Proof. Without loss of generality, we can assume that \(K = \gamma\) is an ordinal.

For \(i\in K\), \(\mathbf{R}_i\leq_{\mathrm{T}}\bigvee \vec{\mathbf{R}}\) is obtained by the maps \(x\mapsto (i,x)\) and \(y\mapsto y(i)\).

If, for each \(i\in K\), \((\Psi^i_-,\Psi^i_+)\) is a Tukey connection for \(\mathbf{R}_i\leq_{\mathrm{T}}\mathbf{R}'\), then \(\bigvee \vec{\mathbf{R}}\leq_{\mathrm{T}}\mathbf{R}'\) is obtained by the maps \((i,x)\mapsto \Psi^i_-(x)\) and \(y'\mapsto \langle\Psi^i_+(y') \colon i\in K\rangle\). This shows [a251a].

It is easy to show that \(\mathbf{R}_i\leq_{\mathrm{T}}\prod \vec{\mathbf{R}}\) for all \(i\in K\), so \(\bigvee \vec{\mathbf{R}}\leq_{\mathrm{T}}\prod \vec{\mathbf{R}}\) by [a251a]. On the other hand, \(\prod \vec{\mathbf{R}}\leq_{\mathrm{T}}\vec{\mathbf{R}}^;\) via the maps \(x \mapsto \langle f^x_\alpha \colon \alpha<\gamma\rangle\) such that each \(f^x_\alpha\) is the constant map with value \(x_\alpha\), and \(y\mapsto y\). This shows [a251b].

By [a251a] it is clear that \(\mathfrak{b}(\bigvee \vec{\mathbf{R}}) \leq\mathfrak{b}(\mathbf{R}_i)\) and \(\mathfrak{d}(\mathbf{R}_i) \leq \mathfrak{d}(\bigvee \vec{\mathbf{R}})\) for all \(i\in K\), so \(\mathfrak{b}(\bigvee \vec{\mathbf{R}}) \leq\min_{i\in K}\mathfrak{b}(\mathbf{R}_i)\) and \(\sup_{i\in K}\mathfrak{d}(\mathbf{R}_i) \leq \mathfrak{d}(\bigvee \vec{\mathbf{R}})\).

For each \(i\in K\), pick some \(\mathbf{R}_i\)-dominating \(D^i:=\{y^i_\xi \colon \xi<\mathfrak{d}(\mathbf{R}_i)\}\subseteq Y_i\). When \(\mathfrak{d}(\mathbf{R}_i)\leq \xi<\sup_{i\in K}\mathfrak{d}(\mathbf{R}_i)\), set \(y^i_\xi:= y^i_0\). So define \(y_\xi:= \langle y^i_\xi \colon i\in K\rangle\) for \(\xi<\sup_{i\in K}\mathfrak{d}(\mathbf{R}_i)\). Then the set \(\{y_\xi \colon \xi<\sup_{i\in K}\mathfrak{d}(\mathbf{R}_i)\}\) is \(\bigvee \vec{\mathbf{R}}\)-dominating, so \(\mathfrak{d}(\bigvee \vec{\mathbf{R}})\leq \sup_{i\in K}\mathfrak{d}(\mathbf{R}_i)\). On the other hand, it is easy to see that \(\prod_{i\in K}D_i\) is dominating for \(\prod \vec{\mathbf{R}}\) and \(\vec{\mathbf{R}}^;\), so \(\mathfrak{d}(\prod \vec{\mathbf{R}}) \leq \mathfrak{d}(\vec{\mathbf{R}}^;)\leq \prod_{i\in K}\mathfrak{d}(\mathbf{R}_i)\).

By [a251b] and what we have proved so far, \(\mathfrak{b}(\vec{\mathbf{R}}^;)\leq \mathfrak{b}(\prod \vec{\mathbf{R}}) \leq \mathfrak{b}(\bigvee \vec{\mathbf{R}}) \leq \min_{i\in K}\mathfrak{b}(\mathbf{R}_i)\) and \(\sup_{i\in K}\mathfrak{d}(\mathbf{R}_i) = \mathfrak{d}(\bigvee \vec{\mathbf{R}}) \leq \mathfrak{d}(\prod \vec{\mathbf{R}}) \leq \mathfrak{d}(\vec{\mathbf{R}}^;)\leq \prod_{i\in K}\mathfrak{d}(\mathbf{R}_i)\). So, to conclude [a251c]–[a251d] and part of [a251e], it is enough to show that \(\min_{i\in K}\mathfrak{b}(\mathbf{R}_i)\leq \mathfrak{b}(\vec{\mathbf{R}}^;)\). Let \(F\subseteq \prod_{\alpha<\gamma}{}^{\prod_{i<\alpha}Y_i}X_\alpha\) be a set of size \({<}\min_{\alpha<\gamma}\mathfrak{b}(\mathbf{R}_\alpha)\). Find \(y_\alpha\in Y_\alpha\) by recursion on \(\alpha<\gamma\) such that \(f_\alpha(y{\upharpoonright}\alpha)\sqsubset_\alpha y_\alpha\) for all \(\bar f = \langle f_\alpha \colon \alpha<\gamma\rangle \in F\). This is fine because \(|F|< \min_{\alpha<\gamma}\mathfrak{b}(\mathbf{R}_\alpha)\). Therefore, \(F\) is \(\vec{\mathbf{R}}^;\)-bounded by \(y:=\langle y_\alpha \colon \alpha<\gamma\rangle\).

