Comparing the closed almost disjointness and dominating numbers

Recall that two infinite subsets $a$ and $b$ of $\omega$ are almost disjoint or a.d. if $a\cap b$ is finite. A family ${\mathscr{A}}$ of infinite subsets of $\omega$ is said to be almost disjoint or a.d. in ${\left[\omega\right]}^{\omega}$ if its members are pairwise almost disjoint. A Maximal Almost Disjoint family, or MAD family in ${\left[\omega\right]}^{\omega}$ is an infinite a.d. family in ${\left[\omega\right]}^{\omega}$ that is not properly contained in a larger a.d. family.

Two functions $f$ and $g$ in ${\omega}^{\omega}$ are said to be almost disjoint or a.d. if they agree in only finitely many places. We say that a family ${\mathscr{A}}\subset{\omega}^{\omega}$ is a.d. in ${\omega}^{\omega}$ if its members are pairwise a.d., and we say that an a.d. family ${\mathscr{A}}\subset{\omega}^{\omega}$ is MAD in ${\omega}^{\omega}$ if $\forall f\in{\omega}^{\omega}\exists h\in{\mathscr{A}}\left[\left|f\cap h\right|={\aleph}_{0}\right]$. Identifying functions with their graphs, every a.d. family in ${\omega}^{\omega}$ is also an a.d. family in ${\left[\omega\times\omega\right]}^{\omega}$; however, it is never MAD in ${\left[\omega\times\omega\right]}^{\omega}$ because any function is a.d. from the vertical columns of $\omega\times\omega$. MAD families in ${\omega}^{\omega}$ that become MAD in ${\left[\omega\times\omega\right]}^{\omega}$ when the vertical columns of $\omega\times\omega$ are thrown in were considered by Van Douwen.

We say that $p\subset\omega\times\omega$ is an infinite partial function if it is a function from some infinite set $A\subset\omega$ to $\omega$. An a.d. family ${\mathscr{A}}\subset{\omega}^{\omega}$ is said to be Van Douwen if for any infinite partial function $p$ there is $h\in{\mathscr{A}}$ such that $\left|h\cap p\right|={\aleph}_{0}$. ${\mathscr{A}}$ is Van Douwen iff ${\mathscr{A}}\cup\{{c}_{n}:n\in\omega\}$ is a MAD family in ${\left[\omega\times\omega\right]}^{\omega}$, where ${c}_{n}$ is the $n$th vertical column of $\omega\times\omega$. The first author showed in [3] that Van Douwen MAD families always exist.

Recall that $\mathfrak{b}$ is the least size of an unbounded family in ${\omega}^{\omega}$, $\mathfrak{d}$ is the least size of a dominating family in ${\omega}^{\omega}$, and ${\mathfrak{a}}$ is the least size of a MAD family in ${\left[\omega\right]}^{\omega}$. It is well known that $\mathfrak{b}\leq{\mathfrak{a}}$. Whether ${\mathfrak{a}}$ could consistently be larger than $\mathfrak{d}$ was an open question for a long time, until Shelah achieved a breakthrough in [4] by producing a model where $\mathfrak{d}={\aleph}_{2}$ and ${\mathfrak{a}}={\aleph}_{3}$. However, it is not known whether ${\mathfrak{a}}$ can be larger than $\mathfrak{d}$ when $\mathfrak{d}={\aleph}_{1}$; this is one of the few major remaining open problems in the theory of cardinal invariants posed during the earliest days of the subject (see [5] and [2]). In this note we take a small step towards resolving this question by showing that if $\mathfrak{d}={\aleph}_{1}$, then there is a MAD family in ${\left[\omega\right]}^{\omega}$ which is the union of ${\aleph}_{1}$ compact subsets of ${\left[\omega\right]}^{\omega}$. More precisely, we will establish the following:

The cardinal invariant ${\mathfrak{a}}_{closed}$ was recently introduced and studied by Brendle and Khomskii [1] in connection with the possible descriptive complexities of MAD families in certain forcing extensions of ${\mathbf{L}}$.

