Proof.

1. (1) We prove that $\mathcal {F}$ is the intersection of $\omega $ many ultrafilters. Indeed, for any $B\in \mathcal {B}$ , let $\mathcal {U}_B$ be an ultrafilter on $\mathbb {Q}$ extending $\mathcal {F}\cup \{B\}$ . Then ${\mathcal {F}=\bigcap _{B\in \mathcal {B}}\mathcal {U}_B}$ : it is clear that $\mathcal {F}\subseteq \bigcap _{B\in \mathcal {B}}\mathcal { U}_B$ . Pick $X\in \bigcap _{B\in \mathcal {B}}\mathcal {U}_B$ . Then for all $B\in \mathcal {B}$ , $X\cap B$ is not empty, so X is a dense subset of $\mathbb {Q}$ . If $X\notin \mathcal {F}$ , then there are $A\in \mathcal {F}$ and $B\in \mathcal {B}$ such that $X\cap B\cap A=\emptyset $ . This implies that $X\notin \mathcal {U}_B$ , which is a contradiction. Since the intersection of less than continuum many ultrafilters is a non-meager filter, then $\mathcal {F}$ is non-meager.

2. (2) Recall that by the Talagrand–Jalaili–Naini theorem, a filter is non-meager if and only if for any interval partition $\langle I_k:k\in \omega \rangle $ of $\omega $ , there is $X\in \mathcal {F}$ which has empty intersection with infinitely many intervals $I_k$ . Fix a $\subseteq $ -decreasing $\langle B_n:n\in \omega \rangle \subseteq \mathcal {F}$ and let $X\in \mathcal {F}$ be a pseudointersection of such sequence. Define an increasing sequence of natural numbers $\langle n_k:k\in \omega \rangle $ as $n_0=0$ , and for all $k\in \omega $ , $X\setminus n_{k+1}\subseteq B_{n_k}$ . Let $Y\in \mathcal {F}$ such that $Y\cap [n_k,n_{k+1})$ is empty for infinitely many $k\in \omega $ . Define $Z=X\cap Y$ . Let $k\in \omega $ be such that $Y\cap [n_k,n_{k+1})=\emptyset $ . Then $Z\setminus n_{k}=Z\setminus n_{k+1}\subseteq B_{n_k}$ .

3. (3) Let $\langle I_k:k\in \omega \rangle $ be an interval partition of $\omega $ and $l\in \omega $ a positive natural number. For $k\in \omega $ , define $J_k=\bigcup \{I_j:lk\leq j<lk+l\}$ , $J^0_k=J_{2k}\cup J_{2k+1}$ , and $J_k^1=J_{2k+1}\cup J_{2k+2}$ . By hypothesis, we can find $X_0,X_1\in \mathcal {F}$ such that for all $k\in \omega $ , $\vert J_k^0\cap X_0\vert ,\vert J_k^1\cap X_1\vert \leq 1$ . Let $Y=X_0\cap X_1$ . Then Y is as required.

Proof. By recursion construct a family $\{A_\eta :\eta \in \omega _1^{<\omega _1}\}$ such that:

1. (1) $A_\emptyset =\mathbb {Q}$ .

2. (2) For all $\alpha \in \omega _1$ , $\{A_f:f\in \omega _1^{\alpha }\}\subseteq Dense(\mathbb {Q})$ is an almost disjoint family.

3. (3) For any interval partition of $\omega $ , $\langle I_n:n\in \omega \rangle $ , there is $\alpha \in \omega _1$ such that for all $f\in \omega _1^\alpha $ , $A_f$ is a partial selector of $\langle I_n:n\in \omega \rangle $ .

4. (4) For all $X\in Dense(\mathbb {Q})$ , there is $\alpha \in \omega _1$ such that for all $f\in \omega _1^\alpha $ , either ${A_f\subseteq ^*X}$ or $A_f\cap X\notin Dense(\mathbb {Q})$ .

5. (5) For $\beta <\alpha $ , $f\in \omega _1^\beta $ , and $g\in \omega _1^\alpha $ such that $f\subseteq g$ , we have $A_{g}\subseteq ^* A_f$ .

For $f\in \omega _1^{\omega _1}$ , define $\mathcal {U}_f$ to be the filter generated by the family $\{A_{f\upharpoonright \alpha }:\alpha \in \omega _1\}$ , and then $\mathscr {F}=\{\mathcal {U}_f:f\in \omega _1^{\omega _1}\}$ . It is easy to see that $\mathcal {U}_f$ is a selective rational filter for all $f\in \omega _1^{\omega _1}$ .

Now assume W is a forcing extension of V having the same cardinals. Let $\{f_\alpha :\alpha \in \omega _1\}\subseteq \omega _1^{\omega _1}\cap V$ be a set of cardinality $\omega _1$ . For $h\in \bigcup _{\alpha \in \omega _1}\omega _1^\alpha $ , define $\langle h\rangle =\{g\in \omega _1^{\omega _1}:h\subseteq g\}$ . There are two cases:

Case 1. There is $h\in \omega _1^{\omega _1}$ such that for all $\alpha \in \omega _1$ , $\langle h\upharpoonright \alpha \rangle \cap \{ f_\beta :\beta \in \omega _1\}$ has cardinality $\omega _1$ . Then by recursion construct $L\in [\omega _1]^{\omega _1}$ and two strictly increasing sequences $\{\beta _\alpha ,\gamma _\alpha \in \omega _1:\alpha \in L\}$ such that $f_{\beta _\alpha }\upharpoonright \gamma _\alpha =h\upharpoonright \gamma _\alpha $ and $f_{\beta _\alpha }(\gamma _\alpha )\neq h(\gamma _\alpha )$ . Now for $\alpha \in L$ define $A_\alpha =A_{f_{\beta _\alpha }\upharpoonright (\alpha +1)}$ . The families $\mathcal {A}=\{A_\alpha :\alpha \in L\}$ and $\{\mathcal {U}_{f_{\beta _\alpha }}:\alpha \in L\}$ are as required.

Case 2. For all $h\in \omega _1^{\omega _1}$ there is $\alpha \in \omega _1$ such that $\langle h\upharpoonright \alpha \rangle \cap \{f_\beta :\beta \in \omega \}$ is at most countable. Note that in this case, given $h\in \omega _1^{\omega _1}$ , we can find $\alpha \in \omega _1$ such that $\langle h\upharpoonright \alpha \rangle \cap \{f_\beta :\beta \in \omega _1\}$ has cardinality at most $1$ . Thus, for each $\beta \in \omega _1$ , we can find $\alpha _\beta \in \omega _1$ such that $\langle f_\beta \upharpoonright \alpha _\beta \rangle \cap \{f_\gamma :\gamma \in \omega _1\}$ has exactly one element. Let $A_\beta =A_{f_\beta \upharpoonright \alpha _\beta }$ . Then $\{f_\beta :\beta \in \omega _1\}$ and $\{A_\beta :\beta \in \omega _1\}$ work.