It remains to show the equality of the dominating numbers in [a251e] when \(K\) is finite. Since sequential composition is associative (modulo Tukey-equivalence), it is enough to assume that \(K=\{0,1\}\). So let \(D\subseteq Y_0\times Y_1\) of size \({<}\mathfrak{d}(\mathbf{R}_0)\cdot\mathfrak{d}(\mathbf{R}_1)\). Set \(D_0\) as the set of \(y_0\in Y_0\) such that \(D_1^{y_0}:=\{y_1 \colon (y_0,y_1)\in D\}\) is \(\mathbf{R}_1\)-dominating. Then \(D_0\) is not \(\mathbf{R}_0\)-dominating, otherwise \(|D_0|\geq \mathfrak{d}(\mathbf{R}_0)\) and, since \(|D^{y_0}_1|\geq \mathfrak{d}(\mathbf{R}_1)\) for \(y_0\in D_0\), we would get that \(|D|\geq \mathfrak{d}(\mathbf{R}_0)\cdot\mathfrak{d}(\mathbf{R}_1)\), a contradiction.

Hence, there is some \(x_0\in X_0\) such that \(x_0\not\sqsubset_0 y_0\) for all \(y_0\in D_0\). Define \(f_0\colon\{\emptyset\}\to X_0\) such that \(f_0(\emptyset):=x_0\) and let \(f_1\colon Y_0\to X_1\) be any map such that, whenever \(y_0\notin D_0\), \(f_1(y_0)\not\sqsubset_1 y_1\) for any \(y_1\in D^{y_0}_1\). We then get that \(\bar f:=(f_0,f_1) \not\sqsubset^; (y_0,y_1)\) for all \((y_0,y_1)\in D\). ◻

Proof. For \(i\in\{0,1\}\), consider the Tukey connections \(\Psi^i_-\colon X_i\to S\) and \(\Psi^i_+\colon Y\to Y_i\) for \(\mathbf{R}_i\leq_{\mathrm{T}}\mathbf{R}\). Define \(\Psi_-\colon X_0\times X_1\to S\) such that \(\Psi_-(x_0,x_1)\) is above \(\Psi^0_-(x_0)\) and \(\Psi^1_-(x_1)\) in \(S\), and define \(\Psi_+\colon Y\to Y_0\times Y_1\) by \(\Psi_+(y):=(\Psi^0_+(y_0),\Psi^1_+(y_1))\). ◻

Proof. [lem:c0:1] follows directly by 1 [ex:c4:0], and [lem:c0:2] is a consequence of [lem:c0:1], [lem:c0:1465] and 1 [a251d]. So it remains to show [lem:c0:1465]. Define the functions \(\Psi_-:X\to\prod_{s\in D}X\) and \(\Psi_+:\prod_{s\in D}\mathcal{I}_s\to\mathcal{I}^*\) where \(\Psi_-(x)\) is the constant sequence with value \(x\), and \(\Psi_+(\langle A_s \colon s\in D\rangle):=\bigcap_{s\in D}A_s\), which is in \(\mathcal{I}^*\) because \(D\) is cofinal in \(S\). These maps form the desired Tukey connection. ◻

Proof. The inequality \(\mathcal{I}^*\leq_{\mathrm{T}}\prod_{s\in D}\mathcal{I}_s\) is easy to show: define \(\Psi_-:\mathcal{I}^*\to\prod_{s\in D}\mathcal{I}_s\) such that \(\Psi_-(A)\) is the constant sequence with value \(A\), and define \(\Psi_+:\prod_{s\in D}\mathcal{I}_s\to\mathcal{I}^*\) such that \(\Psi_+(\langle A_s \colon s\in D\rangle):=\bigcap_{s\in D}A_s\), which is in \(\mathcal{I}^*\) because \(D\) is cofinal in \(S\). Note that \((\Psi_-,\Psi_+)\) is the required Tukey connection.

