Proof $(\mathrm {(a)}\Rightarrow \mathrm {(c)})$ Let $p\in \mathbb {Q}(\mathcal {C})$ . p is a perfect tree by the definition. Thus arrange $\operatorname {\mathrm {split}}(p)$ and assign by induction to each splitting node s a real x from $[p]$ extending s which was either already considered or belongs to a different set from $\mathcal {C}$ than all previously selected reals. This is possible since any $s\in \operatorname {\mathrm {split}}(p)$ may be extended to $t\in \operatorname {\mathrm {split}}(p)$ with $[p(t)]$ being disjoint with finitely many sets from $\mathcal {C}$ containing all previously selected reals. The set of all assigned branches is the required dense set.

$(\mathrm {(c)}\Rightarrow \mathrm {(b)})$ Trivial.

$(\mathrm {(b)}\Rightarrow \mathrm {(a)})$ Let $\beta <\omega _1$ and $s\in p$ . There are $x,y\in [p]$ such that $s\subseteq x,y$ and $\alpha _x\neq \alpha _y$ . We take $z\in \{x,y\}$ such that $\alpha _z\neq \beta $ . Since $z\in [p]\setminus C_{\beta }$ and $C_{\beta }$ is closed, there is $s\subseteq t\subseteq z$ such that $[p(t)]\cap C_{\beta }=\emptyset $ .

Proof We denote $p=\bigcap \{p_n\colon n\in \omega \}$ , $X=\bigcup \{X_n: n\in \omega \}$ , and we assume that we have $s\in p$ . We take $n\in \omega $ and $t\in \operatorname {\mathrm {split}}_n(p)$ such that t extends s. Since $\operatorname {\mathrm {split}}_n(p)=\operatorname {\mathrm {split}}_n(p_{n+1})$ , the set $X_{n+1}$ contains $\mathcal {C}$ -different branches extending t. Hence, X is dense in $[p]$ . One can easily see that X is contained in $[p]$ . Finally, by Lemma 2.4 we conclude that $p\in \mathbb {Q}(\mathcal {C})$ .

Proof Let $p\in \mathbb {Q}(\mathcal {C})$ . One can construct a decreasing sequence $\{q_i\}_{i\in \omega }$ of extensions of p with strictly increasing stems such that $q_n\Vdash \dot {f}\upharpoonright h(n)=f_n$ for some $f_n\in {}^{h(n)}\omega $ . We denote $s_n=\operatorname {\mathrm {stem}}(q_n)$ and we set $x=\bigcup _{i\in \omega }s_i$ . Finally, we take the amalgamation $q=\bigcup _{i\in \omega }q_i(s_i^{\smallfrown } \langle 1-x(|s_i|)\rangle )$ .

Proof Let $p\in \mathbb {Q}(\mathcal {C})$ . We build a fusion sequence $\{(q_n,X_n)\}_{n\in \omega }$ with $q_0\leq p$ such that its fusion q has the required property. Let the condition $q_0$ , branch x, and sequence $\{f_s\}_{s\in x\restriction \operatorname {\mathrm {split}}(q_0)}$ be obtained from Lemma 2.7 for p and $h(n)=n+1$ . We set $X_0=\{x\}$ .

Let $0\leq n<\omega $ . Suppose we have defined $q_n\in \mathbb {Q}(\mathcal {C})$ and finite $X_n\subseteq [q_n]$ . Let $s\in \operatorname {\mathrm {split}}_n(q_n)$ . Take the unique branch $x\in X_n$ extending s, node $r=x\restriction \operatorname {\mathrm {split}}_{n+1}(q_n)$ , and number $i=x(|s|)$ in $\{0,1\}$ . We set $q(s,i)=q_n(r)$ . Let $t\supseteq s^{\smallfrown } \langle 1-i\rangle $ be such that $[q_n(t)]\cap C_{\alpha _x}=\emptyset $ for all already considered branches x (i.e., all branches in $X_n$ and those assigned to previous nodes in some order of $\operatorname {\mathrm {split}}_n(q_n)$ ). Use Lemma 2.7 for $q_n(t)$ and $h(j)=n+j+2$ to obtain $q(s,1-i)\leq q_n(t)$ , branch x, and sequence $\{f_s\}_{s\in x\restriction \operatorname {\mathrm {split}}(q_n)}$ .

Finally, let $X_{n+1}$ be the set of all considered branches in this step, and

$$\begin{align*}q_{n+1}=\bigcup\{q(s,i)\colon s\in\operatorname{\mathrm{split}}_n(q_n), i\in\{0,1\}\}. \end{align*}$$

One can verify that the sequence $\{(q_n,X_n)\}_{n\in \omega }$ is a fusion sequence with witnesses.

Proof As explained above, by Miller’s result it is sufficient to show that $\mathbb {Q}(\mathcal {C})$ is $^{\omega }\omega $ -bounding. Let $\dot {f}$ be a $\mathbb {Q}(\mathcal {C})$ -name for a function in $^{\omega }\omega $ and let $p\in \mathbb {Q}(\mathcal {C})$ . We will show that there is $q\leq p$ and $g\in V\cap {{}^{\omega }\omega }$ such that $q\Vdash \dot {f}\leq ^* \check {g}$ .

By Lemma 2.8 we can assume that there is $q\leq p$ such that for all $m\in \omega $ , for all $t\in \operatorname {\mathrm {split}}_m(q)$ there is $f_t\in {^{m+1}\omega }$ such that $q(t)\Vdash \dot {f}\upharpoonright (m+1)=\check {f}_t$ . Define $g\in {{}^{\omega }\omega }$ as follows:

$$ \begin{align*}g(n)=\max\{f_s(n)+1\colon s\in\operatorname{\mathrm{split}}_n(q)\}.\end{align*} $$

Then $q\Vdash \forall n (\dot {f}(n)<g(n))$ .

Proof We prove just the first part. The second claim follows from the first one and the fact that the forcing notion $\mathbb {Q}(\mathcal {C})$ is ${}^{\omega }\omega $ -bounding (see [Reference Halbeisen26, Lemma 21.12]). Note that a filter base $\mathcal {G}$ generates an ultrafilter on $\omega $ if and only if $\mathcal {P}(\omega )=\langle \mathcal {G}\rangle _{{\mathrm{up}}}\cup \langle \mathcal {G}^{\ast }\rangle _{{\mathrm{dn}}}$ .

Let $\mathcal {U}$ be a P-point in V. We shall prove that the family $\mathcal {U}$ generates an ultrafilter in $V^{\mathbb {Q}(\mathcal {C})}$ , i.e., $V^{\mathbb {Q}(\mathcal {C})}\vDash \mathcal {P}(\omega )=\langle \mathcal {U}\rangle _{{\mathrm{up}}}\cup \langle \mathcal {U}^{\ast }\rangle _{{\mathrm{dn}}}$ . Fix $p\in \mathbb {Q}(\mathcal {C})$ and a $\mathbb {Q}(\mathcal {C})$ -name $\dot {Y}$ such that $p\Vdash \dot {Y}\subseteq \omega $ . By Lemma 2.8 we can assume that for all $m\in \omega $ , for all $t\in \mbox {split}_m(p)$ there is $u_t\in {^{m+1}2}$ such that

$$ \begin{align*}p(t)\Vdash\dot{Y}\upharpoonright(m+1)=\check{u}_t.\end{align*} $$

Note that the latter property remains true for any stronger condition q, since t in the m-th level of q is an extension of some s in the m-th level of p. Let $\{x_t\colon t\in p\}\subseteq [p]$ be a dense set in $[p]$ containing $\mathcal {C}$ -different elements (enumerated such that $s\subseteq x_s$ , and if $s\subseteq t\subseteq x_s$ then $x_t=x_s$ ). Let $Y_t=\bigcup \{u_s\colon s\subseteq x_t\}$ .

