Proof. See [Reference Bartoszyński and Judah1, Theorem 3.2] and [Reference Bartoszyński and Judah2, Theorem 2.7.18], but we provide an argument in connection with Theorem 1.10(1) (which is this theorem for $h = \mathrm {id}_\omega $ ). It is enough to show that, for any $h,h'\in \omega ^{\omega }$ , if h diverges to infinity, then implies . We use an argument similar to [Reference Cardona and Mejía17, Theorem 3.16]. Fix $I\in \mathbb {I}$ . Pick an increasing function $g\in \omega ^{\omega }$ with $g(0)=0$ such that, for $0<k<\omega $ , $h(n)\geq h'(k)$ for all $n\geq g(k)$ . Refine I to $I'\in \mathbb {I}$ such that $I^{\prime }_k:=\bigcup \{I_n \mid \, g(k)\leq n <g(k+1)\}$ . So, by , there is some $\varphi '\in \prod _{k\in \omega }\mathcal {P}(2^{I^{\prime }_k})$ such that $|\varphi '(k)|\leq h'(k)$ for all $k<\omega $ , and $X\subseteq H_{\varphi '}$ . Set

$$\begin{align*}\varphi(n):= \begin{cases} \{s\!\!\upharpoonright\!\! I_n \mid\, s\in\varphi'(k)\} & \text{if } g(k)\leq n<g(k+1) \text{ for some (unique) } k\geq 1,\\ \emptyset & \text{if } n<g(1). \end{cases}\end{align*}$$

It is clear that $\varphi \in \prod _{n\in \omega }\mathcal {P}(2^{I_n})$ and $|\varphi (n)| = 0 \leq h(n)$ whenever $n<g(1)$ ; in the case that n is as in the first case above, $|\varphi (n)|\leq |\varphi '(k)| \leq h'(k)\leq h(n)$ . Moreover, $X\subseteq H_{\varphi '}\subseteq H_\varphi $ , which finishes the proof.

Proof of Theorem A

It just suffices to prove that $\mathrm {non}(\mathcal {N}\!\mathcal {A})\leq \mathrm {add}(\mathcal {N}\!\mathcal {A})$ . Suppose that $\{X_\alpha \mid \, \alpha <\kappa \}\subseteq \mathcal {N}\!\mathcal {A}$ and $\kappa <\mathrm {non}(\mathcal {N}\!\mathcal {A})$ . We prove that $\bigcup _{\alpha <\kappa }X_\alpha \in \mathcal {N}\!\mathcal {A}$ . Let $I\in \mathbb {I}$ . Then by using Theorem 1.10(1), for each $\alpha <\kappa $ choose $\varphi ^I_\alpha \in \mathcal {S}(2^{I},\mathrm {id}_\omega )$ such that $X_\alpha \subseteq H_{\varphi ^I_\alpha }$ . Let $b^I:=\langle [2^{I_n}]^{\leq n} \mid \, n\in \omega \rangle $ . Since $\langle \varphi ^I_\alpha \mid \, \alpha <\kappa \rangle \subseteq \prod b^I$ , by applying Theorem 1.5 there is some $\psi ^I\in \mathcal {S}(b^I,\mathrm {id}_\omega )$ such that $\forall \alpha <\kappa \colon \varphi ^I_\alpha \in ^{*}\psi ^I$ . Set $\varphi ^I\in \mathcal {S}(2^{I},\mathrm {id}_\omega ^2)$ by $\varphi ^I(n):=\bigcup \psi ^I(n)$ . As a result, we get that $\forall \alpha <\kappa \,\forall ^\infty n\colon \varphi ^I_\alpha (n) \subseteq \varphi ^I(n)$ . This implies that $\bigcup _{\alpha <\kappa }X_\alpha \subseteq H_{\varphi ^I}.$ Thus, $\bigcup _{\alpha <\kappa }X_\alpha \in \mathcal {N}\!\mathcal {A}$ by Theorem 2.1 (applied to $h(n):= n^2$ ).

