Notation 3.3. For $\langle T,\vec {f} \rangle \in \mathbb S^\kappa _{\mathbf X}$ :

• • $\eta _T$ denotes the unique ordinal to satisfy $\operatorname {\mathrm {ht}}(T)=\eta _T+1$ ;

• • $\nu _T$ denotes the cardinality of the quotient .

Proof. Set $T':=T\cup \{ t{}^\smallfrown \langle \iota \rangle \mid t\in T_{\eta _T},~\iota <|\nu \cup \theta |\}$ , and then define $\vec {f'}:\eta _{T'}+1\rightarrow {}^\theta ({}^{<\kappa }\kappa )$ by letting for all $\alpha \le \eta _{T'}$ and $\tau <\theta $ ,

$$ \begin{align*}\vec{f'}(\alpha)(\tau):=\begin{cases} \vec{f}(\alpha)(\tau),&\text{if }\alpha\le\eta_{T};\\ (\vec{f}(\beta)(\tau)*\vec{f}(\eta_T)(\tau)){}^\smallfrown\langle \tau\rangle,&\text{otherwise}. \end{cases}\end{align*} $$

It is clear that $\langle T',\vec {f'}\rangle $ is an element of $\mathbb S^\kappa _{\mathbf X}$ as sought.

Proof. Set $\eta :=\sup _{\beta \in \Gamma }\eta _{T^\beta }$ and $\vec {f}:=\bigcup _{\beta \in \Gamma }\vec {f}^{\beta }$ . Define $\vec {f^\gamma }:\eta +1\rightarrow {}^\theta ({}^{<\kappa }\kappa )$ by letting for all $\alpha \le \eta $ and $\tau <\theta $ ,

$$ \begin{align*}\vec{f^\gamma}(\alpha)(\tau):=\begin{cases} \vec{f}(\alpha)(\tau),&\text{if }\alpha<\eta;\\ \bigcup_{\beta\in\Gamma}\vec{f}(\eta_{T^\beta})(\tau),&\text{if }\alpha=\eta. \end{cases}\end{align*} $$

Define $\vec {z^\gamma }:(\gamma ,\delta )\rightarrow T^\gamma $ via

$$ \begin{align*}\vec{z^\gamma}(i):=\bigcup_{\beta\in\Gamma}\vec{z^\beta}(i).\end{align*} $$

Finally, consider $T:=\bigcup _{\beta \in \Gamma }T^\beta $ , and then let

$$ \begin{align*}T^\gamma:=T \cup \bigcup \{ x*\vec{f^\gamma}(\eta)(\tau), x*\vec{z^\gamma}(i)\mid x \in T, \tau<\theta, \gamma<i<\delta \}.\end{align*} $$

To see that $\langle T^\gamma ,\vec {f}^{\gamma }\rangle $ is as sought note that

$$ \begin{align*}y:=\bigcup_{\beta\in\Gamma}\vec{z^\beta}(\gamma)\end{align*} $$

determines a vanishing $\eta $ -branch of $T^\gamma $ and hence $\eta \in V^-(T)=V(T)$ . Also, for every $i \in (\gamma ,\delta )$ , $\vec {z^\gamma }(i)$ and $\vec {f}^\gamma (\eta )(\tau )$ are mutually exclusive for every nonzero $\tau <\theta $ . So, for every $t\in (T^\gamma )_\eta $ , an intersection of two filter sets demonstrates that $\{\tau <\theta \mid t \text { and }\vec f^\gamma (\eta )(\tau )\text { are mutually exclusive}\}$ is in $\mathcal {F}$ .

Proof. By Lemma 3.6. See the proof of the upcoming Lemma 3.10 for a stronger result.

Proof. Let $\xi <\zeta $ , and we shall prove that $\mathbb S^\kappa _{\mathbf X}* \mathbb {\dot {A}}^\kappa _{\mathbf X,\xi }$ has a dense subset that is ${<}\kappa $ -strategically closed.

Let D be the collection of all pairs $(p,\dot q)$ in $\mathbb S^\kappa _{\mathbf X}*\dot {\mathbb A}^\kappa _{\mathbf X,\xi }$ such that if $p=\langle T,\vec {f} \rangle $ , then $\dot q=\check {\eta _T}$ . By Lemma 3.5, D is dense in $\mathbb S^\kappa _{\mathbf X}*\dot {\mathbb A}^\kappa _{\mathbf X,\xi }$ . For simplicity, we shall identify D with the collection of all triples $(T,\vec {f},\eta )$ such that $\langle T,\vec {f}\rangle \in \mathbb S^\kappa _{\mathbf X}$ and $\eta =\eta _T$ , where a triple $(T',\vec {f'} ,\eta ')$ extends $( T,\vec {f} ,\eta )$ iff all of the following hold:

• • $T'\cap {}^{\le \eta }\kappa =T$ ,

• • $\vec {f'}\restriction (\eta +1)=\vec f$ , and

• • $X_\xi \subseteq \circledcirc (\vec f(\eta ),\vec {f'}(\eta '))$ .

To prove that D is ${<}\kappa $ -strategically closed, let $\mu <\kappa $ be arbitrary, and we shall describe a winning strategy for ${\mathbf {II}}$ in the game $\Game _\mu (D)$ . Playing the game will yield a decreasing sequence of conditions in D, $\langle (T^{\beta },\vec {f}^{\beta },\eta _{\beta })\mid \beta <\mu \rangle $ , along with an auxiliary sequence $\langle \vec {z}^{\beta }\mid \beta <\mu \text { nonzero even ordinal}\rangle $ .

