Proof In Proposition 1 of [Reference Brendle and Montoya17], it is proved that nwd and $tr\left ( \mathcal {M}\left ( 2^{\omega }\right ) \right ) $ are Katětov–Blass equivalent. We now show that nwd and $tr\left ( \mathcal {M}\left ( \omega ^{\omega }\right ) \right ) $ are equivalent. For this proof, given $t\in \omega ^{<\omega }$ , define $\left \langle t\right \rangle _{<\omega }=\left \{ s\in \omega ^{<\omega }\mid t\sqsubseteq s\right \} $ and $\left \langle t\right \rangle =\left \{ f\in \omega ^{\omega }\mid t\subseteq f\right \} .$ Fix a well-order $\preceq $ of the rational numbers. We now recursively define the sets $\left \{ U_{s}\mid s\in \omega ^{<\omega }\right \} $ and $\left \{ q_{s}\mid s\in \omega ^{<\omega }\right \} \subseteq \mathbb{Q}$ as follows:

1. (1) Each $U_{s}$ is a clopen interval of the rational numbers and $q_{s}=min_{\preceq }\left ( U_{s}\right ) .$

2. (2) $U_{\emptyset }=\mathbb{Q}.$

3. (3) $\left \{ U_{s^{\frown }i}\mid i\in \omega \right \} $ is a partition (into clopen sets) of $U_{s}\setminus \left \{ q_{s}\right \} .$

4. (4) $\mathbb{Q}=\left \{ q_{s}\mid s\in \omega ^{<\omega }\right \} $ and $\left \{ U_{s}\mid s\in \omega ^{<\omega }\right \} $ is a $\pi $ -base of open sets.

Define $f:\omega ^{<\omega }\longrightarrow \mathbb{Q}$ given by $f\left ( s\right ) =q_{s}.$ We will prove that f is a Katětov-function from $\left ( \omega ^{<\omega },tr\left ( \mathcal {M}\left ( \omega ^{\omega }\right ) \right ) \right ) $ to $(\mathbb{Q},$ nwd $).$ Let $N\subseteq \mathbb{Q}$ be a nowhere dense set, we show that $\pi \left ( f^{-1}\left ( N\right ) \right ) $ is a nowhere dense set of $\omega ^{\omega }.$ Let $s\in \omega ^{<\omega }$ and since N is nowhere dense and $\left \{ U_{s}\mid s\in \omega ^{<\omega }\right \} $ is a $\pi $ -base of open sets, we can find $t\in \omega ^{<\omega }$ extending s such that $U_{t}\cap N=\emptyset .$ It then follows that $\left \langle t\right \rangle \cap \pi \left ( f^{-1}\left ( N\right ) \right ) =\emptyset .$ Now, we will prove that $f^{-1}$ is a Katětov-function from $(\mathbb{Q},$ nwd $)$ to $\left ( \omega ^{<\omega },tr\left ( \mathcal {M}\left ( \omega ^{\omega }\right ) \right ) \right ) .$ It is enough to prove that if $a\in tr\left ( \mathcal {M}\left ( \omega ^{\omega }\right ) \right ) ,$ then $f\left [ a\right ] \in $ nwd $.$ Let $s\in \omega ^{<\omega }$ and since $\pi \left ( a\right ) \cap \left \langle s\right \rangle $ is a $G_{\delta }$ meager set in $\left \langle s\right \rangle $ , it cannot be dense, so there is $t\in \omega ^{<\omega }$ extending s such that $\left \langle t\right \rangle \cap \pi \left ( a\right ) =\emptyset ,$ which implies that $a\cap \left \langle t\right \rangle _{<\omega }$ has finite intersection with every branch, so we can then find an extension r of t such that $a\cap \left \langle r\right \rangle _{<\omega }=\emptyset ,$ which then implies $f\left [ a\right ] \cap U_{r}=\emptyset .$

Proof Assume $\mathcal {I}$ is $\omega $ -hitting and $f:\mathbb{Q}\longrightarrow \omega .$ We will prove that f is not a Katětov function from ( $\mathbb{Q},$ nwd) to $\left ( \omega ,\mathcal {I}\right ) .$ If there is an open non-empty set of $\mathbb{Q}$ whose image is finite, then the preimage of that finite set is not in nwd. Assume that each non-empty open set has infinite image. Let $\left \{ U_{n}\mid n\in \omega \right \} $ be a base of open sets of $\mathbb{Q}.$ Since each $f\left [ U_{n}\right ] $ is infinite and $\mathcal {I}$ is $\omega $ -hitting, we find $A\in \mathcal {I}$ that has infinite intersection with all of them. Hence, $f^{-1}\left ( A\right ) $ is dense, in particular not nowhere dense.

