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\date{}
 
\title{Nice Hamel bases under the Covering Property Axiom}

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\markboth{K.~Ciesielski and J.~Pawlikowski}{Nice Hamel bases under CPA
\ \ \ \ \ \UpdateDate
}

 
\author{
Krzysztof Ciesielski%
\thanks{AMS classification numbers: Primary 26A15, 03E65;
Secondary  26A30, 03E35. \endgraf
\  Key words and phrases: Hamel base, CPA, rigid set, difference property.  \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, when 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{\forces}{\mathrel{\|}\joinrel\mathrel{-}}
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\newcommand{\ignore}[1]{}
\newcommand{\IntTh}{{\rm IntTh}}
\newcommand{\Implies}{\Longrightarrow}
\newcommand{\SoIC}{{s_0^{\rm prism}}}
\newcommand{\SoST}{{s_0^{\rm cube}}}
\newcommand{\psm}{{\rm CPA}}
\newcommand{\psmC}{\mbox{{\rm CPA$_{\rm cube}$}}}
\newcommand{\psmP}{\mbox{{\rm CPA$_{\rm prism}$}}}
\newcommand{\psmCsec}{\mbox{{\rm CPA$_{\rm cube}^{\rm sec}$}}}
\newcommand{\psmPsec}{\mbox{{\rm CPA$_{\rm prism}^{\rm sec}$}}}
\newcommand{\psmPLUS}{{\rm CPA$_{\rm cube}^+$}}
\newcommand{\psmPrPLUS}{{\rm CPA$_{\rm prism}^+$}}
\newcommand{\psmPrGame}{{\rm CPA$_{\rm prism}^{\rm game}$}}
\newcommand{\psmCGame}{{\rm CPA$_{\rm cube}^{\rm game}$}}
\newcommand{\cpa}{{\rm CPA}}
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% characteristic function
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%\newcommand{\proj}{\operatorname{pr}}
%\newcommand{\proj}{\mathop{\rm pr}}
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\def\lin{{\rm LIN}}

\def\cof{{\rm cof}}
\def\cl{{\rm cl}}
\def\diam{{\rm diam}}
\def\dist{{\rm dist}}
\def\inter{{\rm int}}
\def\bd{{\rm bd}}
\def\continuum{{\mathfrak c}}
\def\co{\continuum}
\def\Cantor{{\mathfrak C}}
\def\la{\langle}
\def\ra{\rangle}

\def\dom{{\rm dom}}
\def\range{{\rm range}}

\def\proof{\noindent {\sc Proof. }}
\def\qed{\hfill\vrule height6pt width6pt depth1pt\medskip}
\def\endproof{{\qed}}
\def\AA{{\cal A}}
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\newcommand{\ex}[2]{\begin{example}\label{#1}{\sl #2}\end{example}}
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\newcommand{\rem}[2]{\begin{remark}\label{#1}{\rm #2}\end{remark}} 
\newcommand{\fact}[2]{\begin{Fact}\label{#1}{\sl #2}\end{Fact}} 
\newcommand{\claim}[2]{\begin{Claim}\label{#1}{\sl #2}\end{Claim}}

 
\begin{document}
 
\maketitle



\begin{abstract}
In the paper we prove that axiom \psmPrGame,
which follows from the Covering Property Axiom CPA
and holds in the iterated perfect set model, 
implies that there exists a Hamel basis
which is a union of 
less than continuum many 
pairwise disjoint perfect sets. 
We will also give two consequences of this last fact. 
\end{abstract}



\section{The result and its consequences}

In this paper we will use standard set theoretic terminology as in~\cite{CiBook}. 
We will consider the real line $\real$ as a linear 
space over the rationals $\rational$. Any linear base 
of this space will be referred to as a {\em Hamel base}. 
For $A\subset\real$ we will write $\lin(A)$ 
to denote the linear subspace of $\real$ spanned by $A$. 

Axiom \psmPrGame\ was introduced by the authors in~\cite{CP83},
where it is shown that it holds in the iterated perfect set model. 
Also, \psmPrGame\ is a version of the axiom
\psm\ which is described in a monograph~\cite{CPAbook}. 

It is known that \psmPrGame\
captures, to a big extend, the essence of the iterated perfect set model.
This follows from a resent result of 
J.~Zapletal~\cite{Zap}
who proved that
for a ``nice'' cardinal invariant $\kappa$ if $\kappa<\continuum$
holds in any forcing extension than $\kappa<\continuum$
follows already from \psmPrGame. 

For the reader convenience, we will restate \psmPrGame,
along with necessary definitions, in the next section.

The main result of this paper is the following theorem. 

\thm{th:DisjHamelNew}{
\psmPrGame\ implies that 
there exists a family $\HH$ of $\omega_1$ 
pairwise disjoint perfect subsets of $\real$ such that 
$H=\bigcup\HH$ is a Hamel basis. 
}

This theorem will be proved in the following sections.
For the rest of this section we will discuss its two consequences.

Let $\I$ be a
translation 
invariant ideal on $\real$.
We say that a subset $X$ of $\real$ is {\em $\I$-rigid}\/
provided $X,\real\setminus X\notin\I$ but 
$X\triangle(r+X)\in\I$ for every $r\in\real$.
An easy inductive construction gives a non-measurable 
subset $X$ of $\real$ without the Baire property 
which is $[\real]^{<\continuum}$-rigid.
(First such a construction, under CH, comes from Sierpi\'nski~\cite{Si2c}.
Compare also~\cite{Hulanicki}.)
Thus, under CH or MA there are $\NN\cap\M$-rigid sets,
where $\NN$ and $\M$ stand for the ideals of 
measure zero and of the ideal meager subsets of $\real$, respectively. 
Recently these sets have been studied by 
Laczkovich \cite{Lacz2}
and Cicho\'n, Jasi\'nski, Kamburelis, and Szczepaniak~\cite{CJKS}.
In particular, Laczkovich \cite[Theorem 2]{Lacz2} 
implies that there is no $\NN\cap\M$-rigid set
in the random and Cohen models. 
The next corollary shows that 
the existence of such sets follows from \psmPrGame. 

\cor{cor:diffset}{\psmPrGame\ implies there exists 
an $\NN\cap\M$-rigid set $X$ 
which is neither measurable nor has it the Baire 
property.}

\proof Let $\HH=\{Q_\xi\colon \xi<\omega_1\}$ be from 
Theorem~\ref{th:DisjHamelNew}
and for every $\xi<\omega_1$ 
let $L_\xi=\lin\left(\bigcup_{\eta<\xi}Q_\eta\right)$. 
Then $\real$ is an increasing union of $L_\xi$'s
and each $L_\xi$ belongs to $\NN\cap\M$,
since it is a proper Borel subgroup of $\real$. 

