Proof. We verify this only for Laver trees and for the homogeneous property as the proof for the other cases is similar. Let then $T\in \mathbb {L}_{\mathcal {I}}$ and for each $\sigma \in split(T)$ fix a function $\pi _{\sigma }:\omega \rightarrow succ_{T}(\sigma )$ such that $\pi _{\sigma }^{-1}[X]$ is $\mathcal {I}$ -positive iff $X\subseteq succ_{T}(\sigma )$ is. Define inductively a homeomorphism $h:\omega ^{\omega }\rightarrow [T]$ such that $h(x)=stem(T) ^{\frown } \pi _{\sigma _{0}}(x(0))^{\frown } \pi _{\sigma _{1}}(x(1))^{\frown }...$ where $\sigma _{n}=h(x)|_{|stem(T)|+n}$ . It is easy to see that h satisfies the required properties.

Proof. Let $t^{0}_{\mathcal {B}}$ be any of these two ideals. If $\{X_{n}:n\in \omega \}\subseteq t^{0}$ and $T\in \mathbb {T}$ we construct a fusion sequence $\{T_{n}:n\in \omega \}$ such that $[T_{n}]\cap X_{n}=\emptyset $ . To construct $T_{n+1}$ out of $T_{n}$ in the case of Laver, we use pure decision property, while for Miller, this can be done directly. Then, if S is a fusion of $T_{n}$ ’s we clearly have $[S]\cap \bigcup _{n}X_{n}=\emptyset $ .

Proof. Assume towards a contradiction that there is no such $S\leq _{0}T$ and let ${T_{0}=T}$ . Inductively we construct a sequence $\{T_{n}:n\in \omega \}\subseteq \mathbb {L}_{\mathcal {I}}$ such that for every $n\in \omega $ we have $T_{n+1}\leq _{n}T_{n}$ and for every $\sigma \in split_{n}(T),$ there is no $S\leq T_{n}|{\sigma }$ deciding the value of $\dot {x}$ . Assume that $T_{n}$ has been constructed. For each $\sigma \in split_{n}(T),$ the set $A_{\sigma }=:\{i\in succ_{T_{n}}(\sigma ): \exists S\leq _{0}T_{n}|_{\sigma ^{\frown }i} \text { deciding }\dot {x}\}$ is in $\mathcal {I}$ . We then construct $T_{n+1}$ by letting $succ_{T_{n+1}}(\sigma )=succ_{T_{n}}(\sigma )\setminus A_{\sigma }$ for $\sigma \in split_{n}(T)$ . Finally, if S is the fusion of $T_{n}$ ’s then there is no $R\leq S$ deciding $\dot {x}$ , a contradiction.

Proof. Let $T\Vdash "\dot {x}\in 2^{\omega }$ ”. Inductively we construct a sequence $\{T_{n}:n\in \omega \}\subseteq \mathbb {L}_{\mathcal {I}}$ such that $T_{0}=T$ , for every $n\in \omega T_{n+1}\leq _{n}T_{n}$ and for every $\sigma \in split_{n}(T)$ , $T_{n}|{\sigma }\Vdash "\dot {x}|_{n}=x_{\sigma }"$ . Then, if S is the fusion of $T_{n}$ ’s we define $f:[S]\rightarrow 2^{\omega }$ as $f(y)=\bigcup x_{y|_{n}}$ . Clearly f is continuous and maps generic real into $\dot {x}$ .

Proof. Suppose that $\pi :X\rightarrow 2^{<\omega }$ is such that $\pi ^{-1}[Y]\in \mathcal {I}$ for every $nwd$ -positive $Y\subseteq 2^{<\omega }$ . The real $\dot {c}$ given by $\dot {c}=\pi (\dot {r}_{gen}(0))^{\frown }\pi (\dot {r}_{gen}(1))^{\frown }...$ is a Cohen real.

Proof. Assume $\mathbb {L}_{\mathcal {I}}$ has the constant or 1–1 property, $T\in \mathbb {M}_{\mathcal {I}}$ and $f:[T]\rightarrow 2^{\omega }$ is continuous. Let $T'\in \mathbb {L}_{\mathcal {I}}$ be the unique tree which is constructed from T by “removing the parts in between the splinodes”, i.e., $stem(T')=stem(T)$ and if for any tree T, if $s^{T}:\omega ^{<\omega }\rightarrow split(T)$ is the unique $\subseteq $ -isomorphism which preserves lexicographical order then $succ_{T'}(s^{T'}(stem^{\frown }\tau ))=succ_{T}(s^{T}(stem^{\frown }\tau ))$ for each $\tau \in \omega ^{<\omega }$ . Next let $f':[T']\rightarrow 2^{\omega }$ be a continuous function defined for $x\in [T']$ as $f'(x)=f(\bigcup _{k}s^{T}(\tau _{k})),$ where $\{\tau _{k}:k\in \omega \}\subseteq \omega ^{<\omega }$ is such that $\{x_{n}: n\in \omega \}\cap split(T')=\{s^{T'}(\tau _{k}):k\in \omega \}$ . Now clearly any restriction $S'\in \mathbb {L}_{\mathcal {I}}$ , $S'\leq T'$ with $f'|_{[S']}$ being constant or 1–1 produce $S\in \mathbb {M}_{\mathcal {I}}$ , $S\leq T$ with constant or 1–1 restriction $f|_{[S]}$ .

