Fact 2.9 (Hyttinen–Tuuri [Reference Hyttinen and Tuuri8, Lemma 8.17]).

• • The linear order $\mathcal {R}$ is $(<\kappa , bs)$ -stable and $(\kappa , bs,bs)$ -nice.

• • There is a $\kappa $ -representation $\langle R_\alpha \mathrel {|} \alpha <\kappa \rangle $ and a club $C\subseteq \kappa $ such that for all $\delta \in C$ and $\nu \in \mathcal {R}$ there is $\beta <\delta $ which satisfies the following:

  $$ \begin{align*}\forall\sigma\in R_\delta [\sigma>\nu \Rightarrow \exists\sigma'\in R_\beta\ (\sigma\ge\sigma'\ge\nu)].\end{align*} $$

Proof Let us start by defining the representation $\kappa $ -representation $\langle I^0_\alpha \mathrel {|} \alpha <\kappa \rangle $ . For all $\gamma <\kappa $ , let us define $\langle I^0_\alpha \mathrel {|} \alpha <\kappa \rangle $ by

$$ \begin{align*}I^0_\gamma=\{\nu\in I^0\mathrel{|} \nu_1(n)<\gamma\ \textit{for all } n<\omega\}\end{align*} $$

it is clear that $\langle I^0_\alpha \mathrel {|} \alpha <\kappa \rangle $ is a $\kappa $ -representation.

Let us show item II, i.e., $\nu \notin I^0_\delta $ .

Suppose $\nu \notin I^0_\delta $ . Let $\beta <\delta $ be $max\{\nu _1(i)\mathrel {|} i<n\}$ , where n is the least number such that $\nu _1(n)\ge \delta $ .

Claim 2.12.1. $\beta $ is as wanted, i.e.,

$$ \begin{align*}\forall\sigma\in I^0_\delta\ [\sigma>\nu \Rightarrow \exists\sigma'\in I^0_\beta\ (\sigma>\sigma'>\nu)].\end{align*} $$

Proof Let us suppose $\sigma \in I^0_\delta $ is such that $\sigma \ge \nu $ . By the definition of $I^0$ , there is $n<\omega $ such that $\sigma (n)>\nu (n)$ and n is the minimum number such that $\sigma (n)\neq \nu (n)$ . Since $\sigma \in I^0_\delta $ , for all $m\leq n$ , $\nu _1(m)\leq \sigma _1(m)<\delta $ . Thus for all $m\leq n$ , $\nu _1(m)<\beta $ .

Let us divide the proof in two cases, $\sigma _1(n)=\nu _1(n)$ and $\sigma _1(n)>\nu _1(n)$ .

Case 1. $\sigma _1(n)=\nu _1(n)$ .

By the density of $\mathbb {Q}$ there is r such that $\sigma _2(n)>r>\nu _2(n)$ . Let us define $\sigma '$ by

$$ \begin{align*}\sigma'(m)=\begin{cases} \nu(m), &\mbox{if } m<n,\\ (\nu_1(n), r), & \mbox{if } m=n,\\ 0, & \mbox{otherwise. } \end{cases}\end{align*} $$

Clearly $\sigma>\sigma '>\nu $ . Since $\nu _1(m)<\beta $ for all $m\leq n$ , $\sigma '\in I^0_\beta $ .

Case 2. $\sigma _1(n)>\nu _1(n)$ .

Let us define $\sigma '$ by

$$ \begin{align*}\sigma'(m)=\begin{cases} \nu(m), &\mbox{if } m<n,\\ (\nu_1(n), \nu_2(n)+1), & \mbox{if } m=n,\\ 0, & \mbox{otherwise. } \end{cases}\end{align*} $$

Clearly $\sigma>\sigma '>\nu $ . Since $\nu _1(m)<\beta $ for all $m\leq n$ , $\sigma '\in I^0_\beta $ .

The previous claim proves item II. From the proof of this claim we can see that $\sigma \neq \sigma '$ .

To prove item I, it is enough to prove the case $\nu \in I^0$ .

Suppose $\delta <\kappa $ is a limit and $\nu \in I^0_\delta $ . It is clear that there is $\beta <\delta $ such that $\nu \in I^0_\beta $ and the result follows.

Fact 2.14. Suppose $\nu \in I$ . For all $\sigma \in Gen(\nu )$ , $\sigma \neq \nu $ , there is $n<\omega $ and a sequence $\{\sigma _i\}_{i\leq n}$ such that the following holds:

• • $\sigma _0=\nu $ .

