Proof. (1) Obvious.

(2) Let $\mathcal {F}\subset \mathcal {D}_{\mathcal {K}}$ be an $\geq _{\mathcal {I}}\cap (\mathcal {D}_{\mathcal {K}}\times \mathcal {D}_{\mathcal {J}})$ -unbounded family of cardinality $\mathfrak {b}(\geq _{\mathcal {I}}\cap (\mathcal {D}_{\mathcal {K}}\times \mathcal {D}_{\mathcal {J}}))$ and $\mathcal {F} = \bigcup _{\alpha <\operatorname {\mathrm {cf}}(\mathfrak {b}(\geq _{\mathcal {I}}\cap (\mathcal {D}_{\mathcal {K}}\times \mathcal {D}_{\mathcal {J}}))}\mathcal {F}_\alpha $ with $|\mathcal {F}_\alpha |<\mathfrak {b}(\geq _{\mathcal {I}}\cap (\mathcal {D}_{\mathcal {K}}\times \mathcal {D}_{\mathcal {J}}))$ for every $\alpha $ .

For every $\alpha $ we find $g_\alpha \in \mathcal {D}_{\mathcal {J}}$ with $f\geq _{\mathcal {I}} g_\alpha $ for every $f\in \mathcal {F}_\alpha $ . Let $\mathcal {G}=\{g_\alpha :\alpha <\operatorname {\mathrm {cf}}(\mathfrak {b}(\geq _{\mathcal {I}}\cap (\mathcal {D}_{\mathcal {K}}\times \mathcal {D}_{\mathcal {J}}))\}$ .

Suppose to the contrary that $\operatorname {\mathrm {cf}}(\mathfrak {b}(\geq _{\mathcal {I}}\cap (\mathcal {D}_{\mathcal {K}}\times \mathcal {D}_{\mathcal {J}})))<\mathfrak {b}(\geq _{\mathcal {I}}\cap (\mathcal {D}_{\mathcal {J}}\times \mathcal {D}_{\mathcal {J}}))$ .

Since $\mathcal {G}\subseteq \mathcal {D}_{\mathcal {J}}$ and $|\mathcal {G}|\leq \operatorname {\mathrm {cf}}(\mathfrak {b}(\geq _{\mathcal {I}}\cap (\mathcal {D}_{\mathcal {K}}\times \mathcal {D}_{\mathcal {J}}))<\mathfrak {b}(\geq _{\mathcal {I}}\cap (\mathcal {D}_{\mathcal {J}}\times \mathcal {D}_{\mathcal {J}})$ , there is $g\in \mathcal {D}_{\mathcal {J}}$ with $g_\alpha \geq _{\mathcal {I}} g$ for every $\alpha $ . Consequently, $f\geq _{\mathcal {I}} g$ for every $f\in \mathcal {F}$ , a contradiction.

(3) This can be proved in a similar way as item 2.

(4) If $\mathcal {J}\subseteq \mathcal {K}$ , then $\mathfrak {b}(\geq _{\mathcal {I}}\cap (\mathcal {D}_{\mathcal {K}}\times \mathcal {D}_{\mathcal {J}})) \leq \mathfrak {b}(\geq _{\mathcal {I}}\cap (\mathcal {D}_{\mathcal {J}}\times \mathcal {D}_{\mathcal {J}}))$ and $\mathfrak {b}(>_{\mathcal {I}}\cap (\mathcal {D}_{\mathcal {K}}\times \mathcal {D}_{\mathcal {J}})) \leq \mathfrak {b}(>_{\mathcal {I}}\cap (\mathcal {D}_{\mathcal {J}}\times \mathcal {D}_{\mathcal {J}}))$ . Now using (2) and (3), we get $\mathfrak {b}(\geq _{\mathcal {I}}\cap (\mathcal {D}_{\mathcal {K}}\times \mathcal {D}_{\mathcal {J}})) \leq \mathfrak {b}(\geq _{\mathcal {I}}\cap (\mathcal {D}_{\mathcal {J}}\times \mathcal {D}_{\mathcal {J}}))\leq \operatorname {\mathrm {cf}}(\mathfrak {b}(\geq _{\mathcal {I}}\cap (\mathcal {D}_{\mathcal {K}}\times \mathcal {D}_{\mathcal {J}})))$ and $\mathfrak {b}(>_{\mathcal {I}}\cap (\mathcal {D}_{\mathcal {K}}\times \mathcal {D}_{\mathcal {J}})) \leq \mathfrak {b}(>_{\mathcal {I}}\cap (\mathcal {D}_{\mathcal {J}}\times \mathcal {D}_{\mathcal {J}}))\leq \operatorname {\mathrm {cf}}(\mathfrak {b}(>_{\mathcal {I}}\cap (\mathcal {D}_{\mathcal {K}}\times \mathcal {D}_{\mathcal {J}})))$ .⊣

Proof. (1) Obvious.

(2) It is enough to show $\mathfrak {b}(\mathcal {I},\mathcal {J},\mathcal {K})\geq \mathfrak {b}_s(\mathcal {I},\mathcal {J},\mathcal {K})$ .

Let $\mathcal {E}\subseteq \widehat {\mathcal {P}}_{\mathcal {K}}$ be such that for every $\langle A_n:n\in \omega \rangle \in \mathcal {P}_{\mathcal {J}}$ there is $\langle E_n\rangle \in \mathcal {E}$ with $\bigcup _{n\in \omega }(A_n\cap \bigcup _{i\leq n}E_i)\notin \mathcal {I}$ .

For every $E = \langle E_n\rangle \in \mathcal {E}$ we define $F^E_0=E_0\cup E_1$ and $F^E_n=E_{n+1}$ for $n\geq 1$ . Then $\mathcal {F}=\{\langle F^E_n\rangle :E\in \mathcal {E}\}\subseteq \widehat {\mathcal {P}}_{\mathcal {K}}$ and $|\mathcal {F}|\leq |\mathcal {E}|$ .

Let $\langle A_n\rangle \in \mathcal {P}_{\mathcal {J}}$ . Then there is $\langle E_n\rangle \in \mathcal {E}$ with $B = \bigcup _{n\in \omega }(A_n\cap \bigcup _{i\leq n}E_i)\notin \mathcal {I}$ .

Since $A_0\cap E_0\in \mathcal {J}\cap \mathcal {K}\subseteq \mathcal {I}$ and $ \bigcup _{n\in \omega }(A_{n+1}\cap \bigcup _{i\leq n}F^E_i) = (\bigcup _{n\in \omega }(A_{n}\cap \bigcup _{i\leq n}E_i))\setminus (A_0\cap E_0) = B\setminus (A_0\cap E_0)$ , we get $\bigcup _{n\in \omega }(A_{n+1}\cap \bigcup _{i\leq n}F^E_i) \notin \mathcal {I}$ , and the proof is finished.