Proof. Assume that $T\in \mathbb {L}_{\mathcal {I}}$ and that $f:[T]\rightarrow 2^{\omega }$ is continuous. Using the pure decision property, we construct a sequence $\{T_{n}:n\in \omega \}\subseteq \mathbb {L}_{\mathcal {I}}$ such that $T_{0}=T$ , $T_{n+1}\leq _{0} T_{n}$ and that $T_{n}$ decides the value of $f(\dot {r}_{gen})|_{n}$ to be $x_{n}$ . Then, as $\mathcal {I}$ is $P^+$ , there is a pseudointersection $A\in \mathcal {I}^{+}$ of the family $\{succ_{T_{n}}(stem(T)):n\in \omega \}$ . Let ${x=\bigcup _{n}x_{n}\in 2^{\omega }}$ and let $S\in \mathbb {L}_{\mathcal {I}}$ be such that $stem(S)=stem(T)$ , $succ_{S}(stem(S))=A$ and for $n\in A$ , $S|_{stem(S)^{\frown }n}=T_{k}|_{stem(S)^{\frown }n,}$ where $k\in \omega $ is the last such that $n\in succ_{T_{k}}(stem(S))$ .

Proof. Using the P $^+$ property, it is possible to restrict to some $A_n' \subseteq A_n$ such that $f_n[A_n']$ converges to z, and therefore $f_n[A_n']$ is discrete. Let

$$ \begin{align*}B_n = A_n' \setminus f_n^{-1}(\langle z \restriction n \rangle) \end{align*} $$

then $B_n \subseteq A_n$ , $B_n \in \mathcal {I}^+$ and, since $\bigcup _{n \in \omega } f_n[B_n] \setminus f_n^{-1}(\langle z \restriction n \rangle )$ is finite for every $k \in \omega $ , then $\bigcup _{n \in \omega } f_n[B_n]$ is discrete.

Proof. Let $F: A \rightarrow \omega be$ defined by $F(n) = k$ if and only if $f(n) \restriction k = z \restriction k$ but $f(n) \restriction k + 1 \neq z \restriction k + 1$ . Define $G : B \rightarrow \omega $ in a similar way, using g instead of f. Notice that both F and G are finite to one functions, so there must be $A", B" \subseteq \omega $ such that $A_0 = F^{-1}(A"), A_1 = F^{-1}(\omega \setminus A")$ , $B_0 = G^{-1}(B"), B_1 = G^{-1}(\omega \setminus B")$ are all in $\mathcal {I}^+$ . Then it is easy to show that, either $A_0 \cap B_0$ and $A_1 \cap B_1$ are in $\mathcal {I}^+$ , or $A_1 \cap B_0$ and $A_0 \cap B_1$ are in $\mathcal {I}^+$ .

Proof. By Lemma 2.10, we may assume that $\bigcup _{n \in \omega } f_n[B_n]$ is discrete. Recursively, we are going to construct a family of sequences $\{ S^i_j : i \leq j, j \in \omega \}$ such that, for each $ i \leq j, j \in \omega $ ,

1. (1) $S^i_j \subseteq B_j$ is $\mathcal {I}^+$ ,

2. (2) $S^j_j \subseteq S^i_j$ ,

3. (3) for $m \geq i$ , $f_m[S^i_m] \cap f_{i}[S^i_{i}]$ is finite.

Assume we were able to carry out the construction of $\langle S^i_j : i \leq j, j \in \omega \rangle $ for all $i < k$ , we are going to construct $\langle S^k_j : k \leq j, j \in \omega \rangle $ : Let $S_k' = \bigcap _{i < k}S^i_k$ and recursively apply the previous remark to get two families $\langle S^{\prime }_\ell : k < \ell \rangle $ and $\langle S^k_\ell : k < \ell \rangle $ such that:

• • $S^{\prime }_\ell , S^k_\ell \in \mathcal {I}^+$ ,

• • $S^{\prime }_{\ell + 1} \subseteq S^{\prime }_\ell \subseteq S_k'$ ,

• • for all $i < k$ , $S^k_\ell \subseteq S^i_\ell $ ,

• • $f_k [S^{\prime }_\ell ] \cap f_\ell [S^k_\ell ] = \emptyset $ .

Finally let $S^k_k$ be a positive pseudo-intersection of the family $\langle S^{\prime }_\ell : k < \ell \rangle $ . It is routine to check that $\langle S^k_j : k \leq j, j \in \omega \rangle $ satisfies the properties we are looking for.

Finally, let $C_n = S^n_n \setminus f_n^{-1}(\bigcup _{i < n}f_i [S^i_i])$ . It is easy to see that the family ${\langle C_n : n \in \omega \rangle }$ satisfies the desired properties.