On the other hand, using  [ex:c4:2] and [ex:c4:4] of 1, we obtain \[\prod_{s\in D}\mathcal{I}_s\leq_{\mathrm{T}}\prod_{s\in D}\mathrm{Cv}_{[\mathop{\mathrm{cof}}(\mathcal{I}_s)]^{<\mathop{\mathrm{add}}(\mathcal{I}_s)}}\leq_{\mathrm{T}}\mathrm{Cv}^{D}_{[\mathop{\mathrm{supcof}}(\vec{\mathcal{I}})]^{<\mathop{\mathrm{minadd}}(\vec{\mathcal{I}})}}.\qedhere\] ◻

Proof. It is easy to show that \(\langle\mathbb{I}\times{}^{\omega}2,\sqsubseteq^I\rangle\) is a preorder, so it is enough to prove that \(\mathfrak{b}(\mathbb{I}\times{}^{\omega}2,\sqsubseteq^I)\) is uncountable (which implies directedness). Let \((J^m, y^m)\in \mathbb{I}\times{}^{\omega}2\) for \(m<\omega\) and, wlog, assume that \(I\sqsubseteq J^m\) for all \(m<\omega\). Construct a \(J=\langle J_n \colon n<\omega\rangle\in\mathbb{I}\) such that each \(J_n\) contains one interval from \(J^m\), say \(J^m_{\ell_{n,m}}\), for each \(m\leq n\), also demanding that \(\{J^m_{\ell_{n,m}} \colon m\leq n\}\) is pairwise disjoint and that each member of this set contains an interval from \(I\). So we can define a \(y\in{}^{\omega}2\) such that \(y{\restriction}J_{\ell_{m,n}}^m=y^m{\restriction}J_{\ell_{m,n}}^m\) for all \(m\leq n<\omega\). Then \((J^m,y^m)\sqsubset^I(J,y)\) for all \(m<\omega\). ◻

Proof. \(``\Leftarrow"\) Assume that \(x\in A_{J,y}^{I}\), so \(\forall^\infty n<\omega\, \exists k<\omega\colon I_k\subseteq J_n\) and \(x{\upharpoonright}I_k=y{\upharpoonright}I_k\). Since \((J,y)\sqsubset^I (J',y')\), for large enough \(n\) we can choose an \(\ell<\omega\) such that

1. \(\exists k<\omega\colon I_k\subseteq J_\ell\) and \(x{\restriction}I_k=y{\restriction}I_k\), and

2. \(\forall k<\omega\colon I_k\subseteq J_\ell\Rightarrow I_k\subseteq J'_n\) and \(y{\restriction}I_k=y'{\restriction}I_k\).

Observe that [lem:b2:b] implies \(I_k\subseteq J'_n\) and \(x{\restriction}I_k=y'{\restriction}I_k\) for a \(k\) as in [lem:b2:a]. Hence, \(x\in A_{J',y'}^I\).

\(``\Rightarrow"\) We prove the contrapositive. Assume \((J,y)\not\sqsubset^I (J',y')\), so \(I\sqsubseteq J\), i.e.there is some \(\ell_0<\omega\) such that \(J_\ell\) contains some \(I_k\) for all \(\ell\geq \ell_0\), and there are infinitely many \(n<\omega\) such that:

1. for all \(\ell\geq \ell_0\) there is some \(k<\omega\) such that \(I_k\subseteq J_\ell\) and either \(I_k\not\subseteq J'_n\) or \(y{\restriction}I_{k}\neq y'{\restriction}I_k\).

Let \(w\) be an infinite set of \(n<\omega\) satisfying [stpr] and such that \(|\{n\in w \colon J_\ell\cap J'_n\neq\emptyset\}|\leq1\) for any \(\ell<\omega\). Denote by \(n_\ell\) the unique member of this set, in case it exists.

Define \(x\in{}^{\omega}2\) at each \(I_k\) as follows: First consider the case when \(I_k\subseteq J_\ell\) for some \(\ell\geq \ell_0\) (which is unique). In the case \(n_\ell\) exists, there is some \(k_\ell<\omega\) such that \(I_{k_\ell}\subseteq J_\ell\) and either \(I_{k_\ell}\not\subseteq J'_{n_\ell}\) or \(y{\restriction}I_{k_\ell}\neq y'{\restriction}I_{k_\ell}\). Define \(x{\restriction}I_{k}:=y{\restriction}I_{k}\) when \(k=k_\ell\), otherwise \(x{\restriction}I_{k}\) can be anything different from \(y'{\restriction}I_k\); in the case that \(n_\ell\) does not exist, set \(x{\restriction}I_k:=y{\restriction}I_k\). On the other hand, in the case that \(I_k\) is not contained in any \(J_\ell\) for \(\ell\geq \ell_0\), let \(x{\restriction}I_k\) be anything different from \(y'{\restriction} I_k\).

