Proof. To show that $\aleph _1\leq \operatorname {\mathrm {add}}'(I)$ , fix any countable $\mathcal {A}\subseteq \overline {I}$ . Then $\mathcal {A}=\{B_n: n\in \omega \}$ . Consider now the sequence $\langle B_n:n\in \omega \rangle \in \overline {I}^{\omega }$ . Clearly, for each $A\in \mathcal {A}$ there is $n\in \omega $ with $A=B_n$ , so $A\setminus B_n=\emptyset $ . Thus, $\aleph _1\leq \operatorname {\mathrm {add}}'(I)$ .

Now we show that $\operatorname {\mathrm {add}}'(I)\leq \operatorname {\mathrm {add}}(I)$ . Since I is generated by some family of compact sets, we have

$$ \begin{align*} \operatorname{\mathrm{add}}(I) & =\min\left\{|\mathcal{A}|:\ \mathcal{A}\subseteq I\land\bigcup\mathcal{A}\notin I\right\}\\ & =\min\left\{|\mathcal{A}|:\ \mathcal{A}\subseteq \overline{I}\land\bigcup\mathcal{A}\notin I\right\}\\ & =\min\left\{|\mathcal{A}|:\ \mathcal{A}\subseteq\overline{I}\land\forall_{\langle B_n\rangle\in\overline{I}^{\omega}}\, \bigcup\mathcal{A}\setminus\bigcup_n B_n\neq\emptyset\right\}\\ & =\min\left\{|\mathcal{A}|:\ \mathcal{A}\subseteq\overline{I}\land\forall_{\langle B_n\rangle\in\overline{I}^{\omega}}\, \exists_{A\in\mathcal{A}}\, A\setminus \bigcup_n B_n\neq\emptyset\right\}. \end{align*} $$

Now it is easy to see that each family $\mathcal {A}\subseteq \overline {I}$ witnessing $\operatorname {\mathrm {add}}(I)$ is also a witness for $\operatorname {\mathrm {add}}'(I)$ . Hence, $\operatorname {\mathrm {add}}'(I)\leq \operatorname {\mathrm {add}}(I)$ .

Proof. (a): The inequality $\operatorname {\mathrm {add}}'(\mathcal {M})\leq \operatorname {\mathrm {add}}(\mathcal {M})$ follows from Proposition 3.2, while $\operatorname {\mathrm {add}}'(\mathcal {M})\geq \operatorname {\mathrm {add}}(\mathcal {M})$ is [Reference Kankaanpää21, Lemma 3.5].

(b): Since $\operatorname {\mathrm {add}}'(I)$ depends only on compact members of I, $\operatorname {\mathrm {add}}'(\mathcal {N})=\operatorname {\mathrm {add}}'(\sigma \overline {\mathcal {N}})$ . From Proposition 3.2 and [Reference Bartoszyński and Shelah3, Theorem 3.1] we get $\operatorname {\mathrm {add}}'(\sigma \overline {\mathcal {N}})\leq \operatorname {\mathrm {add}}(\sigma \overline {\mathcal {N}})=\operatorname {\mathrm {add}}(\mathcal {M})$ , while $\mathfrak {p}\leq \operatorname {\mathrm {add}}'(\mathcal {N})$ is shown in [Reference Fremlin19, Theorem 22G].

(c): It follows from Proposition 3.2 and the fact that $\operatorname {\mathrm {add}}([2^{\omega }]^{\leq \omega })=\omega _1$ .

Proof. In the case of $[2^{\omega }]^{<\omega }$ , it suffices to observe that each $f_s$ is a bijection (so $|A|=|f_s[A]|$ ). Since each $f_s$ is also a homeomorphism, $\mathcal {M}$ is invariant over basic sets. Finally, since for each measurable $A\subseteq 2^{\omega }$ the measure of $f_s[A]$ is equal to the measure of A multiplied by $\frac {1}{2^{\operatorname {\mathrm {lh}}(s)}}$ , $\mathcal {N}$ , and $\sigma \overline {\mathcal {N}}$ are invariant over basic sets.

Proof. The case of analytic ideals will follow from the general case. Indeed, each analytic ideal with topological representation $\mathcal {J}_{I}$ is $\mathbf {{\Pi ^0_3}}$ (by [Reference Kwela and Sabok27, Theorem 1.4]), so using [Reference Filipów and Kwela17, Corollary 6.9] we get $\mathfrak {b}(\mathcal {J}_{I})\leq \mathfrak {b}$ .

By [Reference Kwela and Sabok27, Proposition 2.1], without loss of generality we may assume that $\mathcal {J}_I$ is an ideal on $D=\{x\in 2^{\omega }: \exists _{k\in \omega }\, \forall _{n\geq k}\, x(n)=0\}$ . Note that $f_s[D]=D\cap V_s$ , where $V_s$ and $f_s$ are as in Definition 3.4.

We start with $\mathfrak {b}(\mathcal {J}_{I})\geq \min \{\operatorname {\mathrm {add}}'(I),\mathfrak {b}\}$ . Fix any $\kappa <\min \{\operatorname {\mathrm {add}}'(I),\mathfrak {b}\}$ . We need to show that $\kappa <\mathfrak {b}(\mathcal {J}_{I})$ . Let $\{\langle E^\alpha _n:n\in \omega \rangle : \alpha <\kappa \}\subseteq \widehat {\mathcal {P}}_{\mathcal {J}_I}$ be arbitrary. Since $\omega \cdot \kappa <\operatorname {\mathrm {add}}'(I)$ (by Proposition 3.2), there is $\langle B_n:n\in \omega \rangle \in \overline {I}^{\omega }$ such that for each $n\in \omega $ and $\alpha <\kappa $ there exists $f_\alpha (n)\in \omega $ with $\operatorname {\mathrm {cl}}_{2^{\omega }}(E^\alpha _n)\subseteq B_{f_\alpha (n)}$ .

Since $\kappa <\mathfrak {b}$ , there exists $g\in \omega ^{\omega }$ such that $f_\alpha \leq ^\star g$ , for all $\alpha <\kappa $ . Without loss of generality, we may assume that g is strictly increasing.