Proof. Assume that $T\in \mathbb {L}_{\mathcal {I}}$ and $ f:[T]\rightarrow 2^{\omega }$ is continuous. Using a standard fusion argument and Lemma 2.9, we can get a tree $S\leq _{0}T$ and a collection $\{z_{\sigma }:\sigma \in split(S)\}\subseteq 2^{\omega }$ such that, for each $\sigma \in split(S),$ we have

(1) $$ \begin{align} \forall N\in\omega \{i\in succ_{S}(\sigma): f[[S|_{\sigma^{\frown}i}]]\nsubseteq [z_{\sigma}|_{N}]\}\in\mathcal{F}in. \end{align} $$

This, in particular, means that the sequence $\{z_{\sigma ^{\frown }n}:n\in \omega \}$ converges to $z_{\sigma }$ . To decide whether the restriction we are looking for is constant or 1–1, we will use a rank argument. Define $rank:split(S)\rightarrow \omega _{1}$ as:

$$ \begin{align*}rank(\sigma) = \begin{cases} 0 & \text{ if } \{n\in\omega: z_{\sigma^{\frown}n}\neq z_{\sigma}\}\in\mathcal{I}^{+}, \\ \alpha & \text{ if } \alpha \text{ is the smallest such that }\{n\in\omega: rank(\sigma^{\frown}n)<\alpha \}\in\mathcal{I}^{+}. \end{cases} \end{align*} $$

Now we consider two cases.

Case 1: For some $\sigma \in split(S),$ the rank is not defined. This implies that $\{n\in \omega : rank(\sigma ^{\frown }n)$ is defined $\}\in \mathcal {I}$ . Using this, together with a standard fusion argument, we trim S to get an $R\leq _{0} S_\sigma $ such that for all $\tau \in split(R),$ we have $z_{\tau }=z_{\sigma }$ . We conclude that the restriction $f|_{[R]}$ is constant with value $z_{\sigma }$ .

Case 2: The rank is defined for all $\sigma \in split(S)$ . We will inductively construct an increasing sequence $\{F_{n}:n\in \omega \}$ of well-founded subtrees of S and disjoint clopen sets $\{V_{\sigma |_{|\sigma |-1}}:\sigma \in term(F_{n})\}$ such that for all $\sigma \in \omega ^{<\omega }$ and $n\in \omega ,$ we have:

1. A. if $\sigma \in F_{n}$ is not a terminal node of $F_{n}$ then $succ_{F_{n}}(\sigma )\in \mathcal {I}^+$ ,

2. B. if $\sigma \in F_{n+1}\setminus F_{n}$ , then $\sigma $ extends a terminal node of $F_{n}$ ,

3. C. for each $\sigma \in term(F_{n}),$ we have $f[[S_{\sigma }]]\subseteq V_{\sigma |_{|\sigma |-1}}$ .

Assume for a moment that such a sequence has been constructed and let $R=\bigcup _{n}F_{n}$ . Clearly $R\leq S$ , and the condition A implies $R\in \mathbb {L}_{\mathcal {I}}$ . We claim that the restriction $f|_{[R]}$ is 1–1. Let $x,y\in [R]$ , and $x\neq y$ . Then there are $\sigma ,\tau \in \omega ^{<\omega }$ , $\sigma \neq \tau $ and $n\in \omega $ such that $\sigma \subseteq x, \tau \subseteq y$ and $\sigma ,\tau $ are terminal nodes of $F_{n}$ . The condition C guarantees that $f(x)\in V_{\sigma |_{|\sigma |-1}}$ and $f(y)\in V_{\tau |_{|\tau |-1}}$ so $f(x)\neq f(y)$ .

To finish the proof, we will show how to construct such sequence: Let $F_{0}=\{stem(S)\}$ and assume that $F_{n}$ has been built. Fix a terminal node $\sigma $ of $F_{n}$ . As the rank of $\sigma $ is defined, there is a well-founded tree $G_{\sigma }\subseteq \{\tau \in \omega ^{<\omega }:\tau \subseteq \sigma $ or $\sigma \subseteq \tau \}$ such that all terminal nodes of $G_\sigma $ have rank $0$ and all non-terminal nodes of $G_\sigma $ split into $\mathcal {I}$ -positive sets. We then apply Lemma 2.12 to the family $\{A_{\tau }:\tau \in term(G_{\sigma })\}\subseteq \mathcal {I}^{+}$ , $A_{\tau }=\{n\in \omega : z_{\tau ^{\frown }n}\neq z_{\tau }\}$ , and the functions $f_{\tau }:A_{\tau }\rightarrow 2^{\omega }$ defined as $f_{\tau }(n)=z_{\tau ^{\frown }n}$ . We obtain $B_{\tau }\subseteq A_{\tau }$ , $B_{\tau }\in \mathcal {I}^{+}$ such that $D=\bigcup \{f_{\tau }[B_{\tau }]:\tau \in term(G_{\sigma })\}$ is a discrete subset of $2^{\omega }$ and for $\tau \neq \tau '$ , we have that $f_{\tau }[B_{\tau }] \cap f_{\tau '}[B_{\tau '}] =\emptyset $ . Let then $\{U_{z}:z\in D\}$ be a collection of disjoint clopen sets such that $z\in U_{z}$ for each $z\in D$ . By condition (1), we can remove $\mathcal {I}$ -many points from each $B_{\tau }$ to obtain $\mathcal {I}$ -positive sets $C_{\tau }\subseteq B_{\tau }$ such that for every $m\in C_{\tau }$ , we have $f[[S|_{\tau ^{\frown }m}]]\subseteq U_{z_{\tau ^{\frown }m}}$ . Let $H_{\sigma }$ be the tree generated by the set $\{\tau ^{\frown }m:\tau \in term(G_{\sigma })$ and $m\in C_{\tau }\}$ . Finally, we unfix $\sigma $ and let $F_{n+1}$ be the union of all $H_{\sigma }$ ’s, where $\sigma $ is a terminal node of $F_{n}$ . This finishes the construction.

