# Filters, ideal independence and ideal Mrówka spaces

S. Bardyla  Institute of Mathematics, Univeristy of Vienna, Kolingasse 14-16,
1090 Vienna, Austria<EMAIL_ADDRESS>,  J. Cancino-Manríquez  Institute of
Mathematics, Czech Academy of Sciences, Žitná 25, Praha 1, Praha, Czech
Republic<EMAIL_ADDRESS>,  V. Fischer  Institute of Mathematics,
Univeristy of Vienna, Kolingasse 14-16, 1090 Vienna, Austria
<EMAIL_ADDRESS>and  C. Bacal Switzer  Institute of Mathematics,
Univeristy of Vienna, Kolingasse 14-16, 1090 Vienna, Austria
<EMAIL_ADDRESS>
###### Abstract.

A family $\mathcal{A}\subseteq[\omega]^{\omega}$ such that for all finite
$\\{X_{i}\\}_{i\in n}\subseteq\mathcal{A}$ and
$A\in\mathcal{A}\setminus\\{X_{i}\\}_{i\in n}$, the set
$A\setminus\bigcup_{i\in n}X_{i}$ is infinite, is said to be ideal
independent.

We prove that an ideal independent family $\mathcal{A}$ is maximal if and only
if $\mathcal{A}$ is $\mathcal{J}$-completely separable and maximal
$\mathcal{J}$-almost disjoint for a particular ideal $\mathcal{J}$ on
$\omega$. We show that $\mathfrak{u}\leq\mathfrak{s}_{mm}$, where
$\mathfrak{s}_{mm}$ is the minimal cardinality of maximal ideal independent
family. This, in particular, establishes the independence of
$\mathfrak{s}_{mm}$ and $\mathfrak{i}$. Given an arbitrary set $C$ of
uncountable cardinals, we show how to simultaneously adjoin via forcing
maximal ideal independent families of cardinality $\lambda$ for each
$\lambda\in C$, thus establishing the consistency of
$C\subseteq\hbox{spec}(\mathfrak{s}_{mm})$. Assuming $\mathsf{CH}$, we
construct a maximal ideal independent family, which remains maximal after
forcing with any proper, ${}^{\omega}\omega$-bounding, $p$-point preserving
forcing notion and evaluate $\mathfrak{s}_{mm}$ in several well studied
forcing extensions.

We also study natural filters associated with ideal independence and introduce
an analog of Mrówka spaces for ideal independent families.

###### Key words and phrases:

Ideal independent family, ideal, filter, ultrafilter, cardinal characteristic,
Mrowka space

###### 2000 Mathematics Subject Classification:

03E35, 03E17, 54A35

_Acknowledgments._ : The first author was supported by the Austrian Science
Fund FWF (Grant ESP 399). The second, third and fourth authors would like to
thank the Austrian Science Fund (FWF) for the generous support through START
Grant Y1012-N35.

##  1\. Introduction

Given a family $\mathcal{A}\subseteq[\omega]^{\omega}$, the ideal generated by
$\mathcal{A}$ is the collection of all $X\subseteq\omega$ so that
$X\subseteq^{*}\bigcup_{i<n}A_{i}$ for some $\\{A_{i}\\}_{i\in
n}\subseteq\mathcal{A}$ where $\subseteq^{*}$ means inclusion mod finite. A
family $\mathcal{A}$ is ideal independent if no $A\in\mathcal{A}$ is in the
ideal generated by $\mathcal{A}\setminus\\{A\\}$. Almost disjoint families and
independent families are both examples of ideal independent families. An easy
application of Zorn’s lemma shows that there are maximal ideal independent
families, however it is not always the case that a maximal almost disjoint or
a maximal independent family is maximal ideal independent. In this paper to
each ideal independent family $\mathcal{A}$ we correspond an ideal
$\mathcal{J}(\mathcal{A})$ (see Definition 3.1) which allows us to
characterize maximal ideal independent families as follows (see Theorem 3.3):

######  Theorem.

An ideal independent family $\mathcal{A}$ is maximal if and only if
$\mathcal{A}$ is maximal $\mathcal{J}(\mathcal{A})$\- almost disjoint and
$\mathcal{J}(\mathcal{A})$-completely separable.

By $\mathfrak{s}_{mm}$ we denote the least cardinality of a maximal ideal
independent family. An earlier investigation of $\mathfrak{s}_{mm}$ can be
found [5], where it is shown that ${\rm
max}\\{\mathfrak{d},\mathfrak{r}\\}\leq\mathfrak{s}_{mm}$ and that each of the
following inequalities $\mathfrak{u}<\mathfrak{s}_{mm}$,
$\mathfrak{s}_{mm}<\mathfrak{i}$, $\mathfrak{s}_{mm}<\mathfrak{c}$ is
consistent. Here $\mathfrak{d}$, $\mathfrak{u}$, $\mathfrak{r}$,
$\mathfrak{i}$ denote the dominating number, the ultrafilter number, the
reaping and independence numbers, respectively. We refer the reader to [1] for
definitions and basic properties of the combinatorial cardinal
characteristics, which are not stated here. Strengthening and complementing
the above results, in Section 2, we establish the following $\mathsf{ZFC}$
inequality, which also answers Question 17 of [5], see Theorem 3.7:

######  Theorem.

$\mathfrak{u}\leq\mathfrak{s}_{mm}$.

Consequently, we obtain the independence of $\mathfrak{s}_{mm}$ and
$\mathfrak{i}$, as the consistency of $\mathfrak{s}_{mm}<\mathfrak{i}$ is
shown in [5, Theorem 16], while the consistency of $\mathfrak{i}<\mathfrak{u}$
is established in Shelah’s [17] and hence by the above theorem,
$\mathfrak{i}<\mathfrak{s}_{mm}$ holds in the latter model.

######  Corollary.

$\mathfrak{s}_{mm}$ and $\mathfrak{i}$ are independent.

A key role in our investigations is taken by specific filters, which are
naturally associated to a given ideal independent family. On one side, these
are filters to which we refer as complemented filters, see Definition 3.1 and
on the other side, filters resembling the notion of a diagonalization filter
for an independent family, see for example [8, Definition 1]. In difference
with earlier instances of diagonalization reals, associated to say almost
disjoint families, towers, or cofinitary groups, the existence of a
diagonalization real for a given ideal independent family, employs a Cohen
real (see Lemma 4.1). Adjoining diagonalization reals for ideal independent
families cofinally along an appropriate finite support iteration, as well as
building on and modifying earlier forcing constructions used to control for
example the spectrum of independence (see in particular [8, 9]) we establish
the following (see Theorem 4.4):

######  Theorem.

(GCH) Let $R$ be a set of regular uncountable cardinals. Then, there is a ccc
generic extension in which for every $\lambda\in R$ there is a maximal ideal
independent family of cardinality $\lambda$.

Moreover, we look at the preservation of small witnesses to
$\mathfrak{s}_{mm}$. The preservation of the maximality of extremal sets of
reals, like mad families, maximal eventually different families of reals, or
maximal independent families, under forcing iterations is usually a non-
trivial task and often involves the construction of a combinatorial object
which is maximal in a strong sense, examples given by tight almost disjoint
and selective independent families. Partially inspired by the notion of a
$\mathcal{U}$-supported maximal independent family in the higher Baire spaces
given in [7], in Definition 5.1 we introduce the notion of an _$\mathcal{U}$
-encompassing ideal independent family_ and establish the following powerful
preservation result (see Theorems 5.2 and 5.3).

######  Theorem.

(CH) There is a maximal ideal independent family $\mathcal{A}$ which remains
maximal, and so a witness to $\mathfrak{s}_{mm}=\aleph_{1}$, in any generic
extension obtained by a proper, $\omega^{\omega}$-bounding, $p$-points
preserving forcing notion.

The above theorem applies to a large class of partial orders and implies that
in many well-studied forcing extensions,
$\mathfrak{s}_{mm}=\max\\{\mathfrak{d},\mathfrak{u}\\}$.

Finally, to each ideal independent family $\mathcal{A}$ we correspond a
topological space $\psi(\mathcal{A})$ which is a generalized version of the
well-known Mrówka space. Mrówka spaces play important role in set-theoretic
topology (see [12] and references therein). In this paper we characterize
ideal independent families whose corresponding spaces are locally compact or
pseudocompact (see Propositions 6.6 and 6.7). Suprisingly this corresponds to
the case of almost disjoint families, thus suggesting that Mrówka spaces of
non-almost disjoint families behave very differently to the classical theory.
Specifically we show the following.

######  Theorem.

For a maximal ideal independent family $\mathcal{A}$ the following are
equivalent.

  1. (1)

$\psi(\mathcal{A})$ is pseudocompact.

  2. (2)

$\mathcal{A}$ is MAD family.

  3. (3)

$\mathcal{A}$ is a completely separable MAD family.

We conclude the paper with a brief discussion of remaining open questions.

##  2\. Preliminaries

Recall that given $f,g\in\omega$ we write $f\leq^{*}g$ ($f$ is eventually
dominated by $g$), provided there is $n\in\omega$ such that for all $m\geq n$,
$f(m)\leq g(m)$. The cardinal $\mathfrak{b}$, the bounding number is the least
size of a family $A\subseteq\omega^{\omega}$ so that no single
$f\in\omega^{\omega}$ eventually dominates every $g\in A$. Dually, the
dominating number, $\mathfrak{d}$, is the least size of a dominating family,
that is a family $D\subseteq\omega^{\omega}$ so that every
$f\in\omega^{\omega}$ is eventually dominated by some $g\in D$. We denote by
$[\omega]^{\omega}$ the Ramsey space, that is the Polish space of infinite
subsets of $\omega$. Often it is convenient to quotient this space by the
ideal $\mathsf{Fin}$ consisting of finite subsets of $\omega$. For instance,
if $A,B$ in $[\omega]^{\omega}$ then we write $A\subseteq^{*}B$, read “$A$ is
almost contained in $B$” if $A\setminus B$ is finite. Similarly we say that
$A$ and $B$ are almost equal, denoted $A=^{*}B$, if their symmetric difference
is finite, and we say that $A$ and $B$ are almost disjoint, denoted $A\cap
B=^{*}\emptyset$, if their intersection is finite.

