Fact 2.2 [Reference Bartoszyński and Judah4, Lemma 2.3.10].

Let $\varepsilon :\omega \to (0,\infty )$ , and assume that $\sum _{i\in \omega }\varepsilon _i<\infty $ . Then:

1. (a) For any $\bar {c}\in \Omega ^*_\varepsilon $ , $N(\bar {c})\in \mathcal {N}$ .

2. (b) For any $X\subseteq {}^\omega 2$ , $X\in \mathcal {N}$ iff $(\exists \, \bar {c}\in \Omega ^*_\varepsilon )\ X\subseteq N(\bar {c})$ .

Fact 2.3. Let $\varepsilon :\omega \to (0,\infty )$ and assume that $\liminf _{i\to \infty }\varepsilon _i=0$ . Then:

1. (a) For any $\bar {c}\in \Omega ^*_\varepsilon $ , $N^*(\bar {c})\in \mathcal {E}$ .

2. (b) For any $X\subseteq {}^\omega 2$ , $X\in \mathcal {E}$ iff $(\exists \, \bar {c}\in \Omega ^*_\varepsilon )\ X\subseteq N^*(\bar {c})$ .

Proof Item (a) is clear because, for any $n\in \omega $ , $\bigcap _{m\geq n}c_m$ is closed and, since $\liminf \limits _{i\to \infty }\varepsilon _i=0$ and $\bar c\in \Omega ^*_\varepsilon $ , it has measure zero.

For (b), the implication $\Leftarrow $ is clear by (a). To see $\Rightarrow $ , let $X\in {\mathcal {E}}$ , i.e., $X\subseteq \bigcup _{n<\omega }F_n$ for some increasing sequence $\langle F_n:\, n<\omega \rangle $ of closed measure zero sets. For each $n<\omega $ , we can cover $F_n$ with countably many basic clopen sets $[s_{n,k}]$ ( $k<\omega $ ) such that $\sum \limits _{k<\omega }\mu ([s_{n,k}])<\varepsilon _n$ , but by compactness only finitely many of them cover $F_n$ , so $F_n\subseteq c_n:=\bigcup _{k<m_n}[s_{n,k}]$ for some $m_n<\omega $ , and $\mu (c_n)<\varepsilon _n$ . Then $\bar c:=\langle c_n:\, n<\omega \rangle $ is as required.

Fact 2.5. For any $X\subseteq {{}^\omega 2}$ , $X\in {\mathcal {N}}$ iff $X\subseteq N(\bar c)$ for some $\bar c\in {{\overline {\Omega }}}$ .

Proof $\mathrm{(i)} \mathrel \Rightarrow \mathrm{(ii)}$ : Let $\varepsilon $ be a positive rational number. Since $\bar c\in {{\overline {\Omega }}}$ , $\mu (N(\bar c))=0$ , so $\lim \limits _{n\to \infty }\mu \left (\bigcup _{i\geq n}c_i\right )=0$ . Then $\mu \left (\bigcup _{i\geq N}c_i\right )<\varepsilon $ for some $N<\omega $ , which clearly implies that $\mu \left (\bigcup _{N\leq i<n}c_i\right )<\varepsilon $ for all $n\geq N$ .

$\mathrm{(ii)}\mathrel {\Rightarrow }\mathrm{(iii)}$ : Let $\varepsilon \colon \omega \to (0,\infty )$ . Using (ii), by recursion on $n<\omega $ we define an increasing sequence $\langle m_n:\, n<\omega \rangle $ with $m_0=0$ such that $\mu \left (\bigcup _{m_{n+1}\leq i<k}c_i\right )<\varepsilon _{n+1}$ for all $k\geq m_{n+1}$ . Then $I_n:=[m_n,m_{n+1})$ is as required.

$\mathrm{(iii)}\mathrel {\Rightarrow }\mathrm{(iv)}$ : Apply (iii) to $\varepsilon _n:=2^{-n}$ .

$\mathrm{(iv)}\mathrel {\Rightarrow }\mathrm{(i)}$ : Choose $\bar I$ as in (iv), and let $c^{\prime }_n:=\bigcup _{i\in I_n}c_i$ . It is clear that $N(\bar c)=N(\bar c')$ and $\bar c'\in \Omega ^*_\varepsilon $ where $\varepsilon \colon \omega \to (0,\infty )$ , $\varepsilon _n:=\mu (c^{\prime }_n)+2^{-n}$ . Since $\sum _{n<\omega }\mu (c^{\prime }_n)<\infty $ , by Fact 2.2 we obtain that $N(\bar c)=N(\bar c')\in {\mathcal {N}}$ . Thus $\bar c\in {{\overline {\Omega }}}$ .

Proof (a) is proved similarly as Fact 2.3(a), noting that $\bar c\in {{\overline {\Omega }}}$ implies that $\lim _{i\to \infty }\mu (c_i)=0$ (by Lemma 2.6(ii)). (b) follows by (a) and Fact 2.3.

Proof Thanks to (2.4), both ${\mathcal {N}_{{J}}}$ and ${\mathcal {N}^*_{J}}$ contain all finite subsets of ${{}^\omega 2}$ and the whole space ${{}^\omega 2}$ does not belong to them. It is also clear that both families are downwards closed under $\subseteq $ , so it is enough to verify that both ${\mathcal {N}_{{J}}}$ and ${\mathcal {N}^*_{J}}$ are closed under countable unions.

