# P-measures in models without P-points

Piotr Borodulin–Nadzieja  Mathematical Institute, University of Wrocław
pl. Grunwaldzki 2, 50-384 Wrocław, Poland<EMAIL_ADDRESS>,  Jonathan
Cancino-Manríquez  Czech Academy of Sciences
Žitn’a 25, Praha 1, Czech Republic<EMAIL_ADDRESS>and  Adam Morawski
Mathematical Institute, University of Wrocław
pl. Grunwaldzki 2, 50-384 Wrocław, Poland<EMAIL_ADDRESS>
###### Abstract.

We answer in negative the problem if the existence of a P-measure implies the
existence of a P-point. Namely, we show that if we add random reals to a
certain ’unique P-point’ model, then in the resulting model we will have a
P-measure but not P-points. Also, we investigate the question if there is a
P-measure in the Silver model. We show that rapid filters cannot be extended
to a P-measure in the extension by $\omega$ product of Silver forcings and
that in the model obtained by the product of $\omega_{2}$ many Silver forcings
there are no P-measures of countable Maharam type.

###### Key words and phrases:

P-points, P-measures, Additive Property, Silver forcing, random forcing, rapid
filters

###### 2010 Mathematics Subject Classification:

Primary: 03E05, 03E35. Secondary: 03E75, 28E15.

The first author was supported by the Polish National Science Center under the
Weave-UNISONO call in the Weave programme, no. 2021/03/Y/ST1/00124.
The second author was supported by the "Programme to support prospective human
resources – post Ph.D. candidates" of the Czech Academy of Sciences, project
L100192251.

## Introduction

We call an ultrafilter $\mathcal{U}$ on $\omega$ a P-point if every decreasing
sequence of its elements has a pseudointersection in $\mathcal{U}$. Notice
that non-principal ultrafilters on $\omega$ cannot be closed under countable
intersections. In a sense, a non-principal P-point is an ultrafilter as close
to being closed under countable intersections as possible. There is a huge
literature about P-points: they proved their importance in infinitary
combinatorics, topology, and forcing.

The existence of P-points is independent of the usual axioms of set theory.
They exist under Continuum Hypothesis and under many other axioms. However,
Shelah showed that consistently there are no P-points (his proof can be now
found in [Wim82]). Quite recently Chodounský and Guzmán [CG19] showed that
there are no P-points in the classical Silver model. It is still unknown if
P-points exist in the classical random model (see also remarks at the end of
the article).

In a similar fashion to P-points one can define P-measures (note that they are
usually called ’measures with additive property’ in the literature): say that
a (finitely additive) measure on $\omega$ is a P-measure if every decreasing
sequence of subsets of $\omega$ has a pseudointersection whose measure is
equal to the limit of measures of the elements of the sequence. If a measure
on $\omega$ vanishes on points, then it cannot be $\sigma$-additive but
P-measures are as close to being $\sigma$-additive as possible.

If $\mathcal{U}$ is a P-point, then the measure $\delta_{\mathcal{U}}$, i.e.
the 0-1 measure concentrated at $\mathcal{U}$, is a P-measure. It is also
known that if there is a P-point, then there is an atomless P-measure (see
[BFPRN01]) but the following problem remained open.

######  Problem 0.1.

Does the existence of P-measures imply the existence of P-points?

This problem was investigated e.g. in [Mek84], [BFPRN01], [Gre19]. In [Mek84]
Mekler showed that in the Shelah’s model witnessing the lack of P-points there
are no P-measures. In the light of Chodounský-Guzmán theorem it is natural to
ask the following.

######  Problem 0.2.

Is there a P-measure in the Silver model?

Problem 0.2 was the initial motivation for our article and we hoped that the
solution would either bring us the full answer to Problem 0.1 (in case of the
positive answer) or at least it would be a strong indication that Problem 0.1
has a negative solution. Ironically, we have not been able to solve Problem
0.2 but we have solved Problem 0.1.

In [BNS23] it is proved that every P-point can be extended to a P-measure in
the random extensions. The first author believed that this theorem can be used
to prove that there is a P-point in the classical random model, if one manages
to answer Problem 0.1 in positive. To his surprise, the second author used it
to answer Problem 0.1 in negative.

######  Theorem 0.3.

It is consistent with $\mathsf{ZFC}$ that there is a P-measure but there is no
P-point.

The main idea behind the proof is the following. A well-known theorem of Kunen
says that no selective ultrafilter can be extended to a P-point in any random
extension. Shelah proved that there is a model where all P-points are
selective. The model witnessing the negative answer to Problem 0.1 is the
random extension of a model obtained similarly to the Shelah’s model. There
are P-measures there because of the theorem mentioned in the previous
paragraph. The proof that there are no P-points required an upgrade of both
Kunen’s theorem and Shelah’s construction.

Theorem 0.3 made Problem 0.2 slightly less compelling. However, we think that
the question about the existence of P-measures in the Silver model is still
interesting (and it is still open). Moreover, considerations around Problem
0.2 led us to some interesting notions in the theory of measures on $\omega$.

The natural approach toward a solution of Problem 0.2 is to analyze which
filters from the ground model can be extended to a P-measure in the Silver
extension, i.e. for which filters $\mathcal{F}$ there is a measure $\mu$ in
the extension such that $\mu(F)=1$ for each $F\in\mathcal{F}$. (By the Silver
extension we mean the model obtained by side-by-side products of Silver
forcings). E.g. it is not hard to see that every such filter would have to be
a P-filter. Our first achievement in this direction was the following.

######  Theorem 0.4.

No rapid filter can be extended to a P-measure in the Silver extension.

One can also analyze which measures from the ground model can be extended to a
P-measure in the extension (again such measures would also have to be
P-measures). As we have already pointed out, the basic examples of measures
are those concentrated on an ultrafilter. Ultrafilters can be used to define
also non-atomic measures, so called ultrafilter densities (see Section 2 for
the precise definitions). We prove that no such measure can be extended to a
P-measure in the Silver extension.

######  Theorem 0.5.

If there is an ultrafilter $\mathcal{U}$ such that $\delta_{\mathcal{U}}$ is
Rudin-Blass below a measure $\mu$, then $\mu$ cannot be extended to a
P-measure in the Silver model. In particular, no ultrafilter and no
ultrafilter density can be extended to a P-measure in the Silver model.

When we proved the above theorem, in fact, we did not know any example of a
P-measure (under $\mathsf{CH}$) that does not satisfy its assumptions, so we
did not have any candidate for a which can be extended to a P-measure in the
Silver model. Then, we managed to show that under $\mathsf{CH}$ there is a
P-measure $\lambda$ which is in a sense the Lebesgue measure on $2^{\omega}$
in disguise. This measure is basically different from all the examples of
P-measures known before: it is non-atomic but its Maharam type is countable.
Recall that the Maharam type of a measure $\mu$ on $\omega$ is the density of
the pseudo-metric on $\mathcal{P}(\omega)$ defined by
$d_{\mu}(A,B)=\mu(A\triangle B)$.

Additionally, we constructed $\lambda$ in a way that it does not satisfy the
assumption of Theorem 0.5, so it looked like a promising candidate for a
measure that can be extended to a P-measure in the Silver extension. However,
we managed to upgrade the above theorem (to Theorem 4.3) so that as a
corollary we get the following.

######  Theorem 0.6.

In the Silver model there are no P-measures of countable Maharam type.

Since P-points are in a sense measures of Maharam type $2$, the above theorem
can be seen as a generalization of Chodounský-Guzmán theorem. It also says
that $\lambda$, as a measure of countable Maharam type, cannot be extended to
a P-measure in the Silver extension.

Finally, we are again in a situation in which we do not know any example of a
measure that has a chance to be extended to a P-measure in the Silver
extension. It might mean that there are no P-measures in the Silver model. Or
there are still some interesting P-measures to discover in the models
satisfying $\mathsf{CH}$.

###### Contents

  1. 1 P-points and P-measures
  2. 2 Measures on $\omega$
  3. 3 Silver forcing
  4. 4 Measures which cannot be extended to P-measures in the Silver model
  5. 5 Rapid Filters and P-measures
  6. 6 More on P-measures under CH
  7. 7 A model with P-measures and without P-points

##  1\. P-points and P-measures

We start by recalling the basic definitions.

A filter over a set $X$ is a family of subsets of $X$ closed under
intersections and supersets. We will consider only proper filters, i.e. those
which do not contain $\emptyset$. An ultrafilter is a $\subseteq$-maximal
proper filter or, equivalently, a filter such that for every $A\subseteq X$
either it contains $A$ or it contains $A^{c}$. An ultrafilter is called non-
principal if it does not contain any singleton (and thus any finite set).

A finitely additive measure on $X$ is a function
$\mu\colon\mathcal{P}(X)\to\mathbb{R}_{\geq 0}$ such that

  1. (1)

$\mu(\emptyset)=0$,

  2. (2)

for $A\cap B=\emptyset$ we have $\mu(A\cup B)=\mu(A)+\mu(B)$.

We say that a measure vanishes on points if no singleton has a positive
measure (and so all finite sets have measure 0). A probability measure is a
measure with $\mu(X)=1$. A probability measure is said to be atomless if for
any $\varepsilon>0$ the set $X$ can be divided into finitely many sets of
measure not bigger than $\varepsilon$.

Further on, in this paper all ultrafilters will be non-principal ultrafilters
on $\omega$ and all measures will be vanishing on points, finitely additive
probability measures on $\omega$.

We say that $A$ is almost contained in $B$ (denoted $A\subseteq^{*}B$) if
$A\setminus B$ is finite. For a family $\mathcal{F}$ we say that $Z$ is a
pseudointersection of $\mathcal{F}$ if $Z\subseteq^{*}F$ for each
$F\in\mathcal{F}$. Notice that $Z$ is a pseudointersection of a countable
family $\\{F_{n}\colon n\in\omega\\}\subseteq\mathcal{P}(\omega)$ if and only
if there is a function $f\colon\omega\to\omega$ such that $Z\setminus
f(n)\subseteq F_{n}$ for each $n$.

Finally, the following two notions are of main interest. An ultrafilter
$\mathcal{U}$ is called a _P-point_ if for every decreasing sequence
$\\{U_{n}\colon n\in\omega\\}$ of elements of $\mathcal{U}$ there is
$Z\in\mathcal{U}$ which is a pseudointersection of $\\{U_{n}\colon
n\in\omega\\}$. A measure $\mu$ is called a P-measure if for every decreasing
sequence $\\{A_{n}\colon n\in\omega\\}$ there is $Z$, a pseudointersection of
$\\{A_{n}\colon n\in\omega\\}$, with $\mu(Z)=\inf\mu(A_{n})$
($=\lim\mu(A_{n})$). Such measures are also known as measures with the
additive property(*) or AP-measures.

We will formulate now two simple facts concerning P-measures leaving the
proofs for the reader.

######  Proposition 1.1.

The following are equivalent:

  1. (1)

$\mu$ is a P-measure,

  2. (2)

for every increasing sequence $(B_{n})$ there is a set $Z$ with
$\mu(Z)=\sup_{n}(B_{n})$ and $Z\supseteq^{*}B_{n}$ for each $n$,

  3. (3)

for every pairwise disjoint family $(C_{n})$ there is $Z$ st.
$\mu(Z)=\sum_{n}\mu(C_{n})$ and $Z\supseteq^{*}C_{n}$ for each $n$.

######  Proposition 1.2.

A measure $\mu$ is a P-measure if and only if for any decreasing sequence
$(A_{n})$ there is an increasing function $f\colon\omega\to\omega$ such that

$\mu\left(\bigcup_{n\in\omega}\left(A_{n}\cap[f(n),f(n+1))\,\right)\right)=\lim_{n\to\infty}\mu(A_{n}).$

Proposition 1.1 (3) justifies the statement that P-measures are, in some
sense, as close to being $\sigma$-additive as a (vanishing on points) measure
possibly can. Proposition 1.2 carries a valuable intuition that, for the
purposes of P-points and P-measures, any pseudointersection is tightly
associated with a function.

##  2\. Measures on $\omega$

Note that every (finitely additive) measure $\mu$ on $\omega$ can be seen as a
measure on the clopen subsets of $\omega^{*}$ (as, according to our promise
from the previous section, $\mu$ vanishes on points) and then extended
uniquely to a Borel (countably additive) measure on $\omega^{*}$. So, measures
on $\omega$ can be identified with countably additive measures on
$\omega^{*}$.

We will give now some examples of measures on $\omega$.

######  Example 2.1.

Let $\mathcal{U}$ be an ultrafilter on $\omega$. We may produce a measure by
assigning value 1 to all sets in $\mathcal{U}$ and value 0 to their
complements. We will denote such measure by $\delta_{\mathcal{U}}$ (and call
it _the Dirac delta at_ $\mathcal{U}$). Clearly, every Dirac delta is atomic.

There are many atomless measures on $\omega^{*}$, as $\omega^{*}$ is not
scattered (see [Rud57]). Here we present a quite general way to produce
atomless measures extending the asymptotic density.