• • For all $j<n$ ,

  $$ \begin{align*}\sigma_{j+1}=(\sigma_j)^{o(\sigma_{j+1})}.\end{align*} $$

• • $\sigma =\sigma _n=(\sigma _{n-1})^{o(\sigma )}.$

Proof Let $\sigma \neq \nu $ be such that $\sigma \in Gen(\nu )$ . From Definition 2.13, we know that $\sigma \in Gen^{o(\sigma )}(\nu )$ and $Gen^{o(\sigma )}(\nu )\subseteq I^{o(\sigma )}$ . Let us proceed by induction on $o(\sigma )$ . Notice that $o(\nu )<o(\sigma )$ , so the induction starts with the case $o(\sigma )=o(\nu )+1$ . Since $o(\sigma )$ is a successor ordinal, the limit step of the induction is not required.

For the case $o(\sigma )=o(\nu )+1$ it is easy to see from Definition 2.13 that $\sigma =\nu ^{o(\nu )+1}$ . Thus $\sigma _0=\nu $ and $\sigma =\sigma _1=(\sigma _0)^{o(\sigma _{1})}$ is the desire sequence, and $n=1$ .

Let $o(\sigma )=i+1>o(\nu )+1$ be such that for any $\tau \in Gen^i(\nu )$ there are $n<\omega $ and a sequence $\{\tau _j\}_{j\leq n}$ such that the following holds:

• • $\tau _0=\nu $ ;

• • for all $j<n$ ,

  $$ \begin{align*}\tau_{j+1}=(\tau_j)^{o(\tau_{j+1})};\end{align*} $$

• • $\tau =\tau _n=(\tau _{n-1})^{o(\tau )}$ .

We know that $\sigma \in Gen^{o(\sigma )}(\nu )=Gen^{i+1}(\nu )$ . By Definition 2.13, there is $\tau \in Gen^i(\nu )$ such that $\tau ^{i+1}=\sigma $ . We conclude that $n+1<\omega $ and the sequence $\{\sigma _i\}_{i\leq n+1}$ defined by:

• • $\sigma _0=\tau _0=\nu $ ;

• • for all $j\leq n$ , $\sigma _j=\tau _j$ ;

• • $\sigma =\sigma _{n+1}=(\tau _{n})^{i+1}$

are as wanted.

Fact 2.16. Let $i,\delta , \nu $ be such that $\nu \in I^i_\delta $ . Then for all $\sigma \in Gen(\nu )$ , $\sigma \in I^{o(\sigma )}_\delta $ . In particular for all $j<\kappa $

$$ \begin{align*}\sigma\notin I^j_\delta\Rightarrow \sigma\notin I^j.\end{align*} $$

Proof It follows from the construction of $I^{o(\sigma )}$ and the $\kappa $ -representation $\langle I^{o(\sigma )}_\alpha \mathrel {|} \alpha <\kappa \rangle $ .

Fact 2.17. For all $\nu ,\sigma \in I$ , $\sigma \in Gen(\nu )$ , if $\sigma '\in I$ is such that $\nu \ge \sigma '\ge \sigma $ , then $\sigma '\in Gen(\nu )$ .

Proof If $\nu =\sigma $ , the result follows. Thus we only need to prove the case $\nu \neq \sigma $ . Let us suppose towards contradiction that $\sigma '\notin Gen(\nu )$ .

Case $o(\nu )=o(\sigma ')$ . Since $\nu $ and $\sigma $ have the same type of basic formulas over $I^{o(\nu )}\backslash \{\nu \}$ , $\nu $ and $\sigma $ have the same type of basic formulas over $I^{o(\sigma ')}\backslash \{\nu \}$ . Since $\nu \ge \sigma '\ge \sigma $ , $\nu =\sigma '$ a contradiction.

Case $o(\sigma ')<o(\nu )$ . Since $\nu \ge \sigma '$ , there is $\nu '\neq \sigma '$ such that $\nu '>\nu $ , $o(\nu ')=o(\sigma ')$ and $\nu \in Gen(\nu ')$ . Thus $\nu '$ , $\sigma '$ , and $\sigma $ satisfy $\nu '\ge \sigma '\ge \sigma $ , $o(\nu ')=o(\sigma ')$ , and $\sigma \in Gen(\nu ')$ . The result follows from the previous case.