Obviously, ${\mathfrak{a}}_{closed}\leq{\mathfrak{a}}$. Brendle and Khomskii showed in [1] that ${\mathfrak{a}}_{closed}$ behaves differently from ${\mathfrak{a}}$ by producing a model where ${\mathfrak{a}}_{closed}={\aleph}_{1}<{\aleph}_{2}=\mathfrak{b}$. They asked whether $\mathfrak{s}={\aleph}_{1}$ implies that ${\mathfrak{a}}_{closed}={\aleph}_{1}$. As $\mathfrak{s}\leq\mathfrak{d}$, our result in this paper provides a partial positive answer to their question.

Assume $\mathfrak{d}=\aleph_{1}$ in this section. We will build $\aleph_{1}$ many compact subsets of ${\omega}^{\omega}$ whose union is a Van Douwen MAD family. To this end, we will construct a sequence $\langle{T}_{\alpha}:\alpha<\omega_{1}\rangle$ of finitely branching subtrees of ${\omega}^{<\omega}$ such that ${\bigcup}_{\alpha<\omega_{1}}{\left[{T}_{\alpha}\right]}$ has the required properties. Henceforth, $T\subset{\omega}^{<\omega}$ will mean $T$ is a subtree of ${\omega}^{<\omega}$.

Note that if $\eta\leq\xi$ and ${\operatorname{rk}}_{T,p}(\sigma)\geq\xi$, then ${\operatorname{rk}}_{T,p}(\sigma)\geq\eta$, and that for a limit ordinal $\xi$, if $\forall\zeta<\xi\left[{\operatorname{rk}}_{T,p}(\sigma)\geq\zeta\right]$, then ${\operatorname{rk}}_{T,p}(\sigma)\geq\xi$. Also, for any $\sigma,\tau\in T$, if $\sigma\subset\tau$ and ${\operatorname{rk}}_{T,p}(\tau)\geq\xi$, then ${\operatorname{rk}}_{T,p}(\sigma)\geq\xi$. Moreover, if ${\operatorname{rk}}_{T,p}(\sigma)\not\geq\xi$ and if $\tau\in T$ and $l\in A$ are such that $\tau\supset\sigma$, $\left|\sigma\right|\leq l<\left|\tau\right|$, and $p(l)=\tau(l)$, then there is $\zeta<\xi$ such that ${\operatorname{rk}}_{T,p}(\tau)\not\geq\zeta$. Therefore, if there is $f\in\left[T\right]$ with $\left|f\cap p\right|={\aleph}_{0}$, and $\sigma\subset f$ and there is some ordinal $\xi$ such that ${\operatorname{rk}}_{T,p}(\sigma)\not\geq\xi$, then is some $\sigma\subset\tau\subset f$ and some ordinal $\zeta<\xi$ such that ${\operatorname{rk}}_{T,p}(\tau)\not\geq\zeta$, thus allowing us to construct an infinite, strictly descending sequence of ordinals. So if $f\in\left[T\right]$ with $\left|f\cap p\right|={\aleph}_{0}$, then for any $\sigma\subset f$ and any ordinal $\xi$, ${\operatorname{rk}}_{T,p}(\sigma)\geq\xi$. On the other hand, suppose that $\sigma\in T$ with ${\operatorname{rk}}_{T,p}(\sigma)\geq{\omega}_{1}$. Then there is $\tau\in T$ with $\tau\supset\sigma$ and $l\in A$ such that $\left|\sigma\right|\leq l<\left|\tau\right|$, $p(l)=\tau(l)$, and ${\operatorname{rk}}_{T,p}(\tau)\geq{\omega}_{1}$, allowing us to construct $f\in\left[T\right]$ with $\sigma\subset f$ such that $\left|f\cap p\right|={\aleph}_{0}$.