######  Example 2.2.

Let $\mu$ be a measure on $\omega$. Let $d_{A}\colon\omega\to[0,1]$ be defined
by $d_{A}(n)=|A\cap n|/n$. Let

$\nu(A)=\int d_{A}\ d\mu.$

Formally, the above expression does not make much sense, so let us comment on
how to interpret it. First, lift $\mu$ to a Borel measure $\mu^{\prime}$ on
$\omega^{*}$. The function $d_{A}$ is bounded so it induces a continuous
function $d^{\prime}_{A}\colon\omega^{*}\to[0,1]$. Let $\nu^{\prime}(A)=\int
d^{\prime}_{A}\ d\mu^{\prime}$. This is a (countably additive) measure on
$\omega^{*}$ and so it generates a measure $\nu$ on $\omega$. Such a measure
extends the asymptotic density.111Note that it is also possible to define
Lebesgue integral directly for finitely additive measures.

We will consider now measures like the above of quite a specific form.

######  Example 2.3.

Suppose that $\mu$ is a Dirac delta concentrated at an ultrafilter
$\mathcal{U}$. Let $d_{A}$ be defined as in Example 2.2. Then

$\nu(A)=\int d_{A}\ d\mu=\lim_{n\to\mathcal{U}}d_{A}(n).$

This is an atomless measure extending the asymptotic density. We will call
such measures _ultrafilter densities_. Note that if $\mathcal{U}$ is a
P-point, then the resulting measure is a P-measure (see e.g. [BFPRN01]).

Such measures were studied e.g. in [Mek84] or [BFPRN01]. Notice that two
different ultrafilters may induce (in the sense of Example 2.3) the same
measure (see [Gre19], [Kun22]).

It seems to be not easy to define a measure on $\omega$ ’directly’ (even using
ultrafilter as a parameter) in a basically different way than in the above
examples. In Section 6 we will present one more example, which is defined
without any (direct) use of ultrafilters.

Now we will investigate relations between measures on $\omega$. Recall the
definition of Rudin-Blass ordering of filters.

######  Definition 2.4.

Suppose $\mathcal{F}$, $\mathcal{G}$ are filters on $\omega$. We say that
$\mathcal{G}$ is _Rudin-Blass reducible to_ $\mathcal{F}$,
$\mathcal{G}\leq_{RB}\mathcal{F}$ in short, if there is a finite-to-one
function $f\colon\omega\to\omega$ such that $X\in\mathcal{G}$ if and only if
$f^{-1}[X]\in\mathcal{F}$.

We say that a filter $\mathcal{F}$ is _nearly ultra_ if there is an
ultrafilter $\mathcal{U}$ such that $\mathcal{U}\leq_{RB}\mathcal{F}$.

The definition of Rudin-Blass ordering can be naturally re-used to introduce
an ordering on measures on $\omega$.

######  Definition 2.5.

Suppose $\mu$, $\nu$ are measures on $\omega$. We say that $\nu$ is _Rudin-
Blass reducible to_ $\mu$, $\nu\leq_{RB}\mu$ in short, if there is a finite-
to-one function $f\colon\omega\to\omega$ such that for every
$X\subseteq\omega$ we have $\nu(X)=\mu(f^{-1}[X])$.

Similarly, we will say that a measure $\mu$ is _nearly Dirac_ if there is an
ultrafilter $\mathcal{U}$ such that $\delta_{\mathcal{U}}\leq_{RB}\mu$. Notice
that a measure is nearly Dirac if and only if its filter of measure 1 sets is
nearly ultra.

######  Proposition 2.6.

Every measure that is an ultrafilter density is nearly Dirac.

###### Proof.

Let $\mu$ be an ultrafilter extension of density. Fix an ultrafilter
$\mathcal{U}$ such that $\mu(A)=\lim_{n\to\mathcal{U}}|A\cap n|/n$.

Let $g\colon\omega\to\omega$ be an increasing function such that
$g(k+1)/g(k)\to\infty$. Define $L_{i}=\bigcup_{k}[g(2k+i),g(2k+i+1))$ for
$i\in\\{0,1\\}$ and notice that $L_{0},L_{1}$ is a partition of $\omega$ so
exactly one of them has to be in $\mathcal{U}$. Let $j$ be such that $L_{j}$
is in $\mathcal{U}$. In what follows, for simplicity, we will assume that
$j=1$; the other case is analogous.

Let $f\colon\omega\to\omega$ be defined by $f(n)=k$ if
$n\in[g(2k+j),g(2k+j+2))$. Define

$\mathcal{V}=\\{X\subseteq\omega\colon f^{-1}[X]\in\mathcal{U}\\}$

and notice that $\mathcal{V}$ is an ultrafilter. We will show that $f$
witnesses that $\delta_{\mathcal{V}}\leq_{RB}\mu$.

Let $X\in\mathcal{V}$. Fix $\varepsilon>0$. Denote

$A_{\varepsilon}=\\{n\colon|\frac{|f^{-1}[X]\cap n|}{n}-1|<\varepsilon\\}.$

Suppose that $n\in[g(2k+1),g(2k+2))$. Then

$|f^{-1}[X]\cap n|\geq n-g(2k)\geq g(2k+1)-g(2k)>(1-\varepsilon)n$

for $n$ big enough (since $n\geq g(2k+1)$ and $g(2k)/g(2k+1)<\varepsilon$ for
$k$ big enough).

It follows that $f^{-1}[X]\cap L_{1}\subseteq^{*}A_{\varepsilon}$. But
$f^{-1}[X]\in\mathcal{U}$ (as $X\in\mathcal{V}$) and $L_{1}\in\mathcal{U}$.
Thus, $A_{\varepsilon}\in\mathcal{U}$ and, since $\varepsilon$ was arbitrary,
$\mu(X)=1$.

Analogously, if $X\notin\mathcal{V}$, then $X^{c}\in\mathcal{V}$,
$\mu(f^{-1}[X]^{c})=1$ and so $\mu(f^{-1}[X])=0$. Hence,
$\delta_{\mathcal{V}}\leq_{RB}\mu$. ∎

######  Proposition 2.7.

Being a P-filter or a P-measure is hereditary with respect to Rudin-Blass
ordering.

###### Proof.

We only provide a short proof for the case of P-measures. Consider a P-measure
$\mu$, a finite-to-one function $g\colon\omega\to\omega$ and an
$\subseteq^{*}$-decreasing sequence $(A_{n})_{n}$. Let $B$ be a
pseudointersection of the sequence $(g^{-1}[A_{n}])_{n}$ suitable for the
measure $\mu$. Since $\mu(g^{-1}[g[B]])\geq\mu(B)$, for all $n\in\omega$,
$g^{-1}[g[B]]\subseteq^{*}A_{n}$, and we know that $g[B]$ is a
pseudointersection of $(A_{n})_{n}$ suitable for $\mu g^{-1}$. ∎

Hence the ultrafilter $\mathcal{V}$ from the proof of Proposition 2.6 is a
P-point whenever $\mu$ is a P-measure (even if the ultrafilter used in the
definition of $\mu$ is not a P-point). So we have the following corollary
(which was proved also by Grebik in [Gre19]).

######  Corollary 2.8.

If there is a P-measure, which is an ultrafilter density, then there is a
P-point.

Perhaps it is quite surprising that consistently all measures on $\omega$
(vanishing on points) are nearly Dirac. Recall that the principle of filter
dichotomy implies that all filters that are not meager are nearly ultra. But
the filters of measure $1$ sets are always non-meager:

######  Proposition 2.9.

If $\mu$ is a measure on $\omega$, then the filter
$\mathcal{F}=\\{F\colon\mu(F)=1\\}$ is non-meager.

###### Proof.

Suppose that $\mathcal{F}$ is meager. Then, by Talagrand’s characterization of
meager filters (see e.g. [Bla10]), there is an interval partition $(I_{n})$
such that no infinite union of $I_{n}$’s is in the dual ideal. Let
$(A_{\alpha})_{\alpha<\omega_{1}}$ be an almost disjoint family (of subsets of
$\omega$) and for $\alpha<\omega_{1}$ let $N_{\alpha}=\bigcup_{n\in
A_{\alpha}}I_{n}$. Then $\mu(A_{\alpha})>0$ for each $\alpha$ and
$\mu(A_{\alpha}\cap A_{\beta})=0$ if $\alpha\neq\beta$, a contradiction. ∎

Notice that if $\mathcal{U}\leq_{RB}\mathcal{F}$ for a filter of the form
$\\{F\colon\mu(F)=1\\}$ for some measure $\mu$ and an ultrafilter
$\mathcal{U}$, then $\delta_{\mathcal{U}}\leq_{RB}\mu$. So, we have the
following.

######  Proposition 2.10.

Under the principle of filter dichotomy, all measures on $\omega$ are nearly
Dirac.

However, it is not true that all measures on $\omega$ are nearly Dirac in
$\mathsf{ZFC}$. We have the following example.

######  Proposition 2.11.

In the random model there is a measure on $\omega$ which is not nearly Dirac.

###### Proof.

In [BNS23, Theorem 5.6] the authors showed that in the random model there is a
measure supported by a filter $\mathcal{F}$ which is generated by a family
$\mathcal{A}$ of size $\omega_{1}$. Suppose that $\mathcal{G}$ is a filter
such that $\mathcal{G}\leq_{RB}\mathcal{F}$. Then $\mathcal{G}$ is generated
by at most $\omega_{1}$ sets (namely, $\\{f[A]\colon A\in\mathcal{A}\\}$,
where $f$ is a witness for $\mathcal{G}\leq_{RB}\mathcal{F}$). But, in the
random model the ultrafilter number $\mathfrak{u}=\omega_{2}$, so
$\mathcal{G}$ cannot be an ultrafilter. ∎

In Section 6 we will present another example of a measure that is not nearly
Dirac, this time under $\mathsf{CH}$.

##  3\. Silver forcing

By $\mathbb{S}i$ we will denote the set of partial functions from $\omega$ to
$\\{0,1\\}$ with co-infinite domains ordered by the reverse inclusion. Such
partial order is called the Silver forcing. We will use standard forcing
notation. In particular, let $V$ be the ground model, let $G$ be a
$\mathbb{S}i$-generic over $V$. We will use the following abbreviations and
provide the necessary context to conclude whether an object is in $V$ or
$V[G]$:

  * •

instead of $\dot{x}[G]$, the evaluation of a name by a generic, we will simply
write $x$,

  * •

instead of $\check{x}$, the standard name for $x\in V$, we will also write
$x$.

###  3.1. $\omega^{\omega}$-bounding.

We will need a few properties of the Silver forcing. The following facts hold
for $\mathbb{S}i$.

######  Proposition 3.1 ($\omega^{\omega}$-bounding).

If $f\in V[G]$ is a function from $\omega$ to $\omega$, then there is a
function $g\in V$ with $\forall n\ g(n)\geqslant f(n)$.

A proof of Proposition 3.1 can be found in [Hal17].

We can extend our definition of the $\omega^{\omega}$-bounding property to the
relation between models. We say that $W\supseteq V$ has the
$\omega^{\omega}$-bounding property (over $V$) if for any
$f\colon\omega\to\omega$ form $W$ there is $g\colon\omega\to\omega$ from $V$
with $f(n)\leqslant g(n)$.

######  Proposition 3.2.

For $W\supseteq V$ (models of ZFC) and $W$ having the
$\omega^{\omega}$-bounding property over $V$, if $\mu\in V$ is a measure that
can be extended to a P-measure in $W$, then $\mu$ is a P-measure.

###### Proof.

Take any decreasing sequence $\langle\,A_{n}:n\in\omega\,\rangle\in V$. If
$\mu^{\prime}$ is a P-measure extending $\mu$ in $V[G]$ then there is some
function $f\in V[G]$ such that, if

$A=\bigcup_{n\in\omega}\left(A_{n}\cap[f(n),f(n+1))\,\right),$

then $\mu^{\prime}(A)=\lim_{n}\mu^{\prime}(A_{n})$, by Proposition 1.2.

Now by $\omega^{\omega}$-bounding property we can take $\tilde{f}\in V$ with
$\tilde{f}(n)\geqslant f(n)$.

If we put

$A^{\prime}=\bigcup_{n\in\omega}\left(U_{n}\cap[\tilde{f}(n),\tilde{f}(n+1))\,\right),$

then $A^{\prime}\supseteq A$ and $A^{\prime}$ is still a pseudointersection of
$\left\\{\,A_{n}:n\in\omega\,\right\\}$ and

$\lim_{n}\mu^{\prime}(A_{n})\geq\mu^{\prime}(A^{\prime})\geq\mu^{\prime}(A)=\lim_{n}\mu^{\prime}(A_{n})=\lim_{n}\mu(A_{n}).$

But now $A^{\prime}\in V$ and so
$\mu(A^{\prime})=\mu^{\prime}(A^{\prime})=\lim_{n}\mu(A_{n})$. ∎

As we know that a P-point can be viewed as a P-measure we get the following
remark:

######  Remark 3.3.

For $W\supseteq V$ and $W$ having the $\omega^{\omega}$-bounding property over
$V$, if $\mathcal{U}\in V$ is an ultrafilter that can be extended to a P-point
(or, more generally, a P-measure) in $W$, then $\mathcal{U}$ is a P-point.

