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\title[Spaces with pseudo-normally convergent function series]
{Spaces on which every pointwise convergent series of continuous functions 
converges pseudo-normally}
%On $\Sigma\Sigma^*$ subsets of $\real$}
\date{}


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%%%%%%



\author[Lev Bukovsk\'y]{Lev Bukovsk\'y}
\address{Institute of Mathematics, Faculty of Sciences, 
P. J. \v Saf\'arik University, Jesenn\'a 5, 040~01~Ko\v sice, Slovakia}
\email{bukovsky@kosice.upjs.sk}

\author[K. Ciesielski]{Krzysztof Ciesielski}
\address{Department of Mathematics, West Virginia University, Morgantown, WV 26506-6310, USA}
\email{K\_Cies@math.wvu.edu; web page: {\tt http://www.math.wvu.edu/\~{}kcies}}


\thanks{This work was partially supported by NATO Grant PST.CLG.977652.
The second author was also supported by 
2002/03 West Virginia University Senate Research Grant.}

\keywords{$\Sigma\Sigma^*$-space, Rothberger property, 
quasinormal convergence, pseudonormal convergence.}
\subjclass{Primary 54G99, 03E35; Secondary  54A35, 54C30.}


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\begin{document}

\maketitle

\begin{abstract}
A topological space $X$ is a $\Sigma\Sigma^*$-space
provided for every sequence $\la f_n\ra_{n=0}^\infty$ of continuous functions
from $X$ to $\real$, if the series $\sum_{n=0}^\infty|f_n|$ converges 
pointwise then it converges pseudo-normally. 
We show that every regular Lindel\"of
$\Sigma\Sigma^*$-space has Rothberger property. We also construct, 
under the continuum hypothesis, a $\Sigma\Sigma^*$-subset of $\real$ of cardinality
continuum.
\end{abstract}

\section{Introduction}

We will use standard set theoretical notation as in~\cite{CiBook}.
In particular, the ordinal numbers will be identified with the sets of 
their predecessors and cardinals with the initial ordinals. Thus, 
$2=\{0,1\}$ and the first infinite ordinal number $\omega$
is equal to the set of all natural numbers $\{0,1,2,\ldots\}$. 
The family of all functions from a
set $X$ into $Y$ is denoted by $Y^X$. Thus, for $n<\omega$ symbol 
$2^n$ will stand for the set of all sequences
$s\colon \{0,1,2,\ldots,n-1\}\to\{0,1\}$, while
$2^{<\omega}=\bigcup_{n<\omega}2^n$ is the set of all finite sequences 
into $2$. For a set $X$ and a cardinal number $\kappa$ we define $[X]^\kappa$
as the family of all subsets of $X$ of cardinality $\kappa$.
Families $[X]^{<\kappa}$ and $[X]^{\leq\kappa}$ are defined similarly. 
We let $\Cantor\subset[0,1]$ be the classical Cantor middle-thirds set. 
For a topological space $X$ we write  
$\C(X)$ for the family of continuous functions from $X$ into the set $\real$ of real numbers. 
For a sequence $\la f_n\ra_{n<\omega}$ of 
continuous functions from $X$ into $\real$ we
let 
\[
S(\la f_n\ra)=\left\{x\in X\colon\sum_{n<\omega}|f_n(x)|<\infty\right\}.
\]
A series $\sum_{n<\omega}f_n$ of functions from a set 
$X$ into $\real$ {\em converges normally}\/ 
(respectively, {\em pseudo-normally}\/) on a set $A\subset X$
if there exists a sequence $\la \e_n\colon n<\omega\ra$
of positive reals such that $\sum_{n<\omega}\e_n<\infty$ and for every $x\in A$
we have $|f_n(x)|\leq\e_n$  for all (respectively, all but finitely many) $n<\omega$. 
Thus, a topological space $X$ belongs to $\Sigma\Sigma^*$ (see \cite{BRR2})
provided for any sequence $\la f_n\in\C(X)\colon n<\omega\ra$
with $S(\la f_n\ra)=X$ the series $\sum_{n<\omega}f_n$
converges pseudo-normally on $X$.

The class $\Sigma\Sigma^*$ has been studied, in a context of several similar classes
of sets,  by Bukovsk\'y, Rec\l aw, and Repick\'y in~\cite{BRR2}.
In particular, it is known (see \cite[Diagram~2]{BRR2})
that every $\Sigma\Sigma^*$-space $X$
is a {\em $\sigma$-space}, that is, 
every $G_\delta$ subset of $X$ is also an $F_\sigma$ subset of $X$. 
One of the main results of this paper says 
that Lindel\"of  $\Sigma\Sigma^*$-spaces are small also in 
a different sense. 

