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\markboth{\psmPrGame\ and ultrafilters on $\omega$ and $\Q$ \ \ \ \ \
\UpdateDate
}{\psmPrGame\ and ultrafilters on $\omega$ and $\Q$ \ \ \ \ \ \UpdateDate}





\title{\psmPrGame\/ and ultrafilters on $\Q$ and $\omega$}

\author{Andres Mill\'an\thanks{This work is a part of author's
Ph.D. thesis written at West Virginia University under the supervision
of Professor Krzysztof Ciesielski. The author wishes to thank Professor
Ciesielski for
his guidance, patience, and encouragement.}
\\
{\footnotesize Department of Mathematics,}
{\footnotesize West Virginia University,} \\
{\footnotesize Morgantown, WV 26506-6310, USA};
{\footnotesize e-mail: amillan@math.wvu.edu}}

\begin{document}

\maketitle


\begin{abstract}

In this paper we use the version \psmPrGame\ of
the Covering Property Axiom, which has been formulated by Ciesielski
and Pawlikowski and holds in the iterated perfect set model,
to study the relations between different kinds of ultrafilters on $\omega$
and $\Q$.
In particular, we will give a full account for the logical relations
between the properties of being a selective ultrafilter, a $P$-point,
a $Q$-point, and an $\omega_1$-OK point.
\end{abstract}


\section{Introduction}

We use standard set theoretical notation and terminology as in \cite{Jech}.
In particular, if $A$ is a set $|A|$ denotes its cardinality and
$\P(A)$ the set of all its subsets. Lower case Greek letters denote ordinal
numbers. The first infinite cardinal is $\omega$ and $\omega_1$ is the first
uncountable cardinal. The cardinality of $\R$ is denoted by $\continuum$.
We also use the letter $\kappa$ to denote any unespecified uncountable cardinal.
If $A$ and $B$ are arbitrary sets, then we write
$A\subseteq^*{B}$ provided that $|A\setminus{B}|<\omega$.

Let $\U$ be a nonprincipal ultrafilter on an infinite countable set $X$.
(We will use for $X$ either $\omega$ or $\Q$.)
We say that:
\begin{itemize}
\item $\U$ is a \emph{$P$-point}\/ if for every partition
$\P$ of $X$ either $\U\cap{\P}\neq{\emptyset}$ or
there exists a $U\in{\U}$ such that $U\cap{P}$ is finite for each
$P\in{\P}$.

\item
$\U$ is a \emph{$Q$-point}\/ if for every partition
$\P$ of $X$ into finite sets there exists a
$U\in{\U}$ such that $|U\cap{P}|\leq{1}$ for each $P\in{\P}$.

\item
$\U$ is \emph{selective}\/ if for every partition
$\P$ of $X$ either $\U\cap{\P}\neq{\emptyset}$ or
there exists a $U\in{\U}$ such that $|U\cap{P}|\leq{1}$ for each
$P\in{\P}$.

\item $\U$ is a $\kappa$-OK point, where $\kappa$ is an infinite cardinal
number, provided for every
$\la V_n\in\U \colon n<\omega\ra$ there exists a
$\la U_\alpha\in\U \colon \alpha<\kappa\ra$ such that
$\bigcap_{i=1}^n{U_{\alpha_i}}\subseteq^*{V_n}$
for every $n<\omega$ and $\alpha_0<\dots<\alpha_n<\kappa$.
Sequence $\la U_\alpha\in\U \colon \alpha<\kappa\ra$ will be referred
to as OK for $\la V_n\in\U \colon n<\omega\ra$.
\end{itemize}
It is obvious from the definitions that

\FACT{factObvious}{$\U$ is a selective ultrafilter if and only if
$\U$ is simultaneously a $P$-point and a $Q$-point.}

$P$-points have been studied extensively by many people in connection
with the remainder $\omega^*$ of
the \v{C}ech-Stone compactification
of the integers and the problem of its homogeneity.
The existence of $P$-points cannot be proven or refuted
in the usual framework of set theory
ZFC (see, e.g., ~\cite{Wimmers} or ~\cite{BartJudah}) but they do exist under
several additional set theoretical assumptions like the
\emph{Continuum Hypothesis}\/ CH or \emph{Martin's Axiom}\/ MA.

Given a nonprincipal ultrafilter $\U$ on $X$ we say that $\B\subseteq{\U}$
is a basis for $\U$ if for every $U\in{\U}$ there exists a $B\in{\B}$ such
that
$B\subseteq U$. We define the \emph{character} of $\U$ as
$\chi(\U)=\text{min}\{|\B| \colon \B \text{ is a basis for } \U\}$.
If $\kappa=\chi(\U)$ then we say that the ultrafilter $\U$ is
{\em $\kappa$-generated.}

In \cite{Kunen}, K. Kunen introduced $\kappa$-OK
points to give a proof of the nonhomogeneity of
$\omega^*$ without any extra assumption
beyond ZFC. The following results are relevant to this paper.

\prop{prop:prop1}{\rm{(Kunen \cite{Kunen})} Every $P$-point is $\kappa$-OK
for every $\kappa$.}

\prop{prop:prop2}{\rm{(Kunen \cite{Kunen})} There are $2^{\continuum}$ many
distinct $\continuum$-OK points on~$\omega$.
Moreover, these ultrafilters can be made $\continuum$-generated.}

Consider $\Q$ with the subspace topology induced by the usual topology
on $\real$ and denote by $\perf{(\Q)}$ the family of its perfect subsets
(i.e., closed subsets with no isolated points).
\begin{itemize}
\item A nonprincipal filter $\U$ on $\Q$ is \emph{crowded}\/ if the family
$\perf{(\Q)}\cap{\U}$ forms a basis for $\U$.
\end{itemize}

The crowded ultrafilters have
been studied in connection with the remainder of the \v{C}ech-Stone
compactification of $\Q$ and their existence follows from the Continuum
Hypothesis,
Martin's Axiom for countable posets \cite{vanDouwen}, or from
the equality
$\mathfrak{b}=\continuum$ \cite{CoHart}.

In \cite[thm.~4.8 and cor.~4.14]{Crowded} Ciesielski and Pawlikowski showed that
\psmPrGame\ implies that there exist $\omega_1$-generated
selective ultrafilters
as well as $\omega_1$-generated nonselective $P$-points.
Since a nonselective $P$-point cannot be a $Q$-point (see
Fact~\ref{factObvious}),
this second result shows that
\psmPrGame\ implies that there exists a $P$-point which is not a $Q$-point.
In the same paper, \cite[thm.~4.22]{Crowded}, the authors also
established the existence of an
$\omega_1$-generated
crowded ultrafilter on $\Q$ under \psmPrGame.
They also proved, \cite[prop.~4.25]{Crowded}, that a crowded ultrafilter
cannot be a $P$-point.

In this paper we establish, under \psmPrGame, the existence of a nonselective
$Q$-point (i.e., a $Q$-point which is not a $P$-point) by constructing an
$\omega_1$-generated crowded $Q$-point which
is also an $\omega_1$-OK point (Corollary~\ref{cor:corW}).
This improves our construction from~\cite{Millan}
of an $\omega_1$-generated crowded $Q$-point on~$\Q$.
We also prove, under \psmPrGame, that there exist
crowded $\omega_1$-generated $Q$-points that
are not $\omega_1$-OK points (Corollary~\ref{cor:cor5.5}),
crowded $\omega_1$-generated
$\omega_1$-OK points which are neither $P$-points nor
$Q$-points (Theorem~\ref{thm:thm4}),
and crowded $\omega_1$-generated
ultrafilters on $\omega$ that are neither $Q$-points
nor $\omega_1$-OK points (Theorem~\ref{thm:thm1}).
These complete all the logical implications between being a $P$-point, a
$Q$-point, or
an $\omega_1$-OK point as Table~\ref{Ta:1} shows.

\begin{table}
\begin{center}
\begin{tabular}{|c|c|c|c|c|}
\hline
$P$-point & $Q$-point & $\omega_1$-OK point & Existence & Reference\\ \hline
$-$ & $-$ & $-$ & under \psmPrGame & Theorem \ref{thm:thm1} \\ \hline
$-$ & $-$ & + & under \psmPrGame  & Theorem \ref{thm:thm4} \\ \hline
$-$ & + & $-$ & under \psmPrGame  & Corollary~\ref{cor:cor5.5} \\ \hline
$-$ & + & + & under \psmPrGame  & Corollary~\ref{cor:corW} \\ \hline
+ & $-$ & $-$ & No, in ZFC  & Proposition \ref{prop:prop1} \\ \hline
+ & $-$ & + & under \psmPrGame  & \cite{Crowded} or \cite{Book} \\ \hline
+ & + & $-$ & No, in ZFC & Proposition \ref{prop:prop1} \\ \hline
+ & + & + & under \psmPrGame  & \cite{Crowded} or \cite{Book} \\ \hline
\end{tabular}
\end{center}
\vspace{-.5pc}
\caption{Existence of different ultrafilters.
All constructed
ultrafilters are nonprincipal and $\omega_1$-generated. Moreover, the first
four examples can be made also crowded.}\label{Ta:1}
\end{table}



Besides the properties explicitly listed in
Table~\ref{Ta:1} we consider also two other properties:
being $\omega_1$-generated (with $\omega_1<\continuum$) and
being crowded.


As mentioned above, the first four examples
from Table~\ref{Ta:1} are also crowded.
On the other hand that no other example from Table~\ref{Ta:1} can be crowded,
since a crowded ultrafilter cannot be a $P$-point \cite[prop.~4.25]{Crowded}.
It is also easy to see that we can destroy the property of being crowded without
changing any of the remaining properties. To see this, note that if
$\U$ is an ultrafilter on $\Q$ and $f$ is a bijection between $\Q$
and a scattered subset $S$ of $\Q$, then
$\V=\{A\subseteq \Q\colon f^{-1}(A)\in\U\}$
is a noncrowded ultrafilter that has the remaining properties
identical to that of $\U$.

One of the key features of our examples is that they are all
$\omega_1$-generated with $\omega_1<\continuum$.
This cannot be achieved in ZFC, since in many models of ZFC, for example
under MA, every nonprincipal ultrafilter on a countable set
has character $\continuum$. On the other hand, every example cited in
Table~\ref{Ta:1} can be constructed under MA if we are willing
to settle for $\continuum$-generated filters.
An interesting issue is whether under \psmPrGame\ the
examples from Table~\ref{Ta:1} must be $\omega_1$-generated.
The answer is positive for the last example from the table, since
Ciesielski and Pawlikowski proved
(see \cite[cor.~2.7]{Crowded} or \cite[cor.~1.5.4]{Book})
that under \psmPrGame\ every selective ultrafilter
is $\omega_1$-generated.
There is some indication suggesting that
\psmPrGame\ implies that every $P$-point is $\omega_1$-generated.
This would take care of the bottom half of the table.
Recently, the autor have constructed,
under \psmPrGame, a crowded $Q$-point of
character $\continuum$. This will appear in a forthcoming paper.
This particular example is not a weak $P$-point so
it cannot be an $\omega_1$-OK point.
(See \cite[Lemma 1.3]{Kunen}.)
The existence of an example of character $\continuum$
as in the fourth row in the table is left open.
The first two examples from Table~\ref{Ta:1} do not need to be
$\omega_1$-generated.
By Proposition~\ref{prop:prop8} the Fubini product $\U\otimes\U$,
where $\U$ is a Kunen's example
from Proposition~\ref{prop:prop2}, is as the first ultrafilter from
Table~\ref{Ta:1}.
The second of these is justified by a slight modification\footnote{Let
$\F_0$ be the dual filter of the ideal
$\I_0=\{A\subseteq{\omega} \colon
\overline{\lim}_{n\to{\infty}}|A\cap{P_n}|<+\infty\}$,
where $\{P_n \colon n<\omega\}$ is the partition of $\omega$ such that
$P_n=\{m<\omega \colon 2^n-1\leq{m}<2^{n+1}-1\}$. Construct a $\continuum$
by $\continuum$
independent linked family w.r.t $\F_0$ and follow the argument from
\cite{Kunen}.}
of Kunen's example from Proposition~\ref{prop:prop2}.

Finally,  let us address a question, whether any of the examples
from Table~\ref{Ta:1} can be constructed in ZFC.
The answer is clearly no for all but the first two examples, since there are
models of ZFC with no $P$-points (see ~\cite{Wimmers}) as well as models
of ZFC with no $Q$-points (see ~\cite{Miller}).
There are, however, a ZFC examples for the first two entries
of Table~\ref{Ta:1} as mentioned above.
These need not be $\omega_1$-generated, as we already noted.
Whether they can be crowded remains unclear, since
it is an open problem if there exists a crowded
ultrafilter in ZFC.





\section{Axiom \psmPrGame\ and other preliminaries}


The framework of \cpa\ rests on the concept of a
\emph{prism}. If $\Cantor$ denotes the space $2^{\omega}$ with its usual
product topology then we define for a Polish space $\X$
$$\perf{(\X)}=\{C\subseteq{\X} \colon C \text{ is homeomorphic to }
\Cantor\}.$$
If $0<\alpha<\omega_1$ is an ordinal let $\projkeep$
be the set of all continuous injections $f \colon \Cantor^{\alpha} \to
\Cantor^{\alpha}$
with the property that
$$f(x)\restriction{\xi}=f(y)\restriction{\xi}\ \
\Longleftrightarrow\ \ {x\restriction{\xi}=y\restriction{\xi}}
\text{\qquad for all  $\xi<\alpha$ and $x,y\in{\Cantor^{\alpha}}$}.$$
Then we define
$\mathPerf_{\alpha}=\{\text{range}(f) \colon f\in{\projkeep}\}$ and
$\mathPerf_{\omega_1}=\bigcup_{0<\alpha<\omega_1}\mathPerf_{\alpha}.$
The elements of $\mathPerf_{\omega_1}$ are called the \emph{iterated
perfect sets}.
The simplest elements of $\mathPerf_{\alpha}$ are of the form
$C=\prod_{\xi<\alpha}C_\xi$, where
$C_\xi\in{\perf{(\Cantor)}}$ for every $\xi<\alpha$.
We refer to them as  \emph{perfect cubes}.




If $\X$ is a Polish space, then a \emph{prism}\/ in $\X$ is a pair $\la{f,P}\ra$
where $f \colon E \to \X$ is injective and continuous,
$E\in{\mathPerf_{\omega_1}}$, and $P=f[E]$.
Function $f$ can be considered as a coordinate system imposed on $P$.
We will usually abuse this terminology and refer to $P$ itself as a prism.
In this case function $f$, given only implicitly, will be referred
to as a {\em witness function}\/ for $P$.
If the domain of the witness function of a prism $P$ happens to be
a perfect cube, we will sometimes refer also to $P$ as a {\em cube}\/ in $\X$.


If $\la f,P\ra$ is a prism, then we say that $Q$ is its {\em subprism}\/
provided
there exists an iterated perfect set $E\subseteq{\text{dom}(f)}$ such that
$Q=f[E]$.
We will refer to $Q$ as a {\em subcube}\/ of $P$ when $E$ is a perfect cube.
Notice that

\rem{rem1}{If we need to prove that a prism $P$ contains a subprism $Q$
with some ``nice property,'' we can always assume that the witness function
$f$ for $P$ is defined
on the entire set $\Cantor^\alpha$.}

\proof Indeed, assume that we can find a desired subprism $Q$ of a prism
$P$ as long as its witness function $f$ is defined
on the entire set $\Cantor^\alpha$.

Next, take an arbitrary witness function $g$ from
$E\in{\mathPerf_{\alpha}}$ onto $P$
and let $h\in\projkeep$ be onto $E$. Then $f=g\circ h$ is a continuous
injection from $\Cantor^\alpha$ onto $P$, so by the above assumption we
can find a subprism $Q$ of $\la f,P\ra$ with the ``nice property'' we are after.
To finish the argument it is enough to note that $Q$ is also a subprism of
$\la g,P\ra$.
Indeed, since $Q=f[E_0]$ for some $E_0\in {\mathPerf_{\alpha}}$,
there exists
an $h_0\in\projkeep$ onto $E_0$.
But then $h\circ h_0\in\projkeep$ and $Q=f[E_0]=(g\circ h)[h_0[\Cantor^\alpha]]
=g[h\circ h_0[\Cantor^\alpha]]$
is a subprism of $\la g,P\ra$ as
$h\circ h_0[\Cantor^\alpha]\in{\mathPerf_{\alpha}}$. \qed


Since in the game defined below we will need to consider singletons
in the same position as prisms as defined above,
in what follows {\em singletons will be considered as prisms.}
If $P$ is a singleton in $\X$ then its only subprism is $P$ itself.


The following theorem is one of the principal tools for
finding subprism of a prism, so also for using \psm.
This result is a refinement of a theorem proved independently
by H.G. Eggleston~\cite{Eg} and M.L. Brodski\u{\i}~\cite{Br}.