In this article, we will use side-by-side products of Silver forcings. Denote
$\mathbb{P}_{\kappa}=\prod_{\alpha<\kappa}\mathbb{S}i$

######  Theorem 3.4.

Suppose $\mu$ is a measure on $\omega$ in $V^{\mathbb{P}_{\kappa}}$ for
$\kappa$ of uncountable cofinality. Then there is $\alpha<\kappa$ and a
measure $\nu$ in $V^{\mathbb{P}_{\alpha}}$ such that $\mu$ extends $\nu$.

###### Proof.

Suppose that $\dot{\mu}$ is a $\mathbb{P}_{\kappa}$-name for a measure on
$\omega$, for $\kappa$ of uncountable cofinality. For $\alpha\in\kappa$ let
$\dot{\mu}_{\alpha}$ be a $\mathbb{P}_{\kappa}$-name for
$\dot{\mu}\upharpoonright(\mathcal{P}(\omega)\cap V^{\mathbb{P}_{\alpha}})$.

Then the set

$C=\\{\alpha<\kappa\colon\mbox{ there is a }\mathbb{P}_{\alpha}\mbox{-name
}\dot{\nu}_{\alpha}\mbox{ such that
}\mathbb{P}_{\kappa}\Vdash\dot{\mu}_{\alpha}=\dot{\nu}_{\alpha}\\}.$

is a club.

Take $\alpha\in C$ and let $\nu$ be any interpretation of
$\dot{\nu}_{\alpha}$. ∎

###  3.2. Interval partitions

Forcing with $\mathbb{S}i$ adds a new real. Namely, if $G$ is a generic, then
we can define $\dot{s}=\bigcup\\{\,p:p\in G\,\\}$ (in $V[G]$). We will call
this a Silver real, or a generic real. A name for such real can be easily
procured:

$\dot{s}=\\{\,\langle\,\hat{\langle n,i\rangle}\,,\,\\{\,\langle
n,i\rangle\,\\}\,\rangle:n\in\omega,i\in 2\,\\}$

i.e. we use functions with singleton domains to set every element of $s$.

A standard way of creating a set from a real (a 0-1-sequence) is to treat the
real as a characteristic function. While considering ultrafilters this
approach applied to Silver reals yields poor results due to the following
fact:

######  Proposition 3.5.

If $\mathcal{U}$ is an ultrafilter in $V$ and $s$ is a Silver real understood
as above, then (in $V[G]$) for some $U\in\mathcal{U}$ either
$s^{-1}[1]\subseteq U$ or $s^{-1}[1]\cap U=\emptyset$.

###### Proof.

Fix an ultrafilter $\mathcal{U}\in V$. Let $\dot{s}$ be a name for Silver real
and take any $p$ with $\operatorname{Dom}p\in\mathcal{U}$. Notice that the set
of possible $p$’s is dense in $\mathbb{S}i$ as any partial function (of co-
infinite domain) can be extended to have co-infinite domain that is in
$\mathcal{U}$.

Now either $p^{-1}[1]\in\mathcal{U}$ and for $U=\operatorname{Cod}p\cup
p^{-1}[1]$ we have $p\Vdash\dot{s}^{-1}[1]\subseteq U$ or
$p^{-1}[0]\in\mathcal{U}$ and taking $U=p^{-1}[0]$ we have
$p\Vdash\dot{s}^{-1}[1]\cap U=\emptyset$. ∎

In other words, if we want to construct an ultrafilter in $V[G]$ extending
$\mathcal{U}$, we do not have much options for its behavior on $s^{-1}[1]$.

Therefore from this point on, following [CG19], we will rather turn reals into
interval partitions. Namely let $pos(q,k)$ denote the index $m$ for which
$q(m)=1$ for the $k$-th time (more precisely
$pos(q,k)=\min\\{\,m:|q^{-1}[1]\cap[0,m]|=k\,\\}$) and we set $pos(q,0)=0$.
Then we define

$I_{k}(r)=[pos(q,k),pos(q,k+1))$

and $\mathcal{I}(q)=\\{I_{k}(q)\colon k\in\omega\\}$. Notice that such
definitions do not require $q$ to be a function with the full domain.

##  4\. Measures which cannot be extended to P-measures in the Silver model

In this section, we prove the main result concerning P-measures in the Silver
model. We will need the following definition. The name “silver” comes from the
fact that, as we will see, this definition is closely related to the
combinatorics of $\mathcal{I}(q)$, for $q\in\mathbb{S}i$.

######  Definition 4.1.

Let $\mu$ be a measure on $\omega$. We say that it is _silver_ if for each
$N\subseteq\omega$, $r>0$ and each interval partition $(I_{n})$ there is
$\epsilon\in\\{-1,1\\}^{\omega}$ such that

$\mu\big{(}\bigcup_{n}N^{\epsilon(n)}\cap I_{n}\big{)}<r$

(Here we follow the convention that that $N^{1}=N$ and $N^{-1}=N^{c}$.)

If $\mu$ extends asymptotic density, then $\mu$ is not silver (which is
witnessed e.g. by $N$ being the set of odd numbers, $r=1/2$ and
$I_{n}=[2^{n},2^{n+1})$ for each $n$). Every Dirac delta is silver.

######  Lemma 4.2.

Suppose that $\mu$ is a measure which is silver and that $p\in\mathbb{S}i$.
Then for every $i\in 2$, $r>0$ we may find $q\leq p$ such that
$\mu(D_{i}(q))>1-r$.

###### Proof.

Suppose that $\mu$ is silver and so for every interval partition $(I_{n})$,
$N\subseteq\omega$ and every $r>0$ there is $\epsilon\in\\{-1,1\\}^{\omega}$
such that

$\mu\big{(}\bigcup_{n}N^{\epsilon(n)}\cap I_{n}\big{)}>1-r.$

Notice that we may additionally assume that and
$\epsilon(n)\cdot\epsilon(n+1)=-1$ for infinitely many $n$’s. It follows from
the general fact that for every measure $\nu$ on $\omega$ and for every
interval partition $(J_{n})$ we may find an infinite set $X\subseteq\omega$
such that $\nu(\bigcup_{n\in X}J_{n})=0$ (see also Proposition 2.9).

Now, fix $i\in 2$, $p\in\mathbb{S}i$ and $r>0$. Suppose (strengthening $p$ if
necessary) that $\mu(\operatorname{Cod}p)=0$, and let $(a_{n})$ be the
increasing enumeration of $\operatorname{Cod}p$. For each $n\in\omega$ let
$I_{n}=[a_{n},a_{n+1})$. Applying the above remark for $(I_{n})$222$(I_{n})$
may not be a partition but as we deal with a measure vanishing on points, we
can forget some initial segment of $\omega$,
$N=D_{i}(p)\cap\operatorname{Dom}p$ and $r$ we may find
$\epsilon\in\\{-1,1\\}^{\omega}$ such that $\epsilon(n)\cdot\epsilon(n+1)=-1$
for infinitely many $n$’s and

$\mu\big{(}\bigcup_{n}N^{\epsilon(n)}\cap I_{n}\big{)}>1-r.$

Let $q\leq p$ be such that

  * •

$q(a_{0})=1$ iff $\epsilon(0)=-1$

  * •

$a_{n+1}\in\operatorname{Dom}q$ iff $\epsilon(n)\epsilon(n+1)=-1$ and in this
case let $q(a_{n+1})=1$.

Then $q\in\mathbb{S}i$ and $\mu(D_{i}(q))>1-r$. ∎

######  Theorem 4.3.

Suppose $\mu$ is silver. Then $\mu$ cannot be extended to a P-measure in
$V^{\mathbb{S}i_{\omega}}$.

###### Proof.

Suppose $\mu$ is silver. Let $p\in\mathbb{S}i_{\omega}$.

Case 1. There is $q\leq p$ such that

$q\Vdash\forall n\ \forall\varepsilon>0\ \exists m>n\
\mu(\bigcup_{k\in[n,m)}\dot{D}_{0}(\dot{s}_{k}))>1-\varepsilon.$

Let $q$ be as above. Using the fact that $\mathbb{S}i$ is
$\omega^{\omega}$-bounding (to get $h^{\prime}$ and then
$h(n)={h^{\prime}}^{(n)}(0)=h^{\prime}(h^{\prime}(...(0)...))$ ), we may
assume (taking stronger condition if necessary) that there is an increasing
function $h\colon\omega\to\omega$ such that

$q\Vdash\forall n\
\mu(\bigcup_{k\in[h(n),h(n+1))}\dot{D}(\dot{s}_{k}))>1-2^{-(n+2)}.$

Let

$\dot{W}_{n}=\bigcup_{k\in[h(n),h(n+1))}\dot{D}(\dot{s}_{k})$

for each $n$ and let $\dot{Z}_{n}=\bigcap_{m\leq n}\dot{W}_{m}$. Notice that

$q\Vdash\dot{Z}_{n}\mbox{ is }\subseteq-\mbox{decreasing and }\forall n\
\mu(\dot{Z}_{n})\geq 1/2.$

Suppose that there is a pseudointersection $\dot{Z}$ in
$V^{\mathbb{S}i_{\omega}}$ of $(\dot{Z}_{n})$ such that
$\mu(\dot{Z})=\lim_{n}\mu(\dot{Z}_{n})\geq 1/2$. Again, assume that $q$ forces
it and that there is a function $f\colon\omega\to\omega$ in $V$ such that

$q\Vdash\dot{Z}\setminus f(n)\subseteq\dot{Z}_{n}.$

Now, since $\mu$ is silver, subsequently using Lemma 4.2, for each $k$ we can
find $q^{\prime}_{k}\leq q_{k}$ such that

$\mu\big{(}\bigcap_{k<n}D_{1}(q^{\prime}_{k})\big{)}>3/4$

for every $n\in\omega$. Define

$A_{n}=\bigcap_{k<h(n+1)}D_{1}(q^{\prime}_{k}).$

Since $\mu$ is a P-measure, $(A_{n})$ is decreasing and $\mu(A_{n})>3/4$ for
each $n$ we may find a strictly increasing $i\in\omega^{\omega}\cap V$ such
that

$\mu\big{(}\bigcup_{n}A_{n}\cap[i(n),i(n+1))\big{)}\geq 3/4.$

By taking $f$ and $i$ bigger if necessary we may assume that for each $n$ and
for every $k\in[h(n),h(n+1))$ we have
$\mathrm{stem}(q^{\prime}_{k})<f(n)<i(n)$.

For every $n\in\omega$ and $k\in[h(n),h(n+1))$ extend $q^{\prime}_{k}$ to
$r_{k}$ in such a way that

$F_{n}:=A_{n}\cap[i(n),i(n+1))\subseteq D_{1}(r_{k})\cap\mathrm{stem}(r_{k}).$

It is possible as $F_{n}\subseteq D_{1}(q^{\prime}_{k})$ and
$\mathrm{stem}(q^{\prime}_{k})<f(n)<i(n)$.

Let $r=(r_{0},r_{1},\dots)$. Then

$r\Vdash\forall n\
F_{n}\subseteq\bigcap_{k\in[h(n),h(n+1))}\dot{D_{1}}(\dot{s}_{k})$

and so

$r\Vdash\forall n\ F_{n}\cap\dot{Z}_{n}=\emptyset.$

Since $f(n)\leq\min F_{n}$ for each $n$ we have

$r\Vdash\bigcup_{n}F_{n}\cap\dot{Z}=\emptyset$

But $\mu(\bigcup_{n}F_{n})\geq 3/4$, hence $r\Vdash\mu(\dot{Z})\leq 1/4$, a
contradiction.

Case 2. There is $q\leq p$ such that

$q\Vdash\exists m\ \exists\varepsilon>0\ \forall n>m\
\mu(\bigcap_{k\in[m,n)}\dot{D}_{1}(\dot{s}_{k}))>\varepsilon.$

Without loss of generality, we may assume that $m=0$. Let
$\dot{Z}_{n}=\bigcap_{k\leq n}\dot{D}_{1}(\dot{s}_{k})$. Notice that

$q\Vdash\dot{Z}_{n}\mbox{ is }\subseteq-\mbox{decreasing and }\forall k\
\mu(\dot{Z}_{n})\geq\varepsilon.$

Suppose that there is a pseudointersection $\dot{Z}$ in
$V^{\mathbb{S}i_{\omega}}$ of $(\dot{Z}_{n})$ such that
$\mu(\dot{Z})=\lim_{n}\mu(\dot{Z}_{n})\geq\varepsilon$. Assume that $q$ forces
it and that there is a function $f\colon\omega\to\omega$ in $V$ such that

$q\Vdash\dot{Z}\setminus f(n)\subseteq\dot{Z}_{n}.$

For every $n\in\omega$, using Lemma 4.2 we can find $q^{\prime}_{n}\leq q_{n}$
such that $\mu(D_{0}(q^{\prime}_{n}))>1-\varepsilon/2^{n+2}$. Let

$A_{n}=\bigcap_{m<n}D_{0}(q^{\prime}_{m}).$

The sequence $(A_{n})$ is $\subseteq$-decreasing and $\mu(A_{n})\geq
1-\varepsilon/2$ for each $n$. Since $\mu$ is a P-measure, there is an
increasing $i\in\omega^{\omega}\cap V$ such that

$\mu\big{(}\bigcup_{n}A_{n}\cap[i(n),i(n+1))\big{)}\geq 1-\varepsilon/2.$

By taking $f$ and $i$ bigger if necessary we may assume that for each $n$ we
have $\mathrm{stem}(q^{\prime}_{n})<f(n)<i(n)$. For every $n$ extend
$q^{\prime}_{n}$ to $r_{n}$ in such a way that

$F_{n}:=A_{n}\cap[i(n),i(n+1))\subseteq D_{0}(r_{n})\cap\mathrm{stem}(r_{n}).$

Let $r=(r_{0},r_{1},\dots)$. Then

$r\Vdash\forall n\ F_{n}\subseteq D_{0}(\dot{s}_{n})$

and so

$r\Vdash\forall n\ F_{n}\cap\dot{Z}_{n}=\emptyset.$

Since $f(n)\leq\min F_{n}$ for each $n$ we have

$r\Vdash\bigcup_{n}F_{n}\cap\dot{Z}=\emptyset$

But $\mu(\bigcup_{n}F_{n})\geq 1-\varepsilon/2$, hence
$r\Vdash\mu(\dot{Z})<\varepsilon/2$, a contradiction. ∎

We say that a measure $\mu$ is nearly silver if there is a measure
$\nu\leq_{RB}\mu$ such that $\nu$ is silver.