We assume that $\mathcal {Y}_0\subseteq \mathcal {U}$ , the other case may be handled analogously. We take a pseudointersection Z of $\mathcal {Y}_0$ in $\mathcal {U}$ , with $Z\subseteq Y_{\emptyset }$ . We shall simultaneously build two fusion sequences with witnesses, namely $\{(p^0_n,X^0_n)\}_{n\in \omega }$ , $\{(p^1_n,X^1_n)\}_{n\in \omega }$ , and a partition of Z into two sets $Z_0$ , $Z_1$ such that for their respective fusions $q_0,q_1\leq p$ we obtain $q_0\Vdash \check {Z}_0\subseteq \dot {Y}$ and $q_1\Vdash \check {Z}_1\subseteq \dot {Y}$ .

Let $p^0_0=p^1_0=p$ , $X^0_0=X^1_0=\{Y_{\emptyset }\}$ , and $k_0=0$ , $k_1=2$ . We assume that $p^0_n$ , $p^1_n$ , $k_{2n}$ , and $k_{2n+1}$ are constructed. Let $t\in \operatorname {\mathrm {split}}_{k_{2n}}(p)\cap \operatorname {\mathrm {split}}(p^0_n)$ , and set $w^{\ast }(t)=x_t\restriction \operatorname {\mathrm {split}}_{k_{2n+1}}(p)$ . For each $i\in \{0,1\}$ , we take $w^{\ast }(t,i)\in \operatorname {\mathrm {split}}_{k_{2n+1}+1}(p)$ extending $w^{\ast }(t)^{\smallfrown } i$ . There is $k_{2n+2}>k_{2n+1}+1$ such that

$$ \begin{align*} Z\setminus k_{2n+2}\subseteq\bigcap\{Y_{w^*(t,i)}\colon t\in\operatorname{\mathrm{split}}_{k_{2n}}(p)\cap\operatorname{\mathrm{split}}(p^0_n), i\in\{0,1\}\}. \end{align*} $$

We set $w(t,i)=x_{w^{\ast }(t,i)}\restriction \operatorname {\mathrm {split}}_{k_{2n+2}}(p)$ . Take $p_{n+1}^0=\bigcup \{p(w(t,i))\colon t\in \operatorname {\mathrm {split}}_{k_{2n}}(p)\cap \operatorname {\mathrm {split}}(p^0_n), i\in \{0,1\}\}$ and $X_{n+1}^0=\{x_{w(t,i)}\colon t\in \operatorname {\mathrm {split}}_{k_{2n}}(p)\cap \operatorname {\mathrm {split}}(p^0_n), i\in \{0,1\}\}$ . One can see that $p^0_n\Vdash \check {Z}\cap [k_{2n},k_{2n+1})\subseteq \dot {Y}$ . The construction of condition $p^1_n$ and the choice of number $k_{2n+3}$ are done similarly, and leads to $p^1_n\Vdash \check {Z}\cap [k_{2n+1},k_{2n+2})\subseteq \dot {Y}$ . Finally, we define

$$ \begin{align*} Z_0 = Z\cap\bigcup\{[k_{2n},k_{2n+1})\colon n\in\omega\} \ \mathrm{and} \ Z_1 = Z\cap \bigcup\{[k_{2n+1},k_{2n+2}) \colon n\in\omega\}. \end{align*} $$

Since $Z\in \mathcal {U}$ , $Z_0$ or $Z_1$ is in $\mathcal {U}$ , and so $q_0\Vdash \dot {Y}\in \langle \mathcal {U}\rangle _{{\mathrm{up}}}$ or $q_1\Vdash \dot {Y}\in \langle \mathcal {U}\rangle _{{\mathrm{up}}}$ .

Proof First we show that $(1)$ implies $(2)$ . Let $h\in \operatorname {\mathrm {FF}}(\mathcal {A})$ , let $X\subseteq \mathcal {A}^h$ , and suppose $\mathcal {A}^h\backslash X\notin {\mathrm {id}}(\mathcal {A})$ . Thus, there is $h'\in \operatorname {\mathrm {FF}}(\mathcal {A})$ such that for all $h"\supseteq h'$ the set $\mathcal {A}^{h\prime \prime }\cap (\mathcal {A}^h\backslash X)$ is non-empty. Note that if h and $h'$ are incompatible, then $\mathcal {A}^{h'}\cap (\mathcal {A}^h\backslash X)=\emptyset $ , which is a contradiction. Therefore h and $h'$ are compatible and without loss of generality, we can assume that $h'\supseteq h$ . Thus, we have that for all $h"\supseteq h'$ , the set $\mathcal {A}^{h\prime \prime }\backslash X\neq \emptyset $ . Now, since $(1)$ holds, there is $h"\supseteq h'$ such that $\mathcal {A}^{h\prime \prime }\cap X=\emptyset $ . That is, $\mathcal {A}^{h\prime \prime }\subseteq \mathcal {A}^{h}\backslash X$ .

Next, we show that $(2)$ implies $(3)$ . Thus, consider any $X\in \mathcal {P}(\omega )\backslash \operatorname {\mathrm {fil}}(\mathcal {A})$ . Then, in particular $\omega \backslash X\notin {\mathrm {id}}(\mathcal {A})$ and so there is $h\in \operatorname {\mathrm {FF}}(\mathcal {A})$ such that for all $h'\supseteq h$ , $|\mathcal {A}^{h'}\cap (\omega \backslash X)|=|\mathcal {A}^{h'}\backslash X|=\omega $ . Let $Y=\mathcal {A}^h\backslash X$ . Thus, $Y\subseteq \mathcal {A}^h$ . By part $(2)$ either $\mathcal {A}^h\backslash Y\in {\mathrm {id}}(\mathcal {A})$ or there is $h'\supseteq h$ such that $\mathcal {A}^{h'}\subseteq \mathcal {A}^h\backslash Y$ . Suppose $\mathcal {A}^h\backslash Y\in {\mathrm {id}}(\mathcal {A})$ . Then, there is $h'\supseteq h$ such that $\mathcal {A}^{h'}\cap (\mathcal {A}^h\backslash Y)=\mathcal {A}^{h'}\backslash Y=\emptyset $ . However $\mathcal {A}^{h'}\backslash Y=\mathcal {A}^{h'}\cap X=\emptyset $ and so $X\subseteq \omega \backslash \mathcal {A}^{h'}$ and we are done. If there is $h'\supseteq h$ such that $\mathcal {A}^{h'}\subseteq \mathcal {A}^h\backslash Y=\mathcal {A}^h\cap X$ , then $\mathcal {A}^{h'}\cap (\omega \backslash X)=\mathcal {A}^{h'}\backslash X=\emptyset $ , contradicting the choice of h.