Let $\{q_n:\ n\in \omega \}$ be an enumeration of D and define $C_0=\bigcup _{i\leq g(1)}(B_i\cap D)\cup \{q_0\}$ and $C_{n}=(\bigcup _{i\leq g(n+1)}(B_{i}\cap D)\cup \{q_{n}\})\setminus \bigcup _{i<n}C_i$ . Then $\langle C_n: n\in \omega \rangle \in \mathcal {P}_{\mathcal {J}_I}$ , since

$$ \begin{align*}\operatorname{\mathrm{cl}}_{2^{\omega}}(C_n)\subseteq \operatorname{\mathrm{cl}}_{2^{\omega}}\left(\bigcup_{i\leq g(n+1)}B_i\cup\{q_n\}\right)=\bigcup_{i\leq g(n+1)}B_i\cup\{q_n\}\in I.\end{align*} $$

Fix now $\alpha <\kappa $ . We need to show that $\bigcup _{n\in \omega }\left (C_n\cap \bigcup _{i\leq n}E^\alpha _i\right )\in \mathcal {J}_I$ .

Let $k_\alpha \in \omega $ be such that $f_{\alpha }(m)\leq g(m)$ , for every $m>k_\alpha $ . Observe that for each $m>k_\alpha $ we get

$$ \begin{align*}E^\alpha_m\subseteq \operatorname{\mathrm{cl}}_{2^{\omega}}(E^\alpha_m)\cap D\subseteq B_{f_\alpha(m)}\cap D\subseteq \bigcup_{n\leq g(m)}B_n\cap D\subseteq\bigcup_{n\leq m-1}C_n\end{align*} $$

and consequently:

$$ \begin{align*}E^\alpha_m\cap\bigcup_{n\in\omega}\left(C_n\cap\bigcup_{i\leq n}E^\alpha_i\right) \subseteq E^\alpha_m\cap\bigcup_{n\geq m}C_n=\emptyset,\end{align*} $$

since $\langle E^\alpha _n:n\in \omega \rangle \in \widehat {\mathcal {P}}_{\mathcal {J}_I}$ . Hence,

$$ \begin{align*}\bigcup_{n\in\omega}\left(C_n\cap\bigcup_{i\leq n}E^\alpha_i\right)\subseteq\left(\bigcup_{m\in\omega}E^\alpha_m\right)\cap\left(\bigcup_{n\in\omega}\left(C_n\cap\bigcup_{i\leq n}E^\alpha_i\right)\right)\subseteq\bigcup_{m\leq k_\alpha}E^\alpha_m\in\mathcal{J}_I.\end{align*} $$

Now we move to $\mathfrak {b}(\mathcal {J}_I)\leq \operatorname {\mathrm {add}}'(I)$ . Let $\mathcal {A}\subseteq \overline {I}$ be the family of cardinality $\operatorname {\mathrm {add}}'(I)$ such that for each $\langle B_n:n\in \omega \rangle \in \overline {I}^{\omega }$ one can find $A\in \mathcal {A}$ with $A\setminus B_n\neq \emptyset $ , for all n.

Without loss of generality we may assume that $\operatorname {\mathrm {cl}}_{2^{\omega }}(A\cap D)=A$ for each $A\in \mathcal {A}$ . Indeed, given $A\in \mathcal {A}$ , fix any countable $E\subseteq A$ such that $\operatorname {\mathrm {cl}}_{2^{\omega }}(E)=A$ (this is possible as A is a closed subset of $2^{\omega }$ ) and enumerate $E=\{e_i: i\in \omega \}$ . For each $i\in \omega $ find a sequence $\langle x_{i,j}:j\in \omega \rangle \in [D\cap B(e_i,\frac {1}{2^i})]^{\omega }$ converging to $e_i$ (here $B(e_i,\frac {1}{2^i})$ denotes the open ball in $2^{\omega }$ of radius $\frac {1}{2^i}$ centered at $e_i$ ). By defining $\hat {A}=A\cup \{x_{i,j}: i,j\in \omega \}$ we get a closed set such that $\hat {A}=\operatorname {\mathrm {cl}}_{2^{\omega }}(\hat {A}\cap D)$ . Moreover, as $A\subseteq \hat {A}$ , the family $\{\hat {A}: A\in \mathcal {A}\}$ satisfies the same property as $\mathcal {A}$ . Thus, we will assume from now on that $\operatorname {\mathrm {cl}}_{2^{\omega }}(A\cap D)=A$ , for every $A\in \mathcal {A}$ .

Fix any bijection $h:\omega \to 2^{<\omega }$ and define $E^A_n=f_{h(n)}[A\cap D]$ for each $n\in \omega $ and $A\in \mathcal {A}$ . Observe that

$$ \begin{align*}\operatorname{\mathrm{cl}}_{2^{\omega}}(E^A_n)=\operatorname{\mathrm{cl}}_{2^{\omega}}(f_{h(n)}[A\cap D])\subseteq f_{h(n)}[\operatorname{\mathrm{cl}}_{2^{\omega}}(A\cap D)]=f_{h(n)}[A]\in I,\end{align*} $$

since $f_{h(n)}$ is a homeomorphism and I is invariant over basic sets. Hence, each $E^A_n$ belongs to $\mathcal {J}_I$ . We claim that the family $\{\langle \hat {E}^A_n: n\in \omega \rangle :A\in \mathcal {A}\}\subseteq \widehat {\mathcal {P}}_{\mathcal {J}_I}$ , where $\hat {E}^A_n=E^A_n\setminus \bigcup _{i<n}E^A_i$ for all $n\in \omega $ and $A\in \mathcal {A}$ , witnesses $\mathfrak {b}(\mathcal {J}_I)\leq \operatorname {\mathrm {add}}'(I)$ .

Fix any partition $\langle B_n:n\in \omega \rangle \in \mathcal {P}_{\mathcal {J}_I}$ of D.