Case $o(\nu )<o(\sigma ')$ . There is $\sigma ^0\in I$ such that $\sigma ^0>\sigma '$ , $o(\sigma ^0)=o(\nu )$ and $\sigma '\in Gen(\sigma ^0)$ . If $\nu \ge \sigma ^0\ge \sigma $ , then the result follows from the previous cases. Therefore, we are only missing the case $\sigma ^0\ge \nu \ge \sigma '\ge \sigma $ . Since $\sigma ^0$ and $\sigma '$ have the same type of basic formulas of basic formulas over $I^{o(\sigma ^0)}\backslash \{\sigma ^0\}$ , $\sigma ^0=\nu $ and $\sigma '\in Gen(\nu )$ a contradiction.

Proof Notice that if $\nu \in I^{i}_\delta $ , then there is $\theta <\delta $ such that $\nu \in I^{i}_\theta $ and the result follows for $\beta =\theta $ . So we only have to prove the lemma when $\nu \in I^i\backslash I^{i}_\delta $ (the second part of the lemma).

We will proceed by induction over i. The case $i=0$ is precisely Lemma 2.12 II. Let us suppose $i<\kappa $ is such that for all limit ordinal $\delta <\kappa $ and $\nu \in I^i\backslash I^{i}_\delta $ , there is $\beta <\delta $ that satisfies II. Let $\delta <\kappa $ be a limit ordinal and $\nu \in I^{i+1}\backslash I^{i+1}_\delta $ . We have two cases, $\nu \in I^i$ and $\nu \in I^{i+1}\backslash I^i$ .

Case $\nu \in I^i$ . By the induction hypothesis, we know that there is $\beta <\delta $ such that II holds. Let us prove that this $\beta $ is the one we are looking for. Let $\sigma \in I^{i+1}_\delta $ be such that $\sigma>\nu $ . The subcase $\sigma \in I^i_\delta $ follows from the way $\beta $ was chosen.

Subcase $\sigma \in I^{i+1}_\delta \backslash I^i_\delta $ . By the construction of $I^{i+1}$ , there is $\sigma _0\in I^i_\delta $ such that $\sigma =(\sigma _0)^{i+1}$ (so $\sigma _0>\sigma $ ). Thus $\sigma _0>\sigma >\nu $ , and by the way $\beta $ was chosen, there is $\sigma '\in I^0_\beta $ such that $\sigma _0> \sigma '> \nu $ . Since $\sigma _0$ and $\sigma $ have the same type of basic formulas over $I^i\backslash \{\sigma _0\}$ , $\sigma> \sigma '> \nu $ as we wanted.

Case $\nu \in I^{i+1}\backslash I^i$ . By the construction of $I^{i+1}$ , there is $\nu _0\in I^i$ such that $(\nu _0)^{i+1}=\nu $ . Since $\nu \in Gen(\nu _0)$ and $\nu \in I^{i+1}\backslash I^{i+1}_\delta $ , by Fact 2.16 $\nu _0\in I^i\backslash I^i_\delta $ . Thus, by the previous case, there is $\beta <\delta $ such that for all $\sigma \in I^{i+1}_\delta $ ,

$$ \begin{align*}\sigma>\nu_0 \Rightarrow \exists\sigma'\in I^0_\beta\ (\sigma>\sigma'>\nu_0).\end{align*} $$

Let us show that this $\beta $ is as wanted.

From the previous claim, we know that for all $\sigma \in I^{i+1}_\delta $ , $\sigma>\nu $ implies $\sigma>\nu _0$ . By the way $\beta $ was chosen we conclude that for all $\sigma \in I^{i+1}_\delta $ , $\sigma>\nu $ implies the existence of $\sigma '\in I^0_\beta $ such that $\sigma>\sigma '>\nu _0>\nu $ , as we wanted.

Let us proceed with the limit case. Suppose $i<\kappa $ is a limit ordinal such that for all $j<i$ , for all limit ordinal $\delta <\kappa $ , and $\nu \in I^j\backslash I^j_\delta $ , there is $\beta <\delta $ such that II holds for j. Let $\delta <\kappa $ be a limit ordinal and $\nu \in I^i\backslash I^i_\delta $ . Since i is a limit, $o(\nu )<i$ , by the induction hypothesis, there is $\beta $ such that II holds for $o(\nu )$ .