Note the following features of this definition

1. (${\ast}_{1}$)

   $\forall\sigma,\tau\in T\left[\sigma\subset\tau\implies{H}_{T,p}(\sigma)\geq{H}_{T,p}(\tau)\right]$

2. (${\ast}_{2}$)

   for all $\sigma,\tau\in T$ with $\sigma\subset\tau$, if there exists $l\in A$ such that $\left|\sigma\right|\leq l<\left|\tau\right|$ and $p(l)=\tau(l)$, then ${H}_{T,p}(\tau)<{H}_{T,p}(\sigma)$.

On the other hand, notice that if there is a function $H:T\rightarrow{\omega}_{1}$ such that (${\ast}_{1}$) and (${\ast}_{2}$) hold when ${H}_{T,p}$ is replaced with $H$, then $p$ must be a.d. from $[T]$.

It is well known that $\mathfrak{d}$ is also the size of the smallest family of interval partitions dominating any interval partition. So fix a sequence $\langle{I}^{\alpha}:\alpha<{\omega}_{1}\rangle$ of interval partitions such that

1. (1)

   $\forall\alpha\leq\beta<{\omega}_{1}\left[{I}^{\alpha}{\leq}^{\ast}{I}^{\beta}\right]$

2. (2)

   for any interval partition $J$, there exists $\alpha<{\omega}_{1}$ such that $J{\leq}^{\ast}{I}^{\alpha}$.

Fix an $\omega_{1}$-scale $\langle{f}_{\alpha}:\alpha<\omega_{1}\rangle$ such that $\forall\alpha<\omega_{1}\forall n\in\omega\left[f_{\alpha}(n)<f_{\alpha}(n+1)\right]$. For each $\alpha\geq 1$, define ${e}_{\alpha}$ and $g_{\alpha}$ by induction on $\alpha$ as follows. If $\alpha$ is a successor, then $e_{\alpha}:\omega\rightarrow\alpha$ is any onto function, and ${g}_{\alpha}=f_{\alpha}$. If $\alpha$ is a limit, then let $\{e_{n}:n\in\omega\}$ enumerate $\{{e}_{\xi}:\xi<\alpha\}$. Now, define ${e}_{\alpha}:\omega\rightarrow\alpha$ and $g_{\alpha}\in{\omega}^{\omega}$ such that

1. (3)

   $\forall n\in\omega\left[g_{\alpha}(n)\leq g_{\alpha}(n+1)\right]$

2. (4)

   $\forall n\in\omega\forall i\leq n\forall j\leq f_{\alpha}(n)\exists k<g_{\alpha}(n)\left[e_{\alpha}(k)=e_{i}(j)\right]$.

Observe that such an $e_{\alpha}$ must be a surjection. For each $n\in\omega$, put ${w}_{\alpha}(n)=\{{e}_{\alpha}(i):i\leq{g}_{\alpha}(n)\}$.

Now fix $\alpha<\omega_{1}$ and assume that ${T}_{\epsilon}\subset{\omega}^{<\omega}$ has been defined for each $\epsilon<\alpha$ such that each ${T}_{\epsilon}$ is finitely branching and ${\bigcup}_{\epsilon<\alpha}{\left[{T}_{\epsilon}\right]}$ is an a.d. family in ${\omega}^{\omega}$. Let $\langle{\epsilon}_{n}:n\in\omega\rangle$ enumerate $\alpha$, possibly with repetitions. For a tree $T\subset{\omega}^{<\omega}$ and $l\in\omega$, $T\upharpoonright l$ denotes $\{\sigma\in T:\left|\sigma\right|\leq l\}$, and $T(l)$ denotes $\{\sigma\in T:\left|\sigma\right|=l\}$. We will define a sequence of natural numbers $0={l}_{0}<{l}_{1}<\dotsb$ and determine ${T}_{\alpha}\upharpoonright{l}_{n}$ by induction on $n$. ${T}_{\alpha}\upharpoonright{l}_{0}=\{0\}$. Assume that ${l}_{n}$ and ${T}_{\alpha}\upharpoonright{l}_{n}$ are given. Suppose also that we are given a sequence of natural numbers $\langle{k}_{i}:i<n\rangle$ such that

1. (5)

   $\forall i<i+1<n\left[{k}_{i}<{k}_{i+1}\right]$

2. (6)

   ${I}^{\alpha}_{{k}_{i}}\subset[0,{l}_{n})$.