Note that given any $n\in \omega $ , we have $V_{h(n)}\cap \bigcup _{i<n}\operatorname {\mathrm {cl}}_{2^{\omega }}(B_i)\subseteq \bigcup _{i<n}\operatorname {\mathrm {cl}}_{2^{\omega }}(B_i)\in I$ . Since I is invariant over basic sets and $f_{h(n)}$ is a bijection, the set $f_{h(n)}^{-1}[V_{h(n)}\cap \bigcup _{i<n}\operatorname {\mathrm {cl}}_{2^{\omega }}(B_i)]$ belongs to $\overline {I}$ . Hence, there is $A\in \mathcal {A}$ such that

$$ \begin{align*}A\setminus \left(f_{h(n)}^{-1}\left[V_{h(n)}\cap\bigcup_{i<n}\operatorname{\mathrm{cl}}_{2^{\omega}}(B_i)\right]\right)\neq\emptyset\end{align*} $$

for all $n\in \omega $ . To finish the proof, we will show that $B=\bigcup _{n\in \omega }\left (B_n\cap \bigcup _{i\leq n}\hat {E}^A_n\right )$ is dense in $2^{\omega }$ (hence, $\operatorname {\mathrm {cl}}_{2^{\omega }}(B)=2^{\omega }\notin I$ and $B\notin \mathcal {J}_I$ ).

Fix any $s\in 2^{<\omega }$ . We need to find an element of B belonging to $V_s$ . Since

$$ \begin{align*}\operatorname{\mathrm{cl}}_{2^{\omega}}(A\cap D)\setminus f_s^{-1}\left[\!V_s\cap\bigcup_{i<h^{-1}(s)}\operatorname{\mathrm{cl}}_{2^{\omega}}(B_i)\!\right]{\kern-2pt}={\kern-1pt}A\setminus{\kern-2pt} \left(\!f_{s}^{-1}{\kern-2pt}\left[\!V_s\cap\bigcup_{i<h^{-1}(s)}\operatorname{\mathrm{cl}}_{2^{\omega}}(B_i)\!\right]\right)\neq\emptyset,\end{align*} $$

and $f_s^{-1}\left [V_s\cap \bigcup _{i<h^{-1}(s)}\operatorname {\mathrm {cl}}_{2^{\omega }}(B_i)\right ]$ is closed (as $f_s$ is a homeomorphism), we can find $x_s\in (A\cap D)\setminus \left (f_s^{-1}[V_s\cap \bigcup _{i<h^{-1}(s)}B_i]\right )$ . Then, using Remark 2.1, we get

$$ \begin{align*} f_s(x_s) & \in V_s\cap \left(E^A_{h^{-1}(s)}\setminus \bigcup_{i<h^{-1}(s)}B_i\right) \subseteq V_s\cap\bigcup_{j\leq h^{-1}(s)}\left(\hat{E}^A_j\setminus \bigcup_{i<h^{-1}(s)}B_i\right)\\ & \subseteq V_s\cap\bigcup_{j\leq h^{-1}(s)}\left(\hat{E}^A_j\setminus \bigcup_{i<j}B_i\right)\subseteq V_s\cap B. \end{align*} $$

This finishes the proof.

Proof. Follows from Theorem 3.6, Proposition 3.3 and the fact that $\aleph _1\leq \operatorname {\mathrm {add}}(\mathcal {M})\leq \mathfrak {b}$ (by [Reference Bartoszyński, Matthew and Akihiro2, Theorem 3.11]).

Proof. By [Reference Filipów and Staniszewski18, Proposition 4.1], $\mathfrak {b}(\mathcal {I})\geq \aleph _1$ for every ideal $\mathcal {I}$ , so we only need to show that $\mathfrak {b}(\mathcal {I})\leq \aleph _1$ .

Fix any family $\{A^\alpha _n: n\in \omega ,\alpha <\aleph _1\}\subseteq \mathrm {CONV}$ such that:

• • each $A^\alpha _n$ is a sequence in $\mathbb {Q}\cap [0,1]$ convergent in $[0,1]$ ;

• • $\lim A^\alpha _n\neq \lim A^\beta _m$ , for all $(\alpha ,n),(\beta ,m)\in \aleph _1\times \omega $ such that $(\alpha ,n)\neq (\beta ,m)$ ;

• • $A^\alpha _n\cap A^\alpha _m=\emptyset $ for all $\alpha <\aleph _1$ and $n,m\in \omega $ such that $n\neq m$ .

We will show that this family witnesses $\mathfrak {b}(\mathrm {CONV})\leq \aleph _1$ .

Let $\langle B_n:n\in \omega \rangle \in \mathcal {P}_{\mathrm {CONV}}$ . Since $\bigcup _{n\in \omega }\operatorname {\mathrm {cl}}_{[0,1]}(B_n)$ is countable, there is $\alpha <\aleph _1$ such that $\lim A^\alpha _m\notin \bigcup _{n\in \omega }\operatorname {\mathrm {cl}}_{[0,1]}(B_n)$ , for all $m\in \omega $ . In particular, $A^\alpha _m\cap B_n$ is finite for all $n,m\in \omega $ .

Suppose to the contrary that $B=\bigcup _{n\in \omega }\left (B_n\cap \bigcup _{i\leq n}A^\alpha _i\right )\in \mathcal {I}$ , i.e., there are sequences $D_0,\ldots ,D_m\subseteq \mathbb {Q}\cap [0,1]$ convergent in $[0,1]$ and such that $B\subseteq ^\star \bigcup _{i\leq m}D_i$ . Find $j\in \omega $ such that $\lim A^\alpha _j\neq \lim D_i$ for all $i\leq m$ . Observe that:

$$ \begin{align*}A^\alpha_j\setminus\bigcup_{n<j}B_n=A^\alpha_j\cap\bigcup_{n\geq j}B_n\subseteq \bigcup_{n\geq j}\left(B_n\cap\bigcup_{i\leq n}A^\alpha_i\right)\subseteq B.\end{align*} $$

Now, since $A^\alpha _j\cap \bigcup _{n<j}B_n$ is finite, we get $A^\alpha _j\subseteq ^\star B\subseteq ^\star \bigcup _{i\leq m}D_i$ . Thus, there is $i\leq m$ with $A^\alpha _j\cap D_i$ infinite. This contradicts $\lim A^\alpha _j\neq \lim D_i$ and finishes the proof.