Consider $\bar {c}^k \in {{\overline {\Omega }}}$ for each $k\in \omega $ . By recursion, using Lemma 2.6(ii), we define an increasing sequence ${\left \langle {n_l} :\, {l<\omega }\right \rangle }$ of natural numbers such that $\mu \left (\bigcup _{n\geq n_l}c^k_n\right )<\frac {1}{(l+1)2^l}$ for all $k\leq l$ . Let $I_0:=[0,n_1)$ and $I_l:=[n_l,n_{l+1})$ for $l>0$ . Then, we define the sequence $\bar {c}$ by

$$ \begin{align*} c_n = \begin{cases} \bigcup\limits_{k\leq l}c^k_n, & \mbox{if } n\in[n_l,n_{l+1}),\\ \emptyset, & \mbox{if }n<n_0. \end{cases} \end{align*} $$

Finally,

$$ \begin{align*}\sum_{l<\omega}\mu\left(\bigcup_{n\in I_l}c_n\right)\leq \sum_{l<\omega}\sum_{k\leq l}\mu\left(\bigcup_{n\geq n_l}c^k_n\right)\leq \sum_{l<\omega}\sum_{k\leq l}\frac{1}{(l+1)2^l}=\sum_{l<\omega}\frac{1}{2^l}<\infty,\end{align*} $$

so $\bar c\in {{\overline {\Omega }}}$ by Lemma 2.6(iv). It is clear that $\bigcup _{k\in \omega }N_{J}(\bar {c}^k)\subseteq N_{J}(\bar {c})$ and $\bigcup _{k\in \omega }N^*_{J}(\bar {c}^k)\subseteq N^*_{J}(\bar {c})$ .

Proof Without loss of generality, we may assume ${\mathbb {N}}_1={\mathbb {N}}_2=\omega $ in this proof. Fix a finite-to-one function $f\colon \omega \to \omega $ and let $I_n:=f^{-1}[\{n\}]$ for any $n\in \omega $ . Given $\bar c\in {{\overline {\Omega }}}$ we define sequences $\bar c'$ and $\bar c^-$ by $c^{\prime }_n:=\bigcup _{k\in I_n}c_k$ and $c^-_k:= c_{f(k)}$ . Since $N(\bar c')= N(\bar c)$ and $N(\bar c^-)\subseteq N(\bar c)$ , we have that $\bar c',\bar c^- \in {{\overline {\Omega }}}$ .

It is enough to show:

1. (i) $K\subseteq f^\to (J)$ implies $N^*_K(\bar c) \subseteq N^*_J(\bar c^-)$ and $N_J(\bar c) \subseteq N_K(\bar c')$ , and

2. (ii) $f^\to (K)\subseteq J$ implies $N^*_K(\bar c)\subseteq N^*_J(\bar c')$ and $N_J(\bar c)\subseteq N_K(\bar c^-)$ .

(i): Assume $K\subseteq f^\to (J)$ . If $x\in N^*_{K}(\bar c)$ then $\{n<\omega :\, x\notin c_n\}\in {K}$ , so

$$ \begin{align*}\{k<\omega:\, x\notin c^-_k\}=f^{-1}[{\left\{ {n<\omega} :\, {x\notin c_n} \right\}}]\in{J},\end{align*} $$

i.e., $x\in N^*_{J}(\bar c^-)$ ; and if $x\in N_{J}(\bar c)$ , i.e., $\{k<\omega :\, x\in c_k\}\notin {J}$ , then

$$ \begin{align*} f^{-1}[\{n<\omega:\, x\in c^{\prime}_n\}] = \{k<\omega:\, x\in c^{\prime}_{f(k)}\}\supseteq\{k<\omega:\, x\in c_k\}, \end{align*} $$

so $f^{-1}[\{n<\omega :\, x\in c^{\prime }_n\}]\notin {J}$ , i.e., $\{n<\omega :\, x\in c^{\prime }_n\}\notin {K}$ , which means that $x\in N_{K}(\bar c')$ .

(ii) Assume $f^\to (K)\subseteq J$ . If $x\in N^*_K(\bar c)$ , i.e., $\{k<\omega :\, x\notin c_k\}\in K$ , then

$$ \begin{align*} f^{-1}[\{n<\omega:\, x\notin c^{\prime}_n\}] = \{k<\omega:\, x\notin c^{\prime}_{f(k)}\}\subseteq\{k<\omega:\, x\notin c_k\}, \end{align*} $$

so $f^{-1}[\{n<\omega :\, x\notin c^{\prime }_n\}]\in K$ , which implies that $\{n<\omega :\, x\notin c^{\prime }_n\}\in J$ , i.e., $x\in N^*_J(\bar c')$ ; and if $x\in N_J(\bar c)$ then $\{n<\omega :\, x\in c_n\}\notin J$ , so

$$ \begin{align*}\{k<\omega:\, x\in c^-_k\}=f^{-1}[{\left\{ {n<\omega} :\, {x\in c_n} \right\}}]\notin K,\end{align*} $$

i.e., $x\in N_K(\bar c^-)$ .

Main Lemma 3.3. Let $C\subseteq {}^\omega 2$ be a self-supported closed set. If ${J}$ is not meager then $\displaystyle N_{J}(\bar {c})\cap C$ is not co-meager in C for each $\bar {c}\in {{\overline {\Omega }}}$ . As a consequence, $Z\cap C$ is not co-meager in C for any $Z\in {\mathcal {N}}_{J}$ .

