending q, $r^\ast $ extending r, and $s^\ast $ extending p, as illustrated in Figure 3.

We begin by using an automorphism of $\mathbb {P}$ to avoid conflict between $f^q$ and $f^r$ . Consider a permutation $\pi $ which fixes F and moves $\mathrm {dom}(f^q)\backslash F$ to some set which is disjoint with $\mathrm {dom}(f^r)$ . Since $F = \mathrm {dom}(f^p)$ , we have $\hat {\pi } q \leq p$ . It is straightforward to check that $\hat {\pi } q$ witnesses $d \in D$ . Therefore, by replacing q with $\hat {\pi } q$ , we may assume that $\mathrm {dom}(f^q)\backslash F$ and $\mathrm {dom}(f^r)$ are disjoint. It follows that $\mathrm {dom}(f^q) \cap \mathrm {dom}(f^r) = F$ .

Next, fix finite partial functions $g_{q^\ast }$ and $g_{r^\ast }$ on $\mathbb {N}$ such that:

1. – $\mathrm {dom}(g_{q^\ast })$ , $\mathrm {dom}(g_{r^\ast })$ , and $\mathrm {dom}(f^q) \cup \mathrm {dom}(f^r)$ are pairwise disjoint;

2. – $\mathrm {range}(g_{q^\ast }) = \{\sigma \in T^r\backslash T^q: |\sigma | = 1\}$ ;

3. – $\mathrm {range}(g_{r^\ast }) = \{\sigma \in T^q\backslash T^r: |\sigma | = 1\}$ .

We may now define $q^\ast $ as follows:

1. – $T^{q^\ast } = T^q \cup T^r$ ;

2. – $f^{q^\ast }(i) = \begin {cases} f^q(i), & \text {if }i \in \mathrm {dom}(f^q)\backslash F; \\ f^G_i \cap T^{q^\ast }, & \text {if }i \in F; \\ g_{q^\ast }(i), &\text {if }i \in \mathrm {dom}(g_{q^\ast }); \end {cases}$

3. – $h^{q^\ast }$ is defined by cases:

   $$\begin{align*}h^{q^\ast}(\tau) = \begin{cases} h^q(\tau), & \text{if }\tau \in T^q, \\ \infty, & \text{if }\exists i(\tau \subseteq f^{q^\ast}(i)), \\ h^r(\tau), & \text{if }h^r(\tau) < \omega\nu+\omega^2, \\ \omega\nu+\omega^2+|\tau|_Q, &\text{otherwise}, \end{cases} \end{align*}$$
   where $Q = \{\sigma \in T^r\backslash T^q: h^r(\sigma ) \geq \omega \nu +\omega ^2\}$ .
   

   Figure 3 p and r lie in G, while q “looks like” it lies in G.

Next, define $r^\ast $ as follows:

1. – $T^{r^\ast } = T^q \cup T^r$ ;

2. – $f^{r^\ast }(i) = \begin {cases} f^r(i), & \text {if }i \in \mathrm {dom}(f^r)\backslash F; \\ f^G_i \cap T^{r^\ast }, & \text {if }i \in F; \\ g_{r^\ast }(i), & \text {if } i \in \mathrm {dom}(g_{r^\ast }); \end {cases}$

3. – $h^{r^\ast } = h^G\restriction T^{r^\ast }$ .

Finally, define $s^\ast $ as follows:

1. – $T^{s^\ast } = T^q \cup T^r$ ;

2. – $f^{s^\ast } = f^{q^\ast } \cup f^{r^\ast }$ ;

3. – $h^{s^\ast }(\sigma ) = \begin {cases} h^G(\sigma ), &\text {if } h^G(\sigma ) < \omega \nu + \omega ^2; \\ \infty , &\text {otherwise.} \end {cases}$

One may verify that:

1. – $q^\ast $ is a condition which extends q;

2. – $r^\ast $ is a condition which extends r;

3. – $s^\ast $ is a condition which extends p;

4. – $q^\ast $ and $s^\ast $ are $(\omega \nu +\omega ^2)$ - $\mathrm {dom}(f^{q^\ast })$ -retaggings;

5. – $r^\ast $ and $s^\ast $ are $(\omega \nu +\omega ^2)$ - $\mathrm {dom}(f^{r^\ast })$ -retaggings.

The verification is the same as that for ordinary Steel forcing, with minor additions in order to handle the new paths contributed by $g_{q^\ast }$ and $g_{r^\ast }$ . We only discuss $g_{q^\ast }$ here; the argument for $g_{r^\ast }$ is similar. First, note that $g_{q^\ast }$ was defined to ensure that $q^\ast $ satisfies (C2). Second, (E3) is satisfied for $q^\ast \leq q$ because for each $i \in \mathrm {dom}(f^{q^\ast })\backslash \mathrm {dom}(f^q)$ ( $=\mathrm {dom}(g_{q^\ast })$ ), we have $g_{q^\ast } \notin T^q$ and $|g_{q^\ast }(i)| = 1$ .

With the above facts, we may complete the proof that if $M_\infty \models \varphi (\textbf {d})$ , then $d \notin D$ . Since $q^\ast \leq q$ and $q \Vdash \psi ^\mu (\textbf {d})$ , we have $q^\ast \Vdash \psi ^\mu (\textbf {d})$ . It follows from Lemma 3.11 that $s^\ast \Vdash \psi ^\mu (\textbf {d})$ (note $F \subseteq \mathrm {dom}(f^{q^\ast })$ so $\psi ^\mu (\mathbf {d})$ is $\Sigma $ -over- $\mathcal {L}_{\mathrm {dom}(f^{q^\ast })}$ ). Similarly, we have $s^\ast \Vdash \varphi ^\mu (\mathbf {d})$ . But since $M_\infty \models \forall n(\psi ^\mu (n) \to \psi (n))$ and $M_\infty \models \forall n(\varphi ^\mu (n) \to \varphi (n))$ , it follows that $s^\ast \Vdash \psi (\mathbf {d}) \land \varphi (\mathbf {d})$ . This contradicts the fact that $p \Vdash \forall n(\psi (n) \leftrightarrow \neg \varphi (n))$ and $s^\ast \leq p$ .

This completes the proof that $M_\infty $ satisfies $\Delta ^1_1$ - $\mathsf {CA}_0$ .