Given a family $\mathcal{A}$ of subsets of $\omega$ by $\mathcal{A}^{+}$ we
denote the collection of all subsets of $\omega$ which are not in the ideal
generated by $\mathcal{A}$. For an ideal $\mathcal{J}$ on $\omega$ which
contains the ideal $\mathsf{Fin}$ and subsets $A,B$ of $\omega$ we write
$A\subseteq^{\mathcal{J}}B$ if $A\setminus B\in\mathcal{J}$. A family
$\mathcal{A}\subseteq\mathcal{J}^{+}$ is called $\mathcal{J}$-almost disjoint
if $A_{1}\cap A_{2}\in\mathcal{J}$ for any distinct
$A_{1},A_{2}\in\mathcal{A}$. Given a family
$\mathcal{B}\subseteq[\omega]^{\omega}\setminus\mathcal{J}$, let

$\mathcal{B}^{+_{\mathcal{J}}}=[\omega]^{\omega}\setminus\\{X\in[\omega]^{\omega}:X\subseteq^{\mathcal{J}}\bigcup\mathcal{Y},\hbox{
where }\mathcal{Y}\in[\mathcal{B}]^{<\omega}\\}$

and

$\mathcal{B}^{++_{\mathcal{J}}}=\\{X\in[\omega]^{\omega}:\exists\\{B_{i}:i\in\omega\\}\in[\mathcal{B}]^{\omega}\hbox{
such that }X\cap B_{i}\notin\mathcal{J}\quad\forall i\in\omega\\}.$

Elements of the set $\mathcal{B}^{+_{\mathcal{J}}}$ are called positive with
respect to the ideal $\mathcal{J}$. If $\mathcal{J}=\mathsf{Fin}$, then we
write $\mathcal{B}^{+}$ ($\mathcal{B}^{++}$, resp.) instead of
$\mathcal{B}^{+_{\mathcal{J}}}$ ($\mathcal{B}^{++_{\mathcal{J}}}$, resp.).

Given a family $\mathcal{A}\subseteq[\omega]^{\omega}$ we say that
$\mathcal{A}$ has the finite intersection property if any finite subfamily has
infinite intersection. An infinite subset $B\subseteq\omega$ is called a
pseudo-intersection of a family $\mathcal{A}\subseteq[\omega]^{\omega}$ if
$B\subseteq^{*}A$ for every $A\in\mathcal{A}$. The cardinal characteristic
$\mathfrak{p}$ is the least size of a family with the finite intersection
property with no pseudo-intersection. An ultrafilter is said to be principal
if it contains a singleton and non-principal otherwise. Unless otherwise
stated we will assume all ultrafilters to be non-principal. If $\mathscr{U}$
is an ultrafilter then a base for $\mathscr{U}$ is a subset
$\mathcal{B}\subseteq\mathscr{U}$ so that every element $A\in\mathscr{U}$
almost contains some $B\in\mathcal{B}$. In this case, we say that
$\mathcal{B}$ generates $\mathscr{U}$, sometimes denoted
$\langle\mathcal{B}\rangle$. The ultrafilter number $\mathfrak{u}$, is the
least size of a base of a non-principal ultrafilter. For an ultrafilter
$\mathscr{U}$ we say that

  1. (1)

$\mathscr{U}$ is a $p$-point if every countable subfamily of $\mathscr{U}$ has
a pseudo-intersection in $\mathscr{U}$.

  2. (2)

$\mathscr{U}$ is a q-point if every partition of $\omega$ into finite sets
$\\{I_{n}\\}_{n<\omega}$ there is a $U\in\mathscr{U}$ so that $|U\cap
I_{n}|=1$ for each $n<\omega$.

  3. (3)

$\mathscr{U}$ is Ramsey, or, selective if it is a $p$-point and a q-point.

  4. (4)

$\mathscr{U}$ is a $p_{\mathfrak{c}}$-point if any
$\mathscr{F}\subseteq\mathscr{U}$, $|\mathscr{F}|<\mathfrak{c}$ has a pseudo-
intersection in $\mathscr{U}$.

Finally, a family $\mathcal{I}\subseteq[\omega]^{\omega}$ is said to be
independent if whenever $\mathcal{A},\mathcal{B}$ are finite, disjoint, non-
empty subfamilies of $\mathcal{I}$, the set $\bigcap
A\setminus\bigcup\mathcal{B}$ is infinite. While this notion technically makes
sense for finite families it is somewhat degenerate in this case thus we will
implicitly mean by an independent family an infinite independent family. The
least size of a maximal (infinite) independent family is denoted
$\mathfrak{i}$.

##  3\. Filters, Almost Disjointness and Ideal Independence

The following definition plays a central role in this section.

######  Definition 3.1.

Let $\mathcal{A}$ be an ideal independent family and $A\in\mathcal{A}$.

  1. (1)

By $\mathcal{F}(\mathcal{A},A)$ we denote the filter generated by the family

$\\{A\setminus\bigcup\mathcal{B}:A\notin\mathcal{B}\in[\mathcal{A}]^{<\omega}\\}.$

  2. (2)

By $\mathcal{J}(\mathcal{A},A)$ we denote the ideal generated by the family

$\\{A\cap B:B\in\mathcal{A}\setminus\\{A\\}\\}\cup\mathsf{Fin}.$

  3. (3)

By $\mathcal{J}(\mathcal{A})$ we denote the ideal generated by the family
$\bigcup_{A\in\mathcal{A}}\mathcal{J}(\mathcal{A},A)$.

We refer to the filters of the form $\mathcal{F}(\mathcal{A},A)$ as the
complemented filters of $\mathcal{A}$, while for a fixed $A\in\mathcal{A}$ we
say that $\mathcal{F}(\mathcal{A},A)$ is the complemented filter (of
$\mathcal{A}$) corresponding to $A$. Observe that for any $A\in\mathcal{A}$ a
subset $B\subseteq A$ belongs to the ideal $\mathcal{J}(\mathcal{A},A)$ if and
only if $A\setminus B\in\mathcal{F}(\mathcal{A},A)$.

Note that an ideal independent family $\mathcal{A}$ is maximal if and only if
every $X\in[\omega]^{\omega}$ is either in the ideal generated by
$\mathcal{A}$ or belongs to at least one of the filters
$\mathcal{F}(\mathcal{A},A)$ (these two possibilities are not mutually
exclusive). The name “complemented” comes from this observation: under
maximality every element of the complement of the ideal generated by
$\mathcal{A}$ is in some complemented filter.

The proof of the following lemma is straightforward.

######  Lemma 3.2.

For a family $\mathcal{A}\subseteq[\omega]^{\omega}$ the following statements
hold:

  1. (1)

if for some ideal $\mathcal{J}$ on $\omega$ the family $\mathcal{A}$ is
$\mathcal{J}$-almost disjoint, then $\mathcal{A}$ is ideal independent;

  2. (2)

if the family $\mathcal{A}$ is ideal independent, then $\mathcal{A}$ is
$\mathcal{J}(\mathcal{A})$-almost disjoint.

######  Definition.

A $\mathcal{J}$-almost disjoint family $\mathcal{A}$ is called
$\mathcal{J}$-completely separable if for any
$B\in\mathcal{A}^{++_{\mathcal{J}}}$ there exists $A\in\mathcal{A}$ such that
$A\subseteq^{\mathcal{J}}B$.

One can easily check that
$\mathcal{A}^{+_{\mathcal{J}}}=\mathcal{A}^{++_{\mathcal{J}}}$ for any maximal
$\mathcal{J}$-almost disjoint family $\mathcal{A}$. In case
$\mathcal{J}=\mathsf{Fin}$, we refer to $\mathcal{J}$-completely separable
family simply as completely separable. The existence of a completely separable
maximal almost disjoint family in ZFC is still an open problem (see [13] for
the history thereof). Completely separable maximal almost disjoint families
are intensively studied in the literature (see the survey [11] and references
therein). $\mathcal{J}$-almost disjoint families were recently investigated in
[16]. The following theorem establishes a connection between
$\mathcal{J}$-completely separable $\mathcal{J}$-almost disjoint families and
maximal ideal independent families.

######  Theorem 3.3.

An ideal independent family $\mathcal{A}$ is maximal if and only if
$\mathcal{A}$ is maximal $\mathcal{J}(\mathcal{A})$-almost disjoint and
$\mathcal{J}(\mathcal{A})$-completely separable.

###### Proof.

Let $\mathcal{A}$ be a maximal ideal independent family. By Lemma 3.2, the
family $\mathcal{A}$ is $\mathcal{J}(\mathcal{A})$-almost disjoint. To derive
a contradiction assume that there exists
$B\in\mathcal{J}(\mathcal{A})^{+}\setminus\mathcal{A}$ such that $B\cap
A\in\mathcal{J}(\mathcal{A})$ for any $A\in\mathcal{A}$. Then
$\mathcal{A}\cup\\{B\\}$ is ideal independent, which contradicts the
maximality of $\mathcal{A}$. Thus, the family $\mathcal{A}$ is maximal
$\mathcal{J}(\mathcal{A})$-almost disjoint. To show that the family
$\mathcal{A}$ is $\mathcal{J}(\mathcal{A})$-completely separable fix any
$C\in\mathcal{A}^{++_{\mathcal{J}(\mathcal{A})}}$. Note that
$C\in\mathcal{A}^{+}\setminus\mathcal{A}$. By the maximality of $\mathcal{A}$
there exist $A\in\mathcal{A}$ and $\\{A_{i}:i\in
n\\}\in[\mathcal{A}\setminus\\{A\\}]^{<\omega}$ such that
$A\subseteq^{*}\bigcup_{i\in n}A_{i}\cup C$. Hence

$A=A\cap A\subseteq^{*}A\cap(\bigcup_{i\in n}A_{i}\cup
C)\subseteq\big{(}\bigcup_{i\in n}(A\cap A_{i})\big{)}\cup C.$

Then $A\subseteq^{\mathcal{J}(\mathcal{A})}C$, as $\bigcup_{i\in n}(A\cap
A_{i})\in\mathcal{J}(\mathcal{A},A)\subseteq\mathcal{J}(\mathcal{A})$. It
follows that the family $\mathcal{A}$ is $\mathcal{J}(\mathcal{A})$-completely
separable.