Question 2.15. Does $\mathbb {L}_{\mathcal {Z}}$ or $\mathbb {L}_{\mathcal {F}in\times \mathcal {F}in}$ have the constant or 1–1 property?

Proof. To prove $\mathrm {add}_{\omega }^*(\mathcal {I})\leq \mathrm {add}^*(\mathcal {I})$ let $\{X_{\alpha }:\alpha <\kappa \}\subseteq \mathcal {I}$ with $\kappa <\mathrm {add}_{\omega }^*(\mathcal {I})$ . There are then $\{Y_{n}: n\in \omega \}\subseteq \mathcal {I}$ such that each $X_{\alpha }$ is almost-contained in one of the $Y_{n}$ ’s. Then, any pseudo-union $Y\in \mathcal {I}$ of the $Y_{n}$ ’s is a pseudo-union of the $X_{\alpha }$ ’s. This finishes the proof.

Proof. Let , $T\in \mathbb {L}_{\mathcal {I}}$ and suppose we are given a collection $\{L_{\phi _{\alpha }}:\alpha <\kappa \}$ such that $\phi _{\alpha }:\omega ^{<\omega }\rightarrow \omega $ . For each $\sigma \in \omega ^{<\omega }$ , as $\kappa <\mathrm {add}^{*}_{\omega }(\mathcal {I}),$ there is an increasing sequence $\{C^{\sigma }_{n}:n\in \omega \}\subseteq \mathcal {I}$ such that for each $\alpha <\kappa ,$ there is $n_{\alpha }\in \omega $ with $\phi _{\alpha }(\sigma )\subseteq ^{*}C^{\sigma }_{n_{\alpha }}$ . For each $\alpha <\kappa ,$ let $g_{\alpha }:\omega ^{<\omega }\rightarrow \omega $ be defined as $g_{\alpha }(\sigma )=n_{\alpha }$ and $f_{\alpha }:\omega ^{<\omega }\rightarrow \omega $ be defined such that $\phi _{\alpha }(\sigma )\subseteq C^{\sigma }_{n_{\alpha }}\cup f_{\alpha }(\sigma )$ . Now let $g:\omega ^{<\omega }\rightarrow \omega $ and $f:\omega ^{<\omega }\rightarrow \omega $ be such that g dominates all $g_{\alpha }$ ’s and f dominates all $f_{\alpha }$ ’s. Let $\phi :\omega ^{<\omega }\rightarrow \mathcal {I}$ be defined as $\phi (\sigma )=C^{\sigma }_{g(\sigma )}\cup f(\sigma )$ . We claim that $\bigcup _{\alpha }L_{\phi _{\alpha }}\subseteq L_{\phi }$ . This will finish the proof. Let $x\in L_{\phi _{\alpha }}$ . Then for infinitely many n’s, we have $x(n)\in \phi _{\alpha }(x|_{n})\subseteq C^{x|_{n}}_{n_{\alpha }}\cup f_{\alpha }(x|_{n})\subseteq C^{x|_{n}}_{g(x|_{n})}\cup f(x|_{n})=\phi (x|_{n})$ as the sequence of the $C^{x|_{n}}_{k}$ ’s, $k\in \omega $ is increasing.

Proof. We will focus only on $\mathbb {M}_{\mathcal {I}}$ as the proof for $\mathbb {FM}$ is similar. By Theorem 3.3, we only have to show that, for any $T \in \mathbb {M}_{\mathcal {I}}$ , the inequalities

hold. Assume that and a family $\{\phi _{\alpha }:\alpha <\kappa \},$ where $\phi _{\alpha }:\omega ^{<\omega }\rightarrow \mathcal {I}$ . Let $M^{\prime }_{\phi _{\alpha }}=\{x\in \omega ^{<\omega }:\forall n\in \omega x(n)\notin \phi _{\alpha }(x|_{n})\}$ . Inductively we will construct a sequence $\{F_{n}:n\in \omega \}$ of well-founded subtrees of T such that:

1. (1) $F_{n+1}$ is an end-extension of $F_{n}$ and $term(F_{n})\subseteq split(T)$ ,

2. (2) for every $\sigma \in term(F_{n}),$ if $n\in succ_{T}(\sigma )$ then there is unique $\tau \in term(F_{n+1})$ such that $\sigma ^{\frown }n \subseteq \tau $ ,

3. (3) for every $\sigma \in term(F_{n}),$ there are only finitely many $\tau \in term(F_{n+1})$ such that $[\tau ]\cap M^{\prime }_{\phi _{\alpha }}=\emptyset $ .