To see that $(3)$ implies $(2)$ , consider any $h\in \operatorname {\mathrm {FF}}(\mathcal {A})$ and $X\subseteq \mathcal {A}^h$ . Let $Y=\mathcal {A}^h\backslash X$ . If $\omega \backslash Y\in \operatorname {\mathrm {fil}}(\mathcal {A})$ , then $Y=\mathcal {A}^h\backslash X\in {\mathrm {id}}(\mathcal {A})$ . Otherwise, there is $h^*$ such that $\omega \backslash Y\subseteq \omega \backslash \mathcal {A}^{h^*}$ , which implies that $\mathcal {A}^{h^*}\subseteq Y=\mathcal {A}^h\backslash X$ and so $\mathcal {A}^{h^*\cup h}\subseteq \mathcal {A}^{h^*}\subseteq \mathcal {A}^h\backslash X$ .

To see that $(2)$ implies $(1)$ consider any $X\in [\omega ]^{\omega }$ and let $h\in \operatorname {\mathrm {FF}}(\mathcal {A})$ . We want to show that there is $h'\supseteq h$ such that either $\mathcal {A}^{h'}\cap X=\emptyset $ , or $\mathcal {A}^{h'}\backslash X=\emptyset $ . Let $Y=X\cap \mathcal {A}^h$ . Thus, $Y\subseteq \mathcal {A}^h$ . If $\mathcal {A}^h\backslash Y\in {\mathrm {id}}(\mathcal {A})$ , then $\mathcal {A}^h\backslash X\in {\mathrm {id}}(\mathcal {A})$ and so there is $h'\supseteq h$ such that $\mathcal {A}^{h'}\cap (\mathcal {A}^h\backslash X)=\mathcal {A}^{h'}\backslash X=\emptyset $ . Otherwise, there is $h'\supseteq h$ such that $\mathcal {A}^{h'}\subseteq \mathcal {A}^h\backslash Y$ and so $\mathcal {A}^{h'}\cap Y=\emptyset $ . However, $\mathcal {A}^{h'}\cap Y=\mathcal {A}^{h'}\cap (X\cap \mathcal {A}^h)=\mathcal {A}^{h'}\cap X=\emptyset $ .

Proof $(\mathrm {(a)}\Rightarrow \mathrm {(b)})$ Note that the partition relations occurring in this proof, and so implicitly the proof itself, can be found in [Reference Shelah46]. Inductively, choose a sequence $\{n(l)\}_{l\in \omega }$ such that $n(0)=0$ and

$$ \begin{align*}n(l+1)=\min\{n\colon n(l)< n\mbox{ and }\forall m\leq n(l)(f(m) \leq n)\}.\end{align*} $$

We consider the partition $\mathcal {E}_0=\{[n(3l), n(3l+3))\}_{l\in \omega }$ . There is $C_1\in \mathcal {F}$ such that $C_1$ is a selector for $\mathcal {E}_0$ . Now, consider the relation $\mathcal {E}_1$ on $C_1$ defined as follows:

$$ \begin{align*} m\sim_{\mathcal{E}_1} k\mbox{ iff }m=k\lor m<k\leq f(m)\lor k<m\leq f(k). \end{align*} $$

Note that $\mathcal {E}_1$ is clearly reflexive and symmetric. Transitivity holds, since no three pairwise distinct elements of $C_1$ are $\mathcal {E}_1$ -equivalent: Indeed, suppose $m_1<m_2<m_3$ are such that $m_1\mathcal {E}_1 m_2$ and $m_2\mathcal {E}_1 m_3$ . Then $m_1<m_2<m_3\leq f(m_1)$ . There are $l_1<l_2<l_3$ such that $m_i\in [n(3l_i), n(3l_i+3))$ . Then $m_1<n(3l_2)\leq m_2<n(3l_3)\leq m_3\leq f(m_1)$ . However, on the other hand by the definition of sequence $\{n(l)\}_{l\in \omega }$ we have $f(m_1)\leq n(3l_2+1)<n(3l_3)$ , a contradiction. Thus, $\mathcal {E}_1$ is an equivalence relation on $C_1$ and each $\mathcal {E}_1$ -equivalence class has at most two elements.

Extend $\mathcal {E}_1$ to an equivalence relation $\mathcal {E}_2$ on $\omega $ by defining

$$ \begin{align*} m\sim_{\mathcal{E}_2} k\mbox{ iff }m=k\lor m\sim_{\mathcal{E}_1} k. \end{align*} $$

There is $C_2$ in $\mathcal {F}$ such that $C_2$ is a selector for $\mathcal {E}_2$ . Without loss of generality $C_2\subseteq C_1$ and $0\in C_2$ . Let $\{k(n)\}_{n\in \omega }$ enumerate in increasing order $C_2$ . Thus for all $n,n'$ we have that $k(n)\not \sim _{\mathcal {E}_2} k(n')$ . Thus, if $n<n'$ then $k(n')\not \leq f(k(n))$ and so for all $n\in \omega $ , $f(k(n))<k(n+1)$ .

$(\mathrm {(b)}\Rightarrow \mathrm {(a)})$ Let $\mathcal {E}$ be a partition of $\omega $ into finite sets. We set

$$\begin{align*}f(n)=\max\bigcup\{E\in\mathcal{E}\colon (\exists i\leq n)\ i\in E\}. \end{align*}$$

There is $\{k(n)\colon n\in \omega \}\in \mathcal {F}$ such that $f(k(n))<k(n+1)$ for each $n\in \omega $ . The set $\{k(n)\colon n\in \omega \}$ is a selector for $\mathcal {E}$ . Indeed, $k(n)\leq f(k(n))<k(n+1)$ and therefore $k(n+1)$ is from a different set of partition $\mathcal {E}$ than all $k(i)$ for $i\leq n$ .

Proof If $\mathbb {P}$ is an ${}^{\omega }\omega $ -bounding forcing notion, $\mathcal {F}$ a Q-filter in V, then we use part (2) of Lemma 3.7 for $f\in V\cap {{}^{\omega }\omega }$ dominating function $g\in V^{\mathbb {P}}\cap {{}^{\omega }\omega }$ .

Proof $(\mathrm {(a)}\Rightarrow \mathrm {(b)}) \mathcal {F}$ is a P-set and therefore there is $C_0\in \mathcal {F}$ such that $C_0\subseteq ^* G$ for each $G\in \bigcup \{\mathcal {G}_n\colon n\in \omega \}$ . Thus, for some function $f\in {{}^{\omega }\omega }$

$$ \begin{align*} (\forall n\in\omega)\ C_0\setminus f(n)\subseteq \bigcap\mathcal{G}_n. \end{align*} $$

Let us take $\{k(n)\colon n\in \omega \}\in \mathcal {F}$ from Lemma 3.7 such that $C=\{k(n+1)\colon n\in \omega \}\subseteq C_0$ . Hence, we have $k(n+1)\in C_0\setminus f(k(n))$ , and so $k(n+1)\in \bigcap \mathcal {G}_{k(n)}$ .

$(\mathrm {(b)}\Rightarrow \mathrm {(a)})$ First we shall show that $\mathcal {F}$ is a P-set. Let $\{G_i\}_{i\in \omega }$ be a sequence in $\mathcal {F}$ . We set $\mathcal {G}_i=\{G_0\cap G_1\cap \dots \cap G_i\}$ , and we take $a\in \mathcal {F}$ such that $a(n+1)\in \bigcap \mathcal {G}_{a(n)}$ . The set a is a pseudointersection of $\{G_i\}_{i\in \omega }$ . Indeed, if $G_j$ is such that $j\leq a(n)$ then $\{a(k)\colon k\geq n+1\}\subseteq G_j$ .