Proof Let $\mathcal {I}$ be a tall P-ideal and $\left \{ A_{n}\mid n\in \omega \right \} \subseteq \left [ \omega \right ] ^{\omega }.$ Since $\mathcal {I}$ is tall, for every $n\in \omega ,$ we choose $B_{n}\subseteq A_{n}$ with $B_{n}\in \mathcal {I}.$ Using that $\mathcal {I}$ is a P-ideal, we find $B\in \mathcal {I}$ that almost contain every $B_{n}.$ It follows that B has infinite intersection with each $A_{n}.$

Proof We start with the first point. Assume that $\mathcal {I}$ is weakly $\omega $ -hitting and $f:\omega \longrightarrow \omega $ is a Katětov morphism from $\left ( \omega ,\mathcal {J}\right ) $ to $\left ( \omega ,\mathcal {I}\right ) $ . Let $\mathcal {A=}\left \{ A_{n}\mid n\in \omega \right \} \subseteq [\omega ]^\omega $ . We want to find an element of $\mathcal {J}$ that has infinite intersection with infinitely many of them. Consider the set $S=\left \{ n\mid \left \vert f\left [ A_{n}\right ] \right \vert <\omega \right\}$ . We proceed by cases, first assume that S is infinite. For every $n\in S,$ we can find $b_{n}\in \omega $ such that $A_{n}\cap f^{-1}\left ( \left \{ b_{n}\right \} \right ) $ is infinite. If there is a $b\in \omega $ such that there are infinitely many $n\in S$ for which $b_{n}=b,$ then $f^{-1}\left ( \left \{ b\right \} \right ) \in \mathcal {J}$ and has infinite intersection with infinitely many elements of $\mathcal {A}.$ If there is no such $b,$ then the set $B=\left \{ b_{n}\mid n\in S\right \} $ is infinite. Since $\mathcal {I}$ is tall, we find $X\in \mathcal {I}$ such that $X\cap B$ is infinite. It follows that $f^{-1}\left ( X\right ) \in \mathcal {J}$ and has infinite intersection with infinitely many elements of $\mathcal {A}.$ We are now in the case that S is finite. We look at $B=\left \{ f\left [ A_{n}\right ] \mid n\notin S\right \} \subseteq \left [ \omega \right ] ^{\omega }.$ Since $\mathcal {I}$ is weakly $\omega $ -hitting, we find $X\in \mathcal {I}$ such that $X\cap b$ is infinite for every $b\in B$ . Once again, we get that $f^{-1}\left ( X\right ) \in \mathcal {J}$ and has infinite intersection with infinitely many elements of $\mathcal {A}.$ For the second point, note that the same proof as above works, since now S is the empty set, because f is finite to one.

The third point is trivial since no set in fin $\times $ fin can have an infinite intersection with infinitely many of the columns of $\omega ^{2}.$ We move to the fourth point. We will prove that $\mathcal {I}$ is not weakly $\omega $ -hitting if and only if $\mathcal {I}\leq _{ {\textsf {K} }}$ fin $\times $ fin. It is enough to prove that if $\mathcal {I}$ is not weakly $\omega $ -hitting, then $\mathcal {I}\leq _{ {\textsf {K}}}$ fin $\times $ fin since the other implication follows by the previous points. If $\mathcal {I}$ is not tall, there is nothing to do, so assume $\mathcal {I}$ is tall but not weakly $\omega $ -hitting. Thus, we find a partition $\mathcal {P} =\left \{ X_{n}\mid n\in \omega \right \} \subseteq \mathcal {I}$ of $\omega $ into infinite sets witnessing that $\mathcal I$ is not weakly $\omega $ -hitting. Let $f:\omega ^{2}\longrightarrow \omega $ be any bijection such that $f^{-1}\left ( X_{n}\right ) =\left \{ \left ( n,m\right ) \mid m\in \omega \right \} .$ It is easy to see that f is a Katětov morphism from $(\omega ^{2},$ fin $\times $ fin $)$ to $\left ( \omega ,\mathcal {I}\right ) .$