Assume that the family $\mathcal{A}$ is maximal
$\mathcal{J}(\mathcal{A})$-almost disjoint and
$\mathcal{J}(\mathcal{A})$-completely separable. By Lemma 3.2, the family
$\mathcal{A}$ is ideal independent. Fix any $C\in\mathcal{A}^{+}$. Let us show
that $C\in\mathcal{A}^{+_{\mathcal{J}(\mathcal{A})}}$. Assuming the contrary
we can find $\\{A_{i}:i\in n\\}\in[\mathcal{A}]^{<\omega}$ and
$D\in\mathcal{J}(\mathcal{A})$ such that $C\subseteq\bigcup_{i\in n}A_{i}\cup
D$. Then there exists $\\{D_{i}:i\in m\\}\in[\mathcal{A}]^{<\omega}$ such that
$D\subseteq^{*}\bigcup_{i\in m}D_{i}$. But then $C\subseteq^{*}\bigcup_{i\in
n}A_{i}\cup(\bigcup_{i\in m}D_{i})$, witnessing that $C\notin\mathcal{A}^{+}$.
The obtained contradiction implies that
$C\in\mathcal{A}^{+_{\mathcal{J}(\mathcal{A})}}$. Taking into account that
$\mathcal{A}^{+_{\mathcal{J}(\mathcal{A})}}=\mathcal{A}^{++_{\mathcal{J}(\mathcal{A})}}$,
the $\mathcal{J}(\mathcal{A})$-complete separability implies the existence of
$A\in\mathcal{A}$ such that $A\subseteq^{\mathcal{J}(\mathcal{A})}C$. By the
definition of $\mathcal{J}(\mathcal{A})$, the set $A\setminus C$ belongs to
$\mathcal{J}(\mathcal{A})$ and thus it can be covered mod finite by the union
of finitely many elements $E_{1},\ldots,E_{m}\in\mathcal{A}\setminus\\{A\\}$,
witnessing that $C\in\mathcal{F}(\mathcal{A},A)$. Hence the family
$\mathcal{A}$ is maximal ideal independent. ∎

The following corollary of Theorem 3.3 describes almost disjoint families
which are maximal ideal independent.

######  Corollary 3.4.

An almost disjoint family $\mathcal{A}$ is maximal ideal independent if and
only if $\mathcal{A}$ is a completely separable maximal almost disjoint
family.

Note that, independent families are never maximal ideal independent. Indeed,
for every element $A$ of an independent family $\mathcal{A}$ the family
$\mathcal{A}\cup\\{\omega\setminus A\\}$ is ideal independent.

For a filter $\mathcal{F}$ let
$\chi(\mathcal{F})=\min\\{|\mathcal{B}|:\mathcal{B}$ is a base of
$\mathcal{F}\\}$.

######  Theorem 3.5.

Let $\mathcal{A}=\\{A_{\alpha}:\alpha\in\kappa\\}$ be a (maximal) ideal
independent family and for any $\alpha\in\kappa$, $\mathcal{G}_{\alpha}$ be a
filter on $\omega$ such that
$\mathcal{F}(\mathcal{A},A_{\alpha})\subseteq\mathcal{G}_{\alpha}$ and
$\chi(\mathcal{G}_{\alpha})\leq\kappa$. Then there exists a (maximal) ideal
independent family
$\mathcal{A}^{\prime}=\\{A^{\prime}_{\alpha}:\alpha\in\kappa\\}$ such that
$\mathcal{F}(\mathcal{A}^{\prime},A^{\prime}_{\alpha})=\mathcal{G}_{\alpha}$
for all $\alpha\in\kappa$.

###### Proof.

Fix an ideal independent family $\mathcal{A}=\\{A_{\alpha}:\alpha\in\kappa\\}$
and filters $\mathcal{G}_{\alpha}$, $\alpha\in\kappa$ which satisfy the
conditions above. For each $\alpha\in\kappa$ let $\mathcal{C}_{\alpha}$ be a
base of $\mathcal{G}_{\alpha}$ of cardinality $\leq\kappa$. Moreover assume
that for every $\alpha\in\kappa$ and $C\in\mathcal{C}_{\alpha}$ we have that
$C\subseteq A_{\alpha}$. The latter assumption implies that
$\mathcal{C}_{\alpha}\cap\mathcal{C}_{\beta}=\emptyset$ for every
$\alpha\neq\beta$. Enumerate the set
$\mathcal{D}=\bigcup_{\alpha\in\kappa}\mathcal{C}_{\alpha}$ as
$\\{D_{\alpha}:\alpha\in\kappa\\}$, additionally assuming that
$D_{\alpha}\notin\mathcal{C}_{\alpha}$ for every $\alpha\in\kappa$. For each
$\alpha\in\kappa$ let
$A^{\prime}_{\alpha}=A_{\alpha}\cup(A_{\sigma(\alpha)}\setminus D_{\alpha})$,
where $D_{\alpha}\in\mathcal{C}_{\sigma(\alpha)}$. Since for every
$\alpha\neq\beta$ the set $\mathcal{C}_{\alpha}\cap\mathcal{C}_{\beta}$ is
empty, the ordinal $\sigma(\alpha)$ is unique and, thus, well-defined. Let us
show that the family
$\mathcal{A}^{\prime}=\\{A_{\alpha}^{\prime}:\alpha\in\kappa\\}$ is ideal
independent. Fix any $\alpha\in\kappa$ and $\\{\beta_{i}:i\in
n\\}\in[\kappa\setminus\\{\alpha\\}]^{<\omega}$. We need to show that
$A_{\alpha}^{\prime}\setminus\bigcup_{i\in
n}A_{\beta_{i}}^{\prime}\neq^{*}\emptyset$. There are two cases:

  1. (1)

$\alpha\notin\\{\sigma(\beta_{i}):i\in n\\}$;

  2. (2)

$\alpha\in\\{\sigma(\beta_{i}):i\in n\\}$.

_Case 1)_ Observe that $A_{\alpha}\subseteq A_{\alpha}^{\prime}$ and

$\bigcup_{i\in n}A_{\beta_{i}}^{\prime}=\bigcup_{i\in
n}(A_{\beta_{i}}\cup(A_{\sigma(\beta_{i})}\setminus
D_{\beta_{i}}))\subseteq\bigcup_{i\in n}(A_{\beta_{i}}\cup
A_{\sigma(\beta_{i})}).$

Therefore, $A_{\alpha}^{\prime}\setminus\bigcup_{i\in
n}A_{\beta_{i}}^{\prime}\subseteq A_{\alpha}\setminus\bigcup_{i\in
n}(A_{\beta_{i}}\cup A_{\sigma(\beta_{i})})\neq^{*}\emptyset$, as $\alpha$
does not belong to $\\{\beta_{i}:i\in n\\}\cup\\{\sigma(\beta_{i}):i\in n\\}$
and the family $\mathcal{A}$ is ideal independent.

_Case 2)_ Let $\Gamma=\\{i\in n:\alpha=\sigma(\beta_{i})\\}$. Since
$A_{\alpha}\cap\bigcap_{i\in\Gamma}D_{\beta_{i}}\in\mathcal{G}_{\alpha}$ we
obtain:

$\displaystyle A_{\alpha}^{\prime}\setminus\bigcup_{i\in
n}(A_{\beta_{i}}\cup(A_{\sigma(\beta_{i})}\setminus D_{\beta_{i}}))\supseteq$
$\displaystyle\big{(}A_{\alpha}\setminus(\bigcup_{i\in
n}A_{\beta_{i}})\big{)}\cap(A_{\alpha}\cap\bigcap_{i\in\Gamma}D_{\beta_{i}})\cap(A_{\alpha}\setminus(\bigcup_{i\notin\Gamma}A_{\sigma(\beta_{i})}))\in\mathcal{G}_{\alpha}.$

Hence $A_{\alpha}^{\prime}\setminus\bigcup_{i\in
n}A_{\beta_{i}}^{\prime}\neq^{*}\emptyset$, witnessing that the family
$\mathcal{A}_{\alpha}^{\prime}$ is ideal independent.

Fix any $\alpha\in\kappa$ and $C\in\mathcal{C}_{\alpha}$. There is
$\xi\in\kappa\setminus\\{\alpha\\}$ such that $C=D_{\xi}\subseteq A_{\xi}$.
Observe that $\sigma(\xi)=\alpha$. Then

$\displaystyle A_{\alpha}^{\prime}\setminus(A^{\prime}_{\sigma(\alpha)}\cup
A^{\prime}_{\xi})$
$\displaystyle\subseteq(A_{\alpha}\cup(A_{\sigma(\alpha)}\setminus
D_{\alpha}))\setminus\big{(}A_{\sigma(\alpha)}\cup(A_{\sigma(\xi)}\setminus
D_{\xi})\big{)}$ $\displaystyle\subseteq
A_{\alpha}\setminus(A_{\alpha}\setminus D_{\xi})=C,$

witnessing that
$\mathcal{G}_{\alpha}\subseteq\mathcal{F}(\mathcal{A}^{\prime},A^{\prime}_{\alpha})$.
To show the converse inclusion fix any
$F\in\mathcal{F}(\mathcal{A}^{\prime},A^{\prime}_{\alpha})$. Then there exists
$\\{\alpha_{i}:i\in n\\}\in[\kappa\setminus\\{\alpha\\}]^{<\omega}$ such that

$A_{\alpha}^{\prime}\setminus\bigcup_{i\in
n}A_{\alpha_{i}}^{\prime}=\big{(}A_{\alpha}\cup(A_{\sigma(\alpha)}\setminus
D_{\alpha})\big{)}\setminus\bigcup_{i\in
n}\big{(}A_{\alpha_{i}}\cup(A_{\sigma(\alpha_{i})}\setminus
D_{\alpha_{i}})\big{)}\subseteq^{*}F.$

It is straightforward to check that if $\alpha$ is not in
$\\{\sigma(\alpha_{i}):i\in n\\}$, then

$A_{\alpha}^{\prime}\setminus\bigcup_{i\in
n}A_{\alpha_{i}}^{\prime}\in\mathcal{F}(\mathcal{A},A_{\alpha})\subseteq\mathcal{G}_{\alpha},$

implying that $F\in\mathcal{G}_{\alpha}$. Assume that the set $\Gamma=\\{i\in
n:\sigma(\alpha_{i})=\alpha\\}$ is not empty. Then

$F\supseteq^{*}A_{\alpha}^{\prime}\setminus\bigcup_{i\in
n}A_{\alpha_{i}}^{\prime}\supseteq\big{(}A_{\alpha}\setminus\bigcup_{i\in
n}A_{\alpha_{i}}\big{)}\cap(A_{\alpha}\cap\bigcap_{i\in\Gamma}D_{\alpha_{i}})\cap(A_{\alpha}\setminus\bigcup_{i\notin\Gamma}A_{\sigma(\alpha_{i})}\big{)}\in\mathcal{G}_{\alpha}.$

Hence
$\mathcal{G}_{\alpha}=\mathcal{F}(\mathcal{A}^{\prime},B_{\alpha}^{\prime})$
for every $\alpha\in\kappa$.