We shall show that $\mathcal {F}$ is a Q-set using Lemma 3.7. Indeed, let the function $f\in {{}^{\omega }\omega }$ be increasing. We consider sets $\mathcal {G}_i=\{(f(i),+\infty )\}$ , and we take $a\in \mathcal {F}$ such that $a(n+1)\in \bigcap \mathcal {G}_{a(n)}$ . Hence, $a(n+1)>f(a(n))$ .

Proof Let V denote the ground model. We assume that $\mathcal {A}$ is a selective independent family in V, $\mathcal {U}$ is a P-point in V, and $\mathcal {E}$ is a tight MAD family in V (according to [Reference Guzmán, Hrušák and Téllez25]). Using an appropriate bookkeeping device define a countable support iteration $\langle \mathbb {P}_{\alpha },\dot {\mathbb {Q}}_{\beta }\colon \alpha \leq \omega _2,\beta <\omega _2\rangle $ of posets such that for each $\alpha $ , $\mathbb {P}_{\alpha }$ forces that $\mathbb {Q}_{\alpha }=\mathbb {Q}(\mathcal {C})$ for some uncountable partition $\mathcal {C}$ of $2^{\omega }$ into compact sets and such that $V^{\mathbb {P}_{\omega _2}}\vDash \mathfrak {a}_T=\omega _2$ . ${\mathbb {P}_{\omega _2}}$ has the Sacks property and therefore $\operatorname {\mathrm {cof}}(\mathcal {N})=\omega _1$ . By Shelah’s preservation Theorem 3.10 and Corollary 4.10, or alternatively by Theorem 4.1, the family $\mathcal {A}$ remains selective independent in $V^{\mathbb {P}_{\omega _2}}$ and so a witness to $\mathfrak {i}=\omega _1$ . Similarly, $\mathcal {U}$ generates a P-point in $V^{\mathbb {P}_{\omega _2}}$ , so $\mathfrak {u}=\omega _1$ as well. And finally, $\mathfrak {a}=\omega _1$ since $\mathcal {E}$ is a tight MAD family (see [Reference Guzmán, Hrušák and Téllez25]).

Proof Recall that by Lemmas 3.4 and 3.8, $\operatorname {\mathrm {fil}}(\mathcal {A})$ remains a selective filter. Thus, it is sufficient to show that $\mathcal {A}$ remains densely maximal in the generic extension. In $V^{\mathbb {Q}(\mathcal {C})}$ , take any $Y\in \mathcal {P}(\omega )\backslash \langle \operatorname {\mathrm {fil}}(\mathcal {A})\cap V\rangle _{{\mathrm{up}}}$ . Suppose $Y\notin \langle \{\omega \backslash \mathcal {A}^h:h\in \operatorname {\mathrm {FF}}(\mathcal {A})\}\rangle _{{\mathrm{dn}}}$ . Thus, for all $h\in \operatorname {\mathrm {FF}}(\mathcal {A})$ , $Y\not \subseteq \omega \backslash \mathcal {A}^h$ and so for all $h\in \operatorname {\mathrm {FF}}(\mathcal {A})$ , $|Y\cap \mathcal {A}^h|=\omega $ . Therefore in V we can fix $p\in \mathbb {Q}(\mathcal {C})$ and a $\mathbb {Q}(\mathcal {C})$ -name $\dot {Y}$ for Y such that for all $h\in \operatorname {\mathrm {FF}}(\mathcal {A})$ , $p\Vdash |\dot {Y}\cap \mathcal {A}^h|=\infty $ .

By Lemma 2.8 we can assume that for all $m\in \omega $ , for all $t\in \mbox {split}_m(p)$ there is $u_t\in {^{m+1}2}$ such that $p(t)\Vdash \dot {Y}\upharpoonright (m+1)=\check {u}_t$ . Now, in V for each $t\in p$ , let

$$ \begin{align*} Y_t=\{m\in\omega: p(t)\not\Vdash\check{m}\notin\dot{Y}\}. \end{align*} $$

Proof By Claim 4.2(iii), $Y_t\in \operatorname {\mathrm {fil}}(\mathcal {A})\cap V$ for each $t\in \operatorname {\mathrm {split}}(p)$ . By Lemma 3.9 for $\mathcal {G}_n$ being the family of all $Y_t$ ’s with $t\in \operatorname {\mathrm {split}}_{\leq n}(p)$ , we obtain $\{k(n)\colon n\in \omega \}\in \operatorname {\mathrm {fil}}(\mathcal {A})$ such that

$$ \begin{align*} k(n+1)\in\bigcap\{Y_t\colon t\in\operatorname{\mathrm{split}}_{\leq k(n)}(p)\}. \end{align*} $$

Moreover, by part (iv) of Claim 4.2 for any $s\in \operatorname {\mathrm {split}}_{k(n)}(p)$ there is $t\in \operatorname {\mathrm {split}}_{k(n+1)}(p)$ extending s such that $p(t)\Vdash \check {k}(n+1)\in \dot {Y}$ . For each branch $x\in [p]$ we consider the set

$$\begin{align*}i(x)=\{i\colon p(t)\Vdash\check{k}(i+1)\in\dot{Y}\text{ for }t=x\restriction\operatorname{\mathrm{split}}_{k(i+1)}(p)\}. \end{align*}$$

We say that $x\in [p]$ is an acceptable branch if $i(x)$ is cofinite. The smallest n with $i(x)\supseteq [n,\infty )$ is called the degree of acceptability of x. Note that for each acceptable branch x, $y_x=\{k(i+1)\colon i\in i(x)\}\in \operatorname {\mathrm {fil}}(\mathcal {A})$ . Moreover, each $s\in p$ can be extended to an acceptable branch. Indeed, if $s\in \operatorname {\mathrm {split}}_l(p)$ , take j such that $l\leq k(j)$ . Thus, we assume that $k(j+1)\in Y_s$ . By part (iv) of Claim 4.2 there is $t\in \operatorname {\mathrm {split}}_{k(j+1)}(p)$ extending s such that $p(t)\Vdash \check {k}(j+1)\in \dot {Y}$ . Repeat the procedure recursively to build an acceptable branch extending s. In the following, we continue using a fusion argument. We build a fusion sequence $\{(p_n,X_n)\}_{n\in \omega }$ .

To define $p_0$ , take some acceptable branch x extending some node in $\operatorname {\mathrm {split}}_{k(0)+1}(p)$ with degree of acceptability at most $0$ , and a node $s=x\restriction \operatorname {\mathrm {split}}_{k(1)}(p)$ . We set $p_0=p(s)$ and $X_0=\{x\}$ .