Proof In [Reference Hernández-Hernández and Hrušák36], it is proved that $tr\left ( \mathcal {N}\right ) \leq _{ {\textsf {K}}}\mathcal {Z}.$ Since $tr\left ( \mathcal {N}\right ) $ is HL and $\mathcal {Z}$ is not, it follows that $tr\left ( \mathcal {N}\right ) $ is strictly below $\mathcal {Z}.$ By the same reason, $\mathcal {Z}$ cannot be below any $F_{\sigma }$ ideal.

By Lemmas 3.13 and 3.14, we conclude that $tr\left ( \mathcal {N}\right ) \nleqslant _{ {\textsf {K}} }$ nwd. Now, assume that nwd $\leq _{ {\textsf {K}}} tr\left ( \mathcal {N}\right ) ,$ then cov $\left ( \mathcal {N}\right ) \leq $ cov $^{\ast }(tr\left ( \mathcal {N} \right ) )\leq $ cov $^{\ast }($ nwd $)=$ cov $\left ( \mathcal {M}\right ) $ (see Proposition 3.3), which we know is not true in ZFC. Since the Katětov order is absolute between Borel ideals (see [Reference Hrušák40]), we get a contradiction. This shows (3).

Katětov incomparability of fin $\times $ fin and nwd is proved in [Reference Hrušák40]. Since $tr($ ctble $) \subseteq tr({\mathcal M})$ and nwd and $tr\left (\mathcal {M}\right ) $ are Katětov equivalent (Lemma 3.10), we see $tr($ ctble $) \leq _{ {\textsf {K}}}$ nwd and fin $\times $ fin $\nleqslant _{ {\textsf {K}} } tr($ ctble $)$ . On the other hand, in Proposition 76 of [Reference Guzmán31], it was proved that $tr($ ctble $)$ is weakly $\omega $ -hitting, and $tr($ ctble $)\nleqslant _{ {\textsf {K}} }$ fin $\times $ fin follows by Proposition 3.15. This shows (4).

In order to prove that conv $\leq _{ {\textsf {K}}}$ fin $\times $ fin, by Proposition 3.15, it is enough to prove that conv is not weakly $\omega $ -hitting. This is easy, just take countably many convergent sequences converging to different points.

We now prove that conv $\leq _{ {\textsf {K}}}tr($ ctble). For this proof, $\mathbb{Q}_{0}$ will denote the set of all functions $f\in 2^{\omega }$ that are eventually constant $0.$ It is easy to see that $\mathbb{Q}_{0}$ is homeomorphic to the rational numbers. Let conv $(\mathbb{Q}_{0})$ be the ideal generated by all the convergent sequences of $\mathbb{Q}_{0}$ (the point of convergence does not need to be in $\mathbb{Q}_{0}$ ). As discussed before, conv $(\mathbb{Q}_{0})$ and conv are Katětov–Blass equivalent. Define $f:2^{<\omega }\longrightarrow \mathbb{Q}_{0}$ by $f\left ( s\right ) =s^{\frown }1^{\frown }\overline {0}$ (where $\overline {0}$ denotes the constant $0$ function and $^{\frown }$ is the concatenation operation). It is easy to see that f is a Katětov-function from $(2^{<\omega },tr($ ctble)) to $(\mathbb{Q}_{0} ,$ conv $(\mathbb{Q}_{0})).$

Since fin $\times $ fin and $tr($ ctble $) $ are Katětov incomparable, it follows that conv is strictly below both of them. Finally, since nwd and $tr\left ( \mathcal {N}\right ) $ are Katětov incomparable, it follows that $tr($ ctble $\mathsf{)}$ is strictly below both of them.

Proof Let $f:\omega ^{<\omega }\longrightarrow \mathbb{Q},$ we will prove that f is not a Katětov function from $(\omega ^{<\omega },tr(\mathcal {K}_{\sigma }))$ to $(\mathbb{Q},$ nwd $).$ We need the following.

