Claim 3.6. If for every $\bar \alpha \in [B]^k$ , $p_{\bar {\alpha }} \Vdash \bar z_{\bar \alpha } \in \dot {\mathcal {A}}$ , then (1) of the main lemma holds true.

Proof. Let $\bar x_0, \dots , \bar x_{n-1}$ be arbitrary mCg over M. Say $\{x_i(l) : i < n\}$ is enumerated by $\langle y^i_l : i < K_l \rangle $ for every $l<k$ . Now find

$$ \begin{align*}\alpha^0_0 < \dots < \alpha^{K_0 -1}_0 < \dots < \alpha^{0}_{k-1} < \dots < \alpha^{K_{k-1} -1}_{k-1}\end{align*} $$

in $M \cap B$ . Then there is a $\mathbb {Q}$ -generic G over M so that for any $\bar j \in \prod _{l < k} K_l$ ,

$$ \begin{align*}\bar z_{\bar \beta}[G] = (y^{j_0}_0, \dots, y^{j_{k-1}}_{k-1}),\end{align*} $$

where $\bar \beta = (\alpha ^{j_0}_0, \dots , \alpha ^{j_{k-1}}_{k-1})$ . In particular, for each $i<n$ , there is $\bar \beta _i \in [B \cap M]^k$ so that $\bar z_{\bar \beta _i}[G]= \bar x_{i}$ . Since for every $\bar \beta _i$ we have that

$$ \begin{align*}M[G] \models \bar x_i \in \dot{\mathcal{A}}[G]\end{align*} $$

for every $i< n$ and in particular

$$ \begin{align*}M[G] \models \{ \bar x_i : i < n \} \text{ is } E \text{-independent}.\end{align*} $$

By absoluteness $\{ \bar x_i : i < n \}$ is indeed E-independent.⊣

Claim 3.7. For any $\bar \alpha \in [B]^{2k}$ and $\bar \gamma , \bar \delta \in \prod _{l<k} \{\alpha _{2l}, \alpha _{2l +1} \}$ ,

$$ \begin{align*}p_{\bar\gamma} \restriction (\operatorname{\mathrm{dom}} p_{\bar\gamma} \cap \operatorname{\mathrm{dom}} p_{\bar\delta}) = p_{\bar\delta} \restriction (\operatorname{\mathrm{dom}} p_{\bar\gamma} \cap \operatorname{\mathrm{dom}} p_{\bar\delta}).\end{align*} $$

Proof. Suppose not. By homogeneity we find a counterexample $\bar \alpha $ , $\bar \gamma $ , $\bar \delta $ where $B \cap (\alpha _{2l'},\alpha _{2l' +1})$ is non-empty for every $l' <k$ . So let $(l,\beta ) \in \operatorname {\mathrm {dom}} p_{\bar \gamma } \cap \operatorname {\mathrm {dom}} p_{\bar \delta }$ such that $p_{\bar \gamma }(l,\beta ) = u \neq v = p_{\bar \delta }(l,\beta )$ . Let $\bar \rho \in [B]^k$ be such that for every $l' < k$ ,

$$ \begin{align*}\begin{cases} \rho_{l'} \in (\gamma_{l'}, \delta_{l'} ), & \text{if } \gamma_{l'} < \delta_{l'}, \\ \rho_{l'} \in (\delta_{l'}, \gamma_{l'} ), & \text{if } \delta_{l'} < \gamma_{l'}, \\ \rho_{l'} = \gamma_{l'}, & \text{if } \gamma_{l'} = \delta_{l'}. \end{cases}\end{align*} $$

Now note that $\bar \rho $ ’s relative position to $\bar \gamma $ is the same as that of $\bar \delta $ to $\bar \gamma $ . More precisely, let $\bar j, \bar j' \in \prod _{l'<k} \{2l', 2l'+1 \}$ so that $\bar \gamma = (\alpha _{j_0}, \dots , \alpha _{j_{k-1}})$ and $\bar \delta = (\alpha _{j^{\prime }_0}, \dots , \alpha _{j^{\prime }_{k-1}})$ . Then there is $\bar \beta \in [B]^{2k}$ so that $\bar \gamma = (\beta _{j_0}, \dots , \beta _{j_{k-1}})$ and $\bar \rho = (\beta _{j^{\prime }_0}, \dots , \beta _{j^{\prime }_{k-1}})$ . Thus by homogeneity of $[B]^{2k}$ via c, $p_{\bar \rho }(l,\beta ) = v$ . Similarly $\bar \delta $ is in the same position relative to $\bar \rho $ as to $\bar \gamma $ . Thus also $p_{\bar \rho }(l,\beta ) = u$ and we find that $v = u$ —we get a contradiction.⊣

Claim 3.8. For any $l<k$ and $\bar j,\bar j' \in \prod _{l'<k} \{2l', 2l'+1 \}$ , $e_{l,\bar j, \bar j'}(m) = m$ for every $m \in \operatorname {\mathrm {dom}} e_{l,\bar j, \bar j'} $ .