Since, under \psmPrGame, the cofinalities of the ideals 
$\NN$ and $\M$ is equal to $\omega_1$ 
(see \cite{CP88} or \cite{CPAbook}),
there is a family $\{C_\xi\colon \xi<\omega_1\}$
such that every $S\in\M\cup\NN$ is a subset of some $C_\xi$.
By induction choose $X_0=\{x_\xi\colon\xi<\omega_1\}\subset\real$
such that
\[
x_\xi\notin C_\xi\cup\lin(L_\xi\cup\{x_\zeta\colon\zeta<\xi\}).
\]
Then $X_0$ intersects the complement of every set from 
$\M\cup\NN$. Define 
\[
X=\bigcup_{\xi<\omega_1}(x_\xi+L_\xi)
\]
and notice that $X_0\subset X$ and $2\, X_0\subset\real\setminus X$. 
Thus, both $X$ and $\real\setminus X$ 
intersects the complement of every set from $\M\cup\NN$.
In particular, $X,\real\setminus X\notin\M\cup\NN$.

Next notice
that for every $r\in L_\zeta$
\[
X\triangle(r+X)\subset\bigcup_{\xi<\zeta}[(x_\xi+L_\xi)\cup(r+x_\xi+L_\xi)]
\in \NN\cap\M. 
\]
Thus, $X$ is $\NN\cap\M$-rigid, but also $\NN$-rigid
and $\M$-rigid. These last two facts imply that 
$X$ is neither measurable nor does it have the Baire property.
\qed

Our second application of 
Theorem~\ref{th:DisjHamelNew} is the following result. 


\cor{cor:diffFunc}{\psmPrGame\ implies there exists 
a function $f\colon\real\to\real$ such that for every $h\in\real$ 
the difference function $\Delta_h(x)=f(x+h)-f(x)$ is Borel; 
however, for every $\alpha<\omega_1$ there is an $h\in\real$ such that
$\Delta_h$ is not of Borel class~$\alpha$. 
}

Note that answering a question of Laczkovich~\cite{Lacz} 
Filip\'ow and Rec\l aw~\cite{FR} 
gave an example of such an $f$ under CH. 
Rec\l aw also asked (private communication) whether such a function
can be constructed in absence of CH.
Corollary~\ref{cor:diffFunc} gives an affirmative answer to this question. 
It is an open question whether such a function exists in ZFC. 

\medskip

\proof The proof is quite similar to that for Corollary~\ref{cor:diffset}.

Let $\HH=\{Q_\xi\colon \xi<\omega_1\}$ be from
Theorem~\ref{th:DisjHamelNew}.  For every $\xi<\omega_1$ 
define 
$L_\xi=\lin\left(\bigcup_{\eta<\xi}Q_\eta\right)$
and choose a Borel subset $B_\xi$ of $Q_\xi$ 
of Borel class greater than $\xi$. 
Define 
\[
X=\bigcup_{\xi<\omega_1}(B_\xi+L_\xi)
\]
and let $f$ be the characteristic function $\charf{X}$ of $X$. 

To see that $f$ is as required note that 
\[
\Delta_{-h}(x)=
\left[\charf{(h+X)\setminus X} - \charf{X\setminus(h+X)}\right](x).
\]
So, it is enough to show that each of the sets 
$(h+X)\setminus X$ and $X\setminus(h+X)$ 
is Borel, though they can be of arbitrary high class. 
For this, notice that for every $h\in L_{\alpha+1}\setminus L_\alpha$ we have 
\[
h+X=h+\bigcup_{\xi<\omega_1}(B_\xi+L_\xi)
=\bigcup_{\xi\leq\alpha}(h+B_\xi+L_\xi)\cup
\bigcup_{\alpha<\xi<\omega_1}(B_\xi+L_\xi)
\]
and that the sets
$\bigcup_{\xi\leq\alpha}(h+B_\xi+L_\xi)\subset L_{\alpha+1}$
and $\bigcup_{\alpha<\xi<\omega_1}(B_\xi+L_\xi)$ are disjoint. 
So
\[
(h+X)\setminus X=\bigcup_{\xi\leq\alpha}(h+B_\xi+L_\xi)\setminus X
=\bigcup_{\xi\leq\alpha}(h+B_\xi+L_\xi)\setminus 
\bigcup_{\xi\leq\alpha}(B_\xi+L_\xi)
\]
is Borel, since each set $B_\xi+L_\xi$ is Borel. 
(It is a subset of $Q_\xi+L_\xi$, which is homeomorphic
to $Q_\xi\times L_\xi$ via addition function.) 
Similarly, set $X\setminus(h+X)$ is Borel. 

Finally notice that for $h\in Q_\alpha\setminus B_\alpha$
the set
\[
(h+X)\setminus X=\bigcup_{\xi\leq\alpha}(h+B_\xi+L_\xi) 
\]
is of Borel class greater than $\alpha$, since so is 
$(h+Q_\alpha)\cap[(h+X)\setminus X]=h+B_\alpha$.
Thus, $\Delta_h(x)$ can be of an arbitrarily high Borel class.  \qed

\section{\psmPrGame\ and how it implies the theorem}

In what follows the Cantor set $2^\omega$ will 
be denoted by a symbol $\Cantor$.
For a Polish space  $X$ (i.e., a complete separable metric space)
$\perf(X)$ will stand for a 
collection of all subsets of $X$ homeomorphic to the Cantor set $\Cantor$.

The main notion behind a formulation of a \psmPrGame\
is that of a prism in a Polish space  $X$ and of its subprism.
A {\em prism}\/ in $X$ is a perfect set $P\in\perf(X)$
which comes with (implicitly given) 
coordinate system, that is, 
a homeomorphism from $\Cantor^\alpha$, $0<\alpha<\omega_1$, onto $P$.
If $P$ is a prism with a coordinate function $f\colon\Cantor^\alpha\to P$
then its {\em subprism}\/ is any set of the form
$f[E]$, where $E$ is an {\em iterated perfect set}, that is, 
it belongs to the family $\mathPerf_\alpha$ to be defined latter. 

In addition, we consider every singleton 
as a (trivial) prism, whose only subprism is itself. 
We also define $\perf^*(X)$ as a family 
of all sets $P$ such that either $P\in\perf(X)$ or $P$ is a singleton.

\psmPrGame\ is expressed in terms of the 
following game ${\rm GAME}_{\rm prism}(X)$ 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 plays 
a prism $P_\xi\in\perf^*(X)$
and Player~II must respond with a subprism $Q_\xi$ of $P_\xi$. 
The game $\la\la P_\xi,Q_\xi\ra\colon \xi<\omega_1\ra$
is won by Player~I provided
\[
\bigcup_{\xi<\omega_1}Q_\xi=X;
\]
otherwise the game is won by Player~II. 

By a strategy for Player~II
we will understand any function $S$ 
such that
$S(\la\la P_\eta,Q_\eta\ra\colon \eta<\xi\ra,P_\xi)$
is a subprism of $P_\xi$, where 
$\la\la P_\eta,Q_\eta\ra\colon \eta<\xi\ra$ is any partial game. 
(We abuse here slightly the notation, since function $S$ 
depends also on the implicitly given coordinate functions 
making each $P_\eta$ a prism.)
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 $S$. 