\prop{prop:prop3}{
{\rm (K.~Ciesielski and J.~Pawlikowski, \cite[claim~1.1.5]{Book})}
Let $0<\alpha<\omega_1$ and
consider $\Cantor^\alpha$ with its usual topology and its usual product
measure.
If $G$ is a Borel subset of $\Cantor^\alpha$ which
is either of second category or of positive measure, then
$G$ contains a perfect cube $E$. In particular $E\in{\mathPerf_{\alpha}}$.}


Strictly speaking, in \cite[claim~1.1.5]{Book} (see also
\cite[claim~2.3]{Crowded})
the result is proved only for $\alpha=\omega$.
But this easily implies the above version.

We will need also the following fusion lemma, which is an easy compilation
of Lammas~3.1.1 and~3.1.2 from~\cite{Book}. The proof of the compilation
is identical to that of~\cite[cor.~3.1.3]{Book}.


\prop{prop:fusion}{{\rm (K.~Ciesielski and J.~Pawlikowski \cite{Book})}
Let $0<\alpha<\omega_1$ and for every $n<\omega$
let $\D_n\subseteq[\mathPerf_\alpha]^{<\omega}$ be a family
of pairwise disjoint sets such that
$\emptyset\in\D_n$, $\D_n$ is
closed under refinements, and
\begin{itemize}
\item[\rm ($\dagger$)] for every
$\E\in\D_n$ and $E\in\mathPerf_\alpha$ which is disjoint with
$\bigcup\E$ there exists an $E'\in\mathPerf_\alpha\cap\P(E)$ such that
$\{E'\}\cup\E\in\D_n$.
\end{itemize}
Then for every $n<\omega$ there is a
family $\E_n=\{E_k\colon k<2^n\}\in\D_n$ of pairwise disjoint sets such that
$E=\bigcap_{n<\omega}\bigcup\E_n\in\mathPerf_\alpha$.}


For a Polish space $\X$ consider the following game $\gameprism$ of length
$\omega_1$ played by two players, Player I and Player II. At each stage
$\xi<\omega_1$ of the game Player I can play an arbitrary prism $P_{\xi}$
in $\X$
(i.e., $P_{\xi}$ either is a singleton in $\X$ or
it belongs to $\perf(\X)$ and comes with a witness function)
and Player II must respond by playing
a subprism $Q_{\xi}$ of $P_{\xi}$.
The game $\la{\la{P_{\xi},Q_{\xi}}\ra \colon \xi<\omega_1}\ra$ is won by
Player I provided $$\X=\bigcup_{\xi<\omega_1}Q_{\xi};$$
otherwise Player II wins.
A strategy for Player II is any function $S$ such that
$S(\la{\la{P_{\eta},Q_{\eta}}\ra \colon \eta<\xi}\ra, P_{\xi})$
is a subprism of $P_{\xi}$ for every partial game
$\la{\la{P_{\eta},Q_{\eta}}\ra \colon \eta<\xi}\ra$.
We say that a game $\la{\la{P_{\xi},Q_{\xi}}\ra \colon \xi<\omega_1}\ra$
is played according to a strategy $S$ for Player II provided
$Q_{\xi}=S(\la{\la{P_{\eta},Q_{\eta}}\ra \colon \eta<\xi}\ra, P_{\xi})$
for every $\xi<\omega_1$. A strategy $S$ for Player II is a \emph{winning
strategy}
provided Player II wins any game played according the strategy $S$.

The following principle captures a combinatorial
core of the iterated Sacks model.

\begin{description}
\item[\psmPrGame:] $\continuum=\omega_2$ and for any Polish space $\X$ Player II
                               has no winning
%                               \linebreak
                               strategy in the game $\gameprism$.
\end{description}
%
\noindent

\bigskip

\noindent
The axiom is consequence of a slightly more general principle, similar in
spirit,
called  \cpa, see \cite{Book}. Its importance comes from the following theorem.

\prop{prop:prop4}{{\rm (K.~Ciesielski and J.~Pawlikowski
\cite[thm.~7.2.1]{Book})} \cpa\ holds in the iterated
perfect set model. In particular, \psm\ is consistent with ZFC set
theory.}
The proof of the consistency of \psmPrGame\ can be also found in
\cite[thm.~5.3]{Crowded}.


A set $B\subseteq{\Q}$ is {\em scattered}\/ if every nonempty subset of $B$
has isolated points.
It is easy to see that the scattered subsets of $\Q$ form
an ideal, which we will denote by $\I_S$.
The following facts will be used in what follows.
For the proofs see \cite{Crowded} or \cite[Fact~5.5.1]{Book}.

\FACT{fact:nonscat}{Every nonscattered set $B\subseteq{\Q}$ contains a subset
from $\perf(\Q)$.}

Let $\J$ be an ideal on a countable set $X$. Then we define
$\J^+=\P(X)\setminus{\J}$. We say that $\J$ is {\em weakly
selective}\/  if for every $A\in{\J^+}$ and $f\colon A \to X$ there exists
a $B\in{\P(A)\cap{\J^+}}$ such that $f\restriction{B}$ is
either one-to-one or constant.

\FACT{fact:scat}{The ideals $[\omega]^{<\omega}$ and  $\I_S$ are weakly
selective.}

The proof of the following result can be found in
\cite[lem.~4.9(b)]{Crowded} or in
\cite[lem.~5.3.4(b)]{Book}, where $[X]^{\omega}$ comes with a subspace
topology of $\P(X)$, with $\P(X)$ being identified with
$2^X$ via characteristic function.

\prop{prop:prop7}{{\rm (K. Ciesielski, J. Pawlikowski \cite{Crowded, Book})}
Let $X$ be countably infinite and let
$\J\subseteq{\P(X)}$  be a weakly selective ideal. For every prism
$P\subseteq{[X]^{\omega}}$
and every $A\in{\J^+}$ there exist a subprism $Q$ of $P$, a
$B\in{\P(A)\cap{\J^+}}$, and an
$i<2$ such that $g\restriction{B}$ is constant equal to $i$ for every
$g\in{Q}$.}

\section{Some important lemmas.}

Let $X$ be a countably infinite set.
If $\F\subseteq{[X]^{\omega}}$ is nonempty, we say that $\F$ has the
\emph{strong finite intersection property, SFIP}, provided that
$|\bigcap{F}|=\omega$ for every nonempty $F\in{[\F]^{<\omega}}$.
The following is a very well known and easy fact.


\lem{lem:lem1}{If $\F\subseteq [X]^{\omega}$ is nonempty, countable,
and has the SFIP, then there exists a $C(\F)\in{[X]^{\omega}}$
such that $C(\F)\subseteq^*{B}$ for every $B\in{\F}$.}

\proof If $\F$ is finite, we can put $C(\F)=\bigcap{\F}$; otherwise
$\F=\{B_n \colon n<\omega\}$ and we can pick inductively
$b_n\in{\bigcap_{k\leq{n}}B_k}$ such that $b_n\notin{\{b_k \colon k<n\}}$.
The set $C(\F)=\{b_n \colon n<\omega\}$ works.
\qed


Let $X$ be a countably infinite set.
If the set $Z_X=[X]^{<\omega}\setminus{\{\emptyset\}}$
has the discrete topology then the product space
$\Z_X=(Z_X)^{\omega}$ is a Polish space and the sets
$U_{\la{n,a}\ra}=\{z\in{\Z} \colon z(n)=a\}$,
where $a\in{[\omega]^{<\omega}}$ and $n<\omega$, constitute
a subbasis for the product topology.
Consider the set
$$
\Part_X=\{z\in{\Z_X} \colon \mbox{ $\{z(k) \colon k<\omega\}$
is a partition of $\omega$}\}.
$$
If $X=\omega$ we will drop the indexes, that is, $\Z=\Z_{\omega}$ and
$\Part=\Part_{\omega}$.

\lem{lem:lem2}{$\Part_X$ is a $G_{\delta}$ subset of $\Z_X$. Therefore
$\Part_X$ is a
Polish space with the relative topology inherited from $\Z_X$.}

\proof
We can assume that $X=\omega$.
If
$A=\{z\in{\Z} \colon \bigcup_{n<\omega}z(n)=\omega\}$ and
$B=\{z\in{\Z} \colon \{z(n) \colon n<\omega\} \mbox{ is pairwise disjoint} \}$
then $\Part=A\cap{B}$.
The set $A$ is $G_{\delta}$ because we have
$A=\bigcap_{k\in{\omega}}\bigcup_{n<\omega}{\bigcup{\{U_{\la{n,a}\ra} \colon
a\in{[\omega]^{<\omega}}\text{ \& } k\in{a} \}}}$.
The set $B$ is $G_\delta$ since it can be written as
$\bigcap_{m<n<\omega}\bigcup{\{U_{\la{m,a}\ra }\cap{U_{\la{n,b}\ra }}
\colon a\cap{b}=\emptyset \}}$.
Thus, $\Part$ is $G_{\delta}$ in~$\Z$.
\qed

\defi{defi:defiQlike}{Let $X$ be a countably infinite set and
let $\J\subseteq{\P(X)}$ be an ideal on $X$ containing all
the singletons. We say that $\J$ is \emph{$Q$-like} provided that
for every $A\in{\J^+}$ there exists a countable indexed family
$\{A_n\in{[A]^{\omega}} \colon n<\omega\}$ such that
no set $\{b_n \colon n<\omega\}$ belongs to $\J$ provided
$b_n\in{A_n}$ for every $n<\omega$.}

\lem{lem:lem3}{Let $X$ be a countably infinite set, let $\J$ be a
$Q$-like ideal on $X$ and let $A\in{\J^+}$ be arbitrary.
If $P$ is a prism on $\Part_X$, then there exist
a subprism $Q$ of $P$ and a $B\in{\P(A)\cap{\J^+}}$ such that
$|z(k)\cap{B}|\leq{1}$ for every $z\in{Q}$ and $k<\omega$.
Moreover, if $P$ is a cube than $Q$ can
be chosen as a subcube of $P$.}


\proof
We can suppose that $X=\omega$.
Let $\la A_n\in{[A]^{\omega}} \colon n<\omega\ra$ be the sequence
associated to $A$ in the definition of $Q$-like.

Case (a): If $P=\{z\}$ then, define a sequence $\la{b_n\in{\omega} \colon
n<\omega}\ra$
inductively such that
$b_n\in{A_n\setminus{\bigcup\{z(k) \colon k<\omega\ \&\ z(k)
\cap{\{b_0,\dots,b_{n-1}\}\neq{\emptyset}}\}}}$ for every $n<\omega$.
It is easy to see that $B=\{b_n \colon n<\omega\}$ works.

Case (b): If $P\in{\perf(\Part_{\omega})}$, let $f$ be a witness function
for $P$.
By Remark~\ref{rem1} we can assume that $f$ acts from $\Cantor^\alpha$ onto $P$.
Thus, $P$ is a cube. It is enough to find its subcube with the desired
properties.

Let $\mu$ be the standard product probability measure on
$\Cantor^\alpha$.
We construct, by induction on $n<\omega$, a sequence
$\la{K_n\colon n<\omega}\ra$
of open subsets of $\Cantor^\alpha$ and two sequences, $\la{b_n\in{A_n}}
\colon n<\omega\ra$
and $\la{B_n\in{[\omega]^{<\omega}}} \colon n<\omega\ra$,
such that for every $n<\omega$:
\begin{itemize}
\item[(i)] $b_n>\max\left(\{b_i \colon i<n\}\cup{\bigcup_{j<n}B_j}\right)$,
\item[(ii)] $\mu{(K_n)}\geq 1-2^{-(n+2)}$, and
\item[(iii)] $f(h)(k)\subseteq B_n$ for every $h\in K_n$ and $k<\omega$ for
which $b_n\in f(h)(k)$.
\end{itemize}

If this construction is possible, put $B=\{b_n \colon n<\omega\}$.
Then, clearly  $B\in{\P(A)\cap{\J^+}}$ since that $\J$ is $Q$-like
and $b_n\in A_n$ for every $n<\omega$.
Condition (ii) implies
that $\mu\left(\bigcap_{n<\omega}K_n\right)\geq{\frac{1}{2}}$.
Hence, by Proposition~\ref{prop:prop3}, there exists a perfect cube
$C\subseteq{\bigcap_{n<\omega}K_n}$. Then $Q=f[C]$ is a subcube of $P$
and the pair $\la{Q,B}\ra$ is as required.
To see this, it is enough to show that $|z(k)\cap B|\leq 1$
for every $z\in{Q}$ and $k<\omega$.
Let $z=f(h)$ for some $h\in C$.
By conditions (i) and (iii), for every $b_j\in z(k)=f(h)(k)$ and $n>j$
we have that $b_n\notin z(k)$.
Therefore, no two elements of $B$ are in the same $z(k)$ or, in
other words, $|z(k)\cap{B}|\leq{1}$ for every $k<\omega$.

Next, we show that the inductive construction is possible.
Let $n<\omega$ be such that the appropriate $b_i$, $K_i$, and $B_i$ are already
constructed for every $i<n$. We will construct $b_n$, $K_n$, and  $B_n$
satisfying (i)--(iii).
We pick an $b_n$ as an arbitrary element of $A_n$ satisfying condition (i).
Next, we define  $L=\{a\in[\omega]^{<\omega}\colon b_n\in a\}$
and note that $\left\{f^{-1}\left(U_{\la m,a\ra}\right)\colon
\la m,a\ra\in\omega\times L\right\}$
is a partition of $\Cantor^\alpha$ into clopen sets.
Thus, we can find a finite set $S\subseteq \omega\times L$
such that
$K_n=\bigcup\left\{f^{-1}\left(U_{\la m,a\ra}\right)\colon
\la m,a\ra\in S\right\}$
satisfies condition (ii).
Let
$B_n=\bigcup\{a\colon \la m,a\ra\in S\mbox{ for some }m<\omega\}$.
Then clearly, $B_n$ is finite. To see that it satisfies (iii),
take an $h\in K_n$.
Then $f(h)\in U_{\la m,a\ra}$ for some $\la m,a\ra\in S$.
Let $k<\omega$ be such that $b_n\in f(h)(k)$.
Since we have also $b_n\in a=f(h)(m)$,
we conclude that $k=m$. So, $f(h)(k)=f(h)(m)=a\subseteq B_n$.
\qed


\defi{defi:defi1A}{Let $X$ be a countably infinite set. We say that an ideal
$\J$ on $X$ is \emph{prism-friendly}\/ provided that
it contains all singletons and
\begin{itemize}
      \item [($\bullet$)] given a prism $P$ in $2^X$ and an $A\in \J^+$ there
                     exists a subprism $Q$ of $P$, a $B\in{\P(A)\cap{\J^+}}$,
                     and an $i<2$
                     such that $g\restriction{B}$ is constant equal $i$ for
every
                     $g\in{Q}$.

\end{itemize}}

\defi{defi:defi1B}{Let $X$ be a countably infinite set.
We say that an ideal $\J$ on $X$ is \emph{rich} if it is prism-friendly and
\begin{itemize}
\item [(\#)] given an $A\in{\J^+}$ there exists a family
$\A\subseteq{\P(A)\cap{\J^+}}$ of cardinality $\continuum$ which is almost
disjoint, that is,
such that $|A\cap B|<\omega$ for every distinct $A,B\in\A$.
\end{itemize}}

Also, notice that, in ZFC, condition $(\bullet)$ does not imply condition
$(\#)$. Indeed, if $\U$ is a selective ultrafilter, then its dual
ideal $\I_{\U}$ is weakly selective. So, see \cite{Book}, $\I_{\U}$ is
prism-friendly.
However, $\I_{\U}^+=\U$ and no two members in $\U$ can be almost disjoint.

\lem{lem:lem4}{The ideals $[\omega]^{<\omega}$ and $\I_S$ are $Q$-like and
rich.}


\proof
It is easy to see that $[\omega]^{<\omega}$ is $Q$-like.
To see that $\I_S$ is also $Q$-like pick any $A\in{\I_S^+}$.
By Fact~\ref{fact:scat} we can assume that $A\in{\perf(\Q)}$.
Let $\B$ be a countable basis for the topology on $\Q$
and let $\{A_n \colon n<\omega\}$ be an enumeration of the set
$\{S\cap{A} \colon S\in{\B} \; \& \; |S\cap{A}|=\omega\}$.
If $b_n\in{A_n}$ for every $n<\omega$ then
$B=\{b_n \colon n<\omega\}$ is
dense in $A$ and in consequence, it is in $\I_S^+$.

By Fact~\ref{fact:scat}, the ideals $[\omega]^{<\omega}$ and $\I_S$
are weakly selective so, by Proposition~\ref{prop:prop7},
they are prism-friendly. Thus, we need only to check that each
of these ideals satisfies the condition (\#) from the definition of
rich ideal.