Firstly, it is clear that \(x\in A_{J, y}^I\). On the other hand, \(\forall n\in w\,\forall k<\omega\colon I_k\subseteq J'_n\Rightarrow x{\restriction}I_k\neq y'{\restriction}I_k\), i.e, \(x\not\in A_{J', y'}^I\). Indeed, for \(n\in w\) and \(k<\omega\) such that \(I_k\subseteq J'_n\), in case \(I_k\subseteq J_\ell\) for some \(\ell\geq\ell_0\), we have \(n=n_\ell\) and, if \(k=k_\ell\), then \(x{\restriction}I_{k_\ell}=y{\restriction}I_{k_\ell}\neq y'{\restriction}I_{k_\ell}\), else \(x{\restriction}I_k\neq y'{\restriction}I_k\) by the definition of \(x{\restriction}I_k\); otherwise, \(I_k\) is not contained in any \(J_\ell\) for \(\ell\geq \ell_0\), in which case \(x{\restriction}I_k\neq y'{\restriction}I_k\). ◻

Proof. We just prove [lem:b3:1] because [lem:b3:2] is consequence of [lem:b3:1], 1 and 1. For \(x\in{}^{\omega}2\), \(x\in A_{J,y}^{I^1}\) iff \(\forall^\infty n\,\exists k\in J_n\colon x(k)=y(k)\), which is equivalent to \(\forall^\infty n\colon x{\restriction}J_n\neq y'{\restriction}J_n\) where \(y':=y+1 \mod{2}\). This actually shows \(A_{J,y}^{I^1}=B_{y',J}\) (see 1). Therefore, \(\mathcal{M}_{I^1}=\mathcal{M}\). ◻

Proof. For \(X\subseteq{}^{\omega}2\), \[\begin{align} X\in\mathcal{M}\mathcal{A}& \text{\;iff\;} \forall A\in\mathcal{M}\colon X+A\in\mathcal{M}\\ & \text{\;iff\;}\forall (I,y)\in\mathbb{I}\times{}^{\omega}2\,\exists (J,y')\in\mathbb{I}\times {}^{\omega}2\colon X+A_{I,y}^{I^1}=\underbrace{\bigcup_{x\in X}x+A_{I,y}^{I^1}}_{\bigcup_{x\in X}A_{I,x+y}^{I^1}}\subseteq A_{J, y'}^{I^1}\\ & \text{\;iff\;} \forall (I,y)\in\mathbb{I}\times{}^{\omega}2\,\exists (J,y')\in\mathbb{I}\times {}^{\omega}2\,\forall x\in X\colon A_{I,x+y}^{I^1}\subseteq A_{J,y'}^{I^1}\\ & \text{\;iff\;} \forall I\in\mathbb{I}\,\exists(J,y)\in\mathbb{I}\times{}^{\omega}2\,\forall x\in X\colon A_{I,x}^{I^1}\subseteq A_{J,y}^{I^1} \text{ (using a translation)}\\ & \text{\;iff\;} \forall I\in\mathbb{I}\,\exists(J,y)\in\mathbb{I}\times{}^{\omega}2\,\forall x\in X\,\forall^\infty n\in\omega\, \exists\ell<\omega\colon I_\ell\subseteq J_n\,\wedge\,x{\restriction}I_\ell=y{\restriction}I_\ell\\ & \text{\;iff\;} \forall I\in\mathbb{I}\,\exists(J,y)\in\mathbb{I}\times{}^{\omega}2\colon X\subseteq A_{J,y}^I\\ & \text{\;iff\;} \forall I\in\mathbb{I}\colon X\in\mathcal{M}_I\\ & \text{\;iff\;} X\in\bigcap_{I\in\mathbb{I}}\mathcal{M}_I. \qedhere \end{align}\] ◻

Proof. It suffices to show that \(\mathbb{I}\leq_{\mathrm{T}}\langle\mathbb{I}\times {}^{\omega}2, \sqsubset^I\rangle\) because \(\mathbb{I}\cong_{\mathrm{T}}{}^{\omega}\omega\) (see 1). Define the functions \(\Psi_-\colon \mathbb{I}\to\mathbb{I}\times{}^{\omega}2\) and \(\Psi_+\colon \mathbb{I}\times{}^{\omega}2\to \mathbb{I}\) as follows. For \(J\in\mathbb{I}\), pick some \(J^+\in\mathbb{I}\), coarser that \(I\), such that \(J\sqsubseteq J^+\).

So define \[\begin{align} \Psi_-(J) & := (J^+,\bar0),\\ \Psi_+(J',y') & := J', \end{align}\] where \(\bar0\) is the constant sequence with value \(0\).

The pair \((\Psi_-,\Psi_+)\) is the desired Tukey connection: Assume \((J^+,\bar{0})\sqsubset^I(J',y')\). Since \(J^+\) is coarser than \(I\), we obtain \(J^+\sqsubseteq J'\), so \(J\sqsubseteq J'\). ◻

Proof. Find \(I''\in\mathbb{I}\) coarser than \(I'\) such that \(I\sqsubseteq I''\). Using 1 and 1 [lem:b6:1], we get \(\mathrm{Cv}_{\mathcal{M}_I}\leq_{\mathrm{T}}\mathrm{Cv}_{\mathcal{M}_{I''}}\leq_{\mathrm{T}}\mathcal{M}_{I''}\leq_{\mathrm{T}}\mathcal{M}_{I'}\) ◻

Proof. Thanks to 1, it is enough to prove the Tukey connections. Since we already have \(\mathcal{M}_I\leq_{\mathrm{T}}\mathcal{M}\) by 1, the following inequalities remain.