Now, we can formulate the axiom. 

\begin{description}\label{PageCPAGamePrism}
\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}

Now, Theorem~\ref{th:DisjHamelNew} follows quite easily 
form the axiom and the following lemma, which 
proof will take the reminder of this paper. 

\lem{LemRelLinInd}{
Let $M\subset\real$ 
be a sigma-compact and linearly independent.
Then for every prism $P$ in $\real$ there exist a subprism $Q$ of $P$
and a compact subset $R$ of 
$P\setminus M$ such that 
$M\cup R$ is a maximal  linearly independent subset of $M\cup Q$. 
}


\medskip 

\noindent{\sc Proof of Theorem~\ref{th:DisjHamelNew}.}
For a linearly independent sigma-compact set $M\subset\real$
and a prism $P$ in $\real$ let 
$Q(M,P)=Q$ and 
$R(M,P)=R\subset P\setminus M$ 
be as in Lemma~\ref{LemRelLinInd}. 
Consider Player~II strategy $S$ given by
\[
S(\la\la P_\eta,Q_\eta\ra\colon \eta<\xi\ra,P_\xi)=
Q\left(\bigcup\{R_\eta\colon\eta<\xi\}, P_\xi \right),
\]
where $R_\eta$'s are defined inductively by
$R_\eta=R(\bigcup\{R_\zeta\colon\zeta<\eta\}, P_\eta)$.

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, 
$\real=\bigcup_{\xi<\omega_1}Q_\xi$. 

Let $\HH=\{R_\xi\colon \xi<\omega_1\}$ and notice
that $\bigcup\HH$ is a Hamel basis. 
Indeed, clearly $\bigcup\HH$ is linearly independent.
To see that it spans $\real$ it is enough to notice that
$\lin(\bigcup_{\eta<\xi}R_\eta)=\lin(\bigcup_{\eta<\xi}Q_\eta)$
for every $\xi<\omega_1$. 

Although sets in $\HH$ need not to be perfect,
they are clearly pairwise disjoint and compact.
Thus, the theorem follows immediately from the following remark. 
\qed

\rem{rem:DisjHamelNew}{
If there exists a family $\HH$ of $\omega_1$ 
pairwise disjoint compact subsets of $\real$ such that 
$\bigcup\HH$ is a Hamel basis
then there exists such an $\HH$ with $\HH\subset\perf(\real)$. 
}

\proof Let $\HH_0$ be a family 
of $\omega_1$  pairwise disjoint compact subsets of $\real$ such that 
$\bigcup\HH_0$ is a Hamel basis.
Partitioning each $H\in\HH_0$
into its perfect part and singletons from scattered part
we can assume that $\HH_0$ contains only perfect sets and singletons. 
To get $\HH$ as required 
fix a perfect set $P_0\in\HH_0$ and 
an $x\in P_0$ and notice that if 
we replace each $P\in\HH_0\setminus\{P_0\}$ 
with $px+qP$ for some $p,q\in\rational\setminus\{0\}$
then the resulting family will still be pairwise disjoint 
with union being a Hamel basis.
Thus, without loss of generality, we can assume that 
every open interval in $\real$ contains $\omega_1$ perfect 
sets from $\HH_0$. 
Now, for every singleton $\{x\}$ in $\HH_0$
we can choose a sequence $P_1^x>P_2^x>P_3^x>\cdots$ from $\HH_0$
converging to $x$,
and replace a family $\{x\}\cup\{P^x_n\colon n<\omega\}$ with its union. 
(We assume that we choose different sets $P_n^x$ for different singletons.)
If $\HH$ is such a modification of $\HH_0$ then $\HH$ is as desired. 
\qed








\section{Iterated perfect sets and fusion lemmas for prisms}

Let $0<\alpha<\omega_1$. To define 
$\mathPerf_\alpha$ we need to consider 
the family $\Phi_{\rm prism}(\alpha)$
of all continuous injections
$f\colon\Cantor^\alpha\to\Cantor^\alpha$ 
with the property that
\begin{equation}\label{con:PrKeep}
f(x)\restriction\beta=f(y)\restriction\beta\ \Equi\ 
x\restriction\beta=y\restriction\beta
\ \ \ \ \ \mbox{ for all $\beta<\alpha$ and $x,y\in \Cantor^\alpha$}
\end{equation}
or, equivalently, such that for every $\beta<\alpha$
\[
f\restriction\restriction\beta\stackrel{\rm def}{=}
\{\la x\restriction\beta,y\restriction\beta\ra\colon\la x,y\ra\in f\}
\]
is a one-to-one function from $\Cantor^{\beta}$
into $\Cantor^{\beta}$. 
For example, if $\alpha=3$ then $f\in\Phi_{\rm prism}(\alpha)$
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 $\Phi_{\rm prism}(\alpha)$ 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{$\Phi_{\rm prism}(\alpha)$ is closed under the compositions}
\end{equation}
and that for every $0<\beta<\alpha$
\begin{equation}\label{eq:restr}
\mbox{if $f\in\Phi_{\rm prism}(\alpha)$ then
$f\restriction\restriction\beta\in\Phi_{\rm prism}(\beta)$.}
\end{equation}
We define $\mathPerf_\alpha$ as 
\[
\mathPerf_\alpha=\{\range(f)\colon f\in \Phi_{\rm prism}(\alpha)\}.
\]
The simplest possible elements of $\mathPerf_\alpha$ are the {\em perfect cubes},
that is, the sets of the form $\prod_{\beta<\alpha}C_\beta$, where 
$C_\beta\in\Cantor$ for every $\beta<\alpha$. 
(If $f_\beta$ is a continuous injection from $\Cantor$ onto $P_\beta$
and $f\colon\Cantor^\alpha\to\Cantor^\alpha$ is
given by $f(x)(\beta)=f_\beta(x_\beta)$ then
$f\in\Phi_{\rm prism}(\alpha)$ and $\range(f)=\prod_{\beta<\alpha}C_\beta$.)

Note also that 
\begin{equation}\label{eq17}
\mbox{
if $f\in\Phi_{\rm prism}(\alpha)$ and $P\in\mathPerf_\alpha$ then 
$f[P]\in\mathPerf_\alpha$.
}
\end{equation}
Indeed, if $P=g[\Cantor^\alpha]$ for some $g\in\Phi_{\rm prism}(\alpha)$
then, by condition (\ref{eq:compPr}), we have
$f[P]=f[g[\Cantor^\alpha]]=(f\circ g)[\Cantor^\alpha]\in\mathPerf_\alpha$.

In what follows for a fixed $0<\alpha<\omega_1$ and $0<\beta\leq\alpha$
the symbol $\pi_\beta$\label{PageProjPi} 
will stand for the projection  
from $\Cantor^{\alpha}$ onto $\Cantor^\beta$.
We will always consider $\Cantor^\alpha$
with the following standard metric $\rho$:\label{PageRho}
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 $B_\alpha(z,\e)$.\label{PageBallB} 
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(x,\e)]=\pi_\gamma[B_\alpha(y,\e)] \ \ 
\mbox{ for every $x,y\in\Cantor^\alpha$ with 
$x\restriction \gamma=y\restriction \gamma$. }
\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 $\mathPerf_\alpha$. 

