must be the constant function $1$, and this entails the desired conclusion.

We can now prove the following result, which is a combination of proposition
55 and proposition 65:

###### Proposition 70

If $\mathcal{W}$ is a $P$-point and $\mathcal{A}$ is a Laflamme MAD family,
then $\mathbb{F}_{\sigma}\left(\mathcal{A}\right)$ forces that
$\mathcal{\dot{U}}_{gen}\left(\mathcal{A}\right)$ is $\mathcal{W}$-Canjar.

Proof. We will prove the proposition by contradiction. Assume there is
$\mathcal{F}\in\mathbb{F}_{\sigma}\left(\mathcal{A}\right)$ and $\sigma$ such
that $\mathcal{F}$ forces that $\sigma$ is a winning strategy for player
$\mathsf{I}$ in
$\mathcal{H\mathbb{(}W},\mathcal{\dot{U}}_{gen}\left(\mathcal{A}\right))$.
Since strategies for player $\mathsf{I}$ are countable objects and since
$\mathbb{F}_{\sigma}\left(\mathcal{A}\right)$ is $\sigma$-closed, it is enough
to consider ground model strategies. Let
$\mathcal{F}={\textstyle\bigcup\limits_{n\in\omega}}\mathcal{C}_{n}$ where
$\left\langle\mathcal{C}_{n}\right\rangle_{n\in\omega}$ is an increasing
sequence of compact sets. We will use $\sigma$ to construct a winning strategy
for player $\mathsf{I}$ in the game
$\mathcal{G}_{P\text{-}point}\left(\mathcal{W}\right)$, which will be a
contradiction.

Let $L$ be the collection of all $p_{s}$ such that $p$ is a possible response
of player $\mathsf{I}$ according to $\sigma$ and $s\in split\left(p\right).$
In the same way as in the proof of proposition 55, if $p_{s}\in L,$ then
$\mathcal{F}\Vdash``spsuc_{p}\left(s\right)\in(\mathcal{\dot{U}}_{gen}^{<\omega}\mathcal{)}^{+}\textquotedblright,$
hence in particular
$\mathcal{C}\left(spsuc_{p}\left(s\right)\right)\subseteq\left\langle\mathcal{F\cup
I}\left(\mathcal{A}\right)^{\ast}\right\rangle.$ Since $L$ is countable we may
assume (by extending $\mathcal{F}$ if necessary) that
$\mathcal{C}\left(spsuc_{p}\left(s\right)\right)\subseteq\mathcal{F}$ for
every $p_{s}\in L.$ By 61, we can find a family $\left\\{A_{n}\mid
n\in\omega\right\\}\subseteq\mathcal{A}$ such that
$\bigstar\left(A_{n},\mathcal{F},\mathcal{C}\left(spsuc_{p}\left(s\right)\right)\right)$
holds for every $n\in\omega$ and $p_{s}\in L.$

For every
$\mathcal{X}=\left\\{X_{1},...,X_{n}\right\\}\in\left[\left\\{spsuc_{p}\left(s\right)\mid
p\in L\wedge s\in split\left(p\right)\right\\}\right]^{<\omega}$ and for every
$k\in\omega,$ fix a function
$F_{\left(\mathcal{X},k\right)}:\mathcal{X\longrightarrow}$
$\left[\left[\omega^{<\omega}\right]\right]^{<\omega}$ with the following
properties:

  1. 1.

$Y_{i}=F_{\left(\mathcal{X},k\right)}\left(X_{i}\right)\in\left[X_{i}\right]^{<\omega}$
for every $i\leq n.$

  2. 2.

For every $B\in\mathcal{C}_{k}$, for every
$C_{i}^{1},...,C_{i}^{n}\in\mathcal{C}\left(Y_{i}\right)$ (with $i\leq n$) and
for every $k_{1},k_{2}\leq n$, we have that
$B\cap{\textstyle\bigcap\limits_{i,j\leq n}}C_{i}^{j}$ contains an element of
$Y_{k_{1}}\cap\left[A_{k_{2}}\right]^{<\omega}.$

We know such $F_{\left(\mathcal{X},k\right)}$ exists by the previous lemma.
The proof now proceeds in a very similar way as the proof of 55. We define
$\pi$ a strategy for player $\mathsf{I}$ in
$\mathcal{G}_{P\text{-}point}\left(\mathcal{W}\right)$ as follows:

  1. 1.

Player $\mathsf{I}$ starts by playing $W_{0}=\sigma\left(\emptyset\right)$.

  2. 2.