It is well known that (\#) holds for $[\omega]^{<\omega}$.
To check that (\#) also holds for $\I_S$, fix a countable basis $\B$ for
the topology on $\Q$ and pick an $A\in{\I^+_S}$.
By Fact~\ref{fact:nonscat}, we can assume that $A\in\perf(\Q)$.
Let $\{B_n\colon n<\omega\}$ be an enumeration of
$\B_A=\{B\in{\B} \colon |B\cap{A}|=\omega\}$
and construct $\{a_s \colon s\in{2^{<\omega}}\}$ by induction on the length
of $s$ in such a way that
$\{a_s \colon s\in{2^n}\}\in{[A\cap{B_n}]^{2^n}}$  and that
$\{a_s \colon s\in{2^n}\}\cap{\bigcup\{a_t \colon t\in{2^{<n}}\}}=\emptyset$
for every $n<\omega$.
If for $x\in{2^{\omega}}$ we put $A_x=\{a_{x\restriction{n}} \colon n<\omega\}$,
then $A_x\in{\I_S^+}$ for every $x\in{2^{\omega}}$,
since $A_x$ is dense in $A$.
Then  $\A=\{A_x \colon x\in{2^{\omega}}\}$
is almost disjoint and satisfies (\#).
\qed

\defi{defi:Fubini}{Let $X$ be a countably infinite set and let
$\J\subseteq{\P(X)}$
be an ideal on $X$ containing all singletons. The \emph{Fubini product} of
the ideals $[\omega]^{<\omega}$ and $\J$ is the ideal $\K$ on
$\omega\times{X}$  denoted
$[\omega]^{<\omega}\otimes{\J}$ and defined as the family of all subsets $A$ of
$\omega\times{X}$ such that
$${\rm supp}(A)\stackrel{\rm def}{=}\{n<\omega \colon (A)_n\in{\J^+}\}
\text{ is finite},$$
where $(A)_n=\{x\in{X} \colon \la{n,x}\ra\in{A}\}$.}

\lem{lem:Qlike}{If $\J$ is a $Q$-like ideal, then
$\K=[\omega]^{<\omega}\otimes{\J}$ is also $Q$-like.}

\proof
Let $A\in\K^+$. For each $n\in{\rm supp}(A)$ let
$\{A_n^m\in{[(A)_n]^{\omega}} \colon m<\omega\}$
be a family from the definition of $Q$-like for $(A)_n\in\J^+$.
Then the family
$\{\{n\}\times A_n^m\colon n\in{\rm supp}(A)\ \&\ m<\omega\}$
satisfies the definition of $Q$-like for the set $A$.
\qed




\lem{lem:Fubini}{Let $X$ be a countably infinite set, $\J$
a prism-friendly ideal on $X$, $P$ a prism in $2^{\omega\times{X}}$,
$I\in{[\omega]^{\omega}}$, and
let $\la{A_n\in{\J^+} \colon n\in{I}}\ra$ be arbitrary.
Then, there exist a subprism $Q$ of $P$, a set $J\in{[I]^{\omega}}$,
a sequence $\la{B_n\in{\P(A_n)\cap{\J^+}} \colon n\in{J}}\ra$, and an
$i<2$, such that
$g\restriction{B}$ is constant equal $i$ for every $g\in{Q}$ provided
that $B=\bigcup\{\{n\}\times{B_n} \colon n\in{J}\}$.

In particular, if $\J$ is prism-friendly, then so is
$\K=[\omega]^{<\omega}\otimes{\J}$.}

\proof
We can suppose that $I=\omega$. If $P$ is a singleton the lemma
follows easily from the fact that $\J$ is an ideal containing
the singletons and the pigeon hole principle.
So, suppose that $P\in{\perf(2^{\omega\times{X}})}$.
Let $f$ be a function witnessing that $P$ is a prism.
By Remark~\ref{rem1} we can assume that $f$ is defined on $\Cantor^{\alpha}$
for some $0<\alpha<\omega_1$.
We will construct a subprism $Q_0$ of $P$ and  a sequence
$\la{B_n\in{[A_n]^{\omega}\cap{\J^+}} \colon n<\omega}\ra$
such that for every $n<\omega$
\begin{equation}\label{E:1}
g\restriction{\{n\}\times{B_n}} \mbox{ is constant for every } g\in{Q_0}.
\end{equation}
This will be done using Proposition~\ref{prop:fusion}.

For each $n<\omega$ let $\D_n$ be the collection of all pairwise disjoint
families $\E\in{[\mathPerf_{\alpha}]^{<\omega}}$ such that there exists an
$A_{\la{\E,n}\ra}\in{[A_n]^{\omega}\cap{\I^+}}$
with the property that for every $E\in{\E}$
\begin{equation}\label{E:2}
f(h)\restriction{\{n\}\times{A_{\la{\E,n}\ra}}}=f(h^{'})\restriction{\{n\}
\times{A_{\la{\E,n}\ra}}}\mbox{ for all } h, h^{'}\in{E}.
\end{equation}



Clearly, each $\D_n$ is closed under refinaments. To see that $\D_n$ satisfies
the condition $(\dagger)$ from Proposition~\ref{prop:fusion}
pick $\E\in{\D_n}$ and $E\in{\mathPerf_{\alpha}}$ such that
$E\cap{\bigcup{\E}}=\emptyset$.
Decreasing $A_{\la\E,n\ra}$, if necessary, we can assume
that $X\setminus A_{\la\E,n\ra}$ is infinite. 
%
%If this set is finite, function $b_n$cannot be defined. 
%
Let $b_n \colon \omega\times{X} \to X$ be any bijection such that
$b_n(n,a)=a$
for every $a\in A_{\la\E,n\ra}$. 
%
%b_n needs to fix  A_{\la\E,n\ra} for A' to be as desired. 
%
This
bijection induces a homeomorphism  
$f_n \colon 2^{\omega\times{X}} \to 2^X$ defined by
$f_n(g)(x)=g(b_n^{-1}(x))$ for every $g\in{2^{\omega\times{X}}}$ and $x\in{X}$.
Clearly, $f_n$ is continuous and injective. Hence, $Q^*=(f_n\circ{f})[E]$
is a prism in $2^X$.
Since $\J$ is prism-friendly, we can find a
subprism $Q^{**}$ of $Q^*$, an
$A^{'}\in{[A_{\la{\E,n}\ra}]^{\omega}\cap{\J^+}}$, and an $i<2$ such that
$g[A^{'}]=\{i\}$ for every $g\in Q^{**}$.
But $Q^{**}=f_n[E^{'}]$ for some $E^{'}\in{\mathPerf_{\alpha}\cap{\P(E)}}$.
So, if
we put $\E^{'}=\E\cup{\{E^{'}\}}$ and $A_{\la{\E^{'},n}\ra}=A^{'}$ we get that
$\E^{'}\in{\D_n}$ and the condition $(\dagger)$ is satisfied.
Thus, by Proposition~\ref{prop:fusion},
for every $n<\omega$ there exists a
family $\E_n=\{E_k\colon k<2^n\}\in\D_n$ of pairwise disjoint sets with
$E^0=\bigcap_{n<\omega}\bigcup\E_n\in\mathPerf_\alpha$.
We will prove that $Q_0=f[E^0]$ satisfies \eqref{E:1} with
some sequence $\la B_n\colon n<\omega\ra$.

To see this fix an $n<\omega$, for each $k<2^n$
pick an $h_k\in{E_k}$, and define $\varphi_n\colon A_{\la{\E_n,n}\ra}\to
2^{2^n}$ by
$\varphi_n(p)(k)=f(h_k)(n,p)$. Since $A_{\la{\E_n,n}\ra}\in{\I^+}$ and
$\J$ is an ideal, we can find an $s_n\in{2^{2^n}}$ such that
$B_n=\varphi_n^{-1}(s_n)\in\J^+$.
To see that $B_n$ satisfies \eqref{E:1}, pick a $g\in Q_0$.
Then there exists a $k<2^n$ and an $h\in E_k$ such that
$g=f(h)$. Since $B_n\subseteq A_{\la{\E_n,n}\ra}$,  by \eqref{E:2}
we have that $g\restriction \{n\}\times{B_n}=f(h_k)\restriction
\{n\}\times{B_n}$.
In particular, $g(p)=f(h_k)(n,p)=\varphi_n(p)(k)=s_n(k)$
for every $p\in B_n$.
So,
$g\restriction{\{n\}\times{B_n}}$ is constant equal to
$s_n(k)$ and \eqref{E:1} holds.

To finish the proof of the lemma pick a $b_n\in{B_n}$ for each $n<\omega$.
Then, the set $S=\{\la{n,b_n}\ra\in \{n\}\times{B_n}\colon n<\omega\}$ is a
selector
for $\{\{n\}\times{B_n} \colon n<\omega\}$.
Let $\I=[\omega\times X]^{<\omega}$. Then $\I$ is weakly selective and
$S\in\I^+$.
If we identify $2^{\omega\times X}$ with $\P(\omega\times X)$,
then $Q_0$ can be treated as a prism in $\P(\omega\times X)$.
Since $[\omega\times X]^{\omega}$ is residual in $\P(\omega\times X)$,
by Proposition~\ref{prop:prop3} we can assume that
$Q_0$ is a prism in $[\omega\times X]^{\omega}$.
So, by Proposition~\ref{prop:prop7},
there exist a subprism $Q$ of $Q_0$, a set $S_0\in{[S]^{\omega}}$, and
an $i<2$ such that $g[S_0]={\{i\}}$ for every $g\in{Q}$. Define
$J=\{n<\omega \colon \la{n,b_n}\ra\in{S_0}\}$.

To see that the conclusion of the lemma holds take a $g\in Q$ and an
$\la{n,b}\ra\in B$.
Then $n\in J$ and $b\in B_n$.
So, by \eqref{E:1}, $g(n,b)=g(n,b_n)=i$, since $\la{n,b_n}\ra\in S_0$.
\qed

\lem{lem:Rich}{Let $X$ be a countably infinite set and let $\J\subseteq{\P(X)}$
be an ideal containing all singletons and satisfying condition $(\#)$ from the
definition of a rich ideal. Then the ideal $\K=[\omega]^{<\omega}\otimes{\J}$
also satisfies $(\#)$.

In particular, if $\J$ is rich, then so is $\K$.}


\proof Let $A\in{\K^+}$. Then ${\rm supp}(A)$ is infinite.
Let $\A=\{A_{\xi} \colon \xi<\continuum\}\subseteq{[{\rm supp}(A)]^{\omega}}$
be an almost disjoint family. Since $\J$ satisfies $(\#)$, for every $n<\omega$
there exists an almost disjoint family
$\B_n=\{B_{\xi}^n \colon \xi<\continuum\}\subseteq{\P((A)_n)\cap{\J^+}}$.
If for every $\xi<\continuum$ we define
$$U_{\xi}=\bigcup\{\{n\}\times{B_{\xi}^n} \colon n\in{A_{\xi}}\},$$
then the family $\{U_{\xi} \colon \xi<\continuum\}\subseteq{\P(A)\cap{\K^+}}$
works. The other part of the lemma is consequence of this and of Lemma
\ref{lem:Fubini}.
\qed





\section{An $\omega_1$-generated crowded bad point.}


\defi{defi:defi3}{If $\U$ and $\V$ are ultrafilters then, the
\emph{Fubini product}\/ of $\U$ and $\V$ is defined as
$$\U\otimes{\V}=\{A\subseteq{\omega\times{\omega \colon
\{n \colon (A)_n\in{\V}\}\in{\U}}}\}.$$}

\prop{prop:prop8}{{\rm (Folklore)} If $\U$ and $\V$ are nonprincipal
ultrafilters in $\omega$ then $\U\otimes{\V}$ is a nonprincipal
ultrafilter which is not a $P$-point, a $Q$-point,
or even an $\omega_1$-OK point.}

\proof It is easy to see that $\U\otimes{\V}$ is a nonprincipal
ultrafilter.
To see that $\U\otimes{\V}$ cannot be a $P$-point observe that
the set $\{L_m \colon m<\omega\}$ of all sections $L_m=\{\la m,n\ra\colon
n\in{\omega}\}$
is a partition of $\omega\times{\omega}$ into infinite pieces not in
$\U\otimes{\V}$ and that every $X\in{\U\otimes{\V}}$ intersects infiniteley
many $L_m$'s on an infinite set.

To see that $\U\otimes{\V}$ cannot be a $Q$-point consider  the
\emph{partial partition} $\{P_n \colon n<\omega\}$ of $\omega\times{\omega}$
where $P_n=\{\la m,n\ra\colon m\leq{n} \}$ for every $n<\omega$.
Notice that $\bigcup_{n<\omega}P_n\in \U\otimes{\V}$.
Let $\P\subseteq[\omega\times\omega]^{<\omega}$ be a partition of
$\omega\times{\omega}$ such that
$\{P_n \colon n<\omega\}\subseteq{\P}$.
It is easy to see that there is no $X\in{\U\otimes{\V}}$
such that $|X\cap{P}|\leq{1}$ for every $P\in{\P}$.

To see that $\U\otimes{\V}$ is not an $\omega_1$-OK point consider
$\{V_n \colon n<\omega\}\subseteq{\U\otimes{\V}}$, where
$V_n=\bigcup_{m>n}L_m$. By the way of contradiction, suppose that
the sequence $\bar\U=\la U_{\xi}\in\U\otimes{\V}\colon \xi<\omega_1\ra$ is OK
for $\{V_n \colon n<\omega\}$. Then, by the pigeon hole principle, there
exist an $m<\omega$ and an $X\in{[\omega_1]^{\omega_1}}$ such that
$(U_{\xi})_m\in{\V}$ for every $\xi\in{X}$.
Pick ordinals $\xi_1<\xi_2<\dots<\xi_{m}$ in $X$. Since
$\bar\U$ is OK
for $\{V_n \colon n<\omega\}$ we have that
$\bigcap_{i=1}^{m}U_{\xi_i}\subseteq^*{V_{m}}
\subseteq\omega\times\omega\setminus L_m$.
Therefore,
$|\bigcap_{i=1}^{m}U_{\xi_i}\cap{L_m}|<\omega$.
But also,
$(\bigcap_{i=1}^{m}U_{\xi_i})_m=\bigcap_{i=1}^{m}(U_{\xi_i})_m\in{\V}$.
This implies that $|(\bigcap_{i=1}^{m}U_{\xi_i})\cap{L_m}|=\omega$,
which is a contradiction.
\qed

Given $f,g\in{\omega^{\omega}}$ we write $g\leq^*{f}$ provided
that $g(n)\leq{f(n)}$ for all but finitely many $n<\omega$.
We say that an $F\subseteq{\omega^{\omega}}$ is \emph{dominating}\/
provided that for every $g\in{\omega^{\omega}}$ there exists an
$f\in{F}$ such that $g\leq^*{f}$.
The \emph{dominating number} $\mathfrak{d}$ is defined as
the minimum cardinality of a dominating family in $\omega^{\omega}$.
This and other cardinal invariants have been studied extensively
in the literature. See for example \cite{Blass} or \cite{BartJudah}.
It is easy to show that $\omega_1\leq{\mathfrak{d}}\leq{\continuum}$
and that this is all that can be said in ZFC about the value of $\mathfrak{d}$.
For instance, the continuum hipothesis implies that
$\mathfrak{d}=\omega_1=\continuum$, while Martin's Axiom + $\continuum>\omega_1$
imply that $\mathfrak{d}=\continuum>\omega_1$. See, for example \cite{Jech}.

In \cite[sec.~1.3]{Book} Ciesielski and Pawlikowski proved that
a weak version of \psmPrGame, called \psmC, implies that
$\rm{cof}(\mathcal{N})=\omega_1<\continuum$.\footnote{$\rm{cof}(\mathcal{N})=
\min\{|\A| \colon \A\subseteq{\mathcal{N}}\;\forall{X\in{\mathcal{N}}}\;
\exists{Y\in{\A}}\;(X\subseteq{Y})\}$, where $\mathcal{N}$ is the null ideal
on~$\Cantor$.}
It is known that this fact implies that $\mathfrak{d}=\omega_1$.

It is not difficult to prove that
$\mathfrak{d}=\omega_1$
implies that for every countable infinite set $X$
there is an $F\subseteq{([\omega_1]^{<\omega})^X}$ of cardinality
$\omega_1$ which is $\subseteq$-dominant, that is, such that
\begin{center}
for every $g\in ([\omega_1]^{<\omega})^X$ there is an $f\in F$ with
$g(x)\subseteq f(x)$ for all $x\in X$.

\end{center}
This follows from the fact that
$([\omega_1]^{<\omega})^X=\bigcup_{\alpha<\omega_1}([\alpha]^{<\omega})^X$.
This is the form of $\mathfrak{d}=\omega_1$
which we will use in the next proposition.