\medskip











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}
In what follows we will need the following simple fusion lemma,
which can be found in~\cite{CP83}. 
For reader convenience we include here its short proof. 

\lem{FusionSequenceLemma}{
Let $0<\alpha<\omega_1$ and for $k<\omega$ let
$\E_k=\left\{E_s\in\mathPerf_\alpha\colon s\in 2^{A_k}\right\}$.  
Assume that for every $k<\omega$, $s,t\in 2^{A_k}$, 
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 $i<k$ then $E_s\subset E_{s\restriction i}$, 
\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 what follows we will also need the following simple fact,
which follows from the fact that every 
dense $G_\delta$ subset of a Polish space $X\times X$
contains a product $G\times P$, where $G$ is dense $G_\delta$
in $X$ and $P\in\perf(X)$. For the proof see e.g. \cite{CP88} or \cite{CPAbook}. 


\claim{claim1}{Let $0<\alpha<\omega_1$.
If $G$ is a second category Borel subset of $\Cantor^\alpha$ 
then $G$ contains a perfect cube $\prod_{\beta<\alpha}P_\beta$. 
}

We will also use the 
following variant of Kuratowski-Ulam theorem, which can be deduced
from the classical Kuratowski-Ulam theorem via a simple closure argument.
Its proof can be found in~	\cite{CM98} or~\cite{CPAbook}. 




\lem{lem:Gdelta00}{Let $0<\alpha<\omega_1$. For every comeager set $H\subset\Cantor^\alpha$
there exists a comeager set $G\subset H$
such that for every $x\in G$ and $\beta<\alpha$ the set
\[
G_{x\restriction\beta}=\left\{y\in\Cantor^{\alpha\setminus\beta}\colon (x\restriction\beta)\cup y\in G\right\}
\]
is comeager in $\Cantor^{\alpha\setminus\beta}$.} 













\section{Proof of Lemma~\ref{LemRelLinInd}}


Let $X$ be a Polish space,
$0<n<\omega$, and $F\subset X^n$ be an $n$-ary relation.
We say that a set $S\subset X$ is {\em $F$-independent}\/
provided $F(x(0),\ldots,x(n-1))$
does not hold for any one-to-one $x\colon n\to S$. 
For a family $\F$ of finitary relations\index{relation!finitary}  
on $X$ (i.e., relations $F\subset X^n$ where $0<n<\omega$)
we say that 
$S\subset X$ is $\F$-independent
provided $S$ is $F$-independent for every $F\in\F$. 
We will use 
the term\index{relation!unary} 
{\em unary relation}\/ for any $1$-ary relation. 
\index{independent!set $\F$-}\index{set!Findep@{$\F$-independent}}


\prop{prop:Pindep}{Let $0<\alpha<\omega_1$ 
and $\F$ be a countable family of 
closed finitary relations on $\Cantor^\alpha$. 
Assume that every unary relation in $\F$ is 
nowhere dense in $\Cantor^\alpha$ and that 
for every $F\in\F$ there exists  
a comeager subset $G_F$ of $\Cantor^\alpha$ 
such that
\begin{itemize}
\item[\rm (ex)] for every $F$-independent finite set $S\subset G_F$, $x\in S$,
      and $\beta<\alpha$ the set
\[
\left\{z\in\Cantor^{\alpha\setminus\beta}\colon 
\mbox{ $S\cup\{z\cup x\restriction\beta\}\subset G_F$ is $F$-independent}\right\}
\]
      is dense in $\Cantor^{\alpha\setminus\beta}$.
\end{itemize}
Then there is an $E\in\mathPerf_\alpha$ which is $\F$-independent. 
}\index{fusion theorems}

Note that without the assumption that the unary relations in $\F$ are nowhere dense 
the proposition is false:
the unary relation $F=\Cantor^\alpha$ satisfies the condition (ex) 
(with $G_F=\Cantor^\alpha$) and no non-empty set is $F$-independent. 
On the other hand, for any $n$-ary relation $F\in\F$ with $n>1$ condition (ex)
implies that $F$ is nowhere dense in $\left(\Cantor^\alpha\right)^n$. 
However, not every nowhere dense binary relation satisfies (ex).
For example $F=\{\la x,y\ra\colon x(0)=y(0)\}$
is nowhere dense 
and it does not satisfy (ex) if $\alpha>1$. 

\medskip

\proof First notice that applying Lemma~\ref{lem:Gdelta00}, if necessary,
we can assume that for every $F\in\F$, $x\in G_F$, and $\beta<\alpha$ the set
$(G_F)_{x\restriction\beta}$
is comeager in $\Cantor^{\alpha\setminus\beta}$.
But this implies that each set from the condition (ex) is comeager 
in $\Cantor^{\alpha\setminus\beta}$ since it is an intersection 
of $(G_F)_{x\restriction\beta}$ and an open set 
$\left\{z\in\Cantor^{\alpha\setminus\beta}\colon 
\mbox{ $S\cup\{z\cup x\restriction\beta\}$ is $F$-independent}\right\}$.
In particular, if we put $G=\bigcap_{F\in\F}G_F$ 
then $G$ is comeager in $\Cantor^\alpha$ 
and it is easy to see that it satisfies the following condition.
\begin{itemize}
\item[\rm (EX)] For every $\F$-independent finite set $S\subset G$, $x\in S$,
      and $\beta<\alpha$ the set
\[
\left\{z\in\Cantor^{\alpha\setminus\beta}\colon 
\mbox{ $S\cup\{z\cup x\restriction\beta\}\subset G$ is $\F$-independent}\right\}
\]
      is dense in $\Cantor^{\alpha\setminus\beta}$.
\end{itemize}

Let $\{F_k\colon k<\omega\}$ be an enumeration of $\F$
with infinite repetitions. Also, for 
$k<\omega$ let $A_k=\{\la \beta_i,n_i\ra\colon i< k\}$
be as in the condition (\ref{NiceEnum}). 
By induction on $k<\omega$ we will construct
two sequences:  $\la \e_k>0\colon k<\omega\ra$ converging to $0$ 
and $\left\la\left\{x_s\in G\colon s\in 2^{A_k}\right\}\colon k<\omega\right\ra$ 
of $\F$-independent sets
such that for every $\beta<\alpha$, $k<\omega$, 
and $s,t\in 2^{A_k}$
\begin{itemize}
\item[(a)] $x_s\restriction\beta=x_t\restriction\beta$
           if and only if 
           $s\restriction\beta\times\omega=t\restriction\beta\times\omega$;
\item[(b)] if $E_s=B_\alpha(x_s,\e_k)$
           and $\E_k=\{E_s\colon s\in 2^{A_k}\}$ then
           $\E_k$'s satisfy (ii), (ag), and (sp)
           from Lemma~\ref{FusionSequenceLemma};
\item[(c)] if $F_k$ is an $n$-ary relation then $F_k(z_0,\ldots,z_{n-1})$ 
           does not hold provided each $z_i$ is chosen from a different 
           ball from $\E_k$.
\end{itemize}

Before we construct such sequences, let us first
note that $E=\bigcap_{k<\omega}\bigcup\E_k$ is as desired. 
Indeed, $E\in\mathPerf_\alpha$ by Lemma~\ref{FusionSequenceLemma}.
To see that $E$ is 
$\F$-independent
pick an $n$-ary relation $F\in\F$, 
$\{z_0,\ldots,z_{n-1}\}\in[E]^n$,
and find a $k<\omega$ with $F_k=F$ which is big enough 
so that $\e_k$ is smaller than the 
distance between $z_i$ and $z_j$ for all $i<j<n$. 
Then $z_i$'s must belong to distinct 
elements of $\E_k$ so, by (c), 
$F(z_0,\ldots,z_{n-1})$ does not hold.