\thm{th:SMZ}{Every regular and Lindel\"of\/ $\Sigma\Sigma^*$-space 
has Rothberger property.}

\noindent
Recall here that a topological space $X$
{\em has Rothberger property}\/ (called sometimes {\em property $C''$}) provided
for every sequence $\la\G_n\colon n<\omega\ra$ of open covers of $X$
we can find $U_n\in\G_n$ for every $n<\omega$ such that $\{U_n\colon n<\omega\}$
covers~$X$. 

Certainly the most interesting case of 
Theorem~\ref{th:SMZ} is when the space is separable and metric. 
It should be stressed that although many classes 
from~\cite[Diagram~2]{BRR2} are contained
in the class of $\sigma$-spaces, within the class of separable metric spaces 
only $\Sigma\Sigma^*$-spaces have Rothberger property.
Recall also that every metric space $X$ with Rothberger property
is of strong measure zero. (See e.g. \cite[thm.~8.1.11]{BJ}.)
Since there are models of ZFC in which 
every strong measure zero subset of $\real$ 
is countable~\cite[thm.~3.2]{Miller},
an uncountable $\Sigma\Sigma^*$-subset of $\real$ cannot be constructed in ZFC.
Moreover, so far there has been no consistent example of 
a $\Sigma\Sigma^*$-subset of $\real$ of cardinality continuum.
This state changes with the following theorem.

\thm{th:bigSS}{If the continuum hypothesis holds then there exists a set $X\subset[0,1]$
of~cardinality continuum which belongs to $\Sigma\Sigma^*$.}

Note that the existence of such a set does not seem to follow
from any other known constructions, since $\Sigma\Sigma^*$ does
not contain any other known (to us) class for which the analogous result is known. 
Indeed, one of the smallest classes of sets that
admits the consistent examples of cardinality continuum subsets of $\real$ 
is the class of strong $\gamma$-sets.\footnote{Every 
strong $\gamma$-set has $\gamma$-property, 
which is the strongest property in Scheepers' diagram~\cite{coc2}
admiting consistent examples of cardinality continuum.}
(See~\cite{GM}.)
However, there are consistent examples of strong $\gamma$-sets
which are not in $\Sigma\Sigma^*$. 
Actually, slightly modifying the proof of \cite[thm.~8]{GM} 
it is easy to construct, 
under Martin's axiom, a strong $\gamma$-subset of $\real$ of cardinality continuum
which is continuum-concentrated on a countable subset. 
By theorems 3.12 and 4.1 from \cite{BRR2} such a set is not 
in $\Sigma\Sigma^*$. 



\section{Every regular Lindel\"of $\Sigma\Sigma^*$-space has Rothberger property}

We start with the following variation of a well known characterization of Rothberger
property.

\lem{lem:RttChar}{Let $\la m_n\ra_{n<\omega}$ be a sequence of positive integers.
Then a  topological space $X$ has Rothberger property if and only if
\begin{itemize}
\item[($*$)] for every sequence $\la\U_n\ra_{n<\omega}$ of open covers of $X$
there exist an increasing sequence $\la n_i<\omega\colon i<\omega\ra$
and sets $\V_{n_i}\subset\U_{n_i}$ with $|\V_{n_i}|\leq m_{n_i}$ such that 
$X=\bigcup_{i<\omega}\bigcup\mathcal{V}_{n_i}$.
\end{itemize}
}

\proof Clearly every space with Rothberger property satisfies ($*$). 

To see the other implication let $\la\G_n\colon n<\omega\ra$ be a sequence of 
open covers of~$X$. For 
$n<\omega$ put $p_n=\sum_{k=0}^n m_k$ and 
let $\U_n$ be an open cover of $X$ refining covers
$\G_k$  for $k<p_n$. Let also $p_{-1}=0$. 
Applying ($*$) to 
$\la\U_n\ra_{n<\omega}$ we can find 
appropriate sequences $\la n_i\ra_{i<\omega}$ and 
$\la\V_{n_i}\ra_{i<\omega}$.
For $n<\omega$ let $\W_n=\{W_k\in\U_n\colon p_{n-1}\leq k<p_n\}$ 
be such that $\V_{n_i}\subset\W_{n_i}$ for every $i<\omega$. 
(Thus, for $n$ not equal to any $n_i$ the sets in $\W_n$ are arbitrary.)
Then we have $\bigcup_{k<\omega}W_k\supset \bigcup_{i<\omega}\bigcup\V_{n_i}=X$. 
But for $p_{n-1}\leq k<p_n$ the set $W_k$ belongs to $\U_n$, which is a refinement of 
$\G_k$. So, we can find $U_k\in\G_k$ for which $W_k\subset U_k$.
Therefore, $X=\bigcup_{k<\omega}U_k$. \qed


\medskip 

\noindent{\sc Proof of Theorem~\ref{th:SMZ}.} 
Since $X$ is regular and Lindel\"of, it is completely regular.
Thus, being $\Sigma\Sigma^*$-space, 
$X$ must be zero-dimensional --- see \cite[cor.~4.5]{BRR1}.
Let $\la\U_n\ra_{n<\omega}$ be a sequence of open covers of $X$.
We will show that the condition ($*$) from Lemma~\ref{lem:RttChar}
is fulfilled. 