Assume the ideal independent family $\mathcal{A}$ is maximal and fix any
$C\in\mathcal{A}^{\prime+}$. It is easy to see that
$\mathcal{A}^{+}=\mathcal{A}^{\prime+}$. By maximality of $\mathcal{A}$ there
is $\alpha\in\kappa$ such that
$C\in\mathcal{F}(\mathcal{A},A_{\alpha})\subseteq\mathcal{G}_{\alpha}=\mathcal{F}(\mathcal{A}^{\prime},A_{\alpha}^{\prime})$,
witnessing that the family $\mathcal{A}^{\prime}$ is maximal as well. ∎

In particular we obtain:

######  Corollary 3.6.

There exists a maximal ideal independent family $\mathcal{A}$ such that
$\mathcal{F}(\mathcal{A},A)$ is an ultrafilter for all $A\in\mathcal{A}$.

###### Proof.

Let $\mathcal{D}$ be an almost disjoint family of size $\mathfrak{c}$. Enlarge
it to a maximal ideal independent family $\mathcal{A}$. For any
$A\in\mathcal{A}$ fix any ultrafilter $\mathcal{G}_{A}$ which contains the
filter $\mathcal{F}(\mathcal{A},A)$. Theorem 3.5 implies that there exists a
maximal ideal independent family $\mathcal{A}^{\prime}$ such that for any
$A^{\prime}\in\mathcal{A}^{\prime}$ the filter
$\mathcal{F}(\mathcal{A}^{\prime},A^{\prime})$ coincides with the ultrafilter
$\mathcal{G}_{A}$. ∎

We want to point out, that under the assumption $\mathfrak{p}=\mathfrak{c}$
one can obtain in addition the existence of a maximal ideal independent family
$\mathcal{A}$ with the property that all complemented filters of $\mathcal{A}$
are $p_{\mathfrak{c}}$-points.

The relation between ideal independent families and ultrafilters can be
formalized by the following $\mathsf{ZFC}$ theorem.

######  Theorem 3.7.

$\mathfrak{u}\leq\mathfrak{s}_{mm}$.

###### Proof.

Assume otherwise $\mathfrak{s}_{mm}<\mathfrak{u}$ and let $\mathcal{A}$ be a
maximal ideal independent family of minimal cardinality. Maximality implies
that there is a countable subfamily $\\{A_{n}\\}_{n\in\omega}$ of
$\mathcal{A}$ whose union is almost equal to $\omega$. Define $B_{0}=A_{0}$
and for $n>0$, $B_{n}=A_{n}\setminus\bigcup_{i<n}A_{i}$. For each
$n\in\omega$, let $\mathcal{F}_{n}$ be the filter
$\mathcal{F}(\mathcal{A},A_{n})\upharpoonright B_{n}$. Since $\mathcal{F}_{n}$
is not an ultrafilter (by assumption on the size of $\mathfrak{u}$), for any
$s\in 2^{n}$ there is $D_{s}\in\mathcal{F}_{n}^{+}\setminus\mathcal{F}_{n}$
such that for different $s,r\in 2^{n}$, $D_{r}\cap D_{s}=\emptyset$. Now, for
each $f\in 2^{\omega}$, define $D^{f}=\bigcup_{n\in\omega}D_{f\upharpoonright
n}$.

######  Claim.

For each $f\in 2^{\omega}$ and each $\mathcal{F}\in[\mathcal{A}]^{<\omega}$,
$D^{f}\not\subseteq\bigcup\mathcal{F}$.

###### Proof.

Let $\mathcal{F}\in[\mathcal{A}]^{<\omega}$ and let $n\in\omega$ be such that
$A_{n}\notin\mathcal{F}$. We can assume that
$\\{A_{i}:i<n\\}\subseteq\mathcal{F}$. Since $D_{f\upharpoonright n}$ is
$\mathcal{F}_{n}$-positive, it can not be covered by $\bigcup F$ and so
$D^{f}$ can not be covered by $\bigcup\mathcal{F}$ either. ∎

By maximality of $\mathcal{A}$, for any $f\in 2^{\omega}$, there are

$A_{f}\in\mathcal{A}\hbox{ and
}\mathcal{F}_{f}\in[\mathcal{A}\setminus\\{A_{f}\\}]^{<\omega}$

such that $A_{f}\setminus\bigcup\mathcal{F}_{f}\subseteq^{*}D^{f}$. Note that
for no $n\in\omega$ we have $A_{n}=A_{f}$, as otherwise, for some $n\in\omega$
we would have $A_{n}\setminus\mathcal{F}_{f}\subseteq^{*}D^{f}$, which implies

$B_{n}\setminus\bigcup\mathcal{F}_{f}=A_{n}\setminus\left(\bigcup\mathcal{F}_{f}\cup\bigcup_{i<n}A_{i}\right)\subseteq^{*}D_{f\upharpoonright
n},$

contradicting the choice of the set $D_{f\upharpoonright n}$. Since
$\mathfrak{s}_{mm}<\mathfrak{c}$, there are different $f,g\in 2^{\omega}$ such
that $A_{f}=A_{g}$ and $\mathcal{F}_{f}=\mathcal{F}_{g}$. By construction, we
have

$A_{f}\setminus\mathcal{F}_{f}\subseteq^{*}D^{f}\cap
D^{g}\subseteq\bigcup_{i\leq n_{0}}B_{i},$

where $n_{0}\in\omega$ is the maximal natural number such that
$f\upharpoonright n_{0}=g\upharpoonright n_{0}$. But $\bigcup_{i\leq
n_{0}}B_{i}=\bigcup_{i\leq n_{0}}A_{i}$, which means that
$A_{f}\setminus\bigcup\mathcal{F}_{f}\subseteq^{*}\bigcup_{i\leq
n_{0}}A_{n_{0}}$, and so
$A_{f}\subseteq^{*}\bigcup\mathcal{F}_{f}\cup\bigcup_{i\leq n_{0}}A_{i}$, a
contradiction. ∎

Examining the proof of Theorem 3.7 one observes that in fact we only needed
$\mathfrak{s}_{mm}<\mathfrak{c}$ alongside the existence of countably
infinitely many elements $A_{n}\in\mathcal{A}$ whose associated complemented
filters are not ultrafilters to obtain the contradiction. It follows that the
same proof can be used to derive the following proposition which complements
Corollary 3.6.

######  Proposition 3.8.

If $\mathcal{A}$ is a maximal ideal independent family of size
$<\mathfrak{c}$, then there are at most finitely many $A\in\mathcal{A}$ for
which the corresponding complemented filter is not an ultrafilter.

Note, this is not necessarily the case for maximal ideal independent families
with cardinality $\mathfrak{c}$, as any completely separable maximal almost
disjoint family $\mathcal{A}$ is maximal ideal independent (see Corollary
3.4), and for any $A\in\mathcal{A}$ the corresponding complemented filter
$\mathcal{F}(\mathcal{A},A)$ is not an ultrafilter, as it is generated by
cofinite subsets of $A$.

##  4\. Arbitrarily Large Maximal Ideal Independent Families

In this section we examine the question of how to adjoin via forcing a maximal
ideal independent family of desired cardinality and thus begin an
investigation of the spectrum of such families. The spectrum of maximal ideal
independent families, denoted ${\rm spec}(\mathfrak{s}_{mm})$, is defined as
the set of all cardinalities of maximal ideal independent families. Throughout
$V$ denotes the ground model and $\mathbb{C}$ denotes the poset for adding a
single Cohen real.

######  Lemma 4.1.

Let $\mathcal{A}$ be an ideal independent family. There is a $ccc$ forcing
$\mathbb{P}(\mathcal{A})$ which adds a set $z$ such that in
$V^{\mathbb{P}(\mathcal{A})}$:

  1. (1)

$\mathcal{A}\cup\\{z\\}$ is an ideal independent family, and

  2. (2)

for each $y\in V\cap([\omega]^{\omega}\backslash\mathcal{A})$ the family
$\mathscr{A}\cup\\{z,y\\}$ is not ideal independent.

###### Proof.

Add a Cohen real to $V$ and consider a filter $\mathcal{U}$ which contains the
Cohen real and is maximal with respect to the following property:

$(\ast)$ For any $X\in\mathcal{U}$, any $A\in\mathcal{A}$ and finite
$\mathcal{F}\subseteq\mathcal{A}\setminus\\{A\\}$, the set
$X\cap(A\setminus\bigcup\mathcal{F})$ is infinite.

Let $\mathbb{M}(\mathcal{U})$ be Mathias forcing relativized to $\mathcal{U}$,
let $x$ be the generic real added by $\mathbb{M}(\mathcal{U})$ over
$V^{\mathbb{C}}$, let $\dot{x}$ be $\mathbb{M}(\mathcal{U})$-name for $x$ (in
$V^{\mathbb{C}}$) and let
$\mathbb{P}(\mathcal{A})=\mathbb{C}*\mathbb{M}(\dot{\mathcal{U}})$.