\smallskip

For $k=0$ we pick an arbitrary $\F$-independent $x_\emptyset\in G$
by choosing an arbitrary element of $G$ which does not belong to any
nowhere dense unary relation from $\F$. Also, we choose 
an $\e_0\in(0,1]$
ensuring (c), which can be done since 
$F_0$ is closed. 
(This is a non-trivial requirement only when 
$F_0$ is an unary relation.)
Clearly (a)-(c) are satisfied. 


Assume that for some $k<\omega$ the construction is done
up to the level $k$. 
For $s\in 2^{A_{k}}$ and $j<2$ let
$s\hat{\ }j=s\cup\{\la \la\beta_{k},n_{k}\ra,j\ra\}\in 2^{A_{k+1}}$
and define $x_{s\hat{\ }0}=x_s$. Let $\left\{s_i\colon i<2^k\right\}$ be an enumeration of 
$2^{A_k}$ and put $S=\left\{x_{s\hat{\ }0}\colon s\in 2^{A_{k}}\right\}$. 
Points $x_{s_i\hat{\ }1}\in G\cap E_{s_i}$ will be chosen by induction on $i\leq 2^k$
such that the set
$S_i=S\cup\left\{x_{s_j\hat{\ }1}\colon j<i\right\}$ is $\F$-independent and 
the condition (a) is satisfied for the elements of $S_i$. 
Clearly, by the inductive assumption (a) is satisfied for the elements of $S_0=S$. 
So, assume that for some $i\leq 2^k$ the set $S_i$ is already constructed.
We need to find an appropriate $x_{s_i\hat{\ }1}\in G\cap E_{s_i}$. 
Let $\beta<\alpha$ be maximal such that there is an 
$s\in\left\{s\hat{\ }0\colon s\in 2^{A_k}\right\}\cup\left\{s_j\hat{\ }1\colon j<i\right\}$ 
with $s\restriction\beta\times\omega=(s_i\hat{\ }1)\restriction\beta\times\omega$
and let $x=x_s\restriction\beta$. We will choose $x_{s_i\hat{\ }1}$ 
extending $x$ and such that $x_{s_i\hat{\ }1}(\beta)\neq x_t(\beta)$ for all $x_t\in S_i$.
Notice that this will ensure that the condition (a) is satisfied for 
the elements of $S_{i+1}$. Surprisingly, more difficult condition to
insure will be that $x_{s_i\hat{\ }1}\in E_{s_i}=B_\alpha(x_{s_i\hat{\ }0},\e_k)$,
since at the first glance it is not even obvious that 
\begin{equation}\label{eq:indepLem1}
B_\alpha(x_{s_i\hat{\ }0},\e_k) \mbox{ contains an extension of $x$}.
\end{equation}

To argue for this first notice that maximality of $\beta$ insures that 
$\beta\geq\beta_k$, since $s_i\hat{\ }0\in S_i$ and 
$(s_i\hat{\ }0)\restriction\beta_k\times\omega=(s_i\hat{\ }1)\restriction\beta_k\times\omega$.
If $\beta=\beta_k$ we have $x=x_{s_i\hat{\ }0}\restriction\beta$ and (\ref{eq:indepLem1}) is obvious.
So, assume that $\beta>\beta_k$. 
Then there is a $j<i$ such that $s=s_j\hat{\ }1$. 
We also have
$s_j\restriction\beta\times\omega=s_i\restriction\beta\times\omega$
so, by the inductive assumption,
$x_{s_j}\restriction\beta=x_{s_i}\restriction\beta$. 

Now, let $n<\omega$ be the smallest such that $2^{-n}<\e_k$. 
Then, 
by the definition of the metric on $\Cantor^\alpha$, 
the fact that 
$x_s=x_{s_j\hat{\ }1}\in E_{s_j}=B_\alpha(x_{s_j},\e_k)$
means that $x_s(\gamma)(m)=x_{s_j}(\gamma)(m)$ for every 
$\la\gamma,m\ra\in A_n$. Therefore, we have 
$x(\gamma)(m)=x_s(\gamma)(m)=x_{s_j}(\gamma)(m)=x_{s_i}(\gamma)(m)$ for every 
$\la\gamma,m\ra\in A_n$ with $\gamma<\beta$. 
Thus, we can extend $x$ to an element $y\in\Cantor^\alpha$ 
for which 
$y(\gamma)(m)=x_{s_i}(\gamma)(m)$ for every 
$\la\gamma,m\ra\in A_n$. But this $y$ witnesses (\ref{eq:indepLem1}). 

To finish the construction of $x_{s_i\hat{\ }1}$ notice that by (\ref{eq:indepLem1})
we can find an open ball $B$ in $\Cantor^{\alpha\setminus\beta}$
such that $\{x\}\times B\subset B_\alpha(x_{s_i\hat{\ }0},\e_k)$.
Decreasing $B$, if necessary, we can also insure that 
$y(\beta)\neq x_t(\beta)$ for every $t\in S_i$ and $y\in\{x\}\times B$. 
By condition (EX) 
we can find a $z\in B$ such that
$S_i\cup\{x\cup z\}\subset G$ is $\F$-independent.
We put $x_{s_i\hat{\ }1}=x\cup z$.

