%Version of December 9, 2002

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\date{}
 
\title{Crowded and selective ultrafilters under the Covering Property Axiom
% OLD TITLE
%On numbers $\indN$, $\ultraN$, $\reap$, $\madN$, 
%and the existence of selective and crowded ultrafilters
%under the Covering Property Axiom
}


\pagestyle{myheadings}

\newcommand{\reap}{{\mathfrak r}}
\newcommand{\ultraN}{{\mathfrak u}}
\newcommand{\spleat}{{\mathfrak s}}
\newcommand{\madN}{{\mathfrak a}}
\newcommand{\indN}{{\mathfrak i}}


\markboth{K.~Ciesielski and J.~Pawlikowski
}{Crowded and selective ultrafilters under CPA
\ \ \ \ \ \UpdateDate
}

 
\author{
Krzysztof Ciesielski%
\thanks{AMS classification numbers: Primary 54D80, 03E17;
Secondary  03E35. \endgraf
\ \  Key words and phrases: ultrafilter, selective, crowded,
$P$-point, rational numbers, 
independence, reaping, almost disjoint, covering.  \endgraf
\ \ The authors thank Dr. Elliott Pearl for proofreading a version 
of this paper. \endgraf
\ \  The work of the first author was partially supported by 
NATO Grant PST.CLG.977652.}
\\
{\footnotesize Department of Mathematics,}
{\footnotesize West Virginia University,} \\
{\footnotesize Morgantown, WV 26506-6310, USA}\\
{\footnotesize e-mail: K\_Cies@math.wvu.edu}; 
{\footnotesize web page: {\tt http://www.math.wvu.edu/\~{}kcies}}
\and
Janusz Pawlikowski\thanks{
The second
author wishes to thank West Virginia University for its hospitality
during 1998--2001, where the results presented here were obtained. 
}\\
{\footnotesize Department of Mathematics,}
{\footnotesize University of  Wroc\l aw,} \\
{\footnotesize pl. Grunwaldzki 2/4, 50-384 Wroc\l aw, Poland;} 
{\footnotesize e-mail: pawlikow@math.uni.wroc.pl}
}

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\newcommand{\ignore}[1]{}
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\newcommand{\SoIC}{{s_0^{\rm prism}}}
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\newcommand{\psm}{{\rm CPA}}
\newcommand{\psmC}{\mbox{{\rm CPA$_{\rm cube}$}}}
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\newcommand{\psmCsec}{\mbox{{\rm CPA$_{\rm cube}^{\rm sec}$}}}
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\newcommand{\psmPLUS}{{\rm CPA$_{\rm cube}^+$}}
\newcommand{\psmPrPLUS}{{\rm CPA$_{\rm prism}^+$}}
\newcommand{\psmPrGame}{{\rm CPA$_{\rm prism}^{\rm game}$}}
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\newcommand{\Cpr}{{\cal C}_{\rm prism}}

% characteristic function
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\def\lin{{\rm LIN}}

\def\cof{{\rm cof}}
\def\cl{{\rm cl}}
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\def\dist{{\rm dist}}
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\def\co{\continuum}
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\def\proof{\noindent {\sc Proof. }}
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\begin{document}
 
\maketitle



\begin{abstract}
In the paper we formulate an axiom \psmPrGame,
which is the most prominent 
version of the Covering Property Axiom CPA,
and discuss several of its implications. 
In particular, we show that it implies that 
the following cardinal characteristics of
continuum are equal to $\omega_1$, while $\continuum=\omega_2$:
the independence number $\indN$, the reaping number $\reap$, 
the almost disjoint number $\madN$, and the ultrafilter base number
$\ultraN$. We will also show that
\psmPrGame\ implies the existence of 
crowded and selective ultrafilters as well as nonselective $P$-points.
In addition we prove that under \psmPrGame\ 
every selective ultrafilter is $\omega_1$-generated. The paper 
finishes with the proof that \psmPrGame\ 
holds in the iterated perfect set model. 
\end{abstract}



\section{Introduction and preliminaries}

The Covering Property Axiom, CPA, 
constitutes an attempt 
to axiomatize the iterated perfect set (Sacks) model.
In this paper we will consider its prominent 
version, \psmPrGame, as well its three weaker variations: 
\psmP, \psmCGame, and \psmC. They are related to each other 
by the following implications.


\hspace{1.1in}
\begin{picture}(0,100)
\put(0,52){\makebox(0,0){\psmPrGame}}
\put(25,55){\vector(3,1){40}}
\put(25,50){\vector(3,-1){40}}
\put(92,70){\makebox(0,0){\psmCGame}}
\put(92,35){\makebox(0,0){\psmP}}
\put(120,68){\vector(3,-1){40}}
\put(120,37){\vector(3,1){40}}
\put(185,52){\makebox(0,0){\psmC}}
\end{picture}
\vspace{-24pt}
\begin{center}
Chart \Ccounter.\label{count1}
\end{center}
Although in some cases the stronger versions of CPA 
are useful (see e.g.~\cite[chapter~6]{CPAbook}),
it is known that the axiom \psmPrGame\
captures the essence of the Sacks model concerning
the standard cardinal characteristics of continuum.
This follows from a resent result of V.~Kanovei and 
J.~Zapletal~\cite{KZ} (see also~\cite{Zap})
who proved that
for a ``nice'' cardinal invariant $\kappa$ if $\kappa<\continuum$
holds in any forcing extension then $\kappa<\continuum$
follows already from \psmPrGame.

The Covering Property Axiom is 
quite simple in formulation and use, 
nevertheless it requires some new concepts.
To facilitate the absorption of these concepts 
we decided to introduce the axiom in three steps,
beginning with its simplest form. 
(More on CPA can be found in \cite{CPAbook}.)
Thus, in Section~\ref{sectReap} we formulate 
the simplest version of the  
axiom, \psmC, and show that it implies that 
every selective ultrafilter is generated by $\omega_1$ sets 
and that the reaping number $\reap$ is equal to $\omega_1$.
In Section~\ref{secMAD} we will formulate axiom 
\psmCGame\ and show that it 
implies \psmC\ as well as the existence of 
a family $\F\subset[\omega]^\omega$
of cardinality $\omega_1$ which is simultaneously 
maximal almost disjoint, MAD, and reaping.
In particular, \psmCGame\ implies that $\madN=\omega_1<\continuum$. 
In Section~\ref{sec:SelUltra}  we will formulate 
axioms \psmPrGame\ and \psmP\ and show that \psmPrGame\
implies all other versions of the axiom. 
We will also show there that \psmPrGame\ implies the existence
of selective and crowded ultrafilters
as well as nonselective $P$-points. 
In addition we prove there that 
\psmPrGame\ implies the existence
of a family $\F\subset[\omega]^\omega$
of cardinality $\omega_1$ which is simultaneously 
independent and splitting. In particular, under \psmPrGame\ we have 
$\spleat=\indN=\ultraN=\omega_1<\continuum$.
In the last section of the paper we will prove 
the prism fusion lemma, which has been used in 
Section~\ref{sec:SelUltra}, and show that
\psmPrGame\ holds in the iterated perfect set model. 

\medskip

Our set theoretic terminology is standard and follows 
that of~\cite{CiBook}. 
In particular, $|X|$ stands for the cardinality of a set $X$
and $\continuum=|\real|$. 
A Cantor set $2^\omega$ will be denoted by a symbol $\Cantor$. 
We use the term {\em Polish space}\/
for a complete separable metric space {\tt without
isolated points}. 
For a Polish space $X$, the symbol $\perf(X)$
will stand for the 
collection of all subsets of $X$ homeomorphic to a Cantor set $\Cantor$.
For a function $f\colon X\to\real$ and $A\subset X$ an image
of $A$ under $f$ is denoted by $f[A]$, that is,
$f[A]=\{f(x)\colon x\in A\}$. 

For an ideal $\I$ on $\omega$ containing all finite subsets of $\omega$ 
we will use the following generalized selectivity terminology.
We say (see Farah~\cite{Far})
that an ideal $\I$ is {\em selective}\/ 
provided for every sequence $F_0\supset F_1\supset \cdots$
of sets from $\I^+\stackrel{\rm def}{=}\P(\omega)\setminus\I$ 
there exists an $F_\infty\in\I^+$
(called a {\em diagonalization}\/ of this sequence)
such that $F_\infty\setminus \{0,\ldots,n\}\subset F_n$ 
for all $n\in F_\infty$.
Notice that this definition
agrees with the definition of selectivity given by 
Grigorieff in~\cite[p. 365]{Grig}.
(The ideals selective in the above sense
Grigorieff calls {\it inductive}\/ but he also proves 
\cite[cor.~1.15]{Grig} that the inductive ideals
and the ideals selective in his sense are the same notions.)

For $A,B\subset\omega$ we will write $A\subseteq^*B$ when 
$|A\setminus B|<\omega$. 
A set $\D\subset\I^+$ is {\em dense}\/ in $\I^+$
provided for every $B\in\I^+$ there exists an $A\in\D$ such that
$A\subseteq^* B$; set $\D$ is {\em open}\/ in $\I^+$
if $B\in\D$ provided there is an $A\in\D$ such that $B\subseteq^* A$.
For $\bar\D=\la\D_n\subset\I^+\colon n<\omega\ra$ we say 
that $F_\infty\in\I^+$ is a {\em diagonalization}\/ of $\bar\D$
provided $F_\infty\setminus\{0,\ldots,n\}\in\D_n$
for every $n<\omega$.
Following Farah~\cite{Far} we say that an ideal 
$\I$ on $\omega$ is {\em semiselective}\/ provided for every
sequence $\bar\D=\la\D_n\subset\I^+\colon n<\omega\ra$
of dense and open subsets of $\I^+$
the family of all diagonalizations of $\bar\D$ 
is dense in $\I^+$. 

Following Grigorieff~\cite[p. 390]{Grig} 
we say that $\I$ is 
{\em weakly selective}\/ (or {\em weak selective})
provided for every $A\in\I^+$ and $f\colon A\to\omega$ 
there exists a $B\in\I^+$ such that
$f\restriction B$ is either one-to-one or constant.
(Farah in~\cite[sec.~2]{Far} 
terms such ideals as {\it having the $Q^+$-property}.
Note also that Baumgartner and Laver in~\cite{BL} 
call such ideals selective, despite the fact that they 
claim to use Grigorieff's terminology from~\cite{Grig}.)

We have the following implications between these notions.
(See Farah \cite[sec.~2]{Far}.)
\begin{center}
$\I$ is selective \ \ $\Implies$  \ \ 
$\I$ is semiselective \ \ $\Implies$  \ \ 
$\I$ is weakly selective
\end{center}
All these notions represent different generalizations
of the properties of the ideal $[\omega]^{<\omega}$.
In particular, it is easy to see that 
$[\omega]^{<\omega}$ is selective. 

We say that an ideal $\I$ on a countable set $X$
is selective (weakly selective)
provided it is such upon an identification 
of $X$ with $\omega$ via an arbitrary bijection. 
A filter $\F$ on a countable set $X$ 
is selective (semiselective, weakly selective)
provided so is its dual ideal 
$\I=\{X\setminus F\colon F\in\F\}$. 

It is important to note that 
a maximal ideal (or an ultrafilter) 
is selective if and only if it is weakly selective.
This follows, for example, directly from the definitions
of these notions as in Grigorieff~\cite{Grig}.
Recall also that the existence of selective ultrafilters 
cannot be proved in ZFC. 
(Kunen \cite{Kun} proved that there are no selective ultrafilters 
in the random real model. This also follows from the fact that 
every selective ultrafilter is a $P$-point, while Shelah proved that 
there are models with no $P$-points,
see e.g.~\cite[thm.~4.4.7]{BJ}.)




\section{Axiom \psmC\ and its consequences}\label{sectReap}

For a Polish space $X$ we 
will consider $\perf(X)$ as ordered by inclusion.
Thus, a family $\E\subset\perf(X)$ is {\em dense in\/ $\perf(X)$}
provided for every $P\in\perf(X)$ there exists a $Q\in\E$
such that $Q\subset P$. 

Axiom \psmC\ will be of the form
\begin{quote}
if $\E\subset\perf(X)$ is {\em appropriately}\/ dense in $\perf(X)$
then some portion $\E_0$ of 
$\E$ covers almost all of $X$ in a sense that
$|X\setminus\bigcup\E_0|<\continuum$. 
\end{quote}
If the word ``appropriately'' in the above is ignored,
then it implies the following statement.
\begin{description}
\item[{\bf Na\"\i ve-$\cpa$:}]
If $\E$ is dense in $\perf(X)$
then $|X\setminus\bigcup\E|<\continuum$. 
\end{description}
It is a very good candidate for our axiom in the sense that
it implies all the properties we are interested in.
It has, however, one major flaw --- {\em it is false!}\/ 
This is the case since 
$S\subset X\setminus\bigcup\E$ for some dense set $\E$ in $\perf(X)$
provided 
\begin{quote}
for each $P\in\perf(X)$ 
there is a $Q\in\perf(X)$ such that $Q\subset P\setminus S$.
\end{quote}
This means that the family $\G$ of all sets of the form $X\setminus\bigcup\E$,
where $\E$ is dense in $\perf(X)$, coincides with the 
$\sigma$-ideal $s_0$ of Marczewski's sets,
since $\G$ is clearly hereditary. Thus we have 
\begin{equation}\label{eq0}
s_0=\left\{X\setminus\bigcup\E\colon \mbox{ $\E$ is dense in $\perf(X)$}\right\}.
\end{equation}
However, it is well known (see e.g.~\cite[thm.~5.10]{AMillerSubsetsOfR})
that 
there are $s_0$-sets of cardinality~$\continuum$.
Thus, our Na\"\i ve-$\cpa$ ``axiom'' cannot be consistent with ZFC.


In order to formulate the real axiom \psmC\ we need the following 
terminology and notation. 
A subset $C$ of a product $\Cantor^\eta$ of the Cantor set is
said to be a {\em perfect cube}\/ if 
$C=\prod_{n\in\eta} C_n$, where $C_n\in\perf(\Cantor)$ for each $n$.
For a fixed Polish space $X$ 
let $\Fcube$
stand for the family of all continuous 
injections from a perfect cube $C\subset\Cantor^\omega$ onto a set $P$ 
from $\perf(X)$. 
We consider each function $f\in\Fcube$ from $C$ onto $P$ 
as a coordinate system imposed on~$P$.%
\footnote{In a language of forcing a coordinate function $f$ is simply a nice
name for an element from $X$.}
We say that $P\in\perf(X)$ is a {\em cube}\/ if
it is determined by an (implicitly given) 
witness function $f\in\Fcube$ onto $P$,
and $Q$ is a {\em subcube of a cube}\/ $P\in\perf(X)$
provided $Q=f[C]$, where $f\in\Fcube$ is the witness function for $P$ 
and $C$ is a subcube of the domain of $f$. 


We say that a family $\E\subset\perf(X)$ is 
{\em $\Fcube$-dense}\/ (or {\em cube-dense}\/) 
in $\perf(X)$
provided every cube $P\in\perf(X)$ contains a subcube
$Q\in\E$. 
More formally, 
$\E\subset\perf(X)$ is $\Fcube$-dense provided 
\begin{equation}\label{eqDefFdense}
\forall f\in\Fcube\ \exists g\in\Fcube\ 
(g\subset f\ \&\ \range(g)\in\E). 
\end{equation}
It is easy to see that the notion of
$\Fcube$-density is a generalization of 
the notion of density as defined in the first paragraph of this section:
\[
\mbox{if $\E$ is $\Fcube$-dense in $\perf(X)$
then $\E$ is dense in $\perf(X)$.} 
\]
On the other hand, the converse implication is not true, as shown by the 
following simple example. 