######  Corollary 4.4.

If $\mu$ is nearly silver, then it cannot be extended to a P-measure in
$V^{\mathbb{S}i_{\omega}}$.

###### Proof.

Assume $\mu$ can be extended to a P-measure $\mu^{\prime}$ in
$V^{\mathbb{S}i_{\omega}}$. Let $g\in\omega^{\omega}\cap V$ be finite-to-one.
Then by Proposition 2.7 we get that $\mu^{\prime}g^{-1}$ is a P-measure. Since
$\mu^{\prime}g^{-1}$ extends $\mu g^{-1}$, from Theorem 4.3 we get that $\mu
g^{-1}$ cannot be silver.

∎

######  Corollary 4.5.

Suppose that $\mu$ is nearly Dirac. Then $\mu$ cannot be extended to a
P-measure in the Silver model. In particular, no ultrafilter and no
ultrafilter extension of the asymptotic density can be extended to a P-measure
in the Silver model.

######  Problem 4.6.

Assume $\mathsf{CH}$. Is there a measure that is not nearly silver?

If the answer is negative, then Theorem 4.4 and Theorem 3.4 imply that there
are no P-measures in the Silver model. Note that the answer to the question:
“Is there a measure which is not nearly Dirac (under $\mathsf{CH}$)?” is
positive, see Section 6.

Recall that a measure $\mu$ carried by a Boolean algebra $\mathbb{A}$ induces
a pseudo-metric on $\mathbb{A}$: $d_{\mu}(A,B)=\mu(A\triangle B)$. The
_Maharam type_ of $\mu$ is the density of the pseudo-metric space
$(\mathbb{A},d_{\mu})$. In general, the Maharam type of a measure $\mu$ on a
space $K$ is the density of $\mathrm{Bor}(K)_{/\mu=0}$ endowed with the
pseudometric $d_{\mu}$. Notice that if there is $c>0$ and a family
$(B_{\alpha})_{\alpha<\kappa}$ of Borel sets such that
$\mu(B_{\alpha}\triangle B_{\beta})>c$ for each $\alpha\neq\beta<\kappa$, then
$\mu$ has the Maharam type at least $\kappa$.

For a strictly increasing function $g\colon\omega\to\omega$ let
$\widehat{g}\colon\omega\to\omega$ be the finite-to-one function defined by

$\widehat{g}(k)=n\iff g(n)\leq k<g(n+1).$

######  Theorem 4.7.

Suppose that $\mu$ is a measure on $\omega$ which is not nearly silver. Then
$\mu$ has an uncountable Maharam type.

###### Proof.

We will construct sequences $(f_{\alpha})_{\alpha<\omega_{1}}$ and
$(g_{\alpha})_{\alpha<\omega_{1}}$ of strictly increasing functions in
$\omega^{\omega}$, a sequence $(r_{\alpha})_{\alpha<\omega_{1}}$ of positive
reals and a sequence $(N_{\alpha})_{\alpha<\omega_{1}}$ of subsets of
$\omega$. First, we will show how to obtain $f_{\alpha},N_{\alpha},r_{\alpha}$
having $g_{\alpha}$ defined and then how to obtain $g_{\alpha}$ having defined
$f_{\beta}$ for all $\beta<\alpha$.

Suppose that $\alpha<\omega_{1}$ and that $g_{\alpha}$ is defined. Then let
$r_{\alpha},N_{\alpha},f_{\alpha}$ witness that the measure
$\mu\widehat{g}_{\alpha}^{-1}$ is not silver, i.e that for every
$\epsilon\in\\{-1,1\\}^{\omega}$

$\mu\big{(}\widehat{g}_{\alpha}^{-1}\big{[}\bigcup_{n}[f_{\alpha}(n),f_{\alpha}(n+1))\cap
N^{\epsilon(n)}_{\alpha}\big{]}\big{)}>r_{\alpha}.$

Let $g_{0}$ be the identity function. For $\alpha<\omega_{1}$ let
$g_{\alpha+1}=g_{\alpha}\circ f_{\alpha}$. If $\xi<\omega_{1}$ is a limit
ordinal let $g_{\xi}$ be the function such that the interval partition induced
by $g_{\xi}$ dominates all the interval partitions induced by $g_{\alpha}$ for
$\alpha<\xi$ and $g_{\xi}[\omega]$ is a pseudo-intersection of
$g_{\alpha}[\omega]$ for $\alpha<\xi$.

Passing to a subsequence, if needed, we may assume that there is $r$ such that
$r=r_{\alpha}$ for each $\alpha<\omega_{1}$.

Now, for $\alpha<\omega_{1}$ and $\epsilon\in\\{-1,1\\}^{\omega}$ let

$A^{\alpha}_{\epsilon}=\widehat{g}_{\alpha}^{-1}\big{[}\bigcup_{n}[f_{\alpha}(n),f_{\alpha}(n+1))\cap
N_{\alpha}^{\epsilon(n)}\big{]}.$

Then

$\mu((A^{\alpha}_{\epsilon}))>r$

for each $\alpha<\omega_{1}$ and $\epsilon\in\\{-1,1\\}^{\omega}$.

Notice that whenever $\alpha<\beta$ and
$\epsilon,\epsilon^{\prime}\in\\{-1,1\\}^{\omega}$ there is
$\epsilon^{\prime\prime}\in\\{-1,1\\}^{\omega}$ such that

$A^{\alpha}_{\epsilon}\triangle
A^{\beta}_{\epsilon^{\prime}}=^{*}A^{\alpha}_{\epsilon^{\prime\prime}}.$

Denote $A^{\alpha}=A^{\alpha}_{\epsilon}$ if $\epsilon(n)=1$ for each $n$.
Then for each $\alpha<\beta<\omega_{1}$ we have

$\mu(A^{\alpha}\triangle A^{\beta})>r$

and therefore the sequence $(A^{\alpha})$ proves that $\mu$ has uncountable
Maharam type. ∎

Notice that P-points can be treated as P-measures of Maharam type 2 and so
Chodounský-Guzmán theorem can be stated as: there are no P-measures of Maharam
type 2 in the Silver model. Using the above theorem we may strengthen this
fact to the following:

######  Corollary 4.8.

In the Silver model there are no P-measures of countable Maharam type.

###### Proof.

This is a consequence of Theorem 4.7, Theorem 4.3 and Theorem 3.4. ∎

In Section 6 we will show that there exist non-atomic P-measures of countable
Maharam type (under $\mathsf{CH}$).

##  5\. Rapid Filters and P-measures

Let $\mathcal{F}$ be some filter containing all co-finite sets. We will say
that $\mathcal{F}$ is rapid if for any $g\in\omega^{\omega}$ there is an
increasing $f\in\omega^{\omega}$ such that $f>g$ and
$f[\omega]\in\mathcal{F}$.

######  Theorem 5.1.

If $\mathcal{F}$ is a rapid filter then the product $\prod_{\omega}\SS$ forces
that there is no P-measure extending $\mathcal{F}$.

###### Proof.

Suppose that $\mu$ is a measure extending $\mathcal{F}$ in $V[G]$. Let
$D_{i}(s)$ be as above.

There are two cases: either for every $\varepsilon>0$ and every $n$ there is
$m>n$ such that $\mu\left(\bigcap_{n\leqslant
k<m}D_{0}(s_{k})\right)<\varepsilon$ or there are $\varepsilon>0$, $n$ such
that for any $m>n$ we have $\mu\left(\bigcap_{n\leqslant
k<m}D_{0}(s_{k})\right)>\varepsilon$.

Case 1. Using the $\omega^{\omega}$-bounding property we can choose an
increasing function $h$ from $V$ so that $n\in\omega$

$\mu\left(\bigcap_{k\in[h(n),h(n+1))}D_{0}(s_{k})\right)<\frac{1}{2^{n+2}}.$

We put

$M_{n}=\bigcap_{k\in[h(n),h(n+1))}D_{0}(s_{k}).$

Let $p=\langle p_{0},p_{1},...\rangle$ be any condition forcing all the above
and such that

$p\Vdash\forall n\;\dot{Z}\cup g(n)\supseteq\dot{M}_{n}$

for some $\dot{Z},g$ (again $g\in V$ by $\omega^{\omega}$-bounding and perhaps
extending $p$). We want to show that $\mu(Z)=1$.

We can assume (by potentially increasing $g$) that for each $n$ we have

$\forall k\in[h(n),h(n+1))\quad g(n)>\min\operatorname{Cod}(s_{k}).$

Now let $f>g$ be an increasing function such that $f[\omega]\in\mathcal{F}$.
We have $p\Vdash\dot{Z}\supseteq\dot{M}_{n}\setminus f(n)$.

Let $q=\left\langle\,q_{0},q_{1},...\,\right\rangle\leqslant p$ be a condition
such that for any $n$

$\forall k\in[h(n),h(n+1))\;q\Vdash f(n)\in\dot{D}_{0}(\dot{s}_{k}).$

This can be done because $f(n)>\min\operatorname{Cod}(p_{k})$.

Now $q$ forces that $f(n)\in\dot{M}_{n}$ and so $f(n)\in\dot{Z}$. We can
conclude that $\dot{Z}\in\mathcal{F}$ and so

$q\Vdash\dot{\mu}\dot{Z}=1\;\land\;\sum_{n}\dot{\mu}\dot{M}_{n}<\frac{1}{2}$

where $\dot{Z}$ was a name for an arbitrary pseudounion of $M_{n}$.

Case 2. Choose $N$ and $\varepsilon$ such that for every $m>N$ we have

$\mu\left(\bigcap_{k\in[N,m)}D_{0}(s_{k})\right)>\varepsilon.$

Take any $p=\langle p_{0},p_{1},...\rangle$ forcing all the above and such
that (for some $\dot{Z},g$)

$p\Vdash\forall k\geqslant N\;\dot{Z}\setminus
g(k)\subseteq\dot{D}_{0}(\dot{s}_{k}).$

Again assume that $g(k)>\min\operatorname{Cod}(p_{k})$ and let $f>g$ such that
$f[\omega\setminus N]\in\mathcal{F}$.

Now we can choose $q\leqslant p$ such that for any $k\geqslant N$

$q\Vdash f(k)\not\in\dot{D}_{0}(\dot{s}_{k}).$

Now we can see that
$q\Vdash\dot{\mu}\dot{(}Z)=0\;\land\;\lim_{m}\dot{\mu}\left(\bigcup_{k\in[N,m)}\dot{D}_{0}(\dot{s}_{k})\right)\geqslant\varepsilon.$
∎

Finally, we can arrive at the following corollary.

######  Corollary 5.2.

If $\mathcal{F}$ is a rapid filter in $V$, $V[G]$ is a model obtained by
forcing with $\prod_{\omega}\SS$ and $W\supseteq V[G]$ is a model with the
$\omega^{\omega}$-bounding property over $V[G]$ then $\mathcal{F}$ cannot be
extended to a P-measure in $W$.

##  6\. More on P-measures under CH

In this section, we enclose several remarks on measures on $\omega$.

######  Example 6.1.

Let $\mathbb{B}$ be the measure algebra (i.e.
$\mathbb{B}=\mathrm{Bor}(2^{\omega})_{/\lambda=0}$) and let $K$ be its Stone
space. Then, by Sikorski extension theorem (see e.g. [Kun76, Lemma 1.1]), $K$
can be embedded homeomorphically into $\omega^{*}$. Of course, $K$ supports
the Lebesgue measure, and this measure induces naturally a measure $\lambda$
on $\omega$.

What is interesting about the measure $\lambda$ from Example 6.1 is that it is
combinatorially far from the density measures, in particular from the atomless
measures described in Section 2. The reason lies in the Maharam type of this
measure.

Namely, the measure from Example 6.1 has the same Maharam type as the Lebesgue
measure, i.e. countable, whereas density measures have uncountable Maharam
types. The latter fact follows from [FT79, 1J], but we will give a more direct
argument here, due to Grzegorz Plebanek.

######  Theorem 6.2.

Every density measure $\mu$ has the Maharam type $\mathfrak{c}$.