Thus, we constructed an $\F$-independent set
$\{x_{s\hat{\ }j}\colon s\in 2^{A_{k}}\ \&\ j<2\}\subset G$
satisfying (a) and 
such that $x_{s\hat{\ }0},x_{s\hat{\ }1}\in E_s$ for every $s\in 2^{A_{k}}$.
To finish the construction insuring (a)-(c)
we need to choose an $\e_{k+1}\leq 2^{-(k+1)}$ small enough 
to guarantee the following properties. 
\begin{itemize}
\item $E_{s\hat{\ }j}=B_\alpha(x_{s\hat{\ }0},\e_{k_1})\subset E_s$ for every $s\in 2^{A_{k}}$ and $j<2$.
      This will ensure condition (ii).
\item Condition (sp) holds. This can be done, since (a) is satisfied. 
\item Condition (c) is satisfied. This can be done since
      $\left\{x_{s}\colon s\in 2^{A_{k+1}}\right\}$ is $\F$-independent and $F_{k+1}$ is a closed relation.
\end{itemize}
Note that (ag) is guaranteed by (a) and our definition of $E_s$'s. 
This finishes the proof of Proposition~\ref{prop:Pindep}. \qed


We say that an $n$-ary relation $F$ on a Polish space $X$ is\index{relation!symmetric}  
{\em symmetric}\/
provided for any sequence $\la x_i\in X\colon i<n\ra$
and any permutation $\pi$ of $n$ 
\begin{center}
$F\left(x_0,\ldots,x_{n-1}\right)$ holds if and only if
$F\left(x_{\pi(0)},\ldots,x_{\pi(n-1)}\right)$ holds.
\end{center}
For such an $F$ and $A\subset X$ we put 
\[
F*A=A\cup\left\{x\in X\colon 
\left(\exists a_1,\ldots,a_{n-1}\in A\right)\ F(x,a_1,\ldots,a_{n-1})\right\}.
\]
If $F$ is unary relation we interpret the above as $F*A=A\cup F$.
If $\F$ is a family of symmetric finitary relations 
on $X$ then we put $\F*A=\bigcup_{F\in\F}F*A$. Also, an {\em $\F$-closure of $A$}, denoted by
$\cl_\F(A)$,\label{clF} 
is the least $B\subset X$ containing $A$ such that $\F*B=B$. 
Note that $\cl_\F(A)=\bigcup_{n<\omega}\F^n*A$, where
$\F^0*A=A$ and $\F^{n+1}*A=\F*(\F^n*A)$. 
Thus, if $\F$ is a countable family of closed symmetric finitary relations then
$\cl_\F(A)$ is $F_\sigma$ in $X$ for a sigma-compact $A\subset X$
since $F*K$ is closed for every $F\in\F$ and compact $K\subset X$.   

We are the most interested in these notions when we are concerned
with either linear independence (over $\rational$) or algebraic independence in 
$\real$.\index{independent!linearly}\index{independent!algebraically}
In the first case $\F=\F_{\rm lin}$ is defined as the family of all relations
$F_\ell$ of all $\la x_0,\ldots,x_{n-1}\ra$ for which 
\begin{equation}\label{con:Fl}
\ell(x_{\pi(0)},\ldots,x_{\pi(n-1)})=0
\mbox{ for some permutation $\pi$ of $n$},
\end{equation}
where $\ell$ is a non-zero linear function with rational coefficients. 
In this case $\F$-independence stands for linear independence (over $\rational$)
and $\cl_\F(A)$ is the linear span of $A$. When $\F$ is the family of all
relations $F_\ell$, where 
$\ell$ spans over all non-zero polynomials with rational coefficients,
then 
$\F$-independence stands for algebraic independence, while 
$\cl_\F(A)$ is the algebraic closure of $\rational(A)$. 

We will need also one more notion. 
For a family $\F$ of closed symmetric finitary relations on $X$ 
and an $M\subset X$ we define\label{FsubM}
$\F_M$ as the collection
of all possible projections of the relations
from $\F$ along $M$. In other words,
$\F_M$ is the collection of all (symmetric) relations
\begin{equation}\label{con:FM}
\left\{\la x_0,\ldots,x_{k-1}\ra\colon 
\left(\exists a_k,\ldots,a_{n-1}\in M\right)\ F( x_0,\ldots,x_{k-1},a_k,\ldots,a_{n-1})\right\},
\end{equation}
where $F\in\F$ is an $n$-ary relation and $0<k\leq n$. 
Note that if $M$ is compact then each relation in $\F_M$ is still closed
and for every $A\subset X$ we have
\begin{equation}\label{con:Fclosure}
\cl_\F(M\cup A)=\cl_{\F_M}(A).
\end{equation}
Also, if $M$ is $\F$-independent then 
\begin{equation}\label{con:FMind}
\mbox{$A\cup M$ is $\F$-independent provided $A$ is $\F_M$-independent.}
\end{equation}


\lem{lem:IndMyc}{Let $\F$ be an arbitrary family of closed symmetric finitary 
relations in a Polish space $X$. Then for every prism $P$ in $X$
there exists a subprism $Q$ of $P$ and 
a compact $\F$-independent set $R\subset P$ such that $Q\subset\cl_\F(R)$.
}\index{independent!set $\F$-}\index{set!Findep@{$\F$-independent}}

\proof For $0<\alpha<\omega_1$ let $I_\alpha$ be the statement:
\begin{itemize}
\item[$I_\alpha$:] the lemma holds for any prism $P$ with witness function
                   $f\colon\Cantor^\alpha\to P$. 
\end{itemize}
We will prove $I_\alpha$ by induction on $\alpha$. 


First notice that $I_\alpha$ implies the following:
\begin{itemize}
\item[$I^*_\alpha$:] for every $k<\omega$ and continuous functions $g_0,\ldots,g_k\colon\Cantor^\alpha\to X$
there exist an $E\in\mathP_\alpha$ and a compact 
$\F$-independent set $R\subset\bigcup_{i\leq k}g_i[\Cantor^\alpha]$ 
such that $\bigcup_{i\leq k}g_i[E]\subset\cl_\F(R)$. 
\end{itemize}

To see that $I^*_\alpha$ holds true for $k=0$, for every $n$-ary relation $F\in\F$ define 
$F^0=\{\la x_0,\ldots,x_{n-1}\ra\in\left(\Cantor^\alpha\right)^n\colon F(g_0(x_0),\ldots,g_0(x_{n-1}))\}$.
By $I_\alpha$ applied to $\F_0=\{F^0\colon F\in\F\}$ 
we can find an $\F_0$-independent set $R_0\subset\Cantor^\alpha$ and an $E\in\mathP_\alpha$ such that
$E\subset\cl_{\F_0}(R)$. But then $R=g_0[R_0]$ is compact, $\F$-independent, and 
$g_0[E]\subset\cl_\F(g_0[R_0])=\cl_\F(R)$. 

To make an inductive step assume that $I^*_\alpha$ holds for some $k<\omega$
and take continuous functions $g_0,\ldots,g_{k+1}\colon\Cantor^\alpha\to X$. 
By the inductive assumption we can find an $E_0\in\mathP_\alpha$ 
and a compact 
$\F$-independent set $R_0\subset\bigcup_{i\leq k}g_i[\Cantor^\alpha]$ 
such that $\bigcup_{i\leq k}g_i[E_0]\subset\cl_\F(R_0)$. 
Let $h\in\Phi_{\rm prism}(\alpha)$ be a mapping 
from $\Cantor^\alpha$ onto $E_0$. 
Using the case $k=0$ to the function $g_{k+1}\circ h$ and the family $\F_{R_0}$ 
we can find 
an $E_1\in\mathP_\alpha$ and a compact 
$\F_{R_0}$-independent set 
$R_1\subset (g_{k+1}\circ h)[\Cantor^\alpha]$ 
such that $(g_{k+1}\circ h)[E_1]\subset\cl_{\F_{R_0}}(R_1)$. 
Then, by (\ref{con:FMind}), we conclude that
$R=R_0\cup R_1$ is $\F$-independent. 
Put $E=h[E_1]\in\mathP_\alpha$. 
Then, by (\ref{con:Fclosure}), we have 
$g_{k+1}[E]\subset\cl_{\F_{R_0}}(R_1)=\cl_\F(R_0\cup R_1)=\cl_\F(R)$,
while clearly $\bigcup_{i\leq k}g_i[E]\subset\bigcup_{i\leq k}g_i[E_0]\subset\cl_\F(R_0)\subset\cl_\F(R)$. 
Thus, $E$ and $R$ satisfy $I^*_\alpha$.