Fact 2.11. For $b\in (\omega +1)^{\omega }$ , $\mathbf {R}_b\cong _{\mathrm {T}}\langle \prod b,\mathbb {I}\times \omega ^{\omega },\sqsubset ^{\bullet }\rangle $ . As a consequence, if $b'\in (\omega +1)^{\omega }$ and $b\leq ^{*} b'$ , then $\mathbf {R}_b\preceq _{\mathrm {T}}\mathbf {R}_{b'}$ . In particular, $\mathfrak {b}(\mathbf {R}_{b'}) \leq \mathfrak {b}(\mathbf {R}_b)$ and $\mathfrak {d}(\mathbf {R}_b)\leq \mathfrak {d}(\mathbf {R}_{b'})$ .

Proof. Given $b\in \omega ^{\omega }$ , thanks to Fact 2.11 we may assume that there is some $I^b\in \mathbb {I}$ such that $b(n)=2^{|I_n^b|}$ . Then, we can identify numbers ${<}b(n)$ with $0$ - $1$ sequences of length $|I_n^b|$ . We have to find maps $\Psi _-\colon \prod b\to 2^{\omega }$ and $\Psi _+\colon \mathcal {MA}\to \mathbb {I}\times \prod b$ such that, for any $ f\in \prod b$ and for any $X\in \mathcal {MA}$ , $\Psi _-(f)\in X$ implies $f\sqsubset ^{\bullet } \Psi _+(X)$ .

For $f\in \prod b$ define $x_f^{I^b}\in 2^{\omega }$ by

$$\begin{align*}x_f^{I^b}={\underbrace{f(0)}_{\text{length }|I_0^b|}}^{\frown} \cdots\cdots{}^{\frown}{\underbrace{f(n)}_{\text{length } |I_n^b|}}^{\frown}\cdots,\end{align*}$$

so put $\Psi _-(f):= x^{I^b}_{f}$ .

For $X\in \mathcal {MA}$ , $X+B_{I^b}\in \mathcal {M}$ . Note that

$$\begin{align*}X+B_{I^b}=\bigcup_{x\in X}B_{x,I^b}.\end{align*}$$

Then, by Theorem 2.3, there are $y\in 2^{\omega }$ and $J\in \mathbb {I}$ such that

$$\begin{align*}\bigcup_{x\in X}B_{x,I^b}\subseteq B_{y, J}.\end{align*}$$

Let $h\in \prod b$ such that $y=x_{h}^{I^b}$ (recall that $b(n)=2^{|I^b_n|}$ ), so put $\Psi _+(X):=(J',h)$ where

$$\begin{align*}k\in J^{\prime}_n \text{ iff } \min J_n<\max I^b_k \leq \max J_n.\end{align*}$$

It remains to prove that, for any $ f\in \prod b$ and for any $X\in \mathcal {MA}$ , $\Psi _-(f)\in X$ implies $ f\sqsubset ^{\bullet }\Psi _+(X)$ . Suppose that $x_{f}^{I_b}\in X$ and $\Psi _+(X) = (J',h)$ . Then $B_{x_{f}^{I_b},I^b}\subseteq B_{x_{h}^{I^b},J}$ . Hence, by using Lemma 2.4,

$$\begin{align*}\forall^\infty n\,\exists k\colon I_k^b\subseteq J_n\text{ and }x_f^{I^b}{\upharpoonright}I_k^b=x_{h}^{I^b}{\upharpoonright}I_k^b.\end{align*}$$

Since $I^b_k\subseteq J_n$ implies $k\in J^{\prime }_n$ , the equation above implies that $f\sqsubset ^{\bullet }(J',h)$ .

Fact 2.15 [Reference Cardona and Mejía18]

Let $\overline {\mathbf {R}}$ be as in Definition 2.14. Then $\sup _{i\in K}\mathfrak {d}(\mathbf {R}_i)\leq \mathfrak {d}(\prod \overline {\mathbf {R}})\leq \prod _{i\in K}\mathfrak {d}(\mathbf {R}_i)$ and $\mathfrak {b}(\prod \overline {\mathbf {R}})=\min _{i\in K}\mathfrak {b}(\mathbf {R}_i)$ .