\prop{prop:prop9}{Assume $\mathfrak{d}=\omega_1$ and
let $X$ and $Y$ be countably infinte sets.
If $\U$ and $\V$ are $\omega_1$-generated ultrafilters on $X$ and $Y$,
respectively,
then their Fubini product $\U\otimes{\V}$ is also $\omega_1$-generated.}

\proof
Let $\{U_{\alpha} \colon \alpha<\omega_1\}$ and
$\{V_{\beta} \colon \beta<\omega_1\}$ be the bases for $\U$
and $\V$, respectively.
Since $\mathfrak{d}=\omega_1$, there exists a $\subseteq$-dominant family
$\la{f_{\gamma} \colon
\gamma<\omega_1}\ra\subseteq{([\omega_1]^{<\omega})^X}$.
We claim that the family
$\{W_{\alpha,\gamma} \colon \la{\alpha,\gamma}\ra\in{\omega_1\times{\omega_1}}\}
\subseteq{\U\otimes{\V}}$,
where
$W_{\alpha,\gamma}=\bigcup{\{\{x\}\times{\bigcap_{\beta\in{f_{\gamma}(x)}}{V
_{\beta}}}
\colon x\in{U_{\alpha}}\}}$, is a basis for $\U\otimes{\V}$.
To check this, pick an $A\in{\U\otimes{\V}}$.
Then, $\{x\in{X} \colon \{y \colon \la x,y\ra\in{A}\}\in{\V}\}\in{\U}$.
Pick an $\alpha<\omega_1$ such that
$U_{\alpha}\subseteq{\{x\in{X} \colon \{y \colon \la x,y\ra\in{A}\}\in{\V}\}}$.
Then, given an $x\in{U_{\alpha}}$ there exists a $\beta_x<\omega_1$
such that $V_{\beta_x}\subseteq{\{y\in{Y} \colon \la x,y\ra\in{A}\}}$.
This implies that $\{x\}\times{V_{\beta_x}}\subseteq{A}$
for every $x\in{U_{\alpha}}$.

Consider the function $g \colon X \to [\omega_1]^{<\omega}$
defined as
$$
g(x)=
\begin{cases}
\{\beta_x\} & \text{ \rm if } x\in{U_{\alpha}}\\
\emptyset   & \text{ \rm otherwise. }
\end{cases}
$$
Since $\la{f_{\gamma} \colon \gamma<\omega_1}\ra$ is a $\subseteq$-dominant
family,
there exists a $\gamma<\omega_1$ such that $g(x)\subseteq{f_{\gamma}(x)}$
for every $x\in{U_{\alpha}}$.
This implies that $\beta_x\in{f_{\gamma}(x)}$
and that $\{x\}\times{\bigcap_{\beta\in{f_{\gamma}(x)}}V_{\beta}}\subseteq{A}$
for every $x\in{U_{\alpha}}$. Hence, $W_{\alpha,\gamma}\subseteq{A}$.
\qed

\thm{thm:thm1}{\psmPrGame \; implies that there exists an
$\omega_1$-generated crowded ultrafilter which is not a $P$-point,
a $Q$-point, or even an $\omega_1$-OK point.}

\proof \psmPrGame \; implies the existence of an $\omega_1$-generated
crowded ultrafilter $\U$ on $\Q$, see \cite[prop.~4.25]{Crowded}.
We will show that $\U\otimes{\U}$ is as desired.

By Proposition~\ref{prop:prop8}, it is not a $P$-point,
a $Q$-point, or an $\omega_1$-OK point. Also,
since \psmPrGame\ implies $\mathfrak{d}=\omega_1$,
by Proposition~\ref{prop:prop9}  the ultrafilter $\U\otimes{\U}$ is
$\omega_1$-generated by some family $\B$.

To see that $\U\otimes{\U}$ can be treated as crowded, consider $\Q\times{\Q}$
as the product of $\la{\Q,\tau_d}\ra$ and $\la{\Q,\tau_s}\ra$, where
$\tau_d$ is the discrete topology and $\tau_s$ is the standard topology.
Then $\Q\times{\Q}$ is homemorphic to $\Q$.

For $B\in\B$ let
$\bar B={\{x\colon (B)_x\in B\}\in\U\}}$.
Using Fact~\ref{fact:nonscat},  for every $x\in\bar B$ we can
choose a subset $B^x\in\perf(\Q)$ of
$(B)_x$.
Let $B^*=\bigcup\{\{x\}\times B^x\colon x\in\bar B\}$. Then $B^*$ is a perfect
subset of $\Q\times\Q$.
Thus, $\{B^*\colon B\in\B\}$ is a basis of $\U\otimes{\U}$
of cardinality $\omega_1$ formed with perfect subsets of $\Q\times\Q$.
\qed










\section{A crowded $Q$-point which is not an $\omega_1$-OK point.}\label{sec5}



\defi{defi:defi5}{Let $X$ be a countably infinite set and let
$\J\subseteq{\P(X)}$ be an ideal on $X$. If $A, B\in{\J^+}$ we write
$A\preceq^{\J}{B}$ if and only if $A\setminus{B}\in{\J}$.}

\defi{defi:defi6}{Let $X$ be a countably infinite set and let
$\J\subseteq{\P(X)}$ be an ideal on $X$. We
say that $\J$ has the \emph{extension property} provided that for every
$\preceq^{\J}$-decreasing sequence $\la{A_n\in{\J^+} \colon n<\omega}\ra$
there exists an $A\in{\J^+}$ such that $A\preceq^{\J}A_n$
for every $n<\omega$.}

Let $X$ and $\J$ be as above and let $\K=[\omega]^{<\omega}\otimes{\J}$.
We will consider a relation $\sqsubseteq$ defined on $\K^+$ as
$$
A\sqsubseteq{B} \Leftrightarrow
{\rm supp}(A)\subseteq^*{\rm supp}(B) \;\; \&  \;\; (A)_n\preceq^{\J}(B)_n \;\;
\forall n\in{{\rm supp}(A)\cap{{\rm supp}(B)}}.
$$
Note that for $A,B\in{\K^+}$
$$A\subseteq{B} \Implies A\sqsubseteq{B} \Implies A\preceq^{\K}B$$
but none of these implications can be reversed.
Also, it is not difficult to see that the relation $\sqsubseteq$ is not
transitive.
Nevertheless, we say that for $\xi<\omega_1$ a sequence
$\la{U_\eta\in{\K^+}\colon\eta<\xi}\ra$
is $\sqsubseteq$-decreasing provided $U_\eta\sqsubseteq U_\zeta$
for every $\zeta<\eta<\xi$.


\lem{lem:lem5.1}{Let $X$ be a countably infinite set, let
$\J\subseteq{\P(X)}$ be
an ideal on $X$ with the extension property, and let
$\K=[\omega]^{<\omega}\otimes{\J}$.
Then, for every $\xi<\omega_1$ and every
$\sqsubseteq$-decreasing sequence $\la{U_\eta\in{\K^+}\colon\eta<\xi}\ra$
there exists a $C\in{\K^+}$ such that $C\sqsubseteq U_\eta$ for every
$\eta<\xi$.
Moreover, the sequence $\la{U_\eta\colon\eta\leq\xi}\ra$ is
$\sqsubseteq$-decreasing for every $U_\xi\in\P(C)\cap\K^+$.
}

\proof
%Let $\la{U_\eta\in{\K^+}\colon\eta<\xi}\ra$ be a $\sqsubseteq$-decreasing
sequence.
Since the sequence $\la{{\rm supp}(U_\eta) \colon \eta<\xi}\ra$ is
$\subseteq^*$-decreasing, we can find an $S\in[\omega]^\omega$
such that $S\subseteq^*{U_\eta}$ for every $\eta<\xi$.
For each $m\in{S}$ consider the set $I_m=\{\eta<\xi\colon m\in{{\rm
supp}(U_\eta)}\}$.
Then, since $\J$ has the extension property, we can find a $C_m\in{\J^+}$
such that
$C_m\preceq^{\J}(U_\eta)_m$ for every $\eta\in{I_m}$.
Put $C=\bigcup\{\{m\}\times{C_m} \colon m\in{S}\}$.
Then clearly $C\sqsubseteq U_\eta$ for every $\eta<\xi$.
The additional part follows from the fact that
$U\subset C\sqsubseteq V$ implies $U\sqsubseteq V$.
\qed


\lem{lem:lem5.2}{Let $X$,
$\J$, and $\K$ be as above.
Let $\la{U_{\xi}\in{\K^+} \colon \xi<\omega_1 }\ra$ be a
$\sqsubseteq$-decreasing
sequence in $\K^+$ such that for every $g\in{\omega\times{X}}$ there exists
a $\xi<\omega_1$ such that $g\restriction{U_{\xi}}$ is constant.
Then, the family $\{U_{\xi} \colon \xi<\omega_1\}$
forms a base for a nonprincipal ultrafilter on $\omega\times{X}$
which is not an $\omega_1$-OK point.}


\proof We check first that the family $\{U_{\xi} \colon \xi<\omega_1\}$
has $SFIP$. So, choose $\xi_0<\dots<\xi_n<\omega_1$.
Since $U_{\xi_n}\sqsubseteq \cdots \sqsubseteq U_{\xi_1}\sqsubseteq U_{\xi_0}$
we can pick an $m\in{\bigcap_{i\leq{n}}{\rm supp}(U_{\xi_i})}$.
If $I_m=\{\xi<\omega_1 \colon m\in{{\rm supp}(A_{\xi})}\}$,
then $\{\xi_i \colon i\leq{n}\}\subseteq{I_m}$.
Therefore, $(U_{\xi_n})_m\preceq^{\J}\dots\preceq^{\J}{(U_{\xi_0})_m}$.
This implies that $(\bigcap_{i\leq{n}}U_{\xi_i})_m\in{\J^+}$.
In particular, $(\bigcap_{i\leq{n}}U_{\xi_i})_m$ is infinite and so
is $\bigcap_{i\leq{n}}U_{\xi_i}$. Let $\U$ be a filter
generated $\{U_{\xi} \colon \xi<\omega_1\}$.

To see that $\U$ is actually an ultrafilter,
pick any $A\subseteq{\omega\times{X}}$.
Then, there exists a $\xi<\omega_1$ and an $i<2$ such that
$\chi_A\restriction{U_{\xi}}$ is constant equal $i$.
If $i=0$ then $U_{\xi}\subseteq{(\omega\times{X})\setminus{A}}$ and
$(\omega\times{X})\setminus{A}\in{\U}$. If $i=1$ then $U_{\xi}\subseteq{A}$
and $A\in{\U}$.
Therefore, $\U$ is an ultrafilter and $\{U_{\xi} \colon \xi<\omega_1\}$
is a base for $\U$.
Observe that $\U$ is nonprincipal because each set in $\U$ contains
an infinite set~$U_{\xi}$.

To see that $\U$ is not an $\omega_1$-OK point consider a sequence
$\la{V_n\in{\U} \colon n<\omega}\ra$, where $V_n=\bigcup_{i>n}(\{i\}\times{X})$.
Suppose that there exists a $\la{W_{\xi}\in{\U} \colon \xi<\omega_1}\ra$
which is OK for $\la{V_n\in{\U} \colon n<\omega}\ra$.
Since $\{U_{\xi} \colon \xi<\omega_1\}$ is a basis for $\U$, for every
for every $\xi<\omega_1$ there exists a
$U_{\alpha_{\xi}}\subseteq{W_{\xi}}$.
This implies that
$$
\la{U_{\alpha_{\xi}} \colon \xi<\omega_1}\ra \;\; \text{ is OK for }
\;\; \la{V_n\in{\U} \colon n<\omega}\ra.
$$
By the pigeon hole principle, there exist a $T\in{[\omega_1]^{\omega_1}}$
and an $m<\omega$ such that $m={\rm min}({\rm supp}(U_{\alpha_{\xi}}))$ for
every
$\xi\in{T}$. Hence, $T\subseteq{I_m}$.
Pick any ordinals $\alpha_{\xi_0}<\dots<\alpha_{\xi_m}$ in $T$.
Since $\la{U_{\alpha_{\xi}} \colon \xi<\omega_1}\ra$ is
OK for $\la{V_n\in{\U} \colon <\omega}\ra$ we have that
$\bigcap_{i\leq{m}}U_{\alpha_{\xi_i}}\subseteq^*{V_m}$.
Hence, $|(\bigcap_{i\leq{m}}U_{\alpha_{\xi_i}})\cap{(\{m\}\times{X})}|<\omega$
by the definition of $V_m$. On the other hand, $\{\alpha_{\xi_i} \colon
i\leq{m}\}\subseteq{I_m}$.
Therefore,
$(U_{\alpha_{\xi_0}})_m\preceq^{\J}\dots\preceq^{\J}{(U_{\alpha_{\xi_m}})_m}
$.
This implies that $|(\bigcap_{i\leq{m}}U_{\alpha_{\xi_i}})_m|=\omega$.
So, $|(\bigcap_{i\leq{m}}U_{\alpha_{\xi_i}})\cap{(\{m\}\times{X})}|=\omega$.
This contradiction indicates that $\U$ cannot be an $\omega_1$-OK point.
\qed

Let $X$, $\J$, and $\K$ be as before and let $\D\subseteq{\J^+}$ be dense
in the sense that for every $A\in{\J^+}$ there exists a $D\in{\D}$ such
that $D\subseteq{A}$. Then, the family $\D^*\subseteq{\K^+}$ consisting of the
sets of the form $\bigcup\{\{n\}\times{D_n}\colon n\in{I}\}$ is dense in
$\K^+$, where
$I\in{[\omega]^{\omega}}$ and $D_n\in{\D}$ for every $n\in{I}$.
Recall also that $\Part_{\omega\times{X}}$ is the space of all partitions of
$\omega\times{X}$ into finite pieces, as defined in Section 3.


\thm{thm:thm5.4}{Let $X$ be a countably infinite set, let
$\J\subseteq{\P(X)}$ be
an ideal with the extension property, and let $\D\subseteq{\J^+}$ be dense.
If $\J$ is prism-friendly and $Q$-like and $\K=[\omega]^{<\omega}\otimes{\J}$,
then \psmPrGame \; implies that there exists an
$\omega_1$-generated $Q$-point $\U$ on $\omega\times{X}$ which is not an
$\omega_1$-OK point and such that $\U\cap{\D^*}$
is a basis for $\U$.}


\proof
We construct a $\sqsubseteq$-decreasing sequence
$\la{U_{\xi}\in{\K^+\cap{\D^*}} \colon \xi<\omega_1}\ra$
such that:
\begin{itemize}
     \item[\rm (i)] For every $g\in{2^{\omega\times X}}$ there exists a
                           $\xi<\omega_1$ such that $g\restriction{U_{\xi}}$
                           is constant.

   \item[\rm (ii)] For every $z\in{\Part_{\omega\times X}}$ there
      exists a
       $\xi<\omega_1$ such that $|z(k)\cap{U_{\xi}}|\leq{1}$ for
                           every $k\in{\omega}$.
\end{itemize}


\noindent
If this construction is possible, then, by Lemma \ref{lem:lem5.1},
$\{U_{\xi}\in{\D^*} \colon \xi<\omega_1\}$ is a basis for a nonprincipal
ultrafilter $\U$ on $\omega\times X$ which is not an $\omega_1$-OK point.
To see that $\U$ is a $Q$-point pick an arbitrary $z\in{\Part_{\omega\times
X}}$.
Then, by condition (ii), there exists a
$\xi<\omega_1$ such that $|z(k)\cap{U_{\xi}}|\leq{1}$ for every $k<\omega$.
Therefore, $\U$ is
an $\omega_1$-generated $Q$-point.


Let $\Y=2^{\omega\times{X}}\cup{\Part_{\omega\times{X}}}$
and consider it with the
topology $\tau$ formed with all sets $A\subseteq{\Y}$ such that
$A\cap{2^{\omega\times{X}}}$ and $A\cap{\Part_{\omega\times{X}}}$ are open in
$2^{\omega\times{X}}$ and $\Part_{\omega\times{X}}$, respectively.
Then $\la{\Y,\tau}\ra$ is a Polish space.
Note
that, by Lemmas~\ref{lem:Qlike} and~\ref{lem:Fubini},
the ideal $\K$ is $Q$-like and prism-friendly.
For a prism $P$ in $\Y$ and $U\in{\K^+}$
we choose a subprism $Q(U,P)$ of $P$ and $B(U,P)\in\P(U)\cap\D^*$
as follows.
\begin{itemize}
\item If $U\cap 2^{\omega\times{X}}\neq\emptyset$, then we can choose
a subprism $P_0\subseteq 2^{\omega\times{X}}$ of $P$. The choice of $P_0$ is
obvious if
$P$ is a
singleton; otherwise it follows from Proposition~\ref{prop:prop3}.
Then $Q(U,P)$ is a subprism of $P_0$
such that $Q(U,P)$
and $B(U,P)\in\P(U)\cap\K^+$
satisfy condition ($\bullet$) from the definition of the
prism-friendly ideal.