(3) It was proved in [Reference Filipów and Staniszewski18, Reference Šupina35].⊣

Proof. Straightforward.⊣

Proof. First we show $\mathfrak {b}(\mathcal {I},\mathcal {J},\mathcal {K})=\mathfrak {b}_1(\mathcal {I},\mathcal {J},\mathcal {K})$ . Since $\mathfrak {b}(\mathcal {I},\mathcal {J},\mathcal {K})\leq \mathfrak {b}_1(\mathcal {I},\mathcal {J},\mathcal {K})$ is obvious, we only show $\mathfrak {b}(\mathcal {I},\mathcal {J},\mathcal {K})\geq \mathfrak {b}_1(\mathcal {I},\mathcal {J},\mathcal {K})$ . Let $\mathcal {E}\subseteq \widehat {\mathcal {P}_{\mathcal {K}}}$ be such that for every $\langle A_n\rangle \in \mathcal {P}_{\mathcal {J}}$ there is $\langle E_n\rangle \in \mathcal {E}$ with $\bigcup _{n\in \omega }(A_n\cap \bigcup _{i\leq n}E_i)\notin \mathcal {I}$ . For every $E=\langle E_n\rangle \in \mathcal {E}$ we define $F^E_n = E_n\cup \{e_n\}$ , where $e_1,e_2,\dots $ is an enumeration of $B=\omega \setminus \bigcup _{n\in \omega } E_n$ (if B is finite, we put $F^E_n = E_n$ for $n>|B|$ ). Let $\mathcal {F}=\{\langle F^E_n\rangle :E\in \mathcal {E}\}$ , and notice that $\mathcal {F}\subseteq \mathcal {P}_{\mathcal {K}}$ and $|\mathcal {F}|\leq |\mathcal {E}|$ . Let $\langle A_n\rangle \in \mathcal {P}_{\mathcal {J}}$ . There is $E=\langle E_n\rangle \in \mathcal {E}$ with $\bigcup _{n\in \omega }(A_n\cap \bigcup _{i\leq n}E_i)\notin \mathcal {I}$ . Then $\bigcup _{n\in \omega }(A_n\cap \bigcup _{i\leq n}E_i) \subseteq \bigcup _{n\in \omega }(A_n\cap \bigcup _{i\leq n}F^E_i)$ , so $\bigcup _{n\in \omega }(A_n\cap \bigcup _{i\leq n}F^E_i)\notin \mathcal {I}$ . Thus $\mathfrak {b}_1(\mathcal {I},\mathcal {J},\mathcal {K})\leq \mathfrak {b}(\mathcal {I},\mathcal {J},\mathcal {K})$ .

Now we show $\mathfrak {b}(\mathcal {I},\mathcal {J},\mathcal {K})\leq \mathfrak {b}_2(\mathcal {I},\mathcal {J},\mathcal {K})$ . Let $\mathcal {G}\subset \mathcal {K}^{\omega }$ such that for every $\langle A_n\rangle \in \mathcal {P}_{\mathcal {J}}$ there is $\langle G_n\rangle \in \mathcal {G}$ with $\bigcup _{n\in \omega }(A_n\cap G_n)\notin \mathcal {I}$ . For every $G=\langle G_n\rangle \in \mathcal {G}$ we define $E^G_0=G_0$ and $E^G_{n}=G_n\setminus \bigcup _{i<n}G_i$ for $n\geq 1$ . Let $\mathcal {E} = \{\langle E^G_n\rangle : G\in \mathcal {G}\}$ , and notice that $\mathcal {E}\subseteq \widehat {\mathcal {P}_{\mathcal {K}}}$ and $|\mathcal {E}|\leq |\mathcal {G}|$ . Let $\langle A_n\rangle \in \mathcal {P}_{\mathcal {J}}$ . Then there is $G=\langle G_n\rangle \in \mathcal {G}$ with $\bigcup _{n\in \omega }(A_n\cap G_n)\notin \mathcal {I}$ . But $\bigcup _{n\in \omega }(A_n\cap G_n)= \bigcup _{n\in \omega }(A_n\cap \bigcup _{i\leq n} E^G_i)$ , so $\mathfrak {b}(\mathcal {I},\mathcal {J},\mathcal {K})\leq |\mathcal {G}|$ and the proof of this case is finished.

Now we show $\mathfrak {b}(\mathcal {I},\mathcal {J},\mathcal {K})\geq \mathfrak {b}_2(\mathcal {I},\mathcal {J},\mathcal {K})$ . Let $\mathcal {E}\subset \widehat {\mathcal {P}_{\mathcal {K}}}$ such that for every $\langle A_n\rangle \in \mathcal {P}_{\mathcal {J}}$ there is $\langle E_n\rangle \in \mathcal {E}$ with $\bigcup _{n\in \omega }(A_n\cap \bigcup _{i\leq n}E_i)\notin \mathcal {I}$ . For every $E=\langle E_n\rangle \in \mathcal {E}$ we define $G^E_{n}=\bigcup _{i\leq n}E_i$ for $n\in \omega $ . Let $\mathcal {G}= \{\langle G^E_n\rangle : E\in \mathcal {E}\}$ , and notice that $\mathcal {G}\subseteq \mathcal {K}^{\omega }$ and $|\mathcal {G}|\leq |\mathcal {E}|$ . Let $\langle A_n\rangle \in \mathcal {P}_{\mathcal {J}}$ . Then there is $E=\langle E_n\rangle \in \mathcal {E}$ with $\bigcup _{n\in \omega }(A_n\cap \bigcup _{i\leq n} E_i)\notin \mathcal {I}$ . But $\bigcup _{n\in \omega }(A_n\cap \bigcup _{i\leq n} E_i) = \bigcup _{n\in \omega }(A_n\cap G_n)$ , so $\mathfrak {b}_2(\mathcal {I},\mathcal {J},\mathcal {K})\leq |\mathcal {E}|$ and the proof of this case is finished.

Finally, we show $\mathfrak {b}(\mathcal {I},\mathcal {J},\mathcal {K})=\mathfrak {b}_3(\mathcal {I},\mathcal {J},\mathcal {K})$ . Actually, it suffices to observe (using the fact that $\langle A_n\rangle $ is a partition of $\omega $ ) that

$$ \begin{align*} \begin{aligned} x\in\bigcup_{n\in\omega}\left(A_n\cap \bigcup_{i\leq n}E_i\right) &\iff (\exists n\in\omega)(\exists k\leq n)(x\in A_n \land x\in E_k)\\[-3pt] & \iff (\exists k\in\omega)(\exists n\geq k)(x\in E_k \land x\in A_n)\\[-3pt] &\iff (\exists k\in\omega)(\forall n<k)(x\in E_k \land x\notin A_n)\\[-3pt] &\iff x\in\bigcup_{k\in\omega}\left(E_k\setminus \bigcup_{n<k}A_n\right).\\[-2pc] \end{aligned} \end{align*} $$

⊣

Proof. (1) First we show $\mathfrak {b}(>_{\mathcal {I}}\cap (\mathcal {D}_{\mathcal {K}}\times \mathcal {D}_{\mathcal {J}}))\leq \mathfrak {b}(\mathcal {I},\mathcal {J},\mathcal {K})$ . Let $\mathcal {E}\subseteq \widehat {\mathcal {P}_{\mathcal {K}}}$ be such that for every $\langle A_n:n\in \omega \rangle \in \mathcal {P}_{\mathcal {J}}$ there is $\langle E_n\rangle \in \mathcal {E}$ with $\bigcup _{n\in \omega }(A_n\cap \bigcup _{i\leq n}E_i)\notin \mathcal {I}$ .