Let ${\sigma}^{\ast}$ denote the member of ${T}_{\alpha}({l}_{n})$ that is right most with respect to the lexicographical ordering on ${\omega}^{{l}_{n}}$. Suppose we are also given ${L}_{n}:{T}_{\alpha}({l}_{n})\setminus\{{\sigma}^{\ast}\}\rightarrow{W}_{n}$, an injection. Here ${W}_{n}$ is the set of all pairs $\langle{p}_{0},\bar{h}\rangle$ such that

1. (7)

   there are $s\in{{\left[\omega\right]}^{<\omega}}$, and numbers ${i}_{0}<{j}_{0}\leq n$ such that

   1. (a)

      $s\subset{\bigcup}_{i\in[{i}_{0},{j}_{0})}{{I}^{\alpha}_{{k}_{i}}}$

   2. (b)

      for each $i\in[{i}_{0},{j}_{0})$, $\left|s\cap{I}^{\alpha}_{{k}_{i}}\right|=1$

   3. (c)

      ${p}_{0}:s\rightarrow\omega$ such that $\forall m\in s\left[{p}_{0}(m)\leq{f}_{\alpha}(m)\right]$

2. (8)

   There is ${j}_{1}<n$ such that $\bar{h}=\langle{h}_{{\epsilon}_{i}}:i\leq{j}_{1}\rangle$ (if $\alpha=0$, this means that $\bar{h}=0$). For each $i\leq{j}_{1}$, ${h}_{{\epsilon}_{i}}:{T}_{{\epsilon}_{i}}\upharpoonright\max{(s)}+1\rightarrow{w}_{\alpha}(\max{(s)}+1)$ such that (${\ast}_{1}$) and (${\ast}_{2}$) hold when $T$ is replaced there with ${T}_{{\epsilon}_{i}}\upharpoonright\max{(s)}+1$, ${H}_{T,p}$ is replaced with ${h}_{{\epsilon}_{i}}$, $A$ with $s$, and $p$ with ${p}_{0}$.

Assume that for each $i<n$, we are also given ${\sigma}_{i}\in{T}_{\alpha}({l}_{i})$, which we will call the active node at stage $i$. Note that ${T}_{\alpha}({l}_{0})=\{0\}$, and so ${\sigma}_{0}=0$. For each $\sigma\in{T}_{\alpha}({l}_{n})$, let $\Delta(\sigma)=\max\left(\{0\}\cup\{i<n:{\sigma}_{i}=\sigma\upharpoonright{l}_{i}\}\right)$. For, $\sigma,\tau\in{T}_{\alpha}({l}_{n})$, say $\sigma\vartriangleleft\tau$ if either $\Delta(\sigma)<\Delta(\tau)$ or $\Delta(\sigma)=\Delta(\tau)$ and $\sigma$ is to the left of $\tau$ in the lexicographic ordering on ${\omega}^{{l}_{n}}$. Let ${\sigma}_{n}$ be the $\vartriangleleft$-minimal member of ${T}_{\alpha}({l}_{n})$. ${\sigma}_{n}$ will be active at stage $n$. The meaning of this is that none of the other nodes in ${T}_{\alpha}({l}_{n})$ will be allowed to branch at stage $n$. Choose ${k}_{n}$ greater than all ${k}_{i}$ for $i<n$ such that ${I}^{\alpha}_{{k}_{n}}\subset[{l}_{n},\infty)$. Let ${V}_{n}$ be the set of all pairs $\langle{p}_{1},\bar{\mathbf{h}}\rangle$ such that