Proof Let $C\subseteq {}^\omega 2$ be a self-supported closed set, and let $G\subseteq C$ be a co-meager subset in C. Then there is a sequence ${\left \langle {D_n} :\, {n\in \omega }\right \rangle }$ of open dense sets in C such that $\bigcap _{n\in \omega }D_n\subseteq G$ . Moreover, since C is closed, there is a tree T such that $C=[T]$ .

Consider $\bar {c}\in {{\overline {\Omega }}}$ and construct the following strategy of Player I for the meager game for ${J}^d$ .

The first move:

Player I picks an $s_0\in T$ such that $[s_0]\cap C\subseteq D_0$ and chooses $n_0<\omega $ such that ${\mu \left ({\bigcup _{n\geq n_0}c_n}\right )}<{\mu \left ({C\cap [s_0]}\right )}$ (which exists because $\bar c\in {{\overline {\Omega }}}$ ). Player I’s move is $n_0$ .

Second move and further:

Player II replies with $B_0\in [\omega ]^{<\omega }$ such that $n_0\cap B_0=\emptyset $ . Since ${\mu \left ({\bigcup _{n\in B_0} c_n}\right )}<{\mu \left ({C\cap [s_0]}\right )}$ , $C\cap [s_0]\nsubseteq \bigcup _{n\in B_0}c_n$ , so there exists an $x_0\in C\cap [s_0]\smallsetminus \bigcup _{n\in B_0} c_n$ . Then Player I finds an $m_0>|s_0|$ such that $[x_0{\upharpoonright }m_0]\cap \bigcup _{n\in B_0} c_n=\emptyset $ . Since $D_1$ is dense in C, Player I can pick an $s_1\in T$ such that $s_0\subseteq x_0{\upharpoonright } m_0\subseteq s_1$ and $[s_1]\cap C\subseteq D_1$ . Then Player I moves with an $n_1\in \omega $ such that ${\mu \left ({\bigcup _{n\geq n_1} c_n}\right )}<{\mu \left ({C\cap [s_1]}\right )}$ . Player II responds with $B_1\in [\omega ]^{<\omega }$ such that $n_1\cap B_1=\emptyset $ , and the game continues in the same way we just described.

Finally, since ${J}$ is not meager, Player I does not have a winning strategy in the meager game for ${J}^d$ . Hence, there is some match ${\left \langle {(n_k, B_k)} :\, {k\in \omega }\right \rangle }$ where Player I uses the aforementioned strategy and Player II wins. Thus, we have $F:=\bigcup _{k\in \omega } B_k\in {J}^d$ . Define $x:=\bigcup _{k\in \omega }s_k$ . Then, $x\in \bigcap _{k\in \omega }D_k$ . On the other hand, $x\notin \bigcup _{n\in F}c_n$ (because $[s_{k+1}]\cap \bigcup _{n\in B_k}c_k=\emptyset $ ) and hence $x\notin N_{{J}}(\bar {c})$ .

Proof For the implication from left to right, apply Main Lemma 3.3 to $C={{}^\omega 2}$ . For the converse, if ${J}$ has the Baire property then ${\mathcal {N}}_{J}={\mathcal {N}}$ by Theorem 2.18, and it is well-known that ${\mathcal {N}}$ contains a co-meager set in ${{}^\omega 2}$ (Rothberger’s Theorem [Reference Rothberger39]).

Proof Assume that $Z\in {\mathcal {N}}_{J}$ has the Baire property. By Main Lemma 3.3, $Z\cap [s]$ is not co-meager in $[s]$ for all $s\in 2^{<\omega }$ . Therefore, by Lemma 3.4, Z is meager in ${{}^\omega 2}$ .

Proof The implication $\mathrm{(i)}\rightarrow \mathrm{(iii)}$ follows directly from Theorem 2.18. On the other hand, since ${\mathcal {M}}\cap {{\mathcal {N}}} \subseteq \mathcal {N}$ the implication $\mathrm{(ii)}\to \mathrm{(i)}$ is obvious.

It remains to show $\mathrm{(iii)}\to \mathrm{(ii)}$ . First, we choose a closed nowhere dense $C\subseteq ^\omega 2$ of positive measure, which can be found self-supported (as in the hypothesis of Main Lemma 3.3). Then, we find $G\subseteq C$ which is co-meager in C and has Lebesgue measure zero. Hence $G\in {\mathcal {M}}\cap {\mathcal {N}}$ , but $G\notin {\mathcal {N}}_{J}$ by Main Lemma 3.3.

Main Lemma 3.9. Let $T\subseteq 2^{<\omega }$ be a non-empty tree without maximal nodes and let ${J}$ be a non-meager ideal on $\omega $ . Then $N^*_{J}(\bar {c}^T)\cap G\neq \emptyset $ for every co-meager set $G\subseteq [T]$ in $[T]$ , i.e., $N^*_{J}(\bar {c}^T)$ is not meager in $[T]$ .

In other words, for any non-empty closed $H\subseteq {{}^\omega 2}$ there is some $\bar {c}^H\in {{\overline {\Omega }}}$ such that, for any non-meager ideal ${J}$ on $\omega $ , $N^*_{J}(\bar c^H)\cap H$ is non-meager in H.

Proof Let $G\subseteq [T]$ be co-meager in $[T]$ . Then, there is a sequence ${\left \langle {D_n} :\, {n\in \omega }\right \rangle }$ of open dense sets in $[T]$ such that $\bigcap _{n\in \omega }D_n\subseteq G$ . We look for an $x\in N^*_{J}(\bar {c}^T)\cap \bigcap _{n\in \omega }D_n$ by using the non-meager game.