\ex{exCubeDens}{Let $X=\Cantor\times\Cantor$ and let $\E$ be the family 
of all $P\in\perf(X)$ such that either
\begin{itemize}
\item all vertical sections of $P$ are countable, or 
\item all horizontal sections of $P$ are countable. 
\end{itemize}
Then $\E$ is dense in $\perf(X)$, but it is not $\Fcube$-dense in $\perf(X)$. 
}


With these notions in hand we are ready to formulate our 
axiom%
\footnote{This version of the axiom, as well as its prism version
\psmP, can be also formulated replacing the inequalities
``$\leq\omega_1$'' with ``$<\continuum$''
and removing the condition ``$\continuum=\omega_2$.''
Such a version of \psmC\ implies $\continuum\geq\omega_2$.
Also all consequences of the axioms \psmC\ and \psmP\ 
presented in this paper follow also from
the modified versions of these axioms. 
However, we do not know if the modified axioms
are consistent with $\continuum>\omega_2$.
We know only that the modified \psmP\ 
implies that $\continuum$ is a successor cardinal. 
(See \cite{CP84} or \cite{CPAbook}.)} 
\psmC.
\begin{description}
\item[{\bf \psmC:}] $\continuum=\omega_2$ and 
for every Polish space $X$ and every $\Fcube$-dense
family $\E\subset\perf(X)$ there is an $\E_0\subset\E$ 
such that $|\E_0|\leq\omega_1$ and $|X\setminus\bigcup\E_0|\leq\omega_1$. 
\end{description}


It is also worth noticing that in order to check
that $\E$ is $\Fcube$-dense it is enough to consider
in condition (\ref{eqDefFdense})
only functions $f$ defined on the entire space $\Cantor^\omega$,
that is

\fact{factFcube}{ 
$\E\subset\perf(X)$ is $\Fcube$-dense if and only if
\begin{equation}\label{eqDefFdenseWeak}
\forall f\in\Fcube,\ 
\dom(f)=\Cantor^\omega,\ 
\exists g\in\Fcube\ 
(g\subset f\ \&\ \range(g)\in\E). 
\end{equation}
}
\proof
To see this, let $\Phi$ be
the family of all bijections 
$h=\la h_n\ra_{n<\omega}$ between the perfect cubes 
$\prod_{n\in\omega} D_n$ and $\prod_{n\in\omega} C_n$
in $\Cantor^\omega$ such that
each $h_n$ is a homeomorphism between $D_n$ and $C_n$. 
Then 
\[
f\circ h\in\Fcube\ \mbox{ for every $f\in\Fcube$ and $h\in\Phi$
with $\range(h)\subset\dom(f)$}.
\]
Now take an arbitrary $f\colon C\to X$ from $\Fcube$
and choose an $h\in\Phi$
mapping $\Cantor^\omega$ onto $C$.
Then $\hat f=f\circ h\in\Fcube$ maps $\Cantor^\omega$ into $X$
and, using (\ref{eqDefFdenseWeak}), we can find a 
$\hat g\in\Fcube$ such that $\hat g\subset\hat f$ and $\range(\hat g)\in\E$.
Then $g=f\restriction h[\dom(\hat g)]$
satisfies (\ref{eqDefFdense}). \qed

One of the most convenient tools for proving 
$\Fcube$-density is the following fact. 


\claim{claim1}{Consider $\Cantor^\omega$ with its usual
topology and its usual product measure.
If $G\subset \Cantor^\omega$ is either 
comeager or of full measure in $\Cantor^\omega$
then it contains a perfect cube $\prod_{i<\omega}P_i$. }

\proof It  follows easily, by induction
on coordinates, from the following well known fact.
\begin{quotation}
\noindent 
For every comeager (full measure)
subset $H$ of $\Cantor\times \Cantor$ there are
a perfect set $P\subset \Cantor$ and a comeager (full measure) subset
$\hat H$ of $\Cantor$ 
such that $P\times \hat H\subset H$. 
\end{quotation}
The category version is easy and can be found in \cite[Exercise~19.3]{Ke}. 
(Its version for $\real^2$ is also proved, for example, 
in \cite[condition ($\star$), p. 416]{CW}.) 
The measure version follows easily from the fact that
\begin{quotation}
\noindent 
for every full measure
subset $H$ of $[0,1]\times[0,1]$ there are
a perfect set $P\subset \Cantor$ and a positive inner measure subset
$\hat H$ of $[0,1]$ such that $P\times \hat H\subset H$ 
\end{quotation}
which is proved by Eggleston \cite{Eg} and, independently, by 
Brodski\v\i~\cite{Br}. 
\qed




Next we will proceed to demonstrate some 
consequences of \psmC. 
The most important combinatorial fact for us concerning 
semiselective ideals is the following property.
(See theorem~2.1 and remark~4.1 in \cite{Far}.)
This is a generalization of a theorem of 
Laver~\cite{Lav2}
who proved this fact for the ideal $\I=[\omega]^{<\omega}$. 

\prop{propFar2}{{\rm (Farah~\cite{Far})} 
Let $\I$ be a semiselective ideal on $\omega$.
For every analytic set $S\subset\Cantor^\omega\times[\omega]^\omega$
and every $A\in\I^+$ there exist a $B\in\I^+\cap\P(A)$
and a perfect cube $C$ in $\Cantor^\omega$ such that
$C\times[B]^\omega$ is either contained in or disjoint with $S$.
}

With this fact in hand we can prove the following theorem.

\thm{thSemiSel}{Assume that \psmC\ holds. If $\I$ is a semiselective
ideal then there is a family $\W\subset\I^+$, $|\W|\leq\omega_1$, such that 
for every analytic set $A\subset[\omega]^{\omega}$
there is a $W\in\W$ for which 
either $[W]^\omega\subset A$ or $[W]^\omega\cap A=\emptyset$.
}

\proof Let $S\subset\Cantor\times[\omega]^{\omega}$
be a universal analytic set, that is such that 
the family $\{S_x\colon x\in\Cantor\}$
(where $S_x=\{y\in[\omega]^{\omega}\colon \la x,y\ra\in S\}$)
contains all analytic subsets of $[\omega]^{\omega}$. 
(See e.g. \cite[lemma 39.4]{Je}.)
In fact, we will take $S$ such that
for any analytic set $A$ in $[\omega]^{\omega}$
\begin{equation}\label{eqSemiSel1}
|\{x\in\Cantor\colon S_x=A\}|=\continuum. 
\end{equation}
(If $U\subset\Cantor\times[\omega]^{\omega}$ is 
a universal analytic set then 
$S=\Cantor\times U\subset \Cantor\times\Cantor\times[\omega]^{\omega}$
satisfies (\ref{eqSemiSel1}), where we identify 
$\Cantor\times\Cantor$  with $\Cantor$.)
For this particular set $S$ consider the family
$\E$ of all $Q\in\perf(\Cantor)$ for which there exists a 
$W_Q\in\I^+$ such that 
\begin{equation}\label{eqSemiSel2}
\mbox{$Q\times[W_Q]^\omega$ is either contained in or disjoint from $S$.}
\end{equation}
Note that, by Proposition~\ref{propFar2},
the family $\E$ is $\Fcube$-dense in $\perf(\Cantor)$. 
So, by \psmC, there exists an $\E_0\subset\E$, $|\E_0|\leq\omega_1$,
such that $|\Cantor\setminus\bigcup\E_0|<\continuum$. Let
\[
\W=\{W_Q\colon Q\in\E_0\}.
\]
It is enough to see that this $\W$ is as required.

Clearly $|\W|\leq\omega_1$. 
Also, by (\ref{eqSemiSel1}),
for an analytic set $A\subset[\omega]^{\omega}$
there exist a $Q\in\E_0$ and an $x\in Q$ such that
$A=S_x$. 
So, by (\ref{eqSemiSel2}),
$\{x\}\times[W_Q]^\omega$ is either contained in or disjoint 
from $\{x\}\times S_x=\{x\}\times A$. \qed

Recall (see e.g. \cite{BJ} or \cite{Vau}) that a family 
$\W\subset[\omega]^\omega$ is a {\em reaping family}\/ 
provided
\[
\forall A\in[\omega]^\omega\ \exists W\in\W\ 
(W\subset A\ \mbox{ or }\ W\subset\omega\setminus A).
\]
The {\em reaping}\/ (or {\em refinement}\/) number 
$\reap$ is defined as 
the minimum cardinality of a reaping family. 
Also, a number $\reap_\sigma$
is defined as the smallest cardinality
of a family $\W\subset[\omega]^\omega$ such that
for every sequence $\la A_n\in[\omega]^\omega\colon n<\omega\ra$
there exists a $W\in\W$ such that
for every $n<\omega$ either 
$W\subseteq^* A_n$ or $W\subseteq^*\omega\setminus A_n$.
(See \cite{BKR} or \cite{Vau}.)
Clearly $\reap\leq\reap_\sigma$. 

\cor{corReap}{If \psmC\ holds then for every semiselective ideal $\I$
there exists a family $\W\subset\I^+$, $|\W|\leq\omega_1$, such that 
for every $A\in[\omega]^{\omega}$
there is a $W\in\W$ for which 
either $W\subseteq^* A$ or $W\subseteq^*\omega\setminus A$.

In particular, \psmC\ implies that $\reap=\omega_1<\continuum$.
}

\proof The family $\W$ from Theorem~\ref{thSemiSel} works:
since $[A]^\omega$ is analytic in $[\omega]^\omega$
there exists a $W\in\W$ such that
either $[W]^\omega\subset[A]^\omega$
or $[W]^\omega\cap[A]^\omega=\emptyset$. \qed

\cor{corSelUltra}{If \psmC\ holds then every selective ultrafilter
$\F$ on $\omega$ is generated by a family of size $\omega_1<\continuum$.
}

\proof If $\F$ is a selective ultrafilter
on $\omega$ then $\I=\P(\omega)\setminus\F$ is a selective ideal
and $\I^+=\F$. Let $\W\subset\I^+=\F$ be as in Corollary~\ref{corReap}.
Then $\W$ generates~$\F$.

Indeed, if $A\in\F$ then there exists a $W\in\W$ such that
either $W\subset A$ or $W\subset\omega\setminus A$.
But it is impossible that $W\subset\omega\setminus A$
since then we would have 
$\emptyset=A\cap W\in\F$. \qed

As mentioned above, in Theorem~\ref{thm:sel} we will prove that
some version of our axiom implies that there exists a
selective ultrafilter on $\omega$.
In particular, the 
assumptions of the next corollary 
are implied by such a version of our axiom.

\cor{corReapSigma}{If \psmC\ holds 
and there exists a selective ultrafilter $\F$ on $\omega$
then $\reap_\sigma=\omega_1<\continuum$.
}

\proof Let $\W\in[\F]^{\leq\omega_1}$ be a generating family 
for $\F$. We will show that it justifies $\reap_\sigma=\omega_1$.
Indeed, take a sequence 
$\la A_n\in[\omega]^\omega\colon n<\omega\ra$.
For every $n<\omega$ let $A_n^*$ belong to 
$\F\cap\{A_n,\omega\setminus A_n\}$. 
Since $\F$ is selective, there exists an $A\in\F$
such that $A\subseteq^* A_n^*$ for every $n<\omega$.
Let $W\in\W$ be such that $W\subset A$. 
Then for every $n<\omega$ either 
$W\subseteq^* A_n$ or $W\subseteq^*\omega\setminus A_n$.
\qed

We are particularly interested in the number $\reap_\sigma$
since it is related to different variants
of sets of uniqueness coming from harmonic analysis, 
as described in the survey paper \cite{BKR}. In particular, from 
\cite[thm. 12.6]{BKR} it follows
that an appropriate version of our axiom
implies that all covering numbers described in the paper
are equal to~$\omega_1$.


\section{\psmCGame\ and numbers $\madN$ and $\reap$}\label{secMAD} 


Before we get to the formulation of our next version of the axiom
it is good to note that in many applications 
we would prefer to have a full covering of 
a Polish  space $X$ rather that the
almost covering as claimed by \psmC. 
To get better access to the missing singletons
we will extend the notion of a cube by
allowing also the {\em constant cubes}:
a family $\Ccube(X)$
of constant ``cubes'' is defined as the family of 
all constant functions from a perfect cube $C\subset\Cantor^\omega$ into $X$.
We define also $\Fcube^*(X)$
as
\begin{equation}\label{eqFcubeSTAR}
\Fcube^*=\Fcube\cup\Ccube.
\end{equation}
Thus, $\Fcube^*$ is the family of all continuous 
functions from a perfect cube $C\subset\Cantor^\omega$ into $X$ which are either
one-to-one or constant. 
Now the range of every $f\in \Fcube^*$ belongs to the family 
$\perf^*(X)$
of
all sets $P$ such that either $P\in\perf(X)$ or $P$ is a singleton.
The terms ``$P\in\perf^*(X)$ is a cube'' and
``$Q$ is a subcube of a cube $P\in\perf^*(X)$''
are defined in a natural way. 

Consider also the following game
${\rm GAME}_{\rm cube}(X)$
of length $\omega_1$. 
The game has two players, Player~I and Player~II. 
At each stage $\xi<\omega_1$ of the game Player~I can play an
arbitrary cube $P_\xi\in\perf^*(X)$
and Player~II must respond with a subcube $Q_\xi$ of $P_\xi$. 
The game $\la\la P_\xi,Q_\xi\ra\colon \xi<\omega_1\ra$
is won by Player~I provided
\[
\bigcup_{\xi<\omega_1}Q_\xi=X;
\]
otherwise the game is won by Player~II. 

Recall also that a strategy for Player~II
is any function $S$ 
with the property that 
$S(\la\la P_\eta,Q_\eta\ra\colon \eta<\xi\ra,P_\xi)$
is a subcube of $P_\xi$, where 
$\la\la P_\eta,Q_\eta\ra\colon \eta<\xi\ra$ is any partial game. 
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 {\em winning strategy}\/
for Player~II provided Player~II wins any game
played according to the strategy $S$. 

Here is our new version of the 
axiom.%
\footnote{Note that if we remove the assumption $\continuum=\omega_2$
from the axiom then the remaining part follows from 
the continuum hypothesis. Thus, for most of our applications the assumption
that $\continuum\geq\omega_2$ is essential. 
On the other hand, the main body of the axiom
and $\continuum\geq\omega_2$ imply that $\continuum=\omega_2$.}

\begin{description}
\item[{\bf \psmCGame:}] $\continuum=\omega_2$ and for any Polish space $X$ 
Player~II has no winning strategy
in the game ${\rm GAME}_{\rm cube}(X)$.
\end{description}

Notice that

\prop{prop:CGImplC}{Axiom \psmCGame\ implies \psmC.}

\proof Let $\E\subset\perf(X)$ be $\Fcube$-dense.
Thus for every cube $P\in\perf(X)$ there exists 
a subcube $s(P)\in\E$ of $P$. 
Now, for a singleton $P\in\perf^*(X)$ put $s(P)=P$
and consider the following strategy $S$ for Player~II:
\[
S(\la\la P_\eta,Q_\eta\ra\colon \eta<\xi\ra,P_\xi)=s(P_\xi).
\]
By \psmCGame\ it is not a winning strategy for Player~II. 
So there exists a game
$\la\la P_\xi,Q_\xi\ra\colon \xi<\omega_1\ra$
in which $Q_\xi=s(P_\xi)$ for every $\xi<\omega_1$
and Player~II loses, that is, 
$X=\bigcup_{\xi<\omega_1}Q_\xi$. 
Now, let 
$\E_0=\{Q_\xi\colon\xi<\omega_1\ \&\ Q_\xi\in\perf(X)\}$.
Then $|X\setminus\bigcup\E_0|\leq\omega_1$,
so \psmC\ is justified. \qed

Recall that a family $\A\subset[\omega]^\omega$
is {\em almost disjoint}\/ provided 
$|A\cap B|<\omega$ and it is
{\em maximal almost disjoint, MAD,} provided 
it is not a proper subfamily of any other almost disjoint family.
The cardinal $\madN$
is defined as 
follows:
\[
\madN=\min\{|\A|\colon \A\ \mbox{is infinite and MAD}\}. 
\]
The fact that $\madN=\omega_1$ holds in the iterated perfect set model
was apparently first noticed by Spinas 
(see Andreas Blass~\cite[sec. 11.5]{Blass}) 
though it seems that the proof of this result was never provided. 

\thm{thm:a}{\psmCGame\ implies that $\madN=\omega_1$.}

Our proof of Theorem~\ref{thm:a} is based on the following lemma. 

\lem{lem:a}{For every countably infinite almost disjoint family
$\W\subset[\omega]^\omega$ and a cube $P\in\perf([\omega]^\omega)$
there exist a $W\in [\omega]^\omega$ and a subcube $Q$ of $P$ 
such that $\W\cup\{W\}$ is almost disjoint
but $\W\cup\{W,x\}$ is not almost disjoint for every $x\in Q$. 
}

\proof 
Let $\W=\{W_i\colon i<\omega\}$. For every $i<\omega$ choose 
sets $V_i\subset W_i$ such that:
the $V_i$'s are pairwise disjoint,
each $W_i\setminus V_i$ is finite, 
but $V_\omega=\omega\setminus \bigcup_{i<\omega}V_i$ is infinite. 
Let
\[
B=\{x\in P\colon (\forall i\leq\omega)\ |x\cap V_i|<\omega\}
\]
and notice that $B$ is a Borel subset of $P$. 
(In fact, $B$ is an $F_{\sigma\delta}$-set.)
So, by Claim~\ref{claim1}, there is a subcube $P^*$ of $P$
such that either $P^*\subset B$ or $P^*\cap B=\emptyset$. 

If $P^*\cap B=\emptyset$ then $W=V_\omega$ and $Q=P^*$ 
satisfy the conclusion of the 
lemma. So, suppose that $P^*\subset B$.
Let $h\colon\Cantor^\omega\to P^*$, 
$h\in\Fcube$, be a coordinate function making $P^*$ a cube, 
let $\lambda$ be the standard product probability measure on $\Cantor^\omega$,
and define a Borel measure $\mu$ on $P^*$
by the formula $\mu(B)=\lambda(h^{-1}(B))$.