Choose any non-empty clopen set $C_{0}\subseteq U$ for which $C_{1}=U\setminus C_{0}\neq \emptyset .$ We introduce some notation, given $q\in \mathbb{PT}$ and $X\subseteq \omega ^{<\omega },$ we will say that X is dense in q if every node of q can be extended to an element of $X.$ Define D as the set of all $t\in $ suc $_{p}\left ( s\right ) $ for which there is $z\in p_{t}$ such that $f^{-1}\left ( C_{0}\right ) \cap a$ is dense in $p_{z}.$ If D is infinite, we can find $b\subseteq a$ and q as above for which $f\left [ b\right ] \subseteq C_{0},$ hence $f\left [ b\right ] \cap C_{1}=\emptyset $ . On the other hand, if D is finite, we can find $b\subseteq a$ and q such that $f\left [ b\right ] \cap C_{0}=\emptyset .$

We want to recursively construct $a\in tr(\mathcal {K} _{\sigma })^{+}$ whose image is a nowhere dense set. We introduce some notation for the rest of the proof. Given $L\subseteq \omega ^{<\omega },$ by $\left \langle L\right \rangle $ we denote the tree closure of $L,$ in other words, $\left \langle L\right \rangle =\left \{ t\in \omega ^{<\omega }\mid \exists s\in L\left ( t\subseteq s\right ) \right \} .$

First, find $U_{0}\subseteq U$ a non-empty clopen set such that $f\left [ L\right ] \cap U_{0}=\emptyset .$ This is easy since L is finite. Let K be the set of all $t\in \omega ^{<\omega }$ such that either $t\in L$ and is maximal with this property or $t\in \left \langle L\right \rangle \cap $ split $\left ( p\right ) .$ For $t\in K,$ choose $p^{t}$ a Miller tree with the following properties:

1. (1) $p^{t}\leq p$ and $t\subseteq $ st $(p^{t}).$

2. (2) For every $z\in \omega ^{<\omega },$ if $z\in p^{t}\cap L,$ then $z\subseteq t.$

3. (3) If $t\in $ split $\left ( p\right ) ,$ then st $(p^{t})=t.$

We can now apply iteratively the previous claim finitely many times and find $C\subseteq U_{0}$ a non-empty clopen set and $\{(b^{t},q^{t})\mid t\in K\}$ with the following properties:

1. (1) $b^{t}\subseteq \left \{ z\in a\mid t\subseteq z\right \} $ and $f[b^{t}]\cap C=\emptyset .$

2. (2) $q^{t}\leq p^{t}$ and st $(q^{t})=$ st $(p^{t}).$

3. (3) $[q^{t}]\subseteq \pi (b^{t}).$

Define $b=L\cup \bigcup \limits _{t\in K}b^{t}$ and $q=\bigcup \limits _{t\in K}q^{t}.$ It is clear that $b,q$ , and C have the desired properties.

We can now construct a set $a\in tr\left ( \mathcal {K}_{\sigma }\right ) ^{+}$ such that $f\left [ a\right ] \in $ nwd $.$ Using the claim, do a Miller type fusion (preserving only finitely many nodes at each step) and build the desired $a.$

Proof We prove the proposition for $tr($ ctble $),$ a similar argument works for $tr\left ( \mathcal {K}_{\sigma }\right ) .$ Let $a\in tr($ ctble $)^{+}.$ Since $\pi \left ( a\right ) $ is an uncountable Borel set, we know that there is $p\in \mathbb{S}$ such that $\left [ p\right ] \subseteq \pi \left ( a\right ) .$ It follows that forcing with Sacks below p destroys $tr($ ctble $)\upharpoonright a.$ Since $\mathbb{S}$ is a homogeneous forcing, we conclude that Sacks forcing destroys $tr($ ctble $)\upharpoonright a.$ By Theorem 4.3, we conclude that $tr($ ctble $)\upharpoonright a\leq _{ {\textsf {K}} }tr($ ctble $).$