We start by letting $(T^0,\vec {f^0},\eta _0)$ be the maximal element of D, that is, $T^0:=\{\emptyset \}$ , $\vec {f^0}:1\rightarrow {}^\theta \{\emptyset \},$ and $\eta _0:=0$ . Next, suppose that $\gamma <\mu $ is a nonzero even ordinal and we have already obtained a decreasing sequence of conditions $\langle (T^{\beta },\vec {f}^{\beta },\eta _{\beta })\mid \beta <\gamma \rangle $ in D, and a sequence $\langle \vec {z}^{\beta }\mid \beta <\gamma \text { nonzero even ordinal}\rangle $ in such a way that all of the following hold:

1. (i) for all $\alpha <\beta <\gamma $ , $X_\xi \subseteq \circledcirc (\vec f^{\alpha }(\eta _\alpha ), \vec f^{\beta }(\eta _\beta ))$ ;

2. (ii) for every nonzero even $\beta <\gamma $ , $\vec {z}^\beta :(\beta ,\mu )\rightarrow T^\beta $ , where $\beta <i<\mu $ implies:

   • • $\vec {z}^\beta (i)$ belongs to the top level of $T^\beta $ ;

   • • $\vec {z}^\beta (i)\not =^*\vec {z}^\beta (i')$ for every $i'\in (\beta ,\mu )\setminus \{i\}$ ;

   • • $\vec {z}^\beta (i)\not =^*\vec {f}^\beta (\eta _\beta )(0)$ ;

   • • $\vec {z}^\beta (i)$ and $\vec {f}^\beta (\eta _\beta )(\tau )$ are mutually exclusive for every nonzero $\tau <\theta $ ;

3. (iii) for every pair $\alpha <\beta $ of even ordinals below $\gamma $ :

   • • $\circledcirc (\vec f^{\alpha }(\eta _\alpha ), \vec f^{\beta }(\eta _\beta ))=\theta $ ;

   • • $\vec z^\alpha (i) \subseteq \vec z^\beta (i)$ provided $0<\alpha <\beta <i<\mu $ .

Our description of the $\gamma ^{\text {th}}$ -move is divided into three cases, as follows.

${\blacktriangleright }$ If $\gamma =2$ , then using Lemma 3.5, fix a one-step $\eta _1$ -extension $\langle T^2,\vec {f}^2\rangle $ of $\langle T^1,\vec f^1\rangle $ such that $\nu _{T^2}\ge \max \{\mu ,\theta ^+\}$ . Letting $\eta _2:=\eta _{T^2}$ , it is clear that $(T^2,\vec f^2,\eta _2)$ belongs to D and extends $(T^1,\vec f^1,\eta _1)$ . Now, as $\nu _{T^2}\ge \max \{\mu ,\theta ^+\}$ , we may fix a map $\vec {z^2}:(2,\mu )\rightarrow T^2$ , where $2<i<\mu $ implies:

• • $\vec {z^2}(i)$ belongs to the top level of $T^2$ ;

• • $\vec {z^2}(i)\not =^*\vec {z^2}(i')$ for every $i'\in (2,\mu )\setminus \{i\}$ ;

• • $\vec {z^2}(i)\not =^*\vec {f}^2(\eta _{2})(0)$ ;

• • $\vec {z^2}(i)$ and $\vec {f}^2(\eta _{2})(\tau )$ are mutually exclusive for every nonzero $\tau <\theta $ .

Indeed, as $\eta _2=\eta _1+1$ , this can be achieved by ensuring that $\langle \vec {z^2}(i)(\eta _1)\mid 2<i<\mu \rangle {}^\smallfrown \langle \vec {f^2}(\tau )(\eta _1)\mid \tau <\theta \rangle $ be an injective sequence, whereas $\langle \vec {z^2}(i)\restriction \eta _1\mid 2<i<\mu \rangle $ be a constant sequence whose sole element is $\vec {f^2}(\eta _2)(0)\restriction \eta _1$ .

${\blacktriangleright }$ If $\gamma $ is a successor ordinal, say, $\gamma =\alpha +2$ , then using Lemma 3.5, fix a one-step $\eta _\alpha $ -extension $\langle T^\gamma ,\vec {f}^\gamma \rangle $ of $\langle T^{\alpha +1},\vec f^{\alpha +1}\rangle $ such that $\nu _{T^\gamma }\ge \max \{\mu ,\theta ^+\}$ . Letting $\eta _\gamma :=\eta _{T^\gamma }$ , it is clear that $(T^\gamma ,\vec f^\gamma ,\eta _\gamma )$ belongs to D, and

$$ \begin{align*}X_\xi\subseteq \circledcirc(\vec f^{\alpha}(\eta_{\alpha}),\vec{f}^{\alpha+1}(\eta_{\alpha+1}))\subseteq \circledcirc(\vec f^{\alpha+1}(\eta_{\alpha+1}),\vec{f}^\gamma(\eta_\gamma)).\end{align*} $$

Next, we fix a map $\vec {z}^\gamma :(\gamma ,\mu )\rightarrow T^\gamma $ , where $\gamma <i<\mu $ implies:

• • $\vec {z}^\gamma (i)$ belongs to the top level of $T^\gamma $ ;

• • $\vec {z}^\gamma (i)\not =^*\vec {z}^\gamma (i')$ for every $i'\in (\gamma ,\mu )\setminus \{i\}$ ;

• • $\vec z^\alpha (i)\subseteq \vec {z}^\gamma (i)$ and $\vec {z}^\gamma (i)\not =^*\vec {f}^\gamma (\eta _\gamma )(0)$ ;

• • $\vec {z}^\gamma (i)$ and $\vec {f}^\gamma (\eta _\gamma )(\tau )$ are mutually exclusive for every nonzero $\tau <\theta $ .