Proof Let $\delta <\kappa $ be a limit ordinal, and $\nu \in I$ . We have three different cases: $\nu \in I_\delta $ , $\nu \in I^{o(\nu )}_\delta \backslash I_\delta $ , and $\nu \notin I^{o(\nu )}_\delta $ .

Case $\nu \in I_\delta $ . Since $\delta $ is a limit, $o(\nu )<\delta $ and there is $\theta <\delta $ such that $\nu \in I^{o(\nu )}_\theta $ . Let $\beta =max\{o(\nu ),\theta \}$ ; it is clear that $\beta $ is as wanted.

Case $\nu \in I^{o(\nu )}_\delta \backslash I_\delta $ . Recall $I_\delta =I^\delta _\delta $ ; clearly $\delta <o(\nu )$ . There is $\nu _0\in I_\delta $ , such that $\nu \in Gen(\nu _0)$ , with the road to $\nu $ equal to $\{\nu _i\}_{i\leq n}$ , and $\nu _1\notin I^\delta $ . Since $\nu _0\in I^\delta _\delta $ and $\delta $ is a limit, $o(\nu _0)<\delta $ and there is $\theta <\delta $ such that $\nu _0\in I^{o(\nu _0)}_\theta $ . Let $\beta =max\{o(\nu _0),\theta \}$ .

Case $\nu \notin I^{o(\nu )}_\delta $ . Let $\theta =max\{o(\nu ), \delta \}$ ; thus $\nu \in I^\theta $ (notice that we are talking about the order $I^\theta $ and not the element $I_\theta $ of the $\kappa $ -representation $\langle I_\alpha \mathrel {|} \alpha <\kappa \rangle $ ) and by Lemma 2.18 there is $\beta <\delta $ which satisfies the following:

$$ \begin{align*}\forall\sigma\in I^\theta_\delta [\sigma>\nu \Rightarrow \exists\sigma'\in I^0_\beta\ (\sigma>\sigma'>\nu)].\end{align*} $$

Fact 2.20 (Hyttinen–Tuuri [Reference Hyttinen and Tuuri8, Lemma 8.12]).

Let A be a linear order of size $\kappa $ and $\langle A_\alpha \mathrel {|} \alpha <\kappa \rangle $ a $\kappa $ -representation. Then the following are equivalent:

1. (1) A is $(\kappa , bs,bs)$ -nice.

2. (2) There is a club $C\subseteq \kappa $ , such that for all limit $\delta \in C$ , for all $x\in A$ there is $\beta <\delta $ such that one of the following holds:

   • • $\forall \sigma \in A_\delta [\sigma \ge x \Rightarrow \exists \sigma '\in A_\beta \ (\sigma \ge \sigma '\ge x)].$

   • • $\forall \sigma \in A_\delta [\sigma \leq x \Rightarrow \exists \sigma '\in A_\beta \ (\sigma \leq \sigma '\leq x)].$

Proof Recall the linear order $\mathcal {R}$ from Definition 2.8. From the general assumption on $\kappa $ , we know that $\lambda ^\omega =\lambda $ .

For all $A\subseteq I^0$ define $Pr(A)$ as the set $\{f_1\mathrel {|} f\in A\}$ . Let $A\subseteq I^0$ be such that $|A|<\kappa $ . Since $|\mathbb {Q}|=\omega $ , $|\{tp_{bs}(a,A,I^0)\mathrel {|} a\in I^0\}|\leq |\{tp_{bs}(a,Pr(A),\mathcal {R})\mathrel {|} a\in \mathcal {R}\}\times 2^\omega |$ . By Fact 2.9 and since $\lambda ^\omega =\lambda $ , $|\{tp_{bs}(a,A,I^0)\mathrel {|} a\in I\}|< \kappa $ .

Proof Let us fix $A\subset I$ such that $|A|<\kappa $ . From Fact 2.17, for all $a\in I$ and $\nu \in I^0$ such that $a\in Gen(\nu )$ the following holds:

$$ \begin{align*}b\models tp_{bs}(a,A,I) \Leftrightarrow b\models tp_{bs}(\nu,A \backslash Gen(\nu),I)\cup tp_{bs}(a,A\cap Gen(\nu), Gen(\nu)).\end{align*} $$

Thus for all $a\in I$ and $\nu \in I^0$ with $a\in Gen(\nu )$ , the type of a is determined by $tp_{bs}(\nu ,A \backslash Gen(\nu ),I)$ and $tp_{bs}(a,A \cap Gen(\nu ), Gen(\nu ))$ . Let $A'\subseteq I^0$ be such that the following hold:

• • For all $x\in A$ there is $y\in A'$ , $x\in Gen(y)$ .