Proof. Without loss of generality we may assume that $\kappa $ is infinite (as for finite $\kappa $ this is trivial). Recall that for $\mathcal {A}\subseteq \mathcal {P}(\omega )$ we write

$$ \begin{align*}\mathcal{I}(\mathcal{A})=\left\{B\subseteq\omega: B\subseteq^\star\bigcup\mathcal{A}'\text{ for some }\mathcal{A}'\in[\mathcal{A}]^{<\omega}\right\}.\end{align*} $$

In particular, $\mathcal {I}(\mathcal {A})$ is an ideal if and only if $\omega \notin \mathcal {I}(\mathcal {A})$ .

Since $|\mathcal {P}(\omega )^{\kappa \times \omega }|=\mathfrak {c}^\kappa =\mathfrak {c}$ , we can fix an enumeration $\{h_\alpha : \alpha <\mathfrak {c}\}$ of $\mathcal {P}(\omega )^{\kappa \times \omega }$ such that for each $h\in \mathcal {P}(\omega )^{\kappa \times \omega }$ and $\alpha <\mathfrak {c}$ there is $\alpha <\beta <\mathfrak {c}$ with $h_\beta =h$ .

We will recursively define a sequence $\{\mathcal {I}_\alpha : \alpha <\mathfrak {c}\}$ of subsets of $\mathcal {P}(\omega )$ such that for each $\alpha <\mathfrak {c}$ :

1. (i) $\mathcal {I}_\alpha $ is an ideal;

2. (ii) $\mathcal {I}_\alpha $ is generated by at most $|\alpha \cdot \kappa |$ sets;

3. (iii) if $\beta <\alpha $ , then $\mathcal {I}_\beta \subseteq \mathcal {I}_\alpha $ ;

4. (iv) if $h_\alpha [\kappa \times \omega ]=\{h_\alpha (\gamma ,n): \gamma <\kappa ,n\in \omega \}\subseteq \bigcup _{\delta <\alpha }\mathcal {I}_\delta $ , then there is $\langle B_n:n\in \omega \rangle \in \mathcal {P}_{\mathcal {I}_{\alpha }}$ such that $\bigcup _{n\in \omega }\left (B_n\cap \bigcup _{i\leq n}h_\alpha (\gamma ,i)\right )\in \mathcal {I}_\alpha $ , for all $\gamma <\kappa $ .

Assume that $\alpha <\mathfrak {c}$ and $\mathcal {I}_\beta $ , for all $\beta <\alpha $ , are already defined. Observe that $\hat {\mathcal {I}}_\alpha =\bigcup _{\delta <\alpha }\mathcal {I}_\delta $ is an ideal (as a union of an increasing sequence of ideals) generated by at most $|\alpha \cdot (\alpha \cdot \kappa )|=|\alpha \cdot \kappa |$ sets (since $\kappa $ is infinite). Hence, if $h_\alpha (\gamma ,n)\notin \hat {\mathcal {I}}_\alpha $ for some $(\gamma ,n)\in \kappa \times \omega $ , then just put $\mathcal {I}_\alpha =\hat {\mathcal {I}}_\alpha $ and observe that $\mathcal {I}_\alpha $ is as needed.

Assume now that $h_\alpha [\kappa \times \omega ]\subseteq \hat {\mathcal {I}}_\alpha $ . Recursively define a sequence $\langle C_n:n\in \omega \rangle $ of subsets of $\omega $ such that $C_n\notin \mathcal {I}(\hat {\mathcal {I}}_\alpha \cup \{C_j: j<n\})$ and $\mathcal {I}(\hat {\mathcal {I}}_\alpha \cup \{C_j: j\leq n\})$ is an ideal. This is possible as $|\alpha \cdot (\alpha \cdot \kappa )+n|<\mathfrak {c}=\mathfrak {u}$ guarantees that $\mathcal {I}(\hat {\mathcal {I}}_\alpha \cup \{C_j: j<n\})$ is not maximal. Next, for each $n\in \omega $ put $B_n=(\{n\}\cup C_n)\setminus \bigcup _{j<n}B_j$ and note that $\langle B_n:n\in \omega \rangle $ is a partition of $\omega $ and $B_n\notin \mathcal {I}(\hat {\mathcal {I}}_\alpha \cup \{B_j: j<n\})=\mathcal {I}(\hat {\mathcal {I}}_\alpha \cup \{C_j: j<n\})$ , for each n. Moreover, $\mathcal {I}(\hat {\mathcal {I}}_\alpha \cup \{B_k:\ k\in \omega \})$ is an ideal, as otherwise we would have

$$ \begin{align*}\omega{\kern-1pt}\in{\kern-1pt}\mathcal{I}(\hat{\mathcal{I}}_\alpha\cup\{B_k: k\in\omega\}){\kern-1pt}={\kern-1pt}\bigcup_{n\in\omega}\mathcal{I}(\hat{\mathcal{I}}_\alpha\cup\{B_k: k{\kern-1pt}<{\kern-1pt}n\}){\kern-1pt}={\kern-1pt}\bigcup_{n\in\omega}\mathcal{I}(\hat{\mathcal{I}}_\alpha\cup\{C_k: k{\kern-1pt}<{\kern-1pt}n\}),\end{align*} $$

which contradicts the fact that $\omega \notin \mathcal {I}(\hat {\mathcal {I}}_\alpha \cup \{C_k: k\leq n\})$ for each n.

Define

$$ \begin{align*}\mathcal{I}_\alpha=\mathcal{I}\left(\hat{\mathcal{I}}_\alpha\cup\{B_n: n\in\omega\}\cup\left\{\bigcup_{n\in\omega}\left(B_n\cap\bigcup_{i\leq n}h_\alpha(\gamma,i)\right):\gamma<\kappa\right\}\right).\end{align*} $$

Obviously, items (iii) and (iv) are satisfied. What is more, item (ii) is also satisfied as $|\alpha \cdot \kappa +\omega +\kappa |=|\alpha \cdot \kappa |$ . Thus, we only need to check that $\omega \notin \mathcal {I}_\alpha $ .