Proof. Let $\alpha _0 < \dots < \alpha _{2k} \in B$ so that $(\alpha _{2l'},\alpha _{2l'+1}) \cap B \neq \emptyset $ for every $l' <k$ . Consider $\bar \gamma = (\alpha _{j_{l'}})_{l' < k}$ , $\bar \delta = (\alpha _{j^{\prime }_{l'}})_{l' < k}$ and again we find $\bar \rho \in [B]^k$ so that $\rho _{l'}$ is between (possibly equal to) $\alpha _{j_{l'}}$ and $\alpha _{j^{\prime }_{l'}}$ . If $e_{l,\bar j, \bar j'}(m) = m'$ , then if $\beta $ is the m-th element of $H_l(\bar \gamma )$ , then $\beta $ is $m'$ -th element of $H_l(\bar \delta )$ as well as of $H_l(\bar \rho )$ . But also $\beta $ is the m-th element of $H_l(\bar \rho )$ , thus $m =m'$ .⊣

Claim 3.9. $E_{l,m}$ is an equivalence relation.

Proof. The reflexivity and symmetry of $E_{l,m}$ is obvious. Assume that $\bar x_0 E_{l,m} \bar x_1$ and $\bar x_1 E_{l,m} \bar x_2$ , and say $\bar x_0 R_{i_0} \bar x_1$ , $\bar x_1 R_{i_1} \bar x_2$ , and $\bar x_0 R_{i_2} \bar x_2$ . Find $\bar \gamma ^0, \bar \gamma ^1, \bar \gamma ^2 \in [B]^k$ so that

$$ \begin{align*}\{\gamma^i_0 : i <3 \} < \dots < \{\gamma^i_{k-1} : i <3 \}\end{align*} $$

and

$$ \begin{align*}\bar \gamma^0 R_{i_0} \bar \gamma^1, \bar \gamma^1 R_{i_1} \bar \gamma^2, \bar \gamma^0 R_{i_2} \bar \gamma^2.\end{align*} $$

If $\beta $ is the m-th element of $H_l(\bar \gamma ^0)$ , then $\beta $ is also the m-th element of $H_l(\bar \gamma ^1)$ , since we can find an appropriate $\bar \alpha \in [B]^{2k}$ and $\bar j$ , $\bar j'$ so that $\bar \gamma ^0 = (\alpha _{j_l})_{l<k}$ and $\bar \gamma ^1 = (\alpha _{j^{\prime }_l})_{l<k}$ , $\bar j R_{i_0} \bar j'$ and we have that $m \in I_{l,i_0}$ . Similarly $\beta $ is the m-th element of $H_l(\bar \gamma ^2)$ .

But now we find again $\bar \alpha \in [B]^{2k}$ and $\bar j$ , $\bar j'$ so that $\bar \gamma ^0 = (\alpha _{j_l})_{l<k}$ and $\bar \gamma ^2 = (\alpha _{j^{\prime }_l})_{l<k}$ . Thus $m \in I_{l,i_2}$ , as $e_{l,\bar j, \bar j'}(m) = m$ and $\bar x_0 E_{l,m} \bar x_2$ .⊣

Claim 3.10. $E_{l,m}$ is smooth as witnessed by a continuous function, i.e., there is a continuous map $\varphi _{l,m} \colon (2^{\omega })^k \to 2^{\omega }$ so that $\bar x E_{l,m} \bar y$ iff $\varphi _{l,m}(\bar x) = \varphi _{l,m}(\bar y)$ .

Proof. We will check the following:

1. (a) For every open $O \subseteq (2^\omega )^k$ , the $E_{l,m}$ saturation of O is Borel,

2. (b) every $E_{l,m}$ equivalence class is $G_\delta $ .

By a theorem of Srivastava [Reference Srivastava31, Theorem 4.1.], (a) and (b) imply that $E_{l,m}$ is smooth, i.e., we can find $\varphi _{l,m}$ Borel.

