Fact 3. The following properties are very easy to verify.

1. (i) $\psi |_{\mathbb {R}^{\mathsf {R}}}=\psi |_{\mathbb {R}^{\mathsf {U}}}=\psi |_{\mathbb {R}^\xi }=\text {id}$ , and the same for $\psi '$ in place of $\psi $ .

2. (ii) $\psi [\mathbb {U}] = \mathbb {P}^\xi $ , and analogously $\psi '[\mathbb {U}] = \mathbb {P}^{\xi '}$ .

3. (iii) $\varphi |_{\mathbb {R}^{\mathsf {R}}}=\text {id}$ , and the same for $\varphi '$ in place of $\varphi $ .

4. (iv) $\varphi [\mathbb {R}^{\mathsf {U}}] = \mathbb {R}^\xi $ , and analogously $\varphi '[\mathbb {R}^{\mathsf {U}}] = \mathbb {R}^{\xi '}$ .

5. (v) $\psi |_{\mathbb {S}^{\xi '}}=\text {id}$ , and analogously $\psi '|_{\mathbb {S}^{\xi }}=\text {id}$ .

6. (vi) $\varphi |_{\mathbb {S}^{\xi '}}=\text {id}$ , and analogously $\varphi '|_{\mathbb {S}^{\xi }}=\text {id}$ .

7. (vii) $\varphi |_{\mathbb {P}^\xi }=\text {id}$ , and analogously $\varphi '|_{\mathbb {P}^{\xi '}}=\text {id}$ .

Fact 4. For every $\theta \in \{\psi , \varphi , \psi ', \varphi '\}: \theta (p \oplus p')= \theta (p)\oplus \theta (p')$ .

Proof. Let $\alpha \in \kappa $ and $q \in A_\alpha (\dot z)$ . We decompose $q=q^{\mathsf {R}} \oplus q^{\mathsf {U}} \oplus u$ as described above. From Fact 4 it follows that $\bar \chi (q)=\bar \chi (q^{\mathsf {R}}) \oplus \bar \chi (q^{\mathsf {U}}) \oplus \bar \chi (u)$ , and it therefore suffices to show that $q^{\mathsf {R}}$ , $q^{\mathsf {U}}$ and u are fixed by $\bar \chi $ . We use Fact 3.

This finishes the proof of Lemma 3.

Proof. We begin with the first equality. We have $\dot a_{\xi }=\bigcup _{\alpha \in \kappa } \{\check \alpha \}\times A_\alpha (\dot a_\xi )$ and thus, $\chi (\dot a_\xi )= \bigcup _{\alpha \in \kappa } \{\check \alpha \}\times \chi [A_\alpha (\dot a_\xi )]$ . Therefore, we must show that for every $\alpha \in \kappa : \chi [A_\alpha (\dot a_\xi )]=A_\alpha (\dot a_{\xi '})$ .

Let $\alpha \in \kappa $ . First, we deal with $\chi [A_\alpha (\dot a_\xi )] \subseteq A_\alpha (\dot a_{\xi '})$ . Thus, let $p\in A_\alpha (\dot a_\xi )$ . We know that $\varphi \circ \psi (\dot a_\xi )= \dot z$ , which implies that $q:=\varphi \circ \psi (p)\in A_\alpha (\dot z)$ . We also know that $\psi ' \circ \varphi ' (\dot z)=\dot a_{\xi '}$ , and therefore $\psi ' \circ \varphi ' (q) \in A_\alpha (\dot a_{\xi '})$ . Since $\bar \chi (q)=q$ by Lemma 3, we indeed obtain

$$\begin{align*}\chi(p)=\psi' \circ \varphi' \circ \bar \chi \circ \varphi \circ \psi (p)\in A_\alpha(\dot a_{\xi'}).\end{align*}$$

The reverse inclusion $A_\alpha (\dot a_\xi ) \supseteq \chi ^{-1}[A_\alpha (\dot a_{\xi '})]$ follows from essentially the same proof: Note that $\chi ^{-1}=\psi \circ \varphi \circ \bar \chi ^{-1} \circ \varphi ' \circ \psi '$ , and by Lemma 3, $\bar \chi ^{-1}$ is the identity on $A_\alpha (\dot z)$ as well.