This is possible because the combination of $\nu _{T^\gamma }\ge \max \{\mu ,\theta ^+\}$ with $\eta _\gamma =\eta _{\alpha +1}+1$ and $\circledcirc (\vec f^{\alpha }(\eta _\alpha ), \vec f^{\gamma }(\eta _\gamma ))=\theta $ enables us to ensure that $\langle \vec {z^\gamma }(i)(\eta _{\alpha +1})\mid \gamma <i<\mu \rangle {}^\smallfrown \langle \vec {f^\gamma }(\tau )(\eta _{\alpha +1})\mid \tau <\theta \rangle $ be an injective sequence, and $\langle \vec {z^\gamma }(i)\restriction \eta _{\alpha +1}\mid \gamma <i<\mu \rangle $ be equal to $\langle \vec z^\alpha (i)*(\vec {f^\gamma }(\eta _\gamma )(0)\restriction \eta _{\alpha +1})\mid \gamma <i<\mu \rangle $ .

${\blacktriangleright }$ If $\gamma $ is a limit ordinal, then obtain $\langle T^\gamma , \vec {f}^\gamma \rangle $ and $\vec z^\gamma $ by appealing to Lemma 3.6 with the sequence $\langle (T^\beta ,\vec {f^\beta }, \vec {z}^\beta ) \mid \beta <\gamma \text { nonzero even ordinal}\rangle $ . Then let $\eta _\gamma :=\eta _{T^\gamma }$ .

In all cases, $( T^{\gamma },\vec {f}^{\gamma },\eta _\gamma )$ belongs to D and requirements (i)–(iii) are satisfied for $\gamma $ by the induction and the construction of $( T^{\gamma },\vec {f}^{\gamma },\eta _\gamma )$ . This completes the description of our winning strategy for ${\mathbf {II}}$ .

Proof. The “in particular” part follows from the fact that $\zeta <\theta ^+<\kappa $ . Now, fix a $\xi <\zeta $ , and let $\dot {A}$ be an $\mathbb S^\kappa _{\mathbf X}$ -name for a maximal antichain in $\mathbb {\dot {A}}^\kappa _{\mathbf X,\xi }$ . We shall need the following key claim concerning the poset $\mathbb S^\kappa _{\mathbf X}$ .

Proof. By our assumptions, $\mu :=\max \{|T_{\eta _{T}}|,\theta \}^\theta $ is smaller than $\kappa $ . Fix a bijection h from the set of all even ordinals in $\mu $ to the set of all triples $(\vec {x},Y,\pi )$ satisfying (A) and (B). We shall construct a sequence $\langle (T^{j},\vec {f}^{j},\vec {z}^j)\mid j<\mu \rangle $ in such a way that all of the following hold:

1. (i) $\langle T^{0},\vec {f^0}\rangle =\langle T,\vec f\rangle $ ;

2. (ii) for all $i<j<\mu $ , $\langle T^{j},\vec {f}^{j}\rangle \le \langle T^i,\vec {f}^i\rangle ,$ and $X_\xi \subseteq \circledcirc (\vec f^{i}(\eta _{T^i}), \vec f^{j}(\eta _{T^j}))$ ;

3. (iii) for every pair $i<j$ of even ordinals below $\mu $ , $\circledcirc (\vec f^{i}(\eta _{T^i}), \vec f^{j}(\eta _{T^j}))=\theta $ ;

4. (iv) for every nonzero even $j<\mu $ , $\vec {z}^j:(j,\mu ]\rightarrow T^j$ , where $j<\iota <\mu $ implies:

   • • $\vec {z}^j(\iota )$ belongs to the top level of $T^j$ ;

   • • $\vec {z}^j(\iota )\not =^*\vec {z}^j(\iota ')$ for every $\iota '\in (j,\mu ]\setminus \{\iota \}$ ;

   • • $\vec {z}^j(\iota )\not =^*\vec {f}^j(\eta _{T^j})(0)$ ;

   • • $\vec {z}^j(\iota )$ and $\vec {f}^j(\eta _{T^j})(\tau )$ are mutually exclusive for every nonzero $\tau <\theta $ ;

   • • $\vec z^i(\iota ) \subseteq \vec z^j(\iota )$ for every nonzero even $i<j$ and every $\iota \in (j,\mu ]$ .

5. (v) for every even $i<\mu $ , letting $(\langle x_\tau \mid \tau <\theta \rangle ,Y,\pi ):=h(i)$ , there exists an $\alpha \le \eta _{T^{i+1}}$ such that:

   1. (a) $\langle T^{i+1},\vec {f}^{i+1}\rangle \Vdash _{\mathbb S^\kappa _{\mathbf X}}\alpha \in \dot A$ ;

   2. (b) $X_\xi \setminus Y\subseteq \circledcirc (\vec {f}^{i+1}(\alpha ),\vec {f}^{i+1}(\eta _{T^{i+1}}))$ ;

   3. (c) $\vec {f}^{i+1}(\alpha )(\tau )\subseteq x_\tau *\vec {f}^{i+1}(\eta _{T^{i+1}})(\pi (\tau ))$ for every $\tau \in X_\xi \cap Y$ .