For $i,n<\omega$ let
\[
P^n_i=\{x\in P^*\colon x\cap V_i\subset n\}.
\]
Then all the sets $P^n_i$ are Borel (in fact, they are closed)
and $P^*=\bigcup_{n<\omega}P^n_i$ for every $i<\omega$. 
Thus for each $i<\omega$ there exists an $n(i)<\omega$ such that
\[
\mu\left(P_i^{n(i)}\right)>1-2^{-i}.
\]
Then the set $T=\bigcup_{j<\omega}\bigcap_{j<i<\omega} P_i^{n(i)}$
has a $\mu$-measure $1$ so, 
by Claim~\ref{claim1}, there is a subcube $Q$ of $P^*$ which is a subset of $T$. 
Let
\[
W=\bigcup_{i<\omega}\left[V_i\cap n(i)\right].
\]
We claim that $W$ and $Q$ satisfy the lemma.

It is obvious that $W$ is almost disjoint with each $W_i$.
So, fix an $x\in Q$. To finish the proof it is enough to show that
\[
x\subseteq^* W. 
\]
But $x\in Q\subset\bigcup_{j<\omega}\bigcap_{j<i<\omega} P_i^{n(i)}$.
Thus, there exists a $j<\omega$ such that
$x\in \bigcap_{j<i<\omega} P_i^{n(i)}$.
So, 
$x\cap\bigcup_{j<i<\omega}V_i=
\bigcup_{j<i<\omega}(x\cap V_i)\subset
\bigcup_{j<i<\omega}(V_i\cap n(i))
\subset W$ and the set
\[
x\setminus W\subset 
x\cap\left(V_\omega\cup\bigcup_{i\leq j}V_i\right)=
(x\cap V_\omega)\cup\bigcup_{i\leq j}(x\cap V_i)
\]
is finite. 
\qed

\noindent{\sc Proof of Theorem~\ref{thm:a}.}
For a countably infinite almost disjoint family 
$\W\subset[\omega]^\omega$ 
and a cube $P\in\perf([\omega]^\omega)$ 
let $W(\W,P)\in [\omega]^\omega$ and a subcube $Q(\W,P)$ of $P$ 
be as in Lemma~\ref{lem:a}.
For $P=\{x\}\in\perf^*([\omega]^\omega)$ 
we put $Q(\W,P)=P$ and define $W(\W,P)$ as some arbitrary $W$ 
almost disjoint with each set from $\W$ and such that 
$A\cap x$ is infinite for some $A\in\W\cup\{W\}$. 

Let $\A_0\subset[\omega]^\omega$ be an arbitrary infinite almost disjoint family 
and consider the following strategy
$S$ for Player~II:
\[
S(\la\la P_\eta,Q_\eta\ra\colon \eta<\xi\ra,P_\xi)=
Q(\A_0\cup\{W_\eta\colon\eta<\xi\},P_\xi),
\]
where the sets $W_\eta$ are defined inductively by
$W_\eta=W(\A_0\cup\{W_\zeta\colon\zeta<\eta\},P_\eta)$. 
In other words, Player~II remembers (recovers) the sets $W_\eta$ associated 
with the sets $P_\eta$ played so far, and he uses them
(and Lemma~\ref{lem:a}) to get the next answer
$Q_\xi=Q(\A_0\cup\{W_\eta\colon\eta<\xi\},P_\xi)$,  while remembering 
(or recovering each time) 
the set $W_\xi=W(\A_0\cup\{W_\eta\colon\eta<\xi\},P_\xi)$. 

By \psmCGame\ 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, that is, 
$[\omega]^\omega=\bigcup_{\xi<\omega_1}Q_\xi$. 

Now, notice that 
the family $\A=\A_0\cup\{W_\xi\colon \xi<\omega_1\}$ is a MAD
family. 
It is clear that $\A$ is almost disjoint, since
every set $W_\xi$ was chosen as almost disjoint with
every set from $\A_0\cup\{W_\zeta\colon \zeta<\xi\}$.
To see that $\A$ is maximal it is enough to note that
every $x\in[\omega]^\omega$ belongs to a $Q_\xi$ for some
$\xi<\omega_1$, and so
there is an $A\in\A_0\cup\{W_\eta\colon \eta\leq\xi\}$
such that $A\cap x$ is infinite. 
\qed

By Theorem~\ref{thm:a} we see that \psmCGame\ implies
the existence of MAD family of size $\omega_1$. 
Next we will show that such a family can be 
simultaneously a reaping family. This result is 
similar in flavor to that from Theorem~\ref{thm:is}.


\thm{thm:ar}{\psmCGame\ implies that there exists a family $\F\subset[\omega]^\omega$
of cardinality $\omega_1$ which is simultaneously MAD and reaping.}

\proof The proof is just a slight modification of that for Theorem~\ref{thm:a}.

For a countably infinite almost disjoint family 
$\W\subset[\omega]^\omega$ 
and a cube $P\in\perf([\omega]^\omega)$ 
let $W_0\in [\omega]^\omega$ and a subcube $Q_0$ of $P$ 
be as in Lemma~\ref{lem:a}.
Let $A\in[\omega]^\omega$ be almost disjoint with every set from
$\W\cup\{W_0\}$. 
By Laver's theorem~\cite{Lav2} we can also find a subcube 
$Q_1$ of $Q_0$ and a $W_1\in[A]^\omega$ such that 
\begin{itemize}
\item either $W_1\cap x=\emptyset$ for every $x\in Q_1$,
\item or else $W_1\subset x$ for every $x\in Q_1$.   
\end{itemize}
Let $Q(\W,P)=Q_1$ and $\W(\W,P)=\{W_0,W_1\}$. 
If $P\in\perf^*([\omega]^\omega)$ is a singleton 
then we put $Q(\W,P)=P$ and we can easy find 
$W_0$ and $W_1$ satisfying the above conditions. 

Let $\A_0\subset[\omega]^\omega$ be an arbitrary infinite almost disjoint family 
and consider the following strategy
$S$ for Player~II:
\[
S(\la\la P_\eta,Q_\eta\ra\colon \eta<\xi\ra,P_\xi)=
Q\left(\A_0\cup\bigcup\{\W_\eta\colon\eta<\xi\},P_\xi\right),
\]
where $\W_\eta$'s are defined inductively by
$\W_\eta=\W(\A_0\cup\bigcup\{\W_\eta\colon\eta<\xi\},P_\eta)$. 

By \psmCGame\ 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 and Player~II loses, that is, 
$[\omega]^\omega=\bigcup_{\xi<\omega_1}Q_\xi$. 
Then the family $\F=\A_0\cup\bigcup\{\W_\xi\colon \xi<\omega_1\}$ is MAD
and reaping. \qed


\section{On \psmPrGame, selective and crowded ultrafilters, 
nonselective $P$-points, and  numbers $\indN$ and $\ultraN$}\label{sec:SelUltra} 

The axioms \psmCGame\ and \psmC\ dealt with 
the notion of $\Fcube$-density, where $\Fcube$ is 
the family of all injections $f\colon C\to X$ with 
$C$ being a perfect cube in $\Cantor^\omega$. 
In the applications of these axioms we were using the facts that
different subfamilies of $\perf(X)$ were $\Fcube$-dense. 
Unfortunately, in many cases the notion of 
$\Fcube$-density is too weak to do the job ---
in the applications that follow the families $\E\subset\perf(X)$
will not be $\Fcube$-dense, but they will be dense in 
a weaker sense defined below.
Luckily, this weaker notion of density still leads to 
consistent axioms.

\subsection{Prisms and \psmPrGame}




To define this weaker notion of density, 
let us first take another look at the notion of cube. 
Let $A$ be a non-empty countable set of ordinal numbers.
The notion of a perfect cube in $\Cantor^A$ can be defined
the same way as it was done for $\Cantor^\omega$. 
However, it will be more convenient for us to define it 
as follows. 
Let $\Phi_{\rm cube}$
be the family of all continuous injection 
$f\colon\Cantor^A\to\Cantor^A$ 
such that
\[
f(x)(\alpha)=f(y)(\alpha)
\ \mbox{ for all $\alpha\in A$ and $x,y\in \Cantor^A$ with 
$x(\alpha)=y(\alpha)$}.
\]
In other words $\Phi_{\rm cube}$ is 
the family of all functions of the
form $f=\la f_\alpha\ra_{\alpha\in A}$, where each $f_\alpha$
is an injection from $\Cantor$ into $\Cantor$.
Then the family of all perfect cubes in $\Cantor^A$ for an appropriate $A$
is equal to
\[
{\rm CUBE}=\{\range(f)\colon f\in \Phi_{\rm cube}\}
\]
and $\Fcube$ is the family 
all continuous injections $f\colon C\to X$ with 
$C\subset\Cantor^\omega$ and $C\in {\rm CUBE}$. 

In the definitions that follow the notion of ``cube'' will be replaced by that
of a ``prism.'' So, let  
$\Phi_{\rm prism}(A)$
be the family of all continuous injection 
$f\colon\Cantor^A\to\Cantor^A$ 
with the property that
\begin{equation}\label{con:PrKeep}
f(x)\restriction\alpha=f(y)\restriction\alpha\ \Equi\ 
x\restriction\alpha=y\restriction\alpha
\ \ \ \ \ \mbox{ for all $\alpha\in A$ and $x,y\in \Cantor^A$}
\end{equation}
or, equivalently, such that for every $\alpha\in A$
\[
f\restriction\restriction\alpha\stackrel{\rm def}{=}
\{\la x\restriction\alpha,y\restriction\alpha\ra\colon\la x,y\ra\in f\}
\]
is a one-to-one function from $\Cantor^{A\cap\alpha}$
into $\Cantor^{A\cap\alpha}$. 
For example, if $A=\{0,1,2\}$ then $f\in\Phi_{\rm prism}(A)$
provided there exist 
continuous functions $f_0\colon\Cantor\to\Cantor$,
$f_1\colon\Cantor^2\to\Cantor$, and $f_2\colon\Cantor^3\to\Cantor$
such that 
$$
f(x_0,x_1,x_2)=\la f_0(x_0),f_1(x_0,x_1),f_2(x_0,x_1,x_2)\ra
$$
for all $x_0,x_1,x_2\in\Cantor$ and
maps $f_0$, $\la f_0,f_1\ra$, and $f$ are one-to-one. 
Functions $f$ from $\Phi_{\rm prism}(A)$ were first introduced, 
in more general setting, in~\cite{Ka} where they are called
{\em projection-keeping homeomorphisms}.
Note that 
\begin{equation}\label{eq:compPr}
\mbox{$\Phi_{\rm prism}(A)$ is closed under compositions}
\end{equation}
and that for every ordinal number $\alpha>0$
\begin{equation}\label{eq:restr}
\mbox{if $f\in\Phi_{\rm prism}(A)$ then
$f\restriction\restriction\alpha\in\Phi_{\rm prism}(A\cap\alpha)$.}
\end{equation}
Let
\[
\mathPerf_A=\{\range(f)\colon f\in \Phi_{\rm prism}(A)\}.
\]
We will write $\Phi_{\rm prism}$ for 
$\bigcup_{0<\alpha<\omega_1}\Phi_{\rm prism}(\alpha)$
and define 
\[
\mbox{$
\mathPerf_{\omega_1}\stackrel{\rm def}{=}
\bigcup_{0<\alpha<\omega_1}\mathPerf_\alpha=
\{\range(f)\colon f\in \Phi_{\rm prism}\} 
$.}
\]
Following~\cite{Ka} we will refer to elements of 
$\mathPerf_{\omega_1}$ as {\em iterated perfect sets}. 
Also let $\Fpr(X)$
(or just $\Fpr$, if $X$ is clear 
from the context) be the family of 
all continuous injections $f\colon P\to X$ 
where  $P\in\mathPerf_{\omega_1}$
and $X$ is a fixed Polish space.  

We say that a family $\E\subset\perf(X)$ is 
{\em $\Fpr$-dense}\/ provided 
\[
\forall f\in\Fpr\ \exists g\in\Fpr\ 
(g\subset f\ \&\ \range(g)\in\E). 
\]
Similarly as in Fact~\ref{factFcube}, using (\ref{eq:compPr})
we can also prove that

\fact{factFprism}{ 
$\E\subset\perf(X)$ is $\Fpr$-dense if and only if
\begin{equation}\label{eqDefFdenseWeakpr}
\!\!\!\!
\forall \alpha<\omega_1\ 
\forall f\in\Fpr,\ 
\dom(f)=\Cantor^\alpha
\ \exists g\in\Fpr\ 
(g\subset f\ \&\ \range(g)\in\E)\!\!\!\!
\end{equation}
}
Notice also that $\Phi_{\rm cube}\subset\Phi_{\rm prism}$, so every cube is also
a prism. 
From this and Fact~\ref{factFprism} it
also easy to see that 
\begin{equation}\label{eqPr4a}
\mbox{if $\E\subset\perf(X)$ is $\Fcube$-dense
then $\E$ is also $\Fpr$-dense.}
\end{equation}
The converse of (\ref{eqPr4a}), however, is false. 
(See \cite[Remark~3.3.2]{CPAbook}.)

We also adopt the shortcuts similar to that for cubes. 
Thus, we say that $P\in\perf(X)$ is a {\em prism}\/ if
we consider it with an (implicitly given) 
witness function $f\in\Fpr$ onto $P$.
By Fact~\ref{factFprism} to establish $\Fpr$-density
we can always assume that the witness 
function $f$ is in a {\em standard form}, that is, 
defined on the entire set $\Cantor^\alpha$
for an appropriate $\alpha<\omega_1$.  
Then $Q$ is a {\em subprism of a prism}\/ $P\in\perf(X)$
provided $Q=f[E]$, where $f\in\Fpr$ is as above 
and $E\in\mathPerf_\alpha$. 
Also singletons $\{x\}$ in $X$ will be identified 
with constant functions from 
$E\in\mathPerf_{\omega_1}$ to $\{x\}$,
and these functions will be considered as elements 
of $\Fpr^*$,
similarly as in (\ref{eqFcubeSTAR}).

Now we are ready to state the next version of our axiom,
in which the game ${\rm GAME}_{\rm prism}(X)$
is an obvious generalization
of ${\rm GAME}_{\rm cube}(X)$. 

\smallskip 

\begin{description}
\item[{\bf \psmPrGame:}] $\continuum=\omega_2$ and for any Polish space $X$ 
Player~II has no winning strategy
in the game ${\rm GAME}_{\rm prism}(X)$.
\end{description}

Notice that if a prism $P\in\perf(X)$ is considered with 
a witness function $f\in\Fpr$
in a standard form (i.e., $f$ is from $\Cantor^\alpha$ onto $P$) 
then $P$ is also a cube and any subcube
of $P$ is also a subprism of $P$. Thus, any Player~II strategy 
in a game ${\rm GAME}_{\rm cube}(X)$ can be translated 
to a strategy in a game ${\rm GAME}_{\rm prism}(X)$.
(You need to identify appropriately $\Cantor^\alpha$ with
$\Cantor^\omega$: first you identify 
$\Cantor^\alpha$ with $\Cantor^\omega\times\Cantor^{\alpha\setminus\{0\}}$,
which is important for a finite $\alpha$, and then
this second space identify with $\Cantor^\omega$ coordinatewise.)
In particular, \psmPrGame\ implies \psmCGame.
Also, essentially the same argument as used for 
Proposition~\ref{prop:CGImplC} gives also the following. 


\prop{prop:CGImplPr}{Axiom \psmPrGame\ implies the following
prism version of the axiom \psmC:
\begin{description}
\item[{\bf \psmP:}] $\continuum=\omega_2$ and 
for every Polish space $X$ and every $\Fpr$-dense
family $\E\subset\perf(X)$ there is an $\E_0\subset\E$ 
such that $|\E_0|\leq\omega_1$ and $|X\setminus\bigcup\E_0|\leq\omega_1$. 
\end{description}
}
By (\ref{eqPr4a}) it is also obvious that \psmP\ implies \psmC. 
All these implications are summarized in Chart~\ref{count1}.

In what follows for a fixed $0<\alpha<\omega_1$ and $0<\beta\leq\alpha$
the symbol $\pi_\beta$
will stand for the projection  
from $\Cantor^{\alpha}$ onto $\Cantor^\beta$.
We will always consider $\Cantor^\alpha$
with the following standard metric $\rho$:
fix an enumeration
$\{\la \beta_k,n_k\ra\colon k<\omega\}$ of
$\alpha\times\omega$ and 
for distinct $x,y\in\Cantor^\alpha$ define
\begin{equation}\label{eqRHO}
\rho(x,y)=2^{-\min\{k<\omega\colon x(\beta_k)(n_k)\neq y(\beta_k)(n_k)\}}.
\end{equation}
The open ball in $\Cantor^\alpha$ 
with a center at $z\in\Cantor^\alpha$ and radius $\e>0$
will be denoted by $B_\alpha(z,\e)$.
Notice that in this metric any two open balls are either disjoint 
or one is a subset of another. 
Also for every $\gamma<\alpha$ and $\e>0$
\begin{equation}\label{eq18x}
\pi_\gamma[B_\alpha(s,\e)]=\pi_\gamma[B_\alpha(t,\e)] \ \ 
\mbox{ for every $s,t\in\Cantor^\alpha$ with 
$s\restriction \gamma=t\restriction \gamma$. }
\end{equation}
It is also easy to see that any $B_\alpha(z,\e)$
is a clopen set and, in fact, it is a perfect cube in $\Cantor^\alpha$,
so it belongs to $\mathPerf_\alpha$. 
In fact, more can be said:
\begin{equation}\label{eq19}
\mbox{if }\ 
\B_\alpha\stackrel{\rm def}{=}\{B\subset\Cantor^\alpha\colon
\mbox{ $B$ is clopen in $\Cantor^\alpha$\} \ then } \ 
\B_\alpha\subset\mathPerf_\alpha.
\end{equation}
This is the case, since any clopen $E$ in $\Cantor^\alpha$
is a finite union of disjoint open balls, each of which belongs to
$\mathPerf_\alpha$, and it is easy to see that 
$\mathPerf_\alpha$ is closed under finite unions of disjoint sets. 