###### Proof.

Let $\lambda_{\mathfrak{c}}$ be the standard Haar measure on
$2^{\mathfrak{c}}$. By [Fre03, 491G] $\lambda_{\mathfrak{c}}$ has a uniformly
distributed sequence $(x_{n})$, which in particular means that the function
$\varphi\colon\mathrm{Clop}(2^{\mathfrak{c}})\to\mathcal{P}(\omega)$ defined
by $\varphi(C)=\\{n\colon x_{n}\in C\\}$ is a Boolean homomorphism such that
$d(\varphi(A))=\lambda_{\mathfrak{c}}(A)$ for each
$A\in\mathrm{Clop}(2^{\mathfrak{c}})$. For $\alpha<\mathfrak{c}$ let
$C_{\alpha}=\\{x\in 2^{\mathfrak{c}}\colon x(\alpha)=1\\}$ and let
$A_{\alpha}=\varphi(C_{\alpha})$. Then $(A_{\alpha})$ forms a family such that
$d(A_{\alpha})=1/2$ and $d(A_{\alpha}\triangle A_{\beta})=1/4$ for
$\alpha,\beta<\mathfrak{c}$, $\alpha\neq\beta$. If $\mu$ extends $d$, then
$\mu(A_{\alpha}\triangle A_{\beta})>1/4$ for each $\alpha<\mathfrak{c}$ So,
every measure extending the asymptotic density has Maharam type
$\mathfrak{c}$. ∎

The above theorem says, informally, that the asymptotic density has ’Maharam
type‘ $\mathfrak{c}$. It follows that the measure $\lambda$ from Example 6.1
does not extend the asymptotic density.

We will show that under $CH$ we can construct measures as in Example 6.1 with
some additional properties. We start with a theorem by Kunen.

######  Theorem 6.3.

([Kun76, Theorem 1.2]). Assume $\mathsf{CH}$. Then the space
$K=\mathrm{St}(\mathbb{B})$ can be embedded into $\omega^{*}$ as a P-set. In
other words, there is a surjective Boolean homomorphism
$\varphi\colon\mathcal{P}(\omega)/fin\to\mathbb{B}$ whose kernel is a P-ideal.

Now we will show that if $\mathbb{B}$ is embedded as above, then the induced
measure $\lambda$ is a P-measure.

######  Proposition 6.4.

Assume that $\mathbb{B}$ is a Boolean algebra supporting a $\sigma$-additive
measure $\nu$. Suppose that
$\varphi\colon\mathcal{P}(\omega)/fin\to\mathbb{B}$ is a surjective Boolean
homomorphism such that $\mathrm{ker}(\varphi)$ is a P-ideal. Let $\mu$ be a
measure on $\omega$ defined by $\mu(A)=\nu(\varphi(A))$. Then $\mu$ is a
P-measure.

###### Proof.

Let $\mathcal{F}=\\{N\subseteq\omega\colon\varphi(N)=1\\}$. Notice that
$\mathcal{F}$ is a P-filter and that
$\mathcal{F}=\\{N\subseteq\omega\colon\mu(N)=1\\}$. Let $(A_{n})$ be a
decreasing sequence of subsets of $\omega$ and let $B_{n}=\varphi(A_{n})$.
Then $(B_{n})$ is a decreasing sequence of elements of $\mathbb{B}$. There is
$B\in\mathbb{B}$ such that $\lambda(B)=\lim_{n\to\infty}\lambda(B_{n})$.

Let $A\subseteq\omega$ be such that $\varphi(A)=B$. Then $\mu(A\setminus
A_{n})=0$ and so

$A^{\prime}_{n}=(A\cap A_{n})\cup A^{c}\in\mathcal{F}$

for each $n$. As $\mathcal{F}$ is a P-filter, there is $N\in\mathcal{F}$ such
that $N\subseteq^{*}A^{\prime}_{n}$ for every $n$. Then $N\cap
A\subseteq^{*}A_{n}$ for every $n$ and $\mu(N\cap
A)=\lim_{n\to\infty}\mu(A_{n})$. As $(A_{n})$ is arbitrary, $\mu$ is a
P-measure. ∎

As a corollary of the above and of Theorem 6.3 we get the following.

######  Corollary 6.5.

Under $\mathsf{CH}$ there is a P-measure of countable Maharam type.

We will show that we can strengthen Kunen’s construction to obtain a measure
that is not nearly Dirac. To do it we have to take a closer look at the proof
of Theorem 6.3. Let $\mathcal{A}\subseteq P(\omega)$ be a family closed under
finite modifications. For a homomorphism $\psi\colon\mathcal{A}\to\mathbb{B}$
define $\psi^{*}\colon P(\omega)\to\mathbb{B}$ and $\psi_{*}\colon
P(\omega)\to\mathbb{B}$ by

$\psi^{*}(X)=\bigwedge\\{\psi(A)\colon A\in\mathbb{A},X\subseteq A\\}$

and

$\psi_{*}(X)=\bigvee\\{\psi(A)\colon A\in\mathbb{A},A\subseteq X\\}.$

The main tool in the proof of Theorem 6.3 is the following ingredient of the
proof of Sikorski’s theorem. By $\mathbb{A}(X)$ we will denote the Boolean
algebra generated by $\mathbb{A}$ and $X$.

######  Lemma 6.6.

Let $\mathbb{A}\subseteq P(\omega)$ be a countable Boolean algebra (closed
under finite modifications) and let $\psi\colon\mathbb{A}\to\mathbb{B}$ be a
homomorphism. If $X\subseteq\omega$ and $B\in\mathbb{B}$ is such that
$\psi_{*}(X)\subseteq B\subseteq\psi^{*}(X)$, then $\psi$ can be extended to a
homomorphism $\psi^{\prime}\colon\mathbb{A}(X)\to\mathbb{B}$ in such a way
that $\psi^{\prime}(X)=B$.

Theorem 6.3 can be proved by transfinite induction, by taking care of the
following conditions:

  * (1)

for every $B\in\mathbb{B}$ there is $X\subseteq\omega$ such that
$\varphi(X)=B$.

  * (2)

for every countable $\mathcal{A}\subseteq\mathcal{P}(\omega)$, such that
$\lambda(\varphi(A_{\alpha}))=1$ for each $\alpha<\beta$, there is a pseudo-
intersection $X\subseteq\omega$ of $\mathcal{A}$ such that
$\lambda(\varphi(X))=1$.

######  Proposition 6.7.

Assume $\mathsf{CH}$. There is a P-measure that is not nearly Dirac.

###### Proof.

We proceed as explained above, adding one more condition on $\varphi$.

  * (3)

for every finite-to-1 function $f\colon\omega\to\omega$ there is
$N\subseteq\omega$ such that

$\lambda(\varphi(f^{-1}[N]))=\frac{1}{2}.$

It is immediate that (3) implies that the measure $\mu$ defined by
$\mu(X)=\lambda(\varphi(X))$ is not nearly Dirac.

To achieve (3) suppose that $\mathbb{A}$ is countable and a function
$f\colon\omega\to\omega$ is finite-to-1. By induction we may find
$N\subseteq\omega$ such that

$|A\cap f^{-1}[N]|=|A\setminus f^{-1}[N]|=\omega$

for every infinite $A\in\mathbb{A}$.

It means that $\varphi_{*}(A\cap f^{-1}[N])=0$ and $\varphi^{*}(A\cap
f^{-1}[N])=1$ and we can use Lemma 6.6. ∎

##  7\. A model with P-measures and without P-points

In this section, we will prove the following theorem. Denote

$\mathbb{B}_{\kappa}=\mathrm{Bor}(2^{\kappa})/_{\lambda_{\kappa}=0},$

where $\lambda_{\kappa}$ is the standard Haar measure on $2^{\kappa}$. Recall
that $\mathbb{B}_{\kappa}$ is the complete Boolean algebra adding $\kappa$
random reals.

######  Theorem 7.1.

It is relatively consistent with $\mathsf{ZFC}$ that there is a P-measure,
there is no P-point and $2^{\aleph_{0}}$ is arbitrarily large.

The proof will rely on two results concerning the random forcing. The first
one is proved in [BNS23, Theorem 4.15].

######  Theorem 7.2 (Borodulin-Nadzieja, Sobota).

If there is a P-point in the ground model, then after forcing with
$\mathbb{B}_{\kappa}$, for any $\kappa$, there is a P-measure.

The following is the main ingredient of the proof that there are no selective
ultrafilters in the random model, see [Kun76, Theorem 5.1] and [BNS23,
Corollary 5.7] for an alternative proof.

######  Theorem 7.3 (Kunen).

No selective ultrafilter from the ground model can be extended to a P-point
after forcing with $\mathbb{B}_{\kappa}$ for $\kappa\geq\omega$.

Having these two theorems and Problem 0.1 in mind, our attention should be
attracted by the following theorem, see [She17, Theorem 5.13 from Chapter VI,
and Theorem 4.1 from Chapter XVIII].

######  Theorem 7.4 (Shelah).

It is consistent that there is a P-point and all P-points are selective.

Let $V$ be a model witnessing the situation as in the above theorem. If we
force over $V$ with a random forcing, then by Theorem 7.3 no ultrafilter from
$V$ can be extended to a P-point and, by Theorem 7.2 there is a P-measure in
the extension. It sounds like we are close to proving Theorem 7.1. However, a
priori it is possible that by forcing with $\mathbb{B}$ we add some new
P-points (we will come back to this issue in the remarks after Question 7.21).
So, to obtain a model in which there are P-measures but no P-points, we have
to work a little bit more.

First, we need a strengthening of Theorem 7.3. We will show that not only
selective ultrafilters cannot be extended to P-points (in the random
extensions) but in fact all ultrafilters that are nearly coherent with
selective ultrafilters.

######  Definition 7.5 (A. Blass, see [Bla86a]).

Let $\mathcal{F}_{0}$ and $\mathcal{F}_{1}$ be filters on $\omega$. We say
that $\mathcal{F}_{0}$ and $\mathcal{F}_{1}$ are nearly coherent if there is a
finite to one function $f\colon\omega\to\omega$ such that
$f[\mathcal{F}_{0}]\cup f[\mathcal{F}_{1}]$ generates a filter.

Note that if $\mathcal{U}$ is an ultrafilter, then $\mathcal{F}$,
$\mathcal{U}$ are nearly coherent if and only if there is a finite to one
function $f\colon\omega\to\omega$ such that $f[\mathcal{F}]\subseteq
f[\mathcal{U}]$.

For a $\mathbb{B}_{\kappa}$-name for an ultrafilter $\dot{\mathcal{V}}$ and
$p\in\mathbb{B}_{\kappa}$ we denote

$\dot{\mathcal{V}}[p]=\\{A\subseteq\omega\colon p\Vdash
A\in\dot{\mathcal{V}}\\}.$

######  Theorem 7.6.

Let $\mathcal{U}$ be a selective ultrafilter and $\dot{\mathcal{V}}$ be a
$\mathbb{B}_{\kappa}$-name such that
$\mathbb{B}_{\kappa}\Vdash\dot{\mathcal{V}}\text{ is a P-point}$. Then for all
$p\in\mathbb{B}_{\kappa}$, the filters $\mathcal{U}$ and
$\dot{\mathcal{V}}[p]$ are not nearly coherent.

###### Proof.

First note that if $\mathcal{F}$ is a filter which is Rudin-Blass above
$\mathcal{U}$, which is witnessed by $f\colon\omega\to\omega$, then
$\mathcal{F}$ can not be extended to a P-point after forcing with
$\mathbb{B}_{\kappa}$. Indeed, suppose that $\dot{\mathcal{G}}$ is a P-point
extending $\mathcal{F}$. Then $\mathcal{U}\subseteq f[\dot{\mathcal{G}}]$, in
other words $\mathcal{U}$ can be extended to $f[\dot{\mathcal{G}}]$ which is a
P-point. This is impossible by Theorem 7.3.