• • For all $y\in A'$ there is $x\in A$ , $x\in Gen(y)$ .

Clearly $|A'|\leq |A|$ , and by Fact 2.17, for all $\nu \in I^0$ , $tp_{bs}(\nu ,A\backslash Gen(\nu ),I)$ is determined by $tp_{bs}(\nu ,A'\backslash \{\nu \},I^0)$ . So for all $a\in I$ and $\nu \in I^0$ with $a\in Gen(\nu )$ , $tp_{bs}(a,A,I)$ is determined by $tp_{bs}(\nu ,A'\backslash \{\nu \},I^0)$ and $tp_{bs}(a,A\cap Gen(\nu ), Gen(\nu ))$ . Therefore $|\{tp_{bs}(a,A,I)\mathrel {|} a\in I\}|$ is bounded by

$$ \begin{align*}|\{tp_{bs}(\nu,A',I^0)\mathrel{|} \nu\in I^0\}|\ \times\ Sup(\{\alpha_\nu \mathrel{|} \nu \in I^0\}),\end{align*} $$

where

$$ \begin{align*}\alpha_\nu=|\{tp_{bs}(a,A \cap Gen(\nu), Gen(\nu))\mathrel{|} a\in Gen(\nu)\}|.\end{align*} $$

Proof Recall the order $(Gen, <_{Gen})$ . By the non-inductive construction of I, it is enough to show that $(Gen, <_{Gen})$ is $({<}\kappa , bs)$ -stable.

Let $D\subseteq Gen$ be such that $|D|<\kappa $ , and let

$$ \begin{align*}\beta=sup\{f(n)+1\mathrel{|} f\in D\ \&\ n<\omega\}.\end{align*} $$

Since for all $f\in A$ , f is constant to $0$ starting at some m, $\beta <\kappa $ . On the other hand, for all $f,g\in Gen$ , f and g eventually become constants to $0$ , and the order $f<_{Gen}g$ (or $g<_{Gen}f$ ) is determined by the values of $f(i)$ and $g(i)$ , where i is the least ordinal such that $f(i)\neq g(i)$ . Therefore, for all $f\in Gen$ , $tp_{bs}(f,D,Gen)$ is entirely determined by the coordinates n of f in which $f(n)$ is smaller than $\beta +1$ . Since $\lambda ^{\omega }=\lambda $ , and $\beta <\kappa $ , $|\{tp_{bs}(f,D,Gen)\mathrel {|} f\in Gen\}|\leq |\beta ^{<\omega }|\leq \lambda < \kappa $ .

From the previous claim, we conclude that for all $\nu \in I^0$ , $\alpha _\nu <\kappa $ . Since $\kappa =\lambda ^+$ , $Sup(\{\alpha _\nu \mathrel {|} \nu \in I^0\})\leq \lambda $ . From Lemma 2.22 we know that $|\{tp_{bs}(\nu ,A',I^0)\mathrel {|} \nu \in I^0\}|<\kappa $ , so $|\{tp_{bs}(\nu ,A',I^0)\mathrel {|} \nu \in I^0\}|\leq \lambda $ . We conclude $|\{tp_{bs}(a,A,I)\mathrel {|} a\in I\}|<\kappa $ .

Proof From Corollary 2.21 and Lemma 2.23, we only need to show that I is $\kappa $ -colorable. For all $\nu \in I$ let us define $Succ_I(\nu )$ as follows:

$$ \begin{align*} Succ_I(\nu)=\{\sigma\in I\mathrel{|} \sigma=\nu^{o(\sigma)}\}. \end{align*} $$

We use the same notation of ordered trees because I can be seen as an ordered tree. Notice that for all $\nu \in I$ , $|Succ_I(\nu )|=\kappa $ and either $o(\nu )=0$ , or there is a unique $\nu '\in I$ such that $\nu =(\nu ')^{o(\nu )}$ (i.e. $\nu \in Succ_I(\nu ')$ ).