Proof. Assume otherwise $\Sigma $ is a winning strategy for Player I. We will produce an uncountable family of $\mathscr {I}$ -positive sets, each two of which are incompatible in $\mathcal {P}(\omega )/\mathscr {I}$ . Let $\langle s_n:n\in \omega \rangle $ be the enumeration of $2^{<\omega }$ defined by the ordering $s\sqsubset t$ if and only if, either $\vert s\vert <\vert t\vert $ or $\vert s\vert =\vert t\vert $ and $s<_{lex}t$ , where $<_{lex}$ is the lexicographical order ( $\sqsubset $ has order type $\omega $ ). We construct a sequence $\langle p_s:s\in 2^{<\omega }\rangle $ such that:

1. (1) $p_{\emptyset }=(\Sigma (\emptyset ))=(\vec {E}_\emptyset )$ .

2. (2) For $s\in 2^{<\omega }\setminus \{ \emptyset \}$ , $p_s=(\vec {F}_s,\vec {E}_s)$ .

3. (3) $\vec {F}_{s_1}=\vec {E}_\emptyset $ and $\vec {E}_{s_1}=\Sigma (\vec {E}_{\emptyset },\vec {F}_{s_1})$ .

4. (4) For $n>1$ :

   1. (a) If $\vec {E}_{s_n}$ is defined, then define $\vec {F}_{s_{n+1}}=\vec {E}_{s_n}$ .

   2. (b) Now, let

      $$ \begin{align*} P(s_{n+1})=(\vec{E}_\emptyset,\vec{F}_{s_{n+1}\upharpoonright 1},\vec{E}_{s_{n+1}\upharpoonright 1},\ldots,\vec{F}_{s_{n+1}\upharpoonright l},\vec{E}_{s_{n+1}\upharpoonright l},\ldots,\vec{E}_{s_{n+1}\upharpoonright(\vert s_{n+1}\vert-1)}, \vec{F}_{s_{n+1}}). \end{align*} $$

      Then make $\vec {E}_{s_{n+1}}=\Sigma (P(s_{n+1}))$ .

Now, for $f\in 2^\omega $ , define $br(f)$ as

$$ \begin{align*} br(f)=(\vec{E}_\emptyset,\vec{F}_{f\upharpoonright 1},\vec{E}_{f\upharpoonright 1},\ldots,\vec{F}_{f\upharpoonright l},\vec{E}_{f\upharpoonright l},\ldots)_{l\in\omega.} \end{align*} $$

By the construction, $br(f)$ is a run of the game $\mathscr {G}(\mathscr {I})$ on which Player I followed the strategy $\Sigma $ , so we have that $D(f)=\bigcup _{n\ge 1}dom(\vec {F}_{f\upharpoonright n})\setminus dom(\vec {E}_{f\upharpoonright n})\in \mathscr {I}^+$ . However, for different $f,g\in 2^{\omega }$ we have that $D(f)\cap D(g)\in \mathscr {I}$ : first note that the construction of the family $\{p_s:s\in 2^{<\omega }\}$ implies that for all $n\in \omega $ , the relations $dom(\vec {F}_{s_{n+1}})\subseteq dom(\vec {E}_{s_n})\subseteq dom(\vec {F}_{s_n})$ hold, and this last property implies that for any pair of distinct functions $f,g\in 2^\omega $ and $l,k>\vert f\cap g\vert $ ,

$$ \begin{align*}(dom(\vec{F}_{f\upharpoonright k})\setminus dom(\vec{E}_{f\upharpoonright k}))\cap(dom(\vec{F}_{g\upharpoonright l})\setminus dom(\vec{E}_{g\upharpoonright l}))=\emptyset,\end{align*} $$

which in turn implies that $D(f)\cap D(g)\subseteq \bigcup _{i\leq \vert f\cap g\vert }dom(\vec {F}_{f\upharpoonright n})\setminus dom(\vec {E}_{f\upharpoonright n})$ , and this last union of sets is in the ideal $\mathscr {I}$ . Therefore, we have that $\{D(f):f\in 2^{\omega }\}$ is an uncountable antichain in $\mathscr {I}^+$ , which is a contradiction.

Proof. Let $\{s_i:i< 2^n\}$ be an enumeration of all the sequences $s:n\to 2$ . Define a decreasing sequence of conditions $\langle q_i:i\in \{-1\}\cup 2^n\rangle $ such that:

1. (1) $q_{-1}=p$ .

2. (2) $q_{i+1}\leq _{n-1} q_i$ .

3. (3) $q_{i}^{s_i}$ decides the value of $\dot {x}$ .

The construction is done as follows:

1. (1) Assume $q_i$ has been defined. Find $r_{i+1}\leq q_i^{s_{i+1}}$ which decides the value of $\dot {x}$ . Then define $q_{i+1}$ as:

   1. (a) $dom(q_{i+1})=\bigcup _{j<n}a_{j}^{p}/\vec {E}_p\cup dom(\vec {E}_{r_{i+1}})$ .

   2. (b) $\vec {E}_{q_{i+1}}=\{a_j^p/\vec {E}_p:j<n\}\cup \vec {E}_{r_{i+1}}$ .

   3. (c) If $m\in a_{j}^p/\vec {E}_{q_{i+1}}$ for some $j<n$ , then $H_{q_{i+1}}(j)=H_{p}(j)$ .

   4. (d) For all $m\in dom(\vec {E}_{r_{i+1}})$ and $x\in 2^{A(\vec {E}_{q_{i+1}})\cap (m+1)}$ ,

      $$ \begin{align*}H_{q_{i+1}}(m)(x)=H_{r_{i+1}}(m)(x\upharpoonright A(\vec{E}_{r_{i+1}})).\end{align*} $$

   5. (e) For $m\notin dom(\vec {E}_{q_i})$ and $x\in 2^{A(\vec {E}_{q_{i+1}})\cap (m+1)}$ ,

      $$ \begin{align*}H_{q_{i+1}}(m)(x)=H_{q_i}(m)(x_{m,q_{i+1}\to q_i}).\end{align*} $$

   It follows directly from the definitions that $q_{i+1}\leq _{n-1} q_i$ .

Finally, note that by construction, for all $i<2^n$ we have $q_{2^n-1}^{s_i}\leq q_{i}^{s_i}=r_{i}$ , and $r_{i}$ decides the value of $\dot {x}$ , so $q_{2^n-1}^{s_i}$ decides the value of $\dot {x}$ .

Proof. Let $\dot {f}$ be a $\mathbb {Q}(\mathscr {I})$ -name for a real and $p\in \mathbb {Q}(\mathscr {I})$ a condition. We construct a decreasing sequence $\langle q_n:n\in \omega \rangle $ , and on the side give a strategy to Player I in the game $\mathscr {G(I)}$ . Since this is not a winning strategy, there is a run of the game in which Player II wins. The existence of such a run of the game will give us the desired condition.