We start by setting $(T^{0},\vec f^0,\vec z^0):=(T,\vec f,\emptyset )$ . Next, suppose that $j\in (0,\mu ]$ is such that $\langle (T^{i},\vec {f}^{i},\vec {z}^i)\mid i<j \rangle $ has already been defined.

Case 1: $j=i+1$ for an even ordinal i: Let $(\langle x_\tau \mid \tau <\theta \rangle ,Y,\pi ):=h(i)$ . By (B), $\langle \vec f(\eta _T)(\tau )\mid \tau \in \theta \setminus Y\rangle {}^\smallfrown \langle x_\tau \mid \tau \in Y\rangle $ coincides with $\langle x_\tau \mid \tau <\theta \rangle $ . So, by (A), it consists of mutually exclusive nodes. By Clauses (i) and (ii), $\circledcirc (\vec {f}(\eta _{T}),\vec {f^i}(\eta _{T^i}))=\theta $ . So, since $Y\in \mathcal I$ , we can take a one-step extension $\langle S,\vec g\rangle $ of $\langle T^{i}, \vec {f}^{i} \rangle $ such that:

• • $\theta \setminus Y\subseteq \circledcirc (\vec {f}^{i}(\eta _{T^i}),\vec g(\eta _S))$ , and

• • for every $\tau \in Y$ , $x_\tau *\vec {f}^{i}(\eta _{T^i})(\pi (\tau ))\subseteq \vec g(\eta _S)(\tau )$ .

Recalling Definition 3.9, and as A is a maximal antichain, we may find a condition $\langle S', \vec {g'}\rangle \le \langle S,\vec g\rangle $ such that, for some $\alpha \le \eta _{S'}$ ,

• • $\langle S',\vec {g'}\rangle \Vdash _{\mathbb S^\kappa _{\mathbf X}}\alpha \in \dot A$ , and

• • $X_\xi \subseteq \circledcirc (\vec {g}(\eta _S),\vec {g'}(\alpha ))$ .

Fix an injection $\psi :\theta \setminus ((X_\xi \setminus Y)\cup (X_\xi \cap \operatorname {\mathrm {Im}}(\pi )))\rightarrow \theta \setminus X_\xi $ .Footnote ⁸ Denote $\alpha ':=\max \{\eta _S,\alpha \}$ . Pick a one-step extension $\langle T^j,\vec {f^j}\rangle $ of $\langle S',\vec {g'}\rangle $ such that, for every $\tau <\theta $ ,

$$ \begin{align*}\vec{f}^{j}(\eta_{T^j})(\tau)=\begin{cases} (\vec{g'}(\alpha')(\tau)*\vec{g'}(\eta_{S'})(\tau)){}^\smallfrown\langle \tau\rangle,&\text{if }\tau\in X_\xi\setminus Y;\\ (\vec{g'}(\alpha')(\pi^{-1}(\tau))*\vec{g'}(\eta_{S'})(\tau)){}^{\smallfrown\langle\pi^{-1}}(\tau)\rangle,&\text{if }\tau\in X_\xi\cap \operatorname{\mathrm{Im}}(\pi);\\ (\vec{g'}(\alpha')(\psi(\tau))*\vec{g'}(\eta_{S'})(\tau)){}^\smallfrown\langle\psi(\tau)\rangle,&\text{otherwise}. \end{cases}\end{align*} $$

This ensures that the following two hold (Figure 2):

• • $X_\xi \setminus Y\subseteq \circledcirc (\vec {g'}(\alpha ),\vec {f}^{j}(\eta _{T^j}))$ ;

• • $\vec {g'}(\alpha )(\tau )\subseteq x_\tau *\vec {f}^{j}(\eta _{T^j})(\pi (\tau ))=\vec {f}^{j}(\eta _{T^j})(\pi (\tau ))$ for every $\tau \in X_\xi \cap Y$ .

Figure 2.

Case: $\tau \in X_\xi \cap Y$ .

Case 2: $j=2$ : Using Lemma 3.5, fix a one-step extension $\langle T^2,\vec {f}^2\rangle $ of $\langle T^1,\vec f^1\rangle $ such that $\nu _{T^2}\ge \mu $ . In particular, we may fix a map $\vec {z^2}:(2,\mu ]\rightarrow T^2$ , where $2<\iota <\mu $ implies:

• • $\vec {z^2}(\iota )$ belongs to the top level of $T^2$ ;

• • $\vec {z^2}(\iota )\not =^*\vec {z^2}(\iota ')$ for every $\iota '\in (2,\mu ]\setminus \{\iota \}$ ;

• • $\vec {z^2}(\iota )\not =^*\vec {f}^2(\eta _{T^2})(0)$ ;

• • $\vec {z^2}(\iota )$ and $\vec {f}^2(\eta _{T^2})(\tau )$ are mutually exclusive for every nonzero $\tau <\theta $ .