1. (a) The $E_{l,m}$ saturation of O is the set $\{\bar x : \exists \bar y \in O (\bar x E_{l,m} \bar y) \}$ . It suffices to check for each $g \in \{-1,0,1\}^k$ that the set $X = \{\bar x : \exists \bar y \in O (\bar x \tilde R_g \bar y) \}$ is Borel. Let $\mathcal {S} = \{\bar \sigma \in (2^{<\omega })^k : [\sigma _0] \times \dots \times [\sigma _{k-1}] \subseteq O \}$ . Consider

   $$ \begin{align*}\varphi(\bar x) :\leftrightarrow \exists \bar \sigma \in \mathcal{S} \forall l' < k \begin{cases} x_{l'} <_{\operatorname{\mathrm{lex}}} {\sigma_{l'}}^{\frown} 0^\omega, &\text{ if } g(l') = {-1}, \\ x_{l'} \in [\sigma_{l'}], &\text{ if } g(l') = 0, \\ {\sigma_{l'}}^{\frown} 1^\omega <_{\operatorname{\mathrm{lex}}} x_{l'}, &\text{ if } g(l') = 1.\end{cases}\end{align*} $$
   If $\varphi (\bar x)$ holds true then let $\bar \sigma $ witness this. We then see that there is $\bar y \in [\sigma _0] \times \dots \times [\sigma _{k-1}]$ with $\bar x \tilde R_g \bar y$ . On the other hand, if $\bar y \in O$ is such that $\bar x R_g \bar y$ , then we find $\bar \sigma \in \mathcal {S}$ defining a neighborhood of $\bar y$ witnessing $\varphi (\bar x)$ . Thus X is defined by $\varphi $ and is thus Borel.

2. (b) Since finite unions of $G_\delta $ ’s are $G_\delta $ it suffices to check that $\{ \bar x : \bar x \tilde R_g \bar y \}$ is $G_\delta $ for every $\bar y$ and $g \in \{-1,0,1\}^k$ . But this is obvious from the definition.

Now note that given $\varphi _{l,m}$ Borel, we can find perfect $X_0,\dots ,X_{k-1} \subseteq 2^{\omega }$ so that $\varphi _{l,m}$ is continuous on $X_0 \times \dots \times X_{k-1}$ ( $\varphi _{l,m}$ is continuous on a dense $G_\delta $ ). But there is a $<_{\operatorname {\mathrm {lex}}}$ preserving homeomorphism from $X_l$ to $2^{\omega }$ for each $l<k$ so we may simply assume $X_l = 2^\omega $ .⊣

Claim 3.11. (2) of the main lemma holds true with M, $\bar s$ , and $\phi _i$ , $i <N$ , that we just defined.

Proof. Let $\bar x_0, \dots , \bar x_{n-1} \in [\bar s]$ be $\langle 2^\omega : l < k\rangle $ -mCg over M. Let us write $\{ \bar x_i(l) : i <n \} = \{ y_l^i : i < K_l \}$ for every $l < k$ , where $y_l^0 <_{\operatorname {\mathrm {lex}}} \dots <_{\operatorname {\mathrm {lex}}} y_l^{K_l -1} $ . Now find

$$ \begin{align*}\alpha^0_0 < \dots < \alpha_0^{K_0 -1} < \dots < \alpha_{k-1}^0 < \dots < \alpha_{k-1}^{K_{k-1}-1}\end{align*} $$

in $B \cap M$ . For every $\bar j \in \prod _{l < k} K_l$ , define $\bar y_{\bar j} := (y_0^{j(0)}, \dots , y_{k-1}^{j(k-1)})$ and $\bar \alpha _{\bar j} := (\alpha _0^{j(0)}, \dots , \alpha _{k-1}^{j(k-1)})$ . Then, for each $i < n$ , we have $\bar j_i \in \prod _{l < k} K_l$ so that $\bar x_i = \bar y_{\bar j_i}$ . For each $i < n$ define the function $g_i \colon \bigcup _{l < k} \{ l\} \times H_l(\bar \alpha _{\bar j_i}) \to 2^\omega $ , setting

$$ \begin{align*}g_i(l,\beta) = \chi_{l,m}(\varphi_{l,m}(\bar x_{i})),\end{align*} $$

whenever $\beta $ is the m-th element of $H_l(\bar \alpha _{\bar j_i})$ .