The construction of the sequence is done as follows:

1. (1) Start by choosing $q_0\leq p$ which decides $\dot {f}(0)$ ; then Player I plays the $\mathscr {I}$ -partition $\vec {E}_{q_0}*1$

2. (2) Suppose $q_n$ is constructed such that for each $s\in 2^n$ , $q_n^s$ decides $\dot {f}(n)$ , and Player I has played $\vec {E}^1_n=\vec {E}_{q_n}*(n+1)$ . Then Player II has answered with $\vec {E}^2_n$ . Define $\vec {G}_n=\{dom(\vec {E}_n^1)\setminus dom(\vec {E}_n^2)\}\cup \vec {E}_n^2$ . Clearly $\vec {G}_n$ is an $\mathscr {I}$ -partition. Then find $q_{n+1}\leq _{n} q_n*\vec {G}_n$ and such that for all $s:n+1\to 2$ , $q_{n+1}^s$ decides the value of $\dot {f}(n+1)$ . Then Player I plays $\vec {E}^1_{n+1}=\vec {E}_{q_{n+1}}*(n+2)$ .

Since this is not a winning strategy for Player I, there is a run of the game in which $\bigcup _{n\in \omega }dom(\vec {E}_n^1)\setminus dom(\vec {E}_n^2)\in \mathscr {I}^*$ . Suppose $\langle q_n:n\in \omega \rangle $ is the sequence constructed along such run of the game. Let $F_n=dom(E_n^1)\setminus dom(E_n^2)$ and define $\vec {F}=\langle F_n:n\in \omega \rangle $ . Clearly, $\vec {F}$ is an $\mathscr {I}$ -partition. Note that for any $n\in \omega $ , $F_n=E_{n}^{q_{n+1}}$ , and actually, for any $k\ge n+1$ we have that $F_n=E_{n}^{q_{k+1}}$ . Also note that for any $n\in \omega $ and $x\in 2^{A(\vec {F})\cap (n+1)}$ , the sequence $\langle H_{q_k}(n)(x):k\in \omega \rangle $ is eventually constant (it may be not defined on an initial segment, but at some point it gets well defined, and even eventually constant).

Define a condition $q_\omega $ as follows:

1. (I) $\vec {E}_{q_\omega }=\vec {F}$ .

2. (II) For all $n\in \omega $ and $x\in 2^{A(E_{q_\omega })\cap (n+1)}$ , $H_{q_\omega }(n)(x)=\lim _{k\in \omega } H_{q_k}(n)(x)$ .

We need to prove that $q_\omega $ is a condition in $\mathbb {Q}(\mathscr {I})$ and a lower bound of the sequence $\langle q_n:n\in \omega \rangle $ . This clearly is sufficient to finish the proof.

Clauses i) and ii) from Definition 4 are immediate (clause ii) holds because the limit of clause (II) above is well defined). For clause iii), pick $a_k^{q_\omega }\in A(E_{q_\omega })$ , note that $a_k^{q_\omega }=a_{k}^{q_{k+1}}$ . Moreover, for all $l\ge k+1$ , and any $j\leq k+1 a_j^{q_\omega }/E_{q_\omega }=a_{j}^{q_{\omega }}/E_{q_{l}}$ . This in turn implies that for $l\ge k+1$ , $dom(H_{q_l}(a_k^{q_\omega }))=dom(H_{q_{k+1}}(a_k^{q_\omega }))=2^{A(E_{q_\omega })\cap (a_{k}^{q_\omega }+1)}$ . From this it follows that for $l\ge k+1$ ,

$$ \begin{align*} H_{q_l}(a_{k}^{q_\omega})(x)=e_{{a_k^{q_{k+1}}}}(x)=x(a_{k}^{q_{k+1}}). \end{align*} $$

By definition (II) of $H_{q_\omega }(a_k^{q_\omega })$ , we have $H_{q_\omega }(a_k^{q_\omega })=x(a_k^{q_{\omega }})=e_{a_k}^{q_\omega }(x)$ . It is also easy to see that this implies that clause iv) from Definition 4 is also satisfied.

Fix a condition $q_n$ , we want to prove that $q_\omega \leq q_n$ . Note that $\vec {E}_{q_\omega }\leq \vec {E}_{q_n}$ follows the definition of $\vec {E}_{q_\omega }$ . To check clause (2) from Definition 4, pick $l\in dom(E_{q_\omega })$ and note that for big enough $k\ge \max (n+1,l+1)$ , $A(E_{q_\omega })\cap (l+1)=A(E_{q_k})\cap (l+1)$ , which implies that for big enough $k\ge \max (n+1,l+1),$

$$ \begin{align*} H_{q_\omega}(l)(x)=H_{q_k}(l)(x). \end{align*} $$

It is not hard to see that this implies that the complete requirement of clause (2) holds. To check that clause (3) from Definition 4 holds, pick $l\notin dom(\vec {E}_{q_n})$ , and note in addition that for big enough $k\ge \max (n+1,l+1)$ , it holds that

$$ \begin{align*} H_{q_\omega}(l)(x)=H_{q_k}(l)(x)=H_{q_n}(l)(x_{l,q_k\to q_n}), \end{align*} $$

where the last inequality follows from $q_l\leq q_n$ . Therefore,

$$ \begin{align*}H_{q_\omega}(l)(x)=H_{q_n}(l)(x_{l,q_{\omega}\to q_n})\end{align*} $$

since $x_{l,q_{\omega }\to q_n}=x_{l,q_{k}\to q_n}$ for big enough $k\in \omega $ .

Finally, note that for each $s\in 2^n$ , $q_\omega ^s\leq q_n^s$ decides a value for $\dot {f}(n)$ , so $q_\omega $ bounds the possible values of $\dot {f}(n)$ to a set of size at most $2^n$ .

Proof. The argument completely follows similar lines to the proof of Proposition 1, with small modifications. Let $\mathcal {M}\prec H(\theta )$ be a countable elementary submodel such that $\mathbb {Q}(\mathscr {I})\in \mathcal {M}$ , and pick a condition $p\in \mathcal {M}\cap \mathbb {Q}(\mathscr {I})$ . Let $\langle \dot {\alpha }_n:n\in \omega \rangle $ be an enumeration of all the $\mathbb {Q}(\mathscr {I})$ -names in $\mathcal {M}$ for ordinals. Then apply the construction of Lemma 1, with the modification that instead of guessing $\dot {f}(n)$ at step n, we decide the possible values of the name $\dot {\alpha }_n$ , plus the fact that all the steps of the recursion can be done inside $\mathcal {M}$ , that is, the $\mathscr {I}$ -partitions played by both players are in $\mathcal {M}$ ; the same proof of Lemma 3 shows that in this variation of the game $\mathscr {G}(\mathscr {I})$ Player I has no winning strategy (the only modification is that now the strategy $\sigma $ shoots $\mathscr {I}$ -partitions living in $\mathcal {M}$ ), so we can get the condition $q_\omega $ of Proposition 1. Then the condition $q_\omega $ is an $(\mathcal {M},\mathbb {Q}(\mathscr {I}))$ -generic condition.