Proof Let $\mathcal {I}$ be a non-meager ideal and $X\in \left ( \mathcal {I}^{<\omega }\right ) ^{+}.$ Note that since $X\in \left ( \mathcal {I}^{<\omega }\right ) ^{+}$ (and $\mathcal {I}$ contains every finite set) for every $n\in \omega ,$ there is $s\in X$ such that $s\cap n=\emptyset .$ Thus, we can find $Z=\left \{ s_{n}\mid n\in \omega \right \} \subseteq X$ such that if $n\neq m,$ then $s_{n}\cap s_{m}=\emptyset .$ We then define $M=\left \{ A\subseteq \omega \mid \forall ^{\infty }n\left ( s_{n}\nsubseteq A\right ) \right \} $ which is clearly a meager set and thus there must be $A\in \mathcal {I}$ such that $A\notin M$ . Hence, there is $Y\in \left [ X\right ] ^{\omega }$ such that $\bigcup Y\subseteq A\in \mathcal {I}.$

Proof It is easy to see that if $X\in (($ fin $\times $ fin $)^{<\omega })^{+}$ , then there must be infinitely many elements of X that are below the graph of a function, so there must be $Y\in \left [ X\right ] ^{\omega }$ such that $\bigcup Y\in $ fin $\times $ fin.

Proof Let $f:\omega \longrightarrow \omega $ be a Katětov-morphism from $\left ( \omega ,\mathcal {J}\right ) $ to $\left ( \omega ,\mathcal {I}\right ) .$ Letting $X\in \left ( \mathcal {J}^{<\omega }\right ) ^{+}$ , we must find $Y\in \left [ X\right ] ^{\omega }$ such that $\bigcup Y\in \mathcal {J}.$ Defining $X_{1}=\left \{ f\left [ s\right ] \mid s\in X\right \} ,$ we will first argue that $X_{1}\in \left ( \mathcal {I}^{<\omega }\right ) ^{+}.$ To prove this fact, let $A\in \mathcal {I}.$ Since f is a Katětov-morphism, $f^{-1}\left ( A\right ) \in \mathcal {J}$ so there is $s\in X$ for which $s\cap f^{-1}\left ( A\right ) =\emptyset $ and then $f\left [ s\right ] \cap A=\emptyset .$ Since $\mathcal {I}$ is Shelah–Steprāns, there is $Y_{1} \in \left [ X_{1}\right ] ^{\omega }$ such that $\bigcup Y_{1}\in \mathcal {I}.$ Finally, if $Y\in \left [ X\right ] ^{\omega }$ is such that $Y_{1}=\left \{ f\left [ s\right ] \mid s\in Y\right \} $ , then $\bigcup Y\in \mathcal {J}.$

Proof We start with the first point. The equivalence of $a)$ and $c)$ is proved in [Reference Laflamme53]. The equivalence of $b)$ and $c)$ is claimed in that same paper without proof (see the definition of weak P-filter in [Reference Laflamme53]). We provide a sketch of the proof. It is easy to see that $b)$ implies $c).$ We now prove the other implication.

Let $\left \{ F_{s}\mid s\in \omega ^{<\omega }\right \} \subseteq \mathcal {I} ^{\ast }.$ Without loss of generality, we may assume that for every $s,t\in \omega ^{<\omega },$ if $t\subseteq s,$ then $F_{s}\subseteq F_{t}.$ For every $n\in \omega ,$ define:

$$ \begin{align*} \begin{array}[c]{l} H_{n}= {\textstyle\bigcap} \{F_{s}\mid s\in (n+1)^{\leq n} \}. \end{array} \end{align*} $$

Note that $H_{n}\in \mathcal {I}^{\ast }$ for every $n\in \omega .$ By point (c), there is $X\in \mathcal {I}^{+}$ a pseudointersection of $\left \{ H_{n}\mid n\in \omega \right \} .$ We now find a sequence $\left \langle n_{i}\right \rangle _{i\in \omega }$ such that for every $i\in \omega ,$ the following holds:

1. (1) $n_{i}<n_{i+1}.$

2. (2) $X\setminus n_{i+1}\subseteq H_{n_{i}}.$

We now define the following sets:

$$ \begin{align*} \begin{array}[c]{c} Y_{0}=X\cap\left( {\textstyle\bigcup} \{[n_{i},n_{i+1})\mid i\text{ is even}\}\right)\\ Y_{1}=X\cap\left( {\textstyle\bigcup} \{[n_{i},n_{i+1})\mid i\text{ is odd}\}\right). \end{array} \end{align*} $$