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

Define $\Psi _-\colon 2^{\omega }\to \prod _{I\in D_0} 2^I$ by $\Psi _-(x)(I):=\langle x\!\!\upharpoonright \!\! I_n \mid \, n<\omega \rangle $ . And define $\Psi _+\colon \prod _{I\in D_0}\mathbb {I}\times 2^I\to \mathcal {MA}$ such that, for $z=\langle (J^I,z^I) \mid \, I\in \mathbb {I}\rangle $ ,

$$\begin{align*}\Psi_+(z) := \{x\in 2^{\omega} \mid\, \forall I\in D_0\, \forall^\infty n<\omega\, \exists k\in J^I_k\colon x\!\!\upharpoonright\!\! I_k = z^I(k)\}.\end{align*}$$

For each $I\in D_0$ let $I^{\prime }_n:=\bigcup _{k\in J^I_n}I_k$ and $y^I\in 2^{\omega }$ the concatenation of all the $z^I(k)\in 2^{I_k}$ for $k<\omega $ , i.e., $y^I\!\!\upharpoonright \!\! I_k = z^I(k)$ . Then $I':=\langle I^{\prime }_n \mid \, n<\omega \rangle \in \mathbb {I}$ , $I\sqsubseteq I'$ and

$$\begin{align*}\forall x \in \Psi_+(z)\, \forall^\infty n<\omega\, \exists k<\omega\colon I_k\subseteq I^{\prime}_n \text{ and }x\!\!\upharpoonright\!\! I_k = y^I\!\!\upharpoonright\!\! I_k.\end{align*}$$

Therefore, by Theorem 1.10, $\Psi _+(z)\in \mathcal {MA}$ .

It is clear that $(\Psi _-,\Psi _+)$ is the required Tukey connection.

Fact 2.18. Let $\mathbf {R}_i$ be a relational system for $i<3$ . If $\mathbf {R}_0\preceq _{\mathrm {T}}\mathbf {R}_1$ , then $\mathbf {R}_0\preceq _{\mathrm {T}} \mathbf {R}_1\times \mathbf {R}_2 \preceq _{\mathrm {T}} (\mathbf {R}_1;\mathbf {R}_2)$ and $\mathbf {R}_1\times \mathbf {R}_2 \cong _{\mathrm {T}} \mathbf {R}_2\times \mathbf {R}_1$ .

Proof. The Tukey-inequality $\mathbf {Ed}_b^\perp \preceq _{\mathrm {T}}\mathbf {R}_b$ is immediate from the definitions, so we focus on the second one. Define $\Psi _-\colon \prod b \to \prod b \times \mathbb {I}^{\prod b}$ by $\Psi _-(x):= (x,F_x)$ where, for $y\in \prod b$ , if $y =^\infty x$ then $F_x(y):= I^y_x \in \mathbb {I}$ is chosen such that $\forall k<\omega \, \exists i\in I^y_{x,k}\colon y(i) = x(i)$ ; otherwise, $F_x(y)$ can be anything (in $\mathbb {I}$ ).

Define $\Psi _+\colon \prod b\times \mathbb {I}\to \mathbb {I}\times \prod b$ by $\Psi _+(y,J) = (J,y)$ . We check that $(\Psi _-,\Psi _+)$ is a Tukey connection. Assume that $x,y\in \prod b$ , $J\in \mathbb {I}$ and that $\Psi _-(x)\sqsubset _* (y,J)$ , i.e., $x =^\infty y$ and $I^y_x \sqsubseteq J$ . Since each $I^y_{x,k}$ contains a point where x and y coincide, $I^y_x \sqsubseteq J$ implies that, for all but finitely many $n<\omega $ , $J_n$ contains a point where x and y coincide, which means that $x\sqsubset ^{\bullet } (J,y) = \Psi _+(y,J)$ .