Assume that $F_{n}$ has been constructed and fix $\sigma \in term(F_{n})$ . For each $n\in succ_{T}(\sigma ),$ we can choose a point $x^{\sigma }_{n}\in [\sigma ^{\frown }n]\setminus \bigcup _{\alpha } M^{\prime }_{\phi _{\alpha }}$ as $\kappa <\mathrm {cov}(m^{0}_{\mathcal {B}}(\mathcal {I}) |_{[T]})$ (and clearly this cardinal is the same when restricted to basic clopen sets). For every $\alpha <\kappa ,$ there is $f_{\alpha }\in \omega ^{\omega }$ such that

$$ \begin{align*} \forall n\in succ_{T}(\sigma) x^{\sigma}_{n}|_{f_{\alpha}(n)}\notin M^{\prime}_{\phi_{\alpha}}. \end{align*} $$

As , there must be a $g\in \omega ^{\omega }$ such that it dominates all $f_{\alpha }$ ’s. For $n\in \omega \land \sigma \in term(F_{n}),$ let $\rho ^{n}_{\sigma }$ be the shortest splitnode of T which extends $x^{\sigma }_{n}|_{g(n)}$ . Let then $F_{n+1}$ be the tree generated by the set $\{\rho ^{n}_{\sigma }:n\in \omega \land \sigma \in term(F_{n})\}$ . Let $S=\bigcup _{n}F_{n}\in \mathbb {M}_{\mathcal {I}}$ . The condition (3) implies that $[S]\cap M^{\prime }_{\phi _{\alpha }}$ is compact and therefore bounded by some $h_{\alpha }\in \omega ^\omega $ . Let $h\in \omega ^\omega $ be such that it dominates all $h_{\alpha }$ ’s and let $R\in \mathbb {M}_{\mathcal {I}}$ be such that $[R]$ consists of branches of $[S]$ which are not dominated by h. Then, clearly $[R]\cap \bigcup _{\alpha }M^{\prime }_{\alpha }=\emptyset $ for all $\alpha <\kappa $ .

Proof. Let $S \leq T$ . Find $\sigma \in \omega ^{<\omega }$ and $S' \leq S$ such that $S' \Vdash "\sigma \subseteq f(\dot {r}_{gen}) \text { and } \langle \sigma \rangle \cap N = \emptyset "$ . Use the continuity of f to find $\tau \in S'$ such that $f[\langle \tau \rangle ] \subseteq \langle \sigma \rangle $ . If $S" = S'|_\tau $ , then $S" \cap f^{-1}[N] = \emptyset $ follows easily.

Proof. Let $T \in \mathbb {T}$ and let $f : [T] \rightarrow \omega ^\omega $ be a continuous function such that $T \Vdash "f(\dot {r}_{gen}) \text { is }\text {Cohen}".$ We are going to prove that $\mathrm {cov}(t_0|_{[T]}) \leq \mathrm {cov}(\mathcal {M})$ and ${\mathrm {is}(\mathbb {T|_{T}}) \leq \mathfrak {b}}$ . The first inequality follows directly from Proposition 3.9. To prove the second one, assume towards a contradiction that $\mathrm {is}(\mathbb {T|_{T}})> \mathfrak {b}$ : Let $\mathbf {M}$ be an elementary substructure of some $H(\kappa )$ such that $f, T, \mathbb {T} \in \mathbf {M}$ , $\mathbf {M}\cap \omega ^\omega $ is unbounded and $|\mathbf {M}| = \mathfrak {b}$ . Then, since $|\mathbf {M}| < \mathrm {is}(\mathbb {T}|_T)$ there must be an $R \in \mathbb {T}$ , $R \leq T$ such that ${[R] \cap X = \emptyset }$ for every $X \in t^0_{\mathcal {B}} \cap \mathbf {M}$ . By Proposition 3.9, if $N \subseteq \omega ^\omega $ is a nowhere-dense set and $N\in \mathbf {M}$ , then $[R] \cap f^{-1}[N] = \emptyset $ and therefore $f[[R]] \cap N = \emptyset $ . But then, this means that $f[[R]]$ contains exclusively Cohen reals over $\mathbf {M}$ . Since $\sigma $ -compact sets are meager, and $R \Vdash "f(\dot {r}_{gen}) \text { is Cohen}"$ , then $f[[R]]$ cannot be $\sigma $ -compact, so by Theorem 3.3 for Miller trees, $f[[R]]$ must contain the branches of a Miller tree. Therefore, by Theorem 3.8, there must be a real $d \in \omega ^\omega $ such that d dominates every $y \in \omega ^\omega \cap \mathbf {M}$ , which is impossible since $\mathbf {M}$ contains an unbounded family.