Let us fix $G:\kappa \rightarrow \kappa \times \kappa $ a bijection, and $G_1$ , $G_2$ be the functions such that $G(\alpha )=(G_1(\alpha ),G_2(\alpha ))$ . For all $\nu \in I$ let us fix a bijection $g_\nu :Succ_I(\nu )\rightarrow \kappa $ . Let us define $F:I\rightarrow \kappa $ by

$$ \begin{align*} F(\nu)=\begin{cases} 0, &\mbox{if } o(\nu)=0,\\ G_1(g_{\nu'}(\nu)), & \mbox{where }(\nu')^{o(\nu)}=\nu. \end{cases} \end{align*} $$

Fact 3.4 (Hyttinen–Kulikov [Reference Hyttinen and Kulikov5, Lemma 2.5]; Hyttinen–Moreno [Reference Hyttinen and Moreno7, Lemma 4.7]).

Suppose $\kappa $ is such that for all $\gamma <\kappa $ , $\gamma ^\omega <\kappa $ . For every $f,g\in \beta ^\kappa $ the following holds:

$$ \begin{align*} f\ =^\beta_\omega\ g \Leftrightarrow J_f\cong_{ct} J_g, \end{align*} $$

where $\cong _{ct}$ is the isomorphism of coloured trees.

Proof Let $f\in \beta ^\kappa $ , such that $\eta \in (J_g)_{<\omega }$ , and $n=dom(\eta )$ .

Let us construct $\xi $ , $\eta \prec \xi $ and $c_f(\xi )=\theta $ .

• • $\xi {\mathbin \upharpoonright } n=\eta $ .

• • If $n\leq m<\omega $ ,

  1. – $\xi _1(m)=\xi _1(n-1)+1$ .

  2. – $\xi _2(m)=\xi _3(n-1)+\xi _4(n-1)$ .

  3. – $\xi _3(m)=\xi _2(n)+\omega $ .

  4. – Let $\gamma $ and $\zeta $ be such that $dom(\zeta )=[n,\omega )$ , $\zeta \in Z^{\xi _2(n),\xi _3(n)}_{\gamma }$ with $c(\zeta )=\theta $ . Such $\gamma $ and $\zeta $ exist by Definition 3.2.

  5. – $\xi _4(m)=\gamma $ .

  6. – $\xi _5{\mathbin \upharpoonright }_{[n,\omega )}=\zeta $ .

By the way we defined $\xi $ , we know that $\xi \in J_f$ and $\eta \prec \xi $ . By the item (8)(a) on the construction of $J_f$ , we know that $c_f(\xi )=c(\xi _5{\mathbin \upharpoonright }_{[n,\omega )})=\theta $ .

Fact 3.8. Suppose $\eta \in A^f$ has domain $m<\omega $ and $\eta _3(m-1)=\theta $ . Then

$<{\mathbin \upharpoonright } Suc_{A^f}(\eta )$ satisfies ( $*$ ).

Proof Let $b\in Suc_{A^f}(\eta )$ , $(\theta _2,\theta _2)\in \omega \times \operatorname {\mathrm {acc}}(\kappa )$ such that $\theta _3\ge \eta _3(m-1)=\theta $ , and $B\subseteq Suc_{A^f}(\eta )$ have size less than $\kappa $ . Let us denote by q the type

$$ \begin{align*}tp_{bs}(\tilde{\mathcal{G}}_\theta(b_1,b_2,b_3), \tilde{\mathcal{G}}_\theta (B\cap(\kappa\times\omega\times \operatorname{\mathrm{acc}}(\kappa))), I).\end{align*} $$

By the construction of $\tilde {G}_\theta $ , since F is a $\kappa $ -coloration of I,

$$ \begin{align*}|\{a\in I\mathrel{|} a\models q \ \&\ F(a)=\tilde{G}_\theta(\theta_2,\theta_3)\}|=\kappa.\end{align*} $$

Therefore, for all a such that $a\models q$ and $F(a)=\tilde {G}_\theta (\theta _2,\theta _3)$ ,

$$ \begin{align*}\eta^\frown (\tilde{H}_{\tilde{G}_\theta((\theta_2,\theta_3))}(a),\theta_2,\theta_3,\sigma^{\theta_3}_{\overline{\eta}}(\theta_2))\models p.\\[-34pt] \end{align*} $$