For every $E=\langle E_n\rangle \in \mathcal {E}$ let $f_E\in \omega ^{\omega }$ be such that $f_E^{-1}[\{n\}] = E_n$ for every $n\in \omega $ . Let $\mathcal {F} = \{f_E:E\in \mathcal {E}\}$ . Then $\mathcal {F}\subseteq \mathcal {D}_{\mathcal {K}}$ and $|\mathcal {F}|=|\mathcal {E}|$ .

Let $g\in \mathcal {D}_{\mathcal {J}}$ and $A_n=g^{-1}[\{n\}]$ for every $n\in \omega $ . Since $\langle A_n:n\in \omega \rangle \in \mathcal {P}_{\mathcal {J}}$ , there is $E=\langle E_n\rangle \in \mathcal {E}$ with $B = \bigcup _{n\in \omega }(A_n\cap \bigcup _{i\leq n}E_i)\notin \mathcal {I}$ .

Once we show $B \subseteq \{k\in \omega : f_E(k)\leq g(k)\}$ , the proof will be finished in this case. Let $k\in B$ . Then $k\in A_n\cap E_i$ for some $n\in \omega $ and $i\leq n$ . Hence $g(k) = n \geq i = f_E(k)$ .

Now we show $\mathfrak {b}(>_{\mathcal {I}}\cap (\mathcal {D}_{\mathcal {K}}\times \mathcal {D}_{\mathcal {J}}))\geq \mathfrak {b}(\mathcal {I},\mathcal {J},\mathcal {K}) $ . Let $\mathcal {F}\subseteq \mathcal {D}_{\mathcal {K}}$ be such that there is no $g\in \mathcal {D}_{\mathcal {J}}$ such that for all $f\in \mathcal {F}$ we have $f>_{\mathcal {I}} g$ .

For every $f\in \mathcal {F}$ and $n\in \omega $ we define $E^f_n = f^{-1}[\{n\}]$ , and we notice that $\mathcal {E}=\{ \langle E^f_n\rangle : f\in \mathcal {F}\}\subseteq \widehat {\mathcal {P}_{\mathcal {K}}}$ and $|\mathcal {E}|=|\mathcal {F}|$ .

Let $\langle A_n:n\in \omega \rangle \in \mathcal {P}_{\mathcal {J}}$ and $g\in \omega ^{\omega }$ be such that $g^{-1}[\{n\}] = A_n$ . Since $g\in \mathcal {D}_{\mathcal {J}}$ , there is $f\in \mathcal {F}$ with $\neg (f>_{\mathcal {I}} g)$ . Thus, $B = \{k\in \omega : f(k)\leq g(k)\}\notin \mathcal {I}$ .

Once we show $B \subseteq \bigcup _{n\in \omega }(A_n\cap \bigcup _{i\leq n}E^f_i)$ , the proof will be finished in this case.

Let $k\in B$ . Since $k\in A_{g(k)}\cap E^f_{f(k)}$ and $f(k)\leq g(k)$ , $k\in A_{g(k)}\cap \bigcup _{i\leq g(k)}E^f_i$ .

(2) First we show $\mathfrak {b}(\geq _{\mathcal {I}}\cap (\mathcal {D}_{\mathcal {K}}\times \mathcal {D}_{\mathcal {J}})) \leq \mathfrak {b}_s(\mathcal {I},\mathcal {J},\mathcal {K})$ . Let $\mathcal {E}\subseteq \widehat {\mathcal {P}_{\mathcal {K}}}$ be such that for every $\langle A_n:n\in \omega \rangle \in \mathcal {P}_{\mathcal {J}}$ there is $\langle E_n\rangle \in \mathcal {E}$ with $\bigcup _{n\in \omega }(A_{n+1}\cap \bigcup _{i\leq n}E_i)\notin \mathcal {I}$ .

For every $E=\langle E_n\rangle \in \mathcal {E}$ let $f_E\in \omega ^{\omega }$ be such that $f_E^{-1}[\{n\}] = E_n$ for every $n\in \omega $ . Let $\mathcal {F} = \{f_E:E\in \mathcal {E}\}$ . Then $\mathcal {F}\subseteq \mathcal {D}_{\mathcal {K}}$ and $|\mathcal {F}|=|\mathcal {E}|$ .

Let $g\in \mathcal {D}_{\mathcal {J}}$ and $A_n=g^{-1}[\{n\}]$ for every $n\in \omega $ . Since $\langle A_n:n\in \omega \rangle \in \mathcal {P}_{\mathcal {J}}$ , there is $E=\langle E_n\rangle \in \mathcal {E}$ with $B = \bigcup _{n\in \omega }(A_{n+1}\cap \bigcup _{i\leq n}E_i)\notin \mathcal {I}$ .

Once we show $B \subseteq \{k\in \omega : f_E(k)<g(k)\}$ , the proof will be finished in this case. Let $k\in B$ . Then $k\in A_{n+1}\cap E_i$ for some $n\in \omega $ and $i\leq n$ . Hence $g(k) = n+1\geq i+1>i = f_E(k)$ .

Now we show $\mathfrak {b}(\geq _{\mathcal {I}}\cap (\mathcal {D}_{\mathcal {K}}\times \mathcal {D}_{\mathcal {J}})) \geq \mathfrak {b}_s(\mathcal {I},\mathcal {J},\mathcal {K})$ . Let $\mathcal {F}\subseteq \mathcal {D}_{\mathcal {K}}$ be such that for every $g\in \mathcal {D}_{\mathcal {J}}$ there is $f\in \mathcal {F}$ with $\neg (f\geq _I g)$ . For every $f\in \mathcal {F}$ and $n\in \omega $ we define $E^f_n=f^{-1}[\{n\}]$ , and we notice that $\mathcal {E}=\{\langle E^f_n\rangle :f\in \mathcal {F}\}\subseteq \widehat {\mathcal {P}_{\mathcal {K}}}$ and $|\mathcal {E}|\leq |\mathcal {F}|$ .