Assume player $\mathsf{II}$ plays $z_{0}\in\left[W_{0}\right]^{<\omega}$. Let
$p_{0}=\sigma\left(W_{0},z_{0}\right)$, $s_{0}$ be the stem of $p_{0}$ and
$\mathcal{X}_{0}=\left\\{spsuc_{p_{0}}\left(s_{0}\right)\right\\}.$ Define
$n_{0}>d^{-1}\left(s_{0}\right)$ to be the least integer such that$\
d^{-1}(s_{0}{}^{\frown}t)<n_{0}$ for all $t\in
F_{\left(\mathcal{X}_{0},0\right)}\left(spsuc_{p_{0}}\left(s_{0}\right)\right)$.
Player $\mathsf{I}$ will play (in
$\mathcal{G}_{P\text{-}point}\left(\mathcal{W}\right)$)
$W_{1}=\sigma\left(W_{0},z_{0},p_{0},n_{0}\right)$.

  3. 3.

In general, assume that it has been played the sequence $\left\langle
W_{0},z_{0},...,W_{m}\right\rangle.$ At the same time, the player $\mathsf{I}$
has secretly been constructing a sequence $\left\langle
W_{0},z_{0},p_{0},n_{0},W_{1},z_{1},p_{1},n_{1}...,W_{m}\right\rangle$ that is
being forced to be a partial play of the game
$\mathcal{H(W},\mathcal{\dot{U}}_{gen}\left(\mathcal{A}\right)\mathcal{)}$
following $\sigma$ such that for every $i<m,$ the integer $n_{i}$ has the
following important property: letting
$\mathcal{X}_{i}=\left\\{spsuc_{p_{i}}\left(u\right)\mid u\in
T\left(p_{i},n_{i-1}\right)\right\\}$ (we define
$n_{-1}=d^{-1}\left(s_{0}\right)$), for every $t\in
F_{\left(\mathcal{X}_{i},i\right)}\left(spsuc_{p_{i}}\left(u\right)\right),$
we have that $d^{-1}(u^{\frown}t)<n_{i}.$ Assume that player $\mathsf{II}$
plays $z_{m}$ as her next response in
$\mathcal{H(W},\mathcal{\dot{U}}_{gen}\left(\mathcal{A}\right)).$ Let $p_{m}$
be the tree $\sigma\left(W_{0},z_{0},n_{0},W_{1},...,W_{m},z_{m}\right)$ and
let $n_{m}>n_{m-1}$ the least integer with the following property: letting
$\mathcal{X}_{m}=\left\\{spsuc_{p_{m}}\left(u\right)\mid u\in
T\left(p_{m},n_{m-1}\right)\right\\}$, for every $t\in
F_{\left(\mathcal{X}_{m},m\right)}\left(spsuc_{p_{m}}\left(u\right)\right),$
we have that $d^{-1}(u^{\frown}t)<n_{m}.$ Player $\mathsf{I}$ will play
$W_{m+1}=\sigma\left(W_{0},z_{0},n_{0},W_{1},...,W_{m},z_{m},p_{m},n_{m}\right).$

The game $\mathcal{G}_{P\text{-}point}\left(\mathcal{W}\right):$

$\mathsf{I}$ | $W_{0}$ |  | $W_{1}$ |  | $...$
---|---|---|---|---|---
$\mathsf{II}$ |  | $z_{0}$ |  | $z_{1}$ |

The game
$\mathcal{H(W},\mathcal{\dot{U}}_{gen}\left(\mathcal{A}\right)\mathcal{)}:$

$\mathsf{I}$ | $W_{0}$ |  | $p_{0}$ |  | $W_{1}$ |  | $p_{1}$ |  | $...$
---|---|---|---|---|---|---|---|---|---
$\mathsf{II}$ |  | $z_{0}$ |  | $n_{0}$ |  | $z_{1}$ |  | $n_{1}$ |