1. (9)

   there exist $s$ and a natural number ${i}_{1}\leq n$ such that

   1. (a)

      $s\subset{\bigcup}_{i\in[{i}_{1},n+1)}{{I}^{\alpha}_{{k}_{i}}}$

   2. (b)

      for each $i\in[{i}_{1},n+1)$, $\left|s\cap{I}^{\alpha}_{{k}_{i}}\right|=1$

   3. (c)

      ${p}_{1}:s\rightarrow\omega$ such that $\forall m\in s\left[{p}_{1}(m)\leq{f}_{\alpha}(m)\right]$

2. (10)

   There is ${j}_{2}\leq n$ such that $\bar{\mathbf{h}}=\langle{\mathbf{h}}_{{\epsilon}_{i}}:i\leq{j}_{2}\rangle$. For each $i\leq{j}_{2}$, ${\mathbf{h}}_{{\epsilon}_{i}}:{T}_{{\epsilon}_{i}}\upharpoonright\max{(s)}+1\rightarrow{w}_{\alpha}(\max{(s)}+1)$ such that (${\ast}_{1}$) and (${\ast}_{2}$) are satisfied when $T$ is replaced with ${T}_{{\epsilon}_{i}}\upharpoonright\max{(s)}+1$, ${H}_{T,p}$ is replaced with ${\mathbf{h}}_{{\epsilon}_{i}}$, $A$ with $s$, and $p$ with ${p}_{1}$.

Note that ${V}_{n}$ is always finite. Now, the construction splits into two cases.

Case I: ${\sigma}_{n}\neq{\sigma}^{\ast}$. Put $\langle{p}_{0},\bar{h}\rangle={L}_{n}({\sigma}_{n})$. Let ${i}_{0}<n$ be as in (7) above, and let ${j}_{1}<n$ be as in (8). Let

Here ${i}_{1}$ is as in (9), and ${j}_{2}$ is as in (10) with respect to $\langle{p}_{1},\bar{\mathbf{h}}\rangle$. Now choose ${l}_{n+1}>{l}_{n}$ large enough so that ${I}^{\alpha}_{{k}_{n}}\subset[{l}_{n},{l}_{n+1})$ and so that it is possible to pick $\{{\tau}_{x}:x\in{U}_{n}\}\subset{\omega}^{{l}_{n+1}}$ and $\{{\tau}_{\sigma}:\sigma\in{T}_{\alpha}({l}_{n})\}\subset{\omega}^{{l}_{n+1}}$ such that the following conditions are satisfied.

1. (11)

   for each $x\in{U}_{n}$, ${\tau}_{x}\supset{\sigma}_{n}$, and for each $\sigma\in{T}_{\alpha}({l}_{n})$, ${\tau}_{\sigma}\supset\sigma$

2. (12)

   for each $x,y\in{U}_{n}$, if $x\neq y$, then there exists $m\in[{l}_{n},{l}_{n+1})$ such that ${\tau}_{x}(m)\neq{\tau}_{y}(m)$. For each $x\in{U}_{n}$, there exists $m\in[{l}_{n},{l}_{n+1})$ such that ${\tau}_{x}(m)\neq{\tau}_{{\sigma}_{n}}(m)$. For $x=\langle{p}_{1},\bar{\mathbf{h}}\rangle\in{U}_{n}$, if $\{{i}^{\ast}\}=\operatorname{dom}{({p}_{1})}\cap{I}^{\alpha}_{{k}_{n}}$, then ${p}_{1}({i}^{\ast})={\tau}_{x}({i}^{\ast})$.

3. (13)

   for each $x\in{U}_{n}$ and $\sigma\in{T}_{\alpha}({l}_{n})$, $\forall m\in[{l}_{n},{l}_{n+1})\left[{\tau}_{x}(m)\neq{\tau}_{\sigma}(m)\right]$. For $\sigma,\eta\in{T}_{\alpha}({l}_{n})$, if $\sigma\neq\eta$, then $\forall m\in[{l}_{n},{l}_{n+1})\left[{\tau}_{\sigma}(m)\neq{\tau}_{\eta}(m)\right]$.