Take any $\langle A_k\rangle \in \mathcal {P}_{\mathcal {J}}$ and define $g\in \omega ^{\omega }$ by $g(n)=k$ for every $n\in A_k$ , $k\in \omega $ . Since $g\in \mathcal {D}_{\mathcal {J}}$ , there is $f\in \mathcal {F}$ with $B = \{n\in \omega :f(n)<g(n)\}\notin \mathcal {I}$ . Once we show that $B\subseteq \bigcup _{k\in \omega }(A_{k+1}\cap \bigcup _{i\leq k}E^f_i)$ , the proof will be finished in this case.

Let $n\in \omega $ be such that $f(n)<g(n)$ and let $k\in \omega $ be such that $n\in A_k$ . Notice that $k\geq 1$ . (Indeed, if $k=0$ then $n\in A_0$ , so $g(n)=0$ , and consequently $f(n)<0$ , a contradiction.) Since $f(n)<g(n)=k$ , $n\in \bigcup _{i<k}E^f_i$ . Let $l=k-1$ . Then $n\in A_{l+1}\cap \bigcup _{i\leq l}E^f_i$ .⊣

Proof. (1) Follows from Theorem 3.10 and Proposition 3.7(2).

(2) Follows from Theorem 3.10 and Proposition 3.7(3).⊣

Proof. Follows from Theorem 3.10 and Proposition 3.4(4).⊣

Proof. (1) Obvious.

(2) By (1) it is enough to show that $\mathfrak {b}(\mathcal {I},\mathcal {J},\mathcal {K})\leq 1$ . Let $E\in \mathcal {K}\setminus \mathcal {I}$ and $\mathcal {E} = \{\langle E,\emptyset ,\emptyset ,\dots \rangle \}$ . Since $E\in \mathcal {K}$ , $\mathcal {E}\in \widehat {\mathcal {P}_{\mathcal {K}}}$ . Let $\langle A_n\rangle \in \mathcal {P}_{\mathcal {J}}$ . Then $ \bigcup _{n\in \omega }(A_n\cap \bigcup _{i\leq n}E_i) = \bigcup _{n\in \omega }(A_n\cap E) = (\bigcup _{n\in \omega }A_n)\cap E = \omega \cap E=E\notin \mathcal {I}$ . Consequently, $\mathfrak {b}(\mathcal {I},\mathcal {J},\mathcal {K})\leq |\mathcal {E}|=1$ .

(3) Follows from (2) and Proposition 3.7(2).

(4) It follows from [Reference Šupina35, Theorem 4.9(5)].

(5) It follows from [Reference Šupina35, Theorem 4.9(2)].

(6) If $\mathcal {I}\not \perp \mathcal {J}$ , it follows from (5). If $\mathcal {I}\perp \mathcal {J}$ , then $\mathcal {K}\not \subseteq \mathcal {I}$ in this case, so (2) finishes the proof.

(7) It follows from [Reference Šupina35, Theorem 4.9(1)].

(8) Follows from (7) and Proposition 3.7(2).⊣

Proof. It follows from Proposition 3.7(2).⊣

Proof. (1) It follows from Proposition 3.13(2).

(2) It follows from [Reference Filipów and Staniszewski18, Proposition 4.5].

(3a) It follows from [Reference Filipów and Staniszewski18, Proposition 4.1].

(3b) It follows from [Reference Filipów and Staniszewski18, Corollary 4.4].

(4a) It follows from [Reference Šupina35, Theorem 4.9(1)]

(4b) It follows from [Reference Kwela22, Corollary 2.10].

(5) It follows from [Reference Filipów and Staniszewski18, Proposition 4.1].

(6) By (2) and Proposition 3.8 we have $\mathfrak {b}=\mathfrak {b}(\mathrm {Fin},\mathrm {Fin},\mathrm {Fin})\leq \mathfrak {b}(\mathcal {I},\mathrm {Fin},\mathrm {Fin})$ . Now we show $\mathfrak {b}(\mathcal {I},\mathrm {Fin},\mathrm {Fin})\leq \mathfrak {d}$ . Let $\mathcal {F}\subset \omega ^{\omega }$ be a dominating family in $(\omega ^{\omega },\leq ^*)$ . Without loss of generality we can assume that every $f\in \mathcal {F}$ is increasing. We define $E^f_n = [f(n-1), f(n))\cap \omega $ for every $f\in \mathcal {F}$ , $n\in \omega $ (here we put $f(-1) = 0$ ).

Notice that $\langle E^f_n:n\in \omega \rangle \in \widehat {\mathcal {P}}_{\mathrm {Fin}}$ for every $f\in \mathcal {F}$ . Let $\langle A_n:n\in \omega \rangle \in \mathcal {P}_{\mathrm {Fin}}$ . Let $g\in \omega ^{\omega }$ be given by $g(n) = \max (\bigcup _{i\leq n} A_i)$ . Note that if $g(n)<f(n)$ , then $A_n \subseteq \bigcup _{i\leq n} A_i \subseteq [0,f(n)) = \bigcup _{i\leq n}E^f_i$ .

Since $\mathcal {F}$ is dominating, there is $f\in \mathcal {F}$ with $g<^* f$ . Let $N\in \omega $ be such that $g(n)<f(n)$ for all $n\geq N$ .

Then $\bigcup _{n\in \omega }(A_n\cap \bigcup _{i\leq n}E^f_i) \supseteq \bigcup _{n\geq N}(A_n\cap \bigcup _{i\leq n}E^f_i) = \bigcup _{n\geq N}(\bigcup _{i\leq n}E^f_i) =\omega \notin \mathcal {I} $ , and the proof is finished.

(7) It follows from Proposition 3.8.

(8) It follows from Theorem 3.10(1, 2) and Proposition 3.4(4).

(9) The second inequality is obvious and the first inequality follows from Theorem 3.10(1) and Proposition 3.4(2).⊣

Proof. By Corollary 4.1 and Theorem 4.2(2), $\mathfrak {b}(\mathrm {Fin},\mathcal {I},\mathrm {Fin})=\mathfrak {b}_s(\mathrm {Fin},\mathcal {I},\mathrm {Fin})=\mathfrak {b}$ , which is a regular cardinal. The remaining cases follow from Corollary 3.12.⊣

Proof. By Proposition 3.4(2) and Theorem 3.10 we have $\mathfrak {b}(\mathcal {I},\,\mathcal {I},\,\mathcal {I})\leq \operatorname {\mathrm {cf}}(\mathfrak {b}(\mathcal {I},\,\mathcal {I},\mathrm {Fin}))$ , $\operatorname {\mathrm {cf}}(\mathfrak {b}(\mathcal {I},\,\mathcal {I},\mathrm {Fin}))=\operatorname {\mathrm {cf}}(\mathfrak {b}_s(\mathcal {I},\,\mathcal {I},\mathrm {Fin}))$ follows from Corollary 4.1 and $\aleph _1\leq \mathfrak {b}(\mathcal {I},\,\mathcal {I},\,\mathcal {I})$ follows from Theorem 4.2(5).⊣

Proof. Taking into account the previous results and the fact that every P-ideal is a weak P-ideal, it is enough to show $\mathfrak {b}(\mathcal {I},\mathrm {Fin},\mathrm {Fin}) \geq \mathfrak {b}(\mathcal {I},\,\mathcal {I},\mathrm {Fin})$ and $\mathfrak {b}(\mathcal {I},\,\mathcal {I},\,\mathcal {I}) \geq \mathfrak {b}(\mathcal {I},\,\mathcal {I},\mathrm {Fin})$ .