Since $X\in \mathcal {I}^{+},$ there is $i\in 2$ such that $Y_{i}\in \mathcal {I}^{+}$ . If $Y_{0}\in \mathcal {I}^{+}$ , define $f\in \omega ^{\omega }$ , where $f\left ( m\right ) =n_{2m+1}$ and if $Y_{1}\in \mathcal {I}^{+},$ define f by $f\left ( m\right ) =n_{2\left ( m+1\right ) }.$ In either case, it is easy to see that $ {\textstyle \bigcup \limits _{m\in \omega }} \left ( F_{f\upharpoonright m}\cap f\left ( m\right ) \right ) \in \mathcal {I}^{+}.$

For the equivalence of $c)$ and $d),$ see [Reference Hrušák40].

We now move to the second point. The equivalence of $a)$ and $c)$ was proved in [Reference Laflamme53]. For the equivalence of items $b)$ and $c)$ , just note that we can add all finite modifications of each $X_{n}.$

Proof Let $\mathcal {I}$ be an ideal for which $\mathsf{II}$ has a winning strategy in $\mathcal {L}\left ( \mathcal {I}\right ) .$ We will prove that $\mathcal {I}$ is not Shelah–Steprāns. Let $\left \{ X_{n}\mid n\in \omega \right \} \subseteq \left ( \mathcal {I}^{<\omega }\right ) ^{+}$ such that for every $A\in \mathcal {I,}$ there is $n\in \omega $ such that A does not contain any element of $X_{n}.$ For every $n\in \omega ,$ enumerate $X_{n}=\left \{ t_{n}^{i}\mid i\in \omega \right \} $ and $\prod \limits _{j<n}X_{j}=\{p_{n} ^{i}\mid i<\omega \}.$

For every $n,m\in \omega $ and every selector $a\in \left [ \omega \right ] ^{<\omega }$ of $\left ( t_{n}^{0},\ldots ,t_{n}^{m}\right ) $ , we define $F_{\left ( n,m,a\right ) }=p_{n}^{m}\left ( 0\right ) \cup \cdots \cup p_{n}^{m}\left ( n-1\right ) \cup a$ (recall $p_{n}^{m}\in \prod \limits _{j<n}X_{j}$ ). Clearly, each $F_{\left ( n,m,a\right ) }$ is a non-empty finite set. Let X be the collection of all the $F_{\left ( n,m,a\right ) };$ we will prove that X witnesses that $\mathcal {I}$ is not Shelah–Steprāns.

We will first prove that $X\in \left ( \mathcal {I}^{<\omega }\right ) ^{+}.$ Letting $A\in \mathcal {I,}$ we first find $n\in \omega $ such that A does not contain any element of $X_{n}.$ Since each $X_{j}\in \left ( \mathcal {I} ^{<\omega }\right ) ^{+}$ , for every $j<\omega ,$ there is $m\in \omega $ such that A is disjoint with $p_{n}^{m}\left ( 0\right ) \cup \cdots \cup p_{n}^{m}\left ( n-1\right ) .$ Finally, by the assumption on $X_{n}$ , we can find a selector b of $\left ( t_{n}^{0},\ldots ,t_{n}^{m}\right ) $ such that $b\cap A=\emptyset $ and therefore $A\cap F_{\left ( n,m,b\right ) }=\emptyset .$

Letting $Y\in \left [ X\right ] ^{\omega }$ , we will show that $B=\bigcup Y\in \mathcal {I}^{+}.$ There are two cases to consider: first assume there is $n\in \omega $ for which there are infinitely many $\left ( m,a\right ) $ such that $F_{\left ( n,m,a\right ) }\in Y.$ In this case, B intersects every element of $X_{n},$ hence $B\in \mathcal {I}^{+}.$ Now assume that for every $n\in \omega ,$ there are only finitely many $\left ( m,a\right ) $ such that $F_{\left ( n,m,a\right ) }\in Y.$ In this case, there must be infinitely many $n\in \omega $ for which there is $\left ( m,a\right ) $ such that $F_{\left ( n,m,a\right ) }\in Y,$ hence B must contain (at least) one element of every $X_{k}.$ We can then conclude that $B\in \mathcal {I}^{+}.$

Proof By the previous result, we know that 2 implies 1. We first prove that 1 implies 3. Assume that $\mathcal {I}$ is not Shelah–Steprāns, so there is $X=\left \{ s_{n}\mid n\in \omega \right \} \in \left ( \mathcal {I}^{<\omega }\right ) ^{+}$ such that ${\textstyle \bigcup } Y\in \mathcal {I}^{+}$ for every $Y\in \left [ X\right ] ^{\omega }.$ We now define the set $F=\left \{ W\subseteq \omega \mid \forall ^{\infty }n \left ( s_{n} \nsubseteq W\right ) \right \} .$ It is easy to see that F has the desired properties.