To show the second equality, i.e., $\chi (\dot a_{\xi '})=\dot a_\xi $ , we again fix $\alpha \in \kappa $ and show $\chi [A_\alpha (\dot a_{\xi '})]=A_\alpha (\dot a_\xi )$ . Here, we have to deal with the entire $\chi $ at once, we again use Fact 3. To verify the direction “ $\subseteq $ ”, let $p' \in A_\alpha (\dot a_{\xi '})$ . Since $p' \in \mathbb {S}^{\xi '}$ , we have $\psi (p')=p'$ by (v) and $\varphi (p')=p'$ by (vi). The next two automorphisms map $p'$ to $\varphi '(\psi '(p'))$ , which is equal to a condition $q \in A_\alpha (\dot z)$ , since $\varphi '(\psi '(\dot a_{\xi '}))=\dot z$ . Then, q is mapped to $\psi (\varphi (q))$ , which is some $p \in A_\alpha (\dot a_{\xi })$ , because $\psi (\varphi (\dot z))=\dot a_\xi $ . The last two automorphisms $\varphi '$ and $\psi '$ fix p, again by (v) and (vi).

Finally, the proof of the reverse inclusion $\chi [A_\alpha (\dot a_{\xi '})] \supseteq A_\alpha (\dot a_\xi )$ is analogous and left as an exercise to the reader.

Proof. The first statement is easy to verify. To show the second statement, let A be a $\kappa ^{+}$ -sized set of $\mathbb {T}^{\leq \kappa }$ -conditions. For each $p\in A$ , let $S_p:=\bigcup \{\Delta ^{p(\delta )}\times \eta ^{p(\delta )}: \delta \in \text {supp}(p)\}$ . Note that $S_p$ has cardinality ${<}\kappa $ . By the $\Delta $ -system Lemma, we obtain some $A'\subseteq A$ of cardinality $\kappa ^{+}$ and for each $\delta \in \text {dom}(E^{\leq \kappa })$ some $R^{\delta }\in [\mathcal {I}^{\delta }]^{{<}\delta }$ and some $r^{\delta } \in \delta $ , such that for all these $\delta $ and all $p \neq p' \in A': (\Delta ^{p(\delta )}\times \eta ^{p(\delta )}) \cap (\Delta ^{p'(\delta )}\times \eta ^{p'(\delta )})= R^{\delta } \times r^{\delta } $ . Note that the set $C:=\{\delta : R^{\delta } \times r^{\delta } \neq \emptyset \}$ has cardinality ${<}\kappa $ . For each $\delta \in C$ , there is at most one $p\in A'$ with $\eta ^{p(\delta )}\neq r^{\delta }$ . By removing these ${<}\kappa $ many conditions, we can assume that no such p exist in $A'$ .

The set $\bigcup \{R^{\delta } \times r^{\delta }: \delta \in C \}$ has cardinality ${<}\kappa $ . By the $\mathsf {GCH}$ , we have $2^{{<}\kappa }=\kappa $ , and we can therefore assume that for all $p,p' \in A'$ and all $\delta \in \text {dom}(E^{\leq \kappa })$ , the functions $p(\delta )$ and $p'(\delta )$ agree on the intersection of their domains. It is now easy to verify that the conditions in $A'$ are pairwise compatible.

Proof. Let $\kappa \in \text {dom}(E)$ and $\lambda \in [\kappa ^{+}, E(\kappa )]$ . As before, decompose $\mathbb {T}$ as $\mathbb {T}^{\leq \kappa } \times \mathbb {T}^{{>}\kappa }$ and $G=G^{\leq \kappa }\times G^{{>}\kappa }$ accordingly. Since $\mathbb {T}^{{>}\kappa }$ is $\kappa ^{+}$ -closed, the $\mathsf {GCH}$ at $\delta \leq \kappa $ still holds in $\mathbf {V}[G ^{> \kappa }]$ and $(\mathbb {T}^{\leq \kappa })^{\mathbf {V}[G ^{> \kappa }]}=(\mathbb {T}^{\leq \kappa })^{\mathbf {V}}$ . We work in $\mathbf {W}:=\mathbf {V}[G ^{> \kappa }]$ .