Assume now that there is $p_{0}\in\mathbb{B}_{\kappa}$ such that $\mathcal{U}$
and $\dot{\mathcal{V}}[p_{0}]$ are nearly coherent, and that this is witnessed
by $h\colon\omega\to\omega$. It means that
$h[\dot{\mathcal{V}}[p_{0}]]\subseteq h[\mathcal{U}]$. By the previous
paragraph and the fact that $h[\mathcal{U}]$ is also selective, we have that
$h[\mathcal{U}]\nleq_{RB}\dot{\mathcal{V}}[p]$ for every
$p\in\mathbb{B}_{\kappa}$. So, for every $p\leq p_{0}$, there is $A_{p}\in
h[\mathcal{U}]$ such that $h^{-1}[A_{p}]\notin\dot{\mathcal{V}}[p]$, which
implies that there is $q\leq p$ such that $q\Vdash
A_{p}\notin\dot{\mathcal{V}}$. Therefore, the set

$D=\\{p\leq p_{0}\colon\exists A\in h[\mathcal{U}]\ p\Vdash
h^{-1}[A]\notin\dot{\mathcal{V}}\\}$

is dense below $p_{0}$. Let $\mathcal{A}\subseteq D$ be a maximal antichain
below $p_{0}$. As $\mathbb{B}_{\kappa}$ is ccc, $\mathcal{A}$ is necessarily
countable, so there is $X\in h[\mathcal{U}]$ such that $X\subseteq^{*}A_{p}$,
for all $p\in\mathcal{A}$. It follows that $p\Vdash
h^{-1}[X]\notin\dot{\mathcal{V}}$ for all $p\in\mathcal{A}$ (otherwise, for
some $p\in\mathcal{A}$, there exist $q\leq p$ such that $q\Vdash
h^{-1}[A_{p}]\in\dot{\mathcal{V}}$, which is a contradiction). Since
$\mathcal{A}$ is a maximal antichain below $p_{0}$, it follows that
$p_{0}\Vdash h^{-1}[X]\notin\dot{\mathcal{V}}$, which implies that
$p_{0}\Vdash\omega\setminus h^{-1}[X]\in\mathcal{\dot{V}}$. Therefore,
$\omega\setminus X\in h[\dot{\mathcal{V}}[p_{0}]]$. Thus, we have $X\in
h[\mathcal{U}]$ and $\omega\setminus X\in h[\dot{\mathcal{V}}[p_{0}]]$, which
contradicts our assumption that $h[\dot{\mathcal{V}}[p_{0}]]\subseteq
h[\mathcal{U}]$. ∎

######  Remark 7.7.

Having Theorem 7.6 and Theorem 7.3, we may face the following temptation.
Recall that the principle of Near Coherence of Filters (NCF) says that any two
different filters are nearly coherent. It is consistent (e.g. it is implied by
$\mathfrak{u}<\mathfrak{g}$, see [Bla10]). So, if we have NCF with the
existence of selective ultrafilter, then by Theorem 7.6 no ultrafilter would
be extendible to a P-point in the random extensions (and by Theorem 7.2 there
are P-measures in the random extensions). So, by forcing with
$\mathbb{B}_{\omega_{2}}$ over such a model, we would obtain a model with
P-measures but without P-points. Unfortunately, this approach does not work:
NCF implies that there are no selective ultrafilters. Indeed, suppose that NCF
holds and that $\mathcal{U}$ is a selective ultrafilter. NCF implies that
$\mathfrak{u}<\mathfrak{d}$ (see [Bla86b]). So, there is an ultrafilter
$\mathcal{V}$ generated by less than $\mathfrak{d}$ sets. By NCF we have
$h[\mathcal{V}]\leq_{RB}h[\mathcal{U}]$. But $h[\mathcal{U}]$ is selective,
and so $h[\mathcal{V}]$ is also selective which is impossible, see [Bla10,
Remark 9.24].

Now, we will prove Theorem 7.4 in a slightly stronger form in a series of
lemmas. We will extensively use filter games, both the classical one
introduced by Laflamme in [Laf96] and its modification considered by Eisworth
in [Eis01].

######  Definition 7.8 (P-filter game).

Let $\mathcal{F}$ be a filter on $\omega$. Consider the game
$\mathcal{G}(\mathcal{F})$ between Adam and Eve defined in the following way:

  * •

At round $n$ Adam plays a set $A_{n}\in\mathcal{F}$ and Eve responds with a
finite set $F_{n}\subseteq A_{n}$.

Eve wins if $\bigcup_{n}F_{n}\in\mathcal{F}$.

Using the above game gives us a handy characterization of the property of
being a P-filter in the realm of non-meager filters, see [Laf96, Theorem 2.15]

######  Theorem 7.9 (Laflamme).

Suppose that $\mathcal{F}$ is non-meager. Then $\mathcal{F}$ is a P-filter if
and only if Adam does not have a winning strategy in
$\mathcal{G}(\mathcal{F})$.

We will also need a game defined in the same spirit by Eisworth in [Eis01,
Definition 3.3].

######  Definition 7.10.

Let $\mathcal{F}_{0},\mathcal{F}_{1}$ be filters on $\omega$. The game
$\mathcal{G}(\mathcal{F}_{0},\mathcal{F}_{1})$ between Adam and Eve is defined
as follows: a complete stage of the game consists of two rounds:

  * •

At round $2n$, Adam plays sets $A_{n}\in\mathcal{F}_{0}$, and Eve answers with
$a_{n}\in A_{n}$.

  * •

At round $2n+1$, Adam plays sets $B_{n}\in\mathcal{F}_{1}$, and Eve answers
with a non-empty $F_{n}\in[B_{n}]^{<\omega}$.

Eve wins if $\\{a_{n}\colon n\in\omega\\}\in\mathcal{F}_{0}$ and
$\bigcup_{n}F_{n}\in\mathcal{F}_{1}$.

For technical reasons, we will ask Eve to play different numbers $a_{n}$ and
pairwise disjoint sets $F_{n}$. Taking into account her aim in the game, she
should not complain about this additional rule.

Our first step is to prove that under suitable assumptions, Adam has no
winning strategy in the previous game. We need the following lemma. The reader
can find similar theorems in [Eis01, Corollary 2.3] and [Cho11].

######  Lemma 7.11 (Eisworth).

Let $\mathcal{F}_{0},\mathcal{F}_{1}$ be filters which are not nearly
coherent. Let $\mathcal{I}=\\{I_{n}\colon n\in\omega\\}$ be an interval
partition of $\omega$. Then there is an interval partition
$\mathcal{J}=\\{J_{n}\colon n\in\omega\\}$ such that each $J_{n}$ is a union
of elements of $\mathcal{I}$, and

$\bigcup_{k\in\omega}J_{4k+1}\in\mathcal{F}_{0}\text{ and
}\bigcup_{k\in\omega}J_{4k+3}\in\mathcal{F}_{1}$

The following lemma is essentially [Eis01, Theorem 1].

######  Lemma 7.12.

Let $\mathcal{U}$ be a selective ultrafilter, let $\mathcal{F}$ be a P-filter,
and suppose that they are not nearly coherent. Then Adam has no winning
strategy in the game $\mathcal{G}(\mathcal{U},\mathcal{F})$.

###### Proof.

Let $\Sigma$ be a strategy for Adam. We will produce a run of the game in
which he uses $\Sigma$ but nevertheless, Eve wins the game. We think of
$\Sigma$ in a natural way as a tree such that each $\sigma\in\Sigma$ is a
sequence of the form $(A_{0},a_{0},B_{0},F_{0},A_{1},\dots,A_{n})$ or
$(A_{0},a_{0},B_{0},F_{0},A_{1},\dots,B_{n})$, where $A_{i}$, $a_{i}$,
$B_{i}$, $F_{i}$ are as described in Definition 7.10.

Note that $\Sigma$ is a countable tree, so there is $A_{\omega}\in\mathcal{U}$
which is a pseudointersection of all the elements of $\mathcal{U}$ appearing
in $\Sigma$ and, similarly, there is $B_{\omega}\in\mathcal{F}$ which is a
pseudointersection of all the elements of $\mathcal{F}$ which appeared in
$\Sigma$. We can assume that $A_{\omega}\subseteq A_{0}$, the first move of
Adam.

Fix $n\in\omega$. Notice that there are finitely many $\sigma\in\Sigma$ such
that every Eve’s move in $\sigma$ is a subset of $n$; we say that such
$\sigma$ has rank $\leq n$. Denote by $\Sigma_{n}$ all the elements of
$\Sigma$ having rank $\leq n$. Furthermore, let $E_{n}$ denote the collection
of sets $A\in\mathcal{U}$ such that there is a $\sigma\in\Sigma_{n}$ in which
the set $A$ appears. Similarly, denote by $O_{n}$ the collection of sets
$B\in\mathcal{F}$ such that there is $\sigma\in\Sigma_{n}$ in which the set
$B$ appears. Notice that both $E_{n}$ and $O_{n}$ are finite.

Now we define a sequence of natural numbers $\\{k_{n}\colon n\in\omega\\}$ as
follows:

  1. (1)

$k_{0}=0$ and $k_{1}$ is such that $A_{0}\cap k_{1}\neq\emptyset$.

  2. (2)

Assume $k_{i}$ has been defined. Then choose $k_{i+1}$ such that
$A_{\omega}\cap[k_{i},k_{i+1})\neq\emptyset$,
$B_{\omega}\cap[k_{i},k_{i+1})\neq\emptyset$ and for all $A\in E_{k_{i}}$,
$A_{\omega}\setminus k_{i+1}\subseteq A$, and for all $B\in O_{k_{i}}$,
$B_{\omega}\setminus k_{i+1}\subseteq B$.

Having defined the sequence, consider the interval partition given by
$I_{n}=[k_{n},k_{n+1})$. Apply Lemma 7.11 to get a partition $\\{J_{n}\colon
n\in\omega\\}$ such that each $J_{n}$ is union of finitely many consecutive
intervals $I_{k}$, and such that

$\bigcup_{k\in\omega}J_{4k+1}\in\mathcal{U}\text{ and
}\bigcup_{k\in\omega}J_{4k+3}\in\mathcal{F}.$

We may assume that $k_{1}<\min(J_{1})$, and so $k_{2}\leq\min(J_{1})$ (the
minimum of each interval $J_{n}$ is of the form $k_{l}$). Since $\mathcal{U}$
is a selective ultrafilter, and $A_{\omega}\cap I_{n}\neq\emptyset$ for each
$n$ we can find $A=\\{a_{n}\colon n\in\omega\\}\in\mathcal{U}$, $A\subseteq
A_{\omega}$ such that $A\cap J_{4n+1}=\\{a_{n}\\}$ for all $n\in\omega$. Let
$B=B_{\omega}\cap\bigcup_{n\in\omega}J_{4n+3}$. We claim that there is a run
of the game in which Adam follows $\Sigma$ and Eve constructs the sets $A$ and
$B$ along. Clearly, this suffices to prove that $\Sigma$ is not a winning
strategy, as $A\in\mathcal{U}$ and $B\in\mathcal{F}$. Eve should play as
follows:

  1. (1)

After Adam has played $A_{0}$, Eve plays $a_{0}$. Notice that it is a legal
move (i.e. $a_{0}\in A_{0}$) as $a_{0}\in A_{\omega}$ and $A_{\omega}\subseteq
A_{0}$.

  2. (2)

Then Adam answers with $B_{0}$. Note that $\sigma=(A_{0},a_{0},B_{0})$ has
rank $\leq\max(J_{1})+1=\min(J_{2})$, so $B_{0}\in O_{\min(J_{2})}$, which
implies that $B_{\omega}\setminus\min(J_{3})\subseteq B_{0}$, so
$B_{\omega}\cap J_{3}\subseteq B_{0}$. Then Eve can play $F_{0}=B_{\omega}\cap
J_{3}=B\cap J_{3}$ which is non-empty by the choice of the sequence $(k_{n})$.

  3. (3)

Let $A_{1}$ be the answer of Adam. The sequence
$(A_{0},a_{0},B_{0},F_{0},A_{1})$ has rank $\leq\min(J_{4})$, so $A_{1}\in
E_{\min(J_{4})}$, which implies that $A_{\omega}\setminus\min(J_{5})\subseteq
A_{1}$, and thus, $a_{1}\in A_{1}$, which makes it a legal (and recommended)
move for Eve.

  4. (4)

Eve continues playing $a_{n}$ and $F_{n}=B_{\omega}\cap J_{4n+3}$ accordingly.

Note that $\bigcup_{n\in\omega}F_{n}=B$, and clearly $A=\\{a_{n}\colon
n\in\omega\\}$. ∎

The following forcing notion was introduced by Shelah to prove Theorem 7.4.

######  Definition 7.13.

For each natural number $n$, let $S_{n}=\Pi_{k\leq n}2^{n}$ (note that here by
$2^{n}$ we mean the set of sequences of bits having length $n$; to avoid
misunderstandings in what follows we will denote the elements of $2^{n}$ by
Roman letters $s,t,\dots$ and the elements of $S_{n}$ by Greek letters
$\sigma,\tau,..$), and $S=\bigcup_{n\in\omega}S_{n}$. A tree $T\subseteq S$ is
a family closed under initial segments. For a tree $T\subseteq S$, $\tau\in
T$, and $n\in\omega$, define

  * •

$T^{[n]}=\\{\sigma\in T\colon|\sigma|=n\\}$,

  * •

$succ_{T}(\sigma)=\\{s\in 2^{|\sigma|}\colon\sigma^{\frown}s\in T\\}$,

  * •

$T\upharpoonright\tau=\\{\sigma\in
T\colon\tau\subseteq\sigma\vee\sigma\subseteq\tau\\}$.

Let $\mathcal{F}$ be a filter. For a tree $T\subseteq S$ and $k\in\omega$,
define

$spt_{k}(T)=\\{n\in\omega\colon\forall\tau\in T^{[n]}\ \forall s\in 2^{k}\
\exists t\in succ_{T}(\tau)\ s\subseteq t\\}.$

The forcing $SP^{*}(\mathcal{F})$ is defined as the collection of all trees
$T\subseteq S$ such that for all $k\in\omega$, $spt_{k}(T)\in\mathcal{F}$. The
order is given by set inclusion.

Notice that if $T\in SP^{*}(\mathcal{F})$ and $\tau\in T$, then
$T\upharpoonright\tau\in SP^{*}(\mathcal{F})$.