First we show $\mathfrak {b}(\mathcal {I},\mathrm {Fin},\mathrm {Fin}) \geq \mathfrak {b}(\mathcal {I},\,\mathcal {I},\mathrm {Fin})$ .

Let $\mathcal {E}\in \mathcal {P}_{\mathrm {Fin}}$ be such that for every $\langle A_n\rangle \in \mathcal {P}_{\mathrm {Fin}}$ there is $\langle E_n\rangle \in \mathcal {E}$ with $\bigcup _{n\in \omega }(A_n\cap \bigcup _{i\leq n}E_i)\notin \mathcal {I}$ .

Let $\langle B_n\rangle \in \mathcal {P}_{\mathcal {I}}$ . Since $\mathcal {I}$ is a P-ideal, there is $B\in \mathcal {I}$ with $B_n\setminus B\in \mathrm {Fin}$ for every $n\in \omega $ . Let $B=\{b_n:n\in \omega \}$ and set $A_n = (B_n\setminus B)\cup \{b_n\}$ for every $n\in \omega $ . Then $\langle A_n\rangle \in \mathcal {P}_{\mathrm {Fin}}$ , so there is $\langle E_n\rangle \in \mathcal {E}$ with $C = \bigcup _{n\in \omega }(A_n\cap \bigcup _{i\leq n}E_i)\notin \mathcal {I}$ .

Once we show $C\setminus B \subseteq \bigcup _{n\in \omega }(B_n\cap \bigcup _{i\leq n}E_i)$ , the proof of this case will be finished.

Let $k\in C\setminus B$ . Then there is $n\in \omega $ with $k\in [A_n\cap \bigcup _{i\leq n}E_i]\setminus B = [((B_n\setminus B)\cup \{b_n\})\cap \bigcup _{i\leq n}E_i]\setminus B = (B_n\setminus B)\cap \bigcup _{i\leq n}E_i \subseteq B_n\cap \bigcup _{i\leq n}E_i $ .

Now we show $\mathfrak {b}(\mathcal {I},\mathcal {I},\mathcal {I}) \geq \mathfrak {b}(\mathcal {I},\mathcal {I},\mathrm {Fin})$ .

Let $\mathcal {E}\in \mathcal {P}_{\mathcal {I}}$ be such that for every $\langle A_n\rangle \in \mathcal {P}_{\mathcal {I}}$ there is $\langle E_n\rangle \in \mathcal {E}$ with $\bigcup _{n\in \omega }(A_n\cap \bigcup _{i\leq n}E_i)\notin \mathcal {I}$ .

Since $\mathcal {I}$ is a P-ideal, for every $E=\langle E_n\rangle \in \mathcal {E}$ there is $B_E\in \mathcal {I}$ such that $D_n^E = E_n\setminus B_E\in \mathrm {Fin}$ for every $n\in \omega $ . Note that $D_n^E\cap D_k^E=\emptyset $ for every $E\in \mathcal {E}$ and $n\neq k$ .

Let $\langle A_n\rangle \in \mathcal {P}_{\mathcal {I}}$ . Then there is $E=\langle E_n\rangle \in \mathcal {E}$ with $C = \bigcup _{n\in \omega }(A_n\cap \bigcup _{i\leq n}E_i)\notin \mathcal {I}$ .

Once we show $C\setminus B_E \subseteq \bigcup _{n\in \omega }(A_n\cap \bigcup _{i\leq n}D^E_i)$ , the proof of this case will be finished.

Let $k\in C\setminus B_E$ . Then there is $n\in \omega $ with $k \in [\bigcup _{n\in \omega }(A_n\cap \bigcup _{i\leq n}E_i)]\setminus B_E = \bigcup _{n\in \omega }(A_n\cap \bigcup _{i\leq n}(E_i\setminus B_E)) = \bigcup _{n\in \omega }(A_n\cap \bigcup _{i\leq n}D_i^E) $ .⊣

Proof. (1) By Theorem 3.13(4), $\mathfrak {b}_s(\mathrm {Fin},\mathcal {I},\mathcal {I})\geq \aleph _0$ . The revers inequality follows from [Reference Šupina35, Theorem 4.9(3)].

(2) The first inequality follows from [Reference Šupina35, Theorem 4.9(4)], whereas the second one follows from Theorem 4.2(2) and the observation that $\mathfrak {b}_s(\mathrm {Fin},\mathcal {I},\mathcal {I})\leq \mathfrak {b}_s(\mathrm {Fin},\mathcal {I},\mathrm {Fin})=\mathfrak {b}(\mathrm {Fin},\mathcal {I},\mathrm {Fin})=\mathfrak {b}$ .⊣

Proof. If $\operatorname {\mathrm {add} ^{*}}(\mathcal {I})=\mathfrak {c}^+$ , we are done. Assume that $\operatorname {\mathrm {add} ^{*}}(\mathcal {I})\leq \mathfrak {c}$ . Then there is a family $\mathcal {F}=\{F_\alpha :\alpha <\operatorname {\mathrm {add} ^{*}}(\mathcal {I})\}\subseteq \mathcal {I}$ such that for every $A\in \mathcal {I}$ there is $\alpha $ with $|F_\alpha \setminus A|=\aleph _0$ . For every $\alpha $ , we define $E^{\alpha }_0=F_\alpha $ and $E^{\alpha }_n=\emptyset $ for $n\geq 1$ . Then $\langle E^{\alpha }_n:n\in \omega \rangle \in \widehat {\mathcal {P}}_{\mathcal {I}}$ .

Take any $\langle A_n\rangle \in \mathcal {P}_{\mathcal {I}}$ . If we show that $\bigcup _{n\in \omega }(A_{n+1}\cap \bigcup _{i\leq n} E^{\alpha }_i)\notin \mathrm {Fin}$ for some $\alpha $ , the proof will be finished.