Now we have that the $g_i$ agree on their common domain. Namely let $i_0, i_1 < n$ and $(l,\beta ) \in \operatorname {\mathrm {dom}} g_{i_0} \cap \operatorname {\mathrm {dom}} g_{i_1}$ . Then if we set i to be so that $\bar x_{i_0} R_i \bar x_{i_1}$ , we have that $m \in I_{l,i}$ , where $\beta $ is the m-th element of $H_{l}(\bar \alpha _{\bar j_{i_0}})$ and of $H_{l}(\bar \alpha _{\bar j_{i_1}})$ . In particular $\bar x_{i_0} E_{l,m} \bar x_{i_1}$ and $\varphi _{l,m}(\bar x_{i_0}) = \varphi _{l,m}(\bar x_{i_1})$ and thus

$$ \begin{align*}g_{i_0}(l,\beta) = \chi_{l,m}(\varphi_{l,m}(\bar x_{i_0})) = \chi_{l,m}(\varphi_{l,m}(\bar x_{i_1})) = g_{i_1}(l,\beta).\end{align*} $$

Let $g := \bigcup _{i < n} g_i$ . Then we see by Lemma 3.3 that g is Cohen generic in $\prod _{(l,\beta ) \in \operatorname {\mathrm {dom}} g} 2^\omega $ over $M_0$ . Namely consider $K = \sum _{l < k} K_l$ and $(y^0_0, \dots , y_{k-1}^{K_{k-1}-1})$ as a $(2^{<\omega })^K$ -generic over M. Then, if $\langle u_i : i < n'\rangle $ enumerates $\{ \varphi _{l,m}(\bar x_i) : i < n, l<k, m < N_l \}$ , we have that every value of g is contained in $\{ \chi _{l,m}(u_i) : i < n', l <k, m<N_l \}$ . Also note that by construction for every $i < n$ , $p_{\bar \alpha _{\bar j_i}} \restriction \operatorname {\mathrm {dom}} g$ is in the generic filter defined by g. Since $\{ p_{\bar \alpha _{\bar j_i}} : i < n \}$ is centered we can extend the generic filter of g to a $\mathbb {Q}$ -generic G over $M_0$ so that $p_{\bar \alpha _{j_i}} \in G$ for every $i < n$ .

Now we have that

$$ \begin{align*}\bar z_{\bar \alpha_{\bar j_i}}[G] = \bar x_i \text{ and } \bar z_{\bar \beta^{j}(\bar \alpha_{\bar j_i})}[G] = \phi_j(\bar x_{i})\end{align*} $$

for every $i < n$ and $j < N$ . Thus we get that

$$ \begin{align*}M_0[G] \models \bigcup_{i < n} \{ \phi_j(\bar x_i) : j < N \} \subseteq \dot{\mathcal{A}}[G] \wedge \{ \bar x_0 \} \cup \{\phi_j(\bar x_0) : j < N \} \in E.\end{align*} $$

Again, by absoluteness, we get the required result.⊣

Claim 3.15. $(1)_{\tilde R, M, 1}$ is satisfied.

Proof. Suppose $\bar y_0, \dots , \bar y_{n-1}$ are pairwise distinct and strongly mCg over M, but $\{(\bar y_0{}, 0), \dots ,(\bar y_{n-1}, 0) \} \in \tilde R$ as witnessed by $p \in \mathbb {Q}$ , $\langle K_i : i < n \rangle $ and $\langle k_{i,j} : i < n, j < K_i \rangle $ , each $k_{i,j} < k$ . More precisely,

(*0) $$\begin{align} p \Vdash_{\mathbb{Q}} \bigcup_{i<n} \{ (\bar y_i{}^{\frown} \dot z_{0,i,j}, k_{i,j}) : j < K_i \} \in R. \end{align}$$

By absoluteness, $(*_0)$ is satisfied in $M[\bar y_0, \dots , \bar y_{n-1}]$ . Thus, let $\langle z_{0,i,j}, z_{1,i,j} : i, j \in \omega \rangle $ be generic over $M[\bar y_0, \dots , \bar y_{n-1}]$ with p in the associated generic filter. Then the $\bar y_i{}^{\frown } z_{(0,i,j)}$ for $i < n, j < K_i$ are pairwise distinct and strongly mCg over M, but

$$ \begin{align*}\bigcup_{i<n} \{ (\bar y_i{}^{\frown} z_{0,i,j}, k_{i,j}) : j < K_i \} \in R.\end{align*} $$

This poses a contradiction to $(1)_{R,M,k}$ .⊣

Claim 3.16. If $(1)_{\tilde R, M, m}$ is satisfied for every $m \in \omega $ , then also $(1)_{R,M,k+1}$ .