  From this we conclude immediately that
\begin{equation}\label{eqClopen1a}
\mbox{a clopen subset of $E\in\mathPerf_\alpha$
belongs to $\mathPerf_\alpha$}
\end{equation}
and
\begin{equation}\label{eqClopen1b}
\mbox{a clopen subset of a prism is its subprism.}
\end{equation}

Notice also that if $P\in\mathPerf_\alpha$ then 
\begin{equation}\label{eq20x} 
\mbox{$P\cap\pi_\beta^{-1}(P')\in \mathPerf_\alpha$ \ \ \ 
for every $P'\in\mathPerf_\beta$ with $P'\subset\pi_\beta[P]$. 
}
\end{equation}
Indeed, let 
$f\in\Phi_{\rm prism}(\beta)$ and 
$g\in\Phi_{\rm prism}(\alpha)$ be such that 
$f[\Cantor^\beta]=P'$ and $g[\Cantor^\alpha]=P$.
Let 
$Q=(g\restriction\beta)^{-1}[P']
=(g\restriction\beta)^{-1} \circ f[\Cantor^\beta]\in\mathPerf_\beta$.
Then $\pi_\beta^{-1}(Q)$ belongs to $\mathPerf_\alpha$ and 
$P\cap\pi_\beta^{-1}(P')=g[\pi_\beta^{-1}(Q)]\in\mathPerf_\alpha$. 



\subsection{Fusion Lemmas}

One of the main technical tools used to prove that 
a family of perfect sets is dense
is the so called fusion lemma. It says that 
for an appropriately chosen decreasing sequence 
$\{P_n\colon n<\omega\}$ of perfect sets its intersection
$P=\bigcap_{n<\omega}P_n$, called the {\em fusion}, is still a perfect set.
The simple structure of perfect cubes makes it quite easy to
formulate a ``cube fusion lemma'' in which 
the fusion set $P$ is also a cube. 
However, 
so far we did not have any need for such a lemma (at least in an explicit form), 
since its use 
was always hidden in the proofs of the results we quoted, like 
Claim~\ref{claim1} or Proposition~\ref{propFar2}. 
On the other hand, the 
new and more complicated structure of prisms does not leave us
the option of avoiding fusion arguments any longer ---  we have to face it
up front.  

For a fixed $0<\alpha<\omega_1$ 
let $\{\la \beta_k,n_k\ra\colon k<\omega\}$ be an enumeration of
$\alpha\times\omega$ used in the definition 
(\ref{eqRHO}) of the  metric $\rho$ and let
\begin{equation}\label{NiceEnum}
A_k=\{\la \beta_i,n_i\ra\colon i< k\}\ \ \ \mbox{ for every $k<\omega$}.
\end{equation}



\lem{FusionSequenceLemma}{{\rm\bf (Fusion Sequence)}
Let $0<\alpha<\omega_1$ and for every $k<\omega$ let
$\E_k=\left\{E_s\in\mathPerf_\alpha\colon s\in 2^{A_k}\right\}$.  
Assume that for every $k<\omega$, $s,t\in 2^{A_k}$, 
$r\in\bigcup_{i<\omega}2^{A_i}$, and $\beta<\alpha$ we have: 
\begin{itemize}
\item[\rm (i)] the diameter of $E_s$ is less than  or equal to $2^{-k}$,
\item[\rm (ii)] if $r\subset s$ then $E_s\subset E_r$, 
\item[\rm (ag)] (agreement) if
  $s\restriction(\beta\times\omega)=t\restriction(\beta\times\omega)$
    then $\pi_{\beta}[E_s]=\pi_{\beta}[E_t]$, 
\item[\rm (sp)] (split) if
  $s\restriction(\beta\times\omega)\neq t\restriction(\beta\times\omega)$
    then $\pi_{\beta}[E_s]\cap\pi_{\beta}[E_t]=\emptyset$.
\end{itemize}
Then $Q=\bigcap_{k<\omega}\bigcup\E_k$ 
belongs to $\mathPerf_\alpha$.
}


\proof For $x\in\Cantor^\alpha$ let $\bar x\in 2^{\alpha\times\omega}$
be defined by $\bar x(\beta,n)=x(\beta)(n)$. 

First note that, by conditions (i) and (sp), for every $k<\omega$ 
the sets
in $\E_k$ are pairwise 
disjoint and each of the diameter at most $2^{-k}$. 
Thus, taking into account (ii), the function 
$h\colon\Cantor^\alpha\to\Cantor^\alpha$ defined by 
\[
h(x)=r\ \ \Longleftrightarrow\ \  
\{r\}=\bigcap_{k<\omega}E_{\bar x\restriction A_k}
\]
is well defined and is one-to-one.  It is also easy
to see that $h$ is continuous and that 
$Q=h\left[\Cantor^\alpha\right]$.
Thus, we need to prove only that $h\in\Phi_{\rm prism}(\alpha)$,
that is, that $h$ is projection-keeping. 

To show this 
fix $\beta<\alpha$, put $S=\bigcup_{i<\omega}2^{A_i}$,
and notice that, by (i) and (ag), for every $x\in\Cantor^\alpha$
we have
\begin{eqnarray*}
\{h(x)\restriction\beta\}
& = & \pi_{\beta}\left[
\bigcap\{E_{\bar x\restriction A_k}\colon k<\omega\}\right]\\
& = & 
\bigcap\{\pi_{\beta}[E_{\bar x\restriction A_k}]\colon k<\omega\}\\
& = & 
\bigcap\{\pi_{\beta}[E_s]\colon s\in S\ \&\ s\subset\bar x\}\\
& = & 
\bigcap\{\pi_{\beta}[E_s]\colon s\in S\ \&\ 
s\restriction(\beta\times\omega)\subset\bar x\}.
\end{eqnarray*}
Now, if $x\restriction\beta=y\restriction\beta$
then for every $s\in S$
\[
s\restriction(\beta\times\omega)\subset\bar x\ \ \Equi\ \ 
s\restriction(\beta\times\omega)\subset\bar y
\]
so
$h(x)\restriction\beta=h(y)\restriction\beta$.

On the other hand, if $x\restriction\beta\neq y\restriction\beta$
then there exists a $k<\omega$ big enough such that for 
$s=\bar x\restriction A_k$ and 
$t=\bar y\restriction A_k$ we have 
$s\restriction (\beta\times\omega)\neq t\restriction (\beta\times\omega)$.
But then  
$\{h(x)\restriction\beta\}$ and $\{h(y)\restriction\beta\}$
are subsets of $\pi_{\beta}[E_s]$ and $\pi_{\beta}[E_t]$,
respectively, which, by (sp), are disjoint. 
So, $h(x)\restriction\beta\neq h(y)\restriction\beta$. \qed


In all of our applications the task of constructing  
sequences $\la\E_k\colon k<\omega\ra$ satisfying specific 
conditions (ag) and (sp) can be reduced to checking
some simple density properties listed in our next lemma.
In its statement we consider 
$\mathPerf_\alpha$ as ordered by inclusion
and use the standard terminology from the theory of
partially ordered sets:
$D\subset\mathPerf_\alpha$ is {\em dense}\/ 
provided for every $E\in\mathPerf_\alpha$ there is 
an $E'\in D$ with $E'\subset E$; it is {\em open}\/
provided for every $E\in D$ if $E'\in\mathPerf_\alpha$
and $E'\subset E$ then $E'\in D$. 
Moreover, for a family $\E$ of pairwise disjoint subsets of
$\mathPerf_\alpha$ we say that $\E'\subset\mathPerf_\alpha$
is a {\em refinement of $\E$}\/ 
provided $\E'=\{P_E\colon E\in\E\}$ where $P_E\subset E$
for all $E\in\E$. 


\lem{ELemmaNew}{Let $0<\alpha<\omega_1$ and 
$k<\omega$. If
$\E_k=\left\{E_s\in\mathPerf_\alpha\colon s\in 2^{A_k}\right\}$
satisfies (ag) and (sp) then
\begin{itemize}
\item[\rm (A)] there exists an 
$\E_{k+1}=\left\{E_s\in\mathPerf_\alpha\colon s\in 2^{A_{k+1}}\right\}$
such that (i), (ii), (ag), and (sp)
hold for all $s,t\in 2^{A_{k+1}}$ and $r\in 2^{A_k}$.
\end{itemize}
Moreover, if 
$\D\subset[\mathPerf_\alpha]^{<\omega}$ is a family 
of pairwise disjoint sets such that
$\emptyset\in\D$,  $\D$ is 
closed under refinements, and
\begin{itemize}
\item[\rm ($\dagger$)] for every 
$\E\in\D$ and $E\in\mathPerf_\alpha$ which is disjoint with
$\bigcup\E$ there exists an $E'\in\mathPerf_\alpha\cap\P(E)$ such that
$\{E'\}\cup\E\in\D$ 
\end{itemize}
then 
\begin{itemize}
\item[\rm (B)] there exists a refinement $\E'_k\in\D$ of $\E_k$
satisfying (ag) and (sp),
\item[\rm (C)] there exists 
     an $\E_{k+1}$ as in (A) such that $\E_{k+1}\in\D$.
\end{itemize}
}

The proof of this lemma will be postponed to the last section
of this paper. 
One of the most important consequences
of Lemma~\ref{ELemmaNew} is the following.

\cor{DiagonalLemma}{Let $0<\alpha<\omega_1$
and let $\{D_k\colon k<\omega\}$ be a collection of 
dense open subsets of $\mathPerf_\alpha$.
If for every $k<\omega$
\[
D_k^*=
\left\{\bigcup\D\colon \D\in[D_k]^{<\omega} 
\mbox{ and the sets in $\D$ are pairwise disjoint}\right\}
\]
then $\bar D=\bigcap_{k<\omega}D_k^*$ is open and dense in $\mathPerf_\alpha$.
}

\proof It is clear that $\bar D$ is open. To see its density
notice that the families 
\[
\D_k=\left\{\D\in[D_k]^{<\omega} 
\mbox{ and sets in $\D$ are pairwise disjoint}\right\}
\]
satisfy condition ($\dagger$). 
Let $E\in \mathPerf_\alpha$, choose 
an $E_\emptyset\in D_0\subset D_0^*$ below
$E$, and put $\E_0=\{E_\emptyset\}$.
Applying (C) from Lemma~\ref{ELemmaNew} by induction we can 
define families $\E_k\in\D_k$, $k<\omega$,
such that conditions (i), (ii), (ag), and (sp)
from Lemma~\ref{FusionSequenceLemma} are satisfied.
But then
$Q=\bigcap_{k<\omega}\bigcup\E_k\subset E$
belongs to $\bar D$. \qed

\subsection{Selective ultrafilters and number $\ultraN$}

Recall that every weakly selective ultrafilter 
is selective and that the ideal $\I=[\omega]^{<\omega}$ is selective.
Another example of a weakly selective ideal which 
we will use in what follows is given below. 

\fact{factWeakSelQ}{The ideal $\I$ of nowhere dense subset of rationals
$\rational$ is weakly selective.}

\proof 
Let $A\in\I^+$ and take an $f\colon A\to\omega$.
If there is a $B\in\I^+\cap\P(A)$ such that
$f\restriction B$ constant then we are done.
So, assume that it is not the case
and let $A_0\subset A$ be dense on some interval.
By induction on $n<\omega$ define a sequence
$\{b_n\in A_0\colon n<\omega\}$ dense in $A_0$
such that $f$ restricted to $B=\{b_n\colon n<\omega\}$
is one-to-one. Then $B$ is as desired. 
\qed



In what follows we will also need the following fact 
about weakly selective ideals, which can be found 
in Grigorieff~\cite[prop.~14]{Grig}.

\prop{propTreeSel}{Let $\I$ be a weakly selective ideal on $\omega$ and 
$A\in\I^+$. If $T\subset A^{<\omega}$ is a tree such that 
\[
A\setminus\{j<\omega\colon s\hat{\ }j\in T\}\in\I\ 
\mbox{ for every $s\in T$}
\] 
then there exists a branch $b$ of $T$ such that
$b[\omega]\in\I^+$. 
}


\thm{thm:sel}{
\psmPrGame\ implies that for every selective ideal $\I$ on $\omega$
there exists a selective ultrafilter $\F$ on~$\omega$
such that $\F\subset\I^+$. 

In particular if \psmPrGame\ holds then 
there is a selective ultrafilter on~$\omega$.
}

The proof is based on the following lemma. 

\lem{lem:sel}{Let $\I$ be a weakly selective ideal 
on $\omega$. 
\begin{itemize}
\item[\rm (a)] For every $A\in\I^+$ and every prism $P$ in $\omega^\omega$
there exist a $B\in\I^+$, $B\subset A$, and a subprism $Q$ of $P$ 
such that  either
\begin{itemize}
\item[\rm (i)] $g\restriction B$ is one-to-one for every $g\in Q$, or else
\item[\rm (ii)] there exists an $n<\omega$ such that
       $g\restriction B$ is constant equal to $n$ for every $g\in Q$.   
\end{itemize}

\item[\rm (b)] For every $A\in\I^+$ and every prism $P$ in $[\omega]^\omega$
there exist a $B\in\I^+$, $B\subset A$, and a subprism $Q$ of $P$ 
such that  either
\begin{itemize}
\item   $x\cap B=\emptyset$ for every $x\in Q$, or else
\item   $B\subset x$ for every $x\in Q$.   
\end{itemize}

\end{itemize}
}

\proof (a) Fix an $A\in\I^+$, an $f\in\Fpr(\omega^\omega)$ from
$\Cantor^\alpha$ onto $P$, and assume that for no subprism 
$Q$ of $P$ and $B\in\I^+\cap\P(A)$ condition (ii) holds. 
We will find $Q$ satisfying (i). 


For $i,n<\omega$ let 
$D(i,n)=\{E_0\in\mathPerf_\alpha\colon (\forall g\in E_0)\ f(g)(i)\neq n\}$
and for $\gamma\leq\alpha$, $E\in\mathPerf_\alpha$, and $A'\subset\omega$ put
$D_\gamma(E,i,n)=\{\pi_\gamma[E_0]\colon E_0\in D(i,n)\cap\P(E)\}$ and
\[
D_\gamma(E,A',n)=\bigcap_{i\in A'}D_\gamma(E,i,n).
\]
Notice that the sets $D_\gamma(E,i,n)$ and $D_\gamma(E,A',n)$ are open in
$\mathPerf_\gamma$. 
By induction on $0<\beta\leq\alpha$ 
we are going 
to prove the following property. 
\begin{itemize}
\item[$\psi_\beta$:]
For all $0<\gamma\leq\beta$, $E\in\mathPerf_\alpha$,
$n<\omega$, and $\hat A\in\I^+\cap\P(A)$
there exists an $A'\in\I^+\cap\P(\hat A)$
such that $D_\gamma(E,A',n)\neq\emptyset$.
\end{itemize}
In what follows for $k<\omega$ and 
$\E_k=\left\{E_s\in\mathPerf_\beta\colon s\in 2^{A_k}\right\}$
satisfying (i), (ag), and (sp)
from Lemma~\ref{FusionSequenceLemma}
and 
$\E_{k+1}=\left\{E_s\in\mathPerf_\beta\colon s\in 2^{A_{k+1}}\right\}$
we will write
\[
\E_{k+1}\prec \E_k
\]
provided (i), (ii), (ag), and (sp)
hold for all $s,t\in 2^{A_{k+1}}$ and $r\in 2^{A_k}$.
One of the main facts used in the proof of 
$\psi_\beta$ is the following property. 
\begin{itemize}
\item[($*$)] If $\psi_\gamma$ holds for all $\gamma<\beta$,
$\bar A\in\I^+\cap\P(A)$, $n<\omega$, $E\in\mathPerf_\alpha$, 
$\E_k$ is as above and such that $\bigcup\E_k\subset\pi_\beta[E]$, and
\[
Z(\bar A,\E_k,n)=\left\{i\in\bar A\colon
(\exists \E_{k+1}\prec \E_k)\ \bigcup\E_{k+1}\in D_\beta(E,i,n)
\right\}
\]
then $\bar A\setminus Z(\bar A,\E_k,n)\in\I$. 
\end{itemize}

In order to prove ($*$) fix an $\hat A\in\P(\bar A)\cap\I^+$
and note that it 
is enough to show that $\hat A\cap Z(\bar A,\E_k,n)\neq\emptyset$.
Fix an $\bar\E_{k+1}=\{\bar E_s\in\mathPerf_\beta\colon s\in 2^{A_{k+1}}\}$ 
such that $\bar\E_{k+1}\prec \E_k$. We can find such an $\bar\E_{k+1}$ by
Lemma~\ref{ELemmaNew}(A). 
Let $\gamma=\max\{\delta\colon \la\delta,m\ra\in A_{k+1}\}<\beta$.