Since $X$ is Lindel\"of and zero-dimensional
we can assume, replacing with a refinement, if necessary, 
that each cover $\U_n$ is countable and consists 
of pairwise disjoint clopen sets. 
For every $n<\omega$ let 
$\{U_k\colon k\in P_n\}$ be an enumeration of $\U_n$,
where $\{P_n\colon n<\omega\}$ is an appropriate 
partition of $\omega$.

Let $\la m_n\ra_{n<\omega}$ be a sequence of positive integers such that
$r=\sum_{n=0}^{\infty}1/m_n<\infty$. 
For every $n<\omega$ and $k\in P_n$ let 
$f_k=\frac{1}{m_n}\charf{U_k}$, where $\charf{U}\colon X\to\{0,1\}$
is the characteristic function of $U$. 
Then for every $x\in X$ 
\[\sum_{k<\omega}f_k(x)=\sum_{n<\omega}\sum_{k\in P_n}f_k(x)=
\sum_{n<\omega}\frac{1}{m_n}=r.
\] 
Since $X$ is a $\Sigma\Sigma^*$-space, 
there exists a sequence $\la \e_n\colon n<\omega\ra$
of positive reals such that $\sum_{n<\omega}\e_n<\infty$ 
and for every $x\in X$
we have $f_n(x)\leq\e_n$ for all but finitely many $n<\omega$. 
For $i,n<\omega$ let
$X_i=\{x\in X\colon f_k(x)\leq\e_k\mbox{ for all }i\leq k<\omega\}$
and put $T_i^n=\{k\in P_n\colon X_i\cap U_k\neq\emptyset\}$.
We claim that for any $i<\omega$ 
\begin{equation}\label{aa}
\mbox{there exists infinitely many
$n\in\omega$ such that $\vert T_i^n\vert\leq m_n$.}
\end{equation}
To see this, by way of contradiction assume that there is an $i<\omega$
for which (\ref{aa}) is false. So, there exists 
an $n_0$ such that $\vert T_i^n\vert>m_n$ for each $n\geq n_0$. 
Moreover, increasing $n_0$, if necessary, 
we can assume that $k\geq i$ for
any $k\in T_i^n$ and $n\geq n_0$. 
Hence for every $k\in T_i^n$ and $n\geq n_0$
there is an $x\in X_i\cap U_k$, and so
$\frac{1}{m_n}=f_k(x)\leq\varepsilon_k$. Consequently
$\sum_{k=0}^{\infty}\varepsilon_k\geq\sum_{n\geq n_0}
\sum_{k\in T_i^n}\varepsilon_k\geq \sum_{n\geq n_0}\vert T_i^n\vert
\cdot  1/m_n=\infty$,
what is a contradiction.
\par
Now let $\la n_i<\omega\colon i<\omega\ra$ be an
increasing sequence such that
$\vert T_i^{n_i}\vert\leq m_{n_i}$. Setting
$\V_{n_i}=\{U_j^{n_i}\colon j\in T_i^{n_i}\}\subset\U_{n_i}$ 
we obtain a cover satisfying ($*$) from Lemma~\ref{lem:RttChar}.
\qed 

\section{$\Sigma\Sigma^*$-subset of $\real$ of cardinality continuum}

We start with recalling Egoroff's theorem (see e.g. \cite[p.~73]{Ru}):
\begin{quote}
{\em Let $\mu$ be a finite countably additive measure on a set $X$ and let $\la g_k\ra_{k<\omega}$
be a pointwise convergent sequence of measurable functions from $X$ into~$\real$. 
Then for every $\e>0$ there exists a measurable 
set $E\subset X$ such that $\mu(X\setminus E)<\e$ and
$\la g_k\ra_{k<\omega}$ converges uniformly on $E$.}
\end{quote}
In the proof of the theorem we will use the following result, that is of
interest  on its own. 