Proof. The proof follows from Theorems 3.7 and 3.10.

Question 3.13. Is is consistent that $\mathrm {cof}(fm^{0})=\mathfrak {c}$ ?

Proof. Clearly $\mathrm {add}(t^{0})\leq \mathrm {is}(\mathbb {T})$ and $\mathrm {cov}(t^{0}) \leq \mathrm {cov}(t_{\mathcal {B}}^{0})$ , so the conclusion follows from Theorem 3.10 and the fact that .

Proof. Let $\{B_\alpha : \alpha \in \kappa \}$ be a witness for $\mathrm {add}^*_{\omega }(\mathcal {I})$ . For each $\alpha \in \kappa ,$ define $\varphi _\alpha : \omega ^{<\omega } \rightarrow \mathcal {I}$ as $\varphi _\alpha (s) = B_\alpha $ . We will see that $\bigcup _{\alpha \in \kappa } L_{\varphi _\alpha } \notin \ell ^{0}_{\mathcal {B}}(\mathcal {I})$ .

Aiming for a contradiction, assume that there is a sequence $\{ \phi _{n}: n\in \omega \}$ such that, for all $n\in \omega , \phi _{n} : \omega ^{<\omega } \rightarrow \mathcal {I}$ and $\bigcup _{\alpha \in \kappa } L_{\varphi _\alpha } \subseteq \bigcup _{i \in \omega } L_{\phi _i}$ . Then find $\alpha \in \kappa $ such that for all $i \in \omega $ and all $s \in \omega ^{<\omega }$ , $B_\alpha \setminus \bigcup _{\langle i,s\rangle \in F} \phi _i(s) \neq \emptyset $ for every finite $F \subseteq \omega \times \omega ^{<\omega }$ . We will show that $L_{\varphi _\alpha } \setminus \bigcup _{i \in \omega } L_{\phi _i} \neq \emptyset $ : Recursively construct $x:\omega \rightarrow \omega $ such that, for all $n \in \omega :$

• • $x(n) \in B_\alpha $ ,

• • $x(n) \notin \phi _i(x | k)$ for every $i,k \leq n$ .

Since $x(n) \in \varphi _\alpha (x | n)$ for all $n\in \omega $ , then $x \in L_{\varphi _\alpha }$ . On the other hand, for all $i\in \omega , x \notin \phi _i (x | n)$ for all $n> i$ , therefore $x \notin L_{\phi _i}$ .

Proof. Notice that an unbounded family of functions witnesses $\mathrm {add}_{\omega }^*(\mathcal {F}in\times \mathcal {F}in)$ and less than sets from $\mathcal {F}in\times \mathcal {F}in$ cannot, therefore .

Proof. We will focus only on $\mathcal {ED}$ as the proof for $\mathcal {ED}_{fin}$ is identical. Let $[N,F]=:\{(n,m):n<N $ or $m\in \{f(n):f\in F\}\}$ for $N\in \omega $ and $F\in [\omega ^\omega ]^{N}$ . Clearly any $A\in \mathcal {ED}$ can be covered by the set of this form. Fix an eventually different family $\{e_{\alpha }:\alpha <\omega _{1}\}\subseteq \omega ^{\omega }$ . We claim that $\{e_{\alpha }:\alpha <\omega _{1}\}$ is a witness for $add_{\omega }^{*}(\mathcal {ED})$ . To show this, assume that we are given $\{A_{n}:n\in \omega \}\subseteq \mathcal {ED,}$ where $A_{n}=[N_{n},F_{n}]\in \mathcal {ED,}$ where $N_{n}\in \omega $ , $F_{n}\in [\omega ^\omega ]^{<\omega }$ . Observe that for each $f\in \omega ^{\omega }$ , the set $\{\alpha \in \omega _{1}: e_{\alpha }\text { is almost contained in }\bigcup F_{n}\}$ is of cardinality at most $|F_{n}|$ . Choose $\alpha _{0}\in \omega _{1}\setminus \bigcup _{n}\{\alpha : e_{\alpha }\text { is almost contained in }\bigcup F_{n}\}$ . Clearly $e_{\alpha _{0}}$ is as desired.

Proof. Let $\{x_{\alpha }:\alpha <\omega _{1}\}\subseteq 2^{\omega }$ . Define $[F]=\{C\in \Omega : C\cap F\neq \emptyset \},$ where ${F\in [2^{\omega }]^{<\omega }}$ . We claim that the collection $\{[\{x_{\alpha }\}]:\alpha <\omega _{1}\}$ is a witness for $\mathrm {add}_{\omega }^{*}(\mathcal {S})$ . To show this assume that we are given $\{A_{n}:n\in \omega \}\subseteq \mathcal {S,}$ where $A_{n}=[F_{n}]$ and $F_{n}\in [2^{\omega }]^{<\omega }$ . Enumerate $\bigcup _{n}F_{n}$ as $\{y_{n}:n\in \omega \}$ and pick $x_{\alpha }\notin \{y_{n}:n\in \omega \}$ . Clearly, for each $n\in \omega $ , there are infinitely many clopen sets C of measure $1/2$ which contain $x_{\alpha }$ but not $y_{n}$ . This finishes the proof.