We finally prove that 3 implies 2. Assume there is an increasing sequence of closed sets $\left \langle C_{n}\mid n\in \omega \right \rangle $ such that $F= {\textstyle \bigcup \limits _{n\in \omega }} C_{n}$ contains $\mathcal {I}$ and is disjoint from $\mathcal {I}^{\ast }.$ We will now describe a winning strategy for Player $\mathsf{II}$ in $\mathcal {L} ({\mathcal I})$ : In the first round, if Player $\mathsf{I}$ plays $A_{0}\in \mathcal {I}$ then Player $\mathsf{II}$ finds an initial segment $s_{0}$ of $\omega \setminus A_{0}$ such that $\left \langle s_{0}\right \rangle =\left \{ Z\mid s_{0}\sqsubseteq Z\right \} $ is disjoint from $C_{0}$ (where $s_{0}\sqsubseteq Z$ means that $s_{0}$ is an initial segment of Z). At round $n+1,$ if Player $\mathsf{I}$ plays $A_{n+1}\in \mathcal {I}$ then Player $\mathsf{II}$ finds $s_{n+1}$ such that $t={\textstyle \bigcup \limits _{i\leq n+1}} s_{i}$ is an initial segment of $\left ( \omega \setminus A_{n+1}\right ) \cup {\textstyle \bigcup \limits _{j<n+1}} s_{j}$ and $\left \langle t\right \rangle $ is disjoint from $C_{n+1}.$ It is easy to see that this is a winning strategy.

Proof To prove the first item, let Y be the set of all sequences $\left \langle A_{0},s_{0},\ldots ,A_{n},s_{n}\right \rangle $ such that $A_{n}\in \mathcal {I}$ and $s_{n}\in \left [ \omega \setminus A_{n}\right ] ^{<\omega }$ and $\max \left ( s_{i}\right ) \subseteq A_{i+1}$ if $i<n.$ Let T be the tree obtained by closing Y under restrictions. We define $f:\left [ T\right ] \longrightarrow \mathcal P\left ( \omega \right ) $ by $f\left ( b\right ) = {\textstyle \bigcup \limits _{n\in \omega }} b\left ( 2n+1\right ) $ , where $b\in \left [ T\right ] .$ Clearly, $\mathcal {L} \left ( \mathcal {I}\right ) $ is a game equivalent to $\mathcal {G}\left ( T,f,\mathcal {I}\right ) $ , so the result follows from the previous results. The second item is a consequence of the first.

Proof Assume $\mathfrak {b} = \mathfrak {c}$ . It is clear that every almost strongly separable a.d. family is Shelah–Steprāns. Suppose $\mathscr {B}$ is an infinite Shelah–Steprāns a.d. family. The following claim is a simple consequence of Proposition 5.10 and Item (2) of Proposition 5.9. However, we include the argument for completeness.

$\left | \mathscr {B} \right | = \mathfrak {c}$ because being an infinite Shelah–Steprāns a.d. family implies being MAD and because $\mathfrak {b} = \mathfrak {c}$ . Let $\{{X}_{\alpha }: \alpha < \mathfrak {c}\}$ enumerate $\mathcal {P}\left ( {\left [\omega \right ]}^{< \omega } \setminus \{\emptyset \} \right )$ . Construct families $\{{A}_{\alpha }: \alpha < \mathfrak {c}\}$ and $\left \{ {\mathcal {A}}_{\alpha }: \alpha < \mathfrak {c} \right \}$ such that the following hold:

1. (1) each ${\mathcal {A}}_{\alpha }$ is a non-empty finite subset of $\mathscr {B}$ and ${A}_{\alpha } = \bigcup {\mathcal {A}}_{\alpha }$ ;

2. (2) ${\mathcal {A}}_{\alpha } \cap {\mathcal {A}}_{\beta } = \emptyset $ , for all $\alpha < \beta < \mathfrak {c}$ ;

3. (3) either ${X}_{\alpha } \in {\mathcal {I}\left ( \left \{{A}_{\xi }: \xi \leq \alpha \right \} \right )}^{< \omega }$ or $\left | {X}_{\alpha } \cap {\left [ {A}_{\alpha } \right ]}^{< \omega } \right | = {\aleph }_{0}$ .