Assume to the contrary that $\omega \in \mathcal {I}_\alpha $ . Then there are $k,m\in \omega $ , $\gamma _0,\ldots ,\gamma _m<\kappa $ and $A\in \hat {\mathcal {I}}_\alpha $ such that

$$ \begin{align*}\omega=A\cup\bigcup_{j<k}B_j\cup\bigcup_{j\leq m}\left(\bigcup_{n\in\omega}\left(B_n\cap\bigcup_{i\leq n}h_\alpha(\gamma_j,i)\right)\right).\end{align*} $$

However, since $\langle B_n:n\in \omega \rangle $ is a partition of $\omega $ , we have

$$ \begin{align*} B_k & =B_k\cap\omega=B_k\cap\left(A\cup\bigcup_{j<k}B_j\cup\bigcup_{j\leq m}\left(\bigcup_{n\in\omega}\left(B_n\cap\bigcup_{i\leq n}h_\alpha(\gamma_j,i)\right)\right)\right) \\ & \subseteq A\cup\bigcup_{j\leq m}\bigcup_{i\leq k}h_\alpha(\gamma_j,i)\in\hat{\mathcal{I}}_\alpha, \end{align*} $$

which contradicts $B_k\notin \mathcal {I}(\hat {\mathcal {I}}_\alpha \cup \{B_j:\ j<k\})$ . Thus, $\omega \notin \mathcal {I}_\alpha $ .

Once the recursive construction of $\{\mathcal {I}_\alpha : \alpha <\mathfrak {c}\}$ is completed, define $\mathcal {I}=\bigcup _{\alpha <\mathfrak {c}}\mathcal {I}_\alpha $ . Clearly, $\mathcal {I}$ is an ideal (by items (i) and (iii)). To finish the proof we need to show that $\mathfrak {b}(\mathcal {I})>\kappa $ . Fix any $\{\langle E^\gamma _n:n\in \omega \rangle : \gamma <\kappa \}\subseteq \widehat {\mathcal {P}}_{\mathcal {I}}$ . Then for each $(\gamma ,n)\in \kappa \times \omega $ there is $\beta (\gamma ,n)<\mathfrak {c}$ such that $E^\gamma _n\in \mathcal {I}_{\beta (\gamma ,n)}$ . Denote $\beta =\sup \{\beta (\gamma ,n): (\gamma ,n)\in \kappa \times \omega \}$ . Since $|\kappa \times \omega |=\kappa <\mathrm {cf}(\mathfrak {c})$ , we get that $\beta <\mathfrak {c}$ . Thus, we can find $\beta <\alpha <\mathfrak {c}$ with $h_\alpha (\gamma ,n)=E^\gamma _n$ , for all $(\gamma ,n)\in \kappa \times \omega $ . By item (iii), $h_\alpha [\kappa \times \omega ]\subseteq \bigcup _{\delta <\alpha }\mathcal {I}_\delta $ . Using item (iv) for $\alpha $ we get $\langle B_n:n\in \omega \rangle \in \mathcal {P}_{\mathcal {I}_{\alpha }}\subseteq \mathcal {P}_{\mathcal {I}}$ such that $\bigcup _{n\in \omega }\left (B_n\cap \bigcup _{i\leq n}E^\gamma _i\right )\in \mathcal {I}_\alpha \subseteq \mathcal {I}$ , for all $\gamma <\kappa $ . Therefore, $\mathfrak {b}(\mathcal {I})>\kappa $ and the proof is completed.

Proof. The first part follows directly from Lemma 4.1.

We will show that the assumptions $\mathfrak {d}<\mathrm {cf}(\mathfrak {c})\leq \mathfrak {u}=\mathfrak {c} = \mathfrak {c}^{\mathfrak {d}}$ are consistent (specifically, that they hold in the generic extension obtained by adding $\aleph _2$ random reals to the model of GCH).

Suppose that GCH is satisfied in the ground model V. Let

$$ \begin{align*}M(\aleph_2) = \{C\subseteq 2^{\aleph_2}:C \text{ is a closed set of positive measure}\}\end{align*} $$

(ordered by the inclusion modulo null sets) be a forcing which adds $\aleph _2$ random reals (here we consider the product measure on $2^{\aleph _2}$ ). Let G denote a $M(\aleph _2)$ -generic filter.

It is known that $\mathfrak {d}=\aleph _1$ and $\mathfrak {u}=\mathfrak {c}=\aleph _2$ in $V[G]$ (see, e.g., [Reference Blass, Matthew and Akihiro4, p. 474]), so $\mathfrak {d}<\mathrm {cf}(\mathfrak {c})=\mathfrak {u}=\mathfrak {c}$ . It remains to show that $\mathfrak {c} = \mathfrak {c}^{\mathfrak {d}}$ in $V[G]$ .

Denote $\kappa = |M(\aleph _2)|$ , $\lambda =\aleph _1$ and $\delta = \kappa ^{\lambda }$ . Since GCH holds in V, we have $\aleph _2 = 2^{\aleph _1}$ , and consequently

$$ \begin{align*}\kappa = |M(\aleph_2)| = |[\aleph_2]^{<\aleph_0}|^{\aleph_0} =\aleph_2^{\aleph_0} = (2^{\aleph_1})^{\aleph_0} = 2^{\aleph_1}=\aleph_2\end{align*} $$

and

$$ \begin{align*}\delta = \aleph_2^{\aleph_1} = (2^{\aleph_1})^{\aleph_1} = \aleph_2.\end{align*} $$

Hence, $\kappa , \lambda ,\delta $ are infinite cardinals. As $M(\aleph _2)$ is ccc, using [Reference Kunen25, Lemma IV.3.11 at p. 267] we obtain in $V[G]$ the inequality $2^\lambda \leq \delta $ which yields:

$$ \begin{align*}\mathfrak{c}^{\mathfrak{d}} = (2^{\aleph_0})^{\mathfrak{d}} = 2^{\mathfrak{d}} = 2^{\aleph_1} = 2^\lambda\leq \delta = \aleph_2 = \mathfrak{c}.\end{align*} $$

Since $\mathfrak {c}\leq \mathfrak {c}^{\mathfrak {d}}$ , we obtain $\mathfrak {c}=\mathfrak {c}^{\mathfrak {d}}$ , and the proof is finished.