Let us assume that $p_n$ and $X_n$ are defined, and consider $s\in \operatorname {\mathrm {split}}_{k(n)}(p)\cap \operatorname {\mathrm {split}}(p_n)$ . Take the unique acceptable branch $x\in X_n$ extending s. Define $i=x(|s|)\in \{0,1\}$ and $s_i=x\restriction \operatorname {\mathrm {split}}_{k(n+1)}(p)$ . Then we set $s_{1-i}$ to be an extension of $s^{\smallfrown }\langle {1-i}\rangle $ such that:

1. (i) $[p(s_{1-i})]\cap C_{\alpha _x}=\emptyset $ for all already considered acceptable branches x (i.e., all branches in $X_n$ and those assigned to previous nodes in some order of $\operatorname {\mathrm {split}}_{k(n)}(p)\cap \operatorname {\mathrm {split}}(p_n)$ ). This can be easily achieved since each $C_{\alpha _x}$ is nowhere dense in $[p]$ .

2. (ii) $s_{1-i}=x\restriction \operatorname {\mathrm {split}}_{k(j+1)}(p)$ with $j\in i(x)$ for some acceptable branch $x\in [p]$ with degree of acceptability at most j. Let us recall that each node can be extended to an acceptable branch.

Finally, let $X_{n+1}$ be the set of all considered acceptable branches in this step, and

$$\begin{align*}p_{n+1}=\bigcup\{p(s_i)\colon s\in\operatorname{\mathrm{split}}_{k(n)}(p)\cap\operatorname{\mathrm{split}}(p_n), i\in\{0,1\}\}. \end{align*}$$

One can see that the sequence $\{(p_n,X_n)\}_{n\in \omega }$ is a fusion sequence with witnesses. Let $q=\bigcap \{p_n\colon n\in \omega \}$ and let $X=\bigcup \{X_n\colon n\in \omega \}$ .

We shall show that the family $\{y_x\colon x\in X\}$ possesses the desired properties. Indeed, let $x\in X$ . For each $n\in \omega $ we have $y_x(n)=k(i(x)(n)+1)$ . Due to construction of q we have $x\restriction \operatorname {\mathrm {split}}_n(q)=x\restriction \operatorname {\mathrm {split}}_{k(j_n+1)}(p)$ for some increasing sequence $\{j_i\}_{i\in \omega }$ , and if $t=x\restriction \operatorname {\mathrm {split}}_n(q)$ then $p(t)\Vdash \check {k}(j_n+1)\in \dot {Y}$ . Thus $k(j_n+1)\in y_x$ and consequently $j_n\geq i(x)(n)$ for each n. Let us now fix n and consider $t=x\restriction \operatorname {\mathrm {split}}_n(q)$ . The definition of $y_x$ guarantees that $p(s)\Vdash \check {k}(i(x)(n)+1)\in \dot {Y}$ for $s=x\restriction \operatorname {\mathrm {split}}_{k(i(x)(n)+1)}(p)$ . Thus we have $q(s)\Vdash \check {y}_x(n)\in \dot {Y}$ . On the other hand, $s=x\restriction \operatorname {\mathrm {split}}_{k(i(x)(n)+1)}(p)\subseteq x\restriction \operatorname {\mathrm {split}}_{k(j_n+1)}(p)=x\restriction \operatorname {\mathrm {split}}_n(q)=t$ .

The last part of our proof resembles the proof of the previous claim. Let $x_s$ for $s\in \operatorname {\mathrm {split}}(p)$ be the branch in X extending s such that if $s\subseteq t\subseteq x_s$ then $x_t=x_s$ . The corresponding $y_{x_s}$ is denoted by $y_s$ . The set $y_s$ belongs to $\operatorname {\mathrm {fil}}(\mathcal {A})\cap V$ . By Lemma 3.9 for $\mathcal {G}_n$ being the family of all $y_t$ ’s with $t\in \operatorname {\mathrm {split}}_{\leq n}(p)$ , we obtain $\{l(n)\colon n\in \omega \}\in \operatorname {\mathrm {fil}}(\mathcal {A})$ such that

$$ \begin{align*} l(n+1)\in\bigcap\{y_t\colon t\in\operatorname{\mathrm{split}}_{\leq l(n)}(p)\}. \end{align*} $$

Let us denote $C=\{l(n+1)\colon n\in \omega \}$ . We shall construct a condition $q^*\leq p$ such that $q^*\Vdash \check {C}\subseteq \dot {Y}$ . Then $q^*\Vdash \dot {Y}\in \operatorname {\mathrm {fil}}(\mathcal {A})$ which is a contradiction.

We build a fusion sequence $\{(p_n,X_n)\}_{n\in \omega }$ . Let $p_0=p$ , $X_0=\{x_t\}$ for $t\in \operatorname {\mathrm {split}}_0(p)$ , and suppose we have defined $p_n$ . For each $t\in \operatorname {\mathrm {split}}_n(p_n)\subseteq \operatorname {\mathrm {split}}_{l(n)}(p)$ and each $i\in \{0,1\}$ take $w^*(t,i)\in \operatorname {\mathrm {split}}_{l(n)+1}(p)$ such that $w^*(t,i)$ end-extends $t^{\smallfrown } i$ . Then

$$ \begin{align*} l(n+1)\in\bigcap\{y_{w^*(t,i)}\colon t\in\operatorname{\mathrm{split}}_n(p_n), i\in\{0,1\}\} \end{align*} $$

and so for each $t,i$ we take $w(t,i)=x_{w^*(t,i)}\restriction \operatorname {\mathrm {split}}_{l(n+1)}(p)$ . Note that by Claim 4.3 and the fact that $l(n+1)\geq j$ for $l(n+1)=y_{w^*(t,i)}(j)$ we obtain

$$ \begin{align*} p(w(t,i))\Vdash \check{l}(n+1)\in\dot{Y}. \end{align*} $$

Take $p_{n+1}\kern1.2pt{=}\kern1.2pt\bigcup \{p(w(t,i))\colon t\kern1.2pt{\in}\kern1.2pt \operatorname {\mathrm {split}}_n(p_n), i\kern1.2pt{\in}\kern1.2pt \{0,1\}\}$ , $X_{n+1}\kern1.2pt{=}\kern1.2pt\{x_{w(t,i)}\colon t\kern1.2pt{\in}\kern1.2pt \operatorname {\mathrm {split}}_n(p_n), i\in \{0,1\}\}$ . Finally, let $q^*$ be the fusion of $\{(p_n,X_n)\}_{n\in \omega }$ . Then, since $q^*\Vdash l(n+1)\in \dot {Y}$ for all $n\in \omega $ , $q^*$ is as required.

Proof For every $q\in \mathbb {S}$ such that $q\subseteq p$ , let $L(q)=\bigcup H\big [[q]\big ]$ . Now consider the game $\mathfrak {G}(\mathcal {F},\omega ,\mathcal {F})$ . Player I will play the following strategy, while constructing a sequence $\{t_{\sigma }\}_{\sigma \in 2^{<\omega }}\subseteq p$ such that:

1. (a) $\forall \sigma \in 2^{<\omega } \forall i\in 2\big (t_{\sigma }\subsetneq t_{\sigma ^{\smallfrown } i}\big ).$

2. (b) $\forall \sigma ,\tau \in 2^n\big (\sigma \not =\tau \rightarrow (t_{\sigma } \text { and } t_{\tau } \text { are incomparable })\big ).$

On the first turn Player I defines $t_{\emptyset }=\emptyset $ and plays $U_0=L(p(t_{\emptyset }))$ . As the rules dictate, Player II responds with some $a_0\in U_0$ . Since $a_0\in \bigcup H\big [[p(t_{\emptyset })]\big ]$ , there is $f\in [p(t_{\emptyset })]$ such that $a_0\in H(f)$ . As H is continuous there is $k\in \omega $ such that for every $g\in [p(f|_k)]$ we have $a_0\in H(g)$ . Now Player I extends $f|_k$ to incomparable $t_0,t_1\in p$ such that $t_{\emptyset }\subsetneq t_0,t_1$ and plays $U_1=L(p(t_0))\cap L(p(t_1))$ . As the rules dictate Player II responds with some $a_1\in U_1$ .