\item If $U\cap 2^{\omega\times{X}}=\emptyset$, then $P$ is
a prism in $\Part_{\omega\times{X}}$.
Then, by Lemma~\ref{lem:lem3},
there exist
a subprism $Q(U,P)$ of $P$ and a $B(U,P)\in{\P(U)\cap{\K^+}}$ such that
$|z(k)\cap B(U,P)|\leq{1}$ for every $z\in Q(U,P)$ and $k<\omega$.
\end{itemize}
We can also assume that $B(U,P)\in\D^*$, since $\D^*$ is dense in $\K^+$.

Also, for $\xi<\omega_1$ and a $\sqsubseteq$-decreasing sequence
$\la{U_{\eta}\in{\K^+} \colon \eta<\xi}\ra$
let $C_\xi=C(\la{U_{\eta} \colon \eta<\xi}\ra)$ be such that
$C_\xi\sqsubseteq U_\eta$ for every $\eta<\xi$.
Its existence follows from Lemma~\ref{lem:lem5.1}.
Consider the following strategy
$S$ for Player~II:
\[
S(\la{\la{P_{\eta},Q_{\eta}}\ra \colon \eta<\xi}\ra,P_{\xi})=
Q(C(\la{U_{\eta} \colon \eta<\xi}\ra),P_{\xi}),
\]
where the $U_\eta$'s are defined inductively by
$U_\eta=B(C(\la{U_{\zeta} \colon \zeta<\eta}\ra),P_\eta)$.

By \psmPrGame,  the strategy $S$ is not a winning strategy for Player II.
So, there exists a game $\la{\la{P_{\xi},Q_{\xi}}\ra \colon \xi<\omega_1}\ra$
played according to $S$
in which Player~II loses. Thus,
$\Y=\bigcup_{\xi<\omega_1}Q_{\xi}$.
Let $\la{U_{\xi}\in{\D^*} \colon \xi<\omega_1}\ra\subseteq{\K^+}$
be the sequence created in this game.
This sequence is $\sqsubseteq$-decreasing by construction
and Lemma~\ref{lem:lem5.1}.
By the observations made before we only need to check that
$\la{U_{\xi} \colon \xi<\omega_1}\ra$
satisfy conditions (i) and (ii).

If $g\in{2^{\omega\times{X}}}$, then there exists a $\xi<\omega_1$ such
that $g\in{Q_{\xi}}$.  So, $Q_{\xi}\subseteq{2^{\omega\times{X}}}$ and,
by the construction, $g\restriction{U_\xi}$ is constant.
This proves (i).
Similarly, if $z\in{\Part_{\omega\times{X}}}$, then there exists a
$\xi<\omega_1$
such that $z\in{Q_{\xi}}$. Hence, $Q_{\xi}\subseteq{\Part_{\omega\times{X}}}$
and, by the construction, $|z(k)\cap{U_{\xi}}|\leq{1}$ for every $k<\omega$.
This proves (ii).
\qed

\cor{cor:cor5.5}{\psmPrGame \; implies that there exists an
$\omega_1$-generated crowded $Q$-point which is not an $\omega_1$-OK point.}

\proof
Consider $X=\omega\times{\Q}$ with the product topology,
where $\omega$ has the discrete topology and $\Q$ has the subspace
topology inherited from $\real$.
Then $X$ is homeomorphic to $\Q$.
We will find an ideal $\J\subseteq{\P(X)}$ to which
we will apply Theorem~\ref{thm:thm5.4}.

Let $\J=[\omega]^{<\omega}\otimes{\I_S}$. It is clear that $\J$
contains all singletons. Also, $\J$ is prism-friendly
by Lemmas \ref{lem:lem4} and \ref{lem:Fubini} and $Q$-like
by Lemmas \ref{lem:lem4} and~\ref{lem:Qlike}.
To see that $\J$ has the extension property pick a $\preceq^{\J}$-decreasing
sequence $\la{A_n\in{\J^+} \colon n<\omega}\ra$.
By induction construct an increasing sequence
\mbox{$\la n_k\colon k<\omega\ra$} such that
$n_k\in{\rm supp}(A_k)\setminus {\rm supp}\left(\bigcup_{i<k}(A_k\setminus
A_i)\right)$.
The choice can be made, since the set ${\rm
supp}\left(\bigcup_{i<k}(A_k\setminus A_i)\right)$
is finite, as $\bigcup_{i<k}(A_k\setminus A_i)\in\J$.
The choice of $n_k$ gives also $\left(\bigcup_{i<k}(A_k\setminus
A_i)\right)_{n_k}\in \I_S$.
Thus, $\left(\bigcap_{i\leq{k}} A_i\right)_{n_k}\notin\I_S$.
Put
$B=\bigcup\left\{\{n_k\}\times \left(\bigcap_{i\leq{k}}
A_i\right)_{n_k}\colon k<\omega\right\}$.
Then $B\in{\J^+}$ and $B\preceq^{\J}A_n$ for every $n<\omega$.

Since $\bar\D=\perf(\Q)$ is dense in $(\I_S)^+$, the family
$\D=\bar\D^*$ is dense in $\J^+$.
Applying Theorem~\ref{thm:thm5.4} to $\J$ and $\D$,
we can find an $\omega_1$-generated $Q$-point
$\U$ on $\omega\times X$ which is not an $\omega_1$-OK
point and such that $\U\cap\D^*$ contains a basis
for $\U$. Since $\omega\times X$ is homeomorphic to $\Q$
and $\D^*$ consists of perfect set in $\omega\times X$,
it follows that $\U$ is crowded. \qed














\section{Crowded $\omega_1$-generated $\omega_1$-OK points which are not
$P$-points.}

In this section
we prove that the axiom \psmPrGame\; implies the existence of an $\omega_1$-OK
point which is not a $P$-point. For this, we follow the schema used in
\cite{Hart}
for the construction of such an ultrafilter in the model of ZFC obtained by
adding Sacks reals side-by-side. Since that proof uses CH
in the ground model, we have to
modify things
a bit to make it work in the context of \psmPrGame. One possiblity for avoiding
the use of CH is to replace it with some weaker principle consistent with
\psmPrGame\;
like, for instance, $\mathfrak{d}=\omega_1$.
Let $\Gamma$ denote the set of all nonzero limit ordinals
below $\omega_1$.
The following fact is a simple generalization of the remark above
Proposition~\ref{prop:prop9}.

{\fact{{\rm ($\mathfrak{d}=\omega_1$)}\label{fact61}
There exist
a sequence $\la g_\delta\colon\delta<\omega_1\ra$ of functions from
$\omega$ into $[\omega_1]^{<\omega}$
and a partition $\{S_{\delta}\in{[\omega_1]^{\omega_1}} \colon
\delta<\omega_1\}$
of $\Gamma$ such that:
\begin{itemize}
\item%[\rm (F1)]
For every $h\colon\omega\to\omega_1$ there is a $\delta<\omega_1$
               such that $h(n)\in{g_{\delta}(n)}$ for every $n<\omega$.
\item%[\rm (F2)]
$\bigcup{{\rm rang}(g_{\delta})}={{\rm min}(S_{\delta})}$
            for every $\delta<\omega_1$.
\end{itemize}}}

Fix a countably infinite set $X$ and
put $\P=\{\{m\}\times{X} \colon m<\omega\}$.
Then, $\P$ is a partition
of $\omega\times{X}$ into infinitely many infinite pieces.
The idea of the proof is to find a sequence
$\la{U_{\alpha} \colon\alpha<\omega_1}\ra$
that forms a base for a nonprincipal ultrafilter
$\U$ on $\omega\times X$ such that every $U_{\alpha}$ has infinite
intersection with
infinitely many members of $\P$ and, for each $\delta<\omega_1$,
$$
\la{U_{\alpha} \colon \alpha\in{S_{\delta}}}\ra\;\mbox{ is OK for }\;
\left\{\bigcap_{\eta\in{g_{\delta}(n)}}U_{\eta}\colon  n<\omega\right\}.
$$

To see that such an $\U$ is an $\omega_1$-OK point pick
$\la{V_n \colon n<\omega }\ra\in{(\U)^{\omega}}$.
Since the sequence $\la{U_{\alpha} \colon \alpha<\omega_1}\ra$ is a basis
for $\U$, for every $n<\omega$
there is a $\xi_n<\omega_1$ such that $U_{\xi_n}\subseteq{V_n}$.
Therefore, there exists a $\delta<\omega_1$ such that $\xi_n\in{g_{\delta}(n)}$
for every $n<\omega$. Then $\la{U_{\alpha} \colon \alpha\in{S_{\delta}}}\ra$
is OK for
$\la{V_n \colon n<\omega}\ra$ since for any sequence
$\alpha_0<\dots<\alpha_{n}$ of
elements in $S_{\delta}$ we have:
$$\bigcap_{i\leq{n}}U_{\alpha_i}\subseteq^*\bigcap_{\eta\in{g_{\delta}(n)}}
U_{\eta}\subseteq{U_{\xi_n}}\subseteq{V_n}.$$
Observe that $\U$ cannot be a $P$-point because each $U_{\alpha}$ intersects
infinitely many members of $\P$ on an infinite set.

Let us start with fixing  a rich ideal
$\J\subseteq{\P(X)}$ and a dense
$\D\subseteq{\J^+}$.
We will consider the ideal $\K=[\omega]^{<\omega}\otimes{\J}$ on
$\omega\times{X}$
and the set $\D^*\subseteq{\K^+}$ as defined in Section~\ref{sec5}.
We also fix, for each $\xi\in{\Gamma}$,
an enumeration $\{\xi_i \colon i<\omega\}$ of $\xi$.

Let $\T$ be the set of triples $\la{I,f,B}\ra$ satisfying the following
requirements:
\begin{itemize}

\item{$I$ is an infinite subset of $\omega$,}

\item{$f\in{\prod_{m<\omega}(\P(X)\cap{\D})}$, and}

\item{$B\in{\prod_{m\in{\omega}}(\P(f(m))\cap{\D})^{\omega_1}}$
such that every $B(m)$ is a sequence of almost disjoint sets.}

\end{itemize}
If $\xi\leq{\omega_1}$ and
$\la{\la{I_{\eta},f_{\eta},B_{\eta}}\ra\in{\T} \colon \eta<\xi}\ra$, the
sequence
$\la{U_{\eta} \colon \eta<\xi}\ra$ associated with it is defined by
$$
U_{\eta}=\bigcup{\{\{m\}\times f_{\eta}(m) \colon m\in{I_{\eta}}\}}.
$$
Note that each $U_{\eta}$ is in $\D^*$.

To prove that the
resulting ultrafilter $\U$ in our construction is in fact an $\omega_1$-OK point
we will consider for every $\delta<\omega_1$, $\eta<\xi$, and $m<\omega$
the sets
$K(\eta,m)=\{\zeta<\eta \colon
f_{\eta}(m)\subseteq f_{\zeta}(m)\}$,
the numbers $k_{\delta}(\eta,m)=|K(\eta,m)\cap{S_{\delta}}|$, and the functions
$l_{\delta}$ defined by:
$$
l_{\delta}(\eta,m)=
\begin{cases}
\infty &\text{if } \bigcup{\text{rang}(g_{\delta})}\subseteq{K(\eta,m)}\\
-1 &\text{if } g_{\delta}(0)\not\subseteq{K(\eta,m)}\\
\text{max}\{l<\omega \colon \bigcup{g_{\delta}[l+1]}\subseteq{K(\eta,m)}\}
&\quad\quad
\text{ otherwise}.
\end{cases}
$$

\defi{defi:defi4}{For
$\xi\leq\omega_1$ a sequence
$\la{\la{I_{\eta},f_{\eta},B_{\eta}}\ra\in{\T} \colon \eta<\xi}\ra$
is \emph{good} if:
\begin{itemize}

\item[\rm (a)] For every $\zeta<\eta<\xi$ and $m<\omega$, either
$f_{\zeta}(m)\cap{f_{\eta}(m)}$ is finite, or there
exists a $\gamma\leq{\eta}$ such that
$f_{\eta}(m)\subseteq B_{\zeta}(m)(\gamma)\subset f_\zeta(m)$.


\item[\rm (b)] For every $0<\eta<\xi$ and $m<\omega$ there exists a $\zeta<\eta$
such that
$f_{\eta}(m)\subseteq B_{\zeta}(m)(\eta)$.

\item[\rm (c)] If $\eta<\xi$ is limit and $\{m_i \colon i<\omega\}$ is the
increasing enumeration of $I_{\eta}$, then
$$
m_i\in{\bigcap_{j\leq{i}}I_{\eta_j}} \qquad \text{and}
\qquad f_{\eta}(m_i)\subseteq{\bigcap_{j\leq{i}}{f_{\eta_j}(m_i)}},
$$
where $\{\eta_j \colon j<\omega\}$ is our fixed enumeration of $\eta$.

\item[\rm (d)] $f_{\eta}(m)=B_0(m)(\eta)$ for every
$m\in{\omega\setminus{I_{\eta}}}$.

\item[\rm (e)] If $\eta+1<\xi$, then $I_{\eta+1}\subseteq{I_{\eta}}$ and
$f_{\eta+1}(m)\subseteq{B_{\eta}(m)(\eta+1)}\subseteq{f_{\eta}(m)}$ for every
$m\in{I_{\eta+1}}$.

\item[\rm (f)] If $\delta<\omega_1$, $\eta<\xi$, and $\eta\in{S_{\delta}}$, then
$l_{\delta}(\eta,m)>k_{\delta}(\eta,m)$ for every $m\in{I_{\eta}}$.

\item[\rm (g)] If $\delta<\omega_1$,
$\eta<\xi$, and
$\bigcup{\text{rang}(g_{\delta})}\subseteq{\eta}$, then
$$
\lim_{\substack{m\in{I_{\eta}}\\m\to{\infty}}}(l_{\delta}(\eta,m)-k_{\delta}
(\eta,m))=\infty.
$$
\end{itemize}
}

\rem{rem:rem3}{It follows from %conditions
(c) and
(e) that if $\zeta<\eta<\xi$,
then $I_{\eta}\subseteq^*{I_{\zeta}}$.}

\rem{rem:rem4}{It is also easy to check that if $\xi\leq{\omega_1}$ is a
limit ordinal, then the sequence
$\la{\la{I_{\zeta},f_{\zeta},B_{\zeta}}\ra\in{\T} \colon \zeta<\xi}\ra$
is good if and only if the sequence
$\la{\la{I_{\zeta},f_{\zeta},B_{\zeta}}\ra\in{\T} \colon\zeta<\eta}\ra$
is good for every $\eta<\xi$.}


\rem{rem:rem5}{It is not difficult to see that
%conditions (c) and (e) imply that
\begin{equation}
\text{if $\alpha<\beta<\xi$, then $f_{\beta}(m)\subseteq{f_{\alpha}(m)}$ for
all but
finitely many $m\in{I_{\beta}}$.}
\end{equation}
If $\beta\in{\Gamma}$ this follows from (c).
If $\Gamma\cap{(\alpha,\beta]}=\emptyset$, then it follows from (e).
If $\Gamma\cap{(\alpha,\beta]}\neq\emptyset$ then there exist a maximal
$\gamma\in{\Gamma\cap{(\alpha,\beta]}}$ and, by the above two cases,
$f_{\beta}(m)\subseteq{f_{\gamma}(m)}\subseteq{f_{\alpha}(m)}$ for all but
finitely many $m\in{I_{\beta}}$.}

\rem{rem:rem6}{For
every $0<\eta<\xi$ and $m<\omega$
there exists a $\gamma\leq{\eta}$ such that
$f_{\eta}(m)\subseteq {B_0(m)(\gamma)}$.
This follows from condition (b), since
every strictly decreasing
sequence of ordinals is finite.
%It is worth to note that, since  every strictly decreasing
%sequence of ordinals is finite, if $0<\eta<\xi$ and $m<\omega$, then we can
%apply condition (b) finitely many times to get a $\gamma\leq{\eta}$ such that
%$f_{\eta}(m)\subseteq^*{B_0(m)(\gamma)}$.
}

The dual filter of an ideal $\K$ on
a set $\omega\times{X}$ is the family $\F_{\K}$
defined as $\F_{\K}=\{(\omega\times{X})\setminus{A} \colon A\in{\K}\}$.
The importance of the definition of a good sequence derives from the
following lemma.

\lem{lem:lem11}{Let $X$ be a countably infinite set, $\J\subseteq{\P(X)}$ a
rich ideal in $X$, $\D\subseteq{\J^+}$ a dense family, and
let $\K=[\omega]^{<\omega}\otimes{\J}$.
If $\la{\la{I_{\xi},f_{\xi},B_{\xi}}\ra\in{\T} \colon \xi<\omega_1}\ra$
is a good sequence such that for every $g\in{2^{\omega\times{X}}}$ there
exists a
$\xi<\omega_1$ such
that $g\restriction{U_{\xi}}$ is constant, then
$\la{U_{\xi}\in\D^* \colon\xi<\omega_1}\ra$ forms a
base for a nonprincipal ultrafilter on $\omega\times X$ extending $\F_{\K}$
which is an $\omega_1$-OK point but not a $P$-point.}

\proof The fact that $\{U_{\xi} \colon \xi<\omega_1\}\subseteq{\D^*}$
follows immediately from the definition of $U_{\eta}$ and $\D^*$.