Since $A_0\in \mathcal {I}$ , there is $\alpha $ with $|F_\alpha \setminus A_0|=\aleph _0$ . On the other hand, it is easy to see that $F_\alpha \setminus A_0 = E^{\alpha }_0\setminus A_0 \subseteq \bigcup _{n\in \omega }(A_{n+1}\cap \bigcup _{i\leq n} E^{\alpha }_i)$ , so we are done.⊣

Proof. If $\mathcal {I}$ is not a P-ideal, then $\mathfrak {b}_s(\mathrm {Fin},\mathcal {I},\mathcal {I})=\aleph _0$ (by Lemma 4.6(1)) and $\operatorname {\mathrm {add} ^{*}}(\mathcal {I})=\aleph _0$ . Since $\mathfrak {b}>\aleph _0$ , we obtain $\mathfrak {b}_s(\mathrm {Fin},\mathcal {I},\mathcal {I})=\min \{\mathfrak {b},\operatorname {\mathrm {add} ^{*}}(\mathcal {I})\}$ .

Assume that $\mathcal {I}$ is a P-ideal. The inequality “ $\leq $ ” follows from Lemma 4.6(2) and Lemma 4.7. Below we show the inequality “ $\geq $ .”

Take $\kappa <\min \{\mathfrak {b},\operatorname {\mathrm {add} ^{*}}(\mathcal {I})\}$ . We will show that $\kappa <\mathfrak {b}_s(\mathrm {Fin},\mathcal {I},\mathcal {I})$ .

Let $\langle E^{\alpha }_n:n\in \omega \rangle \in \widehat {\mathcal {P}}_{\mathcal {I}}$ for $\alpha <\kappa $ . Since $\kappa \cdot \aleph _0 <\operatorname {\mathrm {add} ^{*}}(\mathcal {I})$ , there is $A\in \mathcal {I}$ such that $|E^{\alpha }_n\setminus A|<\aleph _0$ for every $\alpha <\kappa $ and $n\in \omega $ .

Let $B = \omega \setminus A$ . Since $\langle E^{\alpha }_n\cap B\rangle \in \widehat {\mathcal {P}}_{\mathrm {Fin}(B)}$ for every $\alpha <\kappa $ and $\kappa <\mathfrak {b}=\mathfrak {b}(\mathrm {Fin}(B),\mathrm {Fin}(B),\mathrm {Fin}(B))$ , there is $\langle B_n\rangle \in \mathcal {P}_{\mathrm {Fin}(B)}$ such that

$$ \begin{align*}\bigcup_{n\in \omega}\left(B_{n}\cap \bigcup_{i\leq n} \left(E^{\alpha}_i\cap B\right)\right)\in \mathrm{Fin}(B)\end{align*} $$

for every $\alpha <\kappa $ .

Let $A_0=A$ and $A_n=B_{n-1}$ for $n\geq 1$ . Then $\langle A_n\rangle \in \mathcal {P}_{\mathcal {I}}$ and for every $\alpha <\kappa $ we have

$$ \begin{align*}\bigcup_{n\in \omega}\left(A_{n+1}\cap \bigcup_{i\leq n} E^{\alpha}_i\right) = \bigcup_{n\in \omega}\left(B_{n}\cap \bigcup_{i\leq n} E^{\alpha}_i\right) = \bigcup_{n\in \omega}\left(B_{n}\cap \bigcup_{i\leq n} (E^{\alpha}_i\cap B)\right)\in \mathrm{Fin}.\end{align*} $$

⊣

Proof. We will only prove the first item. The remaining items can be proved in a similar way.

First we prove that $\mathfrak {b}(\mathcal {I}\oplus \mathcal {J},\mathcal {I}\oplus \mathcal {J},\mathcal {I}\oplus \mathcal {J})\leq \min \{\mathfrak {b}(\mathcal {I},\mathcal {I},\mathcal {I}),\mathfrak {b}(\mathcal {J},\mathcal {J},\mathcal {J})\}$ . We will prove that $\mathfrak {b}(\mathcal {I}\oplus \mathcal {J},\mathcal {I}\oplus \mathcal {J},\mathcal {I}\oplus \mathcal {J})\leq \mathfrak {b}(\mathcal {I},\mathcal {I},\mathcal {I})$ (the proof of $\mathfrak {b}(\mathcal {I}\oplus \mathcal {J},\mathcal {I}\oplus \mathcal {J},\mathcal {I}\oplus \mathcal {J})\leq \mathfrak {b}(\mathcal {J},\mathcal {J},\mathcal {J})$ goes in a similar way).

There is a family $\{\langle E^{\alpha }_n:n\in \omega \rangle : \alpha <\mathfrak {b}(\mathcal {I},\mathcal {I},\mathcal {I})\}\subseteq \widehat {\mathcal {P}}_{\mathcal {I}}$ with the property that for every $\langle A_n:n\in \omega \rangle \in \mathcal {P}_{\mathcal {I}}$ there is $\alpha <\mathfrak {b}(\mathcal {I},\mathcal {I},\mathcal {I})$ such that $\bigcup _{n\in \omega }(A_n\cap \bigcup _{i\leq n}E^{\alpha }_i)\notin \mathcal {I}$ .

Consider the family $\{\langle E^{\alpha }_n\times \{0\}: n\in \omega \rangle :\alpha <\mathfrak {b}(\mathcal {I},\mathcal {I},\mathcal {I})\}\subseteq \widehat {\mathcal {P}}_{\mathcal {I}\oplus \mathcal {J}}$ . Let $\langle B_n\oplus C_n:n\in \omega \rangle \in \mathcal {P}_{\mathcal {I}\oplus \mathcal {J}}$ . Since $\langle B_n:n\in \omega \rangle \in \mathcal {P}_{\mathcal {I}}$ , there is $\alpha <\mathfrak {b}(\mathcal {I},\mathcal {I},\mathcal {I})$ such that $C = \bigcup _{n\in \omega }(B_n\cap \bigcup _{i\leq n}E^{\alpha }_i)\notin \mathcal {I}$ . Then $C\times \{0\}\notin \mathcal {I}\oplus \mathcal {J}$ , and $ C \times \{0\}\subseteq \bigcup _{n\in \omega }(B_n\oplus C_n\cap \bigcup _{i\leq n}(E^{\alpha }_i\times \{0\})) $ . That finishes this part of the proof.