Proof. Let $\Psi _-\colon \omega ^{\omega }\to \omega ^{\omega }$ that sends $x\in \omega ^{\omega }$ to some increasing $x'\in \omega ^{\omega }$ above x (everywhere). Define $\Psi _+\colon \mathbb {I}\times \omega ^{\omega }\to \omega ^{\omega }$ such that, for $(I,h)\in \mathbb {I}\times \omega ^{\omega }$ , $\Psi _+(I,h)$ is the map in $\omega ^{\omega }$ that sends each point in the interval $I_n$ to $\max _{k\in I_{n+1}}h(k)$ . Then $(\Psi _-,\Psi _+)$ is a Tukey connection: if $x\in \omega ^{\omega }$ , $(I,h)\in \mathbb {I}\times \omega ^{\omega }$ and $x'\sqsubset ^\bullet (I,h)$ , i.e., $\exists \, k_n\in I_n\colon x'(k_n)=h(k_n)$ for all but finitely many $n<\omega $ , then $x(j)\leq x'(j) < x'(k_{n+1}) \leq \max _{k\in I_{n+1}}h(k)$ for all $j\in I_n$ , i.e., $x\leq ^{*} \Psi _+(I,h)$ .

Proof. Immediately from Theorem 2.22 and Lemma 2.25.

Proof. For $0<m<\omega $ , consider the uniform measure $\mu _m$ on m, which assigns measure $\frac {1}{m}$ to each singleton. Consider the measure $\mathbf {Lb}_b$ on (the completion of) the Borel $\sigma $ -algebra of $\prod b$ obtained as the product measure of the uniform measures of each $b(i)$ . We can define $\mathcal {E}(\prod b)$ on $\prod b$ similarly, and thanks to the map

$$\begin{align*}x\in \prod b \mapsto \sum_{n<\omega}\frac{x(i)}{\prod_{k\leq n}b(k)},\end{align*}$$

we have that $\mathcal {E}(\prod b)\cong _{\mathrm {T}} \mathcal {E}$ and $\mathbf {C}_{\mathcal {E}(\prod b)}\cong _{\mathrm {T}} \mathbf {C}_{\mathcal {E}}$ , see details in [Reference Gavalová and Mejía22, Section 7.1 (arXiv version)].Footnote ⁵ So, for this proof, we can work with $\mathcal {E} = \mathcal {E}(\prod b)$ .

Let $F\colon \prod b\to \prod b$ the identity function. For $(J,h)\in \mathbb {I}\times \prod b$ , define

$$\begin{align*}G(J,h) := \left\{x\in \prod b \;\middle|\; x \sqsubset^{\bullet} (J,h)\right\}.\end{align*}$$

It is enough to show that $G(J,h)\in \mathcal {E}$ to conclude that $(F,G)$ is the desired Tukey connection. It is clear that $G(J,h)$ is an $F_\sigma $ -set, since

$$\begin{align*}G(J,h) = \bigcup_{m<\omega}\bigcap_{n\geq m}\bigcup_{k\in J_n}A^{h(k)}_k, \text{ where }A^{\ell}_k := \left\{x\in \prod b \;\middle|\; x(k) = \ell\right\} \text{ for }\ell<b(k),\end{align*}$$

and each $A^\ell _k$ is clopen. Since $\mathbf {Lb}_b(A^\ell _k) = \frac {1}{b(k)}$ , we obtain

$$\begin{align*}\mathbf{Lb}_b(G(J,h))) \leq \lim_{m\to \infty}\prod_{n\geq m}\sum_{k\in J_n}\frac{1}{b(k)}.\end{align*}$$

This limit above is $0$ because $\sum _{k<\omega }\frac {1}{b(k)}<\infty $ .