We claim that $\pi$ is a winning strategy for player $\mathsf{I}$ in
$\mathcal{G}_{P\text{-}point}\left(\mathcal{W}\right).$ Consider a run of the
game in which player $\mathsf{I}$ played according to $\pi.$ Let
$Z={\textstyle\bigcup\limits_{n\in\omega}}z_{n},$ we will prove that
$Z\notin\mathcal{U}.$ Let
$q={\textstyle\bigcup\limits_{i\in\omega}}T\left(p_{i},n_{i}\right)$ be the
tree that was constructed by player $\mathsf{I}$ during the play. It is easy
to see that
$\mathcal{F\cup}\left\\{\mathcal{C}\left(spsuc_{q}\left(s\right)\right)\mid
s\in split\left(q\right)\right\\}$ generates a condition of
$\mathbb{F}_{\sigma}\left(\mathcal{A}\right)$, call if $\mathcal{K}.$ Note
that $\mathcal{K\leq F}$ hence $\mathcal{K}$ forces that $\sigma$ is a winning
strategy for player $\mathsf{I}$ in
$\mathcal{H\mathbb{(}W},\mathcal{\dot{U}}_{gen})$. Moreover, $\mathcal{K}$
forces that
$q\in\mathbb{PT(}\mathcal{\dot{U}}_{gen}\left(\mathcal{A}\right)).$ Since
player $\mathsf{I}$ is forced to win in
$\mathcal{H\mathbb{(}W},\mathcal{\dot{U}}_{gen}\left(\mathcal{A}\right))$, it
must be the case that $Z\notin\mathcal{W}.$ This shows that $\pi$ is a winning
strategy for player $\mathsf{I}$ in
$\mathcal{G}_{P\text{-}point}\left(\mathcal{W}\right).$ Since player
$\mathsf{I}$ can not have a winning strategy in the $P$-point game, we get a
contradiction.

We can now answer the questions of Brendle and Shelah:

###### Theorem 71

Every MAD family can be destroyed with a proper forcing that preserves all
$P$-points from the ground model. In particular, it is consistent that
$\omega_{1}=\mathfrak{u<a}=\omega_{2}.$

##  7 MAD families build up from closed sets

The _closed almost disjointness number_ was introduced by Brendle and Khomskii
in [14]. The invariant $\mathfrak{a}_{closed}$ is defined as the smallest
number of closed sets of $\left[\omega\right]^{\omega}$ such that its union is
a MAD family. Since singletons are closed, it follows that
$\mathfrak{a}_{closed}\leq\mathfrak{a}$ and it is uncountable by a result of
Mathias (see [44] or [61]). The following are some known results regarding
this cardinal invariant:

###### Proposition 72

  1. 1.

(Raghavan, Törnquist independently) $\mathfrak{p}\leq\mathfrak{a}_{closed}$
(see [61]).

  2. 2.

(Brendle and Khomskii) It is consistent that
$\mathfrak{a}_{closed}<\mathfrak{b}$ (see [14] and [15]).

  3. 3.

(Brendle and Raghavan) It is consistent that $\mathfrak{b<a}_{closed}$ (see
[15]).

  4. 4.

(Raghavan and Shelah) $\mathfrak{d}=\omega_{1}$ implies
$\mathfrak{a}_{closed}=\omega_{1}$ (see [54] and [15]).

There are still many interesting open questions regarding
$\mathfrak{a}_{closed}.$ The following problems are still open:

######  Problem 73 (Raghavan)

Does $\mathfrak{h}\leq\mathfrak{a}_{closed}?$

######  Problem 74 (Brendle, Khomskii)

Does $\mathfrak{s}=\omega_{1}$ imply $\mathfrak{a}_{closed}=\omega_{1}?$

The reader may consult [14] or [15] for more information and open problems
regarding $\mathfrak{a}_{closed}.$ If
$\mathcal{D}\subseteq\left[\omega\right]^{\omega}$ is an AD family, we denote
its _ortogonal_ $\mathcal{D}^{\perp}$ as the set of all $B\subseteq\omega$
that are almost disjoint with every element of $\mathcal{D}.$

As mentioned before, Brendle and Raghavan proved that it is consistent that
$\mathfrak{b<a}_{closed}.$ In fact, they build two models in which this
inequality holds. One is using the creature forcing of Shelah and another was
using a c.c.c. forcing, similar to the model constructed in [8]. In both
cases, their forcings add Cohen reals. We will combine their results with ours
to build a model of $\mathfrak{b<a}_{closed}$ without adding Cohen reals,
moreover; this inequality holds in the model constructed in the previous
section. The key result is the following proposition, which was implicitly
proved in the Lemma 7 of [15]:

######  Proposition 75 (Brendle, Raghavan)

Let $\mathcal{D}\subseteq\left[\omega\right]^{\omega}$ be a closed AD family.
If $\mathcal{U}$ is a $P$-point such that $\mathcal{D\cap U}=\emptyset,$ then
there is $U\in\mathcal{U}\cap\mathcal{D}^{\perp}.$