Next we prove that $\{U_{\xi} \colon \xi<\omega_1\}$
forms a base for a nonprincipal ultrafilter $\U$ on $\omega\times X$
extending the filter $\F_{\K}$.
Given $\xi_0<\dots<\xi_{n}$
pick a $\gamma\in\Gamma$ with $\gamma>\xi_n$.
By (c), we have that
almost every $m\in{I_{\gamma}}$ is in
$\bigcap_{i\leq{n}}I_{\xi_i}$ and that
$f_{\gamma}(m)\subseteq{\bigcap_{i\leq{n}}f_{\xi_i}(m)}$. Therefore,
$\bigcap_{i\leq{n}}U_{\xi_i}\in{\K^+}$ and
$\{U_{\xi} \colon\xi<\omega_1\}$
can  be extended to a proper filter $\U$ on $\omega\times X$.
If $A\subseteq{\omega\times X}$, then
$\chi_A\in{2^{\omega\times X}}$
and there exist a $\xi<\omega_1$ and an $i<2$ such that
$\chi_A\restriction{U_{\xi}}$
is constant equal $i$.
If $i=1$ then $U_{\xi}\subseteq{A}$ and $A\in{\U}$. If $i=0$ then
$U_{\xi}\subseteq{(\omega\times{X})\setminus{A}}$ so
$(\omega\times{X})\setminus{A}\in{\U}$.
This proves that $\U$ is an
ultrafilter and that $\{U_{\xi} \colon \xi<\omega_1\}$ is a base for $\U$.
Since no $A\in{\K}$ contains any $U_{\xi}$, it follows that $\U$ extends
$\F_{\K}$.
In particular, $\U$ is nonprincipal.

To see that $\U$ is not a $P$-point notice that every $U_{\xi}$ intersects
infinitely
many pieces of the partition
$\P=\{\{m\}\times{X} \colon m<\omega\}$ on an infinite set and so does every
$V\in{\U}$.

To prove that $\U$ is an $\omega_1$-OK point
it is enough to prove that for every $\delta<\omega_1$
\begin{equation}\label{E:3}
\la{U_{\alpha} \colon \alpha\in{S_{\delta}}}\ra\;\mbox{ is OK for }\;
\left\{\bigcap_{\eta\in{g_{\delta}(n)}}U_{\eta}\colon n<\omega\right\}.
\end{equation}
Pick $\delta<\omega_1$ and $\xi_0<\dots<\xi_{n}$ in $S_{\delta}$.
First, we prove that for every $m\in{I_{\xi_{n}}}$
\begin{equation}\label{E:55}
\mbox{either }\ \  \bigcap_{i\leq{n}}f_{\xi_i}(m) \ \ \mbox{ is finite, or }\ \
\bigcap_{i\leq{n}}f_{\xi_i}(m)\subseteq\bigcap_{\eta\in{g_{\delta}(n)}}
f_{\eta}(m).
\end{equation}
Indeed, assume that $\bigcap_{i\leq{n}}f_{\xi_i}(m)$ is infinite. Then, by
part (a) of
Definition~\ref{defi:defi4},
$\xi_i\in{K(\xi_{n},m)\cap{S_{\delta}}}$ for each $i\leq{n-1}$.
Therefore, $k_{\delta}(\xi_{n},m)\geq{n}$ and
$l_{\delta}(\xi_{n},m)\geq{n+1}$ by Definition~\ref{defi:defi4}(f).
In particular, $g_{\delta}(n)\subseteq{K(\xi_{n},m)}$.
Hence, by the definition of $K(\eta,m)$,
$$
\bigcap_{i\leq{n}}f_{\xi_i}(m)\subseteq{f_{\xi_{n}}(m)}
\subseteq \bigcap_{\eta\in K(\xi_{n},m)} f_{\eta}(m)
\subseteq{\bigcap_{\eta\in{g_{\delta}(n)}}f_{\eta}(m)}.
$$
Also,
Definition~\ref{defi:defi4}(c)
implies that
$f_{\xi_{n}}(m)\subseteq{\bigcap_{\eta\in{g_{\delta}(n)}}f_{\eta}(m)}$
for all but finitely many $m\in{I_{\xi_{n}}}$.
Thus, the set
$s=\left\{m\in I_{\xi_n}\colon
\bigcap_{i\leq{n}}f_{\xi_i}(m)\not\subseteq\bigcap_{\eta\in g_{\delta}(n)}
f_{\eta}(m)\right\}$
is finite. Moreover, by (\ref{E:55}),
$\bigcap_{i\leq{n}}f_{\xi_i}(m)$ is finite for every $m\in s$.
So,
\begin{eqnarray*}
\bigcap_{i\leq{n}}{U_{\xi_i}}
& = &
\bigcup_{m\in{\bigcap_{i\leq{n}}{I_{\xi_i}}}}\left({\{m\}\times\bigcap_{i\le
q{n}}
            {f_{\xi_i}(m)}}\right)\\
& \subseteq &
\bigcup_{m\in I_{\xi_n} }\left({\{m\}\times\bigcap_{i\leq{n}}
            {f_{\xi_i}(m)}}\right)\\
%{\bigcup\left\{\{m\}\times\bigcap_{i\leq{n}}f_{\xi_i}(m)
%                 \colon m\in{I_{\xi_n}}\right\}}\\
& = &
\bigcup_{m\in s}
  \left(\{m\}\times \bigcap_{i\leq{n}} f_{\xi_i}(m)\right) \cup
\bigcup_{m\in I_{\xi_n}\setminus s}\left(\{m\}\times \bigcap_{i\leq{n}}
f_{\xi_i}(m)\right)\\
& \subseteq^* &
\bigcup_{m\in I_{\xi_n} \setminus s}\left({\{m\}\times\bigcap_{\eta\in
g_\delta(n)}
            {f_{\xi_i}(m)}}\right)\\
& \subseteq &
\bigcup_{m\in I_{\xi_n}}\left({\{m\}\times\bigcap_{\eta\in g_\delta(n)}
            {f_{\xi_i}(m)}}\right)=
\bigcup_{\eta\in g_\delta(n)} U_{\eta},\\
\end{eqnarray*}
which proves \eqref{E:3}. So $\U$ is an $\omega_1$-OK point.
\qed

\lem{lem:lem12}{Let $\la{\la{I_{\eta},f_{\eta},B_{\eta}}\ra\in{\T} \colon
\eta<\xi}\ra$
be a sequence satisfying condition (a) from Definition~\ref{defi:defi4} and let
$\alpha<\beta<\xi$
and $m<\omega$ be such that
$f_{\beta}(m)\subseteq{B_{\alpha}(m)(\beta)}$.
Then $K(\beta,m)=K(\alpha,m)\cup{\{\alpha\}}$. In particular, if the
sequence satisfies conditions (a) and (b) from Definition~\ref{defi:defi4}, then
the set $K(\eta,m)$ is finite for every $\eta<\xi$ and $m<\omega$.}

\proof If $\eta\in{K(\beta,m)}$, then $\eta<\beta$ and
$f_{\beta}(m)\subseteq{f_{\eta}(m)}$. Also, since
$f_{\beta}(m)\subseteq{B_{\alpha}(m)(\beta)}\subseteq{f_{\alpha}(m)}$
we have that $|f_{\eta}(m)\cap{f_{\alpha}(m)}|=\omega$.
If $\alpha<\eta$, then, by condition (a), there exists a $\gamma\leq{\eta}$
such that
$f_{\eta}(m)\subseteq{B_{\alpha}(m)(\gamma)}$; therefore
$f_{\eta}(m)\subseteq{B_{\alpha}(m)(\gamma)\cap{B_{\alpha}(m)(\beta)}}$,
which is impossible. Thus, $\eta\leq{\alpha}$. If $\eta<\alpha$, then, again
by (a), there is a $\gamma\leq{\alpha}$ such that
$f_{\alpha}(m)\subseteq{B_{\eta}(m)(\gamma)}$. Since
$B_{\eta}(m)(\gamma)\subseteq{f_{\eta}(m)}$,
we conclude that $f_{\alpha}(m)\subseteq{f_{\eta}(m)}$. Therefore,
$\eta\in{K(\alpha,m)}$ and
this proves that $K(\beta,m)\subseteq{K(\alpha,m)\cup{\{\alpha\}}}$.

Since $f_{\beta}(m)\subseteq{B_{\alpha}(m)(\beta)}\subseteq{f_{\alpha}(m)}$
we have that
$\alpha\in{K(\beta,m)}$. If $\eta\in{K(\alpha,m)}$, then
$f_{\alpha}(m)\subseteq{f_{\eta}(m)}$. But since
$f_{\beta}(m)\subseteq{B_{\alpha}(m)(\beta)}\subseteq{f_{\alpha}(m)}$
we have that $f_{\beta}(m)\subseteq{f_{\eta}(m)}$. Therefore,
$\eta\in{K(\beta,m)}$ and this
proves that $K(\alpha,m)\cup{\{\alpha\}}\subseteq{K(\beta,m)}$.
Thus, $K(\beta,m)=K(\alpha,m)\cup{\{\alpha\}}$.

Since condition (b) implies that for every $0<\eta<\xi$ and  $m<\omega$ there
exists a $\zeta<\eta$ such that $f_{\eta}(m)\subseteq{B_{\zeta}(m)(\eta)}$,
we have that for every $0<\eta<\xi$ there exists a $\zeta<\eta$ such that
$K(\eta,m)=K(\zeta,m)\cup{\{\zeta\}}$.
Since $K(0,m)=\emptyset$ for every $m<\omega$, we can prove, by induction on
$\eta$, that
$K(\eta,m)$ is finite for every $\eta<\xi$ and $m<\omega$.
\qed

\lem{lem:lem13}{If $\xi\in{\Gamma}$ and
$\la{\la{I_{\eta},f_{\eta},B_{\eta}}\ra\in{\T} \colon \eta<\xi}\ra$
is good, then there exists an
$\la{I_{\xi},f_{\xi},B_{\xi}}\ra\in{\T}$
such that the sequence $\la{\la{I_{\eta},f_{\eta},B_{\eta}}\ra\in{\T} \colon
\eta\leq{\xi}}\ra$ is good.}

\proof
Let $\{\xi_j \colon j<\omega\}$ be the fixed enumeration of $\xi$.
Since
$S_\delta$'s are pairwise disjoint,
the set $\{\delta<\omega_1 \colon \text{min}(S_{\delta})<\xi\}$ is
countable
and it can be enumerated as $\{\delta_i \colon i<\omega\}$.
Let $\delta^*<\omega_1$ be such that $\xi\in{S_{\delta^*}}$.
We  define $I_{\xi}=\{m_i \colon i<\omega\}$ inductively.
Suppose
that $m_j$ has already been defined for every $j<i$.
Put 
$$\ep_i=\text{max}(g_{\delta^*}(0)\cup{g_{\delta^*}(0)}\cup\{\xi_j\colon j\leq {i}\}
\cup{\{\text{min}(S_{\delta_j}) \colon j\leq{i}\}})+1<\xi.$$
Note that $\ep_i<\xi$ since Remark~\ref{rem:rem3} implies that
$I_{\ep_i}\subseteq^*\bigcap_{j\leq{i}}I_{\xi_j}$.
Thus, we can pick an
$m_i\in{I_{\ep_i}}\cap\bigcap_{j\leq{i}}I_{\xi_j}$
so that:
\begin{itemize}

\item[\rm (i)]{$m_i>m_j$ for every $j<i$, }

\item[\rm (ii)]{$l_{\delta_j}(\ep_i,m_i)-k_{\delta_j}(\ep_i,m_i)>i$ for
every $j\leq{i}$, and }

\item[\rm (iii)]
$f_{\ep_i}(m_i)\subseteq \bigcap_{j\leq i}f_{\xi_j}(m_i)\cap
\bigcap\{f_\eta(m_i)\colon \eta\in g_{\delta^*}(0)\cup g_{\delta^*}(1)\}$.

\end{itemize}
Condition (ii) can be achieved since
$\bigcup\text{rang}(g_{\delta_j})\subseteq{\text{min}(S_{\delta_j})<{\ep_i}<
\xi}$ and
$\{\la{I_{\eta},f_{\eta},B_{\eta}}\ra \colon \eta<\xi\}$ is good, so, by (g),
$$
\lim_{\substack{m\in{I_{\ep_i}}\\m\to{\infty}}}
(l_{\delta_j}(\ep_i,m)-k_{\delta_j}(\ep_i,m))=\infty
$$
for every $j\leq{i}$.
Condition (iii) can be ensured by Remark~\ref{rem:rem5}, since
$\xi_j<\ep_i$ for $j\leq i$ and
$\eta<\ep_i for all \eta\in{g_{\delta^*}(0)\cup g_{\delta^*}(1)}$.
This completes the inductive definition of $I_{\xi}$.
Define $f_{\xi}\colon\omega\to\D$ as
$$
f_{\xi}(m)=
\begin{cases}
B_{\ep_i}(m_i)(\xi)&\text{if } m=m_i\in{I_{\xi}}\\\\
B_0(m)(\xi) &\text{otherwise.}
\end{cases}
$$
The $B_{\xi}$ can be defined by taking for each $m<\omega$ an arbitrary
$\omega_1$-sequence of almost disjoint sets in $\P(f_{\xi}(m))\cap{\D}$.
%
This completes the definition of $\la{I_{\xi},f_{\xi},B_{\xi}}\ra$.

To make sure that (a) holds it is enough to check it only
for the pair $\la\eta,\xi\ra$ in place of $\la\zeta,\eta\ra$.
So, choose an $\eta<\xi$ and $m<\omega$.
We need to show that either
$f_{\xi}(m)\cap{f_{\eta}(m)}$ is finite, or there
exists a $\gamma\leq{\xi}$ such that
$f_{\xi}(m)\subseteq B_{\eta}(m)(\gamma)$.
We will consider several cases.
\begin{description}
\item[$m\notin{I_{\xi}}$:] We will consider here two subcases.
  \begin{itemize}
  \item[$\eta=0$:] Then $f_{\xi}(m)=B_0(m)(\xi)=B_{\eta}(m)(\gamma)$
       for $\gamma=\xi$.

  \item[$\eta>0$:] Apply Remark \ref{rem:rem6} to find
     $\gamma\leq{\eta}$ such that
      $f_{\eta}(m)\subseteq{B_0(m)(\gamma)}$. Since
      $f_{\xi}(m)=B_0(m)(\xi)$,
      we have that $|f_{\xi}(m)\cap{f_{\eta}(m)}|<\omega$.
  \end{itemize}
\item[$m=m_i\in{I_{\xi}}$:] We will consider here three subcases.
\begin{itemize}
\item[$\ep_i<\eta$:] We will show that
      $|f_{\xi}(m)\cap{f_{\eta}(m)}|<\omega$. So, by way of contradiction,
      assume that $|f_{\xi}(m)\cap{f_{\eta}(m)}|=\omega$. Therefore, the sets

$f_{\xi}(m)\cap{f_{\eta}(m)}=B_{\ep_i}(m)(\xi)\cap{f_{\eta}(m)}\subseteq
f_{\ep_i}(m)\cap{f_{\eta}(m)}$
      are infinite.
      So, by (a),
      there exists a $\gamma\leq{\eta}$ such that
     $f_{\eta}(m)\subseteq{B_{\ep_i}(m)(\gamma)}$.
     Thus, $B_{\ep_i}(m)(\xi)\cap{B_{\ep_i}(m)(\gamma)}=
          f_{\xi}(m)\cap{B_{\ep_i}(m)(\gamma)}\supseteq
          f_{\xi}(m)\cap{f_{\eta}(m)}$
     is infinite, which is impossible, as $\gamma\leq{\eta}<\xi$.
     This implies that $|f_{\xi}(m)\cap{f_{\eta}(m)}|<\omega$.

\item[$\ep_i>\eta$:] If $f_{\ep_i}(m)\cap{f_{\eta}(m)}$ is finite then so is
     $f_{\xi}(m)\cap{f_{\eta}(m)}\subseteq f_{\ep_i}(m)\cap{f_{\eta}(m)}$.
     Otherwise, by (a), there is
    a $\gamma\leq{\ep_i}$ such that
    $f_{\ep_i}(m)\subseteq {B_{\eta}(m)(\gamma)}$.
    So, $f_{\xi}(m)=B_{\ep_i}(m)(\xi)\subseteq f_{\ep_i}(m)\subseteq
{B_{\eta}(m)(\gamma)}$.

\item[$\ep_i=\eta$:] Clearly $f_{\xi}(m)=B_{\ep_i}(m)(\xi)=B_{\eta}(m)(\gamma)$
       for $\gamma=\xi$.