Proof For every $f\in \beta ^\kappa $ let us define the $\kappa $ -representation $\mathbb {A}^f=\langle A^f_\alpha \mathrel {|} \alpha <\kappa \rangle $ of $A^f$ ,

$$ \begin{align*}A^f_\alpha=\{\eta\in A^f\mathrel{|} rng(\eta)\subseteq \theta\times\omega\times\theta\times\omega\times\theta^4\text{ for some }\theta<\alpha\}. \end{align*} $$

Let f and g be such that $f\ =^\beta _\omega \ g$ . Thus, there is G a coloured tree isomorphism between $J_f$ and $J_g$ . Let $C\subseteq \kappa $ be a club such that $\{\alpha \in C\mathrel {|} cf(\alpha )=\omega \}\subseteq \{\alpha <\kappa \mathrel {|} f(\alpha )=g(\alpha )\}$ . We will show that there are sequences $\{\alpha _i\}_{i<\kappa }$ and $\{F_i\}_{i<\kappa }$ with the following properties:

• • $\{\alpha _i\}_{i<\kappa }$ is a club.

• • If i is a successor, then there is $\theta \in C$ such that $\alpha _{i-1}<\theta <\alpha _i$ .

• • If $i=\gamma +n$ and n is odd, $F_i$ is a partial isomorphism between $A^f$ and $A^g$ , and $A_{\alpha _i}^f\subseteq dom(F_i)$ .

• • If $i=\gamma +n$ and n is even, $F_i$ is a partial isomorphism between $A^f$ and $A^g$ , and $A_{\alpha _i}^g\subseteq rng(F_i)$ .

• • If i is limit, then $F_i:A_{\alpha _i}^f\rightarrow A_{\alpha _i}^g$ .

• • If $i<j$ , then $F_i\subseteq F_j$ .

• • For all $\eta \in dom(F_i)$ , $G(\overline {\eta })=\overline {F_i(\eta )}$ .

We will proceed by induction over i, for the case $i=0$ , let $\alpha _0=0$ and $F_0(\emptyset )=\emptyset $ . Suppose $i=\gamma +n$ with n even is such that $F_i$ is a partial isomorphism, $A_{\alpha _i}^g\subseteq rng(F_i)$ for all $j<i$ , $F_j\subseteq F_i$ , and $G(\overline {\eta })=\overline {F_i(\eta )}$ for all $\eta \in dom(F_i)$ .

Let us choose $\alpha _{i+1}$ to be a successor ordinal such that $\alpha _{i}<\theta <\alpha _{i+1}$ holds for some $\theta \in C$ and enumerate $A^f_{\alpha _{i}}$ by $\{\eta _j\mathrel {|} j<\Omega \}$ for some $\Omega <\kappa $ . Denote by $B_j$ the set $\{x\in A^f_{\alpha _{i+1}}\backslash dom(F_i)\mathrel {|} \eta _j\prec x\}$ .

By the induction hypothesis, we know that for all $j<\Omega $ , $x\in B_j$ , $\overline {F_i(\eta _j)}\prec G(\overline {x})$ . By Remark 3.9, for all $\eta \in A^f$ and $\xi \in A^g$ , $<{\mathbin \upharpoonright } Suc_{A^f}(\eta )$ and $<{\mathbin \upharpoonright } Suc_{A^g}(\xi )$ are isomorphic. Thus, since $|A^f_{\alpha _{i}}|, |B_0|<\kappa $ , by $(\ast )$ there is an embedding $F^0_i$ from $(A^f_{\alpha _i}\cup B_0,\prec ,<)$ to $(A^g,\prec , <)$ that extends $F_i$ and for all $\eta \in dom(F^0_i)$ , $\overline {F^0_i(\eta )}=G(\overline {\eta })$ .

For the case $B_j$ for $j>0$ , let us suppose that $t<\Omega $ is such that the following hold:

• • There is a sequence of embeddings $\{F_i^j\mathrel {|} j<t\}$ , where $F^j_i$ is an embedding from $(A^f_{\alpha _i}\cup \bigcup _{l\leq j}B_l,\prec ,<)$ into $A^g$ .

• • $F_i^l\subseteq F_i^j$ holds for all $l<j<t$ .

• • For all $\eta \in dom(F^j_i)$ , $\overline {F^j_i(\eta )}=G(\overline {\eta })$ .