Proof. Note that if $\mathcal{D\cap U}=\emptyset$ then
$\mathcal{I}\left(\mathcal{D}\right)\cap\mathcal{U}=\emptyset$ since
$\mathcal{U}$ is an ultrafilter. We argue by contradiction, assume that
$\mathcal{U}\cap\mathcal{D}^{\perp}=\emptyset.$ Note that this implies that
for every $B\in\mathcal{U}$ there are $\left\\{A_{n}\mid
n\in\omega\right\\}\subseteq\mathcal{D}$ such that $\left|B\cap
A_{n}\right|=\omega$ for every $n\in\omega.$ Consider the Laver forcing
$\mathbb{L}\left(\mathcal{U}\right).$ Since
$\mathcal{I}\left(\mathcal{D}\right)$ is an analytic set, by a result of Blass
(see [4]), there is $p\in\mathbb{L}\left(\mathcal{U}\right)$ such that either
$\left[p\right]\subseteq\mathcal{I}\left(\mathcal{D}\right)$ or
$\left[p\right]\cap\mathcal{I}\left(\mathcal{D}\right)=\emptyset.$ Since
$\mathcal{U}$ is a $P$-point, we can find $q\leq p$ and $U\in\mathcal{U}$ such
that $suc_{q}\left(s\right)=^{\ast}U$ for every $s\in q$ such that $s$ extends
the stem of $q.$

Let $\left\\{A_{n}\mid n\in\omega\right\\}\subseteq\mathcal{D}$ such that
$A_{n}\cap U$ is infinite for every $n\in\omega.$ On one hand, we can find a
branch $f_{0}\in\left[q\right]$ such that $im\left(f_{0}\right)\subseteq
A_{0},$ so $im\left(f_{0}\right)\in\mathcal{I}\left(\mathcal{D}\right)$, but
on the other hand, we can find $f_{1}\in\left[q\right]$ such that
$im\left(f_{1}\right)\cap A_{n}$ is infinite for every $n\in\omega,$ so
$im\left(f_{1}\right)\notin\mathcal{I}\left(\mathcal{D}\right).$ These two
statement are in contradiction since we know that either
$\left[q\right]\subseteq\mathcal{I}\left(\mathcal{D}\right)$ or
$\left[q\right]\cap\mathcal{I}\left(\mathcal{D}\right)=\emptyset.$

We can now conclude the following:

###### Corollary 76

Let
$\mathcal{A}={\textstyle\bigcup\limits_{\alpha\in\kappa}}\mathcal{C}_{\alpha}$
be a MAD family such that each $\mathcal{C}_{\alpha}$ is a closed subset of
$\left[\omega\right]^{\omega}.$ Let $\mathcal{U}$ be a $P$-point such that
$\mathcal{A}\cap\mathcal{U}=\emptyset.$ If $\mathbb{P}$ is a forcing that
diagonalizes $\mathcal{U},$ and $G\subseteq\mathbb{P}$ is a generic filter,
then
$V\left[G\right]\models``\mathcal{A}^{V\left[G\right]}={\textstyle\bigcup\limits_{\alpha\in\kappa}}\mathcal{C}_{\alpha}^{V\left[G\right]}$
is not a MAD family$\textquotedblright$ (where
$\mathcal{C}_{\alpha}^{V\left[G\right]}$ is the reinterpretation of
$\mathcal{C}_{\alpha}$ in the model $V\left[G\right]$).

Proof. Let $G\subseteq\mathbb{P}$ be a generic filter. We argue in
$V\left[G\right].$ Let $B\in\left[\omega\right]^{\omega}$ be a
pseudointersection of $\mathcal{U}.$ We claim that $B$ is almost disjoint with
$\mathcal{A}^{V\left[G\right]}.$ By the previous result, for every
$\alpha<\kappa$ there is $U_{\alpha}\in\mathcal{U}$ such that the following
statement holds in $V:$ “$U_{\alpha}$ is almost disjoint with every element of
$\mathcal{C}_{\alpha}$”. Since $\mathcal{C}_{\alpha}$ is a closed set, this is
an absolute statement, so $U_{\alpha}$ is almost disjoint with every element
of $\mathcal{C}_{\alpha}^{V\left[G\right]}.$ Furthermore, since
$B\subseteq^{\ast}U_{\alpha},$ then $B$ is also almost disjoint with every
element of $\mathcal{C}_{\alpha}^{V\left[G\right]}$ for every $\alpha<\kappa,$
hence $B$ is almost disjoint with $\mathcal{A}^{V\left[G\right]}.$

In [18] it was proved that Canjar ultrafilters are $P$-points (moreover,
Canjar ultrafilters are precisely the “strong $P$-points”, see [6] for the
definition of strong $P$-point). In this way, we can conclude the following:

###### Corollary 77

The following statements are consistent with the axioms of ZFC:

  1. 1.