4. (14)

   for each $i\leq n$, $\tau\in{T}_{{\epsilon}_{i}}({l}_{n+1})$, $\sigma\in{T}_{\alpha}({l}_{n})$ and $m\in[{l}_{n},{l}_{n+1})$, $\tau(m)\neq{\tau}_{\sigma}(m)$. For each $x\in{U}_{n}$, $i\leq{j}_{2}$, $\tau\in{T}_{{\epsilon}_{i}}({l}_{n+1})$ and $m\in[{l}_{n},{l}_{n+1})$, if ${\tau}_{x}(m)=\tau(m)$, then $m\in\operatorname{dom}{({p}_{1})}$ and ${p}_{1}(m)={\tau}_{x}(m)$.

Define ${L}_{n+1}$ as follows. For any $x\in{U}_{n}$, ${L}_{n+1}({\tau}_{x})=x$. For any $\sigma\in{T}_{\alpha}({l}_{n})\setminus\{{\sigma}^{\ast}\}$, ${L}_{n+1}({\tau}_{\sigma})={L}_{n}(\sigma)$. This finishes case 1.

Case II: ${\sigma}_{n}={\sigma}^{\ast}$. For each $\sigma\in{T}_{\alpha}({l}_{n})\setminus\{{\sigma}_{n}\}$, let $\langle{p}_{0}(\sigma),\bar{h}(\sigma)\rangle={L}_{n}(\sigma)$. Let ${i}_{0}(\sigma)<n$ witness (7) for ${L}_{n}(\sigma)$ and let ${j}_{1}(\sigma)<n$ witness (8) for ${L}_{n}(\sigma)$. Let ${U}_{n}$ be the set of all $\langle{p}_{1},\bar{\mathbf{h}}\rangle\in{V}_{n}$ such that there is no $\sigma\in{T}_{\alpha}({l}_{n})\setminus\{{\sigma}_{n}\}$ so that

Here ${i}_{1}\leq n$ and ${j}_{2}\leq n$ witness (9) and (10) respectively with respect to $\langle{p}_{1},\bar{\mathbf{h}}\rangle$. Choose ${l}_{n+1}>{l}_{n}$ large enough so that ${I}^{\alpha}_{{k}_{n}}\subset[{l}_{n},{l}_{n+1})$ and so that it is possible to choose $\{{\tau}^{\ast}\}$, $\{{\tau}_{x}:x\in{U}_{n}\}$, and $\{{\tau}_{\sigma}:\sigma\in{T}_{\alpha}({l}_{n})\setminus\{{\sigma}_{n}\}\}$, subsets of ${\omega}^{{l}_{n+1}}$, satisfying the following conditions.

1. (15)

   ${\tau}^{\ast}\supset{\sigma}_{n}$. For each $x\in{U}_{n}$, ${\tau}_{x}\supset{\sigma}_{n}$. For each $\sigma\in{T}_{\alpha}({l}_{n})\setminus\{{\sigma}_{n}\}$, ${\tau}_{\sigma}\supset\sigma$.

2. (16)

   ${\tau}^{\ast}$ is the right most branch of ${T}_{\alpha}({l}_{n+1})$. For each $x\in{U}_{n}$, there exists $m\in[{l}_{n},{l}_{n+1})$ such that ${\tau}^{\ast}(m)\neq{\tau}_{x}(m)$. For each $x,y\in{U}_{n}$, if $x\neq y$, then there is $m\in[{l}_{n},{l}_{n+1})$ so that ${\tau}_{x}(m)\neq{\tau}_{y}(m)$. For each $x=\langle{p}_{1},\bar{\mathbf{h}}\rangle\in{U}_{n}$, if $\{{i}^{\ast}\}={I}^{\alpha}_{{k}_{n}}\cap\operatorname{dom}{({p}_{1})}$, then ${p}_{1}({i}^{\ast})={\tau}_{x}({i}^{\ast})$.