\end{itemize}
\end{description}

Conditions
(b), (c), and (d) are immediate from the definition of $f_{\xi}$.
Condition (e) holds because
$\la{\la{I_{\eta},f_{\eta},B_{\eta}}\ra \colon \eta<\xi}\ra$ is good
and $\xi$ is a limit ordinal.

To prove (f) and (g) first observe that, by Lemma~\ref{lem:lem12},
for every $i<\omega$ we have $K(\xi,m_i)=K(\ep_i,m_i)\cup{\{\ep_i\}}$.
This implies that $l_{\delta_j}(\ep_i,m_i)\leq{l_{\delta_j}(\xi,m_i)}$
and $k_{\delta_j}(\xi,m_i)=k_{\delta_j}(\ep_i,m_i)$
for every $j\leq{i}$, because $\ep_i$ is a succesor ordinal.
In particular, for every $j\leq{i}$ we have
\begin{equation}\label{E:66}
l_{\delta_j}(\xi,m_i)-k_{\delta_j}(\xi,m_i)
\geq{l_{\delta_j}(\ep_i,m_i)-k_{\delta_j}(\ep_i,m_i)}.
\end{equation}


To see (f) fix an $m=m_i\in I_\xi$.
We need to show that $l_{\delta^*}(\xi,m)>k_{\delta^*}(\xi,m)$.
First assume that $\xi=\min(S_{\delta^*})$.
Then, $K(\xi,m)\subseteq \xi$ is disjoint with $S_{\delta^*}$, so
$k_{\delta^*}(\xi,m)=|K(\xi,m)\cap S_{\delta^*}|=0$.
On the other hand, condition (iii) implies that
$\bigcup g_{\delta^*}[2]\subseteq K(\xi,m)$.
So, $l_{\delta^*}(\xi,m)\geq 1>0=k_{\delta^*}(\xi,m)$.
Next, consider the case when $\xi>\min(S_{\delta^*})$.
Then,
$\delta^*=\delta_i$ for some $i<\omega$. Therefore, (ii) and (\ref{E:66})
imply that
$$
l_{\delta^*}(\xi,m_i)-k_{\delta^*}(\xi,m_i)
\geq{l_{\delta_i}(\ep_i,m_i)-k_{\delta_i}(\ep_i,m_i)}>i\geq 0.
$$
Thus (f) holds.

To see (g) fix a $\delta<\omega_1$ such that
$\bigcup{\text{rang}(g_{\delta})}\subseteq{\xi}$.
We need to show that
$\lim_{i\to\infty}(l_{\delta}(\xi,m_i)-k_{\delta}(\xi,m_i))=\infty$.
First assume that  $\xi>\text{min}(S_{\delta})$. Then
$\delta=\delta_j$ for some $j<\omega$.
So, by (ii) and (\ref{E:66}), we have that for all $i\geq{j}$
$$l_{\delta}(\xi,m_i)-k_{\delta}(\xi,m_i)=l_{\delta_j}(\xi,m_i)-k_{\delta_j}
(\xi,m_i)
\geq{l_{\delta_j}(\ep_i,m_i)-k_{\delta_j}(\ep_i,m_i)}\geq{i}.$$
This ensures that (g) holds.
Finally, assume that  $\xi\leq{\text{min}(S_{\delta})}$.
Then, for every $m<\omega$, we have
$K(\xi,m)\cap S_\delta=\emptyset$ and so, $k_{\delta}(\xi,m)=0$.
Thus, in this case it is enough to show that
$\lim_{i\to\infty} l_{\delta}(\xi,m_i)=\infty$.
But for every $l<\omega$ we have
$\bigcup g_\delta[l+1]\subseteq \xi=\{\xi_j \colon j<\omega\}$.
Thus, there exists an $i_0<\omega$  such that
$\bigcup g_\delta[l+1]\subseteq \xi=\{\xi_j \colon j\leq i_0\}$.
Since, by (iii), for every $i\geq{i_0}$
we have
$\{\xi_j \colon j\leq i\}\subseteq K(\xi,m_i)$
we conclude that
$\bigcup{g_{\delta}[l+1]}\subseteq{K(\xi,m_i)}$
for every $i\geq{i_0}$.
Thus, $l_{\delta}(\xi,m_i)\geq{l}$ for every $i\geq{i_0}$ and so
$\lim_{i\to\infty} l_{\delta}(\xi,m_i)=\infty$.\qed


\lem{lem:lem14}{Let $\la{\la{I_{\eta},f_{\eta},B_{\eta}}\ra\in{\T} \colon
\eta\leq{\xi}}\ra$ be a good sequence, $I\in{[I_{\xi}]^{\omega}}$,
and let $\la{D_m\in{\P(B_{\xi}(m)(\xi+1))\cap\D} \colon m\in{I}}\ra$ be
arbitrary.
Then, the sequence
$\la{\la{I_{\eta},f_{\eta},B_{\eta}}\ra\in{\T} \colon \eta\leq{\xi+1}}\ra$
is good, where $\la{I_{\xi+1},f_{\xi+1},B_{\xi+1}}\ra\in{\T}$ is defined as
\begin{itemize}
\item[\rm (i)]{$I_{\xi+1}=I$},
\item[\rm (ii)]{$f_{\xi+1}(m)=
\begin{cases}
D_m&\text{if }
m\in{I_{\xi+1}}\\\\
%
B_0(m)(\xi+1) &\text{otherwise},
\end{cases}$}
\item[\rm(iii)]
{$B_{\xi+1}(m)\in{(\P(f_{\xi+1}(m))\cap{\D})^{\omega_1}}$
is any almost disjoint sequence for every $m<\omega$.}
\end{itemize}}

\proof
To show that (a) holds
it is enough to check it only
for the pair $\la\eta,\xi+1\ra$ in place of $\la\zeta,\eta\ra$.
So, choose an $\eta<\xi+1$ and $m<\omega$.
We need to show that either
$f_{\xi+1}(m)\cap{f_{\eta}(m)}$ is finite, or there
exists a $\gamma\leq{\xi+1}$ such that
$f_{\xi+1}(m)\subseteq B_{\eta}(m)(\gamma)$.
We consider several cases.

\newpage

\begin{description}
\item[$m\notin{I_{\xi+1}}$:] We will consider two subcases.
    \begin{itemize}

         \item[$\eta=0$:] Then $f_{\xi+1}(m)=B_0(m)(\xi+1)=B_{\eta}(m)(\gamma)$
                for $\gamma=\xi+1$.

         \item[$\eta>0$:] Apply Remark~\ref{rem:rem6}
              to find a $\gamma\leq{\eta}$ such that
              $f_{\eta}(m)\subseteq^*{B_0(m)(\gamma)}$.
              Since $f_{\xi+1}(m)=B_0(m)(\xi+1)$, we have that
              $|f_{\xi+1}(m)\cap{f_{\eta}(m)}|<\omega$.

    \end{itemize}
\item[$m\in{I_{\xi+1}}$:] We compare $\eta$ with $\xi$.
\begin{itemize}
    \item[$\eta<\xi$:] By (a) either
             $|f_{\xi}(m)\cap{f_{\eta}(m)}|<\omega$ or there exists
              a $\gamma\leq{\xi}$
               such that $f_{\xi}(m)\subseteq^*{B_{\eta}(m)(\gamma)}$. Since

$f_{\xi+1}(m)\subseteq{B_{\xi}(m)(\xi+1)}\subseteq{f_{\xi}(m)}$ we
               have that
               $|f_{\xi+1}(m)\cap{f_{\eta}(m)}|<\omega$ or
               $f_{\xi+1}(m)\subseteq{B_{\eta}(m)(\gamma)}$.

\item[$\eta=\xi$:] Clearly
          $f_{\xi+1}(m)=D_m\subseteq B_{\xi}(m)(\xi+1)=B_{\eta}(m)(\gamma)$
          for $\gamma=\xi+1$.


\end{itemize}
\end{description}
This proves that (a) holds.

Conditions (b), (d), and (e) are obvious by the definition of $f_{\xi+1}$.
Conditions (c) and (f) hold, since there are no new limit ordinals $\eta<\xi+1$.

To see that $(g)$ holds take a $\delta<\omega_1$ such that
$\bigcup\text{rang}(g_\delta)\subseteq \xi+1$.
Then also $\bigcup\text{rang}(g_\delta)\subseteq \xi$
since, by Fact~\ref{fact61},
$\bigcup\text{rang}(g_\delta)=\min(S_\delta)$ is a limit ordinal.
Thus,
$\lim_{\substack{m\in{I_{\xi}}\\m\to{\infty}}}(l_{\delta}(\xi,m)-k_{\delta}(
\xi,m))=\infty$.
Also, by Lemma~\ref{lem:lem12} and the
definition
of $f_{\xi+1}(m)$ we have
$K(\xi+1,m)=K(\xi,m)\cup{\{\xi\}}$ for every $m\in I_{\xi+1}$.
This implies that
$k_{\delta}(\xi+1,m)\leq{k_{\delta}(\xi,m)+1}$  and
$l_{\delta}(\xi,m)\leq{l_{\delta}(\xi+1,m)}$ for every $m\in I_{\xi+1}$.
So,
$l_{\delta}(\xi+1,m)-k_{\delta}(\xi+1,m)
\geq{l_{\delta}(\xi,m)-k_{\delta}(\xi,m)-1}$.
Since $I_{\xi+1}\subseteq{I_{\xi}}$ is infinite, we have
\[
\lim_{\substack{m\in{I_{\xi+1}}\\m\to{\infty}}}
(l_{\delta}(\xi+1,m)-k_{\delta}(\xi+1,m))\geq
\lim_{\substack{m\in{I_{\xi}}\\m\to{\infty}}}(l_{\delta}(\xi,m)-k_{\delta}(\
xi,m)-1)=\infty.
\]
So, $(h)$ holds.
\qed

\cor{cor:cor2}{Let
$\la{\la{I_{\eta},f_{\eta},B_{\eta}}\ra\in{\T} \colon\eta\leq{\xi}}\ra$
be good.
If $P$ is a prism in $2^{\omega\times{X}}$, then there exists an
$\la{I_{\xi+1},f_{\xi+1},B_{\xi+1}}\ra\in{\T}$,
a suprism $Q$ of $P$, and an $i<2$ such that
\begin{itemize}
    \item[\rm (i)]{$\la{\la{I_{\eta},f_{\eta},B_{\eta}}\ra
\colon\eta\leq{\xi+1}}\ra$
                      is good and}
    \item[\rm (ii)]{$g\restriction{U_{\xi+1}}$ is constant equal to $i$ for
every
                       $g\in{Q}$.}
\end{itemize}}

\proof Apply Lemma \ref{lem:Fubini} to the prism $P$, the set $I_{\xi}$, and
the family $\{B_{\xi}(m)(\xi+1) \colon m\in{I_{\xi}}\}$
to find a subprism $Q$ of $P$, a set $I_{\xi+1}\in{[I_{\xi}]^{\omega}}$,
a sequence
$\la{B_m\in{\P(B(m)(\xi+1))\cap{\J^+}}\colon m\in{I_{\xi+1}}}\ra$,
and an $i<2$ such that $g\restriction{B}$ is constant equal to $i$,
where $B=\bigcup\{\{m\}\times{B_m} \colon m\in{I_{\xi+1}}\}$.
For every $m\in{I_{\xi+1}}$ choose $D_m\in\P(B_m)\cap\D$.
Then, if we define $f_{\xi+1}$ and $B_{\xi+1}$ as in Lemma \ref{lem:lem14},
$\la{\la{I_{\eta},f_{\eta},B_{\eta}}\ra \colon\eta\leq{\xi+1}}\ra$
is good and $g\restriction{U_{\xi+1}}$ is constant equal to $i$.
\qed

\cor{cor:cor3}{Let $X$ be a countably infinite set, $\J\subseteq{\P(X)}$
a $Q$-like ideal on $X$,
$\la{\la{I_{\eta},f_{\eta},B_{\eta}}\ra\in{\T}\colon \eta\leq{\xi}}\ra$
be good, and $P$ be a prism on $\Part_{\omega\times{X}}$.
Then, there exists a $\la{I_{\xi+1},f_{\xi+1},B_{\xi+1}}\ra\in{\T}$ and a
subprism $Q$ of $P$ such that
\begin{itemize}
      \item[\rm (i)]{$\la{\la{I_{\eta},f_{\eta},B_{\eta}}\ra \colon
\eta\leq{\xi+1}}\ra$
                     is good and}
      \item[\rm (ii)]{$|z(k)\cap{U_{\xi+1}}|\leq{1}$ for every $z\in{Q}$
and $k<\omega$.}
\end{itemize}}

\proof Let
$A=\bigcup\{\{m\}\times B_{\xi}(m)(\xi+1) \colon m\in{I_{\xi}}\}$.
Then $A\in\K^+$ and, by Lemma~\ref{lem:Qlike}, $\K$ is $Q$-like.
So, by Lemma~\ref{lem:lem3}, there is a subprism $Q$ of $P$
and a $B\in{\P(A)\cap{\K^+}}$ such that
$|z(k)\cap{B}|\leq{1}$ for every $z\in{Q}$ and $k<\omega$.
Let $I_{\xi+1}={\rm supp}(B)\subseteq I_\xi$
and for every $m\in{I_{\xi+1}}$ choose $D_m\in\P((B)_m)\cap\D$.
Then, if we define $f_{\xi+1}$ and $B_{\xi+1}$ as in Lemma \ref{lem:lem14},
$\la{\la{I_{\eta},f_{\eta},B_{\eta}}\ra \colon\eta\leq{\xi+1}}\ra$
is good and $|z(k)\cap{U_{\xi+1}}|\leq{1}$ for every $z\in{Q}$ and $k<\omega$.
\qed

\thm{thm:thm3}{Let $X$ be a countably infinite set,
$\J\subseteq{\P(X)}$ be a rich ideal, let
$\D\subseteq{\J^+}$ be a dense family, and put $\K=[\omega]^{<\omega}\otimes\J$.
Then, \psmPrGame\ implies that there exists an
$\omega_1$-generated $\omega_1$-OK point extending $\F_{\K}$
with a basis $\{U_{\xi} \colon \xi<\omega_1\}\subseteq{\D^*}$
which is not a $P$-point.}

\proof
To define a triple $\la{I_0,f_0,B_0}\ra$ put $I_0=\omega$,
for every $m<\omega$ define
$f_0(m)=X$, and let $B_0(m)=\la B_0(m)(\gamma) \colon \gamma<\omega_1\ra$
be an arbitrary $\omega_1$-sequence of almost disjoint sets in
$\P(f_0(m))\cap\D$.

For a good sequence $\la\la I_{\eta},f_{\eta},B_{\eta} \ra \colon
\eta\leq{\xi} \ra$
and a prism $P$ in $2^{\omega\times X}$ let us define
a subprism $Q(\la{\la{I_{\eta},f_{\eta},B_{\eta}}\ra \colon
\eta\leq{\xi}}\ra,P)=Q$ of $P$
and the triple $T(\la{\la{I_{\eta},f_{\eta},B_{\eta}}\ra \colon
\eta\leq{\xi}}\ra,P)
=\la I_{\xi+1},f_{\xi+1},B_{\xi+1}\ra\in\T$ as in Corollary~\ref{cor:cor2}.
We define a strategy $S$ for Player II in the game
$\text{GAME}_{\text{prism}}(2^{\omega\times{X}})$ as:
$$
S(\la{\la{P_{\eta},Q_{\eta}}\ra \colon \eta<\xi}\ra, P_{\xi})=
Q(\la{\la{I_{\eta},f_{\eta},B_{\eta}}\ra \colon \eta\leq{\xi}}\ra,P_{\xi}),
$$
where $\la{\la{I_{\eta},f_{\eta},B_{\eta}}\ra \colon \eta\leq{\xi}}\ra$
is a good sequence defined by induction on $\eta\leq{\xi}$ as follows.
Assume that $\la\la I_{\zeta},f_{\zeta},B_{\zeta}\ra \colon \zeta<\eta\ra$
is already defined.

If $\eta=0$, then
$\la{I_\eta,f_\eta,B_\eta}\ra=\la{I_0,f_0,B_0}\ra$ is defined as above.

If $\eta=\zeta+1$, then we put
$\la{I_{\eta},f_{\eta},B_{\eta}}\ra=
T(\la{\la I_\delta,f_\delta,B_\delta\ra \colon \delta\leq\zeta}\ra,P_\zeta)$.

If $\eta\in\Gamma$, then $\la{I_\eta,f_\eta,B_\eta}\ra$ is found
using Lemma~\ref{lem:lem13}.

Notice that the sequence $\la\la I_{\zeta},f_{\zeta},B_{\zeta}\ra \colon
\zeta<\eta\ra$
is good by the inductive hypothesis and Remark~\ref{rem:rem4}.