Proof of Theorem 1.13

We show that, for any fixed $h\in \omega ^{\omega }$ diverging to infinity and any dominating family $D\subseteq \omega ^{\omega }$ , $\mathcal {SN} \preceq _{\mathrm {T}} \prod _{b\in D}\mathbf {Lc}(b,h)$ , so we also have $\mathrm {cof}(\mathcal {SN})\leq \mathfrak {d}\left (\prod _{b\in D}\mathbf {Lc}(b,h)\right )$ .Footnote ⁶ Let $b_-(n):=\lfloor \log _2 b(n) \rfloor $ and fix an injection $s^n\colon 2^{b_-(n)}\to b(n)$ with left inverse $t^n$ . First define $\Psi _-\colon \mathcal {SN}\to \prod _{b\in D}\prod b$ . Let $X\in \mathcal {SN}$ . For $b\in D$ , by Lemma 2.30 pick $\sigma ^b_{X}\in (2^{<\omega })^{\omega }$ such that $\mathrm {ht}_{\sigma ^b_X}=b_-$ and $X\subseteq [\sigma ^b_X]_\infty $ , and set $\Psi _-(X):=\langle \langle s^n(\sigma ^b_X(n)) \mid \, n<\omega \rangle \mid \, b\in D\rangle $ .

Now define $\Psi _+\colon \prod _{b\in D} \mathcal {S}(b,h)\to \mathcal {SN}$ . Partition $\omega $ into intervals $I_n$ of length $h(n)$ . Let $\bar \varphi = \langle \varphi ^b \mid \, b\in D\rangle \in \prod _{b\in D} \mathcal {S}(b,h)$ . For each $b\in D$ pick $\psi ^b\in \mathcal {S}(b,h)$ such that $\varphi ^b(n) \subseteq \psi ^b(n) = \{\ell ^b_i \mid \, i\in I_n\}$ (i.e., $|\psi ^b(n)|=h(n)$ ) for all $n<\omega $ . Set $\tau ^b:=\langle t^n(\ell ^b_i) \mid \, i<\omega \rangle $ and $\Psi _+(\bar \varphi ):=\bigcap _{b\in D}[\tau ^b]_\infty $ . We get that $\{\mathrm {ht}_{\tau ^b} \mid \, b\in D\}$ forms a dominating family, so $\Psi _+(\bar \varphi )\in \mathcal {SN}$ . Indeed, for $x\in \omega ^{\omega }$ , define $x'(n):= 2^{\max _{i\in I_n}x(i)}$ , so $x'\leq ^{*} b$ for some $b\in D$ , i.e., $\max _{i\in I_n}x(i) \leq b_-(n)$ for all but finitely many n, thus $x(i) \leq b_-(n) = \mathrm {ht}_{\tau ^b}(i)$ for $i\in I_n$ .

The pair $(\Psi _-,\Psi _+)$ is a Tukey-connection. For $X\in \mathcal {SN}$ and $\bar \varphi \in \prod _{b\in D}\mathbf {Lc}(b,h)$ , assume $\Psi _-(X)\sqsubset ^\times \bar \varphi $ , i.e., for all $b\in D$ , for all but finitely many $n<\omega $ , $s^n(\sigma ^b_X(n)) \in \varphi ^b(n)$ , which means that $s^n(\sigma ^b_X(n))=\ell ^b_{i_n}$ for some $i_n\in I_n$ . This implies $\sigma ^b_X(n)=t^n(\ell ^b_{i_n})=\tau ^b(i_n)$ . Therefore, $X\subseteq [\sigma ^b_X]_\infty \subseteq [\tau ^b]_\infty $ for any $b\in D$ , so $X\subseteq \Psi _+(\bar \varphi )$ .

Claim 3.22. Assume $M<\omega $ , $\{(s_k,t_k,m_k) \mid \, k<M\}\subseteq \omega ^{<\omega }\times \mathrm {seq}_{<\omega }(b)\times \omega $ , $\{\bar {p}^k \mid \, k<M\}$ such that each $\bar {p}^k=\langle p_{k,n} \mid \, n<\omega \rangle $ is a sequence in $P_{s_k,t_k,m_k}$ , $q_k$ is the D-limit of $\bar {p}^k$ for each $k<M$ , and $q\in \mathbb {P}_b$ is stronger than every $q_k$ . Then, for any $a\in D$ , there are $n\in a$ and $q'\leq q$ stronger than $p_{k,n}$ for all $k<M$ (i.e., $q'$ forces $a\cap \bigcap _{k<M}\dot {W}(\bar {p}^k)\neq \emptyset $ ).