######  Claim 4.2.

In $V^{\mathbb{P}(\mathcal{A})}$ the family $\mathcal{A}\cup\\{\omega\setminus
x\\}$ is ideal independent.

###### Proof.

Let $\mathcal{F}$ be a finite subset of $\mathcal{A}$. First we prove that
$\bigcup\mathcal{F}$ does not almost contain $\omega\setminus x$. Let
$A\in\mathcal{A}\backslash\mathcal{F}$, $(s,B)\in\mathbb{M}(\mathcal{U})$ and
let $n\in\omega$ be arbitrary. Since the Cohen real belongs to $\mathcal{U}$,
we can assume that $B$ is a subset of it, so
$A\setminus(B\cup\bigcup\mathcal{F})$ is infinite. Let $k\in\omega$ be big
enough so $[\max(s),k)\cap(A\setminus(B\cup\bigcup\mathcal{F}))$ has more than
$n$ elements. Then $(s\cup\\{k\\},B)$ forces that
$(\omega\setminus\dot{x})\setminus\bigcup\mathcal{F}$ has more than $n$
elements and since $n$ was arbitrary, it follows that $(\omega\setminus
x)\setminus\bigcup\mathcal{F}$ is infinite. A genericity argument shows that
$x\cap(A\setminus\bigcup\mathcal{F})$ is infinite for any $A\in\mathcal{A}$
and $\mathcal{F}\in[\mathcal{A}\setminus\\{A\\}]^{<\omega}$, which implies
that $(A\setminus\bigcup\mathcal{F})\setminus(\omega\setminus x)$ is infinite.
∎

######  Claim 4.3.

Let $A\in([\omega]^{\omega}\cap V)\backslash\mathcal{A}$. Then in
$V^{\mathbb{P}(\mathcal{A})}$, $\mathcal{A}\cup\\{\omega\setminus x,A\\}$ is
not ideal independent.

###### Proof.

Let $A\in[\omega]^{\omega}\cap V$ be an arbitrary set. If there are
$X\in\mathcal{U}$, $B\in\mathcal{A}$ and
$\mathcal{F}\in[\mathcal{A}\setminus\\{B\\}]^{<\omega}$ such that
$X\cap(B\setminus\bigcup\mathcal{F})\subseteq^{*}A$, then
$x\cap(B\setminus\bigcup\mathcal{F})\subseteq^{*}A$. But
$x\cap(B\setminus\bigcup\mathcal{F})=(B\setminus\bigcup\mathcal{F})\setminus(\omega\setminus
x)$, so $A$ can not be added to $\mathcal{A}\cup\\{\omega\setminus x\\}$, as
witnessed by $B,\mathcal{F}$ and $\omega\setminus x$. On the other hand, if
for all $X\in\mathcal{U}$, $B\in\mathcal{A}$ and
$F\in[\mathcal{A}\setminus\\{B\\}]^{<\omega}$ it happens that
$X\cap(B\setminus\bigcup F)\nsubseteq^{*}A$, then $(\omega\setminus A)\cap
X\cap(B\setminus\bigcup F)$ is infinite. Thus, by maximality of $\mathcal{U}$,
$\omega\setminus A\in\mathcal{U}$, which implies that
$x\subseteq^{*}\omega\setminus A$ and so $A\subseteq^{*}\omega\setminus x$. ∎

This completes the proof of the Lemma. ∎

######  Theorem 4.4.

Assume $GCH$. Let $C$ be a set of uncountable cardinals and let $\kappa$ be a
regular uncountable cardinal such that $\sup C\leq\kappa$. Then there is a
$ccc$ generic extension in which

$C\subseteq{\rm spec}(\mathfrak{s}_{mm}).$

###### Proof.

Add $\kappa$ Cohen reals to the ground model $V$ to obtain a model of
$\mathfrak{c}=\kappa$ and for each $\lambda\in C$ let $\mathcal{A}_{\lambda}$
be an ideal independent family of cardinality $\lambda$. Let
$\langle\lambda_{\beta}:\beta<\gamma\rangle$ be an enumeration of $C$. Proceed
with a finite support iteration
$\langle\mathbb{P}_{\alpha},\dot{\mathbb{Q}}_{\alpha}:\alpha<\omega_{1}\rangle$
where each iterand is a finite support iteration of length, the cardinality of
$C$, as follows:

Let $\mathbb{P}_{0}$ be the finite support iteration
$\langle\mathbb{R}_{\beta}^{0},\dot{\mathbb{S}}_{j}^{0}:j<\gamma\rangle$
defined by $\mathbb{R}_{0}^{0}=\mathbb{P}(\mathcal{A}_{\lambda_{0}})$ and
$\mathbb{R}_{j}^{0}\Vdash\dot{\mathbb{S}}_{j}^{0}=\mathbb{P}(\mathcal{A}_{\lambda_{j}})$.
After forcing with $\mathbb{P}_{0}$, for each $j<\gamma$, define
$\mathcal{A}_{j}^{0}=\mathcal{A}_{\lambda_{j}}\cup\\{x_{j}\\}$, where $x_{j}$
is the real from Lemma 4.1 added by the $j$-th-step of the iteration
$\mathbb{P}_{0}$. Now, assume $\mathbb{P}_{\beta}$ and
$\\{\mathcal{A}_{j}^{\beta}:j<\gamma\\}$ are defined. The next step
$\dot{\mathbb{Q}}_{\beta}$ is the finite support iteration
$\langle\mathbb{R}_{j}^{\beta},\dot{\mathbb{S}}_{j}^{\beta}:j<\gamma\rangle$
such that $\mathbb{R}_{0}^{\beta}=\mathbb{P}(\mathcal{A}_{0}^{\beta})$ and
$\mathbb{R}_{j}^{\beta}\Vdash\dot{\mathbb{S}}_{j}^{\beta}=\mathbb{P}(\mathcal{A}_{j}^{\beta})$.
In $V[G_{\beta+1}]$, after forcing with
$\mathbb{P}_{\beta}*\dot{\mathbb{Q}}_{\beta}$, define
$\mathcal{A}_{j}^{\beta+1}=\mathcal{A}_{j}^{\beta}\cup\\{x_{j}\\}$, where
$x_{j}$ is the real from Lemma 4.1 added by the $j$-th-step of the iteration
$\dot{\mathbb{Q}}_{\beta}$. If $\beta$ is a limit ordinal and
$\mathbb{P}_{\alpha}$, $\mathcal{A}_{j}^{\alpha}$ are defined for all
$\alpha<\beta$ and $j<\gamma$, let $\mathbb{P}_{\beta}$ be the finite support
iteration
$\langle\mathbb{P}_{\alpha},\dot{\mathbb{Q}}_{\alpha}:\alpha<\beta\rangle$ and
for $j<\gamma$,
$\mathcal{A}_{j}^{\beta}=\bigcup_{\alpha<\beta}\mathcal{I}_{j}^{\alpha}$.

Let $\mathbb{P}_{\omega_{1}}$ be the above iteration and for any $j<\gamma$,
let $\mathcal{A}_{j}^{*}=\bigcup_{\alpha<\omega_{1}}\mathcal{A}_{j}^{\alpha}$.
Fix $j$. Let $y\in
V[G_{\omega_{1}}]\cap[\omega]^{\omega}\backslash\mathcal{A}_{j}^{*}$. Then,
there is $\alpha<\omega_{1}$ such that $y$ is added at stage $\alpha$ of the
iteration. Since $y\notin\mathcal{A}_{j}$, it is also the case that
$y\notin\mathcal{A}_{j}^{\alpha+1}$ and so by the diagonalization properties
of the generic $x$ added at by $\mathbb{S}_{j}^{\alpha}$,
$\mathcal{A}^{j}_{\alpha+1}\cup\\{x,y\\}$ is not ideal independent. Since
$x\in\mathcal{A}_{j}$, $\mathcal{A}_{j}\cup{y}$ is not ideal independent. Thus
$\mathcal{A}_{j}^{*}$ is maximal. The poset $\mathbb{P}_{\omega_{1}}$
preserves all cardinals and we only added $\omega_{1}$ sets to the family
$\mathcal{A}_{j}$ to obtain $\mathcal{A}_{j}^{*}$, $\mathcal{A}_{j}^{*}$ has
indeed size $\lambda_{j}$. ∎

######  Remark 1.

The cardinality of a maximal ideal independent family can have countable
cofinality, while the character of any ultrafilter is uncountable. The first
assertion follows from the previous theorem by taking
$\kappa>\aleph_{\omega}$. The second assertion can be found in [3].

##  5\. Forcing Invariant Maximal Ideal Independent Families

In the following, we construct a maximal ideal independent family with strong
combinatorial properties, which guarantee that its maximality is preserved by
a large number of forcing notions.

######  Definition 5.1.

Let $\mathcal{U}$ be an ultrafilter. A maximal ideal independent family
$\mathcal{A}$ is called $\mathcal{U}\hbox{-\em encompassing}$ if the following
conditions hold:

  1. (1)

$\mathcal{U}\cap\mathcal{A}=\emptyset$, i.e. $\mathcal{A}$ is contained in the
dual ideal of $\mathcal{U}$.

  2. (2)

For every $X\in\mathcal{U}$ the set of $A\in\mathcal{A}$ so that
$X\in\mathcal{F}(\mathcal{A},A)$ is co-countable.

######  Theorem 5.2.

Assume $\mathsf{CH}$. For any $p$-point $\mathcal{U}$ there is a
$\mathcal{U}$-encompassing maximal ideal independent family $\mathcal{A}$ such
that for all $A\in\mathcal{A}$, the corresponding complemented filter
$\mathcal{F}(\mathcal{A},A)$ is a $p$-point.