\({}^{\omega}\omega\times\bigvee_{J\in\mathbb{I}}\mathrm{Cv}_{\mathcal{M}_J} \leq_{\mathrm{T}}\mathcal{M}_I\): By 1 and 1, \(\bigvee_{J\in\mathbb{I}}\mathrm{Cv}_{\mathcal{M}_J} \leq_{\mathrm{T}}\mathcal{M}_I\). On the other hand, \({}^{\omega}\omega\leq_{\mathrm{T}}\mathcal{M}_I\) by 1, so we can use 1 to conclude the desired inequality.

\(\mathcal{M}\leq_{\mathrm{T}}\left(\mathbb{I};\bigvee_{J\in\mathbb{I}}\mathrm{Cv}_{\mathcal{M}_J}\right)\): We can use \(\langle\mathbb{I}\times{}^{\omega}2,\sqsubset^{I^1}\rangle\) instead of \(\mathcal{M}\). Define \(\Psi_-\colon \mathbb{I}\times {}^{\omega}2\to \mathbb{I}\times {}^\mathbb{I}(\mathbb{I}\times{}^{\omega}2)\) by \(\Psi_-(I,y):=(I,\bar y)\), where \(\bar y\colon \mathbb{I}\to \mathbb{I}\times{}^{\omega}2\), \(J\mapsto (J,y)\), and define \(\Psi_+\colon \mathbb{I}\times\prod_{J\in\mathbb{I}}\mathcal{M}_J \to \mathbb{I}\times {}^{\omega}2\) by \(\Psi_+(J,\bar B):= (J',y')\), where \((J',y')\) is chosen such that \(B_J\subseteq A^J_{J',y'}\). This is the required Tukey connection: if \(I\sqsubseteq J\) and \(\bar y(J) \sqsubset^\vee \bar B\), i.e.\(y\in B_J\), then \(y\in A^J_{J',y'}\), so \((I,y)\sqsubset^{I^1} (J',y')\). ◻

Proof. Since \(\mathcal{M}_I\) is generated by \(F_\sigma\) sets, [b5i]\({}\Leftrightarrow{}\)[b5ii] is clear. So it suffices to prove [b5i]\({}\Leftrightarrow{}\)[b5iii]. Notice that \[A_{J,y}^I=\bigcup_{m<\omega}\bigcap_{n\geq m}\bigcup_{\substack{k<\omega\\ I_k\subseteq J_n}}[y{\restriction}I_k].\] \(``\Leftarrow"\) Assume that \(\sum_{k<\omega}2^{-|I_k|}<\infty\). Let \(X\in\mathcal{M}_I\), so there are \(J\in\mathbb{I}\) and \(y\in{}^{\omega}2\) such that \(X\subseteq A_{J,y}^I\). To see that \(X\in\mathcal{N}\), it is sufficient to show that \[\mathop{\mathrm{Lb}}\left(\bigcup_{m<\omega}\bigcap_{n\geq m}\bigcup_{\substack{k<\omega\\ I_k\subseteq J_n}}[y{\restriction}I_k]\right)=0.\] Observe that \[\mathop{\mathrm{Lb}}\left(\bigcup_{\substack{k<\omega\\ I_k\subseteq J_n}}[y{\restriction}I_k]\right)\leq \sum_{\substack{k<\omega\\ I_k\subseteq J_n}}2^{-|I_k|}.\] Since \(\sum_{k<\omega}2^{-|I_k|}<\infty\), \(\bigcap_{n\geq m}\bigcup_{\substack{k<\omega\\ I_k\subseteq J_n}}[y{\restriction}I_k]\) has measure zero. Therefore, \(A_{J,y}^I\in\mathcal{N}\).