\medskip 



Now, we are ready to prove $I_\alpha$. 
So, fix $0<\alpha<\omega_1$ and assume that $I_\gamma$ is true for all $0<\gamma<\alpha$.  
Let $P$ be a prism in $X$ with witness function $f\colon\Cantor^\alpha\to P$. 
We need to find appropriate $Q$ and $R$. 


Let $W$ be the set of all $\beta\leq\alpha$
for which there exists an $E\in\mathPerf_\alpha$ and an $F\in\F$ 
such that for every $z\in\pi_\beta[E]$ there is a finite set $R_z\subset P$ for which 
\begin{equation}\label{conFinDimMM}
f[\{x\in E\colon z\subset x\}]\subset F*R_z.
\end{equation}
Notice that $W$ is non-empty since $\alpha\in W$. 
So $\beta=\min W$ is well defined. 
Let $E\in\mathPerf_\alpha$ be such that (\ref{conFinDimMM}) holds for $\beta$. 
Replacing
$f$ with its composition with an appropriate function from
$\Phi_{\rm prism}(\alpha)$ (compare (\ref{eq17})), if necessary, 
we can assume that $E=\Cantor^\alpha$. 


If $\beta=0$ then $f[\Cantor^\alpha]\subset\cl_\F(R_0)$ for some finite set $R_0\subset P$,
and we can find an $\F$-independent finite $R\subset R_0$ with
$f[\Cantor^\alpha]\subset\cl_\F(R)$. 
(Note that if $T$ is $\F$-independent and $x\in X\setminus\cl_\F(T)$ then
$T\cup\{x\}$ is also $\F$-independent.) 
Thus, $Q=f[\Cantor^\alpha]$ and $R$ satisfy $I_\alpha$. 
So, for the rest of the proof we will assume that $\beta>0$. 


Next, assume that $0<\beta<\alpha$. Let $\B_\beta$ be a countable basis 
of $\Cantor^{\alpha\setminus\beta}$ consisting of non-empty clopen sets
and assume that $F$ satisfying (\ref{conFinDimMM}) is $(n+1)$-ary. 
For every $B\in\B_\beta$ consider the set
\[
K_B=\left\{z\in\Cantor^\beta\colon\left(\exists \la x_1,\ldots,x_n\ra\in P^n\right)
\left(\forall y\in B\right)
\; F(f(z\cup y),x_1,\ldots,x_n)\right\}.
\]
It is easy to see that each set $K_B$ is closed. Notice also that
\begin{equation}\label{eq:Kb}
\Cantor^\beta=\bigcup_{B\in\B_\beta}K_B.
\end{equation}
To see this, fix a $z\in \Cantor^\beta$. By (\ref{conFinDimMM}), there exists a finite set 
$S_z\subset\Cantor^\alpha$ such that
$\Cantor^{\alpha\setminus\beta}=
\bigcup_{x_1,\ldots,x_n\in f[S_z]}\{y\in \Cantor^{\alpha\setminus\beta}\colon 
F(f(z\cup y),x_1,\ldots,x_n)\}$. Since each set
$\{y\in \Cantor^{\alpha\setminus\beta}\colon F(f(z\cup y),x_1,\ldots,x_n)\}$ is closed,
one of them must contain a $B\in\B_\beta$, and so $z\in K_B$. 

Thus, by (\ref{eq:Kb}), there exists a $B\in\B_\beta$ such that $K_B$ has a non-empty interior.
In particular, there is a non-empty clopen set $U\subset K_B$. 
But then for every $z\in U$ there exists a
$g(z)=\la g_1(z),\ldots,g_n(z)\ra\in P^n$ such that 
$F(f(z\cup y),g_1(z),\ldots,g_n(z))$ holds for every $y\in B$. 
Now  
\[
T=\{\la z,\bar p\ra\in U\times P^n\colon (\forall y\in B)\ F(f(z\cup y),\bar p)\}
\] 
is a compact subset of $U\times P^n$ and $g$ constitutes a selector of $T$. 
Thus, we can choose $g$ to be Borel. In particular, there is a dense $G_\delta$ 
subset $W$ of $U$ such that $g\restriction W$ is continuous. 
So, by Claim~\ref{claim1}, we can find a perfect cube $C\subset W\subset\Cantor^\beta$. 
Now, identifying $C$ with $\Cantor^\beta$, we conclude that
functions $g_1,\ldots,g_n\colon \Cantor^\beta\to P$ are continuous and that 
$F(f(z\cup y),g_1(z),\ldots,g_n(z))$ holds for every $z\in\Cantor^\beta$ and $y\in B$. 

Since, by the inductive hypothesis, $I_\beta$ is true, condition $I^*_\beta$ holds as well.
Thus, there exist 
an $E\in\mathP_\beta$ and a compact 
$\F$-independent set $R\subset P$ 
such that $\bigcup_{i=1}^n g_i[E]\subset\cl_\F(R)$. 
Since $Q=f[E\times B]$ is a subprism of $P$, we just need to show that
$Q\subset\cl_\F(R)$. To see this it just note
that for every $z\in E$ we have 
$f[\{z\}\times B]\subset F*\{g_1(z),\ldots,g_n(z)\}
\subset\cl_\F\left(\bigcup_{i=1}^n g_i[E]\right)\subset\cl_\F(R)$.
This finishes the proof of the case $0<\beta<\alpha$.

\medskip 

For the reminder of the proof we will assume that $\beta=\alpha$. This means that 
there is no $E\in\mathPerf_\alpha$ such that for some $F\in\F$ and $\beta<\alpha$
\begin{equation}\label{conFinDimNegMM}
\left(\forall z\in\pi_\beta[E]\right)\left(\exists R_z\in[P]^{<\omega}\right)\ 
f[\{x\in E\colon z\subset x\}]\subset F*R_z.
\end{equation}
For every $n$-ary $F\in\F$ let 
$F^*=\{\la x_0,\ldots,x_{n-1}\ra\colon F(f(x_0),\ldots,f(x_{n-1}))\}$
and let $\F^*=\{F^*\colon F\in\F\}$.
We will apply 
Proposition~\ref{prop:Pindep} to find an $\F^*$-independent $E\in\mathP_\alpha$.
Then $Q=f[E]$ is $\F$-independent subprism of $P$ and together with $R=Q$ they
satisfy the lemma. 