Proof. Write $p_{k,n}=(s_k,t_k,F_{k,n})$ , $q_k=(s_k,t_k,F_{k})$ where each $F_k = \{x^k_j \mid \, j<m_{*,k}\}$ is the D-limit of $F_{k,n} = \{x^{k,n}_j \mid \, j<m_{*,k}\}$ (increasing $\lhd $ -enumeration) with $m_{*,k}\leq m_k$ . Assume that $q = (s,t,F)\leq q_k$ in $\mathbb {P}_b$ for all $k<M$ . Let

$$ \begin{align*} U_k & :=\big\{\langle x_j \mid\, j<m_{*,k}\rangle \mid\, \\ &\forall j<m_{*,k}\, \forall \ell\in|s|\smallsetminus|s_k|\, \exists m\in[s(\ell-1),s(\ell))\colon x_j(m)=t(m)\big\}, \end{align*} $$

which is an open neighborhood of $\langle x^k_j \mid \, j<m_{*,k}\rangle $ in $(\prod b)^{m_{*,k}}$ . Then

$$\begin{align*}b_k & := \big\{n<\omega \mid\, \\ & \forall j<m_{*,k}\, \forall \ell\in|s|\smallsetminus|s_k|\, \exists m\in[s(\ell-1),s(\ell))\colon x_{j}^{k,n}(m)=t(m) \big\}\in D.\end{align*}$$

Hence, $a\cap \bigcap _{k<M}b_k\neq \emptyset $ , so choose $n\in a\cap \bigcap _{k<M}b_k$ and put $q'=(s,t,F')$ where $F':=F\cup \bigcup _{k<M}F_{k,n}$ . This is a condition in $\mathbb {P}_b$ because $|F'|\leq |F|+\sum _{k<M}m_{*,k}$ . Furthermore, $q'$ is stronger than q and $p_{n,k}$ for any $k<M$ .

Proof. We show (1) (the other properties follow immediately from this). Let $(\sigma ,N,F)\in \mathbb {Q}_{f}$ and $N'\geq N$ in $\omega $ . We need to extend $\sigma $ to $\sigma '$ to ensure that, for any $i\in N'\smallsetminus N$ and $\tau \in F$ , $\tau (i)$ extends some $\sigma '(n)$ . For this purpose, we aim for $|\sigma '| = |\sigma | + |F|(N'-N)$ . Enumerate $F=\{\tau _i \mid \, i<|F|\}$ . For $d<N'-N$ and $i<|F|$ , we have

$$\begin{align*}f(|\sigma|+d |F|+i) &\leq f(N^2 + d N + i)<f(N(N+d+1)) < f((N+d+1)^2) \\ &= |\tau_i(N+d)|,\end{align*}$$

so it is enough to define $\sigma '(|\sigma |+d |F|+i):= \tau _i(N+d)\!\!\upharpoonright \!\! f(|\sigma |+d |F|+i)$ .

Claim 3.26. Assume $M<\omega $ , $\{(\sigma _k,N_k) \mid \, k<M\}\subseteq (2^{<\omega })^{<\omega }\times \omega $ , $\{\bar {p}^k \mid \, k<M\}$ such that each $\bar {p}^k=\langle p_{k,n} \mid \, n<\omega \rangle $ is a sequence in $Q_{\sigma _k,N_k}$ , $q_k$ is the D-limit of $\bar {p}^k$ for each $k<M$ , and $q\in \mathbb {Q}_{f}$ is stronger than every $q_k$ . If $a\in D$ then there are some $n\in a$ and $q'\leq q$ stronger than $p_{k,n}$ for all $k<M$ .