Proof. Let $\Sigma $ be a strategy for Player I, we need to produce a run of the game in which Player I follows $\Sigma $ but Player II wins. We think of $\Sigma $ as the family sequences of possible and allowed moves of both players, so we identify $\Sigma $ with a tree with the following properties: for $s\in \Sigma $ and $i<\vert s\vert :$

1. (1) It has a root node $\langle m_0^0\rangle $ , which is the starting move of Player I.

2. (2) If $i=4N$ for some N, then $s(i)\in \omega $ is the movement of Player I determined by $\Sigma $ . The successors of $s\upharpoonright (i+1)$ are the set of all the possibilities that Player II is allowed to play, which is the set of all natural numbers bigger than $s(i)$ .

3. (3) If $i=4N+2$ for some N, then $s(i)\in \mathcal {G}$ is the movement of Player I determined by $\Sigma $ . The successors of $s\upharpoonright (i+1)$ are the set of all the possibilities that Player II is allowed to play, which is the set of all natural numbers which are elements of $s(i)$ .

For $n\in \omega $ , define $\Sigma ^{n}=\{s\in \Sigma :\vert s\vert =n+1\}$ . For $j\in 4$ , let $\Sigma ^{[j]}=\bigcup _{k\in \omega }\Sigma ^{4k+j}$ Also, if $s\in \Sigma $ is a state of the game in which last movement was given by Player II, $\Sigma (s)$ denotes the answer of Player I to the last movement of Player II in s, according to $\Sigma $ .

For $n\in \omega $ , define the following sets:

$$ \begin{align*}\mathcal{E}_n=\{s\in\Sigma^{[2]}:(\forall i<\vert s\vert)(\text{if } 4i+1<\vert s\vert\Longrightarrow s(4i+1)\leq n)\}.\end{align*} $$

Each one of these sets is finite. Now, since $\Sigma $ is a countable set, we have that the family of sets $A\in \mathcal {G}$ which appear in some $s\in \Sigma $ is countable, so we can find $A_\omega \in \mathcal {G}$ which is a pseudointersection of all of them. We construct two sequences of natural numbers $\langle l_n:n\in \omega \rangle $ and $\langle \alpha _n:n\ge 2\rangle $ as follows:

1. (1) $l_0=\Sigma (\emptyset )=m_0^0$ and $l_1>l_0$ is such that $A_\omega \cap [l_0,l_1)\neq \emptyset $ .

2. (2) For each $j\in (l_0,l_1]$ , $\langle l_0,j\rangle $ is an allowed state of the game in which last movement was given by Player II, so $\Sigma (\langle l_0,j\rangle )=A_j$ is well defined. Choose $\alpha _2>l_1$ such that for each $j\in (l_0,l_1]$ , $A_j\cap \alpha _2\neq \emptyset $ . Now pick $l_2>\alpha _2$ such that for each $j\in (l_0,l_1]$ and $i\in A_j\cap \alpha _2$ , $\Sigma (l_0,j,A_j,i)\leq l_2$ .

3. (3) If $l_n\ge l_2$ has been defined, define $l_{n+1}>\alpha _{n+1}>l_n$ such that the following holds:

   1. (a) $\alpha _{n+1}>l_n$ is such that for each $s\in \mathcal {E}_{l_n}$ and $i<\vert s\vert $ such that $4i+2<\vert s\vert $ , there is $j\in s(4i+2)\cap [\alpha _n,\alpha _{n+1})$ .

   2. (b) We require $l_{n+1}>\alpha _{n+1}$ to satisfy that for each $s\in \mathcal {E}_{l_{n}}$ and $i<\vert s\vert $ such that $4i+2<\vert s\vert $ , and $j\in s(4i+2)\cap \alpha _{n+1}$ , $\Sigma (s\upharpoonright (4i+3)^\frown j)<l_{n+1}$ .

   3. (c) Define $H_{n+1}=\{s(4i+2):s\in \mathcal {E}_{l_{n}}\land 4i+2<\vert s\vert \}$ . Then, for all $A\in H_{n+1}$ , $A_\omega \setminus l_{n+1}\subseteq A$ .

   4. (d) $A_\omega \cap [l_n,l_{n+1})\neq \emptyset $ .

Then we get an interval partition $\langle I_n=[l_n,l_{n+1}):n\in \omega \rangle $ and by Lemma 6, we can find an interval partition $\mathcal {J}$ such that each element of $\mathcal {J}$ is union of intervals $I_n$ , and,

$$ \begin{align*}\bigcup_{k\in\omega}J_{4k}\in\mathcal{F}\text{ and }\bigcup_{k\in\omega}J_{4k+2}\in\mathcal{G.}\end{align*} $$

Let $B\in \mathcal {G}$ be a selector for $\mathcal {J}$ and define $B_\omega =B\cap A_\omega \cap \bigcup _{k\in \omega }J_{4k+2}$ and let $\{b_n:n\in \omega \}$ be its increasing enumeration. Note that since we assume $A_\omega \cap [l_n,l_{n+1})\neq \emptyset $ for all $n\in \omega $ , we can assume that $B_\omega \cap J_{4n+2}\neq \emptyset $ for all $n\in \omega $ . Let also $\langle \eta _k:k\in \omega \rangle $ be the increasing sequence such that $J_k=[\eta _k,\eta _{k+1})$ , and $\langle \epsilon _k:k\in \omega \rangle $ be such that $\eta _k=l_{\epsilon _k}$ . Now we defined the following run of the game:

1. (1) Player I starts by playing $m_0^0=l_0$ and let Player II play $m_0^1=\eta _1=l_{\epsilon _1}$ . Note that $[m_0^0,m_0^1)=J_0$ . Define $A_0=\Sigma (m_0^0,m_0^1)$ , so $A_0\in H_{{\epsilon _1}+1}$ (since ${m_0^0}^\frown {m_1^0}^\frown A_0\in \mathcal {E}_{l_{\epsilon _1}}$ ), which implies that $A_\omega \setminus l_{\epsilon _1+1}\subseteq A_0$ . Note that $b_0\ge \min (J_2)=l_{\epsilon _2}\ge l_{\epsilon _1+1}$ , so Player II can play $b_0$ . This finishes the first round of the game.