Case 3: $j=i+2$ for an even ordinal i: Using Lemma 3.5, let $\langle T^j,\vec {f^j}\rangle $ be a one-step $\eta _{T^i}$ -extension of $\langle T^{i+1},\vec {f^{i+1}}\rangle $ with $\nu _{T^j}\ge \mu $ . As $\circledcirc (\vec f^{i}(\eta _{T^i}), \vec f^{j}(\eta _{T^j}))=\theta $ , we may then fix a map $\vec {z}^j:(j,\mu ]\rightarrow T^j$ , where $j<\iota <\mu $ implies:

• • $\vec {z}^j(\iota )$ belongs to the top level of $T^j$ ;

• • $\vec {z}^j(\iota )\not =^*\vec {z}^j(\iota ')$ for every $\iota '\in (j,\mu ]\setminus \{\iota \}$ ;

• • $\vec z^i(\iota )\subseteq \vec {z}^j(\iota )$ and $\vec {z}^j(\iota )\not =^*\vec {f}^j(\eta _{T^j})(\tau )$ ;

• • $\vec {z}^j(\iota )$ and $\vec {f}^j(\eta _{T^j})(\tau )$ are mutually exclusive for every nonzero $\tau <\theta $ .

Case 4: $j\in \operatorname {\mathrm {acc}}(\mu +1)$ : Obtain $(T^j,\vec {f^j},\vec z^j)$ by appealing to Lemma 3.6 with the sequence $\langle (T^i,\vec {f^i}, \vec {z}^i) \mid i<j\text { nonzero even ordinal}\rangle $ , using $\delta :=\mu +1$ .

Once we are done with the recursion, it is clear that $\langle T',\vec {f'}\rangle :=\langle T^\mu ,\vec {f^\mu }\rangle $ is as sought.

Let $\langle T,\vec {f}\rangle $ be an arbitrary condition in $\mathbb S^\kappa _{\mathbf X}$ , and we shall find an extension of it forcing that A is bounded in $\kappa $ . To this end, we shall construct a sequence $\langle (T^{j},\vec {f}^{j},\vec {z}^j)\mid j<\theta ^+ \rangle $ in such a way that all of the following hold:

1. (i) $\langle T^{0},\vec {f^0}\rangle \le \langle T,\vec f\rangle $ ;

2. (ii) for all $i<j<\theta ^+$ , $(T^{j},\vec {f}^{j}\rangle \le (T^i,\vec {f}^i\rangle $ and $\circledcirc (\vec f^{i}(\eta _{T^i}), \vec f^{j}(\eta _{T^j}))=\theta $ ;

3. (iii) for every $j<\theta ^+$ , $\vec {z}^j:(j,\theta ^+]\rightarrow T^j$ , where $j<\iota <\theta ^+$ implies:

   • • $\vec {z}^j(\iota )$ belongs to the top level of $T^j$ ;

   • • $\vec {z}^j(\iota )\not =^*\vec {z}^j(\iota ')$ for every $\iota '\in (j,\theta ^+]\setminus \{\iota \}$ ;

   • • $\vec {z}^j(\iota )\not =^*\vec {f}^j(\eta _{T^j})(0)$ ;

   • • $\vec {z}^j(\iota )$ and $\vec {f}^j(\eta _{T^j})(\tau )$ are mutually exclusive for every nonzero $\tau <\theta $ ;

   • • $\vec z^i(\iota ) \subseteq \vec z^j(\iota )$ for every $i<j$ and every $\iota \in (j,\theta ^+]$ .

We start by appealing to Lemma 3.5 to find a condition $\langle T^0,\vec {f}^0\rangle \le \langle T,\vec f\rangle $ with $\nu _{T^0}\ge \theta ^+$ . Then, fix a map $\vec {z}^0:(0,\theta ^+]\rightarrow T^0$ , where $0<\iota <\theta ^+$ implies:

• • $\vec {z}^0(\iota )$ belongs to the top level of $T^0$ ;

• • $\vec {z}^0(\iota )\not =^*\vec {z}^0(\iota ')$ for every $\iota '\in (0,\theta ^+]\setminus \{\iota \}$ ;

• • $\vec {z}^0(\iota )\not =^*\vec {f}^0(\eta _0)(0)$ ;

• • $\vec {z}^0(\iota )$ and $\vec {f}^0(\eta _0)(\tau )$ are mutually exclusive for every nonzero $\tau <\theta $ .

Next, for every $i<\theta ^+$ such that $(T^i,\vec {f}^i,\vec {z}^i)$ has already been determined, obtain $\langle T^{i+1},\vec f^{i+1}\rangle $ by appealing to Claim 3.11.1 with $\langle T^i,\vec {f}^i\rangle $ . By possibly replacing $\langle T^{i+1},\vec {f}^{i+1}\rangle $ with a one-step $\eta _{T^{i+1}}$ -extension of it (using Lemma 3.5), we may also assume that $\eta _{T^{i+1}}$ is a successor ordinal and $\nu _{T^{i+1}}\ge \theta ^+$ . Consequently, we may construct $\vec z^{i+1}$ satisfying requirement (iii) for $j:=i+1$ . At a limit stage i, we obviously use Lemma 3.6 with $\gamma :=i$ and $\delta :=\theta ^++1$ . Hereon, for notational simplicity, denote $\eta _{T^i}$ by $\eta _i$ .