Proof. Let $\bar x_0, \dots , \bar x_{n-1} \in (2^\omega )^\alpha $ be pairwise distinct, strongly mCg over M and let $k_0, \dots , k_{n-1} \leq k$ . Then we may write $\{ (\bar x_0, k_0), \dots , (\bar x_{n-1}, k_{n-1}) \}$ as

(*1) $$\begin{align}\bigcup_{i < n'} \{ (\bar y_i{}^{\frown} z_{0,i,j}, k_{i,j}) : j < K_i \} \cup \{ (\bar y_i{}^{\frown} z_{1,i,j}, k) : j < m_i \},\end{align}$$

for some pairwise distinct $\bar y_0, \dots ,\bar y_{n'-1}$ , $\langle K_i : i < n' \rangle $ , $\langle k_{i,j} : i < n', j<K_i \rangle $ , $\langle m_i : i < n' \rangle $ and $\langle z_{0,i,j} : i, j \in \omega \rangle $ , $\langle z_{1,i,j} : i, j \in \omega \rangle 2^\omega $ -mutually Cohen generic over $M[\bar y_0, \dots , \bar y_{n'-1}]$ . Letting $m = \max _{i < n'} m_i +1$ , we follow the R-independence of the set in $(*_1)$ from $(1)_{\tilde R,M,m}$ .⊣

Claim 3.17. If there is $m \in \omega $ so that $(1)_{\tilde R, M, m}$ fails, then there is a countable model $M^+ \supseteq M$ so that $(2)_{R,M^+,k}$ .

Proof. Let $m\geq 1$ be least so that $(2)_{\tilde R, M_0, m}$ for some countable model $M_0 \supseteq M$ . We know that such m exists, since from $(1)_{\tilde R, M, 1}$ we follow that either $(1)_{\tilde R, M, 2}$ or $(2)_{\tilde R, M_0,1}$ for some $M_0$ , then, if $(1)_{\tilde R, M, 2}$ , either $(1)_{\tilde R, M, 3}$ or $(2)_{\tilde R, M_0, 2}$ for some $M_0$ , and so on. Let $\phi _0, \dots , \phi _{N-1}$ , $m_0, \dots , m_{N-1} \leq m$ and $\bar s \in \bigotimes _{i < \beta }(2^{<\omega })$ witness $(2)_{\tilde R, M_0, m}$ . Let $M_1$ be a countable elementary model such that $\phi _0, \dots , \phi _{N-1}, M_0 \in M_1$ . Then we have that for any $\bar y$ that is Cohen generic in $(2^\omega )^\beta \cap [\bar s]$ over $M_1$ , in particular over $M_0$ , that

$$ \begin{align*}\{(\bar y, m) \} \cup \{ (\phi_j(\bar y), m_j) : j <N \} \in \tilde R,\end{align*} $$

i.e., there is $p \in \mathbb {Q}$ , there are $K_i \in \omega $ , $k_{i,0}, \dots , k_{i,K_i -1} < k$ for every $i \leq N$ , so that

By extending $\bar s$ , we can assume wlog that p, $\langle K_i : i \leq N \rangle $ , $\langle k_{i,j} : i \leq N, j < K_i \rangle $ are the same for each $\bar y \in [\bar s]$ generic over $M_1$ , since $(*_2)$ can be forced over $M_1$ . Also, from the fact that $\phi _j$ is continuous for every $j < N$ , that $\phi _j(\bar y) \neq \bar y$ for every $j < N$ , and that $\phi _{j_0}(\bar y) \neq \phi _{j_1}(\bar y)$ for every $j_0 < j_1 < N$ , we can assume wlog that for any $\bar y_0, \bar y_1 \in [\bar s]$ and $j_0 < j_1 < N$ ,

(*3) $$\begin{align} \phi_{j_0}(\bar y_0) \neq \bar y_1 \text{ and } \phi_{j_0}(\bar y_0) \neq \phi_{j_1}(\bar y_1). \end{align}$$

Let us force in a finite support product over $M_1$ continuous functions $\chi _{0,i,j} \colon (2^\omega )^\beta \to [p(0,i,j)]$ and $\chi _{1,i,j} \colon (2^\omega )^\beta \to [p(1,i,j)]$ for $i,j \in \omega $ and write $M^+ = M_1[\langle \chi _{0,i,j}, \chi _{1,i,j} : i,j \in \omega \rangle ]$ . For every $i < N$ and $j < K_i$ and $\bar x \in (2^\omega )^\alpha $ , define

$$ \begin{align*}\phi_{0,i,j}(\bar x ) := \phi_i(\bar x \restriction \beta){}^{\frown} \chi_{0,i,j}(\phi_i(\bar x \restriction \beta)) \text{ and } k_{0,i,j} = k_{i,j}.\end{align*} $$

For every $i< N$ and $j < m_i$ and $\bar x \in (2^\omega )^\alpha $ , define

$$ \begin{align*}\phi_{1,i,j}(\bar x ) := \phi_i(\bar x \restriction \beta){}^{\frown} \chi_{1,i,j}(\phi_i(\bar x \restriction \beta)) \text{ and } k_{1,i,j} = k.\end{align*} $$