2. (2) Assume we have constructed a partial run of the game s such that $\vert s\vert =4n$ for some positive $n\in \omega $ . Then the last movement was given by Player II and corresponds to Player II playing $b_{n-1}\in J_{4(n-1)+2}$ . Then we have $b_{n-1}\in [l_{j},l_{j+1})$ for some $j+1\leq \epsilon _{4(n-1)+3}$ , which implies that $b_{n-1}<l_{j+1}<\alpha _{j+2}$ , which means that $\Sigma (s)=m_{n}^0<l_{j+2}\leq l_{\epsilon _{4(n-1)+4}}=l_{\epsilon _{4n}}=\eta _{4n}<\eta _{4n+1}$ , and Player II can play $m_n^1=\eta _{4n+1}$ . Define $A_n=\Sigma (s^\frown {m_n^0}^\frown m_n^1)$ , and note that $A_n\in H_{\epsilon _{4n+1}+1}$ (since $s^\frown {m_n^0}^\frown {m_n^1}^\frown A_n\in \mathcal {E}_{l_{\epsilon _{4n+1}}}$ ), which implies that $A_\omega \setminus l_{\epsilon _{4n+1}+1}\subseteq A_n$ . Since $b_n\in J_{4n+2}\cap B_\omega $ and $\min (J_{4n+2})=\eta _{4n+2}=l_{\epsilon _{4n+2}}\ge l_{\epsilon _{4n+1}+1}$ , we have that ${b_n\in A_n}$ , so Player II can play $b_n\in A_n$ . This finishes the definition of the current round, which results in $s^\frown {m_n^0}^\frown {m_n^1}^\frown {A_n}^\frown b_n$ . Note that $J_{4n}\subseteq [m_n^0,m_n^1)$ .

Note that after this run of the game, $\bigcup _{k\in \omega }J_{4k}\subseteq \bigcup _{k\in \omega }[m_k^0,m_k^1)$ and the second set constructed by Player II is exactly $B_\omega $ . Thus, Player II wins.

Proof. If $\Sigma $ is a winning strategy for Player I in the game $\mathscr {G}_{\vec {E}}(\mathcal {F},\mathcal {G})$ , then $\Sigma $ is a winning strategy in the game $\mathscr {G}(h(\mathcal {F}),\mathcal {G})$ . However, $h(\mathcal {F})$ and $\mathcal {G}$ are not nearly coherent.

Proof. Let $\dot {X}$ be a name for a dense subset of the rational numbers and $p\in \mathbb {Q}(\mathscr {I})$ a condition. We have to find a condition $q\leq p$ such that either, there is $Z\in \mathcal {F}$ such that $q\Vdash Z\cap \dot {X}\notin Dense(\mathbb {Q})$ , or $q\Vdash Z\subseteq \dot {X}$ . By Proposition 1 we can assume that for all $n\in \omega $ and $s\in 2^n$ , $p^s$ decides the truth value of $n\in \dot {X}$ .

For each $s\in 2^{<\omega }$ , let $X_s$ be the set $X_s=\{n\in \omega :\exists q\leq p^{s}\ q\Vdash n\in \dot {X}\}$ . If there is $s\in 2^{<\omega }$ such that $X_s\notin \mathcal {F}$ , then there is $Z\in \mathcal {F}$ such that $X_s\cap Z\notin Dense(\mathbb {Q})$ . Since $p^{s}$ forces $\dot {X}$ to be a subset of $X_s$ , then $\dot {X}\cap Z$ is forced by $p^s$ to be a not dense subset of the rational numbers, so we are done with this case and we can assume that $X_s\in \mathcal {F}$ for all $s\in 2^{<\omega }$ .

Define a function $f:dom(\vec {E}_p)\to \omega $ as $f(k)=n$ if and only if $k\in E_n$ . Then $\mathcal {F}\cup f(\mathscr {I}^*)$ does not generate a filter. Moreover, our assumptions on $\mathscr {I}^*$ and $\mathcal {F}$ imply that $\mathcal {F}$ and $f(\mathscr {I}^*)$ are not nearly coherent, so Player I has no winning strategy in the game $\mathscr {G}_{\vec {E}_p}(\mathscr {I}^*,\mathcal {F})$ . We define a strategy for Player I in the game $\mathscr {G}_{\vec {E}_p}(\mathscr {I}^*,\mathcal {F})$ , in which Player I constructs along the game a decreasing sequence of conditions. This is done as follows:

1. (1) Player I starts by playing $m_0^0=0$ . Define $q_{-1}=p$ .

2. (2) After Player II answers with $m^1_0>m_0^0$ , define $A_0=\bigcap _{s\in 2^{m_0^1}}X_s$ , and let Player I play the set $A_0$ . Then Player II answers with $k_0\in A_0$ . This finishes the first round of the game.

3. (3) Assume round number n has finished and Player II has played $k_n\in A_n$ . Then, for each $s\in 2^{m_n^1}$ , pick $r_s\in 2^{<\omega }$ extending s and such that $p^{r_s}\Vdash k_n\in A_n$ . Let $m_{n+1}^0\ge \max \{\vert r_s\vert :s\in 2^{m_n^1}\}$ and let Player I play $m_{n+1}^0$ . Then Player II answers with $m_{n+1}^1>m_{n+1}^0$ . Let Player I play the set $A_{n+1}=\bigcap _{s\in 2^{m_{n+1}^1}}X_s$ . Then Player II answers with $k_{n+1}\in A_{n+1}$ , and the round is finished.

   Now we define condition $q_n$ . For $s\in 2^{m_n^1}$ , pick $\circ s\in 2^{m_{n+1}^0}$ such that $r_s\subseteq \circ s$ . Define $q_{n}$ as follows:

   1. (a) $dom(\vec {E}_{q_n})=\bigcup _{l\leq n}\bigcup _{j\in [m_l^0,m_l^1)}E^{q_{n-1}}_j\cup \bigcup _{j\ge m_{n+1}^0}E_j^{q_{n-1}}$ .

   2. (b) $\vec {E}_{q_n}=\vec {E}_{q_{n-1}}\upharpoonright dom(\vec {E}_{q_n})=\vec {E}_p\upharpoonright dom(\vec {E}_{q_n})$ .

   3. (c) For $j\in [m_{n}^1,m_{n+1}^0)$ and $x\in 2^{A(E_{q_n})\cap (a_{j}^{q_{n-1}}+1)}$ , let $s_x=x\upharpoonright a_{m_n^1}^{q_n}$ and define $H_{q_n}(a_{j}^{q_{n-1}})(x)=\circ s_x(j)$ . This implicitly defines $H_{q_\omega }(j)$ for $j\in \bigcup _{l\in [m_{n}^1,m_{n+1}^0)}E_l^{q_{n-1}}$ (which should satisfy clause (2) from Definition 4).

   4. (d) For $j\in dom(\vec {E}_{q_n})$ , define $H_{q_n}(j)$ in the natural way to satisfy clauses ii)–iv) and (2) from Definition 4.

   5. (e) For $j\notin dom(\vec {E}_{q_{n-1}})$ , $H_{q_n}(j)$ is defined accordingly to clause (3) from Definition 4.