We claim that $\langle T^{\theta ^+},\vec {f}^{\theta ^+}\rangle $ forces that A is bounded in $\kappa $ , and in fact $A\subseteq \eta _{\theta ^+}$ . Suppose not, and pick a condition $\langle S,\vec {g}\rangle \le \langle T^{\theta ^+},\vec {f}^{\theta ^+}\rangle $ that forces some $\beta \geq \eta _{\theta ^+}$ to be in A. Without loss of generality, $\beta \le \eta _S$ .

Consider $Y:=\theta \setminus \circledcirc (\vec {g}(\beta ),\vec {f}^{\theta ^+}(\eta _{\theta ^+}))$ which is an element of $\mathcal I$ . Recalling that we have invoked Lemma 3.6 with $\delta :=\theta ^++1$ , every element of $T_{\eta _{\theta ^+}}$ is equal modulo bounded to $\vec {f}^{\theta ^+}(\eta _{\theta ^+})(\tau )$ for some $\tau <\theta $ . Consequently, we may fix a map $\pi :Y\rightarrow \theta $ such that $\vec {g}(\beta )(\tau )\restriction \eta _{\theta ^+}=^*\vec {f}^{\theta ^+}(\eta _{\theta ^+})(\pi (\tau ))$ for every $\tau \in Y$ . As the elements of $\vec g(\beta )$ are mutually exclusive, it follows that $\pi $ is an injection from Y to Y. To summarize, we have found a $Y\in \mathcal I$ and an injection $\pi :Y\rightarrow Y$ such that the following two hold:

1. (1) $\vec {g}(\beta )(\tau )\restriction \eta _{\theta ^+}=~\vec {f}^{\theta ^+}(\eta _{\theta ^+})(\tau )$ for every $\tau \in \theta \setminus Y$ ;

2. (2) $\vec {g}(\beta )(\tau )\restriction \eta _{\theta ^+}=^*\vec {f}^{\theta ^+}(\eta _{\theta ^+})(\pi (\tau ))$ for every $\tau \in Y$ .

Next, as $\operatorname {\mathrm {dom}}(\vec {g}(\beta ))=\theta $ , we may find a large enough $i<\theta ^+$ such that:

1. (3) $\vec {g}(\beta )(\tau )\restriction \eta _{\theta ^+}=(\vec {g}(\beta )(\tau )\restriction \eta _i)*\vec {f}^{\theta ^+}(\eta _{\theta ^+})(\pi (\tau ))$ for every $\tau \in Y$ .

In particular:

1. (4) $\vec {g}(\beta )(\tau )\restriction \eta _{i+1}=~\vec {f}^{i+1}(\eta _{i+1})(\tau )$ for every $\tau \in \theta \setminus Y$ , and

2. (5) $\vec {g}(\beta )(\tau )\restriction \eta _{i+1}=^*\vec {f}^{i+1}(\eta _{i+1})(\pi (\tau ))$ for every $\tau \in Y$ .

Consider $\vec {x}:=\langle \vec {g}(\beta )(\tau ) \restriction \eta _{i+1} \mid \tau <\theta \rangle $ . Then the triple $(\vec x,Y,\pi )$ satisfies requirements (A) and (B) with respect to $\langle T^{i+1},\vec {f}^{i+1}\rangle $ . Therefore, there exists an $\alpha <\eta _{i+2}$ such that:

1. (6) $\langle T^{i+2},\vec {f}^{i+2}\rangle \Vdash _{\mathbb S^\kappa _{\mathbf X}}\alpha \in \dot A$ ;

2. (7) $X_\xi \setminus Y\subseteq \circledcirc (\vec {f}^{i+2}(\alpha ),\vec {f}^{i+2}(\eta _{i+2}))$ ;

3. (8) $\vec {f}^{i+2}(\alpha )(\tau ) \subseteq x_\tau *\vec {f}^{i+2}(\eta _{i+2})(\pi (\tau ))$ for every $\tau \in X_\xi \cap Y$ .

By Clauses (1) and (7), for every $\tau \in X_\xi \setminus Y$ ,

$$ \begin{align*}\vec{f}^{i+2}(\alpha)(\tau)\subseteq \vec{f}^{i+2}(\eta_{i+2})(\tau)\subseteq \vec{f}^{\theta^+}(\eta_{\theta^+})(\tau)\subseteq \vec{g}(\beta)(\tau).\end{align*} $$

By Clauses (3) and (8), for every $\tau \in X_\xi \cap Y$ , we have

$$ \begin{align*}\begin{aligned} \vec{f}^{i+2}(\alpha)(\tau)&\subseteq x_\tau*\vec{f}^{i+2}(\eta_{i+2})(\pi(\tau))\\ &= (\vec{g}(\beta)(\tau)\restriction \eta_{i+1})*\vec{f}^{i+2}(\eta_{i+2})(\pi(\tau))\\ &\subseteq(\vec{g}(\beta)(\tau)\restriction \eta_{i+1})*\vec{f}^{\theta^+}(\eta_{\theta^+})(\pi(\tau))\\ &=\vec{g}(\beta)(\tau)\restriction \eta_{\theta^+}\subseteq \vec g(\beta)(\tau). \end{aligned}\end{align*} $$

As $\vec {f^{i+2}}\subseteq \vec {f}^{\theta ^+}\subseteq \vec g$ , we infer that $\langle S,\vec {g} \rangle $ forces that $\beta \in A$ is a proper extension of $\alpha \in A$ , contradicting the fact that A is an antichain.