Construct the following strategy of Player I for the non-meager game for ${J}^d$ along with fragments of the desired x.

The first move:

Player I chooses some $s_0\in T$ extending $t^0_0$ such that $[s_0]\cap [T]\subseteq D_0$ . Since $s_0\in T$ , $s_0=t_{n_0}$ for some $n_0\in \omega $ . Player I moves with $A_0:=n_0+1$ .

Second move and further:

Player II replies with $B_0$ . Since $B_0\cap A_0=\emptyset $ , by (ii)–(iii) of Definition 3.8, Player I can extend $s_0$ to some $s^{\prime }_0\in T$ such that, for any $\ell \in B_0$ , $s^{\prime }_0$ extends $t^\ell _{n_0}$ , and further finds $s_1\in T$ extending $s^{\prime }_0$ such that $[s_1]\cap [T]\subseteq D_1$ . Here, $s_1=t_{n_1}$ for some $n_1\in \omega $ , and Player I moves with $A_1:=n_1+1$ . Player II would reply with some $B_1$ not intersecting $A_1$ , and Player I continues playing in the same way.

Now, since ${J}$ is not meager, Player I does not have a winning strategy. In particular, there is some match ${\left \langle {(A_n,B_n)} :\, {n\in \omega }\right \rangle }$ of the game where Player I uses the strategy defined above and Player II wins, i.e., $F:=\bigcup _{n\in \omega }B_n$ is in ${J}^d$ . On the other hand, Player I constructed the increasing sequence ${\left \langle {s_n} :\, {n\in \omega }\right \rangle }$ of members of T, so $x:=\bigcup _{n\in \omega }s_n$ is a branch of the tree. By the definition of the strategy, we have that $x\in c^T_m$ for any $m\in F$ , so $x\in N^*_{J}(\bar {c}^T)$ . Also $x\in D_n$ for every $n\in \omega $ .

Proof $\mathrm{(iv)}\rightarrow \mathrm{(i)}$ follows by Corollary 3.10; $\mathrm{(i)}\rightarrow \mathrm{(ii)}$ is obvious; $\mathrm{(ii)}\rightarrow \mathrm{(iii)}$ is a consequence of ${\mathcal {E}}\subseteq {\mathcal {M}}\cap {\mathcal {N}}$ ; $\mathrm{(iii)}\rightarrow \mathrm{(iv)}$ is a consequence of Theorem 2.18.

Proof Since ${\mathcal {N}}_{J}^*\subseteq {\mathcal {N}}_{J}$ , the conclusion is clear by Theorem 3.11 when ${J}$ is not meager. Otherwise ${\mathcal {N}}_{J}={\mathcal {N}}$ , and ${\mathcal {N}}$ contains a co-meager set.

Main Lemma 4.7. Let $C\subseteq {}^\omega 2$ be a closed self-supported set. If ${J}$ and ${K}$ are not nearly coherent ideals on $\omega $ then $N^*_{J}(\bar {c}^T)\notin {\mathcal {N}}_{K}$ where T is the tree without maximal nodes such that $C=[T]$ , and $\bar c^T$ is as in Definition 3.8.

Proof We use the nearly coherence game $C_{{J}^d,{K}^d}$ and build a strategy for Player I, in order to get an $x\in N^*_{J}(\bar {c}^T)\smallsetminus N_{K}(\bar {c}')$ for any $\bar {c}'\in {{\overline {\Omega }}}$ .

The first move: Player I first moves with $A_0=\{0\}$ , and put $s_0:= t_0$ (which is the empty sequence).

The second and third moves, and further:

After Player II replies with $B_0\subseteq \omega \smallsetminus A_0$ , Player I finds $s_1\supseteq s_0$ in T such that $s_1\supseteq t^n_0$ for every $n\in B_0$ , and further finds $n_1\geq |s_1|$ such that $\mu \left (\bigcup _{\ell \geq n_1}c^{\prime }_\ell \right )<\mu (C\cap [s_1])$ , which implies that there is some $x_1\in C\cap [s_1]\smallsetminus \bigcup _{\ell \geq n_1}c^{\prime }_\ell $ . Player I moves with $A_1=n_1+1$ .

After Player II replies with $B_1\subseteq \omega \smallsetminus A_1$ , Player I finds some $m_2\geq n_1$ such that $[s_2]\cap \bigcup _{n\in B_1}c^{\prime }_n=\emptyset $ where $s_2:=x_1{\upharpoonright }m_2$ . Player I moves $A_2=n_2+1$ where $n_2<\omega $ is such that $s_2=t_{n_2}$ .

Afterwards, Player II moves with $B_2\subseteq \omega \smallsetminus A_2$ , and the same dynamic is repeated: Player I finds an $s_3\supseteq s_2$ in T such that $s_3\supseteq t^n_{n_2}$ for every $n\in B_2$ , and some $n_3\geq |s_3|$ such that $\mu \left ( \bigcup _{\ell \geq n_3} c^{\prime }_\ell \right )<\mu (C\cap [s_3])$ , which implies that there is an $x_3\in C\cap [s_3]\smallsetminus \bigcup _{\ell \geq n_3}c^{\prime }_\ell $ . Player I moves with $A_3=n_3+1$ , and the game continues as described so far.