   Note that for each $s\in 2^{m_n^1}$ , $q_n^s\leq p^{\circ s}\leq p^{r_s}$ , so $q_n^s\Vdash k_n\in \dot {X}$ . Thus, $q_n\Vdash k_n\in \dot {X}$ .

This is not a winning strategy for Player I, so there is a run of the game in which Player II wins, so, for such run, we have $\bigcup _{n\in \omega }\bigcup _{j\in [m_n^0,m_n^1)}E^p_j\in \mathscr {I}^*$ and $\{k_n:n\in \omega \}\in \mathcal {F}$ . Let $\langle m_n^0,m_n^1,k_n, q_n:n\in \omega \rangle $ be the sequences constructed along such run of the game. Define a condition $q_\omega $ as follows:

1. (1) $dom(\vec {E}_{q_\omega })=\bigcup _{n\in \omega }\bigcup _{j\in [m_n^0,m_n^1)}E^p_{j}\in \mathscr {I}^*$ .

2. (2) $\vec {E}_{q_\omega }=\vec {E}_p\upharpoonright dom(\vec {E}_{q_\omega })$ .

3. (3) For any $j\in \omega $ and $x\in 2^{A(\vec {E}_{q_\omega })\cap (j+1)}$ , let $H_{q_\omega }(j)=\lim _{n\to \omega } H_{q_n}(j)(x)$ .

Checking that $q_\omega $ is a condition and a lower bound of $\{q_n:n\in \omega \}$ is quite similar to the case of Proposition 1, so we omit the details. Finally, note that since $q_n\Vdash k_n\in \dot {X}$ and $q_\omega \leq q_n$ for all $n\in \omega $ , we also have that $q_\omega \Vdash \{k_n:n\in \omega \}\subseteq \dot {X}$ .

Proof. Let R be the collection of all open subsets of $\omega $ which are resolvable. Note that if $U\in R$ and $V\subseteq U$ is open, then $V\in R$ . Let $\mathcal {A}\subseteq R$ be a maximal subset of disjoint open sets. Note that $W=\bigcup \mathcal {A}$ is resolvable, and $\omega \setminus W$ has non-empty interior (otherwise $(\omega ,\tau )$ would be resolvable). Define $X=int_{\tau }(\omega \setminus W)$ . Clearly $(X,\tau \upharpoonright X)$ is strongly irresolvable: otherwise, there would be an open set $U\subseteq X$ which is resolvable, and since X is open in $\tau $ , U is open in $\tau $ , so we get $U\in R$ , which implies $X\cap W\neq \emptyset $ , which is a contradiction.

Proof. Clearly the inequality $\leq $ is true. The other inequality follows from the previous lemma and the fact that the $\pi $ -weight of $(X,\tau \upharpoonright X)$ is at most the $\pi $ -weight of $(\omega ,\tau )$ .

Proof. First note that for any $A\in \mathsf {nwd}(\tau )^+$ , A has non-empty interior: assume otherwise and pick some $A\in \mathsf {nwd}(\tau )^+$ with empty interior, and let $U\in \tau $ be such that $U\subseteq \bar {A}$ . Note that $U\setminus A$ and $U\cap A$ are both dense in U, which means that U is resolvable, which is a contradiction. Now, let $\mathcal {A}\subseteq \mathcal {P}(\omega )/\mathsf {nwd}(\tau )$ be a maximal antichain. Then $\{int(A):A\in \mathcal {A}\}$ is also an antichain, and moreover, for any two $A,B\in \mathcal {A}$ , $int(A)\cap int(B)=\emptyset $ , which implies that $\{int(A):A\in \mathcal {A}\}$ is countable (since $\omega $ is countable), which in turn implies that $\mathcal {A}$ is countable.

Proof. Let $p=(H_p,E_p)\in \mathbb {Q}(\mathsf {nwd}(\tau ))$ be a condition. Note that for each $n\in dom(E_p)$ , we can find extensions $q_0,q_1\leq p$ such that for each $i\in 2$ , $H_{q_i}(n)$ has value constant i, and $\min (dom(q_i))>n$ . Thus, if $V\in \tau $ is a basic open set, we can pick distinct $n_0,n_1\in V\cap dom(E_p)$ and find a condition $q\leq p$ such that for each $i\in 2$ , $H_q(n_i)$ has constant value i, and $\min (dom(E_q))>\max \{n_0,n_1\}$ . This means that $q\Vdash n_0\notin \dot {x}_{gen}\cap V,n_1\in \dot {x}_{gen}\cap V$ .

Proof. Let $\mathcal {B}$ be our fixed base for $\mathbb {Q}$ . Let $\mathcal {U}\subseteq Dense(\mathbb {Q})$ be a rational filter. Define a base for a topology as $\mathcal {B}_{\mathcal {U}}=\{B\cap A:B\in \mathcal {B}\land A\in \mathcal {U}\}$ and let $\tau (\mathcal {U})$ be the topology generated by $\mathcal {B}_{\mathcal {U}}$ . Then, $(\mathbb {Q},\tau (\mathcal {U}))$ is a $T_2$ irresolvable space: let $X\subseteq \mathbb {Q}$ be an arbitrary $\tau (\mathcal {U})$ -dense set, which in particular implies that $X\in Dense(\mathbb {Q})$ , and the definition of $\mathcal {B}_{\mathcal {U}}$ implies that the intersection of X with any element of $\mathcal {U}$ is a dense set in $\mathbb {Q}$ , so the maximality of $\mathcal {U}$ implies that $X\in \mathcal {U}$ holds; this clearly implies that the intersection of any two dense sets is not empty. The space $(\mathbb {Q},\tau (\mathcal {U}))$ is $T_2$ because the base $\mathcal {B}$ separates points. It is also clear that the $\pi $ -weight of $(\mathbb {Q},\tau (\mathcal {U}))$ is at most $\chi (\mathcal {U})$ . Thus, $\mathfrak {irr}_2\leq \mathfrak {u}_{\mathbb {Q}}$ follows.

To prove the inequality $\mathfrak {irr}_2\leq \mathfrak {irr}$ , since any $T_3$ irresolvable space is $T_2$ , it suffices to prove that any $T_3$ irresolvable topology on $\omega $ is $\mathbb {Q}$ -compatible. To prove this, let $\mathcal {M}\prec H(\theta )$ be countable such that $(\omega ,\tau )\in \mathcal {M}$ and consider the topology generated by the family $\tau \cap \mathcal {M}$ . Since $\mathcal {M}$ models that $(\omega ,\tau )$ is a $T_3$ space, the topology $\tau '$ generated by $\tau \cap \mathcal {M}$ is $T_3$ : for any $n\in \omega $ and any basic set $U\in \tau \cap \mathcal {M}$ such that $n\in U$ , by elementarity, there is $V\in \tau \cap \mathcal {M}$ such that $n\in \overline {V}^{\tau }\subseteq U$ ; moreover, by elementarity we also have $\overline {V}^\tau =\overline {V}^{\tau '}$ for any set $V\in \tau \cap \mathcal {M}$ . Clearly $(\omega ,\tau ')$ has no isolated points. Therefore, by Sierpiński’s theorem, we have that $(\omega ,\tau ')$ is homeomorphic to the rational numbers with their usual topology.