Proof. Write the forcing conditions as $p_{k,n}=(\sigma _k,N_k, F_{k,n})$ where $F_{k,n}=\{\tau _i^{k,n} \mid \, i<N_0^k\}$ (increasing enumeration) with $N_0^k\leq N_k$ , for all $n\in a$ (wlog), and let $q_k=(\sigma _k,N_k,F_{k})$ be such that each $F_k=\{\tau ^k_i \mid \, i<N_0^k\}$ is the D-limit of $\langle F_{k,n} \mid \, n<\omega \rangle $ , that is, $\tau _i^k=\lim _n^D \tau _i^{k,n}$ for $i< N_0^k$ .

Assume that $q=(\sigma ,N,F) \leq q_k$ for all $k<M$ . By strengthening q if necessary, we assume that $|F|+\sum _{k<M}N_k\leq N$ . Then

$$\begin{align*}\forall i<N_0^k\, \forall j\in N\smallsetminus N_k\,\exists \ell<|\sigma|\colon\sigma(\ell)\subseteq\tau_i^k(j),\end{align*}$$

so $b_k:=\{n<\omega \mid \, \forall i<N_0^k\,\forall j\in N\smallsetminus N_k\,\exists \ell <|\sigma |:\sigma (\ell )\subseteq \tau _i^{k,n}(j)\}\in D$ . Hence $a\cap \bigcap _{k<M}b_k$ is not empty. Choose an n in that set and put $q':=(\sigma ,N,F')$ where $F':=F\cup \bigcup _{k<M} F_{n,k}$ . This is a condition in $\mathbb {Q}_{f}$ because $|F'|\leq |F|+\sum _{k<M}N_k \leq N$ . Thus $q'$ is stronger than q and $p_{n,k}$ for $k<M$ .

Proof. First force with $\mathrm {Fn}_{<\theta }(\nu ,\theta )$ to obtain $\mathfrak {d}_\theta = 2^\theta = \nu $ in its generic extension, where $\mathfrak {d}_\theta := \mathfrak {d}(\langle \theta ,\theta ,\leq \rangle ^\theta )$ (as a relational system), which coincides with the canonical dominating number of $\theta ^\theta $ . Notice that cardinalities (and cofinalities) are preserved, as well as the cardinal arithmetic hypothesis.

Afterwards, perform a FS iteration $\mathbb {P} = \langle \mathbb {P}_\xi ,\dot {\mathbb {Q}}_\xi \mid \, \xi <\lambda \theta \rangle $ where $\mathbb {Q}_\xi $ is a $\mathbb {P}_\xi $ -name of $\mathbb {D}\ast \mathbb {P}_{\dot d_\xi }$ where $\dot d_\xi $ is the name of the dominating real over $V_\xi := V^{\mathbb {P}_\xi }$ added by $\mathbb {D}$ . The iterands of this iteration are $\mathbf {Cn}$ -good (see Example 3.8(1)), so $\mathbb {P}$ forces $\mathbf {C}_{[\lambda ]^{<\aleph _1}} \preceq _{\mathrm {T}} \mathbf {Cn}$ by Theorem 3.10. On the other hand, $\mathbb {P}$ forces $\mathfrak {c}=\lambda $ , so it follows that $\mathrm {cov}(\mathcal {N}) = \aleph _1$ and $\mathrm {non}(\mathcal {N}) = \mathfrak {c}$ . On the other hand, by the cofinaly-many Cohen and dominating reals $\langle \dot d_{\lambda \rho } \mid \, \rho <\theta \rangle $ added along the iteration, since $\mathrm {cf}(\lambda \theta )=\theta $ , we obtain $\mathfrak {b} = \mathrm {non}(\mathcal {M}) = \mathrm {cov}(\mathcal {M}) = \mathfrak {d} = \theta $ . which implies $\mathrm {add}(\mathcal {M}) = \mathrm {cof}(\mathcal {M})=\theta $ . Even more, we obtain $\omega ^{\omega } \cong _{\mathrm {T}} \mathbf {C}_{\mathcal {M}} \cong _{\mathrm {T}} \theta $ .