Since $\kappa $ and $\lambda $ are fixed and since we are only interested in the $\kappa $ -th coordinate of each $\mathbb {T}^{\leq \kappa }$ -condition p, define for notational simplicity for each $p \in \mathbb {T}^{\leq \kappa } $ the following abbreviation $q_p$ :

1. (i) $\forall \xi \in E(\kappa )\;\forall \alpha \in \kappa : q_p(\xi , \alpha ):=p(\kappa )(\langle \kappa , \xi \rangle , \alpha ).$

2. (ii) $\Delta ^{q_p}:= \Delta ^{p(\kappa )}.$

3. (iii) $\eta ^{q_p}:=\eta ^{p(\kappa )}$ .

In $\mathbf {W}[G^{\leq \kappa }]$ , define for each $\xi \in E(\kappa )$ the $\kappa $ -real $g_\xi := \{\alpha \in \kappa : \exists p \in G^{\leq \kappa }: q_p(\xi , \alpha ) =1\}$ . We assume that $\lambda < E(\kappa )$ and define $a_\xi :=\text {cl}(g_\xi \setminus g_\lambda )$ for all $\xi < \lambda $ . We show that the sequence $\langle a_{\xi }: \xi \in \lambda \rangle $ is a $\kappa $ -tower of length $\lambda $ in $\mathbf {W}[G^{\leq \kappa }]$ . If $\lambda = E(\kappa )$ , it follows by a very similar but simplified argument that setting $a_\xi := \text {cl}(g_\xi )$ yields a $\kappa $ -tower of length $E(\kappa )$ .

It is easy to see that $\langle g_{\xi }: \xi \in E(\kappa )\rangle $ is well-ordered by $\supseteq ^{*}$ , and therefore, $\langle a_{\xi }: \xi \in \lambda \rangle $ is as well. In order to show that $\langle a_{\xi }: \xi \in \lambda \rangle $ does not have a pseudo-intersection in $\mathbf {W}[G^{\leq \kappa }]$ , let $\dot x$ be a $\mathbb {T}^{\leq \kappa }$ -name for a subset of $\kappa $ and $p_0 \in G^{\leq \kappa }$ a condition such that $p_0 \Vdash {"}|\dot x|=\kappa "$ . For each $\alpha \in \kappa $ , let $A_{\alpha }$ be a maximal antichain deciding $"\alpha \in \dot x"$ . By the $\kappa ^{+}$ -c.c. of $\mathbb {T}^{\leq \kappa }$ , the set $\Delta := \bigcup \{\Delta ^{q_p}: p \in A_\alpha ,\;\alpha \in \kappa \}$ has cardinality at most $\kappa $ . Thus, by regularity of $\lambda $ , there exists $\langle \kappa , \xi _0\rangle \in \mathcal {I}^{\kappa }$ such that $\xi < \xi _0 < \lambda $ for every $\xi < \lambda $ with $\langle \kappa , \xi \rangle \in \Delta $ . We show that for every $\nu \in \kappa $ , the set of conditions forcing $"\dot x \setminus \nu \nsubseteq \dot {a}_{\xi _0}"$ is dense below $p_0$ .

Let $p \leq p_0$ . By extending p, we can assume that $\langle \kappa , \lambda \rangle \in \Delta ^{q_p}$ . Since $p \Vdash {"}|\dot x|=\kappa " $ , there exists $\alpha _0> \max \{\eta ^{q_p}, \nu \}$ and $\bar p \leq p$ with $\bar p \Vdash \check \alpha _0 \in \dot x$ . Therefore $\bar p$ is compatible with some $r \in A_{\alpha _0}$ via some common extension s. In particular, p and r are compatible via s. Without loss of generality, we can assume that $\langle \xi _0, \alpha _0 \rangle , \langle \lambda , \alpha _0\rangle \in \text {dom}(q_s)$ .

Note that for all $\xi _0 \leq \xi < \lambda $ and all $\max \{\eta ^{q_p}, \nu \}\leq \alpha \leq \alpha _0$ with $\langle \xi , \alpha \rangle \in \text {dom}(q_{s}):\langle \xi , \alpha \rangle \notin \text {dom}(q_p) \cup \text {dom}(q_r)$ , since $\alpha \geq \eta ^{q_p}$ and by the choice of $\xi _0$ . Therefore, we can set $\bar s$ equal to s except that for all such $\xi $ and $\alpha : q_{\bar s}(\xi , \alpha ):=\min \{q_s(\xi , \alpha ),\,q_s(\lambda , \alpha )\}$ . It follows that $\bar s$ is a common extension of p and r, and for every $\max \{\eta ^{q_p}, \nu \}\leq \alpha \leq \alpha _0: \;\bar s \Vdash {"}\check \alpha \in \dot g_{\xi _0} \implies \check \alpha \in \dot g_{\lambda }"$ . Thus, $\bar s \Vdash {"}\check \alpha _0 \in \dot x \setminus \text {cl}(\dot g_{\xi _0}\setminus \dot g_{\lambda })"$ , finishing the proof of the proposition.