In general, suppose that it is the $n+1$ turn and that Player I has constructed $\{t_{\sigma }\}_{\sigma \in 2^{\leq n}}$ and for every $m\leq n$ played $U_m=\bigcap \limits _{\sigma \in 2^{\leq m}} L(p(t_{\sigma }))$ . As $a_n\in \bigcap \limits _{\sigma \in 2^n}\big (\bigcup H\big [[p(t_{\sigma })]\big ]\big ) $ , for every $\sigma \in 2^n$ there is $f_{\sigma }\in [p(t_0)]$ such that $a_n\in H(f_{\sigma }).$ Since H is continuous, there is $k\in \omega $ such that for every $\sigma \in 2^n$ and every $g\in [p(f_{\sigma }|_k)], a_n\in H(g)$ . Now Player I extends each $f_{\sigma }|_k$ to incomparable $t_{\sigma ^{\smallfrown } 0},t_{\sigma ^{\smallfrown } 1}\in p$ such that $t_{\sigma } \subsetneq t_{\sigma ^{\smallfrown } 0},t_{\sigma ^{\smallfrown } 1}$ and plays $U_{n+1}=\bigcap \limits _{\sigma \in 2^{\leq {n+1}}} L(p(t_{\sigma }))$ . As the rules dictate, Player II responds with some $a_{n+1}\in U_{n+1}$ .

Since $\mathcal {F}$ is a selective filter, this is not a winning strategy for Player I. Therefore there is a match where Player I plays by the above strategy, but Player II wins. Let $\{a_n\}_{n\in \omega }$ and $\{t_{\sigma }\}_{\sigma \in 2^{<\omega }}$ be the sequences associated with one of these matches and let $q=\{\tau \in p\;|\;\exists \sigma \in 2^{<\omega }\big (\tau \subseteq t_{\sigma }\big )\}$ . It is straightforward that q and $Y=\{a_n\}_{n\in \omega }$ are the objects we are looking for.

Proof Let $\dot {X}$ be a name for a subset of $\omega $ , $p\in \mathbb {Q}(\mathcal {C})$ , and suppose no condition below p forces $(b)$ . By Lemma 4.4 there is a continuous $H:[p]\longrightarrow P(\omega )$ such that $p\Vdash H(\dot {r}_{gen})=\dot {X}$ . For every $q\in \mathbb {S}$ such that $q\subseteq p$ , let $L(q)=\bigcup H\big [[q]\big ]$ and note that if $q\in \mathbb {Q}(\mathcal {C})$ then $L(q)\in \mathcal {F}$ . This holds, because $q\Vdash \dot {X}\subseteq L(q)$ and q does not force $(b)$ . We say that a condition $q\in \mathbb {S}$ , which is not necessarily in $\mathbb {Q}(\mathcal {C})$ , is special if $q\subseteq p$ and for every $s\in q$ we have that $L(q(s))\in \mathcal {F}$ .

We will make use of the following notion: Given $s\in p$ we say that the pair $(q,T)$ is s-special if $q\in \mathbb {S}$ is special, $T\in \mathcal {C}$ , $[q]\subseteq [p(s)]\cap T$ . We divide the proof in cases.

In this case, consider the game $\mathfrak {G}(\mathcal {F},\omega ,\mathcal {F})$ . Player I will play by the following strategy, while recursively constructing sequences $\{q_{\sigma }\}_{\sigma \in 2^{<\omega }}\subseteq \mathbb {S}$ , $\{s_{\sigma }\}_{\sigma \in 2^{<\omega }}\subseteq p$ , and $\{T_{\sigma }\}_{\sigma \in 2^{<\omega }}\subseteq \mathcal {C}$ such that:

1. (a) $\forall \sigma \in 2^{<\omega }$ the pair $(q_{\sigma },T_{\sigma })$ is $s_{\sigma }$ -special;

2. (b) $\forall \sigma \in 2^{<\omega }\big (q_{\sigma ^{\smallfrown } 0}\subseteq q_{\sigma }\big )$ ;

3. (c) $\forall \sigma \in 2^{<\omega }\big (T_{\sigma ^{\smallfrown } 0}= T_{\sigma }\big )$ ;

4. (d) $\forall \sigma ,\tau \in 2^n\big (\sigma \not =\tau \rightarrow T_{\sigma }\not =T_{\tau }\big )$ ;

5. (e) $\forall \sigma \in 2^{<\omega }\forall i\in 2\big (s_{\sigma }\subsetneq s_{\sigma ^{\smallfrown } i}\big )$ ;

6. (f) $\forall \sigma \in 2^{<\omega }\big ( s_{\sigma ^{\smallfrown } 0}\in q_{\sigma } \wedge [p( s_{\sigma ^{\smallfrown } 1})]\cap T_{\sigma }=\emptyset \big )$ .

On the first turn, Player I defines $s_{\emptyset }=\emptyset $ , an $s_{\emptyset }$ -special pair $(q_{\emptyset },T_{\emptyset })$ and plays $U_0=L(q_{\emptyset })$ . As the rules dictate, Player II responds with some $a_0\in U_0$ . Since $a_0\in \bigcup H\big [[q_{\emptyset }]\big ]$ , there is some $f\in [q_{\emptyset }]$ such that $a_0\in H(f)$ and since H is continuous there is $k\in \omega $ such that for every $g\in [p(f|_k)]$ , $a_0\in H(g)$ . Notice that since $(q_{\emptyset },T_{\emptyset })$ is $s_{\emptyset }$ -special, we have that $f|_k$ is compatible with $s_{\emptyset }$ and moreover we can extend $f|_k$ to incomparable $s_0,s_1\in p$ such that $s_{\emptyset }\subsetneq s_0,s_1$ , $s_0\in q_{\emptyset }$ and $[p(s_1)]\cap T_{\emptyset }=\emptyset $ . Now Player I defines $q_0=q_{\emptyset }(s_0)$ , $T_0=T_{\emptyset }$ , an $s_1$ -special pair $(q_1, T_1)$ and plays $U_1= L(q_0)\cap L(q_1)$ . As the rules dictate, Player II responds with some $a_1\in U_1$ .