Proof. It is known that $\mathcal {R}\leq _{K} conv$ (see [Reference Hrušák11]), so by Proposition 4.6, it is enough to show that $\mathrm {non}_{\omega }^{*}(conv) = \omega _1$ . Let $\{x_{\alpha }:\alpha <\omega _{1}\}\subseteq 2^{\omega }$ , for each $\alpha <\omega _{1}$ choose $A_{\alpha }\subseteq \mathbb {Q}$ to be a sequence converging to $x_{\alpha }$ . We claim that $\{A_{\alpha }:\alpha \in \omega _{1}\}\subseteq conv$ witnesses $\mathrm {non}_{\omega }^{*}(conv)$ . If $\{B_{n}:n\in \omega \}\subseteq conv$ is given, choose $\alpha \in \omega _{1}$ such that $x_{\alpha }$ is not a convergence point of any $B_{n}$ . Clearly then $A_{\alpha }\cap B_{n}$ is finite.

Proof. To prove the second inequality, simply notice that if $\{N_{\alpha }:\alpha <\kappa \}$ is a collection of nowhere dense subsets of $2^{\omega }$ such that $\bigcup _{\alpha }N_{\alpha }$ is not meager then the nowhere dense trees $\{T_{\alpha }:\alpha <\kappa \}\subseteq nwd$ , $[T_{\alpha }]=N_{\alpha }$ witness for $\mathrm {add}_{\omega }^*(nwd)$ .

To prove the first inequality, assume that and that we are given $\{T_{\alpha }:\alpha <\kappa \}$ with $T_{\alpha }\in nwd$ . We define a partial order $\mathbb {P}$ consisting of finite partial functions from $\omega $ to $nwd\times \omega $ . Let $p=\{(A^{p}_{k},i^{p}_{k}):k<n\}$ . The order is given by $p\leq q$ if $dom(q)\subseteq dom(p)$ and for each $k\in dom(q),$ we have $i^{p}_{k}>i^{q}_{k}$ , $A^{q}_{k}\subseteq A^{p}_{k}$ and $A^{p}_{k}\cap 2^{i^{q}_{k}}=A^{p}_{k}\cap 2^{i^{q}_{k}}$ . First, observe that $\mathbb {P}$ is $\sigma $ -centered: if p and q have the same domains, and for each $k\in dom(p)$ , $i^{p}_{k}=i^{q}_{k}=i$ and $A^{p}_{k}\cap 2^{i}=A^{q}_{k}\cap 2^{i}$ , then p and q are compatible. Second, note that for any $\alpha <\kappa ,$ the set $D_{\alpha }=\{ p\in \mathbb {P}: A^{p}_{k}=T_{\alpha } \text { for some } k\in dom(p)\}$ is dense in $\mathbb {P}$ . If $G\subseteq \mathbb {P}$ is generic filter over the family $\{D_{\alpha }:\alpha <\kappa \}$ then, the sequence $\{B_{n}:n\in \omega \}\subseteq nwd$ defined as $B_{n}=\bigcup \{A^{p}_{n}\cap i^{p}_{n}: p\in G\}$ is such that for each $\alpha ,$ there is $n\in \omega $ with $T_{\alpha }\subseteq B_{n}$ which finishes the proof.

Question 4.10. Is $\mathrm {add}_{\omega }^*(nwd)=\mathrm {add}(\mathcal {M})$ true?

Question 4.11. What are the values of $\mathrm {non}_{\omega }^{*}(\mathcal {S})$ , $\mathrm {non}_{\omega }^{*}(\mathcal {ED}),$ and $\mathrm {non}_{\omega }^{*}(\mathcal {ED}_{fin})$ ?

Proof. It follows from Theorems 4.9 and 4.1 together with Propositions 3.6 and 4.3.

Proof. The first inequality follows from the fact that $M_{\phi }$ ’s are meager. For the second inequality, let be a dominating family. If $\phi _{\alpha }$ is such that $\phi _{\alpha }(\sigma )=f_{\alpha }(|\sigma |)$ then the sets $M_{\phi _{\alpha }}$ cover the entire $\omega ^{\omega }$ . For the last inequality, pick an unbounded family instead of a dominating one.