Now we prove $\mathfrak {b}(\mathcal {I}\oplus \mathcal {J},\mathcal {I}\oplus \mathcal {J},\mathcal {I}\oplus \mathcal {J})\geq \min \{\mathfrak {b}(\mathcal {I},\mathcal {I},\mathcal {I}),\mathfrak {b}(\mathcal {J},\mathcal {J},\mathcal {J})\}$ . Let $\kappa <\min \{\mathfrak {b}(\mathcal {I},\mathcal {I},\mathcal {I}),\mathfrak {b}(\mathcal {J},\mathcal {J},\mathcal {J})\}$ and $\{\langle E^{\alpha }_n\oplus F^{\alpha }_n: n\in \omega \rangle : \alpha <\kappa \}\subseteq \widehat {\mathcal {P}}_{\mathcal {I}\oplus \mathcal {J}}$ . Since $\kappa <\mathfrak {b}(\mathcal {I},\mathcal {I},\mathcal {I})$ and $\{\langle E^{\alpha }_n: n\in \omega \rangle :\alpha <\kappa \}\subseteq \widehat {\mathcal {P}}_{\mathcal {I}}$ , there is $\langle B_n:n\in \omega \rangle \in \mathcal {P}_{\mathcal {I}}$ such that $\bigcup _{n\in \omega }(B_n\cap \bigcup _{i\leq n}E^{\alpha }_i)\in \mathcal {I}$ for all $\alpha <\kappa $ . Similarly, there is $\langle C_n:n\in \omega \rangle \in \mathcal {P}_{\mathcal {J}}$ such that $\bigcup _{n\in \omega }(C_n\cap \bigcup _{i\leq n}F^{\alpha }_i)\in \mathcal {J}$ for all $\alpha <\kappa $ . Then $\langle B_n\oplus C_n:n\in \omega \rangle \in \mathcal {P}_{\mathcal {I}\oplus \mathcal {J}}$ and $\bigcup _{n\in \omega }((B_n\oplus C_n)\cap \bigcup _{i\leq n}(E^{\alpha }_i\oplus F^{\alpha }_i))\in \mathcal {I}\oplus \mathcal {J}$ for all $\alpha <\kappa $ . Thus, $\kappa <\mathfrak {b}(\mathcal {I}\oplus \mathcal {J},\mathcal {I}\oplus \mathcal {J},\mathcal {I}\oplus \mathcal {J})$ .⊣

Proof. By Theorem 4.2, $\mathfrak {b}(\mathcal {I},\mathcal {I},\mathcal {I})\leq \mathfrak {b}(\mathcal {I},\mathcal {I},\mathrm {Fin})$ and $\mathfrak {b}\leq \mathfrak {b}(\mathcal {I},\mathrm {Fin},\mathrm {Fin})\leq \mathfrak {b}(\mathcal {I},\mathcal {I},\mathrm {Fin})$ , so it is enough to show $\mathfrak {b}(\mathcal {I},\mathcal {I},\mathrm {Fin})\leq \mathfrak {b}$ .

Let $A\subseteq \omega $ be infinite such that for every $B\subseteq A$ , $B\in \mathcal {I} \iff B\text { is finite}$ . Let $\mathrm {Fin}(A)$ denote the family of all finite subsets of A. Then $\mathcal {I}=\mathrm {Fin}(A)\oplus (\mathcal {I}\restriction (\omega \setminus A))$ and, by Theorem 5.1(2), we have $\mathfrak {b}(\mathcal {I},\mathcal {I},\mathrm {Fin})\leq \mathfrak {b}(\mathrm {Fin}(A),\mathrm {Fin}(A),\mathrm {Fin})=\mathfrak {b}(\mathrm {Fin},\mathrm {Fin},\mathrm {Fin})=\mathfrak {b}$ .⊣

Proof. Follows from Theorems 4.5 and 5.2.⊣

Proof. By Theorem 6.1 there is a sequence $n_0<n_1<\cdots $ such that $\{k: [n_k,n_{k+1})\subseteq A\}$ is finite for every $A\in \mathcal {I}$ .

It is enough to define h such that h is increasing, $g\leq h$ and $[h(n),h(n+1))$ contains at least one interval $[n_k,n_{k+1})$ for every $n\in \omega $ . We can define h inductively in the following manner. Let $h(0)=g(0)$ . Suppose that $h(i)$ has been defined for $i\leq n$ . Let $k_0=\min \{k: h(n)\leq n_k\}$ . We put $h(n+1)=\max \{n_{k_0+1},h(n)+1,g(n+1)\}$ .⊣

Proof. By Theorem 4.2(6) we only have to show that $\mathfrak {b}(\mathcal {I},\mathrm {Fin},\mathrm {Fin})\leq \mathfrak {b}$ . We will show that $\kappa <\mathfrak {b}(\mathcal {I},\mathrm {Fin},\mathrm {Fin})$ implies $\kappa <\mathfrak {b}$ .

Let $\kappa <\mathfrak {b}(\mathcal {I},\mathrm {Fin},\mathrm {Fin})$ and $g_\alpha \in \omega ^{\omega }$ for $\alpha <\kappa $ .

By Corollary 6.2 we can assume that $g_\alpha $ are strictly increasing and

$$ \begin{align*}\bigcup_{n\in A}[g_\alpha(n),g_\alpha(n+1))\notin\mathcal{I}\end{align*} $$

for every infinite $A\subseteq \omega $ and $\alpha <\kappa $ .

For $\alpha <\kappa $ we define $A^{\alpha }_0 = [0,g_\alpha (1))$ and $A^{\alpha }_n = [g_\alpha (n),g_\alpha (n+1))$ for $n\geq 1$ .

Since $\{\langle A^{\alpha }_n:n\in \omega \rangle : \alpha <\kappa \}\subseteq \mathcal {P}_{\mathrm {Fin}}$ and $\kappa <\mathfrak {b}(\mathcal {I},\mathrm {Fin},\mathrm {Fin})$ , there is a partition $\langle B_n: n\in \omega \rangle \in \mathcal {P}_{\mathrm {Fin}}$ such that $\bigcup _{n\in \omega }(B_n\cap \bigcup _{i\leq n}A^{\alpha }_i)\in \mathcal {I}$ for every $\alpha <\kappa $ .

We define $g\in \omega ^{\omega }$ by $g(n) = \max (\bigcup _{i\leq n} B_i)$ . If we show that $g_\alpha \leq ^* g$ for every $\alpha <\kappa $ , the proof will be finished.

Let $\alpha <\kappa $ . Since $B = \bigcup _{n\in \omega }(B_n\cap \bigcup _{i\leq n}A^{\alpha }_i)\in \mathcal {I}$ , the set $A = \{n: [g_\alpha (n), g_\alpha (n+1))\subseteq B\}$ is finite.