Fact 5. If $f: \mathbb {Q} \to \mathbb {Q}'$ is a cofinal embedding, then $\mathfrak {b}(\mathbb {Q}')=\mathfrak {b}(\mathbb {Q})$ and $\mathfrak {d}(\mathbb {Q}')=\mathfrak {d}(\mathbb {Q})$ .

Proof. Let p be a $\mathbb {P}$ -condition iff p is a well-founded subset of $[\mu ]^{{<}\beta }$ of cardinality ${<}\beta $ . The order is given by

$$\begin{align*}q \leq p:\iff p \subseteq q \;\,\text{and}\,\; \forall x \in p \;\forall y \in q \setminus p: y \nsubseteq x.\end{align*}$$

Now, let H be $\mathbb {P}$ -generic over $\mathbf {V}$ and define $\mathbb {Q}^{*}:=\bigcup H$ . By the above claim, cardinalities and cofinalities are preserved in $\mathbf {V}[H]$ and we have $([\mu ]^{<\beta })^{\mathbf {V}[H]}=([\mu ]^{<\beta })^{\mathbf {V}}$ .

It is easy to see that for every $x \in [\mu ]^{{<}\beta }$ , the set $\mathcal {D}_x:=\{p \in \mathbb {P}: \exists y \in p: y \supseteq x\}$ is open dense in $\mathbb {P}$ , by adding $\bigcup p \cup x$ to the p in question. Thus, $\mathbb {Q}^{*}$ is indeed cofinal in $[\mu ]^{{<}\beta }$ . Well-foundedness of $\mathbb {Q}^{*}$ follows from H being directed. By the same reason, we have that for every $x \in \mathbb {Q}^{*}: \{y \in \mathbb {Q}^{*}: y \subseteq x\} \subseteq p$ , where $p \in H$ is any condition containing x. Thus $|\{y \in \mathbb {Q}^{*}: y \subseteq x\}| < \beta $ .

It remains to show (iii) and (iv). In order to verify $\mathfrak {b}(\mathbb {Q}^{*})=\beta $ and $\mathfrak {d}(\mathbb {Q}^{*})=\mu $ , it suffices, by Fact 5, to verify $\mathfrak {b}(([\mu ]^{<\beta }, \subseteq ))= \beta $ and $\mathfrak {d}(([\mu ]^{<\beta }, \subseteq ))=\mu $ in $\mathbf {V}[H]$ . To check the first statement, note that by regularity of $\beta $ , every $B \subseteq [\mu ]^{<\beta }$ of cardinality ${<}\beta $ is bounded. On the other hand, for any $X \in [\mu ]^{\beta }$ , the set $\{\{\eta \}: \eta \in X\}$ is unbounded, which yields $\mathfrak {b}(([\mu ]^{<\beta }, \subseteq ))= \beta $ .

Similarly, any $D \subseteq [\mu ]^{<\beta }$ of cardinality ${<}\mu $ cannot be dominating, since $\bigcup D \neq \mu $ . This gives us $\mathfrak {d}([\mu ]^{<\beta })\geq \mu $ . The reverse inequality holds because $|[\mu ]^{<\beta }|=\mu $ , which follows by the assumption $\text {cf}(\mu )\geq \beta $ and by the $\mathsf {GCH}$ in $\mathbf {V}$ . Since $\mathbb {Q}^{*}$ is itself cofinal, this also yields $|\mathbb {Q}^{*}|=\mu $ .