For every $j < K_{N}$ and $\bar x \in (2^\omega )^\alpha $ , define

$$ \begin{align*}\phi_{0,N,j}(\bar x ) := \bar x \restriction \beta{}^{\frown} \chi_{0,N,j}(\bar x \restriction \beta) \text{ and } k_{0,N,j} = k_{N,j}.\end{align*} $$

At last, define for every $j < m-1$ and $\bar x \in (2^\omega )^\alpha $ ,

$$ \begin{align*}\phi_{1,N,j}(\bar x ) := \bar x \restriction \beta{}^{\frown} \chi_{1,N,j}(\bar x \restriction \beta) \text{ and } k_{1,N,j} = k.\end{align*} $$

Let $\bar t \in \bigotimes _{i < \alpha } 2^{<\alpha }$ be $\bar s$ with $p(1,N,m-1)$ added in coordinate $\beta $ . Now we have that for any $\bar x \in [\bar t]$ that is Cohen generic in $(2^\omega )^\alpha $ over $M^+$ ,

$$ \begin{align*} \{ (\bar x, k) \} &\cup \{ (\phi_{0,i,j}(\bar x), k_{0,i,j}) : i \leq N, j < K_N \} \cup \{ (\phi_{1,i,j}(\bar x), k_{1,i,j}) : i < N, j < m_i\} \\ &\cup \{(\phi_{1,N,j}(\bar x), k_{1,N,j}) : j < m -1 \} \in R. \end{align*} $$

This follows from $(*)_2$ and applying Lemma 3.3 to see that the $\chi _{0,i,j}(\phi _i(\bar x \restriction \beta ))$ , $\chi _{1,i,j}(\phi _i(\bar x \restriction \beta ))$ , $\chi _{0,N,j}(\bar x \restriction \beta )$ , $\chi _{1,N,j}(\bar x \restriction \beta )$ , and $x(\beta )$ are mutually Cohen generic over $M_1[\bar x \restriction \beta ]$ . Moreover they correspond to the reals $z_{0,i,j}$ , $z_{1,i,j}$ added by a $\mathbb {Q}$ -generic over $M_1[\bar x \restriction \beta ]$ , containing p in its generic filter. Also, remember that $(*)_2$ is absolute between models containing the relevant parameters, which $M_1[\bar y]$ is, with $\bar y = {\bar x \restriction \beta }$ .

On the other hand, whenever $\bar x_0, \dots , \bar x_{n-1} \in (2^\omega )^\alpha \cap [\bar t]$ are pairwise distinct and strongly mCg over $M^+$ , letting $\bar y_0, \dots , \bar y_{n'-1}$ enumerate $\{\bar x_i \restriction \beta : i < n \}$ , we have that

(*4) $$\begin{align} \{ (\bar y_i{},m-1) : i < n' \} \cup \{ (\phi_j(\bar y_i), m_j) : i < n', j < N \} \text{ is } \tilde R \text{-independent.} \end{align}$$

According to the definition of $\tilde R$ , $(*_4)$ is saying, e.g., that whenever $A \cup B \subseteq (2^\omega )^\alpha $ is an arbitrary set of strongly mCg reals over $M_1$ , where $A \restriction \beta , B \restriction \beta \subseteq \{ \bar y_i, \phi _j(\bar y_i) : i < n', j < N \}$ and in B, $\bar y_i$ is extended at most $m-1$ many times and $\phi _j(\bar y_i)$ at most $m_j$ many times for every $i < n'$ , $j < N$ , and, assuming for now that $k> 0$ , if $f \colon A \to k$ , then

$$ \begin{align*}\{ (\bar x{}, f(\bar x)) : \bar x \in A \} \cup (B \times \{k\}) \text{ is } R\text{-independent.}\end{align*} $$

As an example for such sets A and B, we have

$$ \begin{align*}A = \{\phi_{0,i,j}(\bar x_l) : l < n, i \leq N, j < K_i \} \cup \{ \bar x_l : l < n' \} \text{ and}\end{align*} $$

$$ \begin{align*}B = \{ \phi_{1,i,j}(\bar x_l) : l < n, i < N, j < m_i \} \cup \{ \phi_{1,N,j}(\bar x_l) : l < n', j < m-1 \}.\end{align*} $$