###### Proof.

Let $\mathcal{U}$ be a $p$-point and $\langle
Y_{\alpha}:\alpha\in\omega_{1}\rangle$ be an $\subseteq^{*}$-decreasing
sequence which generates the $p$-point $\mathcal{U}$. Let $\langle
X_{\alpha}:\alpha\in\omega_{1}\rangle$ be an enumeration of
$[\omega]^{\omega}$. By recursion we construct a sequence
$\langle\mathcal{A}_{\alpha}:\alpha\in[\omega,\omega_{1})\rangle$ such that:

  1. (1)

For all $\alpha$, $\mathcal{A}_{\alpha}\subseteq\mathcal{U}^{*}$ is a
countable ideal independent family.

  2. (2)

For all $\alpha$, if $X_{\alpha}\notin\mathcal{A}_{\alpha+1}$ then
$\mathcal{A}_{\alpha+1}\cup\\{X_{\alpha}\\}$ is not an ideal independent
family.

  3. (3)

For all $\alpha$, $\mathcal{A}_{\alpha+1}=\mathcal{A}_{\alpha}$ or
$\mathcal{A}_{\alpha+1}=\mathcal{A}_{\alpha}\cup\\{A_{0}^{\alpha},A_{1}^{\alpha}\\}$
for some $A_{0}^{\alpha},A_{1}^{\alpha}\in\mathcal{U}^{*}$ and such that
$A_{0}^{\alpha}\setminus A_{1}^{\alpha},A_{1}^{\alpha}\setminus
A_{0}^{\alpha}\subseteq Y_{\alpha}$.

  4. (4)

If $\alpha$ is a limit ordinal, then
$\mathcal{A}_{\alpha}=\bigcup_{\beta<\alpha}\mathcal{A}_{\beta}$.

  5. (5)

If $A_{i}^{\alpha}\in\mathcal{A}$ is added in step $\alpha$ of the iteration,
$\langle\mathcal{P}^{\alpha,i}_{\beta}:\beta\in[\alpha+1,\omega_{1})\rangle$
is the enumeration of all partitions of $A^{\alpha}_{i}$, for $i\in 2$, and
for all $\alpha$ and $\beta>\alpha$, there are a finite
$F\subseteq[\beta\setminus\\{\alpha\\}]^{<\omega}$ and $k\in\omega$, such that
for any $i,j\in 2$ the set
$A^{\alpha}_{i}\setminus\left(A_{j}^{\beta}\cup\bigcup F\right)\setminus k$:

     * •

either is a partial selector of the partition
$\mathcal{P}^{\alpha,i}_{\beta}$,

     * •

or is contained in one element of the partition
$\mathcal{P}^{\alpha,i}_{\beta}$.

After the recursion we define
$\mathcal{A}=\bigcup_{\alpha<\omega_{1}}\mathcal{A}_{\alpha}$. Condition (1)
makes sure that $\mathcal{A}$ is an ideal independent family and (2) makes
sure that $\mathcal{A}$ is maximal. Condition (3) makes sure that
$\mathcal{A}$ is $\mathcal{U}$-encompassing. Condition (5) makes sure that the
filters $\mathcal{F}(\mathcal{A},A)$ are selective ultrafilters for all
$A\in\mathcal{A}$. We start by setting $\mathcal{A}_{\omega}=\langle
A_{n}:n\in\omega\rangle$ be a partition of $\omega$ into infinitely many
infinite sets. For each $n\in\omega$, let $A_{0}^{n}=A_{1}^{n}=A_{n}$ and let
$\langle\mathcal{P}_{\beta}^{n,i}:\beta\in[\omega,\omega_{1})\rangle$
enumerate all partitions of $A_{n}$. Assume $\mathcal{A}_{\alpha}$ has been
constructed. We take care of the set $X_{\alpha}$ and define
$\mathcal{A}_{\alpha+1}$.

If _$X_{\alpha}\in\mathcal{U}$_ , we just define
$\mathcal{A}_{\alpha+1}=\mathcal{A}_{\alpha}$, and condition (2) from
Definition 5.1 will make sure that $X_{\alpha}$ can not be added to the family
$\mathcal{A}$. If _$X_{\alpha}$ is in the ideal generated by the family
$\mathcal{A}_{\alpha}$_ we have nothing to do and we can define
$\mathcal{A}_{\alpha+1}=\mathcal{A}_{\alpha}$ again. _Otherwise_ ,
$X_{\alpha}\notin\mathcal{U}$ and $X_{\alpha}$ is positive relative to the
ideal generated by $\mathcal{A}_{\alpha}$. Let
$e_{\alpha}:\omega\to\mathcal{A}_{\alpha}$ be an enumeration of the elements
of $\mathcal{A}_{\alpha}$, and define
$C_{0}=e_{\alpha}(0),C_{n+1}=e_{\alpha}(n+1)\setminus\bigcup_{i\leq
n}e_{\alpha}(i)$.

If there are $n\in\omega$ and finite
$F\subseteq\mathcal{A}_{\alpha}\setminus\\{e_{\alpha}(0),\ldots,e_{\alpha}(n)\\}$
such that $C_{n}\setminus X_{\alpha}\subseteq^{*}\bigcup F$ and $C_{n}\cap
X_{\alpha}$ is infinite, then we have $C_{n}\setminus\bigcup
F\subseteq^{*}C_{n}\cap X_{\alpha}\subseteq X_{\alpha}$, and we can define
again $\mathcal{A}_{\alpha+1}=\mathcal{A}_{\alpha}$.

So let us assume that for all $n\in\omega$, $C_{n}\setminus X_{\alpha}$ is
finite or $C_{n}\setminus X_{\alpha}$ is not covered by any
$F\subseteq\mathcal{A}_{\alpha}\setminus\\{e_{\alpha}(0),\dots,e_{\alpha}(n)\\}$.
Since $e_{\alpha}(n)$ is not almost contained in the union of finitely many
elements from $\mathcal{A}_{\alpha}\setminus\\{e_{\alpha}(n)\\}$, and
$\mathcal{A}_{\alpha}\setminus\\{e_{\alpha}(n)\\}$ is countable, by recursion
we can construct an infinite set $B_{n}\subseteq e_{\alpha}(n)$, such that:

  1. (1)

$B_{n}\cap X_{\alpha}=\emptyset$,

  2. (2)

for all $Z\in\mathcal{A}_{\alpha}$ different from $e_{\alpha}(n)$, we have
$Z\cap B_{n}=^{*}\emptyset$,

  3. (3)

and moreover, by going to a subset if necessary, $B_{n}$ is a partial selector
of the partition $\mathcal{P}^{\gamma,i}_{\alpha}$ or is completely contained
in one element of the partition $\mathcal{P}^{\gamma,i}_{\alpha}$, where
$\gamma$ and $i$ are such that $A_{i}^{\gamma}=e_{\alpha}(n)$

Also, since for all $n\in\omega$, $C_{n}\notin\mathcal{U}$ and $\mathcal{U}$
is a $p$-point, there is $A\in\mathcal{U}$ such that for all $n\in\omega$,
$C_{n}\cap A$ is finite, $X_{\alpha}\cap A=\emptyset$ and $A\subseteq
Y_{\alpha}$. Let $W_{0},W_{1}$ be infinite disjoint subsets of $A$ which are
not in the ultrafilter $\mathcal{U}$. Define $W$ as,

$W=\left(\bigcup_{n\in\omega}C_{n}\setminus B_{n}\right)\setminus A$

Finally, define $A_{i}^{\alpha}=W\cup W_{i}$, for $i\in 2$, and
$\mathcal{A}_{\alpha+1}=\mathcal{A}_{\alpha}\cup\\{A_{0}^{\alpha},A_{1}^{\alpha}\\}$.
Note that $X_{\alpha}\subseteq A_{i}^{\alpha}$, $A_{i}^{\alpha}\setminus
A_{1-i}^{\alpha}=W_{i}\subseteq Y_{\alpha}$. It remains to observe that by
construction of $A_{0}^{\alpha}$ and $A_{1}^{\alpha}$ the family
$\mathcal{A}_{\alpha+1}$ is ideal independent. ∎

######  Theorem 5.3.

Let $\mathcal{U}$ be a $p$-point and let $\mathbb{P}$ be a proper,
$\omega^{\omega}$-bounding forcing notion which preserves $p$-points. Then
$\mathbb{P}$ preserves the maximality of any $\mathcal{U}$-encompassing
maximal ideal independent family $\mathcal{A}$ such that for all
$A\in\mathcal{A}$, the corresponding complemented filter
$\mathcal{F}(\mathcal{A},A)$ is a $p$-point.

Note that this theorem implies that under $\mathsf{CH}$, in the generic
extension by any proper $\omega^{\omega}$-bounding $p$-point preserving
forcing notion $\mathfrak{s}_{mm}$ is $\aleph_{1}$.

###### Proof.

Fix an ultrafilter $\mathcal{U}$, a $\mathcal{U}$-encompassing maximal ideal
independent family $\mathcal{A}$ with the property that all of the
complemented filters of $\mathcal{A}$ are $p$-points, and a proper,
$\omega^{\omega}$-bounding, $p$-point preserving forcing notion $\mathbb{P}$.
Let $p\in\mathbb{P}$ and let $\dot{X}$ be a name so that
$p\Vdash\dot{X}\in[\omega]^{\omega}$. We need to show that some $q\leq p$
forces that $\dot{X}$ cannot be added to $\mathcal{A}$ without destroying
ideal independence. More precisely, this means that we need to either find a
$q\leq p$ so that $q$ forces that $\dot{X}$ is in the ideal generated by
$\mathcal{A}$ or else find $q\leq p$ and an $A\in\mathcal{A}$ so that $q$
forces that $\dot{X}$ is in the complemented filter corresponding to $A$.

Thus suppose towards a contradiction that $p$ forces that $\dot{X}$ is neither
in the ideal generated by $\mathcal{A}$ nor in any filter
$\mathcal{F}(\mathcal{A},A)$ for any $A\in\mathcal{A}$. Note that this implies
in particular that $\dot{X}$ is not in $\mathcal{U}$ since if it were, then in
would be in some filter $\mathcal{F}(\mathcal{A},A)$ (in fact co-countably
many). Since $\mathbb{P}$ preserves $\mathcal{U}$ being a $p$-point it follows
that $p$ forces that the complement of $\dot{X}$ is in $\mathcal{U}$ and
therefore we can find a $q\leq p$ and a $Z\in\mathcal{U}$ so that
$q\Vdash\dot{X}\cap Z=\emptyset$. Fix such a $q$ and $Z$. To complete the
proof it suffices to therefore show that some $r\leq q$ forces that
$n\in\dot{X}$ for some $n\in Z$.