Now we show that $g_\alpha (n)\leq g(n)$ for every $n\in \omega \setminus A$ (i.e. $g_\alpha \leq ^* g$ ). Let $n\in \omega \setminus A$ . Since $[g_\alpha (n),g_\alpha (n+1))\cap (\omega \setminus B)\neq \emptyset $ , there is $m\in [g_\alpha (n),g_\alpha (n+1))\cap (\omega \setminus B)$ . Then $g_\alpha (n)\leq m$ . On the other hand, $m\notin B = \bigcup _{n\in \omega }(B_n\cap \bigcup _{i\leq n}A^{\alpha }_i)$ and $m\in [g_\alpha (n), g_\alpha (n+1)) = A^{\alpha }_n$ . Thus $m\notin B_i$ for every $i\geq n$ , so $m\in B_i$ for some $i<n$ . Hence $g(n) = \max (\bigcup _{i\leq n} B_i) \geq m \geq g_\alpha (n)$ .⊣

Proof. Follows from Theorems 4.5 and 6.3.⊣

Proof. It was proved by Hernández and Hrušák [Reference Hernández-Hernández and Hrušák20, Theorem 2.2] that $\operatorname {\mathrm {add} ^{*}}(\mathcal {I})=\operatorname {\mathrm {add}}(\mathcal {N})$ in these cases. Since $\operatorname {\mathrm {add}}(\mathcal {N})\leq \mathfrak {b}$ (see e.g., [Reference Bartoszyński and Judah1, p. 35]), we get $\mathfrak {b}_s(\mathrm {Fin},\mathcal {I},\mathcal {I}) = \operatorname {\mathrm {add}}(\mathcal {N})$ . Moreover, since $\mathcal {I}$ is a P-ideal with the Baire property (see e.g., [Reference Farah16, Examples 1.2.3(c,d)]), Corollary 6.4 finishes the proof.⊣

Proof. Since $\mathcal {I}$ in not a P-ideal, we get $\mathfrak {b}_s(\mathrm {Fin},\mathcal {I},\mathcal {I}) = \aleph _0$ . Since $\mathcal {I}$ has the Baire property (see e.g., [Reference Hrušák21]), we can apply Theorem 6.3 to obtain $\mathfrak {b}(\mathcal {I},\mathrm {Fin},\mathrm {Fin}) = \mathfrak {b}$ . Since $\mathcal {I} = \mathrm {Fin}^n = \mathrm {Fin}\otimes \mathrm {Fin}^{n-1}$ , we can apply Theorem 5.13(1) to obtain $\mathfrak {b}(\mathcal {I},\mathcal {I},\mathcal {I}) = \mathfrak {b}(\mathrm {Fin},\mathrm {Fin},\mathrm {Fin}) = \mathfrak {b}$ . Since $\mathcal {I}$ is not a weak P-ideal, we can apply Proposition 4.2(4a) to obtain $\mathfrak {b}(\mathcal {I},\mathcal {I},\mathrm {Fin}) = \mathfrak {c}^+$ .⊣

Proof. By Theorem 4.2, we only need to show $\mathfrak {b}(\mathcal {I},\mathcal {I},\mathrm {Fin})\leq \mathfrak {b}$ .

Let $\{f_\alpha :\alpha <\mathfrak {b}\}$ be an unbounded family on $\omega ^{\omega }$ . Without loss of generality we can assume that each $f_\alpha $ is strictly increasing.

For each $n\in \omega $ and $\alpha <\mathfrak {b}$ define $F^{\alpha }_n=\{i\in \omega :i\leq f_\alpha (n)\}$ . We will show that $\{\langle F^{\alpha }_n:n\in \omega \rangle : \alpha <\mathfrak {b}\}\subseteq \mathrm {Fin}^{\omega }$ witnesses $\mathfrak {b}(\mathcal {I},\mathcal {I},\mathrm {Fin})\leq \mathfrak {b}$ .

Fix $\langle B_n:n\in \omega \rangle \in \mathcal {P}_{\mathcal {I}}$ . Let $\{\mathcal {Z}_k:k\in \omega \}$ be a family of $\mathcal {I}^*$ -universal sets $\omega $ -diagonalizing $\mathcal {I}$ . For each k, let $\mathcal {Z}_k=\{Z_k^n:n\in \omega \}$ .

For each $k,n\in \omega $ find $i(k,n)\in \omega $ such that $Z_k^{i(k,n)}\cap (B_0\cup \cdots \cup B_n)=\emptyset $ and define $g_k\in \omega ^{\omega }$ by $g_k(n)=\max Z_k^{i(k,n)}$ . Let $g\in \omega ^{\omega }$ be given by $g(i)=\max \{g_1(i),\ldots ,g_i(i)\}$ .

There is $\alpha <\mathfrak {b}$ such that $f_\alpha (i)>g(i)$ for infinitely many i. If we show that for each $k\in \omega $ there is $i\in \omega $ with $Z_k^i\subseteq \bigcup _{j\in \omega }(B_j\cap F^{\alpha }_j)$ , then, by $\omega $ -diagonalizability of $\mathcal {I}$ , it will follow that $\bigcup _{j\in \omega }(B_j\cap F^{\alpha }_j)\notin \mathcal {I}$ , and the proof will be finished.

Fix $k\in \omega $ . There is $n>k$ such that $f_\alpha (n)>g(n)\geq g_k(n)=\max Z_k^{i(k,n)}$ . Thus, $Z_k^{i(k,n)}\subseteq F^{\alpha }_n\subseteq F^{\alpha }_{n+1}\subseteq \cdots $ and $Z_k^{i(k,n)}\subseteq B_{n+1}\cup B_{n+2}\cup \cdots $ . Hence, $Z_k^{i(k,n)}\subseteq \bigcup _{j>n}(B_j\cap F^{\alpha }_j)\subseteq \bigcup _{j\in \omega }(B_j\cap F^{\alpha }_j)$ and we are done.⊣

Proof. Consider the game $G\left (\mathcal {I}\right )$ , defined by Laflamme (see [Reference Laflamme27]) as follows: Player I in his n’th move plays an element $C_{n}\in \mathcal {I}$ , and then Player II responses with any $F_{n}\in \left [\omega \right ]^{<\omega }$ such that $F_{n}\cap C_{n}=\emptyset $ . Player I wins if $\bigcup _{n\in \omega } F_{n}\in \mathcal {I}$ . Otherwise, Player II wins.

By [Reference Kwela and Staniszewski25, Theorem 5.1], $G\left (\mathcal {I}\right )$ is determined for coanalytic ideals (see also [Reference Kwela and Sabok24, Theorem 1.6]). Moreover, by [Reference Laflamme27, Theorem 2.16], Player I has a winning strategy in $G\left (\mathcal {I}\right )$ if and only if $\mathcal {I}$ is not a weak P-ideal. Thus, in our case Player II has a winning strategy. Again by [Reference Laflamme27, Theorem 2.16], this is in turn equivalent to $\mathcal {I}$ being $\omega $ -diagonalizable by $\mathcal {I}^*$ -universal sets. Using Theorem 6.7 we get $\aleph _1\leq \mathfrak {b}(\mathcal {I},\mathcal {I},\mathcal {I})\leq \mathfrak {b} = \mathfrak {b}(\mathcal {I},\mathrm {Fin},\mathrm {Fin})=\mathfrak {b}(\mathcal {I},\mathcal {I},\mathrm {Fin})$ and we are done.⊣

Proof. By [Reference Debs and Raymond15, Theorems 7.5 and 9.1], each $\bf {\Pi ^0_4}$ ideal is a weak P-ideal. Thus, the corollary follows from Theorem 6.8.⊣