Again, to see this we apply Lemma 3.3 to show that the relevant reals are mutually generic over the model $M_1[\bar y_0, \dots , \bar y_{n'-1}]$ . Also, remember from the definition of $\phi _{1,i,j}$ for $i < N$ and $j < m_i$ that, if $\phi _i(\bar x_{l_0} \restriction \beta ) = \phi _i(\bar x_{l_1} \restriction \beta )$ , then also $\phi _{1,i,j}(\bar x_{l_0}) = \phi _{1,i,j}(\bar x_{l_1})$ , for all $l_0, l_1 < n$ . Equally, if $\bar x_{l_0} \restriction \beta = \bar x_{l_1} \restriction \beta $ , then $\phi _{1,N,j}(\bar x_{l_0}) = \phi _{1,N,j}(\bar x_{l_1})$ for every $j < m-1$ . Use $(*_3)$ to note that $\{ \bar y_i : i < n'\}$ , $\{\phi _0(\bar y_i) : i < n' \}, \dots $ , ${\{ \phi _{N-1}(\bar y_i) : i < n' \}}$ are pairwise disjoint. From this we can follow that indeed, each $\bar y_i$ is extended at most $m-1$ many times in B and $\phi _j(\bar y_i)$ at most $m_i$ many times. In total, we get that

$$ \begin{align*} &\{(\phi_{0,i,j}(\bar x_l), k_{0,i,j}) : l < n, i \leq N, j < K_i \} \cup \{ (\bar x_l,k-1) : l < n' \} \cup \\ &\{ (\phi_{1,i,j}(\bar x_l), k) : l < n, i < N, j < m_i \} \cup \{ (\phi_{1,N,j}(\bar x_l), k) : l < n', j < m-1 \} \\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\text{ is } R\text{-independent}. \end{align*} $$

It is now easy to check that we have the witnesses required in the statement of $(2)_{R,M^+,k}$ . For example, $\phi _{0,i,j}(\bar x) \neq \bar x$ when $i < N$ follows from $\phi _i(\bar x) \neq \bar x$ . For the values $\phi _{0,N,j}(\bar x)$ we simply have that $\chi _{0,N,j}(\bar x \restriction \beta ) \neq x(\beta )$ , as the two values are mutually generic. Everything else is similar and consists only of a few case distinctions. Also, the continuity of the functions is clear.

If $k = 0$ , then we can simply forget the set A above, since $K_i$ must be $0$ for every $i \leq N$ . In this case we just get that

$$ \begin{align*} \{ (\phi_{1,i,j}(\bar x_l), k) : l < n, i < N, j < m_i \} \cup \{ (\phi_{1,N,j}(\bar x_l), k) &: l < n', j < m-1 \} \\ &\quad\text{ is } R\text{-independent}, \end{align*} $$

which then yields $(2)_{R,M^+,k}$ .⊣

Claim 3.18. For every $\xi < \alpha $ , $(1)_{R_\xi ,M,1}$ .

Claim 3.19. Assume that for every $\xi < \alpha $ , $(1)_{R_\xi ,M,2}$ . Then also $(1)_{R,M,k+1}$ .

Proof. Let $\bar x^0_0, \dots , \bar x^0_{n_0-1}, \bar x^1_0, \dots , \bar x^1_{n_1-1}$ be pairwise distinct and strongly mCg over M and $k_0, \dots , k_{n_0 -1} < k$ . Then there is $\xi < \alpha $ large enough so that $\bar x^0_0 \restriction \xi , \dots , \bar x^0_{n_0-1}\restriction \xi , \bar x^1_0\restriction \xi , \dots , \bar x^1_{n_1-1}\restriction \xi $ are pairwise distinct and in particular, $\bar x^0_0 \restriction [\xi ,\alpha ), \dots , \bar x^0_{n_0-1}\restriction [\xi , \alpha ) , \bar x^1_0\restriction [\xi , \alpha ), \dots , \bar x^1_{n_1-1}\restriction [\xi , \alpha )$ are pairwise different in every coordinate. Let $\xi _0 = \xi $ , $\xi _1 = \alpha $ , $K_i = 1$ for every $i < n_0$ and $t_{0,0}, \dots , t_{n_0 -1,0} \in T \cap \omega ^1$ pairwise distinct. Also, write $k_{0,0} = k_0, \ldots , k_{n_0-1,0} = k_{n_0 -1}$ . Then, from $(1)_{R_\xi ,M,2}$ , we have that

By absoluteness, this holds true in $M[\langle \bar x^0_i \restriction \xi ,\bar x^1_j\restriction \xi : i < n_0, j < n_1\rangle ]$ and we find that

$$ \begin{align*}\{(\bar x^0_i{}, k_i) : i < n_0 \} \cup \{ (\bar x^1_i, k) \} \text{ is } R\text{-independent},\end{align*} $$

as required.⊣

Claim 3.20. If there is $\xi < \alpha $ so that $(1)_{R_\xi ,M,2}\,$ fails, then there is a countable model $M^+ \supseteq M$ so that $(2)_{R,M^+,k}$ .