For any $u\in\mathbb{P}$ let
$X_{u}=\\{n\in\omega\;|\;u\nVdash\check{n}\notin\dot{X}\\}$ be the outer hull
of $\dot{X}$ with respect to $u$, i.e. the collection of $n<\omega$ forced to
be in $\dot{X}$ by some $u^{\prime}\leq u$. Note that
$u\Vdash\dot{X}\subseteq\check{X}_{u}$ for any $u\in\mathbb{P}$. It follows
that for any condition $u$ stronger than $p$, $X_{u}$ is not in the ideal
generated by $\mathcal{A}$. By the maximality of $\mathcal{A}$, moreover we
get that for every $r\leq q$ the set $X_{r}$ is in some complemented filter of
$\mathcal{A}$. Therefore to finish the proof it suffices to show that in fact
any such $X_{r}$ is actually in uncountably many such filters. This suffices
since if this is the case then in particular it applies to $X_{q}$ and, since,
by the definition of $\mathcal{U}$-encompassing, $Z$ is in
$\mathcal{F}(\mathcal{A},A)$ for co-countably many $A\in\mathcal{A}$, there is
some $A\in\mathcal{A}$ so that $Z\cap X_{q}\in\mathcal{F}(\mathcal{A},A)$ and
so $Z\cap X_{q}$ has infinite intersection. Thus, some $r\leq q$ forces that
$n\in\dot{X}$ for some (in fact infinitely many) $n\in Z$. Summing up, it
suffices to show the following claim.

######  Claim 5.4.

For any $u\in\mathbb{P}$ stronger than $p$ the set
$X_{u}\in\mathcal{F}(\mathcal{A},A)$ for uncountably many $A\in\mathcal{A}$.

Fix such a $u\in\mathbb{P}$ and suppose towards a contradiction that there
were only countably many $A\in\mathcal{A}$ with
$X_{u}\in\mathcal{F}(\mathcal{A},A)$. Let $M\prec H_{\theta}$ be a countable
model for $\theta$ sufficiently large with
$\mathbb{P},p,u,\mathcal{A},\mathcal{U}\in M$ containing every $A$ so that
$X_{u}\in\mathcal{F}(\mathcal{A},A)$. Enumerate $\mathcal{A}\cap M$ as
$\\{A_{i}\;|\;i<\omega\\}$. Since $p$, and hence $u$, forces that for each
$n<\omega$ the name $\dot{X}$ is not in $\mathcal{F}(\mathcal{A},A_{n})$, and
each one of such filters is a $p$-point by our assumption and hence an
ultrafilter preserved by $\mathbb{P}$, there is in $M$ a dense set of
conditions below $u$ forcing that $\dot{X}\cap
A_{n}\setminus\bigcup_{i<k_{n},i\neq n}A_{i}$ is finite for some
$k_{n}\in\omega$. Applying $\omega^{\omega}$-boundedness and properness we can
find in the ground model functions $f,g\in\omega^{\omega}$ and a condition
$r\leq u$ which is $(M,\mathbb{P})$-generic so that for each $n<\omega$ we
have

$r\Vdash\dot{X}\cap A_{n}\setminus\bigcup_{i<\check{f}(n),i\neq
n}A_{i}\subseteq\check{g}(n)$

In particular we get that $X_{r}\cap A_{n}\setminus\bigcup_{i<f(n),i\neq
n}A_{i}\subseteq g(n)$ and thus $X_{r}\notin\mathcal{F}(\mathcal{A},A_{n})$
for any $n<\omega$. But then, by applying the same argument to $r$ that we
applied to $u$, we get that $X_{r}$ is in some $\mathcal{F}(\mathcal{A},B)$
for some $B\in\mathcal{A}$ with $B\neq A_{n}$ for any $n<\omega$. This is a
contradiction however since $X_{r}\subseteq X_{u}$ and by definition of the
$A_{n}$’s $X_{u}\notin\mathcal{F}(\mathcal{A},B)$. This contradiction implies
that $X_{q}$ is in uncountably many complemented filters of $\mathcal{A}$ and
hence the proof is complete. ∎

As an straightforward corollary we obtain:

######  Corollary 5.5.

  1. (1)

$\mathfrak{s}_{mm}=\aleph_{1}$ in the Sacks model.

  2. (2)

$\mathfrak{s}_{mm}=\aleph_{1}$ in the Miller partition model and hence
$\mathfrak{s}_{mm}<\mathfrak{a}_{T}$ is consistent.

  3. (3)

$\mathfrak{s}_{mm}=\aleph_{1}$ in the $h$-perfect tree forcing model and hence
$\mathfrak{s}_{mm}<\hbox{non}(\mathcal{N})$ is consistent.

###### Proof.

For (1), it is a standard fact that the iterated Sacks forcing preserves
$p$-points and it is $\omega^{\omega}$-bounding. For (2), in [14], Miller has
constructed a forcing, known as Miller partition forcing, which makes the
cardinal invariant $\mathfrak{a}_{T}$ equal to $\aleph_{2}$, as recently shown
in [4] preserves $p$-points, and as shown in [18]) is
$\omega^{\omega}$-bounding. For (3) recall, that the $h$-perfect tree forcing
is proper, ${}^{\omega}\omega$-bounding, preserves $p$-points and that in the
$h$-perfect tree forcing model $\hbox{non}(\mathcal{N})=\aleph_{2}$, see [10,
Section 2]. ∎

An alternation of Miller partition and $h$-perfect tree forcings will lead to
a model of
$\mathfrak{i}=\mathfrak{s}_{mm}<\hbox{non}(\mathcal{N})=\mathfrak{a}_{T}=\aleph_{2}$
(for the effect of the respective posets on $\mathfrak{i}$ see [4] and [19]).

######  Corollary 5.6.

$\mathfrak{s}_{mm}$ is independent of $\mathfrak{a}_{T}$

###### Proof.

In the Miller partition model, $\mathfrak{s}_{mm}<\mathfrak{a}_{T}$. On the
other hand, it is well known that $\mathfrak{a}_{T}<\mathfrak{u}$ holds in the
Random model and hence $\mathfrak{a}_{T}<\mathfrak{s}_{mm}$ holds in that
model as well. ∎

##  6\. Ideal Mrówka Spaces

To each ideal independent family $\mathcal{A}$ we can associate a topological
space $\psi(\mathcal{A})$ defined as follows.

######  Definition.

Let $\mathcal{A}$ be an ideal independent family. The space
$\psi(\mathcal{A})$ is the set $\omega\cup\mathcal{A}$ endowed with the
topology $\tau$ which satisfies the following condition:

  * •

a set $U\in\tau$ if and only if for any $A\in\mathcal{A}\cap U$ there exists
$F\in\mathcal{F}(\mathcal{A},A)$ such that $F\subseteq U$.

It is easy to see that the set $\omega$ is a dense discrete subset of
$\psi(\mathcal{A})$ and for any $A\in\mathcal{A}$ the family
$\\{F\cup\\{A\\}:F\in\mathcal{F}(\mathcal{A},A)\\}$ forms an open neighborhood
base at $A$ in $\psi(\mathcal{A})$. The set
$\mathcal{A}\subseteq\psi(\mathcal{A})$ is closed and discrete. For any
$A\in\mathcal{A}$ the set $\\{A\\}\cup A$ is a clopen neighborhood of $A$.
Thus, we obtain the following:

######  Lemma 6.1.

For any ideal independent family $\mathcal{A}$ the space $\psi(\mathcal{A})$
is Tychonoff zero-dimensional separable scattered (of height $2$) space.

Observe that for any almost disjoint family $\mathcal{A}$ the space
$\psi(\mathcal{A})$ is the well-known Mrówka space corresponding to the almost
disjoint family $\mathcal{A}$.

######  Lemma 6.2.

Let $\mathcal{A}$ be an ideal independent family. If $\mathcal{A}$ is maximal,
then each $C\in\mathcal{A}^{+}$ has an accumulation point in
$\psi(\mathcal{A})$. Moreover, if all but finitely many
$\mathcal{F}(\mathcal{A},A)$, $A\in\mathcal{A}$ are ultrafilters, then
$\mathcal{A}$ is maximal, whenever each $C\in\mathcal{A}^{+}$ has an
accumulation point.

###### Proof.

Let $\mathcal{A}$ be a maximal ideal independent family and
$C\in\mathcal{A}^{+}$. The maximality of $\mathcal{A}$ implies that there
exists $A\in\mathcal{A}$ such that $C\in\mathcal{F}(\mathcal{A},A)$. Then $A$
is an accumulation point of the set $C$.

Let $\mathcal{A}$ be an ideal independent family such that
$\mathcal{F}(\mathcal{A},A)$ is an ultrafilter for all but finitely many
$A\in\mathcal{A}$, and each $C\in\mathcal{A}^{+}$ has an accumulation point in
$\psi(\mathcal{A})$. Denote
$\mathcal{P}=\\{A\in\mathcal{A}:\mathcal{F}(\mathcal{A},A)$ is not an
ultrafilter$\\}$. Consider any $C\in\mathcal{A}^{+}$. Let
$D=C\setminus\bigcup\mathcal{P}$. Since the family $\mathcal{P}$ is finite,
the set $D$ is positive. Then there exists an accumulation point
$A\in\mathcal{A}$ of $D$ in $\psi(\mathcal{A})$. It follows that $D\cap
F\neq\emptyset$ for any $F\in\mathcal{F}(\mathcal{A},A)$. Clearly,
$A\notin\mathcal{P}$. Then $\mathcal{F}(\mathcal{A},A)$ is an ultrafilter and,
consequently, $C\supseteq D\in\mathcal{F}(\mathcal{A},A)$. Thus, the ideal
independent family $\mathcal{A}$ is maximal. ∎

For an ideal independent family $\mathcal{A}$ by
$\mathsf{Cl}_{\mathcal{A}}(\omega)$ we denote the set of all closed subsets of
$\omega$ in $\psi(\mathcal{A})$. The following lemma topologically describes
the ideal $\mathcal{J}(\mathcal{A})$ corresponding to an ideal independent
family $\mathcal{A}$.