By \psmPrGame\ strategy $S$ is not a winning strategy for Player II.
So, there exists a game $\la{\la{P_{\xi},Q_{\xi}}\ra \colon \xi<\omega_1}\ra$
played according to $S$ for which Player II loses, this is,
$2^{\omega\times{X}}=\bigcup_{\xi<\omega_1}Q_{\xi}$.
If $\la{\la{I_{\xi},f_{\xi},B_{\xi}}\ra\in{\T} \colon\xi<\omega_1}\ra$
is the sequence created when Player II uses strategy $S$,
then this sequence is good by construction.
Application of Lemma~\ref{lem:lem11} to this sequence finishes the proof.
\qed


\thm{thm:thm4}{\psmPrGame \; implies that there exists an
$\omega_1$-generated, crowded $\omega_1$-OK point on $\Q$
which is neither a $P$-point nor a $Q$-point.}

\proof The idea is to apply Theorem \ref{thm:thm3} to an apropriate
ideal to get a crowded ultrafilter which is not a $Q$-point.
Consider $X=\Q\times{\omega}$ with a natural product topology.
Then, $X$ is homeomorphic to $\Q$.
For every $m<\omega$ put
$P_m=\{n<\omega \colon 2^m-1\leq{n}<2^{m+1}-1\}$.
Then $\{P_m \colon m<\omega\}$ is a partition of $\omega$
and $|P_m|=2^m$.
For $A\subset\Q\times\omega$ put
$$
N_A(m)=\max\{k<\omega \colon \exists \; U\in{\I_S^+} \; \exists \;
P\in{[P_m]^k} \;\; U\times{P}\subseteq{A}\}
$$
and define $\J\subseteq{\P(\Q\times{\omega})}$ as
$$
\J=\{A\subseteq{\Q\times{\omega}} \colon
\overline{\lim}_{m\to{\infty}}N_A(m)<\infty\}.
$$
To see that $\J$ is closed under finite unions notice first that
\begin{center}
$N_{A\cup B}(m)\leq N_A(m)+N_B(m)$ \ \ for every $m<\omega$ and
$A,B\subseteq\Q\times\omega$.
\end{center}
Indeed, take a $P\subseteq P_m$ of cardinality $N_{A\cup B}(m)$
and $U\in{\I_S^+}$ such that $U\times{P}\subseteq A\cup B$.
Let $h\colon U\times{P}\to 2$ be a characteristic function of
$A\cap(U\times{P})$ and let
$\varphi\colon U\to 2^P$ be defined by $\varphi(u)(p)=h(u,p)$.
Since $2^P$ is finite, there exists a $g\in 2^P$ such that
$V=\varphi^{-1}(g)$ belongs to $\I_S^+$.
Let $P_A=g^{-1}(1)$ and $P_B=g^{-1}(0)$.
Then $V\times P_A\subseteq A$ and $V\times P_B\subseteq B$.
Therefore,
$N_A(m)\geq |P_A|$ and $N_B(m)\geq |P_B|$.
So,
$N_{A\cup B}(m)=|P|=|P_A|+|P_B|\leq N_A(m)+N_B(m)$.

The above proved inequality easily implies that
\[
 \overline{\lim}_{m\to{\infty}}N_{A\cup B}(m)\leq
 \overline{\lim}_{m\to{\infty}}N_A(m)+
 \overline{\lim}_{m\to{\infty}}N_B(m)
\]
for every and $A,B\subseteq\Q\times\omega$.
Thus, $\J$ is closed under finite unions. Since it clearly is closed
also under subsets, we can conclude that
$\J$ is an ideal on $\Q\times{\omega}$
containing all the singletons.
We will prove that
\begin{equation}\label{eq:Jrich}
\mbox{the ideal $\J$ is rich.}
\end{equation}

First notice how (\ref{eq:Jrich}) implies the theorem.
Since $\perf(\Q)$ is dense in $\I_S^+$,
it is easy to see that $\D=\perf(\Q\times{\omega})$
is dense in $\J^+$.
Let $\U$ be an ultrafilter on $\omega\times X$ from Theorem~\ref{thm:thm3}
applied to $\J$ and $\D$.
Since $X=\Q\times \omega$ is homeomorphic to $\Q$, so is
$\omega\times X$ and $\D^*$ contains only its perfect subsets.
Therefore, $\U$ can be considered as crowded.
Moreover, a partition
$\P=\{\{n\}\times(\{q\}\times{P_m})\colon q\in{\Q}\ \&\ n,m<\omega\}$
of $\omega\times X$ into finite sets
does not admit partial selector in $\U$,
since each such partial selector belongs to $\K=[\omega]^{<\omega}\times \J$.
Thus, $\U$ is not a $Q$-point.

To prove property (\ref{eq:Jrich})
fix an $A\in{\J^+}$.  Then there exist $\la{m_k\in{\omega} \colon k<\omega}\ra$,
$\{U_k\in{\I_S^+} \colon k<\omega\}$, and
$\la{Q_k\subseteq{P_{m_k}} \colon k<\omega}\ra$ such that
$U_k\times{Q_k}\subseteq{A}$ and $|Q_k|>k \cdot 2^{2^k}$ for every $k<\omega$.

First we prove condition (\#) from Definition~\ref{defi:defi1B}.
Since, by Lemma~\ref{lem:lem4},
the ideal $\I_S$ on $\Q$ is rich, for every $k<\omega$ there exists
an almost disjoint family $\{U_f^k \colon f\in
2^\omega\}\subseteq{\P(U_k)\cap{\I_S^+}}$.
Also, for every $k<\omega$ there exists
a pairwise disjoint family $\{A_s \colon s\in{2^k}\}\subseteq{[Q_k]^k}$.
For $f\in 2^\omega$  define
$A_f=\bigcup{\{U_f^k\times{A_{f\restriction{k}}} \colon k<\omega\}}$.
Then, $\{A_f \colon  f\in 2^\omega\}\subseteq{\P(A)\cap{\J^+}}$
is almost disjoint, proving (\#).

To prove that $\J$ is prism-friendly let
$P$ be a prism in $2^X$. If $P$ is singleton then
condition ($\bullet$) is clearly satisfied.
So, assume that $P\in{\perf(2^X)}$
and let $f$ be a witness function for it.
By Remark~\ref{rem1} we can assume that
$f$ is defined on $\Cantor^\alpha$ for some $0<\alpha<\omega_1$.
Our first goal is to find a subprism $Q'$ of $P$ and two sequences
$\{V_k\subseteq{U_k} \colon k<\omega\}\subseteq{\I_S^+}$
and $\{A_k\in{[P_{m_k}]^k} \colon k<\omega\}$ such that
\begin{equation}\label{E:4}
g\restriction{V_k\times{A_k}}\
\mbox{ is constant for every } \  g\in{Q'}.
\end{equation}
For every $k<\omega$ define $\D_k$  as the set of all disjoint collections
$\E\in{[\mathPerf_{\alpha}]^{<\omega}}$ such that there exists a
$V_{\la{\E,k}\ra}\in{\P(U_k)\cap{\I_S^+}}$ such that
for every $q\in{Q_k}$, $E\in{\E}$, and $h,h^{'}\in{E}$, each
$f(h)$ is constant on $V_{\la{\E,k}\ra}\times{\{q\}}$
and
\begin{equation}\label{E:5}
f(h)\restriction{V_{\la{\E,k}\ra}\times{\{q\}}=f(h^{'})
\restriction{V_{\la{\E,k}\ra}\times{\{q\}}}}.
\end{equation}

It is immediate that $\D_k$ is closed under refinaments.
To prove that $\D_k$ satisfies the condition $(\dagger)$ from Proposition
\ref{prop:fusion} let $\E\in{\D_k}$ and $E\in{\mathPerf_{\alpha}}$ be such that
$E\cap{\bigcup{\E}}=\emptyset$.
Let $\{q_i\colon i\leq r\}$ be an enumeration of $Q_k$.
Using Proposition \ref{prop:prop7},
construct inductively
decreasing sequences $\la{E_i\in{\mathPerf_{\alpha}\cap{\P(E)}} \colon
i\leq r}\ra$,
$\la V_i\in{\P(V_{\la{\E,k}\ra})\cap{\I_S^+}} \colon i\leq r\ra$, and
a sequence $\la{j_i<2\colon i\leq r}\ra$ such that for every $i\leq r$
\begin{equation}\label{E:6}
f(h)\restriction{V_i\times{\{q_i\}}} \ \text{ is constant equal to $j_i$
for every }
\ h\in{E_i}.
\end{equation}
Therefore, if we put $E^{'}=E_r$ and $V_{\la{\E\cup{\{E^{'}\}},k}\ra}=V_r$,
then $\E\cup{\{E^{'}\}}\in{\D_k}$ and condition $(\dagger)$ is
satisfied. Thus, by Proposition \ref{prop:fusion}, for every $k<\omega$
there exists a family $\E_k=\{E_i \colon i<2^k\}\in{\D_k}$ of
pairwise disjoint sets with
$E^0=\bigcap_{k<\omega}\bigcup{\E_k}\in{\mathPerf_{\alpha}}$.
We will prove that $Q'=f[E^0]$ satisfies \eqref{E:4} with
$V_k=V_{\la{\E_k,k}\ra}$ and
some sequence $\la{A_k\in{[Q_k]^k} \colon k<\omega}\ra$.

To see this fix $k<\omega$ and $v_0\in V_k={V_{\la{\E_k,k}\ra}}$, and
for each $i<2^k$ pick an $h_i\in{E_i}\in{\E_k}$.
Define $\varphi_k \colon Q_k \to 2^{2^k}$ by $\varphi_k(p)(i)=f(h_i)(v_0,p)$.
Since $|Q_k|>k\cdot{2^{2^k}}$, there exists an $s_k\in{2^{2^k}}$ such that
$|\varphi_k^{-1}\{s_k\}|\geq{k}$. Pick an $A_k\in{[\varphi_k^{-1}\{s_k\}]^k}$.
To see that the pair $\la{V_k,A_k}\ra$ satisfies \eqref{E:4}, pick a $g\in{Q'}$.
Then there exists an $i<2^k$ and an $h\in{E_i}\in{\E_k}$ such that $g=f(h)$.
We will show that $g[V_k\times A_k]=\{s_k(i)\}$.

Let $\la v,q\ra\in V_k\times A_k$.
Since, by (\ref{E:5}), $f(h)$ is constant on $V_k\times{\{q\}}$, we have
$f(h)(v,q)=f(h)(v_0,h)$.
Also, \eqref{E:5} gives
$f(h)(v_0,q)=f(h_i)(v_0,q)$.
Hence, $g(v,q)=f(h_i)(v_0,q)=\varphi_k(q)(i)=s_k(i)$.
So, $g\restriction{V_k\times{A_k}}$ is constant
equal to $s_k(i)$ and \eqref{E:4} holds.



To finish the proof for every $k<\omega$ pick
$\la v_k,a_k\ra\in{V_k\times{A_k}}$
and put $S=\{\la v_k,a_k\ra\colon k<\omega\}$.
Let $\I=[X]^{<\omega}$. Then $\I$ is weakly selective and
$S\in\I^+$.
If we identify $2^{X}$ with $\P(X)$,
then $Q'$ can be treated as a prism in $\P(X)$.
Since $[X]^{\omega}$ is residual in $\P(X)$,
by Proposition~\ref{prop:prop3} we can assume that
$Q'$ is a prism in $[X]^{\omega}$.
So, by Proposition~\ref{prop:prop7},
there exist a subprism $Q$ of $Q'$, a set $S_0\in{[S]^{\omega}}$, and
an $i<2$ such that $g[S_0]={\{i\}}$ for every $g\in{Q}$.
Put $B=\bigcup\{V_k\times{A_k} \colon
\la v_k,a_k\ra\in{S_0}\}$.  Then, $g\restriction{B}$ is constant equal $i$
for every $g\in{Q}$.
It is clear that $B\subseteq{A}$. Since $V_k\times{A_k}\subseteq{B}$ and
$A_k\in{[P_{m_k}]^k}$ we have $N_B(m_k)\geq{k}$. This implies that
$\overline{\lim}_{m\to{\infty}}N_B(m)=\infty$ and that $B\in{\J^+}$.
So, $Q$ and $B$ satisfy ($\bullet$).\qed


\thm{thm:thm5}{Let $X$ be a countably infinite set, $\J\subseteq{\P(X)}$  a
rich and $Q$-like ideal on $X$, and let $\D\subseteq{\J^+}$ be dense.
Then, \psmPrGame \; implies that there exists an
$\omega_1$-generated, crowded $\omega_1$-OK point on $\omega\times{X}$
which is also a $Q$-point but not a $P$-point.}

\proof This proof combines the elements of the proofs of
Theorems~\ref{thm:thm5.4} and~\ref{thm:thm3}.
Let $\Y=\Part_{\omega\times{X}}\cup{2^{\omega\times{X}}}$
be as in Theorem~\ref{thm:thm5.4}.


For a good sequence $\bar G=\la\la I_{\eta},f_{\eta},B_{\eta} \ra \colon
\eta\leq{\xi} \ra$
and a prism $P$ in $\Y$ let us define
a subprism $Q(\bar G,P)$ of $P$ and a triple
$T(\bar G,P)\in\T$ as follows.
\begin{itemize}
\item If $U\cap 2^{\omega\times{X}}\neq\emptyset$, then we can choose
a subprism $P_0\subseteq 2^{\omega\times{X}}$ of $P$. The choice of $P_0$ is
obvious if
$P$ is a singleton, and it follows from Proposition~\ref{prop:prop3},
otherwise. Then we apply Corollary~\ref{cor:cor2}
to $\bar G$ and $P_0$ to find appropriate subprism $Q(\bar G,P)$ of $P_0$
and $\la I_{\xi+1},f_{\xi+1},B_{\xi+1}\ra\in\T$.
We put
$T(\bar G,P)=\la I_{\xi+1},f_{\xi+1},B_{\xi+1}\ra$.

\item If $U\cap 2^{\omega\times{X}}=\emptyset$, then $P$ is
a prism in $\Part_{\omega\times{X}}$.
Then, we can use  Corollary~\ref{cor:cor3}
to find appropriate $\la I_{\xi+1},f_{\xi+1},B_{\xi+1}\ra\in\T$ and
a subprism $Q(\bar G,P)$ of $P_0$.
We put
$T(\bar G,P)=\la I_{\xi+1},f_{\xi+1},B_{\xi+1}\ra$.
\end{itemize}

We define a strategy $S$ for Player II in the game
$\text{GAME}_{\text{prism}}(\Y)$ as:
$$
S(\la{\la{P_{\eta},Q_{\eta}}\ra \colon \eta<\xi}\ra, P_{\xi})=
Q(\la{\la{I_{\eta},f_{\eta},B_{\eta}}\ra \colon \eta\leq{\xi}}\ra,P_{\xi}),
$$
where
the sequence $\la{\la{I_{\eta},f_{\eta},B_{\eta}}\ra \colon \eta\leq{\xi}}\ra$
is defined as in Theorem~\ref{thm:thm3}.

By \psmPrGame\ strategy $S$ is not a winning strategy for Player II.
So, there exists a game $\la{\la{P_{\xi},Q_{\xi}}\ra \colon \xi<\omega_1}\ra$
played according to $S$ for which Player II loses, this is,
$\Y=\bigcup_{\xi<\omega_1}Q_{\xi}$.
If $\la{\la{I_{\xi},f_{\xi},B_{\xi}}\ra\in{\T} \colon\xi<\omega_1}\ra$
is the sequence created when Player II uses strategy $S$,
then, by Remark \ref{rem:rem4}, this sequence is good.





If $g\in{2^{\omega\times{X}}}$, then there exists a $\xi<\omega_1$
such that $g\in{Q_{\xi}}$. Therefore, $Q_{\xi}\subseteq{2^{\omega\times{X}}}$
and $g\restriction{U_{\xi+1}}$ is constant.
Thus,
by Lemma \ref{lem:lem11}, the family $\{U_{\xi} \colon \xi<\omega_1\}$
forms a base for a nonprincipal ultrafilter $\U$ on $\omega\times{X}$ which
is an $\omega_1$-OK point but not a $P$-point.
Note that $\{U_{\xi} \colon \xi<\omega_1\}\subseteq{\D^*}$.
To see that $\U$ is a $Q$-point, take a $z\in{\Part_{\omega\times{X}}}$.
Then, there exists a $\xi<\omega_1$ such that $z\in{Q_{\xi}}$.
This means that $Q_{\xi}\subseteq{\Part_{\omega\times{X}}}$ and that
$|z(k)\cap{U_{\xi+1}}|\leq{1}$ for every $k<\omega$.
Hence, $\U$ is also a $Q$-point.
\qed

\cor{cor:corW}{\psmPrGame\ implies that there is an
$\omega_1$-generated, crowded $\omega_1$-OK point on $\omega\times{X}$
which is also a $Q$-point but not a $P$-point.}
%
\proof
Apply Theorem \ref{thm:thm5} with $X=\Q$, $\J=\I_S$, and $\D=\perf(\Q)$.
\qed




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