\(``\Rightarrow"\) Assume that \(\sum_{k<\omega}2^{-|I_k|}=\infty\). Let \(J\in\mathbb{I}\) be coarser than \(I\) such that \[\sum_{\substack{k<\omega\\ I_k\subseteq J_n}}2^{-|I_k|}\geq n.\] We prove that \(A_{J,y}^I\notin \mathcal{N}\), in fact \(\mathop{\mathrm{Lb}}({}^{\omega}2\smallsetminus A_{J,y}^I)=0\). Notice that \[{}^{\omega}2\smallsetminus A_{J,y}^I=\bigcap_{m<\omega}\bigcup_{n\geq m}\bigcap_{\substack{k<\omega\\ I_k\subseteq J_n}}({}^{\omega}2\smallsetminus[y{\restriction}I_k]).\] To show \(\mathop{\mathrm{Lb}}({}^{\omega}2\smallsetminus A_{J,y}^I)=0\), it is enough to see that \[\lim_{m\to\infty}\mathop{\mathrm{Lb}}\left(\bigcup_{n\geq m}\bigcap_{\substack{k<\omega\\ I_k\subseteq J_n}}{}^{\omega}2\smallsetminus[y{\restriction}I_k]\right)=0.\] Indeed, \[\begin{align} \mathop{\mathrm{Lb}}\left(\bigcap_{\substack{k<\omega\\ I_k\subseteq J_n}}{}^{\omega}2\smallsetminus[y{\restriction}I_k]\right)&=\prod_{\substack{k<\omega\\ I_k\subseteq J_n}}(1-2^{-|I_n|})\\ &\leq e^{-\sum_{\substack{k<\omega\\ I_k\subseteq J_n}}2^{-|I_k|}}\\ &\leq e^{-n} \end{align}\] Since \(\sum_{n<\omega} e^{-n}<\infty\), we conclude the desired limit. ◻

Proof. In the proof of 1, we produced a set \(X\in\mathcal{M}_I\) such that \({}^{\omega}2\smallsetminus X\in\mathcal{N}\). Since both ideals \(\mathcal{M}_I\) and \(\mathcal{N}\) are translation invariant, we obtain \(\mathrm{Cv}_\mathcal{N}\leq_{\mathrm{T}}\mathrm{Cv}_{\mathcal{M}_I}^\perp\) as in Rothberger’s proof of \(\mathrm{Cv}_\mathcal{N}\leq_{\mathrm{T}}\mathrm{Cv}_{\mathcal{M}}^\perp\), namely, using the maps \(x\mapsto x+X\) and \(y\mapsto y+({}^{\omega}2\smallsetminus X)\). ◻

Proof. Assume wlog that, for all \(k\in\omega\), \(b(k)=2^{|I_k|}\), so \(\prod b=2^I := \prod_{k<\omega}{}^{I_k}2\). Define \[\begin{align} f\colon 2^I & \longrightarrow {}^{\omega}2\\ \eta & \longmapsto \bigcup_{n<\omega}\eta(n) \end{align}\] Note that \(f\) is a bijection.

Let \(\mathbb{I}_I\) be the set of \(J\in\mathbb{I}\) coarser than \(I\). Define the maps \[\begin{align} g \colon \mathbb{I}_I & \longrightarrow \mathbb{I}\text{ and} \notag\\ h \colon \mathbb{I}\times 2^I & \longrightarrow\mathbb{I}_I\times{}^{\omega}2\notag \intertext{by the assignments} J & \longmapsto \langle\{k<\omega \colon I_k\subseteq J_n\} \colon n<\omega\rangle \text{ and} \\ (J,y) &\longmapsto \big(g^{-1}(J),f(y)\big), \end{align}\] respectively. These maps are bijections as well.

To define \(\Psi_+\colon\mathcal{M}_I\to\mathbb{I}\times2^I\), for \(X\in \mathcal{M}_I\), pick an \((J,y)\in\mathbb{I}_I\times{}^{\omega}2\) such that \(X\subseteq A_{J,y}^I\) and put \(\Psi_+(X):=h^{-1}(J,y)\). Notice that \(f(\eta)\in A_{J,y}^I\) iff \(\forall^\infty n\,\exists k\colon I_k\subseteq J_n\) and \(\eta(k)=y{\restriction}I_k\), which is equivalent to \(\forall^\infty n\,\exists k\in J_n'\colon\eta(k)=y{\restriction}I_k\) where \(J'=g(J)\), i.e.\(\eta\sqsubset^{\bullet}\Psi_+(X)\). Then \((f,\Psi_+)\) witnesses \(\mathbf{R}_b\leq_{\mathrm{T}}\mathrm{Cv}_{\mathcal{M}_I}\).

To prove \(\mathrm{Cv}_{\mathcal{M}_I}\leq_{\mathrm{T}}\mathbf{R}_b\), define \[\begin{align} \Psi'_+\colon\mathbb{I}\times2^I & \rightarrow \mathcal{M}_I \\ (J',\eta) & \mapsto A_{(g^{-1}(J'),f(\eta))}^I \end{align}\] It is clear that \((f^{-1},\Psi'_+)\) witnesses \(\mathrm{Cv}_{\mathcal{M}_I}\leq_{\mathrm{T}}\mathbf{R}_b\). Indeed, \(f^{-1}(x)\sqsubset^{\bullet}(J',\eta)\) iff \(\forall^{\infty}n\,\exists k\in J'_n\colon x{\restriction} I_k=\eta(k)\), which is equivalent to \(\forall^{\infty}n\,\exists k\colon I_k\subseteq J_n\) and \(x{\restriction}I_k=\eta(k)\) , i.e.\(x\in A_{(g^{-1}(J'),f(\eta))}^I\). ◻

Proof. Under 1 [thm:a0:NA], we only need to prove “\(\Leftarrow\)”, that is, the statement about \(D\) implies the same statement but for \(\mathbb{I}\) in the place of \(D\).