First assume that $\gamma=0$. Then for every $s\in 2^{A_{k+1}}$
the set 
\[
Z_s=\left\{i\in \hat A\colon D_\beta(E,i,n)\cap\P(\bar E_s)=\emptyset\right\}
\]
belongs to $\I$, since otherwise 
$Q=f\left[\pi_\beta^{-1}[\bar E_s]\cap E\right]$ and 
$B=Z_s\in\I^+\cap\P(A)$ would satisfy 
the condition (ii), contradicting our assumption. 
Let us 
define $A'=\hat A\setminus\bigcup\{Z_s\colon s\in 2^{A_{k+1}}\}\in\I^+$
and notice that $A'\subset Z(\bar A,\E_k,n)$. 
Indeed, take an $i\in A'$ and for every $s\in 2^{A_{k+1}}$
choose $E_s\in D_\beta(E,i,n)\cap\P(\bar E_s)$. 
Then $\E_{k+1}\stackrel{\rm def}{=}\{E_s\colon s\in 2^{A_{k+1}}\}\prec\E_k$, since 
(i), (ii), and (sp) hold for $\E_{k+1}$ as they were true for $\bar\E_{k+1}$,
and (ag) is satisfied trivially, by the maximality of $\gamma$. 
Condition $\bigcup\E_{k+1}\in D_\beta(E,i,n)$ is guaranteed by 
the choice of $E_s$'s, so indeed $i\in Z(\bar A,\E_k,n)$. 


Next assume that $\gamma>0$. Let
$B=\{\la\delta,m\ra\in A_{k+1}\colon \delta<\gamma\}$,
and define 
$\E_{k+1}^*=\{E^*_t\in\mathPerf_\gamma\colon t\in 2^B\}$, where
$E^*_t=\pi_\gamma[\bar E_s]$ for any $s\in 2^{A_{k+1}}$ with $t\subset s$.
Note that, by (ag), the definition of $E^*_t$
is independent of the choice of $s$. 
It is easy to see that $\E_{k+1}^*$ satisfies 
(ag) and (sp), where $\alpha$ is replaced by $\gamma$. 
For $\hat A_0\in\P(\hat A)\cap\I^+$ and 
$t\in 2^B$ 
define $D(\hat A_0,t)$ as 
the collection of all $E_0\in\mathPerf_\gamma$
for which there exists an $A'\in\P(\hat A_0)\cap\I^+$ such that 
\[
E_0\in 
\bigcap\left\{
D_\gamma\left(\pi_\beta^{-1}[\bar E_s]\cap E,A',n\right)
\colon {t\subset s\in 2^{A_{k+1}}}
\right\}.
\]
It is clear that each $D(\hat A_0,t)$ is open, since so is 
each $D_\gamma\left(\pi_\beta^{-1}[\bar E_s]\cap E,A',n\right)$.
It is also important to notice that $D(\hat A_0,t)$ is dense below $E^*_t$.
To see it, fix an $E_0\in\mathPerf_\gamma\cap\P(E^*_t)$
and let $\{s_1,\ldots,s_m\}$ be an enumeration of 
$\{s\in 2^{A_{k+1}}\colon t\subset s\}$. 
By induction on $i\leq m$ define two decreasing sequences
$\{E_i\in\mathPerf_\gamma\colon i\leq m\}$ and 
$\{\hat A_i\in\I^+\colon i\leq m\}$ such that
$E_i\in D_\gamma\left(\pi_\gamma^{-1}(E_{i-1})\cap
(\pi_\beta^{-1}[\bar E_{s_i}]\cap E),\hat A_i,n\right)$
provided $0<i\leq m$. The inductive step can be made 
since condition $\psi_\gamma$ holds. Then 
$E_m\in 
\bigcap\left\{
D_\gamma\left(
\pi_\beta^{-1}[\bar E_s]\cap E,\hat A_m,n\right)
\colon {t\subset s\in 2^{A_{k+1}}}
\right\}$ and we have 
$E_m\in D(\hat A_0,t)\cap\P(E_0)$. 

Let $\D$ be the collection of all 
pairwise disjoint families $\E\in[\mathPerf_\gamma]^{<\omega}$ 
for which there exists an $A'\in\P(\hat A)\cap\I^+$
working simultaneously for all $E_0\in\E$, that is, such that 
for all $t\in 2^B$ 
and $E_0\in\E$ if $E_0\subset E^*_t$ then 
\[
E_0\in \bigcap\left\{D_\gamma
\left(\pi_\beta^{-1}[\bar E_s]\cap E,A',n\right)
\colon t\subset s\in 2^{A_{k+1}}\right\}.
\]
Notice that $\D$ satisfies condition 
($\dagger$) from Lemma~\ref{ELemmaNew} used with $\alpha$ 
replaced by $\gamma$.
Indeed, if $\E\in\D$ is witnessed by $A'\in\P(\hat A)\cap\I^+$
and $E\in\mathPerf_\gamma$ is disjoint with
$\bigcup\E$ choose 
$E'\in\mathPerf_\gamma$ below $E$ which is either disjoint with 
$\bigcup\E_{k+1}^*$ or contained in some $E^*_t\in\E_{k+1}^*$. 
If $E'\cap\bigcup\E_{k+1}^*=\emptyset$
then $\{E'\}\cup\E\in\D$ is witnessed by $A'$.
If $E'\subset E^*_t\in\E_{k+1}^*$ 
by the density of $D(A',t)$ below $E^*_t$ we can find 
an $A''\in\P(A')\cap\I^+$ and 
\[
E''\in \P(E')\cap 
\bigcap\left\{
D_\gamma\left(\pi_\beta^{-1}[\bar E_s]\cap E,A'',n\right)
\colon {t\subset s\in 2^{A_{k+1}}}
\right\}.
\]
Then $\{E''\}\cup\E\in\D$ is witnessed by $A''$.

Now, by Lemma~\ref{ELemmaNew}(B), there exists an 
$\hat\E_{k+1}=\left\{\hat E_t\in D\colon t\in 2^B\right\}\in\E$
satisfying (ag) and (sp) such that 
$\hat E_t\subset E^*_t$ for all $t\in 2^B$. 
Let $A'\in\P(\hat A)\cap\I^+$ witness $\hat\E_{k+1}\in\E$.
We will show that $A'\subset Z(\bar A,\E_k,n)$.
So fix an $i\in A'$. Since for every $t\in 2^B$ and $t\subset s\in 2^{A_{k+1}}$
we have 
$\hat E_t\in D_\gamma(\pi_\beta^{-1}[\bar E_s]\cap E,A',n)$
there exists an 
$\hat E_s\in D(\pi_\beta^{-1}[\bar E_s]\cap E,i,n)$
with $\pi_\gamma[\hat E_s]=\hat E_t$. 
Let $E_s=\pi_\beta[\hat E_s]\subset\bar E_s$ and notice that
$\E_{k+1}\stackrel{\rm def}{=}\{E_s\colon s\in 2^{A_{k+1}}\}\prec\E_k$.
Indeed, $\E_{k+1}$ satisfies (i), (ii), and (sp) since 
they were true for $\bar\E_{k+1}$ and 
$\E_{k+1}$ is a refinement of $\bar\E_{k+1}$.
Condition (ag) is satisfied by $\E_{k+1}$
since, by the maximality of $\gamma$, it is non-trivial for $\hat\beta\leq\gamma$ 
and for such $\hat\beta$ it guaranteed by (ag) for $\hat\E_{k+1}$.
Finally, $\bigcup\E_{k+1}\in D_\beta(E,i,n)$ is guaranteed by our definition, so
indeed $i\in Z(\bar A,\E_k,n)$. 
This finishes the proof of ($*$). 

\smallskip

To prove $\psi_\beta$ assume that $\psi_\gamma$ holds for all $\gamma<\beta$.
Fix $E\in\mathPerf_\alpha$,
$n<\omega$, and $\hat A\in\I^+\cap\P(A)$. We need to find an 
$A'\in\I^+\cap\P(\hat A)$
such that $D_\beta(E,A',n)\neq\emptyset$, that is,
$\bigcap_{i\in A'}D_\beta(E,i,n)\neq\emptyset$.
We will construct a tree $T\subset\hat A^{<\omega}$ 
and the mapping $T\ni s\mapsto \E_s\in[\mathPerf_\beta]^{<\omega}$ 
such that $\E_\emptyset=\{E\}$ and 
for every $r\in T$ and $s=r\hat{\ }i\in T$
we have $\E_s\prec\E_r$ and $\bigcup\E_s\in D_\beta(E,i,n)$.
Notice that, by ($*$), for every $r\in T$ 
we can define $\E_{r\hat{\ }i}$ for all $i\in Z(\hat A,\E_r,n)$. 
So we can ensure that $T$ satisfies the assumptions of 
Proposition~\ref{propTreeSel}.
Let $b$ be a branch of $T$ with $A'=b[\omega]\in\I^+$.
By Lemma~\ref{FusionSequenceLemma}
$E_0=\bigcap_{k<\omega}\bigcup\E_{b\restriction k}$ 
belongs to $\mathPerf_\beta$
and $E_0\in \bigcap_{i\in A'}D_\beta(E,i,n)$.
This concludes the proof of $\psi_\beta$.

\medskip 

For the conclusion of the proof we first need to refine the prism $P$. 
For every $i<\omega$ let 
$h_i\colon \Cantor^\alpha\to\omega\subset\real$
be defined by $h_i(g)=f(g)(i)$. 
Clearly each $h_i$ is continuous. 
Hence each set $h_i^{-1}(n)$ is open in $\Cantor^\alpha$, so
\[
D_i=\{E\in\mathPerf_\alpha\colon h_i \mbox{ is constant on } E\}
\]
is dense and open in $\mathPerf_\alpha$. Thus, by 
Corollary~\ref{DiagonalLemma},
there exists an $E\in\bigcap_{i<\omega}D_i^*$, where 
\[
D_i^*=
\left\{\bigcup\D\colon \D\in[D_i]^{2^i} 
\mbox{ and the sets in $\D$ are pairwise disjoint}\right\}.
\]
Let $P_0=f[E]$. Then $P_0$ is a subprism of $P$.
We will find a subprism $Q$ of $P_0$.
Notice also that, by our construction, 
for every $i<\omega$ there is a set $V_i\in[\omega]^{\leq 2^i}$
such that
\begin{center}
$f(g)(i)\in V_i$ for all $g\in E$ and $i<\omega$. 
\end{center}

Notice also that since $\psi_\alpha$ holds, so 
is the conclusion of ($*$) for $\beta=\alpha$.  
In particular, for every $n<\omega$
and $\E_k=\left\{E_s\in\mathPerf_\alpha\colon s\in 2^{A_k}\right\}$
satisfying (i), (ag), and (sp) from Lemma~\ref{FusionSequenceLemma}
and such that $\bigcup\E_k\subset E$ we have 
\[
Z(A,\E_k,n)=\left\{i\in A\colon
(\exists \E_{k+1}\prec \E_k)\ \bigcup\E_{k+1}\in D(i,n)\right\}
\]
and $A\setminus Z(A,\E_k,n)\in\I$. 

\medskip

We will construct a tree $T\subset A^{<\omega}$ 
as in Proposition~\ref{propTreeSel}
and the mapping $T\ni s\mapsto \E_s\in[\mathPerf_\alpha]^{<\omega}$.
The construction is done by induction on the levels of $T$.
We start with 
$\E_\emptyset=\{E\}$ and, for every $r\in T$ 
and $s=r\hat{\ }i\in T$, we ensure that $\E_s\prec\E_r$ and 
\begin{equation}\label{conEs}
\bigcup\E_s\in\bigcap\left\{D(i,n)\colon 
n\in V_j \mbox{ for some } j\in\range(r)\right\}.
\end{equation}
Notice that for every $r\in T$ if 
$Z_r=\bigcap\left\{Z(A,\E_r,n)\colon n\in\bigcup_{j\in\range(r)}V_j\right\}$ 
then $A\setminus Z_r\in\I$.
Moreover, for all $i\in Z_r$ we can find 
$\E_{r\hat{\ }i}$ as in (\ref{conEs}). 
So, $T$ as above can be constructed. 
Take a branch $b$ of $T$ with $B=b[\omega]\in\I^+$ and 
$E_0=\bigcap_{k<\omega}\bigcup\E_{b\restriction k}\in\mathPerf_\beta$.
Then $B$ and $Q=f[E_0]$ satisfy (i). 
This finishes the proof of (a). 

\medskip

(b) Since the characteristic function $\charf{}$ gives an embedding
from $[\omega]^\omega$ into $2^\omega\subset\omega^\omega$
the prism $P$ can be identified with $\charf{}[P]=\{\charf{x}\colon x\in P\}$.
Applying part (a) to $\charf{}[P]$ we can find
a subprism $Q$ of $P$, $k<2$, and $B\in\I^+$, $B\subset A$,
such that $\charf{x}\restriction B\equiv k$ for every $x\in Q$.
If $k=0$ this gives $x\cap B=\emptyset$ for every $x\in Q$.
If $k=1$ we have $B\subset x$ for every $x\in Q$.
\qed



\noindent{\sc Proof of Theorem~\ref{thm:sel}.}
Let $\I$ be a selective ideal on $\omega$. 
For a countable family $\A\subset\I^+$ linearly ordered by $\subset^*$
let $C(\A)\in\I^+$ be such that $C(\A)\subset^* A$ for every $A\in\A$. 

For 
$A\in\I^+$ 
and $f\in\Fpr(\omega^\omega)$ put $P=\range(f)$ and
let $B(A,P)\in[A]^\omega$ and a subprism $Q(A,P)$ of $P$ 
be as in Lemma~\ref{lem:sel}(a).
If $f\in\Cpr(\omega^\omega)$ and $P=\range(f)=\{x\}$ 
then we put $Q(A,P)=P$ and take $B(A,P)\in[A]^\omega$
satisfying the conclusion of Lemma~\ref{lem:sel}(a).

Consider the following strategy $S$ for Player~II:
\[
S(\la\la P_\eta,Q_\eta\ra\colon \eta<\xi\ra,P_\xi)=
Q(C(\{B_\eta\colon\eta<\xi\}),P_\xi),
\]
where the sets $B_\eta$ are defined inductively by
$B_\eta=B(C(\{B_\zeta\colon\zeta<\eta\}),P_\eta)$. 

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 $S$ 
in which and Player~II loses, that is, 
$\omega^\omega=\bigcup_{\xi<\omega_1}Q_\xi$. 

Now, let $\F$ be a filter generated by $\{B_\xi\colon\xi<\omega_1\}$
and notice that $\F$ is a selective ultrafilter.
It is a filter, since $\{B_\xi\colon\xi<\omega_1\}$
is decreasing with respect to $\subset^*$. 
It also easy to see that 
\begin{quote}
for every $f\in\omega^\omega$
there exists a $B\in\F$ such that 
$f\restriction B$ is either one-to-one 
or constant.
\end{quote}
Indeed, if $f\in\omega^\omega$ then there exists a $\xi<\omega_1$ such that
$f\in Q_\xi$. Then $B=B_\xi$ is as desired.

Now, to see that $\F$ is an ultrafilter take an $A\subset\omega$ and let 
$f\in\omega^\omega$ be a characteristic function of $A$.
Then $B\in\F$ as above is a subset of either $A$ 
or its complement. 

It is easy to see that the above two properties imply that
$\F$ is a selective ultrafilter. \qed


Notice that \psmPrGame\ implies also that we have many different 
selective ultrafilters. The consistency of this fact, 
in a model obtained by adding many side-by-side Sacks reals,
was first noticed by Hart in~\cite{Hart}.


\rem{remManySel}{\psmPrGame\ implies that there are 
$\omega_2$ different selective ultrafilters. }

\proof This can be easily deduced by a simple transfinite induction
from 
\begin{itemize}
\item[($*$)] for every family $\U=\{\F_\xi\colon\xi<\omega_1\}$
of ultrafilters on $\omega$ there is a selective
ultrafilter
$\F\notin\U$,.
\end{itemize}
Property ($*$) is proved as above, where we use $\I=[\omega]^{<\omega}$
and the operator $C(\{B_\eta\colon\eta<\xi\})$
is replaced with $C_\xi(\{B_\eta\colon\eta<\xi\})\notin\F_\xi$.
\qed

Now, from the above we obtain that  
``$2^{\omega_1}=\omega_2$''+\psmPrGame\ (which is consistent) 
implies that
there are $2^{\omega_1}$ different selective ultrafilters. 
Since \psm\ is also consistent with $2^{\omega_1}>\omega_2$
it is worth noticing that 
the existence of $2^{\omega_1}$ different selective ultrafilters
can be also deduced from a slightly stronger 
version of \psmPrGame\ also in this case.
This can be found in~\cite{CPAbook}. 