Proof. If $\alpha =1$ then this is true by [Reference Filipów and Kwela17, Theorems 4.2(4) and 4.2(9)].

Let now $1<\alpha <\omega _1$ be any ordinal. We need to standardize the notation so that we will be able to proceed with the proof for both successor and limit ordinals $\alpha $ .

If $\alpha =\beta +1$ is a successor ordinal, put $\gamma =\omega $ and write $\beta _\delta =\beta $ for all $\delta <\gamma $ . Note that $\mathrm {Fin}^\alpha =\mathrm {Fin}\otimes \mathrm {Fin}^\beta =\mathcal {I}_\gamma -\sum _{\delta <\gamma }\mathrm {Fin}^{\beta _\delta }$ .

On the other hand, if $\alpha $ is a limit ordinal, put $\gamma =\alpha $ and write $\beta _\delta =\delta $ for all $\delta <\gamma $ . Note that $\mathrm {Fin}^\alpha =\mathcal {I}_\alpha -\sum _{\delta <\gamma }\mathrm {Fin}^{\delta }=\mathcal {I}_\gamma -\sum _{\delta <\gamma }\mathrm {Fin}^{\beta _\delta }$ .

Now we can proceed with the proof not caring if $\alpha $ is a successor or a limit ordinal.

For each $\delta <\gamma $ let $\mathrm {Fin}^{\beta _\delta }(\omega )$ be any isomorphic copy of $\mathrm {Fin}^{\beta _\delta }$ on $\omega $ . Then $\mathrm {Fin}^\alpha $ and $\mathcal {J}=\mathcal {I}_\gamma -\sum _{\delta <\gamma }\mathrm {Fin}^{\beta _\delta }(\omega )$ are isomorphic, so $\mathcal {J}\sqsubseteq \mathcal {I}$ . Thus, without loss of generality we may assume that $\mathcal {I}$ is an ideal on $\gamma \times \omega $ such that $\mathcal {J}\subseteq \mathcal {I}$ (this can be done by considering the ideal $\{f[A]:A\in \mathcal {I}\}\supseteq \mathcal {J}$ , instead of $\mathcal {I}$ , where f is the bijection witnessing $\mathcal {J}\sqsubseteq \mathcal {I}$ ).

We will show that $\mathfrak {b}(\mathcal {I},\mathcal {I},\mathcal {J})\leq \mathfrak {d}$ . This will finish the proof as $\mathfrak {b}(\mathcal {I})\leq \mathfrak {b}(\mathcal {I},\mathcal {I},\mathcal {J})$ (by $\mathcal {J}\subseteq \mathcal {I}$ ).

Fix a strictly increasing sequence $\langle \gamma _i:i\in \omega \rangle \in \gamma ^{\omega }$ converging to $\gamma $ (in the order topology) such that $\gamma _0=0$ . Let $\{f_\alpha :\alpha <\mathfrak {d}\}\subseteq \omega ^{\omega }$ be a $\leq _{\mathrm {Fin}}$ -dominating family. Without loss of generality we may assume that each $f_\alpha $ is strictly increasing and satisfies $f_\alpha (0)=0$ . Define $A^\alpha _n=(\gamma _{f_\alpha (n+1)}\setminus \gamma _{f_\alpha (n)})\times \omega $ for all $\alpha <\mathfrak {d}$ and $n\in \omega $ . Then $\{\langle A^\alpha _n:n\in \omega \rangle :\alpha <\mathfrak {d}\}\subseteq \widehat {\mathcal {P}}_{\mathcal {J}}$ . We claim that this family witnesses $\mathfrak {b}(\mathcal {I},\mathcal {I},\mathcal {J})\leq \mathfrak {d}$ .

Fix any $\langle B_n:n\in \omega \rangle \in \mathcal {P}_{\mathcal {I}}$ . There are two possibilities.

Assume first that there is $X\notin \mathcal {I}$ such that $(X\cap B_n)_{(\delta )}\in \mathrm {Fin}^{\beta _\delta }(\omega )$ , for all $n\in \omega $ and $\delta <\gamma $ . For $\delta <\gamma $ let $m_\delta \in \omega $ be such that $\gamma _{f_0(m_\delta )}\leq \delta <\gamma _{f_0(m_\delta +1)}$ (i.e., $\{\delta \}\times \omega \subseteq A^0_{m_\delta }$ ). Observe that

$$ \begin{align*} \left(\bigcup_{i\in\omega}\left(A^0_i\setminus \bigcup_{n<i}B_n\right)\right)_{(\delta)} & \supseteq \left(A^0_{m_\delta}\setminus \bigcup_{n<m_\delta}B_n\right)_{(\delta)}\\ & =\omega\setminus\bigcup_{n<m_\delta}(B_n)_{(\delta)}\\ & \supseteq X_{(\delta)}\setminus\bigcup_{n<m_\delta}(X\cap B_n)_{(\delta)}. \end{align*} $$

Define $Y_\delta =\bigcup _{n<m_\delta }(X\cap B_n)_{(\delta )}\in \mathrm {Fin}^{\beta _\delta }(\omega )$ and $Y=\bigcup _{\delta <\gamma }\{\delta \}\times Y_\delta \in \mathcal {J}\subseteq \mathcal {I}$ . Then we have

$$ \begin{align*}\mathcal{I}\not\ni X\setminus Y\subseteq\bigcup_{i\in\omega}\left(A^0_i\setminus \bigcup_{n<i}B_n\right)\subseteq \bigcup_{n\in\omega}\left(B_n\cap\bigcup_{i\leq n}A^0_i\right)\end{align*} $$

(the last inclusion follows from Remark 2.1).