Let \(J\in\mathbb{I}\) be such that \(|J_n|=(n+1)^2\) for all \(n<\omega\), and fix \(I\in \mathbb{I}\). Set \(I_n':=\bigcup_{k\in J_n}I_k\), so \(I'\in\mathbb{I}\). Since \(D\) is \(\mathbb{I}\)-dominating, we can find an \(I^*\in D\) such that, for all but finitely many \(n<\omega\), \(I^*\) contains at least two intervals from \(I'\). Therefore, there is some \(\varphi^*\in\mathcal{S}({}^{I^*}2,\mathop{\mathrm{id}})\) such that \(X\subseteq H_{\varphi^*}\). It is enough to construct a \(\varphi\in\mathcal{S}({}^I2,\mathop{\mathrm{id}})\) such that \(H_{\varphi^*}\subseteq H_\varphi\).

We first show that, for all but finitely many \(n<\omega\) and for any \(k<\omega\), if \(I_k\subseteq I^*_n\) then \(k\geq n(n+1)\): For any \(\ell<\omega\), \(I'_\ell\) is composed of \((\ell+1)^2\)-many intervals from \(I\), so \(I_k\subseteq I'_\ell\) iff \(\frac{\ell(\ell+1)(2\ell+1)}{6} \leq k <\frac{(\ell+1)(\ell+2)(2\ell+3)}{6}\). Now, for all but finitely many \(n\), we have that whenever \(I'_\ell\) intersects \(I^*_n\), \(\ell\geq n\) and \(\ell(\ell+1)< \frac{(\ell-1)\ell(2\ell-1)}{6}\). Therefore, for any \(I_k\subseteq I^*_n\), \(n(n+1)\leq \ell'(\ell'+1)< \frac{(\ell'-1)\ell'(2\ell'-1)}{6} \leq k\), where \(\ell'\) is the smallest \(\ell\) such that \(I'_\ell\cap I^*_n\neq\emptyset\).

Fix an \(n\) as above. Whenever \(I_k\subseteq I^*_n\), define \(\varphi(k):= \{t{\restriction}I_k \colon t\in\varphi^*(n)\}\), so \(|\varphi(k)|\leq|\varphi^*(n)|\leq n\leq n(n+1) \leq k\); and whenever \(I_k\) intersects both \(I^*_n\) and \(I^*_{n+1}\), define \[\varphi(k):=\{s{\restriction}(I_k\cap I_n^*)\cup t{\restriction}(I_{k}\cap I_{n+1}^*) \colon s\in\varphi^*(n), t\in\varphi^*(n+1)\},\] so \(|\varphi(k)|\leq|\varphi^*(n)|\cdot|\varphi^*(n+1)|\leq n(n+1)\leq k\).

So far, we have defined \(\varphi(k)\) for all but finitely many \(k<\omega\), so for the remaining ones set \(\varphi(k):=\emptyset\). Therefore, \(\varphi\in\mathcal{S}({}^I2,\mathop{\mathrm{id}})\).

Lastly, it is clear that \(H_{\varphi^*}\subseteq H_{\varphi}\), as required. ◻

Proof. Without loss of generality, we may assume that there is some \(\mathbb{I}\)-dominating family \(D_0\) such that, for each \(b\in D\), there is some \(I\in D_0\) such that \(b = {}^{I}2\), i.e.\(b(n) = {}^{I_n}2\) for all \(n<\omega\).

We define \(\Psi_-\colon {}^{\omega}2\to\prod_{I\in D_0}{}^{I}2\) and \(\Psi_+\colon\prod_{I\in D_0}\mathcal{S}({}^{I}2,\mathop{\mathrm{id}})\to \mathcal{N}\!\mathcal{A}\) as follows: For \(x\in{}^{\omega}2\) and \(I\in D_0\), set \(x_I:=\langle x{\restriction}I_n \colon n\in\omega\rangle\) and define \(\Psi_-(x):=\langle x_I \colon I\in D_0\rangle\); for \(\bar\varphi=\langle\varphi_I \colon I\in D_0\rangle\in\prod_{I\in D_0}\mathcal{S}({}^{I}2,\mathop{\mathrm{id}})\) define \[\Psi_+(\bar\varphi):=\bigcap_{I\in D_0}\{x\in{}^{\omega}2 \colon \forall^\infty n\in\omega\colon x{\restriction}I_n\in\varphi_I(n)\}.\] Notice that \(\Psi_+(\bar\varphi)\in\mathcal{N}\!\mathcal{A}\) by 1. It is clear that, for any \(x\in{}^{\omega}2\) and \(\bar\varphi=\langle\varphi_I \colon I\in D_0\rangle\in\prod_{I\in D_0}\mathcal{S}({}^{I}2,\mathop{\mathrm{id}})\), \(\Psi_-(x)\sqsubset^\otimes \bar\varphi\) implies \(x\in\Psi_+(\bar\varphi)\). ◻