\medskip 


Recall that the number $\ultraN$ is defined as the smallest 
cardinality of the base for a non-principal ultrafilter 
on $\omega$. Thus Theorem~\ref{thm:sel} and 
Corollaries~\ref{corSelUltra} and~\ref{corReapSigma}
imply that

\cor{cor:numberU}{\psmPrGame\ implies that 
$\ultraN=\reap_\sigma=\omega_1$.}

\subsection{Non-selective $P$-points and number $\indN$}


Recall that an ultrafilter $\F$ on $\omega$ is a 
{\em $P$-point}\/ 
provided for every partition $\P$ of $\omega$ either 
$\P\cap\F\neq\emptyset$ or
there is an $F\in\F$ such that $|F\cap P|<\omega$
for all $P\in\P$. 
Clearly every selective ultrafilter is a $P$-point. 
Thus, \psmPrGame\ implies the existence of
a $P$-point. On the other hand,
Shelah proved that 
there are models with no $P$-points. (See 
e.g.~\cite[thm.~4.4.7]{BJ}.)
Hart in~\cite{Hart} proved that 
in a model obtained by adding many side-by-side Sacks reals
there is a $P$-point which is not selective. 
Next, we will prove that this follows also from
\psmPrGame. The main idea of the proof is 
the same as that used in~\cite{Hart}. 

For $m<\omega$ let $P_m=\{n<\omega\colon 2^m-1\leq n<2^{m+1}-1\}$
and define a partition $\P$ of $\omega$ by
$\P=\{P_m\colon m<\omega\}$. Consider the following ideal
$\bar\I$ on $\omega$
\begin{equation}\label{defI}
\bar\I=\left\{A\subset\omega\colon 
\limsup_{m\to\infty} |A\cap P_m|<\omega\right\}
\end{equation}
and notice the following simple fact.

\fact{factSigmaCl}{If $\A\in [\bar\I^+]^{\leq\omega}$
is linearly ordered by $\subset^*$ then
there is a $C(\A)\in\bar\I^+$ such that 
$C(\A)\subset^* A$ for all $A\in\A$. 
}

\proof Let $\{A_n\colon n<\omega\}\subset\A$
be a $\subset^*$-decreasing sequence coinitial with $\A$.
For every $i<\omega$ choose $m_i<\omega$ and 
$C_i\in [P_{m_i}]^i$ such that 
$C_i\subset \bigcap_{j\leq i}A_j$. Then 
$C(\A)=\bigcup_{i<\omega}C_i$ is as desired. \qed

To construct a nonselective $P$-point we are going 
to prove the following theorem. 


\thm{th:nonselP}{If \psmPrGame\  holds then there exists
a $\subset^*$-decreasing sequence 
$\B=\{B_\xi\in \bar\I^+\colon \xi<\omega_1\}$
such that the filter 
$\F$ generated by $\B$ is an ultrafilter on~$\omega$. 
}

Notice that from this we will immediately deduce
the required result.

\cor{cor:nonselP}{If \psmPrGame\  holds then there exists
a nonselective $P$-point.}

\proof Let $\F$ be as in Theorem~\ref{th:nonselP}. 
Clearly $\F$ is nonselective, since $\P$
is disjoint with $\bar\I^+\supset \F$
and every selector of $\P$ is in
$\bar\I\subset\P(\omega)\setminus\F$.  The fact that $\F$ is a
$P$-point follows from the fact that
$\F$ has a base linearly ordered by $\subset^*$.
Indeed, if $\{S_n\colon n<\omega\}\subset\P(\omega)\setminus\F$ 
is a partition of $\omega$ then for every 
$m<\omega$ there is $\xi_m<\omega_1$
such that $B_{\xi_m}\subset^*\omega\setminus\bigcup_{n\leq m}
S_n$. Let $\beta<\omega_1$ be such that 
$B_\beta\subset^*B_{\xi_m}$ for all $m<\omega$.
Then $F=B_\beta\in\F$ is such that $|F\cap S_n|<\omega$
for all $n<\omega$. \qed

The proof of Theorem~\ref{th:nonselP}
will be based on the following lemma, 
which is analogous to Lemma~\ref{lem:sel}.
Note, that although the statement of this lemma is 
identical to that of Lemma~\ref{lem:sel}(b), we cannot
apply this lemma here, since the ideal $\bar\I$
is not weakly selective. 



\lem{lem:nonselP}{Let $\bar\I$ be as in (\ref{defI}). 
Then for every $A\in \bar\I^+$ and a prism $P$ in $2^\omega$
there exist a $B\in \bar\I^+$, $B\subset A$, a subprism $Q$ of $P$,
and a $j<2$ such that
\begin{itemize}
\item[{\rm ($\circ$)}]
$g\restriction B$ is constant equal to $j$ for every $g\in Q$.   
\end{itemize}
}

\proof 
Fix an $A\in \bar\I^+$ and an $f\in\Fpr(2^\omega)$ from
$\Cantor^\alpha$ onto $P$.
Since $A\in \bar\I^+$ for every $k<\omega$ 
we can find an $m_k<\omega$ such that 
$|A\cap P_{m_k}|\geq k\ 2^{2^k}$. 
First we will construct a subprism $Q_0$ of $P$
and a sequence 
$\la A_k\in[A\cap P_{m_k}]^k\colon k<\omega\ra$
such that for every $k<\omega$
\begin{equation}\label{con:nonselP1}
\mbox{$g\restriction A_k$ is constant for every $g\in Q_0$.}
\end{equation}
This will be done using Lemmas~\ref{ELemmaNew} 
and~\ref{FusionSequenceLemma}. 
So, for each $k<\omega$ let  
$\D_k$ be the collection of all
pairwise disjoint families $\E\in[\mathPerf_\alpha]^{<\omega}$
such that for every $E\in\E$
\begin{equation}\label{con:nonselP1AA}
\mbox{$f(h)\restriction P_{m_k}=f(h')\restriction P_{m_k}$
for all $h,h'\in E$.}
\end{equation}
Clearly each $\D_k$ satisfies condition ($\dagger$)
from Lemma~\ref{ELemmaNew}, so by an easy induction
we can find a sequence $\la\E_k\in\D_k\colon k<\omega\ra$ 
satisfying the assumptions of Lemma~\ref{FusionSequenceLemma}.
Let $E_0=\bigcap_{k<\omega}\bigcup\E_k\in\mathPerf_\alpha$.
We will show that $Q_0=f[E_0]$ satisfies (\ref{con:nonselP1}).


Indeed, fix a $k<\omega$ and notice that 
$\E_k=\{E_i\colon i<2^k\}$. For each $i<2^k$ choose an $h_i\in E_i$
and define $\p\colon A\cap P_{m_k}\to 2^{2^k}$ by 
$\p(p)(i)=f(h_i)(p)$. Since $|A\cap P_{m_k}|\geq k\ 2^{2^k}$,
by the pigeon hole principle
we can find an $s\in 2^{2^k}$ such that $|\p^{-1}(s)|\geq k$.
Choose an $A_k\in[\p^{-1}(s)]^{k}$.
Then for every $i<2^k$ and $p\in A_k$ we have
$f(h_i)(p)=\p(p)(i)=s(i)$. So, $f(h_i)\restriction A_k$
is constant equal to $s(i)$. Combining this with the inclusion 
$A_k\subset P_{m_k}$ and the condition (\ref{con:nonselP1AA})
we obtain (\ref{con:nonselP1}).

To finish the proof fix a selector 
$\bar A$ from the family $\{A_k\colon k<\omega\}$.
Then $\bar A\in\I^+$, where $\I$ is the ideal of finite subsets of
$\omega$. Applying Lemma~\ref{lem:sel}(a) to $\bar A$ and $Q_0$
we can find a $j<2$, an $S\in[\bar A]^\omega$, and a subprism $Q$ of $Q_0$
such that $g\restriction S$ is constant equal to $j$
for every $g\in Q$.
Put $B=\bigcup_{k\in S}A_k$.
Then, by (\ref{con:nonselP1}), 
$g\restriction B$ is constant equal to $j$ for every $g\in Q$.

It is clear that $B\in\bar\I^+\cap\P(A)$, since it is a union of infinitely 
many sets $A_k\in[A\cap P_{m_k}]^k$. 
\qed 

Note also that, similarly as for Remark~\ref{remManySel},
the conclusion of the following fact holds
in a model obtained by adding many side-by-side Sacks reals. 
This was first noticed by Hart in~\cite{Hart}.


\rem{remManyNonselP}{\psmPrGame\ implies that there are 
$\omega_2$ different nonselective $P$-points. }

The existence of $2^{\omega_1}$ different such ultrafilters 
follows also from a slightly stronger 
version of \psmPrGame. This can be found in~\cite{CPAbook}.

\medskip 



Recall also that a family $\J\subset[\omega]^\omega$
is an {\em independent family}\/ provided the set 
\[
\bigcap_{A\in\A}A \cap\bigcap_{B\in\B}(\omega\setminus B)
\]
is infinite for every disjoint finite subsets $\A$ and $\B$ of $\J$. 
It is often convenient to express this definition in a slightly
different notation. Thus, for $W\subset\omega$ let $W^0=W$ 
and $W^1=\omega\setminus W$. A family $\J\subset[\omega]^\omega$
is independent provided the set 
\[
\bigcap_{W\in\J_0}W^{\tau(W)}
\]
is infinite for every finite subset $\J_0$ of $\J$ and $\tau\colon\J_0\to\{0,1\}$.

The {\em independence}\/ cardinal $\indN$ is defined as follows:
\[
\indN=\min\{|\J|\colon \J\ \mbox{is infinite maximal independent family}\}. 
\]
According to Andreas Blass~\cite[sec. 11.5]{Blass}
the fact that the equation $\indN=\omega_1$ 
holds in the iterated perfect set model
was first proved by Eisworth and Shelah (unpublished). 




\thm{thm:i}{\psmPrGame\ implies that $\indN=\omega_1$.}

The proof of the theorem is based on the following lemma.  
We say that a family $\W\subset[\omega]^\omega$ {\em separates points}\/
provided for every $k<\omega$ there are $U,V\in\W$ such that $k\in U\setminus V$. 


\lem{lem:i}{For every countable independent family
$\W\subset[\omega]^\omega$ separating points 
and a prism $P$ in $[\omega]^\omega$
there exist $W\in [\omega]^\omega$ and a subprism $Q$ of $P$ 
such that $\W\cup\{W\}$ is independent 
but $\W\cup\{W,x\}$ is not independent for every $x\in Q$. 
}

\proof Let $\W=\{W_i\colon i<\omega\}$ and let $\p\colon \omega\to 2^\omega$
be a Marczewski function for $\W$, that is,
for $i,k<\omega$
\[
\p(k)(i)=\left\{
\begin{array}{ll}
1 &\mbox{ for } k\in W_i\\
0 &\mbox{ for } k\notin W_i.
\end{array}\right.
\]
Note that $\p$ is one-to-one, since $\W$ separates points. 
Notice also that for every $k,n<\omega$ and $\tau\in 2^n$
\[
k\in \bigcap_{i<n}W_i^{\tau(i)} 
\Equi\ (\forall i<n)\ k\in W_i^{\tau(i)}
\Equi\ (\forall i<n)\ \p(k)(i)=\tau(i)
\Equi\ \tau\subset\p(k).
\]
Now, if $[\tau]=\{t\in 2^\omega\colon \tau\subset t\}$
then sets $\{[\tau]\colon \tau\in 2^{<\omega}\}$ form a base for $2^\omega$
and
\begin{equation}\label{eq:PHI}
k\in \bigcap_{i<n}W_i^{\tau(i)} \Equi\ \p(k)\in[\tau].
\end{equation}
Thus, 
independence of $\W$ implies that $\p[\omega]$ is dense in $2^\omega$,
so it is homeomorphic to the set $\rational$ of rational numbers. 
Note also that from (\ref{eq:PHI}) it follows immediately that
\begin{itemize}
\item[(a)] if $W\subset\omega$ is such that $\p[W]$ 
and $\p[\omega]\setminus\p[W]$
are dense then $\W\cup\{W\}$ is independent;

\item[(b)] if $W,x\subset\omega$ are such that for some $\tau\in 2^{<\omega}$
      either $\p[x\cap W]\cap[\tau]=\emptyset$
      or $\p[W]\cap[\tau]\subset\p[x]$
      then $\W\cup\{W,x\}$ is not independent.
\end{itemize}
Let 
$\I$ be the ideal of nowhere dense subsets of $\p[\omega]$. 
Then, by Fact~\ref{factWeakSelQ}, $\I$ is weakly selective, since 
$\p[\omega]$ is homeomorphic to $\rational$.
So, identifying $\p[\omega]$ with $\omega$ 
and applying Lemma~\ref{lem:sel}(b), we can 
find a subprism $Q$ of $P$ and a 
$V\in[\omega]^\omega\setminus\I$ such that either 
\begin{itemize}
\item  $x\cap V=\emptyset$ for every $x\in Q$, or else
\item  $V\subset x$ for every $x\in Q$.   
\end{itemize}
Since $V\notin\I$, there exists a $\tau\in2^{<\omega}$ such that
$\p[V]$ is dense in $[\tau]$. Trimming $V$, if necessary, we can assume that
$\p[V]\subset[\tau]$ and that 
$\p[\omega\setminus V]$ is also dense in $[\tau]$.
Now let $W\supset V$ be such that $\p[W]\cap[\tau]=\p[V]$
and both $\p[W]$ and $\p[\omega\setminus W]$ are dense in $\p[\omega]$.
Then, by (a) and (b), $\W\cup\{W\}$ is independent
while $\W\cup\{W,x\}$ is not independent for every $x\in Q$. \qed


\noindent{\sc Proof of Theorem~\ref{thm:i}.}
For a countable independent family
$\W\subset[\omega]^\omega$ separating points 
and an $f\in\Fpr([\omega]^\omega)$ put $P=\range(f)$ and
let $W(\W,P)\in [\omega]^\omega$ and a subprism $Q(\W,P)$ of $P$ 
be as in Lemma~\ref{lem:i}.
If $f\in\Cpr([\omega]^\omega)$ and $P=\range(f)=\{x\}$ 
then we put $Q(\W,P)=P$ and define $W(\W,P)$ as an arbitrary $W$ 
such that $\W\cup\{W\}$ is independent while $\W\cup\{W,x\}$ is not. 



Let $\A_0\subset[\omega]^\omega$ be an arbitrary countable
independent family  separating points 
and consider the following strategy
$S$ for Player~II:
\[
S(\la\la P_\eta,Q_\eta\ra\colon \eta<\xi\ra,P_\xi)=
Q(\A_0\cup\{W_\eta\colon\eta<\xi\},P_\xi),
\]
where sets $W_\eta$ are defined inductively by
$W_\eta=W(\A_0\cup\{W_\zeta\colon\zeta<\eta\},P_\eta)$. 

By \psmCGame\ 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 and Player~II loses, that is, 
$[\omega]^\omega=\bigcup_{\xi<\omega_1}Q_\xi$. 

Now, notice that the family $\J=\A_0\cup\{W_\xi\colon \xi<\omega_1\}$ is a 
maximal independent family. 
It is clear that $\J$ is independent, since
every set $W_\xi$ was chosen so that 
$\A_0\cup\{W_\zeta\colon \zeta\leq\xi\}$ is independent. 
To see that $\J$ is maximal it is enough to note that
every $x\in[\omega]^\omega$ belongs to a $Q_\xi$ for some
$\xi<\omega_1$, and so
$\A_0\cup\{W_\zeta\colon \zeta\leq\xi\}\cup\{x\}$ is not independent.
\qed


By Theorem~\ref{thm:i} we see that \psmPrGame\ implies
the existence of an independent family of size $\omega_1$. 
Next, answering a question of Michael Hru\v s\'ak~\cite{Hru}
we show that such a family can be 
simultaneously a splitting family. 
This is similar in flavor to Theorem~\ref{thm:ar}.
In the proof we will use the following lemma.

\lem{lem:is}{For every countable family $\V\subset[\omega]^\omega$ 
and a perfect set $P$ in $[\omega]^\omega$
there exists a $W_1\in[\omega]^\omega$ such that
$\V\cup\{W_1\}$ is independent and $W_1$ splits every $A\in P$.}

\proof We follow the argument from \cite[p.~121]{vD}
that ${\mathfrak s}\leq{\mathfrak d}$. 


For every $A\in[\omega]^\omega$ let $b_A$ be a strictly increasing
bijection from $\omega$ onto $A$. Then $b\colon [\omega]^\omega\to\omega^\omega$
defined by $b(A)=b_A$ is continuous. 
In particular $b[P]=\{b_A\colon A\in P\}$ is compact, so there exists
a strictly increasing $f\in\omega^\omega$ such that
$b_A(n)<f(n)$ for every $A\in P$ and $n<\omega$. 
For $n<\omega$ let $f^n$ denote the $n$-fold composition of $f$
and let $S_n=\{m<\omega\colon f^n(0)\leq m<f^{n+1}(0)\}$.
Then $f^n(0)\leq b_A(f^n(0))<f(f^n(0))=f^{n+1}(0)$ for every 
$A\in P$ and $n<\omega$. In particular, for every $A\in P$
\[
S_n\cap A\neq\emptyset.
\]
So, if $T\subset\omega$ be infinite and co-infinite 
and $W_1=\bigcup_{n\in T}S_n$ then
$W_1$ splits every $A\in P$. 
Thus, it is enough to take infinite and co-infinite
$T\subset\omega$ such that 
$\V\cup\{W_1\}$ is independent. \qed
 





\thm{thm:is}{\psmPrGame\ implies that there exists a family $\F\subset[\omega]^\omega$
of cardinality $\omega_1$ which is simultaneously 
independent and splitting.}

\proof The proof is just a slight modification of that for Theorem~\ref{thm:i}.
(Compare also Theorem~\ref{thm:ar}.)