Since Player I does not have a winning strategy, there is some run as described above where Player II wins. Thus, $F_0:=\bigcup _{n\in \omega }B_{2n}\in {J}^d$ and $F_1:=\bigcup _{n\in \omega }B_{2n+1}\in {K}^d$ . Set $x:=\bigcup _{n\in \omega }s_n$ . Note that $x\in c^T_\ell $ for any $\ell \in F_0$ , and $x\notin c^{\prime }_\ell $ for any $\ell \in F_1$ , which means that $x\in N^*_{J}(\bar {c}^T)$ and $x\notin N_{K}(\bar {c}')$ .

Proof $\mathrm{(i)}\mathrel {\Rightarrow } \mathrm{(ii)}$ is obvious; $\mathrm{(ii)}\mathrel {\Rightarrow } \mathrm{(iii)}$ is immediate from Theorem 2.15; $\mathrm{(iii)}\mathrel {\Rightarrow }\mathrm{(iv)}$ is obvious; and $\mathrm{(iv)}\mathrel {\Rightarrow } \mathrm{(i)}$ follows by (the contrapositive of) Theorem 4.8.

Proof Partition $\omega ={\mathbb {N}}_1\cup {\mathbb {N}}_2$ into two infinite sets, let $g_e\colon \omega \to {\mathbb {N}}_e$ be a bijection for each $e\in \{1,2\}$ , and let $J^{\prime }_e= g_e^\to (J_e)$ , which is an ideal on ${\mathbb {N}}_e$ isomorphic with $J_e$ . Here, our interpretation of $J_1\oplus J_2$ is $J^{\prime }_1\oplus J^{\prime }_2$ . It is clear that ${\mathcal {N}}_{J^{\prime }_1}={\mathcal {N}}_{J_1}$ and ${\mathcal {N}}^*_{J^{\prime }_1}={\mathcal {N}}^*_{J_1}$ (by Theorem 2.15(c)).

Recall that non-nearly coherent ideals must be non-meager (this also follows by Theorem 4.8), so $J^{\prime }_1\times J^{\prime }_2$ is a non-meager subset of ${{}^\omega 2}= {}^{{\mathbb {N}}_1}2 \times {}^{{\mathbb {N}}_2}2$ by Kuratowski–Ulam Theorem, which implies that $J^{\prime }_1\oplus J^{\prime }_2$ is non-meager.

By Example 2.13, ${\mathcal {N}}^*_{J^{\prime }_1\oplus J^{\prime }_2}={\mathcal {N}}^*_{J^{\prime }_1}\cap {\mathcal {N}}^*_{J^{\prime }_2} = {\mathcal {N}}^*_{J_1}\cap {\mathcal {N}}^*_{J_2}$ and ${\mathcal {N}}_{J^{\prime }_1 \oplus J^{\prime }_2}$ contains both ${\mathcal {N}}_{J_1}$ and ${\mathcal {N}}_{J_2}$ , so ${\mathcal {N}}^*_{J^{\prime }_1\oplus J^{\prime }_2}= {\mathcal {N}}_{J^{\prime }_1\oplus J^{\prime }_2}$ would imply that ${\mathcal {N}}_{J_1}\cup {\mathcal {N}}_{J_2}\subseteq {\mathcal {N}}^*_{J_1}\cap {\mathcal {N}}^*_{J_2}$ , and in turn ${\mathcal {N}}_{J_2}\subseteq {\mathcal {N}}^*_{J_1}$ , which implies ${\mathcal {N}}^*_{J_2}\subseteq {\mathcal {N}}_{J_1}$ (because ${\mathcal {N}}^*_{J_e}\subseteq {\mathcal {N}}_{J_e}$ ), contradicting Theorem 4.10 and the fact that $J_1$ and $J_2$ are not nearly-coherent. Therefore, ${\mathcal {N}}^*_{J^{\prime }_1\oplus J^{\prime }_2}\neq {\mathcal {N}}_{J^{\prime }_1\oplus J^{\prime }_2}$

Fact 5.1. If ${\mathcal {I}}$ is an ideal on X, ${\mathcal {I}}\subseteq {\mathcal {I}}'$ and ${\mathcal {J}}\subseteq {\mathcal {P}}(X)\smallsetminus \{X\}$ , then $({\mathbf {c}}_{{\mathcal {I}}'}^{\mathcal {J}})^\perp\ {\mathrel {\preceq _{\mathrm {T}}}} {\mathbf {C}}_{\mathcal {I}}{\mathrel {\preceq _{\mathrm {T}}}}{\mathbf {c}}_{[X]^{<\aleph _0}}^{\mathcal {I}}$ . In particular, ${\mathrm {add}}({\mathcal {I}}',{\mathcal {J}})\leq {\mathrm {cov}}({\mathcal {I}})$ and ${\mathrm {non}}({\mathcal {I}})\leq {\mathrm {cof}}({\mathcal {I}}',{\mathcal {J}})$ .

Proof We only show the first Tukey connection. Define $F\colon {\mathcal {J}}\to X$ such that $F(B)\in X\smallsetminus B$ (which exists because $X\notin {\mathcal {J}}$ ), and define $G\colon {\mathcal {I}}\to {\mathcal {I}}'$ by $G(A):=A$ . Then, for $A\in {\mathcal {I}}$ and $B\in {\mathcal {J}}$ , $F(B)\in A$ implies $B\nsupseteq A$ . Hence, $(F,G)$ witnesses $({\mathbf {c}}_{{\mathcal {I}}'}^{\mathcal {J}})^\perp\ {\mathrel {\preceq _{\mathrm {T}}}}\ {\mathbf {C}}_{\mathcal {I}}$ .