Proof. Perform a FS iteration \(\mathbb{P}= \langle\mathbb{P}_\xi,\dot{\mathbb{Q}}_\xi \colon \xi<\lambda\theta\rangle\) (\(\lambda\theta\) as product of ordinal numbers) where \(\mathbb{Q}_\xi\) is a \(\mathbb{P}_\xi\)-name of \(\mathbb{D}\ast \mathbb{LOC}_{\dot{d}_\xi, \mathop{\mathrm{id}}}\) where \(\dot{d}_\xi\) is the name of the dominating real over \(V_\xi := V^{\mathbb{P}_\xi}\) added by \(\mathbb{D}\). Using the cofinaly-many Cohen and dominating reals \(\langle\dot{d}_{\lambda\rho} \colon \rho<\theta\rangle\) added along the iteration and \(\mathop{\mathrm{cf}}(\lambda\theta)=\theta\), we obtain \(\mathfrak{b}= \mathop{\mathrm{non}}(\mathcal{M}) = \mathop{\mathrm{cov}}(\mathcal{M}) = \mathfrak{d}= \theta\). which implies \(\mathop{\mathrm{add}}(\mathcal{M}) = \mathop{\mathrm{cof}}(\mathcal{M})=\theta\). Even more, we obtain \(\omega^\omega \cong_{\mathrm{T}}\mathrm{Cv}_\mathcal{M}\cong_{\mathrm{T}}\theta\).

In the final generic extension \(V_{\lambda\theta}\), it is clear that \(D:=\{d_{\lambda\rho} \colon \rho<\theta\}\) is \(\leq^*\)-increasing and dominating in \({}^{\omega}\omega\). Denote \(d'_\rho := d_{\lambda\rho}\) for \(\rho<\theta\). We show that \(\mathrm{Lc}(d'_\rho,\mathop{\mathrm{id}}) \cong_{\mathrm{T}}\theta\). On the one hand, \(\theta \cong_{\mathrm{T}}\mathrm{Cv}_\mathcal{M}\leq_{\mathrm{T}}\mathrm{Lc}(d'_\rho,\mathop{\mathrm{id}})\). For the converse, define \(F\colon \prod d'_\rho \to \theta\) such that, for \(x\in \prod d'_\rho\), \(F(x)\) is some ordinal \(\eta>\rho\) such that \(x\in V_{\lambda\eta}\); and define \(F'\colon \theta\to \mathcal{S}(d'_\rho,\mathop{\mathrm{id}})\) such that \(F'(\varrho)\) is the \(\mathbb{LOC}_{d'_\varrho, \mathop{\mathrm{id}}}\)-generic real added by \(\mathbb{Q}_{\lambda\varrho}\) when \(\varrho\geq\rho\), otherwise \(F'(\varrho):= F'(\rho)\). It is clear that \((F,F')\) is the desired Tukey connection.

Since \(\mathrm{Lc}(d'_\rho,\mathop{\mathrm{id}}) \cong_{\mathrm{T}}\theta\), we get \(\mathrm{minLc}=\mathrm{supLc}=\theta\), so by 1 \(\mathop{\mathrm{add}}(\mathcal{N})=\mathop{\mathrm{cof}}(\mathcal{N})=\theta\). On the other hand, given that \(\mathbb{P}\) is ccc, the equality \(\mathfrak{d}_\theta = \nu\) is preserved (see e.g. [6]). Now, by 1, \[\mathrm{Cv}_{\mathcal{N}\!\mathcal{A}} \leq_{\mathrm{T}}\prod_{b\in D}\mathrm{Lc}(b,\mathop{\mathrm{id}}) \cong_{\mathrm{T}}\langle\theta,\leq \rangle^\theta,\] so \(\mathop{\mathrm{cov}}(\mathcal{N}\!\mathcal{A}) \leq \mathfrak{d}_\theta = \nu\). ◻

Proof. Let \(\mathbb{P}\) be the FS iteration of ccc posets constructed in the proof of [1]. It remains to show that \(\mathbb{P}\) forces \(\mathop{\mathrm{cof}}(\mathcal{M}\mathcal{A})\leq \nu\). Since \(\mathbb{P}\) has the ccc, the identity \(\mathfrak{d}_\theta=\nu\) is preserved in the generic extension. Now, in the generic extension, by 1, \(\mathop{\mathrm{cof}}(\mathcal{M}\mathcal{A})\leq \mathop{\mathrm{cov}}\left(([\theta]^{<\theta})^\theta\right) = \mathfrak{d}_\theta=\nu\). ◻