Proof. The argument is essentially the same as for the generic existence of p-points. For a given filter $\mathcal {F}\subseteq Dense(\mathbb {Q})$ and $X\in Dense(\mathbb {Q})$ , let us say that X is $(\mathcal {F},\mathsf {nwd})$ -positive if for all $A\in \mathcal {F}$ , $A\cap X\in Dense(\mathbb {Q})$ . Let $\mathcal {F}\subseteq Dense(\mathbb {Q})$ be a filter of character less than $\mathfrak {c}$ and let $\mathcal {H}\subseteq \mathcal {F}$ a base for $\mathcal {F}$ of minimal cardinality, fix $\{\vec {D}_\alpha :\alpha \in \mathfrak {c}\}$ an enumeration of all $\subseteq $ -decreasing sequences of dense subsets of the rationals. Construct a $\subseteq $ -increasing sequence of families $\{\mathcal {F}_\alpha :\alpha \in \mathfrak {c}\}$ such that:

1. (1) $\mathcal {F}_0=\mathcal {H}$ .

2. (2) For each $\alpha \in \mathfrak {c}$ , if each set in the sequence $\vec {D}_\alpha $ is $(\mathcal {F}_\alpha ,\mathsf {nwd})$ -positive, then there is $X_\alpha \in \mathcal {F}_{\alpha +1}$ which is almost contained in each term of the sequence $\vec {D}_\alpha $ .

3. (3) For all $\alpha \in \mathfrak {c}$ , $\vert \mathcal {F}_\alpha \vert <\mathfrak {c}$ .

Assume $\mathcal {F}_\alpha $ has been constructed and each term of $\vec {D}_\alpha =\langle D_n^\alpha :n\in \omega \rangle $ is $(\mathcal {F}_\alpha ,\mathsf {nwd})$ -positive (in other case define $\mathcal {F}_{\alpha +1}=\mathcal {F}_\alpha $ ). Let $\langle B_n:n\in \omega \rangle $ be an enumeration of a base for the rationals. For each $X\in \mathcal {F}_\alpha $ , define the function

$$ \begin{align*}\varphi_X(n)=\min\{k\in\omega:(\forall i\leq n)(D^\alpha_n\cap B_i\cap X\cap k\neq\emptyset)\}.\end{align*} $$

The family $\{\varphi _X:X\in \mathcal {F}_\alpha \}$ has cardinality less than $\mathfrak {d}$ , so there is a function $f_\alpha $ which is not $\leq ^*$ -dominated by each one of the functions $\varphi _X$ . Let $Z_\alpha =\bigcup _{n\in \omega } D_n^\alpha \cap f(n)$ . It is easy to see that $Z_\alpha $ is a pseudointersection of $\vec {D}_\alpha $ and $(\mathcal {F}_\alpha ,\mathsf {nwd})$ -positive, so we can define $\mathcal {F}_{\alpha +1}=\mathcal {F}_\alpha \cup \{Z_\alpha \}$ . At limit steps just take the union of all the previously constructed families. Now let $\mathcal {G}$ be the filter generated by $\bigcup _{\alpha \in \mathfrak {c}}\mathcal {F}_\alpha $ . Clause (2) makes sure $\mathcal {G}$ is a p-filter. To see that $\mathcal {G}$ is maximal, pick $D\in Dense(\mathbb {Q})$ , and let $\alpha $ be such that $\vec {D}_\alpha $ is the constant sequence with values equal to D, then either, D is not $(\mathcal {F}_\alpha ,\mathsf {nwd})$ -positive, or $Z_\alpha \subseteq ^* D$ .

Proof. It is not hard to see that the functions described below can be chosen to be Borel, so we omit the details of this. First let us recall that if $\diamondsuit (\mathfrak {r}_{\mathbb {Q}}^+)$ holds, then $\mathfrak {r}_{\mathbb {Q}}^+=\omega _1$ , and since $\mathfrak {d}\leq \mathfrak {r}_{\mathbb {Q}}^+$ holds, we get $\mathfrak {d}=\omega _1$ as well. Let $\{\vec {\mathcal {I}}_\alpha :\alpha \in \omega _1\}$ be a dominating family of interval partitions, that is, for any interval partition $\vec {\mathcal {J}}$ , there is $\alpha \in \omega _1$ such that any $I\in \vec {\mathcal {I}}_\alpha $ contains one element from $\vec {\mathcal {J}}$ . For each limit ordinal $\alpha \in \omega _1$ , fix a bijection $e_\alpha :\omega \to \alpha $ . Now, by a suitable coding, we can assume that the domain of the function F is $dom(F)=\bigcup _{\alpha \in \lim (\omega _1)}[\omega ]^{\omega }\times ([\omega ]^\omega )^\alpha $ . Given a sequence $\vec {X}=\langle X_\beta :\beta \in \alpha \rangle $ , let us say that $\vec {X}$ is $\mathbb {Q}$ -centered if the intersection of finitely many terms of the sequence is a dense set. Given a $\mathbb {Q}$ -centered sequence $\vec {X}=\langle X_\beta :\beta <\alpha \rangle $ , let $\varphi _\alpha (\langle X_\beta :\beta <\alpha \rangle )$ be the set $\{k_n:n\in \omega \}$ defined by

$$ \begin{align*} k_n=\min\left(\bigcap_{j\leq n}X_{e_\alpha(j)}\cap B_n\right). \end{align*} $$

Note that the set $\{k_n:n\in \omega \}$ is a dense set and a pseudointersection of $\{X_\beta :\beta <\alpha \}$ . Now define the function F as follows:

1. a) If $(C,\langle X_\beta :\beta <\alpha \rangle )$ is such that $\langle X_\beta :\beta <\alpha \rangle $ is not a $\mathbb {Q}$ -centered sequence, let $F(C,\langle X_\beta :\beta <\alpha \rangle )=\omega $ .

2. b) If $\langle X_\beta :\beta <\alpha \rangle $ is a $\mathbb {Q}$ -centered sequence, but $C\cap \varphi _\alpha (\langle X_\beta :\beta <\alpha \rangle )$ is not a dense set, define $F(C,\langle X_\beta :\beta <\alpha \rangle )=\omega $ .

3. c) In the remaining case, let $\psi _{\varphi _\alpha (\langle X_\beta :\beta \in \alpha \rangle )}:\omega \to \varphi _\alpha (\langle X_\beta :\beta <\alpha \rangle )$ be a homeomorphism between $\omega $ and $\varphi _\alpha (\langle X_\beta :\beta <\alpha \rangle )$ , which is constructed in a recursive way. Then define $F(C,\langle X_\beta :\beta <\alpha \rangle )=\psi _\alpha ^{-1}[C\cap \varphi _\alpha (\langle X_\beta :\beta <\alpha \rangle )]$ .