Proof. (1) $T(G)$ is a uniformly homogeneous $\kappa $ -tree thanks to Lemmas 3.5 and 3.7. Towards a contradiction, suppose that $T(G)$ admits a $\kappa $ -sized antichain. It then follows from the upcoming Clause (4) that there exists an injective sequence $\langle (\alpha _\gamma ,\tau _\gamma )\mid \gamma <\kappa \rangle $ of elements of $\kappa \times X_0$ such that $\{ \vec {f^G}(\alpha _\gamma )(\tau _\gamma )\mid \gamma <\kappa \}$ is an antichain in $T(G)$ . Find a $\tau \in X_0$ for which $\Gamma :=\{\gamma <\kappa \mid \tau _\gamma =\tau \}$ has size $\kappa $ . Then $\{\alpha _\gamma \mid \gamma \in \Gamma \}$ is a $\kappa $ -sized antichain in $\mathbb A^\kappa _{\mathbf X,0}$ , contradicting Lemma 3.11.

(2) By Definition 3.2(3), it suffices to prove that for every $\varepsilon <\kappa $ , there exists a $\langle T,\vec f\rangle \in G$ such that $\eta _T\in \operatorname {\mathrm {acc}}(\kappa \setminus \varepsilon )$ . Now, this follows from Lemma 3.6, as demonstrated by the proof of Lemma 3.10.

(3) By Definition 3.2(2).

(4) By a density argument, using Definition 3.2(4).

Proof. For an ordinal $\tau $ , let $\mathbb P_\tau $ be the Easton-support iteration of length $\tau $ which forces with $\mathbb S^\eta _{\mathbf X}*\mathbb A^\eta _{\mathbf X}$ at every strongly inaccessible cardinal $\eta <\tau $ . As usual, at $\eta <\tau $ that is not strongly inaccessible, we use trivial forcing.

Let $G_\kappa $ be $\mathbb P_\kappa $ -generic over V. Let g be $\mathbb S^\kappa _{\mathbf X}$ -generic over $V[G_\kappa ]$ and let h be $\mathbb A^\kappa _{\mathbf X, 0}$ -generic over $V[G_\kappa ][g]$ . Note that $G_{\kappa +1}:=G_\kappa *g*h$ is $\mathbb P_{\kappa +1}$ -generic over V.

By Corollary 3.12, in $V[G_\kappa ][g]$ , $T(g)$ is a uniformly homogeneous $\kappa $ -Souslin tree admitting an $\mathcal F$ -ascent path $\vec f^g$ , and $V(T(g))$ covers a club in $\kappa $ . By [Reference Rinot, Yadai and You13, Lemma 2.15], then,

$$ \begin{align*}V[G_\kappa][g]\models\text{"there exists a }\kappa\text{-Souslin tree }T\text{ such that }V(T)=\operatorname{\mathrm{acc}}(\kappa)."\end{align*} $$

It thus remains to prove that $\chi (\kappa )\le 1$ holds in $V[G_\kappa ][g]$ . To this end, let $\vec {C}=\langle C_\beta \mid \beta <\kappa \rangle $ be an arbitrary C-sequence in $V[G_\kappa ][g]$ .

Work in V. Let $\dot {C}$ be a $\mathbb P_\kappa *\dot {\mathbb S}^\kappa _{\mathbf X}$ -name for $\vec {C}$ . Take a $\kappa $ -model M containing $\mathbb P_\kappa *\dot {\mathbb S}^\kappa _{\mathbf X}*\dot {\mathbb A}^{\kappa }_{\mathbf X}$ and $\dot {C}$ . As $\kappa $ is weakly compact, we may now pick a $\kappa $ -model N and a nontrivial elementary embedding $j_0:M \rightarrow N$ with critical point $\kappa $ . Hereafter, work in $V[G_{\kappa +1}]$ .

Claim 4.1.1. $j_0$ may be lifted to an elementary embedding

$$ \begin{align*}j_1:M[G_\kappa]\rightarrow N[G_{\kappa+1}*G_{tail}].\end{align*} $$

Claim 4.1.2. $j_1$ may be lifted to an elementary embedding

$$ \begin{align*}j_2:M[G_\kappa*g]\rightarrow N[G_{\kappa+1}*G_{tail}*g_{tail}].\end{align*} $$

Proof. For every $n\in X_0$ , define $b_n:\kappa \rightarrow \kappa $ via:

$$ \begin{align*}b_n:=\bigcup\{\vec f^g(\alpha)(n)\mid \alpha\in h\}.\end{align*} $$

As h is $\mathbb A^\kappa _{\mathbf X, 0}$ -generic over $V[G_\kappa ][g]$ , each such $b_n$ is a cofinal branch through $T(g)$ . Furthermore, the elements of $\langle b_n\mid n\in X_0\rangle $ are mutually exclusive. Take $n_0 \in X_0\setminus X_1$ and then let

$$ \begin{align*}T:=T(g)\cup\{ x*b_n\mid x\in T(g), n\in X_0\ \&\ n \neq n_0\}.\end{align*} $$

Then T is a uniformly homogeneous tree for which $b_{n_0}$ is a vanishing $\kappa $ -branch. Thus, $V(T)=V(T(g))\cup \{\kappa \}$ . Next, we extend the ascent path $\vec {f}^g$ by setting $\vec {f}:=\vec {f}^g\cup \{ (\kappa ,\langle b_{\pi (n)}\mid n<\omega \rangle )\}$ , where $\pi :\omega \rightarrow X_0 \setminus \{n_0\}$ is some bijection such that $\pi \restriction X_1$ is the identity function.