In general, suppose that it is the $n+1$ turn and Player I has constructed $q_{\sigma }, s_{\sigma }$ , and $T_{\sigma }$ for every $\sigma \in 2^{\leq n}$ . Moreover, suppose that for every $m\leq n$ Player I has played $U_m=\bigcap \limits _{\sigma \in 2^m}L(q_{\sigma })$ . Since $a_n \in \bigcap \limits _{\sigma \in 2^n}\big (\bigcup H\big [[q_{\sigma }]\big ]\big )$ , for every $\sigma \in 2^n$ there is $f_{\sigma }\in [q_{\sigma }]$ such that $a_n\in H(f_{\sigma })$ . As H is continuous, there is $k\in \omega $ such that for every $\sigma \in 2^n$ and every $g\in [p(f_{\sigma }|_k)]$ , $a_n\in H(g)$ . As each $(q_{\sigma },T_{\sigma })$ is $s_{\sigma }$ -special, we have that $f_{\sigma }|_k$ is compatible with $s_{\sigma }$ and that $\bigcup \limits _{\sigma \in 2^n} T_{\sigma }\cap [p]$ is nowhere dense in $[p]$ . Then we can extend each $f_{\sigma }|_k$ to incomparable $s_{\sigma ^{\smallfrown } 0},s_{\sigma ^{\smallfrown } 1}\in p$ such that $s_{\sigma }\subsetneq s_{\sigma ^{\smallfrown } 0}$ , $s_{\sigma ^{\smallfrown } 1}$ , $s_{\sigma ^{\smallfrown } 0}\in q_{\sigma }$ and $[p( s_{\sigma ^{\smallfrown } 1})]\cap T_{\sigma }=\emptyset $ . Now Player I defines $q_{\sigma ^{\smallfrown } 0}=q_{\sigma }(s_{\sigma ^{\smallfrown } 0})$ , $T_{\sigma \smallfrown 0}=T_{\sigma }$ , an $s_{\sigma ^{\smallfrown } 1}$ -special pair $(q_{\sigma ^{\smallfrown } 1},T_{\sigma ^{\smallfrown } 1})$ and plays $U_{n+1}=\bigcap \limits _{\sigma \in 2^{n+1}}L(q_{\sigma })$ . As the rules dictate, Player II responds with some $a_{n+1}\in U_{n+1}$ .

Since $\mathcal {F}$ is a selective filter, the above is not a winning strategy for Player I and so, there is a match where Player I follows the strategy, but Player II wins. Let $\{a_n\}_{n\in \omega }$ , $\{q_{\sigma }\}_{\sigma \in 2^{\omega }}$ , $\{s_{\sigma }\}_{\sigma \in 2^{\omega }}$ , and $\{T_{\sigma }\}_{\sigma \in 2^{\omega }}$ be the sequences associated with one of these matches. To finish Case 1, define $q=\{\tau \in p\;|\;\exists \sigma \in 2^{\omega }\big (\tau \subseteq s_{\sigma }\big )\}$ . Moreover, if $c_0$ is the constant $0$ function in $2^{\omega }$ and $\sigma \in 2^{<\omega }$ then $g_{\sigma }=\bigcup \{s_{\tau }\;|\; \tau \subseteq \sigma ^{\smallfrown } c_0\}\in T_{\sigma }$ and the set $Q=\{ g_{\sigma }\;|\;\sigma \in 2^{<\omega }\}$ is dense in $[q]$ . But then, by Lemma 2.4, $q\in \mathbb {Q}(\mathcal {C})$ . Since for every $n\in \omega $ , every $\sigma \in 2^{n+1}$ , and every $g\in [p(s_{\sigma })]$ , we have that $a_n\in H(g)$ and every $g\in [q]$ satisfies this condition for some $\sigma \in 2^{n+1}$ , we obtain that for every $g\in [q]$ , $\{a_n\}_{n\in \omega }\subseteq H(g)$ . In particular we have that $q\Vdash \{a_n\}_{n\in \omega }\subseteq \dot {X}$ . Since $\{a_n\}_{n\in \omega }\in \mathcal {F}$ , we are done.

In this case, we use Lemma 4.7 to find $q\in \mathbb {S}$ and $Y\in \mathcal {F}$ such that $q\subseteq p(s_0)$ and for every $f\in [q]$ , $Y\subseteq H(f)$ . Notice that q is special. Suppose towards a contradiction that $q\notin \mathbb {Q}(\mathcal {C})$ . Since every element of $\mathcal {C}$ is closed, this means that there is some $T\in \mathcal {C}$ such that $T\cap [q]$ has non-empty interior in $[q]$ and so we can find $\tau \in q$ such that $[q(\tau )]\subseteq T$ . Then $(q(\tau ),T)$ is $s_0$ -special, which is a contradiction. Therefore $q\in \mathbb {Q}(\mathcal {C})$ . To finish this case, just note that as before $q\Vdash Y \subseteq \dot {X}$ .

Proof Since $\mathbb {Q}(\mathcal {C})$ is proper and has the Sacks property, $\langle \operatorname {\mathrm {fil}}(\mathcal {A})^V \rangle $ is a selective filter in $V[G]$ , but by Lemma 3.4 we know that $\operatorname {\mathrm {fil}}(\mathcal {A})^{V[G]}=\langle \operatorname {\mathrm {fil}}(\mathcal {A})^V \rangle $ . To show that $\mathcal {A}$ remains densely maximal in $V[G]$ , note that by Theorem 4.8, the family $P(\omega )^V\backslash \operatorname {\mathrm {fil}}(\mathcal {A})^V$ is cofinal in $P(\omega )^{V[G]}\backslash \langle \operatorname {\mathrm {fil}}(\mathcal {A})^V\rangle $ . However, by hypothesis $\{\omega \backslash \mathcal {A}^h:h\in \operatorname {\mathrm {FF}}(\mathcal {A})\}$ is cofinal in $P(\omega )^V\backslash \operatorname {\mathrm {fil}}(\mathcal {A})^V$ and so we are done.

Claim 5.2. Each A-n-determined function is equal to a function $\varphi (g_0,\dots ,g_n)$ which is obtained as an interpretation of a formula $\varphi (a_0,\dots ,a_n)$ of propositional calculus. The symbols $\wedge $ , $\vee $ , $\neg $ are interpreted as a maximum, minimum, and complement (i.e., $1-g_i$ ), respectively. The formula $\varphi (a_0,\dots ,a_n)$ may contain constant symbols $0,1$ which are interpreted as constant functions $0,1$ .

Claim 5.6. Let $p\in \mathbb {Q}_{\mathcal {I}}$ . For an initial segment u of $A^p$ , and $h\colon u\to \{0,1\}$ , let $p^{[h]}$ be the pair $q=(H^q,E^q)$ defined by (i) and (ii) below:

1. (i) $E^q=E^p\restriction \bigcup \{[i]_{E^p}\colon i\in A^p\setminus u\}$ .

2. (ii) If $H^p(n)$ is $\varphi (g_0,\dots ,g_n)$ then $H^q(n)$ is $\varphi (g_0,\dots ,g_i/h(i),\dots ,g_n)$ , where the substitution is done just for $i\in u$ .

Then we have:

1. (1) $p^{[h]}$ is a condition in $\mathbb {Q}_{\mathcal {I}}$ stronger than p.

2. (2) The set $\{p^{[h]}\colon h\in {}^u\{0,1\}\}$ is predense below p.

3. (3) If u is the set of first n elements of $A^p$ , D a dense subset of $\mathbb {Q}_{\mathcal {I}}$ then there is $q\in \mathbb {Q}_{\mathcal {I}}$ such that $q\leq _np$ and $q^{[h]}\in D$ for any $h\in {}^u\{0,1\}$ .