Proof. Parts (i) and (ii) are proved in [Reference Cummings and Shelah3, Claims 1 and 2]. Part (iii) is straightfoward to check. For part (iv), let $|\mathbb {Q}|= \rho $ with $\rho ^\kappa =\rho $ and let $a \in \mathbb {Q}\cup \{\text {top}\}$ . Assume by induction that for all $b < a$ there are at most $\rho $ many nice $\mathbb {D}(\mathbb {Q}_b)$ -names for subsets of $\kappa $ . In particular, there are at most $\rho $ many nice $\mathbb {D}(\mathbb {Q}_b)$ -names for elements of $\kappa ^\kappa $ . Since $\mathbb {D}(\mathbb {Q}_a)$ satisfies the $\kappa ^{+}$ -c.c., the number of nice $\mathbb {D}(\mathbb {Q}_a)$ -names for subsets of $\kappa $ is bounded by $|\mathbb {D}(\mathbb {Q}_a)|^{\kappa }$ . Note that $|\mathbb {D}(\mathbb {Q}_a)| \leq |\mathbb {Q}_a|^{{<}\kappa } \cdot \kappa ^{<\kappa } \cdot \rho ^{{<}\kappa }$ by the induction hypothesis. This is at most $\rho $ , because $\mathbb {Q}_a\subseteq \mathbb {Q}$ and $\rho ^{{<}\kappa }=\rho $ , which finally yields that there are at most $\rho ^\kappa =\rho $ nice $\mathbb {D}(\mathbb {Q}_a)$ -names for subsets of $\kappa $ .

Proof. We have $|\mathbb {Q}^{*}|=\mu $ by Lemma 6 (iv), which implies by Lemma 7 (iv) that there are at most $\mu $ many nice $\mathbb {D}(\mathbb {Q}^{*})$ -names for subsets of $\kappa $ . Thus, $\mathbf {W}[G]\models 2^\kappa \leq \mu $ . In order to verify the remaining claims, it suffices by the above Lemma 8 and by Fact 5 to check that $\mathfrak {b}(\mathbb {Q}^{*})=\beta $ and $\mathfrak {d}(\mathbb {Q}^{*})=\mu $ still holds in $\mathbf {W}[G]$ . However, this very easily follows from $\mathbb {D}(\mathbb {Q}^{*})$ satisfying the $\kappa ^{+}$ -c.c.

Proof. Assume towards a contradiction that $\langle a_\xi : \xi \in \beta ^{+}\rangle $ is a strictly $\supseteq ^{*}$ -descending sequence in $\mathbf {W}[G]$ . For each $\xi \in \beta ^{+}$ , let $\dot {a}_\xi =\bigcup _{\alpha \in \kappa }\{\check {\alpha }\}\times A_\alpha ^\xi $ be a nice $\mathbb {D}(\mathbb {Q}^{*})$ -name for $a_\xi $ . Assume $p_0\in \mathbb {D}(\mathbb {Q}^{*})$ is such that for all $\xi < \xi ' < \beta ^{+}: p_0 \Vdash _{\mathbb {D}(\mathbb {Q}^{*})} \dot {a}_\xi \supsetneq ^{*} \dot {a}_{\xi '}$ .

Define for every $\xi \in \beta ^{+}$ the set

$$\begin{align*}d_\xi:=\bigcup \{x: x \in \text{dom}(p), \;p \in A_\alpha^\xi,\;\alpha \in \kappa\} \;\cup\; \bigcup \{x: x \in \text{dom}(p_0)\},\end{align*}$$

which is a subset of $\mu $ of size ${<}\beta $ . Since $\mathbb {Q}^{*}$ is cofinal in $[\mu ]^{<\beta }$ , we find for each $\xi \in \beta ^{+}$ some $D_\xi \supseteq d_\xi $ in $\mathbb {Q}^{*}$ . As noted before, the $\mathsf {GCH}$ holds in $\mathbf {W}$ below $\beta $ and we may therefore apply the $\Delta $ -system Lemma to the family $\{D_\xi : \xi \in \beta ^{+}\}$ to obtain some $X \subseteq \beta ^{+}$ of cardinality $\beta ^{+}$ and a root R. Set $\mathbb {Q}^{*}_\xi :=\{y \in \mathbb {Q}^{*}: y \subseteq D_\xi \}$ and $\mathbb {R}:=\{y \in \mathbb {Q}^{*}: y \subseteq R\}$ . Note that $\mathbb {R}$ is the root of the $\mathbb {Q}^{*}_\xi $ . By Lemma 6 (ii), we have $| \mathbb {Q}^{*}_\xi |<\beta $ , and we may therefore assume by the pigeonhole principle that $\forall \xi \in X: |\mathbb {Q}^{*}_\xi |=\theta < \beta $ .