Proof. If $(1)_{R_\xi , M, 2}$ fails, then there is a countable model $M_0 \supseteq M$ so that $(2)_{R_\xi , M, 1}$ holds true as witnessed by $\bar s \in \bigotimes _{i < \xi } 2^{<\omega }$ , $\phi ^0_{0}, \dots , \phi ^0_{N_0 -1}, \phi ^1_{0}, \dots , \phi ^1_{N_1 -1} \colon (2^\omega )^\xi \to (2^\omega )^\xi $ such that for any pairwise distinct $\bar y_0, \dots , \bar y_{n-1} \in (2^\omega )^\xi \cap [\bar s]$ that are strongly mCg over $M_0$ ,

(*5) $$\begin{align} \{ (\bar y_i, 0) : i < n\} \cup \{(\phi^0_j(\bar y_i), 0) : i < n, j < N_0\} \cup \{ (\phi^1_j(\bar y_i), 1) : i < n, j < N_1\} \end{align}$$

is $R_\xi $ -independent, but

(*6) $$\begin{align} \{ (\bar y_0, 1) \} \cup \{ (\phi^0_j(\bar y_0), 0) : j < N_0\} \cup \{ (\phi^1_j(\bar y_0), 1) : j < N_1\} \in R_\xi. \end{align}$$

As before, we may pick $M_1 \ni M_0$ elementary containing all relevant information, assume that $(*_6)$ is witnessed by fixed $\xi _0 = \xi < \dots < \xi _{k'} = \alpha $ , $(p,q) \in \mathbb {Q}_{\xi _0, \dots , \xi _{k'}}$ , $K_0, \dots , K_{N_0-1}$ , $k_{i,0}, \dots , k_{i, K_i -1}$ , and $t_{i,0}, \dots , t_{i, K_i-1} \in T \cap \omega ^{k'}$ for every $i < N_0$ , so that for every generic $\bar y_0 \in (2^\omega )^\xi \cap [\bar s]$ over $M_1$ ,

(*7) $$\begin{align} (p,q) \Vdash_{\mathbb{Q}_{\bar \xi}} \{(\bar y_0{}^{\frown} \bar z^1_{N_1}, k) \}\cup \bigcup_{i < N_0}&\{ (\phi^0_i(\bar y_0){}^{\frown} \bar z^0_{t_{i,j}}, k_{i,j}): j < K_i\} \cup \\ &\{ (\phi^1_j(\bar y_0){}^{\frown} \bar z^1_{j}, k): j < N_1 \} \in R.\nonumber \end{align}$$

As before, we may also assume that $\bar y_0 \neq \phi ^{j_0}_{i_0}(\bar y_0) \neq \phi ^{j_1}_{i_1}(\bar y_1)$ for every $\bar y_0, \bar y_1 \in [\bar s]$ and $(j_0,i_1) \neq (j_1,i_1)$ . We let $\bar s' = \bar s{}^{\frown } q(N_1)$ . Now we force continuous functions $\chi ^0_{l,i} \colon (2^\omega )^\xi \to (2^{\omega })^{[\xi _l, \xi _{l+1})} \cap [p(l,i)]$ and $\chi ^1_{i} \colon (2^\omega )^\xi \to (2^{\omega })^{[\xi , \alpha )} \cap [q(i)]$ over $M_1$ for every $i \in \omega $ , $l < k'$ and we let $M^+ = M_1[\langle \chi ^0_{l,i}, \chi ^1_{i} : i \in \omega , l < k'\rangle ]$ . Finally we let

$$ \begin{align*}\phi_{0,i,j}(\bar x) := \phi^0_i(\bar x \restriction \xi){}^{\frown} \chi_{0,t_{i,j}(0)}{}(\phi^0_i(\bar x \restriction \xi)){}^{\frown} \cdots{}^{\frown} \chi_{k'-1,t_{i,j}(k'-1)}(\phi^0_i(\bar x \restriction \xi))\end{align*} $$

for every $i < N_0$ and $j < K_i$ , $\bar x \in (2^\omega )^\alpha $ , and

$$ \begin{align*}\phi_{1,i}(\bar x) := \phi^1_i(\bar x \restriction \xi){}^{\frown} \chi_{1,i}(\phi ^1_i(\bar x \restriction \xi))\end{align*} $$

for every $i < N_1$ , $\bar x \in (2^\omega )^\alpha $ .

We get from $(*_7)$ , and, as usual, applying Lemma 3.3, that for any $\bar x \in (2^\omega )^\alpha \cap [\bar s']$ which is generic over $M^+$ ,