Proof Let $p\in \mathbb {Q}_{\mathcal {I}}$ , M a countable elementary submodel of $H(\kappa )$ such that $\mathcal {I}, \mathcal {A}, p\in M$ and $B\in \mathcal {I}(\mathcal {A})$ for which $|B\cap Y|=\omega $ for every $Y\in \mathcal {I}(\mathcal {A})^+\cap M$ . We fix an enumeration $\{D_n\colon n\in \omega \}$ of all open dense subsets of $\mathbb {Q}_{\mathcal {I}}$ that are in M, and an enumeration $\{\dot {Z}_n\colon n\in \omega \}$ with infinite repetitions of all $\mathbb {Q}_{\mathcal {I}}$ -names for elements of $\mathcal {I}(\mathcal {A})^+$ that are in M.

We define a strategy for the first player in the game GM $_{\mathcal {I}}(E)$ , which cannot be winning in all rounds. Work in M. We set $p_0=q_0=p$ and $u_0=\emptyset $ . We assume that the first player has chosen $E^1_n$ , $q_n$ , $p_n$ , $u_n$ , and the second one has chosen $E^2_n$ . We give instructions to choose $E^1_{n+1}$ , $q_{n+1}$ , $p_{n+1}$ , $u_{n+1}$ . We begin with $q_{n+1}$ :

1. (1) $\operatorname {\mathrm {dom}} E^{q_{n+1}}=\operatorname {\mathrm {dom}} E^{p_n}$ .

2. (2) $xE^{q_{n+1}}y$ iff one of the following holds:

   1. (i) $xE^2_ny$ .

   2. (ii) There is $k\in u_n$ with $x,y\in [k]_{E^{p_n}}$ and $x,y\not \in \operatorname {\mathrm {dom}} E^2_n$ .

   3. (iii) $x,y\notin \bigcup \{[i]_{E^{p_n}}:i\in u_n\}\cup \operatorname {\mathrm {dom}} E^2_n$ .

3. (3) $H^{q_{n+1}}$ is chosen such that:

   1. (i) If $l\in A^{q_{n+1}}$ , then $H^{q_{n+1}}(l)=g_l$ .

   2. (ii) If $l\in A^{p_n}\backslash A^{q_{n+1}}$ , then $H^{q_{n+1}}(l)=g_{\min [l]_{E^{q_{n+1}}}}$ .

   3. (iii) If $l\in \operatorname {\mathrm {dom}} E^{p_n}\setminus A^{p_n}$ , $H^{p_n}(l)=g_i$ then $H^{q_{n+1}}(l)=H^{q_{n+1}}(i)$ .

   4. (iv) If $l\in \operatorname {\mathrm {dom}} E^{p_n}\setminus A^{p_n}$ , $H^{p_n}(l)=1-g_i$ then $H^{q_{n+1}}(l)=1-H^{q_{n+1}}(i)$ .

   5. (v) If $l\in \omega \setminus \operatorname {\mathrm {dom}} E^{p_n}$ then $H^{q_{n+1}}(l)=^{\ast \ast }H^{p_n}(l)$ .

Note that for the already defined condition $q_{n+1}$ we have $q_{n+1}\leq _n p_n$ . Let $z^n$ be the least element of $\operatorname {\mathrm {dom}} E^{q_{n+1}}$ generating the $E^{q_{n+1}}$ -equivalence class of (2)(iii) above and let $u_{n+1}=u_n\cup \{z^n\}$ . By Lemma 5.12, the set $D^{\prime }_n=\{r\in \mathbb {Q}_{\mathcal {I}}\colon r\Vdash \text {"}(\dot {Z}_n\cap B)\setminus n\neq \emptyset \text {"}\}$ is open dense below p (and also below $q_{n+1}$ ). Then $D^{\prime }_n\cap D_n$ is dense below $q_{n+1}$ . Therefore we can apply Claim 5.6(3) to obtain $p_{n+1}\leq _{n+1} q_{n+1}$ such that for each $h\in {}^{u_{n+1}}\{0,1\}$ , the condition $p^{[h]}_{n+1} \in D^{\prime }_n\cap D_n\cap M$ . In particular, if $h\in {}^{u_{n+1}}\{0,1\}$ then $p_{n+1}^{[h]}\Vdash \text {"} (\dot {Z}_n\cap B)\setminus n\neq \emptyset \text {"}$ and $p_{n+1}^{[h]}\in D_n\cap M$ . By Claim 5.6(2) we have $p_{n+1}\Vdash \text {"} (\dot {Z}_n\cap B)\setminus n\neq \emptyset .\text {"}$ Finally, we set

$$\begin{align*}E^1_{n+1}=E^{p_{n+1}}\restriction(\operatorname{\mathrm{dom}} E^{p_{n+1}}\setminus\bigcup\{[i]_{E^{p_{n+1}}}\colon i\in u_{n+1}\}). \end{align*}$$

We define a fusion q of the sequence $\langle p_n\colon n\in \omega \rangle $ . The relation $E^q$ has $\operatorname {\mathrm {dom}} E^q=\bigcap _{n\in \omega }\operatorname {\mathrm {dom}} E^{p_n}$ and $xE^qy$ if for every n large enough, $xE^{p_n}y$ . The function $H^q$ is equal to $H^{p_n}$ for large enough n. In order to guarantee that $\operatorname {\mathrm {dom}} E^q\in \mathcal {I}^*$ , it is sufficient to choose a play with the first player using the described strategy, but loosing, which by Lemma 5.9 exists. The other properties for $q\in \mathbb {Q}_{\mathcal {I}}$ are satisfied by the definition of q.

Finally q is (M, $\mathbb {Q}_{\mathcal {I}}$ )-generic, $q\leq _n p_n$ for all n and so $q\Vdash \text {"}(\forall \dot {Z}\in \mathcal {I}(\mathcal {A})\cap M[\dot {G}])\ |\dot {Z}\cap B|=\omega .\text {"}$

Proof Work over a model of $\textsf {CH}$ . Let $\mathcal {A}_0$ be Shelah’s selective independent family and let $\mathcal {A}_1$ be a tight mad family. Using an appropriate bookkeeping device define a countable support iteration $\langle \mathbb {P}_{\alpha },\dot {\mathbb {Q}}_{\beta }\colon \alpha \leq \omega _2,\beta <\omega _2\rangle $ of posets such that for even $\alpha $ , $\mathbb {P}_{\alpha }$ forces that $\mathbb {Q}_{\alpha }=\mathbb {Q}(\mathcal {C})$ for some uncountable partition $\mathcal {C}$ of $2^{\omega }$ into compact sets, for odd $\alpha $ , $\mathbb {P}_{\alpha }$ forces that $\mathbb {Q}_{\alpha }=\mathbb {Q}_{\mathcal {I}}$ for some maximal ideal $\mathcal {I}$ on $\omega $ , and such that $V^{\mathbb {P}_{\omega _2}}\vDash \mathfrak {a}_T=\mathfrak {u}=\omega _2$ . The iteration ${\mathbb {P}_{\omega _2}}$ has the Sacks property and therefore $\operatorname {\mathrm {cof}}(\mathcal {N})=\omega _1$ . By the indestructibility of selective independence the family $\mathcal {A}_0$ remains maximal independent in $V^{\mathbb {P}_{\omega _2}}$ and so a witness to $\mathfrak {i}=\omega _1$ . Moreover, by the preservation properties of tight MAD families (see [Reference Guzmán, Hrušák and Téllez25]), and the above preservation theorems, $\mathcal {A}_1$ is a witness to $\mathfrak {a}=\omega _1$ in the final model.