For a countable independent family
$\W\subset[\omega]^\omega$ separating points 
and an $f\in\Fpr([\omega]^\omega)$ put $P=\range(f)$ and
let $W_0\in [\omega]^\omega$ and a subprism $Q$ of $P$ 
be as in Lemma~\ref{lem:i}.
Let $W_1$ be as in Lemma~\ref{lem:is} used with $P=Q$ and $\V=\W\cup\{W_0\}$.
We put $Q(\W,P)=Q_1$ and $\W(\W,P)=\{W_0,W_1\}$. 

If $f\in\Ccube([\omega]^\omega)$ and $P=\range(f)=\{x\}$ 
then we put $Q(\W,P)=P$ and 
$\W(\W,P)=\{W_0,W_1\}$, where $W_0$ and $W_1$ are such that
$\W\cup\{W_0,W_1\}$ is independent and $W_1$ splits $P=\{x\}$.


Let $\A_0\subset[\omega]^\omega$ be an arbitrary countable
independent family  separating points 
and consider the following strategy
$S$ for Player~II:
\[
S(\la\la P_\eta,Q_\eta\ra\colon \eta<\xi\ra,P_\xi)=
Q(\A_0\cup\bigcup\{\W_\eta\colon\eta<\xi\},P_\xi),
\]
where $\W_\eta$'s are defined inductively by
$\W_\eta=\W(\A_0\cup\bigcup\{\W_\eta\colon\eta<\xi\},P_\eta)$. 

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 $S$ 
in which and Player~II loses, that is, 
$[\omega]^\omega=\bigcup_{\xi<\omega_1}Q_\xi$. 
Then the family $\F=\A_0\cup\bigcup\{\W_\xi\colon \xi<\omega_1\}$ is 
independent and splitting. \qed

\subsection{Crowded ultrafilters on $\rational$}






Let $\perf(\rational)$ stand for the family of all
closed subsets $A$ of $\rational$ without isolated points, 
that is, such that their closures 
$\cl_\real(A)$ in $\real$ are perfect sets. 
Recall that an ideal $\I$ on $\rational$ of is 
{\em crowded}\/ provided $\I^+=\P(\rational)\setminus\I$ 
is generated by
the sets from $\perf(\rational)$. 
Crowded ultrafilters were studied 
by several authors (see e.g. \cite{vD2,CH})
in connection with the reminder $\beta\,\rational \setminus\rational$ 
of the \v Cech-Stone 
compactification $\beta\,\rational$ of $\rational$. 

In what follows we will also use the following simple fact,
in which a non-scattered subset of $\rational$ 
is understood as a set containing a subset dense in itself. 


\fact{fact:crow}{Every non-scattered set $B\subset\rational$ 
contains a subset from $\perf(\rational)$. 
}

\proof Since $B$ is non-scattered, decreasing it, if necessary, we
can assume that $B$ is dense in itself.
Let $\{k_n\leq n\colon n<\omega\}$ be an
enumeration of $\omega$ with infinitely many repetitions
and let $\rational\setminus B=\{a_n\colon n<\omega\}$.
By induction construct a sequence 
$\la\la p_n,U_n\ra\in B\cap \P(\rational)\colon n<\omega\ra$
such that $p_n\in B\setminus\bigcup_{i<n}U_i$ and 
$|p_n-p_{k_n}|<2^{-n}$
while $U_n\ni a_n$ is a clopen subset of 
$\rational\setminus\{p_i\colon i\leq n\}$. 
Then $\rational\setminus \bigcup_{n<\omega}U_n\subset B$ is as desired.
\qed


The following theorem answers in positive 
a question of M.~Hru\v s\'ak~\cite{Hru}
on whether there exists a crowded ultrafilter in the iterated
perfect set model.


\thm{thm:crow}{\psmPrGame\ implies there exists a non-principal  
ultrafilter on $\rational$ which is crowded.}

Fix a $p\in\real\setminus\rational$ and 
for a family $\D\subset\P(\rational)$
let $F(\D)$ denote a filter on $\rational$ 
generated by the family $\D\cup\{I_n\cap\rational\colon n<\omega\}$,
where $I_n=[p-2^{-n},p+2^{-n}]$. 
The proof of the theorem is based on the following lemma,
in which $[\rational]^\omega$ is considered with the same topology as
$[\omega]^\omega$ upon natural identification. 

\lem{lem:crow}{Let $\D\subset\perf(\rational)$ be a countable family 
such that $F(\D)$ is non-trivial. Then for
every prism $P$ in $[\rational]^\omega$ there exist a 
subprism $Q$ of $P$ and a $Z\in\perf(\rational)$
such that $F(\D\cup\{Z\})$ is non-trivial and either 
\begin{itemize}
\item[(i)] $Z\cap x=\emptyset$ for every $x\in Q$, or else 
\item[(ii)] $Z\subset x$ for every $x\in Q$.   
\end{itemize}
}

\proof In what follows we will identify $[\rational]^\omega$
with $2^\rational$, the identification 
mapping given by the characteristic function. 
Thus, we will consider $P$ as a prism in~$2^\rational$. 
Fix an $f\in\Fpr(2^\rational)$ from $\Cantor^\alpha$ onto $P$.

Let 
$\{D_n\in\perf(\rational)\colon n<\omega\}$ be a cofinal sequence in 
$F(\D)$ with a property 
that $D_{n+1}\subset D_n\subset I_n$ for every $n<\omega$. 
Choosing a subsequence, if necessary, we can find
disjoint intervals $J_n$ such that 
$K_n=D_n\cap J_n\in\perf(\rational)$. 

First we will show that there exist 
a sequence $\la B_n\subset K_n\colon n<\omega\ra$
of non-scattered sets and a subprism $P_0$ of
$P$ such that
\begin{equation}\label{bullet}
\mbox{$g\restriction B_n$ is constant for every $g\in P_0$ and $n<\omega$.} 
\end{equation}

For each $n<\omega$ let 
$\D_n$ be the collection of all
pairwise disjoint families $\E\in[\mathPerf_\alpha]^{<\omega}$
for which there exists a non-scattered set $B_n\subset K_n$
with the property that 
\begin{center}
$g\restriction B_n$ is constant for every $g\in f\left[\bigcup\E\right]$. 
\end{center}
To see that the families $\D_n$ satisfy condition
($\dagger$) from Lemma~\ref{ELemmaNew}
it is enough to notice that for every 
non-scattered set $B\subset\rational$ and every prism 
$P_1$ there is a subprism $Q_1$ of $P_1$ and a 
non-scattered subset $B'$ of $B$
such that $g\restriction B'$ is constant for every $g\in Q_1$.
But $B$ contains a subset $W$ homeomorphic to $\rational$.
So, by Fact~\ref{factWeakSelQ}, 
the ideal $\I$ of nowhere dense subsets of $W$
is weakly selective. So, applying 
Lemma~\ref{lem:sel} to this ideal and the prism $P_1$
we can find a $B'\in\I^+$, which clearly is not scattered, 
and a $Q_1$ as desired. 


Thus, using Lemma~\ref{ELemmaNew}, 
we can find
a sequence $\la\E_n\in\D_n\colon n<\omega\ra$ 
satisfying the assumptions of Lemma~\ref{FusionSequenceLemma}.
Let $E_0=\bigcap_{n<\omega}\bigcup\E_n\in\mathPerf_\alpha$.
It is easy to see that the sets $B_n$ witnessing $\E_n\in\D_n$ 
and $Q_0=f[E_0]$
satisfy (\ref{bullet}).
Notice also that by Fact~\ref{fact:crow}
we can assume that $B_n\in\perf(\rational)$ for every $n<\omega$. 

Now let $A$ be a selector from the family $\{B_n\colon n<\omega\}$.
Then $A\in\I^+$, where $\I$ is the ideal of finite subsets of $\rational$.
Applying Lemma~\ref{lem:sel}(a) to $A$ and $P_0$
we can find $i<2$, $S\in[A]^\omega$, and a subprism $Q$ of $P_0$
such that $g\restriction S$ is constant 
equal to $i$ for every $g\in Q$.
Put $Z=\bigcup_{n\in S}B_n$.
Then, by (\ref{bullet}), 
$g\restriction Z$ is constant for every $g\in Q$.
Finally, note that $Z\in\perf(\rational)$
since $Z$ is closed, as $B_n\to p\notin\rational$. 
\qed 


\noindent{\sc Proof of Theorem~\ref{thm:crow}.}
For a prism $P$ in $[\rational]^\omega$ and 
a countable family $\D\subset\perf(\rational)$
for which 
$F(\D)$ is non-trivial
let $Z(\D,P)\in\perf(\rational)$ and a subprism $Q(\D,P)$ of $P$
be as in Lemma~\ref{lem:crow}. 
Consider the following strategy $S$ for Player~II:
\[
S(\la\la P_\eta,Q_\eta\ra\colon \eta<\xi\ra,P_\xi)=
Q(\{Z_\eta\colon\eta<\xi\},P_\xi),
\]
where sets $Z_\eta$ are defined inductively by
$Z_\eta=Z(\{Z_\zeta\colon\zeta<\eta\},P_\eta)$. 

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 $S$ 
in which Player~II loses, that is, 
$[\rational]^\omega=\bigcup_{\xi<\omega_1}Q_\xi$. 

Now, let $\F=F(\{Z_\xi\colon\xi<\omega_1\})$.
Then clearly $\F$ is a crowded non-principal filter. 
To see that it is maximal, take an $x\in[\rational]^\omega$.
Then there is a $\xi<\omega_1$ such that
$x\in Q_\xi$. Then either 
$Z_\xi\cap x=\emptyset$ or $Z_\xi\subset x$. 
Thus, either $x$ or its complement belong to $\F$. 
\qed


Note also that similarly as for Remarks~\ref{remManySel}
and~\ref{remManyNonselP} we can argue that there are many 
non-principal crowded ultrafilters. 

\rem{remManyCrowded}{\psmPrGame\ implies that there are 
$\omega_2$-many different crowded non-principal ultrafilters. }

The existence of $2^{\omega_1}$ different such ultrafilters 
follows also from a slightly stronger 
version of \psmPrGame. This can be found in~\cite{CPAbook}.

\medskip 

The construction of crowded ultrafilters is quite similar
to that of selective ultrafilters and of nonselective $P$-points.
This similarity suggests that it may be possible to construct
a crowded ultrafilter which is also selective.
This, however, cannot be done: 

\prop{prop:NoCrowSel}{There is 
no non-principal crowded ultrafilter on $\rational$
which is also a $P$-point.}

\proof Let $\F$ be a non-principal crowded ultrafilter on $\rational$
and let $\{x\}=\bigcap_{F\in\F}\cl_\real(F)\in\real\setminus\rational$. 
Then $I_n=(x-2^{-n},x+2^{-n})\cap\rational$ belongs
to $\F$ for every $n<\omega$. 
Let 
$\P=\{I_n\setminus I_{n+1}\colon n<\omega\}\cup\{\rational\setminus I_0\}$.
Then $\P$ is a partition of $\omega$ disjoint with $\F$. 
It is also easy to see that if $F\subset\omega$
is such that $|F\cap P|<\omega$ for every $P\in\P$ then $F\notin\F$. 
\qed

It is also not difficult to show that no 
non-principal crowded ultrafilter on $\rational$
can be a $Q$-point. 

It is worth mentioning that \psmPrGame\ 
implies also the existence of many other 
kinds of ultrafilters, like those constructed in~\cite{Hart}.
In fact, many constructions that are done under 
CH can be carried out also under \psmPrGame. 
However, this always needs some combinatorial
lemma, such as Lemma~\ref{lem:sel},
which allows replacing points with prisms.



\section{Proof of fusion Lemma ~\ref{ELemmaNew}
and of consistency of \psmPrGame}

\noindent{\sc Proof of Lemma~\ref{ELemmaNew}.}
For $s\in 2^{A_{k}}$ and $j<2$ let
$s\hat{\ }j$ stand for 
$s\cup\{\la \la\beta_{k},n_{k}\ra,j\ra\}\in 2^{A_{k+1}}$. 

Let $\{s_i\colon i<2^{k+1}\}$ be an enumeration of
$2^{A_{k+1}}$.
By induction on $i<2^{k+1}$ we will construct 
a sequence $\la x_{s_i}\in\Cantor^\alpha\colon i<2^{k+1}\ra$
such that for every $i<2^{k+1}$
\begin{itemize}
\item[(a)] if $s_i=s\hat{\ }j$, where $s\in 2^{A_{k}}$ and $j<2$,
      then $x_{s_i}\in E_s$, 
\item[(b)] if $m<i$ and 
$\beta=\max\{\bar\beta\colon 
s_i\restriction(\bar\beta\times\omega)=
s_m\restriction(\bar\beta\times\omega)\}$
      then 
\begin{center}
$x_{s_i}\restriction\beta=x_{s_m}\restriction\beta$
      and $x_{s_i}(\beta)\neq x_{s_m}(\beta)$. 
\end{center}
\end{itemize}
The point $x_{s_0}$ is chosen arbitrarily from $E_{s_0\restriction A_k}$.
To make an inductive step, if for some $0<i\leq 2^{k+1}$ points
$\{x_{s_m}\colon m<i\}$  are already constructed
choose an $\bar m<i$ for which  $\beta$ as in (b) is maximal.
Notice that by the inductive assumption and the condition (ag) we have 
$x_{s_{\bar m}}\restriction\beta\in
\pi_\beta[E_{s_{\bar m}\restriction A_k}]=
\pi_\beta[E_{s_i\restriction A_k}]$.
So we can choose 
an $x_{s_i}\in E_{s_i\restriction A_k}$ extending 
$x_{s_{\bar m}}\restriction\beta$
and such that $x_{s_i}(\beta)\neq x_{s_m}(\beta)$ for all $m<i$.
It is easy to see that such $x_{s_i}$ satisfies (a) and 
the condition (b) for $m=\bar m$. For other $m<i$ condition (b) follows
from the maximality of $\beta$ 
and the assumption that $\E_k$
satisfies (ag) and (sp).

Conditions (a) and (b) imply that 
$\E_{k+1}'=\left\{\{x_s\}\colon s\in 2^{A_{k+1}}\right\}$ 
satisfy (A) except for being a subset of $\mathPerf_\alpha$. 
Let $\e\in (0,2^{-(k+1)}]$ be small enough that for every 
$m<i<2^{k+1}$ and $\beta$ as in (b) we have
$\pi_{\beta+1}[B_\alpha(x_{s_i},\e)]
\cap\pi_{\beta+1}[B_\alpha(x_{s_m},\e)]=\emptyset$. 
For $s\in 2^{A_{k}}$ and $j<2$ define 
\[
E_{s\hat{\ }j}=E_{s}\cap B_\alpha(x_{s\hat{\ }j},\e).
\]
Then $\E_{k+1}=\left\{E_s\colon s\in 2^{A_{k+1}}\right\}$ is a subset of 
$\mathPerf_\alpha$ by (\ref{eqClopen1a}). 
Conditions (i) and (ii) are clear from the construction,
while (ag) for $\E_{k+1}$ follows from 
(b) and (\ref{eq18x}).
Property (sp) holds by (b) and the choice of $\e$,  
since (sp) was true for $\E_{k+1}'$. 
We have completed the proof of (A).


To prove condition (B), 
fix an enumeration $\{s_i\colon i<2^{k}\}$ of
$2^{A_{k}}$ and define 
$\gamma=\max\{\beta_0,\ldots,\beta_{k}\}<\alpha$. Also
for $i,m<2^k$ put $E_{s_i}^{-1}=E_{s_i}$ and 
\[
\beta^m_i=\max\{\beta\leq\gamma\colon 
s_i\restriction(\beta\times\omega)=s_m\restriction(\beta\times\omega)\}.
\]
By induction we will construct the sequences
$\la\{E_{s_i}^m\in\mathPerf_\alpha\colon i<2^k\}\colon m< 2^k\ra$ 
and $\la P_m\in\mathPerf_\alpha\colon m<2^k\ra$
such that for every $j,m< 2^k$
\begin{itemize}
\item[(a)] $\E^m=\{E_{s_i}^m\colon i<2^k\}$ satisfies (ag), 
\item[(b)] $E_{s_j}^m\subset E_{s_j}^{m-1}$
           and if $x\in E_{s_j}^{m-1}$ and $\pi_\gamma(x)\in\pi_\gamma[E_{s_j}^m]$
           then $x\in E_{s_j}^m$,
\item[(c)] $\pi_\gamma[P_m]=\pi_\gamma[E_{s_m}^m]$,
\item[(d)] $P_m\subset E_{s_m}^{m-1}$ and 
           $\{P_i\colon i\leq m\}\in\D$.
\end{itemize}
So, assume that for some $m<2^k$ the sequence 
$\la P_i\colon i<m\ra$ and the 
family 
$\E^{m-1}$ satisfying (ag) are already constructed.
Notice that, by (b), sets in $\E^{m-1}$ are pairwise disjoint,
since this was the case for $\E^{-1}=\E_k$.
So, by condition ($\dagger$)
applied to $\E=\{P_i\colon i<m\}$, we can choose 
$P_m\in\mathPerf_\alpha\cap\P(E_{s_m}^{m-1})$
such that $\{P_m\}\cup\{P_i\colon i<m\}\in\D$.
This guarantees (d). 