Since $|A^f_{\alpha _{i}}\cup \bigcup _{j< t}B_j|, |B_t|<\kappa $ , by $(\ast )$ there is an embedding $F^t_i$ from $(A^f_{\alpha _i}\cup \bigcup _{j\leq t}B_j,\prec ,<)$ to $(A^g,\prec , <)$ that extends $\bigcup _{j<t} F_i^j$ and for all $\eta \in dom(F^t_i)$ , $\overline {F^t_i(\eta )}=G(\overline {\eta })$ .

Finally $F_{i+1}=\bigcup _{j<\Omega } F_i^j$ is as wanted.

The case $i=\gamma +n$ with n odd is similar. For i limit, we define $\alpha _i=\bigcup _{j<i} \alpha _j$ and $F_{\alpha _{i}}=\bigcup _{j<i} F_j$ .

It is clear that $F=\bigcup _{j<\kappa } F_j$ witnesses that $A^f$ and $A^g$ are isomorphic as ordered trees. Let us show that $d_f(\eta )=d_g(F(\eta ))$ , and suppose $\eta \in A^f$ is a leaf. Let l be the least ordinal such that $\eta \in A^f_{\alpha _l}$ . If there is $n<\omega $ such that for all $j<l$ , $\eta {\mathbin \upharpoonright } n\notin A^f_{\alpha _j}$ , then by the way F was constructed, $d_f(\eta )=d_g(F(\eta ))$ . On the other hand, if for all $n<\omega $ there is $j<l$ such that $\eta {\mathbin \upharpoonright } n\in A^f_{\alpha _j}$ , then there is an $\omega $ -cofinal ordinal i such that $s_\omega (\eta )=\alpha _i$ and $i+1=l$ .

By the construction of $A^f$ (recall equation (2)) we know that

$$ \begin{align*}d_f(\eta)=\begin{cases} c_f(\overline{\eta}), &\mbox{if } s_1(\eta)< s_\omega(\eta),\\ f(s_1(\eta)), & \mbox{if } s_1(\eta)= s_\omega(\eta). \end{cases} \end{align*} $$

Since $s_\omega (\eta )=\alpha _i$ , either $d_f(\eta )=f(s_1(\eta ))$ (if $s_1(\eta )=\alpha _i$ ) or $d_f(\eta )=c_f(\overline {\eta })$ (if $s_1(\eta )<\alpha _i$ ).

Therefore, if $s_1(\eta )=\alpha _i$ , then $d_f(\eta )=f(\alpha _i)$ .

Let us calculate $d_f(\eta )$ , when $s_1(\eta )< s_\omega (\eta )$ . Notice that $\overline {\eta }\in J_j$ , so there is $\zeta =(\zeta _1,\zeta _2,\zeta _3,\zeta _4,\zeta _5)$ such that $\overline {\eta }=\zeta \in J_j$ .

From Definition 3.3 items (5) and (7), since $\zeta \in (J_f)_\omega \backslash J_f^{\alpha _i}$ and for all $n<\omega $ , $\zeta {\mathbin \upharpoonright } n\in J_f^{\alpha _i}$ holds,

$$ \begin{align*}sup(rng(\zeta_4))\leq sup(rng(\zeta_2))=sup(rng(\zeta_3))=sup(rng(\zeta_5)).\end{align*} $$

Since $\overline {\eta }=\zeta $ ,

$$ \begin{align*}s_7(\eta)\leq s_8(\eta)=s_6(\eta)=s_8(\eta)=sup(rng(\zeta_5).\end{align*} $$

It is easy to see that $s_2(\eta ),s_3(\eta ),s_4(\eta )\leq s_5(\eta )$ .

We conclude that $s_\omega (\eta )=s_8(\eta )=sup(rng(\zeta _5))$ and $\alpha _i=sup(rng(\zeta _5))$ . From Definition 3.3(8),

$$ \begin{align*}c_f(\overline{\eta})=c_f(\zeta)=f(sup(rng(\zeta_5)))=f(\alpha_i).\end{align*} $$

Therefore $d_f(\eta )=f(\alpha _i)$ in both cases ( $s_1(\eta )=s_\omega (\eta )$ and $s_1(\eta )<s_\omega (\eta )$ ).

By the same argument and using the definition of F, we can conclude that $d_g(F(\eta ))=g(\alpha _i)$ . Finally since i is a limit ordinal with cofinality $\omega $ , $\alpha _i$ is an $\omega $ -limit of C. Thus $d_f(\eta )=f(\alpha _i)=g(\alpha _i)=d_g(F(\eta ))$ and F is a coloured tree isomorphism.