3. (17)

   For each $x\in{U}_{n}$ and $m\in[{l}_{n},{l}_{n+1})$, ${\tau}_{x}(m)\neq{\tau}^{\ast}(m)$. For each $\sigma\in{T}_{\alpha}({l}_{n})\setminus\{{\sigma}_{n}\}$ and for each $m\in[{l}_{n},{l}_{n+1})$, ${\tau}^{\ast}(m)\neq{\tau}_{\sigma}(m)$, and for each $x\in{U}_{n}$, ${\tau}_{\sigma}(m)\neq{\tau}_{x}(m)$. For each $\sigma,\eta\in{T}_{\alpha}({l}_{n})\setminus\{{\sigma}_{n}\}$, if $\sigma\neq\eta$, then for all $m\in[{l}_{n},{l}_{n+1})$, ${\tau}_{\sigma}(m)\neq{\tau}_{\eta}(m)$.

4. (18)

   For each $i\leq n$, $\tau\in{T}_{{\epsilon}_{i}}({l}_{n+1})$, $m\in[{l}_{n},{l}_{n+1})$, and $\sigma\in{T}_{\alpha}({l}_{n})\setminus\{{\sigma}_{n}\}$, ${\tau}^{\ast}(m)\neq\tau(m)$ and ${\tau}_{\sigma}(m)\neq\tau(m)$. For each $x=\langle{p}_{1},\bar{\mathbf{h}}\rangle\in{U}_{n}$, $i\leq{j}_{2}$, $\tau\in{\tau}_{{\epsilon}_{i}}({l}_{n+1})$ and $m\in[{l}_{n},{l}_{n+1})$, if ${\tau}_{x}(m)=\tau(m)$, then $m\in\operatorname{dom}{({p}_{1})}$ and ${p}_{1}(m)={\tau}_{x}(m)$.

For each $\sigma\in{T}_{\alpha}({l}_{n})\setminus\{{\sigma}_{n}\}$, define ${L}_{n+1}({\tau}_{\sigma})={L}_{n}(\sigma)$. For each $x\in{U}_{n}$, set ${L}_{n+1}({\tau}_{x})=x$. This completes the construction. We now check that it is as required.

Note that Lemma 6 implies that for any $\sigma\in{T}_{\alpha}$, there is a unique minimal extension of $\sigma$ which is active.

The construction in this paper is very specific to ${\omega}_{1}$; indeed, it is possible to show that $\mathfrak{d}$ is not always an upper bound for ${\mathfrak{a}}_{closed}$. A modification of the methods of Section 4 of [4] shows that if $\kappa$ is a measurable cardinal and if $\lambda=\operatorname{cf}{(\lambda)}={\lambda}^{\kappa}>\mu=\operatorname{cf}{(\mu)}>\kappa$, then there is a c.c.c. poset $\mathbb{P}$ such that $\left|\mathbb{P}\right|=\lambda$, and $\mathbb{P}$ forces that $\mathfrak{b}=\mathfrak{d}=\mu$ and ${\mathfrak{a}}={\mathfrak{a}}_{closed}=\mathfrak{c}=\lambda$.

As mentioned in Section 1, we see the result in this paper as providing a weak positive answer to the following basic question, which has remained open for long.

There are also several open questions about upper and lower bounds for ${\mathfrak{a}}_{closed}$.

Regarding Question 10, it is proved in Brendle and Khomskii [1] that if ${\mathbf{V}}$ is any ground model satisfying $\mathrm{CH}$ and $\mathbb{P}$ is any poset forcing that all splitting families in ${\mathbf{V}}$ remain splitting families in ${{\mathbf{V}}[G]}$, then $\mathbb{P}$ also forces that ${\mathfrak{a}}_{closed}={\aleph}_{1}$. This result suggests that Question 10 should have a positive answer, and showing this would be an improvement of the result in this paper.