Proof. It follows from Proposition 2.7, Theorem 4.1, and Proposition 4.14 that $\mathrm {cov}(\ell ^{0}_{\mathcal {B}}(nwd))\leq \mathrm {add}(\mathcal {M})$ . To prove $\mathrm {add}(\mathcal {M})\leq \mathrm {cov}(\ell ^{0}_{\mathcal {B}}(nwd))$ assume that we are given a family ${\phi _{\alpha }:\alpha <\kappa }$ , $\phi _{\alpha }:\omega ^{<\omega }\rightarrow nwd,$ where $\kappa <\mathrm {add}(\mathcal {M})$ . Without loss of generality, we may assume that $\phi _{\alpha }(\sigma )$ is a nowhere dense subtree of $2^{<\omega }$ . As ${\kappa <\mathrm {cov}(\mathcal {M})}$ , we pick $z\in 2^{\omega }\setminus \bigcup _{\alpha , \sigma }[\phi _{\alpha }(\sigma )]$ . For each $\alpha <\kappa $ define $f_{\alpha }:\omega ^{<\omega }\rightarrow \omega $ such that $f_{\alpha }(\sigma )$ is any natural number $m\in \omega $ such that $x|_{m}\notin \phi _{\alpha }(\sigma )$ . As , there is $g:\omega ^{<\omega }\rightarrow \omega $ which dominates all $f_{\alpha }$ ’s. Construct $x\in \omega ^{\omega }$ inductively in such a way that $x(n)=z|_{g(x|_{n})}$ . It follows that $x\notin \bigcup _{\alpha }L_{\phi _{\alpha }}$ .

Proof. Let $\pi {\kern-1pt}:{\kern-1pt}dom(\mathcal {I}_{1}){\kern-1pt}\rightarrow{\kern-1pt} dom(\mathcal {I}_{0})$ be Katetov reduction and let ${\kappa {\kern-1pt}<{\kern-1pt}\mathrm {cov}(\ell ^{0}_{\mathcal {B}}(\mathcal {I}_{1}))}$ . Given a family $\{\phi _{\alpha }:\alpha \in \kappa \}$ such that $\phi _{\alpha }:dom(\mathcal {I}_{0})^{<\omega }\rightarrow \mathcal {I}_{0}$ , define $\psi _{\alpha }:dom(\mathcal {I}_{1})^{<\omega }\rightarrow \mathcal {I}_{1}$ as $\psi _{\alpha }(\sigma )=\pi ^{-1}[\phi _{\alpha }(\pi (\sigma ))]\in \mathcal {I}_{1}$ . Now, if $x\notin \bigcup _{\alpha }L_{\psi _{\alpha }}$ then $\pi \circ x\notin \bigcup _{\alpha }L_{\phi _{\alpha }}$ which finish the proof.

Proof. For $\mathcal {F}in\times \mathcal {F}in$ , it follows from Corollary 4.13 and Proposition 4.14. For the rest, it follows from Proposition 4.16.

Question 4.18. Do one of the forcings $\mathbb {L}_{\mathcal {S}}$ , $\mathbb {L}_{\mathcal {S}um}$ , $\mathbb {L}_{\mathcal {Z}}$ , or $\mathbb {L}_{\mathcal {ED}_{fin}}$ add a Cohen real?

Question 4.19. Is it possible to consistently distinguish the covering numbers $\mathrm {cov}(m^{0}_{\mathcal {B}}(\mathcal {I}))$ for different $\mathcal {I}$ ’s? In particular, is consistent?

Proof. Let and let $\{\phi _{\alpha }:dom(\mathcal {S})^{<\omega }\rightarrow \mathcal {S}:\alpha <\kappa \}$ . As $\kappa <\mathrm {non}(\mathcal {E}),$ the set $\bigcup _{\alpha <\kappa }\bigcup _{\sigma }\phi _{\alpha }(\sigma )$ can be covered by countable increasing union $\bigcup _{n}N_{n}$ of closed sets of measure zero. Now for each $\alpha <\kappa $ , define a function $f_{\alpha }:\omega ^{<\omega }\rightarrow \omega $ as $f_{\alpha }(\sigma )=\min \{m\in \omega : \phi _{\alpha }(\sigma )\subseteq N_{m}\}$ . The function $f_{\alpha }$ is well defined as each set $\phi _{\alpha }(\sigma )$ is finite and the $N_{n}$ ’s are increasing. Let $g:\omega ^{<\omega }\rightarrow \omega $ be a function dominating all the $f_{\alpha }$ ’s. We define $x\in dom(\mathcal {S})^{\omega }$ such that $x(n)$ is disjoint from $N_{g(x|_{n})}$ . Then clearly $x\notin \bigcup _{\alpha <\kappa }L_{\phi _{\alpha }}$ .

Proof. For (1), it follows from Proposition 3.1 and Corollaries 4.8 and 4.17. For (2), it follows from Proposition 3.1 with Corollary 4.12 and Proposition 4.14. (3): For the Miller case, it follows from Theorems 2.6, 3.7, and 3.11 and the fact that $nwd$ is weakly homogeneous. For the Laver case, it follows from Corollary 4.13 and Proposition 4.15. (4) follows from Proposition 3.1, Corollary 4.13, and Proposition 4.14. For (5), use Proposition 3.1 with Theorem 3.7, Corollary 4.8, and Propositions 4.14 and 4.20.

Question 4.22. What are the values of $\mathrm {is}(\mathbb {M}_{\mathcal {F}in\times \mathcal {F}in})$ , $\mathrm {is}(\mathbb {L}_{nwd})$ , $\mathrm {is}(\mathbb {L}_{\mathcal {S}}),$ and $\mathrm {is}(\mathbb {L}_{\mathcal {ED}_{fin}})$ ?