Define the downward-closed partially ordered set $\mathbb {A}:=\bigcup _{\xi \in X'}\mathbb {Q}^{*}_{\xi }$ . Note that by definition of $D_\xi $ , $\dot a_\xi $ is a nice $\mathbb {D}(\mathbb {Q}^{*}_\xi )$ -name and thus a nice $\mathbb {D}(\mathbb {A})$ -name. Furthermore, $p_0$ is a $\mathbb {D}(\mathbb {R})$ -condition. For a fixed $\xi _0 \in X'$ , the isomorphism $\psi _{\xi , \xi _0}$ extends to an automorphism of order 2 of $\mathbb {A}$ , which we denote by $\psi _{\xi }$ . This automorphism $\psi _{\xi }$ naturally induces an automorphism $\varphi _{\xi }$ of $\mathbb {D}(\mathbb {A})$ in the obvious way: Let ${a \in \mathbb {A}\cup \{\text {top}\}}$ and assume by induction that for every $b< a$ , the isomorphism

$$\begin{align*}\varphi_{\xi}|_{\mathbb{D}(\mathbb{A}_b)}: \mathbb{D}(\mathbb{A}_b) \to \mathbb{D}(\mathbb{A}_{\psi_{\xi}(b)})\end{align*}$$

has been defined (note the abuse of notation). In particular, this isomorphism extends to $\mathbb {D}(\mathbb {A}_b)$ -names. Now let p be any $\mathbb {D}(\mathbb {A}_a)$ -condition. We write for every $b \in \text {dom}(p): p(b)=\langle s(b), \dot f(b)\rangle $ , and define

$$\begin{align*}\varphi_{\xi}|_{\mathbb{D}(\mathbb{A}_a)}(p):=q, \;\text{where}\; \begin{cases} \text{dom}(q):=\psi_{\xi}[\text{dom}(p)] \;\;\text{and}\\ \forall \psi_{\xi}(b) \in \text{dom}(q): q(\psi_{\xi}(b)):= \langle s(b), \varphi_{\xi}|_{\mathbb{D}(\mathbb{A}_b)}(\dot f)\rangle \end{cases} \end{align*}$$

It follows by induction that $\varphi _{\xi }$ is an automorphism and that $\varphi _{\xi }|_{\mathbb {D}(\mathbb {R})}=\text {id}$ .

Note that $\varphi _{\xi }(\dot a_\xi )$ is a nice $\mathbb {D}(\mathbb {Q}^{*}_{\xi _0})$ -name and that by Lemma 7 (iv), there are at most $|\mathbb {Q}^{*}_{\xi _0}|<\beta $ many nice $\mathbb {D}(\mathbb {Q}^{*}_{\xi _0})$ -names for subsets of $\kappa $ . Thus, we can extract $X" \subseteq X'$ of cardinality $\beta ^{+}$ such that $\varphi _{\xi }(\dot a_\xi )$ is the same nice $\mathbb {D}(\mathbb {Q}^{*}_{\xi _0})$ -name for all $\xi \in X"$ .

Fix $\xi < \xi ' \in X"\setminus \{\xi _0\}$ and define the automorphism $\chi _{\xi , \xi '}:=\varphi _{\xi '} \circ \varphi _{\xi } \circ \varphi _{\xi '}$ of $\mathbb {A}$ . By construction, $\chi _{\xi , \xi '}(\dot a_\xi )=\dot a_{\xi '}$ , $\chi _{\xi , \xi '}(\dot a_{\xi '})=\dot a_{\xi }$ and $\chi _{\xi , \xi '}(p_0)=p_0$ . Since $\mathbb {D}(\mathbb {A})$ is a complete suborder of $\mathbb {D}(\mathbb {Q}^{*})$ by Lemma 7 (iii), we have $p_0 \Vdash _{\mathbb {D}(\mathbb {A})} \dot a_\xi \supsetneq ^{*} \dot a_{\xi '} $ , which yields the contradiction $p_0 \Vdash _{\mathbb {D}(\mathbb {A})} \dot a_{\xi '} \supsetneq ^{*} \dot a_{\xi } \land \dot a_{\xi } \supsetneq ^{*} \dot a_{\xi '}$ , just as in the proof of Theorem 1.