Fact 5.2. If ${\mathcal {I}}\subseteq {\mathcal {I}}'$ and ${\mathcal {J}}'\subseteq {\mathcal {J}}$ then ${\mathbf {c}}^{\mathcal {J}}_{\mathcal {I}}{\mathrel {\preceq _{\mathrm {T}}}} {\mathbf {c}}^{{\mathcal {J}}'}_{{\mathcal {I}}'}$ . In particular, ${\mathrm {add}}({\mathcal {I}}',{\mathcal {J}}')\leq {\mathrm {add}}({\mathcal {I}},{\mathcal {J}})$ and ${\mathrm {cof}}({\mathcal {I}},{\mathcal {J}})\leq {\mathrm {cof}}({\mathcal {I}}',{\mathcal {J}}')$ .

Proof Define $F\colon {\mathcal {N}}_J\to {{\overline {\Omega }}}$ such that $X\subseteq N_J(F(X))$ for any $X\in {\mathcal {N}}_J$ , and define $G\colon {{\overline {\Omega }}} \to {\mathcal {N}}_J$ by $G(\bar d):=N_J(\bar d)$ for any $\bar d\in {{\overline {\Omega }}}$ . It is clear that $(F,G)$ is a Tukey connection from ${\mathcal {N}}_J$ into ${\mathbf {S}}_J$ , because $\bar c\subseteq ^J \bar d$ implies $N_J(\bar c)\subseteq N_J(\bar d)$ .

Similarly, we obtain a Tukey connection from ${\mathcal {N}}^*_J$ into ${\mathbf {S}}_J$ via functions $F^*\colon {\mathcal {N}}^*_J\to {{\overline {\Omega }}}$ and $G^*\colon {{\overline {\Omega }}} \to {\mathcal {N}}_J$ such that $X\subseteq N^*_J(F^*(X))$ for $X\in {\mathcal {N}}^*_J$ and $G^*(\bar d):= N^*_J(\bar d)$ .

Proof It is clear that (c) follows from (a) and (b).

Let $f\colon \omega \to \omega $ be a finite-to-one function and denote $I_n:=f^{-1}[\{n\}]$ . Like in the proof of Theorem 2.15, define $F'\colon {{\overline {\Omega }}}\to {{\overline {\Omega }}}$ by $F'(\bar c):=\bar c'$ where $c^{\prime }_n:=\bigcup _{k\in I_n}c_k$ , and define $F^-\colon {{\overline {\Omega }}}\to {{\overline {\Omega }}}$ by $F^-(\bar c):=\bar c^-$ where $c^-_k:=c_{f(k)}$ .

If $K\subseteq f^{\to }(J)$ then $(F',F^-)$ is a Tukey connection from ${\mathbf {S}}_J$ into ${\mathbf {S}}_K$ , which shows (a). To prove this, assume $\bar c,\bar d\in {{\overline {\Omega }}}$ and $\bar c'\subseteq ^K \bar d$ , and we show $\bar c \subseteq ^J \bar d^-$ . The hypothesis indicates that ${\left \{ n<\omega :\, c^{\prime }_n\subseteq d_n \right \}}\in K^d$ , which implies that ${\left \{ {k<\omega } :\, {c^{\prime }_{f(k)}\subseteq d_{f(k)}} \right \}}\in J^d$ . Since $c_k\subseteq c^{\prime }_{f(k)}$ , the previous set is contained in ${\left \{ k<\omega :\, c_k\subseteq d^-_k \right \}}$ , so $\bar c \subseteq ^J \bar d^-$ .

To show (b), we verify that, whenever $f^\to (K)\subseteq J$ , $(F^-,F')$ is a Tukey connection from ${\mathbf {S}}_J$ into ${\mathbf {S}}_K$ . Let $\bar c,\bar d\in {{\overline {\Omega }}}$ and assume that $\bar c^-\subseteq ^K \bar d$ , i.e., ${\left \{ {k<\omega } :\, {c_{f(k)}\subseteq d_k} \right \}}\in K^d$ . Since $d_k\subseteq d^{\prime }_{f(k)}$ , this set is contained in ${\left \{ {k<\omega } :\, {c_{f(k)}\subseteq d^{\prime }_{f(k)}} \right \}}$ , so ${\left \{ n<\omega :\, c_n\subseteq d^{\prime }_n \right \}}\in J^d$ , i.e., $\bar c\subseteq ^J \bar d'$ .

Proof Note that ${\mathcal {N}}={\mathcal {N}}_{\mathrm {Fin}}{\mathrel {\preceq _{\mathrm {T}}}} {\mathbf {S}}_{\mathrm {Fin}}$ by Theorem 5.10, so ${\mathfrak {b}}_{\mathrm {Fin}}({{\overline {\Omega }}})\leq {\mathrm {add}}({\mathcal {N}})$ and ${\mathrm {cof}}({\mathcal {N}})\leq {\mathfrak {d}}_{\mathrm {Fin}}({{\overline {\Omega }}})$ .