Altogether, $\langle T, \vec {f} \rangle $ is a legitimate condition of $\mathbb S^{j_1(\kappa )}_{\mathbf X}$ as computed in $N[G_{\kappa +1}]$ , in particular, it is an element of the last iterand of $j_1(\mathbb P_\kappa *\dot {\mathbb S}^\kappa _{\mathbf X})/ G_{\kappa +1}*G_{tail}$ . In addition, $\langle T, \vec {f} \rangle $ extends every condition lying in $j_1[g]$ .

Since ${}^{<\kappa }N[G_{\kappa +1}] \subseteq N[G_{\kappa +1}]$ and $G_{tail}$ is a generic for a $\kappa $ -strategically closed forcing, ${}^{<\kappa }N[G_{\kappa +1}*G_{tail}] \subseteq N[G_{\kappa +1}*G_{tail}]$ . By Lemma 3.10, $\mathbb S^{j_1(\kappa )}_{\mathbf X}$ is ${<}j_1(\kappa )$ -strategically closed in $N[G_{\kappa +1}*G_{tail}]$ . But ${}^{<\kappa }N[G_{\kappa +1}*G_{tail}] \subseteq N[G_{\kappa +1}*G_{tail}]$ , and hence $\mathbb S^{j_1(\kappa )}_{\mathbf X}$ is $\kappa $ -strategically closed. As N has no more than $\kappa $ many $j_1(\mathbb P_{\kappa })$ -names, the model $N[G_{\kappa +1}*G_{tail}]$ has no more than $\kappa $ many dense sets of $j_1(\mathbb P_\kappa *\dot {\mathbb S}^\kappa _{\mathbf X})/ G_{\kappa +1}*G_{tail}$ . It thus follows that a $j_1(\mathbb P_\kappa *\dot {\mathbb S}^\kappa _{\mathbf X})/ G_{\kappa +1}*G_{tail}$ -generic over $N[G_{\kappa +1}*G_{tail}]$ may be recursively constructed, say $g_{tail}$ , so that $N[G_{\kappa +1}*G_{tail}][g_{tail}]$ is a $j_1(\mathbb P_\kappa *\dot {\mathbb S}^\kappa _{\mathbf X})$ -generic extension of N and $\langle T,\vec f\rangle \in g_{tail}$ . By Silver’s criterion, we may now lift $j_1$ to a $j_2:M[G_\kappa ][g]\rightarrow N[G_{\kappa +1}*G_{tail}][g_{tail}]$ , as sought.

Let $j_2:M[G_\kappa ][g]\rightarrow N[G_{\kappa +1}*G_{tail}*g_{tail}]$ be given by the preceding claim. Clearly, $\vec C$ is in $M[G_\kappa ][g]$ , so that $j_2(\vec C)$ is a C-sequence over $j_2(\kappa )$ . Let $C_\kappa $ denote $j_2(\vec {C})(\kappa )$ .

Proof. Suppose not, and let $\alpha <\kappa $ be a counterexample. Denote $A:=C_\kappa \cap \alpha $ , and define a map $\varphi :\kappa \setminus \alpha \rightarrow \alpha $ via:

$$ \begin{align*}\varphi(\beta):=\min\{\varepsilon<\alpha\mid A\cap\varepsilon\neq C_\beta\cap\varepsilon\}.\end{align*} $$

By elementarity, $j_2(A)\cap j_2(\varphi )(\kappa )\neq C_\kappa \cap j_2(\varphi )(\kappa )$ . However, $j_2(A)=A=C_\kappa \cap \alpha \subseteq \alpha>j_2(\varphi )(\kappa )$ . This is a contradiction.

Recall that $C_\kappa $ is a club in $\kappa $ lying in $V[G_\kappa ][g][h]$ . As h is $\mathbb A^\kappa _{\mathbf X}$ -generic over $V[G_\kappa ][g]$ , and as $\mathbb A^\kappa _{\mathbf X}$ satisfies the $\kappa $ -cc, it follows that there exists a club $D \subseteq C_\kappa $ with $D\in V[G][g]$ . Evidently, for every $\alpha <\kappa $ , there exists a $\beta \ge \alpha $ such that $D \cap \alpha \subseteq C_\beta \cap \alpha $ .

Proof. Work in the model of Theorem 4.1. We have $\chi (\kappa )=1$ which implies that $\chi (\kappa )\le 1$ , i.e., Clause (II) holds. In addition, there exists a $\kappa $ -Souslin tree T such that $V(T)=\operatorname {\mathrm {acc}}(\kappa )$ . An inspection of the proof of [Reference Rinot and Shalev12, Theorem 2.23] makes it clear that there exists a club $D\subseteq \kappa $ such that for every partition $\mathcal S$ of D into stationary sets, $\clubsuit _{\operatorname {\mathrm {AD}}}(\mathcal S,\omega ,{<}\omega )$ holds. In particular, $\clubsuit _{\operatorname {\mathrm {AD}}}(\{D\},1,1)$ holds. An inspection of the proof of [Reference Rinot and Shalev12, Theorem 2.30] then yields that Clause (I) must fail.

Question 4.4. Is Postulate (I) strictly weaker than weak compactness?