(Brendle, Raghavan) $\omega_{1}=\mathfrak{b<a}_{closed}$ $\mathfrak{=}$
cov$\left(\mathcal{M}\right)=\mathfrak{c}=\omega_{2}.$

  2. 2.

$\omega_{1}=\mathfrak{u}$ $\mathfrak{<a}_{closed}$ $\mathfrak{=}$
$\omega_{2}.$

##  8 Open Problems

We will list some open problems the authors do not know how to answer.
Regarding the forcings $\mathbb{Q}\left(\mathcal{F}\right)$ of Sabok and
Zapletal, we know that they might or might not diagonalize $\mathcal{F}.$ It
would be interesting to know the answer of the following:

###### Problem 78

Is there a nice combinatorial characterization for the filters $\mathcal{F}$
for which $\mathbb{Q}\left(\mathcal{F}\right)$ diagonalizes $\mathcal{F}$?

The forcings $\mathbb{M}\left(\mathcal{F}\right)$ and
$\mathbb{L}\left(\mathcal{F}\right)$ have been proven to be very useful (see
for example [38], [53], [21], [36], [22] or [13] for some applications of this
forcings). We expect the forcings $\mathbb{PT}\left(\mathcal{F}\right)$ to
have interesting applications as well. It would then be useful to have a
deeper understanding of such forcings. For example we have the following:

###### Problem 79

Let $\mathcal{F}$ be a filter.

  1. 1.

Is there a combinatorial characterization of when
$\mathbb{PT}\left(\mathcal{F}\right)$ does not add Cohen reals?

  2. 2.

Is there a combinatorial characterization of when
$\mathbb{PT}\left(\mathcal{F}\right)$ does not add dominating reals?

  3. 3.

Is there a filter $\mathcal{F}$ such that
$\mathbb{PT}\left(\mathcal{F}\right)$ does not add dominating reals but
$\mathbb{M}\left(\mathcal{F}\right)$ adds dominating reals?

It would be interesting if the previous properties have characterizations in
terms of the Katětov order, similar to the results for
$\mathbb{Q}\left(\mathcal{F}\right)$ obtained in [55]. Regarding the
preservation of $P$-points, we have the following:

###### Problem 80

  1. 1.

Let $\mathcal{U}$ be a $P$-point and $\mathcal{F}$ a Canjar filter such that
$\mathbb{PT}\left(\mathcal{F}\right)$ forces that $\mathcal{U}$ generates a
non-meager filter. Does $\mathbb{PT}\left(\mathcal{F}\right)$ preserve
$\mathcal{U}$?101010This question was recently answered positively by
Chodounský, Verner and the first author. This result will be published in
another article.

  2. 2.

Assuming CH, is there a Canjar filter $\mathcal{F}$ such that
$\mathbb{PT}\left(\mathcal{F}\right)$ destroys all $P$-points?

Regarding half-Cohen reals, we do not know the answer of the following
questions:

###### Problem 81

If $\mathbb{P}$ does not add half-Cohen reals and
$\mathbb{P\Vdash}``\mathbb{\dot{Q}}$ does not add half-Cohen
reals$\textquotedblright,$ is it true that $\mathbb{P\ast\dot{Q}}$ does not
add half-Cohen reals?

###### Problem 82

If $\delta$ is a limit ordinal,
$\langle\mathbb{P}_{\alpha},\mathbb{\dot{R}}_{\alpha}\mid\alpha<\delta\rangle$
is a countable support iteration of proper forcings such that each
$\mathbb{P}_{\alpha}$ does not add half-Cohen reals, is it true that
$\mathbb{P}_{\delta}$ does not add half-Cohen reals? What if each
$\mathbb{P}_{\alpha}$ does not add dominating reals?

Recall that not adding Cohen reals is not preserved under two step iteration
by the result of Zapletal in [63].

###### Acknowledgement 83

The authors would like to thank Juris Steprāns for valuable comments and hours
of stimulating conversations. We would also like to thank the referee for
her/his generous comments that improved the paper. The research for this paper
was started when the first author was a postdoc at York University, concluded
after he became a postdoc in the University of Toronto, and the final version
was finished while working for the Universidad Autónoma de México (UNAM). He
would like to express his gratitude to the three institutions.

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Osvaldo Guzmán

Centro de Ciencias Matematicas, UNAM.

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Damjan Kalajdzievski

York University

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