To see that the assumptions of Proposition~\ref{prop:Pindep} are satisfied, first 
notice that unary relations in $\F^*$ are nowhere dense.
Indeed, otherwise there is a unary relation $F^*\in\F^*$ and a non-empty clopen
set $E\subset F^*$. But then $E$ contradicts 
(\ref{conFinDimNegMM}), as $f[E]\subset F*\emptyset$. 
Thus, we just need to show that the condition (ex) is satisfied. 

So, fix an $F\in\F$. 
For $0<\beta<\alpha$ and $B\in\B_\beta$ let
\[
K(B)=\left\{z\in\Cantor^\beta\colon 
\left(\exists R_z\in[P]^{<\omega}\right)\ f[\{z\}\times B]\subset F*R_z\right\}.
\]
Clearly $K(B)$ is $F_\sigma$. Notice also that it is meager,
since otherwise there would exist a non-empty clopen $U\subset K(B)$ and 
$E=U\times B$ would contradict (\ref{conFinDimNegMM}). Thus, each set
$K_\beta=\bigcup_{B\in\B_\alpha}K(B)$
is meager. Also, for every $z\in\Cantor^\beta\setminus K_\beta$ and 
for every finite $R\subset P$ the set
$\left\{y\in\Cantor^{\alpha\setminus\beta}\colon f(z\cup y)\notin F*R\right\}$
is dense and open. In particular, if $R$ is a finite $F$-independent subset of $P$
then 
\begin{equation}\label{eq:indW}
W_R=\left\{y\in\Cantor^{\alpha\setminus\beta}\colon R\cup\{f(z\cup y)\} \mbox{ is $F$-independent}\right\}
\end{equation}
is dense and open. 
Let
\[
H=\bigcap_{0<\beta<\alpha}
\left(
\left(\Cantor^\beta\setminus K_\beta\right)\times\Cantor^{\alpha\setminus\beta}
\right)
\]
and notice that $H$ is comeager since each $K_\beta$ is meager in $\Cantor^\beta$. 
By Lemma~\ref{lem:Gdelta00} we can find a comeager set $G\subset H$
such that 
\[
G_{x\restriction\beta}=\left\{y\in\Cantor^{\alpha\setminus\beta}\colon (x\restriction\beta)\cup y\in G\right\}
\]
is comeager every $x\in G$ and $\beta<\alpha$.
To finish the proof it is enough to show that 
$G$ satisfy (ex) for $F^*$.
So, take an $F^*$-independent finite set $S\subset G$, an $x\in S$, and a $\beta<\alpha$.

First let us assume that $\beta>0$. Then $x\in S\subset G\subset H$ implies that 
$z=x\restriction \beta\in\Cantor^\beta\setminus K_\beta$. 
In particular, the set $W_{f[S]}$ from (\ref{eq:indW})
is comeager, and so is $W_{f[S]}\cap G_{x\restriction\beta}$.
To get (ex) it is enough to notice that $W_{f[S]}\cap G_{x\restriction\beta}$
is a subset of 
$\left\{y\in\Cantor^{\alpha\setminus\beta}\colon 
\mbox{ $S\cup\{y\cup z\}\subset G$ is $F^*$-independent}\right\}$.

Finally assume that $\beta=0$. We need to show that
the set 
\[
\{y\in G\colon S\cup\{y\} \mbox{ is $F^*$-independent}\}
\]
is dense. But this set must be comeager, since
otherwise its complement would contain a non-empty clopen set $E$
which wold contradict
(\ref{conFinDimNegMM}) with $\beta=0$. \qed





\bigskip

\noindent{\sc Proof of Lemma~\ref{LemRelLinInd}.}
Let $\F=\F_{\rm lin}$ be the linear independence family defined at (\ref{con:Fl})
and let $\bar M=\la M_n\colon n<\omega\ra$ 
be an increasing family of compact sets such that $M=\bigcup_{n<\omega}M_n$. 
Let $\F_{\bar M}=\bigcup_{n<\omega}\F_{M_n}$, 
where each $\F_{M_n}$ is defined at (\ref{con:FM}), that is, $\F_{M_n}$ is the 
the collection
of all possible projections of the relations from $\F$ along $M_n$.

If $M\cap P$ is of second category in $P$
then we can choose a subprism $Q$ of $P$ with $Q\subset M$.
Then $Q$ and  $R=\emptyset$
have the desired properties.
On the other hand, if $M\cap P$ is of first category in $P$
then, by Claim~\ref{claim1}, we can find a subprism 
$P_1$ of $P$ disjoint with $M$.

Now, applying Lemma~\ref{lem:IndMyc} we can find
a subprism $Q$ of $P_1$ and 
a compact $\F_{\bar M}$-independent set 
$R\subset P_1\subset P\setminus M$ such that $Q\subset\cl_{\F_{\bar M}}(R)$.
But then $M\cup R$ is $\F$-independent, see (\ref{con:FMind}). 
Moreover,
\[
Q\subset\cl_{\F_{\bar M}}(R)=\cl_\F(M\cup R)=\lin(M\cup R).
\]
So, $M\cup Q\subset\lin(M\cup R)$ proving that $Q$ and $R$ are as desired.  \qed


\section{Remarks}

It is worth to notice that in case when $M=\emptyset$
Lemma~\ref{LemRelLinInd} can be proved easier, and in a stronger form.


\prop{pr:linInd}{Every prism $P$ in $\real$ there is a subprism $Q$ 
which is linearly independent. 
}

\proof This follows 
from Proposition~\ref{prop:Pindep}
used with $\F=\F_{\rm lin}$. 
\qed



\rem{rem:DisjHamel1}{
Note that Proposition~\ref{pr:linInd} is false if we 
require $Q$ to be a subcube of prism $P$, that is,
$Q=f[C]$, where $C$ is a perfect cube in $\Cantor^\alpha$
and $f\colon\Cantor^\alpha\to P$ is a coordinate function making $P$ a prism. 
}

\proof Indeed, let $P_1$ and $P_2$ be disjoint perfect subsets of $\real$ 
such that $P_1\cup P_1$ is linearly independent over $\rational$.
Let $f\colon P_1\times P_2\to\real$ be defined by 
a formula $f(x_1,x_2)=x_1+x_2$. 
Identifying $P_1$ and $P_2$ with $\Cantor$ we 
think about $f$ as defined on $\Cantor^2$ and 
treat $P$ as a prism. 
To see that $P$ 
has no linearly independent
subcube let $Q=Q_1\times Q_2$ be a subcube of $P$ and 
choose different $a_1,b_1\in Q_1$ and $a_2,b_2\in Q_2$.
Then $\{a_1+a_2,a_1+b_2,b_1+a_2,b_1+a_2\}\subset Q$ and they 
are clearly linearly dependent. \qed

\rem{rem:DisjHamel2}{In Lemma~\ref{LemRelLinInd} 
we cannot require $R=Q$. 
}

\proof Let $P_1$, $P_2$, and $f$ be as in Remark~\ref{rem:DisjHamel1}.
If $M=P_2$ then $P$ has no subprism $Q$ such that $M\cup Q$
is linearly independent, since any vertical section
of $Q$ is a translation of a portion of $M$. 
\qed



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\end{document}