Let now g be a guessing function for F. Define an increasing sequence of families $\{\mathcal {F}_\alpha :\alpha \in \lim (\omega _1)\}$ as follows:

1. (1) Start with $\mathcal {F}_\omega =\{\omega \setminus k:k\in \omega \}$ .

2. (2) Suppose $\mathcal {F}_\alpha =\{D_\beta :\beta <\alpha \}$ has been defined. Let $D_\alpha \subseteq \psi _{\varphi _\alpha (\langle D_\beta :\beta \in \alpha \rangle )}[g(\alpha )]$ be such that it is a dense set and a partial selector of $\vec {\mathcal {I}}_\alpha $ , and put $\mathcal {F}_{\alpha +1}=\mathcal {F}_\alpha \cup \{D_\alpha \}$ .

It is clear from the construction that $\mathcal {F}_\alpha $ is a family of dense sets generating a filter. Also note that $\mathcal {H}=\{D_\alpha :\alpha \in \omega _1\}$ is a $\subseteq ^*$ -decreasing sequence of dense subsets, so it is enough to prove that the filter generated by $\mathcal {H}$ is maximal and a q-filter. Let $X\in Dense(\mathbb {Q})$ be arbitrary and consider $(X,\langle D_\alpha :\alpha \in \omega _1\rangle )$ . By the guessing property of g, there is $\alpha \in \omega _1$ such that $g(\alpha ) Dense(\mathbb {Q})$ -reaps $F(X,\langle D_\beta :\beta <\alpha \rangle )$ , that is, $g(\alpha ) Dense(\mathbb {Q})^+$ -reaps the set $\psi _{\varphi _\alpha (\langle D_\beta :\beta <\alpha \rangle )}^{-1}[X\cap \varphi _\alpha (\langle D_{\beta }:\beta <\alpha \rangle )]$ , so we have that, either $D_\alpha \subseteq ^*X$ or $D_\alpha \cap X$ is not dense. This shows that the filter generated by $\mathcal {H}$ maximal. Since it is generated by a $\subseteq ^*$ -decreasing sequence, it is also a p-filter. To see that this filter is a q-filter, note that by clause (2) above, $D_\alpha $ is a partial selector of $\vec {\mathcal {I}}_\alpha $ ; this clearly implies that the filter generated by $\mathcal {H}$ has partial selector for all interval partitions.

Proof. We have seen in the paragraph preceding this proposition the existence of a $\tau (\mathcal {F})$ -open set V such that $\mathcal {F}\upharpoonright V$ is strongly irresolvable. A base for the topology $\tau (\mathcal {F})$ was defined as $\mathcal {B}_{\mathcal {F}}=\{B\cap X:B\in \mathcal {B}\land X\in \mathcal {F}\}$ , so we can find $B\in \mathcal {B}$ and $X\in \mathcal {F}$ such that $X\cap B\subseteq V$ . It is clear that $\mathcal {F}\upharpoonright (X\cap B)$ is a p-filter. To prove that it is also a saturated filter, by Lemma 8, it is enough to prove that $\mathsf {nwd}(\tau (\mathcal {F}))\upharpoonright (X\cap B)^*\subseteq \mathcal {F}\upharpoonright (X\cap B)$ . Note that a base for the topology $\tau (\mathcal {F})\upharpoonright (X\cap B)$ is given by $\mathcal {B}_{\mathcal {F}}'=\{W\in \mathcal {B}_{\mathcal {F}}:W\subseteq X\cap B\}$ . For simplicity, let us write $\tilde {\tau }$ instead of $\tau (\mathcal {F})\upharpoonright (X\cap B)$ . Note that since $X\cap B$ is $\tau (\mathcal {F})$ -open, we have that $N\subseteq X\cap B$ is $\tilde {\tau }$ -nowhere dense if and only if it is $\tau (\mathcal {F})$ -nowhere dense, so $\mathsf {nwd}(\tilde {\tau })=\mathsf {nwd}(\tau (\mathcal {F}))\upharpoonright (X\cap B)$ .

Pick $N\in \mathsf {nwd}(\tilde {\tau })$ , we want to show that $(X\cap B)\setminus N\in \mathcal {F}\upharpoonright (X\cap B)$ . Since N is nowhere dense, any open set $W\in \tau (\mathcal {F})\upharpoonright (X\cap V)$ can be shrunk to a basic open set having empty intersection with N. Thus, we can find a maximal antichain of basic open sets $\{B_n\cap X_n:n\in \omega \}\subseteq \mathcal {B}_{\mathcal {F}}'$ , each one of whose elements has empty intersection with N. Note that by the definition of the basis $\mathcal {B}_{\mathcal {F}}$ and $\mathcal {B}_{\mathcal {F}}'$ , we should have $B_n\cap B_m=\emptyset $ whenever $n\neq m$ . Note also that $\bigcup _{n\in \omega }B_n$ is an open dense subset of B (in the usual topology of the rationals). Since $\mathcal {F}$ is a p-filter, there is $X_\omega $ which is almost contained in each one of the sets $X_n$ . Then $F_n=X_\omega \cap B_n\setminus X_n\cap B_n$ is a finite subset of $B_n$ ; this implies that $\bigcup _{n\in \omega }F_n$ is a discrete set in $\bigcup _{n\in \omega }B_n$ , so $Z=X_\omega \setminus \bigcup _{n\in \omega }F_n\in \mathcal {F}$ . Then we have that $Z\cap B_n$ is a basic open set and is contained in $X_n\cap B_n$ , so it has empty intersection with N. On the other hand, $W=\bigcup _{n\in \omega }Z\cap B_n$ is an open $\tilde {\tau }$ -dense set, and it is an element of the filter $\mathcal {F}\upharpoonright (X\cap B)$ : note that $U=(\omega \setminus \overline {B})\cup \bigcup _{n\in \omega }B_n$ is an open dense set of $\mathbb {Q}$ , so we have $U\in \mathcal {F}$ , which implies $Z\cap U\in \mathcal {F}$ , and it is easy to see that $B\cap Z\cap U=W$ . Thus, we conclude that $W\in \mathcal {F}\upharpoonright (X\cap B)$ , which in turn implies $\omega \setminus N\in \mathcal {F}\upharpoonright (X\cap B)$ (because W and N are disjoint). We have proved that $\mathsf {nwd}(\tilde {\tau })^*\subseteq \mathcal {F}\upharpoonright (X\cap B)$ .

Proof. In the model constructed in [Reference Frankiewicz, Shelah and Zbierski5] there are no saturated p-filters, so the previous corollary implies that in such model there is no rational p-filter as well.