We show the converse inequality for ${\mathrm {add}}({\mathcal {N}})$ . It is enough to prove that, whenever $F\subseteq {{\overline {\Omega }}}$ has size ${<}{\mathrm {add}}({\mathcal {N}})$ , it has some upper $\subseteq ^{\mathrm {Fin}}$ -bound. For each $\bar c\in F$ , find a function $f^{\bar c}\in {{}^\omega \omega }$ such that $\mu \big (\bigcup _{k\geq f^{\bar c}(n)}c_k\big )<\frac {1}{(n+1)2^n}$ for all $n<\omega $ . Now $|F|<{\mathrm {add}}({\mathcal {N}})\leq {\mathfrak {b}}$ , so there is some increasing $f\in {{}^\omega \omega }$ with $f(0)=0$ dominating ${\left \{ f^{\bar c} :\, \bar c\in F \right \}}$ , which means that, for any $\bar c\in F$ , $\mu \big (\bigcup _{k\geq f(n)}c_k\big )<\frac {1}{(n+1)2^n}$ for all but finitely many $n<\omega $ . By making finitely many modifications to each $\bar c\in F$ , we can assume that the previous inequality is valid for all $n<\omega $ .

For each $n<\omega $ , let $I_n:=[f(n),f(n+1))$ . Consider the functions $b^f$ and h with domain $\omega $ such that $b^f(n){\kern-1pt}:={\kern-1pt}{\left \{ {s\in {}^{I_n}\Omega } :\, {\mu \left ( \bigcup _{k\in I_n}s_k\right ){\kern-1pt}<{\kern-1pt}\frac {1}{(n+1)2^n}} \right \}}$ and $h(n){\kern-1pt}:={\kern-1pt}n+1$ . Since ${\mathbf {Lc}}(b^f,h){\mathrel {\cong _{\mathrm {T}}}} {\mathbf {Lc}}(\omega ,h){\mathrel {\cong _{\mathrm {T}}}} {\mathcal {N}}$ , we obtain that $\mathfrak {b}^{\mathrm {Lc}}_{b^f,h}={\mathrm {add}}({\mathcal {N}})$ by Theorem 5.8. Since F can be seen as a subset of $\prod b^f$ , there is some $\varphi \in {S}(b^f,h)$ such that, for any $\bar c\in F$ , $\bar c\upharpoonright I_n\in \varphi (n)$ for all but finitely many $n<\omega $ . Define $\bar d\in {}^\omega \Omega $ by $d_k:=\bigcup _{s\in \varphi (n)}s_k$ for $k\in I_n$ . Note that

$$ \begin{align*}\mu\left(\bigcup_{k\in I_n}d_k\right) = \mu\left( \bigcup_{s\in\varphi(n)} \bigcup_{k\in I_n}s_k\right)<\frac{1}{(n+1)2^n}(n+1)=\frac{1}{2^n},\end{align*} $$

thus $\bar d\in {{\overline {\Omega }}}$ . On the other hand, for any $\bar c\in F$ , $c_k\subseteq d_k$ for all but finitely many $k<\omega $ , i.e., $\bar c\subseteq ^{\mathrm {Fin}} \bar d$ .

The proof of ${\mathfrak {d}}_{\mathrm {Fin}}({{\overline {\Omega }}})\leq {\mathrm {cof}}({\mathcal {N}})$ is similar. Fix a dominating family D of size ${\mathfrak {d}}$ formed by increasing functions f such that $f(0)=0$ . For each $f\in D$ , note that $\mathfrak {d}^{\mathrm {Lc}}_{b^f,h}={\mathrm {cof}}({\mathcal {N}})$ by Theorem 5.8, so we can choose some witness $S^f\subseteq {S}(b^f,h)$ and, for each $\varphi \in S^f$ , define $d^{f,\varphi }_k:=\bigcup _{s\in \varphi (n)}s_k$ for $k\in [f(n),f(n+1))$ . Then, $E:={\left \{ \bar d^{f,\varphi } :\, \varphi ^f\in S^f,\ f\in D \right \}}$ has size ${\leq }{\mathrm {cof}}({\mathcal {N}})$ and it is ${\mathbf {S}}_{\mathrm {Fin}}$ -dominating, i.e., any $\bar c\in {{\overline {\Omega }}}$ is $\subseteq ^{\mathrm {Fin}}$ -bounded by some $\bar d\in E$ .

Proof Since ${\mathrm {Fin}}\ {\mathrel {\leq _{\mathrm {KB}}}}\ J$ , by Theorem 5.11 we obtain that ${\mathfrak {b}}_{\mathrm {Fin}}({{\overline {\Omega }}})\leq {\mathfrak {b}}_J({{\overline {\Omega }}})$ and ${\mathfrak {d}}_J({{\overline {\Omega }}})\leq {\mathfrak {d}}_J({{\overline {\Omega }}})$ . Hence, by Theorems 5.10 and 5.12, we obtain

$$ \begin{align*} {\mathrm{add}}({\mathcal{N}})\leq &\ {\mathfrak{b}}_J({{\overline{\Omega}}})\leq \min\{{\mathrm{add}}({\mathcal{N}}_J),{\mathrm{add}}({\mathcal{N}}^*_J)\} \mbox{ and }\\ \max\{{\mathrm{cof}}({\mathcal{N}}_J),{\mathrm{cof}}({\mathcal{N}}^*_J)\}\leq &\ {\mathfrak{d}}_J({{\overline{\Omega}}})\leq {\mathrm{cof}}({\mathcal{N}}). \end{align*} $$

On the other hand, by Fact 5.2 and Theorem 5.4, ${\mathrm {add}}({\mathcal {N}}_J)={\mathrm {add}}({\mathcal {N}}_J,{\mathcal {N}}_J)\leq {\mathrm {add}}({\mathcal {E}},{\mathcal {N}})={\mathrm {cov}}({\mathcal {M}})$ and ${\mathrm {non}}({\mathcal {M}})= {\mathrm {cof}}({\mathcal {E}},{\mathcal {N}}) \leq {\mathrm {cof}}({\mathcal {N}}_J,{\mathcal {N}}_J)={\mathrm {cof}}({\mathcal {N}}_J)$ , likewise for ${\mathcal {N}}^*_J$ .