######  Lemma 6.3.

For any ideal independent family $\mathcal{A}$ we have that
$\mathcal{J}(\mathcal{A})\subseteq\mathsf{Cl}_{\mathcal{A}}(\omega)$.
Moreover, if $\mathcal{A}$ is maximal, then
$\mathcal{J}(\mathcal{A})=\mathsf{Cl}_{\mathcal{A}}(\omega)$.

###### Proof.

Let $A_{1},A_{2}$ be any distinct elements of an ideal independent family
$\mathcal{A}$ and put $C=A_{1}\cap A_{2}$. Observe that for any
$A\in\mathcal{A}\setminus\\{A_{2}\\}$ the set $A\setminus
A_{2}\in\mathcal{F}(\mathcal{A},A)$ and $(A\setminus A_{2})\cap C=\emptyset$.
Similarly, the set $A_{2}\setminus A_{1}\in\mathcal{F}(\mathcal{A},A_{2})$ and
$(A_{2}\setminus A_{1})\cap C=\emptyset$. Thus, the set $C$ is closed and
discrete in $\psi(\mathcal{A})$. At this point it is easy to see that for any
$A\in\mathcal{A}$ the ideal $\mathcal{J}(\mathcal{A})$ consists of closed (in
$\psi(\mathcal{A})$) sets. Since a finite union of closed sets is closed and
any subset of a closed discrete set is closed, we get that
$\mathcal{J}(\mathcal{A})\subseteq\mathsf{Cl}_{\mathcal{A}}(\omega)$.

Let $\mathcal{A}$ be a maximal ideal independent family. Consider an arbitrary
set $C\in\mathsf{Cl}_{\mathcal{A}}(\omega)$. Since the set $C$ is closed we
get that $C\notin\mathcal{F}(\mathcal{A},A)$ for every $A\in\mathcal{A}$. The
maximality of $\mathcal{A}$ implies that $C\notin\mathcal{A}^{+}$. Hence there
exists a finite collection $\\{A_{i}:i\in n\\}\in[\mathcal{A}]^{<\omega}$ such
that $C\subseteq^{*}\bigcup_{i\in n}A_{i}$. Since $C$ is closed, for every
$i\in n$ there exists $F_{i}\in\mathcal{F}(\mathcal{A},A_{i})$ such that
$F_{i}\cap C=\emptyset$. Then for every $i\in n$ there exists $m(i)\in\omega$
and a finite family $\\{D^{i}_{j}:j\in
m(i)\\}\in[\mathcal{A}\setminus\\{A_{i}\\}]^{<\omega}$ such that
$(A_{i}\setminus\bigcup_{j\in m(i)}D^{i}_{j})\cap C=^{*}\emptyset$, witnessing
that $C\subseteq^{*}\bigcup_{i\in n}\bigcup_{j\in m(i)}(A_{i}\cap
D_{j}^{i})\in\mathcal{J}(\mathcal{A})$. Hence
$\mathcal{J}(\mathcal{A})=\mathsf{Cl}_{\mathcal{A}}(\omega)$. ∎

A Tychonoff space $X$ is called pseudocompact if each continuous real-valued
function on $X$ is bounded. The following lemma is a folklore.

######  Lemma 6.4.

Let $X$ be a space with an open discrete dense subspace $D$. Then $X$ is
pseudocompact if and only if each $A\in[D]^{\omega}$ has an accumulation point
in $X$.

######  Lemma 6.5.

Let $\mathcal{A}$ be an ideal independent family. The space
$\psi(\mathcal{A})$ is pseudocompact if and only if $\mathcal{A}$ is a maximal
almost disjoint family.

###### Proof.

If $\mathcal{A}$ is a maximal almost disjoint family, then by [12] the Mrówka
space $\psi(\mathcal{A})$ is pseudocompact.

Assume that the ideal independent family $\mathcal{A}$ is not almost disjoint.
Then there exists $A_{1},A_{2}\in\mathcal{A}$ such that the set $C=A_{1}\cap
A_{2}$ is infinite. By Lemma 6.3, the set $C$ is closed in
$\psi(\mathcal{A})$. Lemma 6.4 implies that the space $\psi(\mathcal{A})$ is
not pseudocompact. ∎

Corollary 3.4 and Lemma 6.5 imply the following.

######  Proposition 6.6.

For a maximal ideal independent family $\mathcal{A}$ the following conditions
are equivalent:

  1. (1)

$\psi(\mathcal{A})$ is pseudocompact;

  2. (2)

$\mathcal{A}$ is a maximal almost disjoint family;

  3. (3)

$\mathcal{A}$ is a completely separable maximal almost disjoint family.

######  Proposition 6.7.

Let $\mathcal{A}$ be an ideal independent family. The space
$\psi(\mathcal{A})$ is locally compact if and only if $\psi(\mathcal{A})$ is a
homeomorphic to $\psi(\mathcal{B})$ for some almost disjoint family
$\mathcal{B}$.

###### Proof.

The sufficiency follows from the fact that the Mrówka space
$\psi(\mathcal{B})$ is locally compact for any almost disjoint family
$\mathcal{B}$.

Assume that $\psi(\mathcal{A})$ is locally compact for some ideal independent
family $\mathcal{A}$. Since the space $\psi(\mathcal{A})$ is zero-dimensional
(see Lemma 6.1), for each $A\in\mathcal{A}$ we can find a compact open
neighborhood $U_{A}$ of $A$. Recall that for every $A\in\mathcal{A}$ the set
$A\cup\\{A\\}$ is a clopen neighborhood of $A$. Then
$V_{A}=U_{A}\cap(A\cup\\{A\\})$ is a compact open neighborhood of $A$. Since
$V_{A}\cap\mathcal{A}=\\{A\\}$ we deduce that $A$ is the only non-isolated
point of $V_{A}$. The compactness of $V_{A}$ implies that
$B_{A}=V_{A}\cap\omega$ is a convergent sequence to $A$ in
$\psi(\mathcal{A})$. Let $\mathcal{B}=\\{B_{A}:A\in\mathcal{A}\\}$. Clearly,
the family $\mathcal{B}$ is almost disjoint. Define
$f:\psi(\mathcal{B})\rightarrow\psi(\mathcal{A})$ by

$f(x)=\begin{cases}n&\hbox{ if }x=n\in\omega;\\\ A&\mbox{ if
}x=B_{A}\in\mathcal{B}.\end{cases}$

A routine verification shows that the map $f$ is a homeomorphism. ∎

##  7\. Conclusion and Open Questions

Of interest remains the following question:

######  Question 7.1.

Is it consistent that for every maximal ideal independent family $\mathcal{A}$
there is an $A\in\mathcal{A}$ so that $\mathcal{F}(\mathcal{A},A)$ is an
ultrafilter?

A positive answer to this question would imply that there are no completely
separable maximal almost disjoint families. It is known that such families
exist under either $\mathfrak{s}\leq\mathfrak{a}$ or
$2^{\aleph_{0}}<\aleph_{\omega}$, see [15] and [12], respectively.

In the context of topology we ask more about Mrówka spaces for ideal
independent families. In this paper we have barely scratched the surface of
what may be possible, for instance the following is a natural question which
is open.

######  Question 7.2.

Does there exist a $\mathsf{ZFC}$ example of an infinite maximal ideal
independent family $\mathcal{A}$ such that $\psi(\mathcal{A})$ is Fréchet-
Urysohn or first-countable?

Clearly, the above question has positive answer if completely separable
maximal almost disjoint families exist in ZFC.

The results of the current paper together with those of [5] give either a
$\mathsf{ZFC}$ relation, or establish the independence between
$\mathfrak{s}_{mm}$ and any other well studied cardinal characteristic, with
the exception of the almost disjointness number $\mathfrak{a}$. The following
remains open.

######  Question 7.3.

Is it consistent that $\mathfrak{s}_{mm}<\mathfrak{a}$?

The corresponding question for $\mathfrak{i}$, i.e. the consistency of
$\mathfrak{i}<\mathfrak{a}$ is one of the most interesting open problems in
cardinal characteristics of the continuum and many of the roadblocks towards
solving that problem are the same as trying to answer the question above. See
the appendix of [4] for an interesting discussion on Vaughan’s problem.

As noted in the introduction, Theorem 5.2 implies that $\mathfrak{s}_{mm}={\rm
max}\\{\mathfrak{d},\mathfrak{u}\\}$ in many standard forcing extensions.
However, this is not a $\mathsf{ZFC}$ equality as $\mathfrak{s}_{mm}>{\rm
max}\\{\mathfrak{d},\mathfrak{u}\\}$ holds in the Boolean ultrapower model,
see for example [2]. That model requires a measurable cardinal and increases
both $\mathfrak{u}$ and $\mathfrak{d}$. As a result the following two
questions remain very interesting:

######  Question 7.4.

Is ${\rm max}\\{\mathfrak{d},\mathfrak{u}\\}<\mathfrak{s}_{mm}$ consistent
with $\mathsf{ZFC}$?

######  Question 7.5.

If $\mathfrak{d}=\mathfrak{u}=\aleph_{1}$ does $\mathfrak{s}_{mm}=\aleph_{1}$?

The later question is an ideal independent version of Roitman’s problem.
Theorem 4.4 opens up the possibility of a maximal ideal independent families
of size $\aleph_{\omega}$. We can therefore ask:

######  Question 7.6.

Is it consistent that $\mathfrak{s}_{mm}=\aleph_{\omega}$? More generally can
$\mathfrak{s}_{mm}$ have countable cofinality?

Finally, we can ask more generally about the spectrum of maximal ideal
independent families:

######  Question 7.7.

What $\mathsf{ZFC}$ restrictions are there on the set ${\rm
spec}(\mathfrak{s}_{mm})$? Can it be equal to any set of regular cardinals
which includes the continuum?

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