Next, for $i<2^k$ define 
\[
E^m_{s_i}=
E^{m-1}_{s_i}\cap\pi_{\beta^m_i}^{-1}(\pi_{\beta^m_i}[P_m])
=\left\{x\in E^{m-1}_{s_i}\colon x\restriction \beta^m_i\in 
\pi_{\beta^m_i}[P_m]\right\}
\]
and notice that
$\pi_{\beta^m_i}[P_m]\subset
\pi_{\beta^m_i}[E^{m-1}_{s_m}]=\pi_{\beta^m_i}[E^{m-1}_{s_i}]$. 
So, by  (\ref{eq20x}), 
$E^{m}_{s_i}\in\mathPerf_\alpha$. Also, the definition ensures (b) since 
$\beta^m_i\leq\gamma$.

Note that, by the inductive assumption (a), for all $i<2^k$ we have 
\[
\pi_{\beta^m_i}[E^m_{s_i}]
=\pi_{\beta^m_i}[E^{m-1}_{s_i}]\cap\pi_{\beta^m_i}[P_m]
=\pi_{\beta^m_i}[E^{m-1}_{s_m}]\cap\pi_{\beta^m_i}[P_m]
=\pi_{\beta^m_i}[P_m].
\]
Since $\beta_m^m=\gamma$, this implies (c). To prove (a) 
pick $\beta<\alpha$ and different $i,j<2^k$
such that 
$s_i\restriction(\beta\times\omega)=s_j\restriction(\beta\times\omega)$.
If $\beta\leq\beta^m_i$ then also $\beta\leq\beta^m_j$
and 
$\pi_{\beta}[E^m_{s_i}]=\pi_{\beta}[P_m]=\pi_{\beta}[E^m_{s_j}]$.
So, assume that 
$\beta>\beta^m_i$ and $\beta>\beta^m_j$. Then
$\beta^m_i=\beta^m_j$ and
\begin{eqnarray*}
\pi_\beta[E^m_{s_i}]
& = &
\left\{\pi_\beta(x)\colon x\in E^{m-1}_{s_i}\ \&\ 
x\restriction \beta^m_i
\in \pi_{\beta^m_i}[P_m]\right\}\\
& = &
\left\{\pi_\beta(x)\colon x\in E^{m-1}_{s_j}\ \&\ 
x\restriction \beta^m_j
\in \pi_{\beta^m_j}[P_m]\right\}\\
& = &
\pi_\beta[E^m_{s_j}].
\end{eqnarray*}
So $\E^m$ satisfies (a). This finishes the construction. 

Notice that by the maximality of $\gamma$ and the properties (a) and (c)
the family 
$\E'_k=\{P_m\colon m<2^k\}$ satisfies
(ag). Since it is a refinement of $\E_k$ it also satisfies (sp).
So (B) is proved. 

To find $\E_{k+1}$ as in (C) first take an 
$\E_{k+1}'$ satisfying (A) and then 
use (B) to find its refinement $\E_{k+1}'\in\D$
satisfying (ag) and (sp). \qed



\bigskip

In what follows we will show that \psmPrGame\ holds in the generic
extension
$V[G]$ of a model $V$ of ZFC+CH, where $G$ is a $V$-generic filter
over $\mathS_{\omega_2}$, 
the $\omega_2$ countable support iteration of Sacks forcing.
We will use here terminology from~\cite{BJ}. 

Let $\mathPerf=\la \perf(\Cantor),\subset\ra$. 
Recall that perfect set (Sacks) forcing 
$\mathS$ is usually defined as the set of all trees
$T(P)=\{x\restriction n\in 2^{<\omega}\colon x\in P\ \&\ n<\omega\}$,
where $P\in\mathPerf$,
and is ordered by inclusion, that is, $s\in \mathS$ is stronger than
$t\in\mathS$, $s\leq t$, if $s\subset t$.
It is important to realize that 
\[
P\subset Q \ \ \mbox{ if and only if } \ \ T(P)\subset T(Q),
\]
so $T\colon\mathPerf\to\mathS$ establishes
isomorphism between forcings $\mathPerf$ and $\mathS$.
Also if for $s\in\mathS$ we define 
$\lim(s)=\{x\in 2^\omega\colon \forall n<\omega\ (x\restriction n\in s)\}$
then $\lim\colon\mathS\to\mathPerf$ is the inverse of $T$. 

Perfect set forcing is usually represented as $\mathS$
rather that in its more natural form $\mathPerf$
since the conditions in $\mathS$ are absolute, unlike those in
$\mathPerf$. 
However, in light of our axiom, it is important to 
think of this forcing in terms of $\mathPerf$. 

Recall also that for a countable $A\subset\omega_2$
we defined $\Phi_{\rm prism}(A)$
as the family of all projection-keeping homeomorphisms
$f\colon\Cantor^A\to\Cantor^A$. For $A\subset\alpha\leq\omega_2$
define 
\[
\mathPerf_A^\alpha=\{[\range(f)]_\alpha\colon f\in \Phi_{\rm prism}(A)\},
\]
where 
$[E]_\alpha=\left\{g\in \Cantor^{\alpha}\colon g\restriction A\in E\right\}$
for every $E\subset \Cantor^A$. Also, we put
\[
\mathPerf_\alpha=
\bigcup\left\{\mathPerf_A^\alpha\colon A\in[\alpha]^{\leq\omega}\right\}
\]
and order it by inclusion. 
(Thus, for a countable $\alpha$ we have two different definitions 
of $\mathPerf_\alpha$. However, it is not difficult to see that they
describe the same family.)

It is known that forcing 
$\mathPerf_{\alpha}$ is equivalent to $\mathS_{\alpha}$,
a countable support iteration 
of $\mathS$ of length $\alpha$.
This fact is stated explicitly by Kanovei in~\cite{Ka},
though it was also used, in less explicit form, 
in earlier papers of Miller~\cite{Mi} and 
Stepr\={a}ns~\cite{St}. 
More precisely, in~\cite{Mi} and~\cite{St}
the authors consider the 
family $\mathS^D_{\alpha}\subset\mathPerf_{\alpha}$ of 
{\em determined conditions}\/ in $\mathPerf_\alpha$,
which form a dense subset of $\mathPerf_\alpha$,
and notice that $\mathS^D_{\alpha}$ is equivalent to 
$\mathS_{\alpha}$. 
This fact is most precisely described by the following fact, 
whose explicit proof can be found in~\cite{CPAbook}. 


\fact{factForcEq}{$\mathS_{\alpha}$ is order isomorphic to 
$\mathS^D_{\alpha}$ for every $0<\alpha\leq\omega_2$.
In particular, forcings 
$\mathS_{\alpha}$ and $\mathPerf_{\alpha}$ are equivalent.}







In the proof of the consistency of \psmPrGame\
we will use the following proposition, which is of interest 
by its own. In its statement 
the symbol \psmPrGame$[X]$ stands for \psmPrGame\ for a fixed
Polish space $X$.


\prop{FixedX}{For any Polish space $X$ axiom
\psmPrGame$[X]$ implies the full axiom \psmPrGame.}

\proof Let $X$ be a Polish space. 
First notice the following two facts.

\begin{itemize}
\item[(F1)] If $Y$ is a Polish subspace of $X$
then \psmPrGame$[X]$ implies \psmPrGame$[Y]$.
\end{itemize}
To see it, by way of contradiction assume that Player~II
has a winning strategy $S$ in ${\rm GAME}_{\rm prism}(Y)$.
For each prism $P$ in $X$ let 
$Q_P$ be its subprism such that
either $Q_P\cap Y=\emptyset$ or $Q_P\subset Y$. 
Such a subprism can be found by Claim~\ref{claim1}
since $Y$ is a $G_\delta$ subset of $X$. 
Define a strategy $\bar S$ for Player~II
in the game ${\rm GAME}_{\rm prism}(X)$ by putting
\[
\bar S(\la\la P_\eta,Q_\eta\ra\colon  \eta<\xi\ra,P)=
S(\la\la P_\eta,Q_\eta\ra\colon  \eta<\xi\ \&\ Q_{P_\eta}\subset Y
\ra,Q_P)
\]
provided $Q_P\subset Y$, and 
$\bar S(\la\la P_\eta,Q_\eta\ra\colon  \eta<\xi\ra,P)=Q_P$
otherwise. It is easy to see that $\bar S$ is 
a winning strategy for  
Player~II in ${\rm GAME}_{\rm prism}(X)$,
contradicting \psmPrGame$[X]$. 
So (F1) is proved. 

\begin{itemize}
\item[(F2)] If a Polish space $Y$ is a $1-1$ continuous
image of $X$ then \psmPrGame$[X]$ implies \psmPrGame$[Y]$.
\end{itemize}
Indeed, let $f$ be a continuous bijection from $X$ onto $Y$
and by way of contradiction assume that Player~II
has a winning strategy $S$ in ${\rm GAME}_{\rm prism}(Y)$.
Define a strategy $\bar S$ for Player~II
in ${\rm GAME}_{\rm prism}(X)$ by putting
\[
\bar S(\la\la P_\eta,Q_\eta\ra\colon \eta<\xi\ra,P)=
\bar S(\la\la f[P_\eta],f[Q_\eta]\ra\colon \eta<\xi\ra,f[P]).
\]
(Here if $h$ is a coordinate function for a prism $P$
then prism $f[P]$ is considered with
a coordinate system $h\circ f$.)
It is easy to see that $\bar S$ is 
a winning strategy for  
Player~II in ${\rm GAME}_{\rm prism}(X)$,
contradicting \psmPrGame$[X]$. 
So (F2) is proved. 

To finish the proof take a Polish space $X$ for which 
\psmPrGame$[X]$ holds and recall that 
the Baire space $\omega^\omega$ is homeomorphic
to a subspace of $X$ (since $X$ contains a copy of 
$\Cantor$ and $\Cantor$ contains a copy of $\omega^\omega$). 
Thus, by (F1), \psmPrGame$[Z]$ holds for an arbitrary
Polish subspace $Z$ of $\omega^\omega$.
Now, if $Y$ is an arbitrary Polish space then there 
exists a closed subset $F$ of $\omega^\omega$ 
such that $Y$ is a one-to-one continuous
image of $F$. (See e.g. \cite[thm.~7.9]{Ke}.)
So, by (F2), \psmPrGame$[Y]$ holds as well. \qed


\thm{th21}{\psmPrGame\ holds in the iterated perfect set model. In
particular, it is consistent with ZFC set theory.}

\proof Start with a model $V$ of ZFC+CH and 
let $V[G]$ be a generic extension of $V$ with respect to
forcing $\mathPerf_{\omega_2}$. By Fact~\ref{factForcEq} forcing 
$\mathPerf_{\omega_2}$ is equivalent to $\mathS_{\omega_2}$,
so it preserves cardinals and $\continuum=\omega_2$ in $V[G]$.
For $\alpha\leq\omega_2$ let $G_\alpha=G\restriction\alpha$.
Then $G_\alpha$ is $V$-generic over $\mathPerf_\alpha$.
By Proposition~\ref{FixedX} it is enough to
prove only \psmPrGame$[X]$ for $X=\Cantor$. 


Let $\{c_\alpha\colon \alpha<\omega_2\}$
be an enumeration, in $V[G]$, of $\Cantor$ 
such that for every $\alpha$-th element $\omega_1\,\alpha$ of $\Gamma$,
$\alpha>0$, we have: 
\begin{itemize}
\item $\{c_\xi\colon \xi<\omega_1\,\alpha\}=\Cantor\cap V[G_\alpha]$;

\item $c_{\omega_1\,\alpha}$  is the Sacks generic 
real in $V[G_{\alpha+1}]$
over $V[G_\alpha]$. 
\end{itemize}
We will show that this sequence satisfies \psmPrGame\ in $V[G]$.

So let $S$ be a strategy for Player~II. 
Thus, $S$ is a function from a subset of 
$D=\bigcup_{\xi<\omega_1}
\left(\Fpr^*(X)\times\Fpr^*(X)\right)^\xi\times\Fpr^*(X)$ 
into $\Fpr^*(X)$. 
(Here each prism is considered with its explicit 
coordinate function from $\Fpr^*(X)$.)
Since $\mathPerf_{\omega_2}$ is $\omega_2$-cc and satisfies
axiom $A$, there is an $\alpha\in\Gamma$ such that 
\begin{equation}\label{eqFor}
\omega_1\,\alpha=\alpha\ \ \&\ \ 
S\cap V[G_\alpha]=S\cap[(D\times\Fpr^*(X))\cap V[G_\alpha]]
\in V[G_\alpha].%
\footnote{Formally no
$f\in\Fpr^*(X)\cap V[G_{\omega_2}]$
belongs to $V[G_\alpha]$ with $\alpha<\omega_2$.
However in this proof the expression ``$f\in\Fpr^*(X)\cap V[G_{\alpha}]$''
will be understood
as ``Code$(f)$ belongs to $V[G_\alpha]$'',
where for $\dom(f)=P\subset\Cantor^\alpha$
with $P=\range(g)$, $g\in\Phi_{\rm prism}(\alpha)$, 
and $D_\alpha\in V$ being a fixed countable 
dense subset of $\Cantor^\alpha$ 
we define Code$(f)=f\restriction g[D_\alpha]$.
}
\end{equation}

Since the quotient forcing $\mathPerf_{\omega_2}/\mathPerf_\alpha$
is equivalent to $\mathPerf_{\omega_2}$
we can assume that $\alpha=0$, that is, that 
$V[G_\alpha]$ is our ground model $V$.

Let $\la\la f_\xi,g_\xi\ra\colon\xi<\omega_1\ra$ be a game played 
according to the strategy $S$ in which Player~I plays 
in such a way that 
$\{f_\xi\colon \xi<\omega_1\}=\Fpr^*(X)\cap V$.
Then $\G=\{g_\xi\colon \xi<\omega_1\}\in V$ is $\Fpr^*$-dense. 
It is enough to show that
\[
X\setminus V\subset
\bigcup_{g_\xi\in\Fpr}
\range(g_\xi).
\]

So, take an $r\in X\setminus V$. 
Then there exists a $\mathPerf_{\omega_2}$-name
$\tau$ for $r$ such that
\[
\mathPerf_{\omega_2}\forces \tau\in X\setminus V.
\]
We can also choose $\tau$ such that it is a $\mathPerf(A)$-name
for some $A\in[\omega_2]^{\leq\omega}$ with $0\in A$.
Then all the information on $r$ is coded by $G_A=G\restriction A$. 
Therefore $r\in V[\{c_{\omega_1\;\xi}\colon \xi\in A\}]$. 
Assume that $A$ has an order type
$\alpha$. Clearly $\alpha<\omega_1$
and $\mathPerf(A)$ is isomorphic to $\mathPerf_\alpha$.
Applying this isomorphism we can assume that 
$\tau$ is a $\mathPerf_\alpha$-name for $r$ and 
$\mathPerf_\alpha\forces \tau\in X\setminus V$. 
Picking the smallest $\alpha$ with this property, we can also assume that
for every 
$\beta<\alpha$ we have
\[
\mathPerf_\alpha\forces \tau\in X\setminus V[G_\beta].
\]
Now, for any such a name $\tau$ and any $R\in\mathPerf_{\alpha}$
there exist $P\in\mathPerf_\alpha$, $P\subset R$, and a continuous 
injection function
$f\colon P\to X$ (so $f\in\Fpr(X)\cap V$) which 
``reads $\tau$ continuously'' in the sense that
\begin{equation}\label{eq:forc1}
Q\forces \tau\in f[Q]
\end{equation}
for every $Q\subset P$, $Q\in\mathPerf_\alpha$.
(See \cite[Lemma 3.1]{St} or \cite[Lemma 6, p. 580]{Mi}.
This also can be deduced 
from~\cite[Lemma~3.2.2]{CPAbook}.)
So, the set 
\[
D=\{Q\in\mathPerf_\alpha\colon (\exists\xi<\omega_1)\ \ 
Q=\dom(g_\xi)\ \&\ Q\forces \tau\in g_\xi[Q]\}\in V
\]
is dense in $\mathPerf_\alpha$.
(For $R\in \mathPerf_\alpha$ take $f$ as in (\ref{eq:forc1}), 
find $\xi<\omega_1$ with $f=f_\xi$, and notice that $Q=\dom(g_\xi)$
justifies the density of $D$.)

Take $Q\in D\cap G_{\omega_1}$ and $\xi<\omega_1$ such that $Q=\dom(g_\xi)$.
Then there is a $z\in Q$ such that $g_\xi(z)=r$. 
This finishes the proof. \qed










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\end{document}