Proof Consider Cohen forcing ${\mathbb {C}}$ as the poset formed by pairs of finite sequences $p=(n^p,c^p)$ such that $n^p=\langle n^p_k:\, k<m \rangle $ is an increasing sequence of natural numbers, $c^p= \langle c^p_i :\, i<n^p_{m-1}\rangle $ is a sequence of clopen subsets of ${{}^\omega 2}$ and

(6.1) $$ \begin{align} \mu\left(\bigcup_{i=n^p_k}^{n^p_{k+1}-1} c^p_i\right)<2^{-(k+1)} \end{align} $$

for any $k<m-1$ . The order is $q\leq p$ iff $n^q$ end-extends $n^p$ and $c^q$ end-extends $c^p$ . If G is ${\mathbb {C}}$ -generic over the ground model V, we define $\bar e$ by $e_k:=c^p_k$ for some $p\in G$ (this value does not depend on such a p). It is easy to show that $\bar e \in {{\overline {\Omega }}}$ .

It is enough to show that, in the Cohen extension,

$$ \begin{align*}J\cup{\left\{ {{\left\{ {k<\omega} :\, {c_k\nsubseteq e_k} \right\}}} :\, {\bar c\in {{\overline{\Omega}}}\cap V} \right\}} \mbox{ has the finite union property,}\end{align*} $$

i.e., $\omega $ cannot be covered by finitely many members of that collection. Then, the promised $J'$ will be the ideal generated by this collection.

So, in the ground model, fix $a\in J$ , $F\subseteq {{\overline {\Omega }}}$ finite and $p\in {\mathbb {C}}$ with $n^p=\langle n^p_k:\, k<m \rangle $ and $c^p= \langle c^p_i :\, i<n^p_{m-1}\rangle $ . We have to show that there is some $q\leq p$ and some $k<\omega $ such that q forces

$$ \begin{align*}k\notin a\cup\bigcup_{\bar c\in F}{\left\{ k<\omega :\, c_k\nsubseteq e_k \right\}},\end{align*} $$

that is, $k\notin a$ and $c_k\subseteq c^q_k$ for any $\bar c\in F$ . To see this, find some $n'\in \omega \smallsetminus a$ larger than $n^p_{m-1}$ such that, for any $\bar c\in F$ ,

$$ \begin{align*}\mu\left(\bigcup_{k\geq n'}c_k\right)<\frac{1}{(|F|+1)2^{m+1}}.\end{align*} $$

Define $q\in {\mathbb {C}}$ such that $n^q$ has length $m+2$ , it extends $n^p$ , $n^q_m:=n'$ and $n^q_{m+1}:=n'+1$ , and such that $c^q$ extends $c^p$ , $c^q_i=\emptyset $ for all $n^p_{m-1}\leq i< n'$ , and $c^q_{n'}:= \bigcup _{\bar c\in F}c_{n'}$ . It is clear that $q\in {\mathbb {C}}$ is stronger than p, and that $c^q_{n'}$ contains $c_{n'}$ for all $\bar c\in F$ . So $k:=n'$ works.

Proof Let $L:=\{0\}\cup {\left \{ {\alpha <\pi } :\, {\alpha \mbox { limit}} \right \}}$ . For each $\alpha \in L$ let $\bar e_\alpha $ be a ${\mathbb {P}}_{\alpha +\omega }$ -name of a Cohen real in ${{\overline {\Omega }}}$ (in the sense of the proof of Lemma 6.1) over the ${\mathbb {P}}_\alpha $ -extension. We construct, by recursion, a sequence $\langle \dot J_\alpha :\, \alpha \in L\cup \{\pi \}\rangle $ such that $\dot J_\alpha $ is a ${\mathbb {P}}_\alpha $ -name of an ideal on $\omega $ and ${\mathbb {P}}$ forces that $\dot J_\alpha \subseteq \dot J_\beta $ when $\alpha <\beta $ . We let $\dot J_0$ be (the ${\mathbb {P}}_0$ -name of) any ideal $J_0$ in the ground model.Footnote ⁷ For the successor step, assume we have constructed $\dot J_\alpha $ at a stage $\alpha \in L$ . By Lemma 6.1, we obtain a ${\mathbb {P}}_{\alpha +\omega }$ -name $\dot J_{\alpha +\omega }$ of an ideal extending $\dot J_{\alpha }$ , such that $\bar e_\alpha \ \subseteq ^{\dot J_{\alpha +\omega }}$ -dominates all the $\bar c\in {{\overline {\Omega }}}$ from the ${\mathbb {P}}_{\alpha }$ -extension. For the limit step, when $\gamma $ is a limit point of L (which includes the case $\gamma =\pi $ ), just let $\dot J_\gamma $ be the ${\mathbb {P}}_\gamma $ -name of $\bigcup _{\alpha <\gamma }\dot J_\alpha $ . This finishes the construction.