######  Remark 7.14.

The conditions of $SP^{*}(\mathcal{F})$ are slightly complicated so maybe this
is a good moment to see how the forcing actually ’works’. Let $G$ be
$SP^{*}(\mathcal{F})$-generic, and define $g\in[S]$ to be such that
$\bigcup\bigcap G=\\{g\\}$. We can see it as a sequence of reals: for each
natural number define $\dot{r}_{n}\colon\omega\setminus(n+1)\to 2$ as

$\dot{r}_{n}(k)=1\mbox{ if and only if }\dot{g}(k)(n)=1.$

Actually, we will rather treat it as a sequence of subsets of $\omega$:

$\dot{R}_{n}=\\{k\in\omega\colon\dot{r}_{n}(k)=1\\}.$

A simple density argument shows that for all $n\in\omega$ we have
$\dot{R}_{n},\omega\setminus\dot{R}_{n}\in\mathcal{F}^{+}$. The family
$\\{\dot{R}_{n}\colon n\in\omega\\}$ will be used in a crucial step of the
proof.

######  Proposition 7.15 (Shelah, see [She17]).

If $\mathcal{F}$ is a non-meager P-filter, the forcing $SP^{*}(\mathcal{F})$
is proper and $\omega^{\omega}$-bounding.

###### Proof.

We only prove that $SP^{*}(\mathcal{F})$ is $\omega^{\omega}$-bounding, the
proof for properness follows similar lines, just choosing an elementary
submodel as usual and taking care that each step of the construction lives
inside the submodel. So let $p\in SP^{*}(\mathcal{F})$ be a condition and let
$\dot{f}$ be a name for a function from $\omega$ to $\omega$. We are going to
find a sequence of finite sets $\\{W_{n}\colon n\in\omega\\}$ and $q\leq p$
such that $q\Vdash\dot{f}(n)\in W_{n}$ for each $n\in\omega$. To do so, we
give a strategy to Adam in the P-filter game $\mathcal{G}(\mathcal{F})$.

  * •

Adam starts by extending $p$ to a condition $p_{0}\leq p$ which decides the
value of $\dot{f}(0)$, and plays the set $spt_{1}(p_{0})$.

  * •

Eve answers with a finite set $F_{0}\subseteq spt_{1}(p_{0})$. Then Adam
chooses a natural number $n_{0}>\max(F_{0})$. Define $W_{0}=\\{l\\}$, where
$l$ is such that $p_{0}\Vdash\dot{f}(0)=l$.

  * •

At round $k$, suppose Eve has previously played a set $F_{k-1}$, Adam has
chosen $n_{k-1}>\max(F_{k-1})$ and defined conditions $\\{p_{i}\colon i<k\\}$
and finite set $\\{W_{i}\colon i<k\\}$ such that $p_{i}\Vdash\dot{f}(i)\in
W_{i}$. Now, for each node $\sigma\in p_{k-1}^{[n_{k-1}]}$, let $q_{s}\leq
p_{k-1}\upharpoonright\sigma$ be a condition which decides the value of
$\dot{f}(k)$, and let Adam play the set
$\bigcap\\{spt_{k+1}(q_{\sigma})\colon{\sigma\in p_{k-1}^{[n_{k-1}]}}\\}$.

  * •

Eve answers with a finite set $F_{k}$. Then Adam chooses a natural number
$n_{k}>\max(F_{k})$, defines $p_{k}=\bigcup\\{q_{\sigma}\colon{\sigma\in
p_{k-1}^{[n_{k-1}]}}\\}$, and

$W_{k}=\\{l\in\omega\colon\exists\sigma\in p_{k-1}^{[n_{k-1}]}\
q_{\sigma}\Vdash\dot{f}(k)=l\\}.$

Notice that $F_{k}\subseteq spt_{k}(p_{k})$ and $p_{k}\Vdash\dot{f}(k)\in
W_{k}$.

This is not a winning strategy for Adam, because of Theorem 7.9. So, there is
a run of the game in which Eve wins. Let $\\{p_{n}\colon n\in\omega\\}$ and
$\\{W_{n}\colon n\in\omega\\}$ be the sequences constructed along such a run.
Define $p_{\omega}=\bigcap_{n\in\omega}p_{n}$. Notice that
$p_{\omega}^{[i]}=p_{n}^{[i]}$ for every $i\leq n$. So, as $n_{k}>\max F_{k}$
for each $k$, we have that $\bigcup_{l\geq k}F_{l}\subseteq
spt_{k}(p_{\omega})$ for each $k$. Hence, $p_{\omega}$ is a condition. Also,
since $p_{\omega}\leq p_{k}$ for each $k$, we have
$p_{\omega}\Vdash\dot{f}(k)\in W_{k}$ and we are done.

∎

The following is essentially a part of the proof of Shelah’s theorem on the
consistency of the existence of only one selective ultrafilter up to
permutation, we include the proof for completeness.

######  Lemma 7.16 (S. Shelah, see [She17]).

Let $\mathcal{U}$ be a selective ultrafilter and $\mathcal{F}$ be a non-meager
P-filter such that $\mathcal{U}$ and $\mathcal{F}$ are not nearly coherent.
Let $\dot{\mathbb{Q}}$ be a $SP^{*}(\mathcal{F})$-name for a proper
$\omega^{\omega}$-bounding forcing. Then:

  1. (1)

$SP^{*}(\mathcal{F})$ preserves $\mathcal{U}$ as a selective ultrafilter.

  2. (2)

$SP^{*}(\mathcal{F})\ast\dot{\mathbb{Q}}$ forces that $\mathcal{F}$ can not be
extended to a P-point.

###### Proof.

We first prove (1). Let $\dot{x}$ be a name for a subset of $\omega$ and let
$p\in SP^{*}(\mathcal{F})$. We are going to show that there is $q\leq p$ such
that

  1. (i)

$q\Vdash\dot{x}$ is disjoint with an element of $\mathcal{U}$ or

  2. (ii)

$q\Vdash\dot{x}$ has a subset belonging to $\mathcal{U}$.

It will mean that $\mathcal{U}$ generates an ultrafilter in the forcing
extension, and we will be done.

For $q\in SP^{*}(\mathcal{F})$ define $\dot{x}[q]=\\{n\in\omega\colon\exists
r\leq q\ r\Vdash n\in\dot{x}\\}$. If there is a condition $q\leq p$ such that
$\dot{x}[q]\notin\mathcal{U}$, then we are in the situation (i). So we may
assume that for all $q\leq p$ $\dot{x}[q]\in\mathcal{U}$. We give a strategy
for Adam in the game $\mathcal{G}(\mathcal{U},\mathcal{F})$. On the side, Adam
will construct a decreasing sequence of conditions $\\{p_{n}\colon
n\in\omega\\}$ and a sequence of natural numbers $\\{a_{n}\colon
n\in\omega\\}=A\in\mathcal{U}$, and after a run of the game in which Eve wins,
there will be a lower bound $p_{\omega}$ of $\\{p_{n}\colon n\in\omega\\}$ and
$p_{\omega}\Vdash A\subseteq\dot{x}$.

We dictate to Adam the following strategy.

  * •

Adam starts by playing $\dot{x}[p]$. Then Eve chooses $a_{0}\in\dot{x}[p]$.
Now Adam extends $p$ to a condition $p_{0}\Vdash a_{0}\in\dot{x}$, and plays
the set $spt_{1}(p_{0})$. Then Eve answers with a finite set $F_{0}\subseteq
spt_{1}(p_{0})$ and Adam chooses a natural number $n_{0}>\max(F_{0})$.

  * •

At stage $k$, after Eve has played $F_{k-1}$, Adam has chosen $n_{k-1}$ and
has defined $p_{k-1}$, we proceed as follows. Adam plays the set

$A_{k}=\bigcap\\{\dot{x}[p_{k-1}\upharpoonright\sigma]\colon\sigma\in
p_{k-1}^{[n_{k-1}]}\\}.$

Then Eve answers with $a_{k}\in A_{k}$. For each $\sigma\in
p_{k-1}^{[n_{k-1}]}$, Adam extends the condition
$p_{k-1}\upharpoonright\sigma$ to a condition $q_{\sigma}$ which forces
$a_{k}\in\dot{x}$. Then he plays the set

$B_{k}=\bigcap\\{spt_{k+1}(q_{\sigma})\colon\sigma\in p_{k-1}^{[n_{k-1}]}\\}$

Then Eve answers with a finite set $F_{k}\subseteq B_{k}$, and Adam chooses
$n_{k}>\max(F_{k})$. Finally, he defines
$p_{k}=\bigcup\\{q_{\sigma}\colon\sigma\in p_{k-1}^{[n_{k-1}]}\\}$. Note that
$p_{k}\Vdash a_{k}\in\dot{x}$.

By Theorem 7.9 the above strategy is not winning. Let $A=\\{a_{k}\colon
k\in\omega\\}$, $\\{F_{k}\colon k\in\omega\\}$ and $\\{p_{k}\colon
k\in\omega\\}$ be the sequences constructed along a run of the game in which
Eve wins. Define $p_{\omega}=\bigcap_{k\in\omega}p_{n}$. Similarly as in the
proof of Proposition 7.15 we have that $p_{\omega}\in SP^{*}(\mathcal{F})$. We
also have that $p_{\omega}\Vdash A\subseteq\dot{x}$, and $A\in\mathcal{U}$.
So, we are in the situation (ii), for $q=p_{\omega}$. This finishes the proof
of (1).

We now prove (2). We will use the terminology of Remark 7.14. Let
$\dot{\mathcal{V}}$ be a $SP^{*}(\mathcal{F})\ast\dot{\mathbb{Q}}$-name for an
ultrafilter extending $\mathcal{F}$, and define $\dot{h}\colon\omega\to 2$ as
$\dot{h}(n)=1$ if and only if $\dot{R}_{n}\in\dot{\mathcal{V}}$. So,

$\dot{R}_{n}^{\dot{h}(n)}\in\dot{\mathcal{V}}$

for every $n$.

Fix a condition $(p,\dot{q})\in SP^{*}(\mathcal{F})\ast\dot{\mathbb{Q}}$. We
have two cases: (a) there is $(p_{0},\dot{q}_{0})\leq(p_{0},\dot{q_{0}})$
which forces $\dot{h}$ to be eventually constant; (b) $(p,\dot{q})$ forces
that $\dot{h}$ is not eventually constant. We only prove (b) since the
argument for (a) is similar and slightly simpler.

Since $\dot{\mathbb{Q}}$ is forced to be $\omega^{\omega}$-bounding, also
$SP^{*}(\mathcal{F})\ast\dot{\mathbb{Q}}$ is $\omega^{\omega}$-bounding, and
so there is a condition $(p_{0},\dot{r}_{0})\leq(p,\dot{q})$ and a function
$f\in\omega^{\omega}$ such that for all $n\in\omega$, $n+1<f(n)$ and

$(p_{0},\dot{r}_{0})\Vdash\exists i,j\in[f(n),f(n+1))\
\dot{h}(i)\neq\dot{h}(j).$

For each $n\in\omega$ define

$\dot{W}_{n}=\bigcap_{j\in[f(n),f(n+1))}\dot{R}_{j}^{\dot{h}(j)}.$

Clearly, $\dot{W}_{n}$ is forced to be in $\dot{\mathcal{V}}$. We will show
that $\\{\dot{W}_{n}\colon n\in\omega\\}$ has no pseudo-intersection in
$\dot{\mathcal{V}}$. Indeed, let $\dot{Z}$ be a name for a pseudointersection
of $\\{\dot{W}_{n}\colon n\in\omega\\}$. We may assume that there is
$g\in\omega^{\omega}$ in the ground model such that $(p_{0},\dot{r}_{0})$
forces $\dot{Z}\setminus g(n)\subseteq\dot{W}_{n}$ for each $n$ (extending and
renaming $(p_{0},\dot{r}_{0})$ if needed).

Again, we will play the P-filter game. We suggest the following strategy to
Adam. Let $q_{0}=p_{0}$ and let $k_{0}>g(0)$. At stage $n$, after defining
$q_{n}$ and $k_{n}>g(n)$, Adam plays the set
$A_{n}=spt_{f(n+1)}(q_{n})\setminus k_{n}$. Eve responds with a finite set
$F_{n}=\\{a_{0}^{n}<\ldots<a_{m_{n}}^{n}\\}\subseteq A_{n}$. We construct a
sequence of decreasing conditions $\\{p_{j}^{n}\colon j\leq m_{n}\\}$ as
follows. For each $\sigma\in q_{n}^{[a_{0}^{n}]}$ and $s\in 2^{n+1}$ using the
fact that $a_{0}^{n}\in spt_{f(n+1)}(q_{n})$ we can find $r_{s}^{\sigma}\in
succ_{q_{n}}(\sigma)$ such that

  * •

$s\subseteq r_{s}^{\sigma}$, and

  * •

for all $l\in[f(n),f(n+1))$ we have $r_{s}^{\sigma}(l)=0$.