Suppose for a moment that this can be done. Let $\mathscr {A} = \{{A}_{\alpha }: \alpha < \mathfrak {c}\}$ . $\mathscr {A}$ is a.d. because of (1), (2), and the fact that $\mathscr {B}$ is a.d. If $X \in {\left ( {\mathcal {I}(\mathscr {A})}^{< \omega } \right )}^{+}$ , then $X = {X}_{\alpha }$ , for some $\alpha < \mathfrak {c}$ , and since $X \notin {\mathcal {I}\left ( \left \{{A}_{\xi }: \xi \leq \alpha \right \} \right )}^{< \omega }$ , $\left | {X}_{\alpha } \cap {\left [ {A}_{\alpha } \right ]}^{< \omega } \right | = {\aleph }_{0}$ by (3). This shows $\mathscr {A}$ is almost strongly separable.

To construct such families, fix $\alpha < \mathfrak {c}$ and assume that $\{{A}_{\xi }: \xi < \alpha \}$ and $\left \{{\mathcal {A}}_{\xi }: \xi < \alpha \right \}$ are given. Define $\mathcal {A}\left [\alpha \right ] = {\bigcup }_{\xi < \alpha }{{\mathcal {A}}_{\xi }}$ and $\mathscr {B}\left [\alpha \right ] = \mathscr {B} \setminus \mathcal {A}\left [\alpha \right ]$ . Note that $\mathcal {A}\left [\alpha \right ]$ and $\mathscr {B}\left [\alpha \right ]$ are subsets of $\mathscr {B}$ , $\left | \mathcal {A}\left [\alpha \right ] \right | < \mathfrak {c}$ , and $\left | \mathscr {B}\left [\alpha \right ] \right | = \mathfrak {c}$ . First, suppose that ${X}_{\alpha } \in {\mathcal {I}(\{{A}_{\xi }: \xi < \alpha \})}^{< \omega }$ . Then, choose any $B \in \mathscr {B}\left [\alpha \right ]$ , define ${\mathcal {A}}_{\alpha } = \{B\}$ , and ${A}_{\alpha } = \bigcup {\mathcal {A}}_{\alpha }$ – i.e., ${A}_{\alpha } = B$ . Then, it is clear that (1)–(3) hold.

Next suppose that ${X}_{\alpha } \in {\left ( {\mathcal {I}(\{{A}_{\xi }: \xi < \alpha \})}^{< \omega } \right )}^{+}$ , but ${X}_{\alpha } \in {\mathcal {I}(\mathscr {B})}^{< \omega }$ . Let $l \in \omega $ and $\mathcal {B} \subseteq \mathscr {B}$ be a finite set such that $\forall s \in {X}_{\alpha }\left [ s \cap \left ( l \cup \left ( \bigcup \mathcal {B} \right ) \right ) \neq \emptyset \right ]$ . Define ${\mathcal {A}}_{\alpha } = \mathcal {B} \setminus \mathcal {A}\left [\alpha \right ]$ and ${A}_{\alpha } = \bigcup {\mathcal {A}}_{\alpha }$ . ${\mathcal {A}}_{\alpha } \neq \emptyset $ because ${X}_{\alpha } \notin {\mathcal {I}\left (\{{A}_{\xi }: \xi < \alpha \}\right )}^{< \omega }$ . Hence, (1) and (2) hold. (3) holds because ${X}_{\alpha } \in {\mathcal {I}(\left \{ {A}_{\xi }: \xi \leq \alpha \right \})}^{< \omega }$ .

Finally, suppose that ${X}_{\alpha } \in {\left ( {\mathcal {I}(\mathscr {B})}^{< \omega } \right )}^{+}$ .