Assume now that if $X\subseteq \gamma \times \omega $ satisfies $(X\cap B_n)_{(\delta )}\in \mathrm {Fin}^{\beta _\delta }(\omega )$ for all $n\in \omega $ and $\delta <\gamma $ then $X\in \mathcal {I}$ . Recall that the ideal $\mathrm {Fin}\otimes \mathcal {J}$ (which is isomorphic to $\mathrm {Fin}^{\alpha +1}$ ) has three kinds of generators:

• • sets of the form $\{n\}\times \gamma \times \omega $ , for $n\in \omega $ (generators of the first type);

• • sets of the form $\bigcup _{n\in \omega }\{n\}\times \gamma _{g(n)}\times \omega $ , for $g\in \omega ^{\omega }$ (generators of the second type);

• • sets $G\subseteq \omega \times \gamma \times \omega $ such that $G_{(n,\delta )}\in \mathrm {Fin}^{\beta _\delta }(\omega )$ , for all $(n,\delta )\in \omega \times \gamma $ (generators of the third type).

Consider the function $h:\gamma \times \omega \to \omega \times \gamma \times \omega $ defined by the formula $h(\delta ,j)=(n(\delta ,j),\delta ,j)$ , where $n(\delta ,j)\in \omega $ is given by $(\delta ,j)\in B_{n(\delta ,j)}$ . Observe that:

• • if $G=\{n\}\times \gamma \times \omega $ is a generator of he first type, then $h^{-1}[G]\subseteq B_n\in \mathcal {I}$ , for every $n\in \omega $ ;

• • if $G\subseteq \omega \times \gamma \times \omega $ is a generator of the third type, then $(h^{-1}[G]\cap B_n)_{(\delta )}\subseteq G_{(n,\delta )}\in \mathrm {Fin}^{\beta _\delta }(\omega )$ , for all $(n,\delta )\in \omega \times \gamma $ , so $h^{-1}[G]\in \mathcal {I}$ by our assumption.

On the other hand, $\mathrm {Fin}^{\alpha +1}\not \leq _K\mathcal {I}$ (by $\mathrm {Fin}^{\alpha +1}\not \sqsubseteq \mathcal {I}$ and [Reference Barbarski, Filipów, Mrożek and Szuca1, Example 4.1]), so there has to be a generator of the second type, $G=\bigcup _{n\in \omega }\{n\}\times \gamma _{g(n)}\times \omega $ for some $g\in \omega ^{\omega }$ , such that

$$ \begin{align*}\mathcal{I}\not\ni h^{-1}\left[G\right]\subseteq \bigcup_{n\in\omega}\left(B_n\cap(\gamma_{g(n)}\times\omega)\right).\end{align*} $$

Find $\alpha <\mathfrak {d}$ with $g\leq ^\star f_\alpha $ and let $m\in \omega $ be such that $g(n)\leq f_\alpha (n)$ for all $n>m$ . Note that

$$ \begin{align*} \bigcup_{n\in\omega}\left(B_n\cap\bigcup_{i\leq n}A^\alpha_i\right) & \supseteq\bigcup_{n\in\omega}\left(B_n\cap(\gamma_{f_\alpha(n)}\times\omega)\right)\\ & \supseteq\bigcup_{n>m}\left(B_n\cap(\gamma_{g(n)}\times\omega)\right)\supseteq h^{-1}[G]\setminus \bigcup_{n\leq m} B_n\notin\mathcal{I} \end{align*} $$

(as $\bigcup _{n\leq m} B_n\in \mathcal {I}$ ). This finishes the proof.

Proof. We will show it inductively.

This is clear for $\alpha =2$ as witnessed by the family:

$$ \begin{align*}\left\{(f(0)\times\omega)\cup\bigcup_{i\in\omega}\{i\}\times f(i):f\in\omega^{\omega}\right\}\subseteq\mathrm{Fin}^2.\end{align*} $$

Fix now any partition $\langle C_n:n\in \omega \rangle $ of $\omega $ into infinite sets and for each n let $h_n:\omega \to C_n$ be the increasing enumeration of $C_n$ .

Assume that for some $1<\alpha <\omega _1$ we have the needed family $\{A_f:f\in \omega ^{\omega }\}\subseteq \mathrm {Fin}^\alpha $ . We need to find the required family for $\alpha +1$ . Given $f\in \omega ^{\omega }$ , define

$$ \begin{align*}B_f=(f(0)\times \bigcup\mathrm{Fin}^\alpha)\cup\bigcup_{n\in\omega}\{n\}\times A_{f\circ h_n}.\end{align*} $$

Again it is clear that $\{B_{f}:f\in \omega ^{\omega }\}\subseteq \mathrm {Fin}^{\alpha +1}$ is as needed.

Finally, assume that $1<\alpha <\omega _1$ is a limit ordinal and for each $\beta <\alpha $ we have the required family $\{A^\beta _{f}:f\in \omega ^{\omega }\}\subseteq \mathrm {Fin}^\beta $ . Find a bijection $g:\omega \to \alpha $ and a strictly increasing sequence $\langle \beta _i:i\in \omega \rangle $ converging to $\alpha $ (in the order topology). For each $f\in \omega ^{\omega }$ , define

$$ \begin{align*}B_f=\left(\sum_{\beta<\beta_{f(0)}}\bigcup\mathrm{Fin}^\beta\right)\cup\left(\bigcup_{n\in\omega}\{g(n)\}\times A^{g(n)}_{f\circ h_n}\right).\end{align*} $$

Once more it is clear that $\{B_{f}:f\in \omega ^{\omega }\}\subseteq \mathrm {Fin}^{\alpha }$ is as needed.

Proof. Find an increasing sequence $\langle \beta _i:i\in \omega \rangle $ converging to $\alpha $ (in the order topology). Since $\mathrm {Fin}^{\beta }\sqsubseteq \mathcal {I}$ for all $\beta <\alpha $ , we can assume that $\mathcal {I}$ is an ideal on $\omega $ and $\mathrm {Fin}^{\beta _i}(\omega )\subseteq \mathcal {I}$ for all i, where $\mathrm {Fin}^{\beta _i}(\omega )$ is some copy of $\mathrm {Fin}^{\beta _i}$ on $\omega $ . For each $i\in \omega $ let $\{B^i_{f}:f\in \omega ^{\omega }\}\subseteq \mathrm {Fin}^{\beta _i}(\omega )$ be the family from Lemma 5.2.