Define

$p_{0}^{n}=\bigcup\\{q_{n}\upharpoonright(\sigma^{\frown}r_{s}^{\sigma})\colon\sigma\in
q_{n}^{[a_{0}^{n}]}\land s\in 2^{n+1}\\}.$

Notice that for each $j\in[f(n),f(n+1))$

$(p_{0}^{n},\dot{r}_{0})\Vdash a^{n}_{0}\notin\dot{R}_{j}$

and since there is $i\in[f(n),f(n+1))$ such that
$(p_{0},\dot{r}_{0})\Vdash\dot{h}(i)=0$ we have also

$(p_{0}^{n},\dot{r}_{0})\Vdash a^{n}_{0}\notin\dot{W}_{n}.$

Now, having defined $p_{j}^{n}$, for each
$\sigma\in(p_{j}^{n})^{[a_{j+1}^{n}]}$ and $s\in 2^{n+1}$, let
$r_{s}^{\sigma}\in succ_{p^{n}_{j}}(\sigma)$ be such that

  * •

$s\subseteq r_{s}^{\sigma}$, and

  * •

for all $l\in[f(n),f(n+1))$, $r_{s}^{\sigma}(l)=0$.

Define

$p_{j+1}^{n}=\bigcup\\{p_{j}^{n}\upharpoonright(\sigma^{\frown}r_{s}^{\sigma})\colon\sigma\in(p^{n}_{j})^{[a_{j+1}^{n}]}\land
s\in 2^{n+1}\\}.$

Finally, let $q_{n+1}=p_{m_{n}}^{n}$, choose
$k_{n+1}>\max\\{g(n+1),\max(F_{n})\\}$, and notice that

$(q_{n+1},\dot{r}_{0})\Vdash F_{n}\cap\dot{W}_{n}=\emptyset.$

According to Theorem 7.9 there is a run of the game in which the above
strategy fails. Let $\\{q_{n}\colon n\in\omega\\}$ and $\\{F_{n}\colon
n\in\omega\\}$ be the sequences constructed along such a run. Then
$X=\bigcup_{n\in\omega}F_{n}\in\mathcal{F}$. Define
$q_{\omega}=\bigcap_{n\in\omega}q_{n}$ and notice that, as in the proof of
Theorem 7.15, $q_{\omega}$ is a condition. Then for each $n$

$(q_{\omega},\dot{r}_{0})\Vdash F_{n}\cap\dot{W}_{n}=\emptyset.$

But, as $\min F_{n}>g(n)$ and $(p_{0},\dot{r}_{0})\Vdash\dot{Z}\setminus
g(n)\subseteq\dot{W}_{n}$ it means that

$(q_{\omega},\dot{r}_{0})\Vdash F_{n}\cap\dot{Z}=\emptyset.$

So, $(q_{\omega},\dot{r}_{0})$ forces $\dot{Z}$ to be outside
$\dot{\mathcal{V}}$. ∎

Before the proof of the main theorem, we need one more fact: if we force with
a random forcing over a model $V$, then every $\mathbb{B}_{\kappa}$-name for a
non-principal ultrafilter on $\omega$ contains a substantial portion of a
ground model ultrafilter.

######  Proposition 7.17.

Suppose that $\dot{\mathcal{V}}$ is a $\mathbb{B}_{\kappa}$-name for an
ultrafilter on $\omega$. Let $p\in\mathbb{B}_{\kappa}$. Then, there is a
measure $\mu$ on $\omega$, vanishing on points, such that

$\\{F\subseteq\omega\colon\mu(F)=1\\}\subseteq\dot{\mathcal{V}}[p]\
(=\\{A\subseteq\omega\colon p\Vdash A\in\dot{\mathcal{V}}\\}).`$

Additionally, if $p$ forces that $\dot{\mathcal{V}}$ is a P-point, then $\mu$
can be assumed to be a P-measure.

###### Proof.

Let $\dot{\mathcal{V}}$ be a name for an ultrafilter. Let $\mu$ be a measure
on $\omega$ defined by

$\mu(A)=\lambda(\llbracket A\in\dot{\mathcal{V}}\rrbracket\cap p)/\lambda(p).$

Clearly, if $\mu(A)=1$, then $p\Vdash A\in\dot{\mathcal{V}}$ and so
$A\in\dot{\mathcal{V}}[p]$.

Now, suppose that $p\Vdash\dot{\mathcal{V}}$ is a P-point. Let $(A_{n})$ be a
decreasing sequence of subsets of $\omega$. Suppose, towards the
contradiction, that $\lim_{n\to\infty}\mu(A_{n})=a>0$ and there is no
pseudointersection $A$ of $(A_{n})$ such that $\mu(A)=a$. In fact, we may
assume that there is no pseudointersection $A$ of $(A_{n})$ such that
$\mu(A)>0$ (see [BNS23, Lemma 3.3]). Let $p=\bigwedge\llbracket
A_{n}\in\dot{\mathcal{U}}\rrbracket$. Then $p\neq 0$ and $p\Vdash\forall n\
A_{n}\in\dot{\mathcal{V}}$. So, there is $q\leq p$ such that

$q\Vdash\exists\dot{B}\in\dot{\mathcal{U}}\ \forall n\
\dot{B}\subseteq^{*}A_{n}.$

By the fact that $\mathbb{B}_{\kappa}$ is $\omega^{\omega}$-bounding, there is
$r\leq q$ and $A\in V$, $p\Vdash\dot{B}\subseteq A$ such that

$r\Vdash\forall n\ A\subseteq^{*}A_{n}.$

Then $r\leq\llbracket A\in\dot{\mathcal{U}}\rrbracket$ and so $\mu(A)>0$, a
contradiction. ∎

In the proof of the main theorem we will only need the first part of the above
proposition, the part about P-measure will be needed for other purposes.
Actually, we will use only the following corollary.

######  Corollary 7.18.

Suppose that $\dot{\mathcal{V}}$ is a $\mathbb{B}_{\kappa}$-name for an
ultrafilter on $\omega$. Then $\dot{\mathcal{V}}[p]$ is non-meager for each
$p\in\mathbb{B}_{\kappa}$.

###### Proof.

Suppose that $\mathcal{F}$ is a filter containing a family
$\\{A\subseteq\omega\colon\mu(A)=1\\}$ for some measure on $\omega$. According
to Proposition 2.9 the filter $\mathcal{F}$ is non-meager and so, thanks to
Proposition 7.17, we are done. ∎

######  Remark 7.19.

Recall that a filter $\mathcal{F}$ is called ccc if
$\mathcal{P}(\omega)/\mathcal{F}^{*}$ is ccc. In the proof of Proposition 2.9
we actually show that every ccc filter is nonmeager. It is not difficult to
prove that if $\mathbb{P}$ is a ccc forcing notion, and $\dot{\mathcal{U}}$ is
a $\mathbb{P}$-name for an ultrafilter on $\omega$, then there is a ground
model ccc filter $\mathcal{F}$ such that $\dot{\mathcal{U}}$ extends
$\mathcal{F}$ (see e.g. [Wim82]). As $\mathbb{B}_{\kappa}$ is obviously ccc,
we could use this fact to prove Corollary 7.18 instead of Proposition 7.17.

Now we are ready to prove the main theorem of this section.

###### Proof.

(of Theorem 7.1) Assume $V$ is a model of $\mathsf{ZFC+CH+\diamondsuit(S)}$,
where $S\subseteq\omega_{2}$ is stationary. Let $(A_{\alpha}:\alpha\in S)$ be
a $\diamondsuit(S)$-guessing sequence. Let $\kappa\geq\omega_{2}$ be an
uncountable regular cardinal. Let $\mathcal{U}$ be a selective ultrafilter and
$\mathcal{V}$ a P-point which is not Rudin-Blass above $\mathcal{U}$. Define a
countable support iteration
$(\mathbb{P}_{\alpha},\dot{\mathbb{Q}}_{\alpha}\colon\alpha<\omega_{2})$ as
follows:

  1. (1)

$\mathbb{P}_{0}=SP^{*}(\mathcal{U})$.

  2. (2)

If $\alpha\notin S$, define $\dot{\mathbb{Q}}_{\alpha}$ to be the trivial
forcing.

  3. (3)

If $\alpha\in S$, and $A_{\alpha}$ codifies a $\mathbb{P}_{\alpha}$-name for a
non-meager P-filter $\dot{\mathcal{F}}$ which is not nearly coherent with
$\mathcal{U}$, define
$\mathbb{P}_{\alpha}\Vdash\dot{\mathbb{Q}}_{\alpha}=SP^{*}(\dot{\mathcal{F}})$;
otherwise, let $\dot{\mathbb{Q}}_{\alpha}$ be the trivial forcing.

Let $\mathbb{P}_{\omega_{2}}$ the resulting forcing, and let
$\dot{\mathbb{B}}_{\kappa}$ be a $\mathbb{P}_{\omega_{2}}$-name for the
forcing adding $\kappa$ random reals. Our working forcing is
$\mathbb{P}_{\omega_{2}}\ast\dot{\mathbb{B}}_{\kappa}$.

Let $G\ast H$ be a
$\mathbb{P}_{\omega_{2}}\ast\dot{\mathbb{B}}_{\kappa}$-generic filter, and for
each $\alpha<\omega_{2}$ let $G_{\alpha}$ be the restriction of $G$ to
$\mathbb{P}_{\alpha}$. Since all steps of the iteration preserve $\mathcal{U}$
(by Lemma 7.16), we have that $\mathcal{U}$ continues to generate a selective
ultrafilter in $V[G]$. So, by Theorem 7.2, there is a P-measure in $V[G\ast
H]$.

Now, suppose for the sake of contradiction, that there is a P-point
$\mathcal{F}$ in $V[G*H]$. By Lemma 7.6 we have that in $V[G]$ the P-filter
$\dot{\mathcal{F}}[1]$ is not nearly coherent with $\mathcal{U}$ and by
Corollary 7.18 it is non-meager. Let $\dot{\mathcal{H}}$ be a
$\mathbb{P}_{\omega_{2}}$-name for the filter $\dot{\mathcal{F}}[1]$. For each
$\alpha\in\omega_{2}$, define
$\dot{\mathcal{H}}_{\alpha}=\dot{\mathcal{H}}\upharpoonright\mathbb{P}_{\alpha}$.
Let

$C_{0}=\\{\alpha\in\omega_{2}\colon\mathbb{P}_{\alpha}\Vdash\dot{\mathcal{H}}_{\alpha}=\dot{\mathcal{H}}\cap
V[G_{\alpha}]\\}$

and let

$C_{1}=\\{\alpha\in\omega_{2}:\mathbb{P}_{\alpha}\Vdash\text{
$\dot{\mathcal{H}}_{\alpha}\in V[G_{\alpha}]$ is not nearly coherent with
}\mathcal{U}\\}.$

The set $C_{0}\cap C_{1}$ is a club set, so there is $\alpha\in S\cap
C_{0}\cap C_{1}$ such that $A_{\alpha}$ encodes a $\mathbb{P}_{\alpha}$-name
for the filter $\dot{\mathcal{H}}_{\alpha}$. Thus, we have
$\mathbb{P}_{\alpha}\Vdash\dot{\mathbb{Q}}_{\alpha}=SP^{*}(\dot{\mathcal{H}}_{\alpha})$.
By Lemma 7.16 we have that $\mathbb{P}_{\alpha+1}$ forces that
$\dot{\mathcal{H}}_{\alpha}$ cannot be extended to a P-point in any further
$\omega^{\omega}$-bounding extensions, so it cannot be extended to a P-point
in $V[G\ast H]$. Since
$\dot{\mathcal{H}}_{\alpha}\subseteq\dot{\mathcal{F}}[1]\subseteq\mathcal{F}$,
we have that $\mathcal{F}$ is not a P-point, which is a contradiction.

Since $\mathbb{B}_{\kappa}$ is ccc and adds $\kappa$ new reals, all cardinal
numbers are preserved and the continuum is at least $\kappa$ in $V[G\ast H]$.

∎

######  Corollary 7.20.

It is relatively consistent with $\mathsf{ZFC}$ that there is a P-measure but
there is no P-point (and so, there is no ultrafilter density that is a
P-measure).

We will finish this section with some remarks on the random forcing.

As we have mentioned at the beginning of this section, the proof of Theorem
7.1 could be substantially shortened if we know that the random forcing cannot
add a ’new’ P-point (i.e. that every P-point in the random extension extends a
ground model P-point). However, we do not know that, nor do we know that this
is not true.

######  Problem 7.21.

Is there a model $V$ such that in the random extension $V^{\mathbb{B}}$ there
is a P-point which does not extend any P-point from $V$?

Recall that it is still an open problem if there is a P-point in the classical
random model. Kunen (see [Bre17]) proved that if we first add $\omega_{1}$
Cohen reals to the ground model and then add random reals, then in the
resulting model there is a P-point. Dow (see [Dow20]) showed that this
assertion holds also if we assume $\square_{\omega_{1}}$ in the ground model.
In both cases, the constructed P-point is an extension of a P-point from the
ground model.

The proof of Theorem 7.1 shows that consistently random forcings do not add
’new’ P-points. Also, (the second part of) Proposition 7.17 says that if we
add random reals to a model without P-measures, then we cannot add any
P-point.

So, there is a chance that Problem 7.21 has a negative solution. On the other
hand, if the answer is positive, the proof could bring us closer to the
solution of the problem of the existence of P-point in the random model.

Finally, notice a peculiar symmetry of Theorem 7.2 and Proposition 7.17: every
P-point in the ground model induces naturally a P-measure in the random
extension, but also every P-point in the random extension induces in a natural
way a P-measure in the ground model.

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