Fix any partition $\langle C_n:n\in \omega \rangle $ of $\omega $ into infinite sets and for each n let $h_n:\omega \to C_n$ be the increasing enumeration of $C_n$ . Let $\{f_\alpha :\alpha <\mathfrak {d}\}\subseteq \omega ^{\omega }$ be a $\leq _{\mathrm {Fin}}$ -dominating family of strictly increasing functions and recursively define $A^\alpha _n=B^n_{f_\alpha \circ h_n}\setminus \bigcup _{i<n}A^\alpha _i$ , for all $\alpha <\mathfrak {d}$ and $n\in \omega $ . Then $\{\langle A^\alpha _n:n\in \omega \rangle :\alpha <\mathfrak {d}\}\subseteq \widehat {\mathcal {P}}_{\mathcal {I}}$ as $\mathrm {Fin}^{\beta _i}(\omega )\subseteq \mathcal {I}$ for all i. We will show that $\{A^\alpha _n:n\in \omega ,\alpha <\mathfrak {d}\}$ witnesses $\mathfrak {b}(\mathcal {I})\leq \mathfrak {d}$ .

Fix any $\langle B_n: n\in \omega \rangle \in \mathcal {P}_{\mathcal {I}}$ and consider the function $h:\omega \to \alpha \times \omega $ defined by $h(i)=(\beta _n,i)$ , where $n\in \omega $ is given by $i\in B_n$ . Since $\mathcal {I}_\alpha -\sum _{\beta <\alpha }\mathrm {Fin}^{\beta }(\omega )$ is isomorphic to $\mathrm {Fin}^\alpha $ and $\mathrm {Fin}^{\alpha }\not \leq _K\mathcal {I}$ (by $\mathrm {Fin}^{\alpha }\not \sqsubseteq \mathcal {I}$ and [Reference Barbarski, Filipów, Mrożek and Szuca1, Example 4.1]), there is $Y\in \mathcal {I}_\alpha -\sum _{\beta <\alpha }\mathrm {Fin}^{\beta }(\omega )$ such that $h^{-1}[Y]\notin \mathcal {I}$ . Actually, for each $\beta <\alpha $ either $h^{-1}[\{\beta \}\times \omega ]=\emptyset \in \mathcal {I}$ (if $\beta \neq \beta _i$ for all $i\in \omega $ ) or $h^{-1}[\{\beta \}\times \omega ]\subseteq B_n\in \mathcal {I}$ for some $n\in \omega $ (if $\beta =\beta _n$ ). Hence, the set

$$ \begin{align*}X=h^{-1}\left[\bigcup\left\{\{\delta\}\times Y_{(\delta)}: Y_{(\delta)}\in\mathrm{Fin}^\delta(\omega)\right\}\right],\end{align*} $$

does not belong to $\mathcal {I}$ , while $h[X]\subseteq h[h^{-1}[Y]]\subseteq Y\in \mathcal {I}_\alpha -\sum _{\beta <\alpha }\mathrm {Fin}^{\beta }(\omega )$ . Observe that $X\cap B_n\subseteq (h[X])_{(\beta _n)}\in \mathrm {Fin}^{\beta _n}$ for all $n\in \omega $ .

Then for each n there is $g_n\in \omega ^{\omega }$ such that $X\cap B_n\subseteq B^n_{g_n}$ . Define $g\in \omega ^{\omega }$ by $g\restriction C_n=g_n\circ h_n^{-1}$ . Find $\alpha <\mathfrak {d}$ with $g\leq ^\star f_\alpha $ and let $m\in \omega $ be such that $g(i)\leq f_\alpha (i)$ whenever $i\geq m$ . Let $k\in \omega $ be maximal such that $C_k\cap m\neq \emptyset $ . Observe that for $n>k$ and any $j\in \omega $ we have

$$ \begin{align*}(f_\alpha\circ h_n)(j)=f_\alpha(h_n(j))\geq g(h_n(j))=g_n(j).\end{align*} $$

Hence, $X\cap B_n\subseteq B^n_{g_n}\subseteq B^n_{f_\alpha \circ h_n}\subseteq \bigcup _{i\leq n}A^\alpha _i$ , whenever $n>k$ . Thus,

$$ \begin{align*}\bigcup_{n\in\omega}\left(B_n\cap\bigcup_{i\leq n}A^\alpha_i\right)\supseteq\bigcup_{n>k}\left(B_n\cap \bigcup_{i\leq n}A^\alpha_i\right)\supseteq\bigcup_{n>k}\left(X\cap B_n\right)\notin\mathcal{I}\end{align*} $$

as $\mathcal {I}\not \ni X=\bigcup _{n\in \omega }\left (X\cap B_n\right )$ and $\bigcup _{n\leq k}B_n\in \mathcal {I}$ .

Proof. Let $\alpha <\omega _1$ be minimal such that $\mathrm {Fin}^{\alpha }\not \sqsubseteq \mathcal {I}$ . If $\alpha $ is a successor ordinal, use Lemma 5.1. On the other hand, if $\alpha $ is a limit ordinal, we can apply Lemma 5.3.

Proof. We claim that there is $\alpha <\omega _1$ such that $\mathrm {Fin}^\alpha \not \sqsubseteq \mathcal {I}$ . Indeed, by [Reference Debs and Raymond15, Theorems 3.2 and 6.5 and Lemma 7.2], if $\mathrm {Fin}^\alpha \sqsubseteq \mathcal {I}$ , then $\mathcal {I}$ is not $\mathbf {\Delta ^0_{1+\alpha }}$ -separated from $\mathcal {I}^\star $ . Thus, the assumption that $\mathrm {Fin}^\alpha \sqsubseteq \mathcal {I}$ , for all $\alpha <\omega _1$ , would contradict the fact that $\mathcal {I}$ is Borel-separated from $\mathcal {I}^\star $ . Now it suffices to apply Theorem 5.4.

For each analytic ideal, its dual filter is also analytic (by [Reference Debs and Raymond15, Section 1.1]), so they can be Borel-separated (by the Lusin separation theorem). This proves the “In particular” part.