\prop{prop1}{Let $P$ be a perfect subset of $\real$ and 
$\la f_n\in\C(P)\colon n<\omega\ra$.
If $S(\la f_n\ra)=P$ then there exists a sequence 
$\la \e_n\colon n<\omega\ra$ of positive numbers
such that $\sum_{n<\omega}\e_n<\infty$ and the closed set
$K=\bigcap_{n<\omega}\{x\in P\colon |f_n(x)|\leq\e_n\}$
has cardinality continuum. 
}

\proof For $k<\omega$ and $x\in P$ define $g_k(x)=\sum_{n=k}^\infty|f_n(x)|$.
Then functions $g_k\colon P\to\real$ are measurable
(in fact, Baire class one) and
they converge pointwise to~$0$. By Egoroff's theorem 
(used with a countably additive Borel probability measure on $P$
which vanishes on points)
there exists 
a perfect subset $Q$ of $P$ on which the sequence $\la g_k\ra$ converges uniformly.
By induction on $k<\omega$ we will construct 
the sequences $\la\e_n^k>0\colon k,n<\omega\ra$ 
and $\la x_s\in Q\colon s\in 2^k\ \&\ k<\omega\ra$ such that
for every $0<k<\omega$, $s\in 2^{k-1}$, $t\in 2^k$, and $n<\omega$
the following holds true. 
\begin{itemize}
\item[(i)]    $x_\emptyset\in Q$ is arbitrary, $\e_n^0=|f_n(x_\emptyset)|+2^{-n}$;
              thus $\sum_{m=0}^\infty \e_m^0<\infty$;
\item[(ii)]   $x_{s\hat{\ }0}=x_s$ and 
    $0<|x_{s\hat{\ }1}-x_s|<4^{-k}\delta_{k-1}$, where number $\delta_{k-1}$
             is defined as 
$\min\left(\{|x_{s_0}-x_{s_1}|>0\colon s_0,s_1\in 2^{k-1}\}\cup\{1\}\right)$;

\item[(iii)]  $|f_n(x_t)|<\e_n^k$ and $\e_n^{k-1}\leq\e_n^k$;
\item[(iv)]   $\sum_{m=0}^\infty \e_m^k\leq 2^{-k}+\sum_{m=0}^\infty\e_m^{k-1}$
              and $\e_i^k=\e_i^i$ for $i<k$.
\end{itemize}

First notice that if such sequences can be constructed 
and we put $\e_n=\e_n^n$ than the sequence $\la \e_n\ra_{n<\omega}$ is as desired. 
Indeed, the series converges, since by (iv) for every $k<\omega$ we have
\[
\sum_{m=0}^k \e_m=
\sum_{m=0}^k \e_m^m=
\sum_{m=0}^k \e_m^k\leq 
\sum_{m=0}^\infty \e_m^k\leq 
\sum_{m=1}^k 2^{-m}+\sum_{m=0}^\infty\e_m^0<
1+\sum_{m=0}^\infty\e_m^0
\]
so $\sum_{m=0}^\infty \e_m\leq 1+\sum_{m=0}^\infty\e_m^0<\infty$.

To see that the set $K=\bigcap_{n<\omega}\{x\in P\colon |f_n(x)|\leq\e_n^n\}$
has cardinality continuum, notice that $K$ is closed and that, by conditions (iii) and (iv),
we have $x_t\in K$ for every $t\in 2^{<\omega}$. 
So $K$ contains the closure of 
the set $\{x_t\colon t\in 2^{<\omega}\}$.
But, by (ii), the mapping associating 
$\lim_{k\to\infty}x_{\varphi\restriction k}\in K$
to each $\varphi\in 2^\omega$ 
is one-to-one. 

To make an inductive step in the construction
take a $k>0$ and assume that $(k-1)$-th step of the construction is already done.
Find a $j<\omega$ such that $j>k$ and $g_j(x)<4^{-k}$ for all $x\in Q$.
Let $0<\delta\leq 4^{-k}\delta_{k-1}$ be such that
for every $s\in 2^{k-1}$, $n\leq j$, and $x\in Q$, if
$|x-x_s|<\delta$ then $|f_n(x)|<\e_n^{k-1}$.
Existence of such a $\delta$ follows from the inductive assumption (iii)
and the continuity of functions~$f_n$. 
For $s\in 2^{k-1}$ pick $x_{s\hat{\ }1}\in Q$ such that 
$0<|x_{s\hat{\ }1}-x_s|<\delta$.
This insures (ii).
For $n< j$ put $\e_n^k=\e_n^{k-1}$
and for $j\leq n<\omega$ define $\e_n^k=\e_n^{k-1}+\sum_{t\in 2^k}|f_n(x_t)|$.
This clearly guarantees (iii) and the second part of (iv). 
To see the first part of (iv) notice that
\begin{eqnarray*}
\sum_{m=0}^\infty \e_m^k
& = &
\sum_{m<j}\e_m^k+\sum_{m=j}^\infty \e_m^k\\
& = &
\sum_{m<j} \e_m^{k-1}+
\sum_{m=j}^\infty \left(\e_m^{k-1}+\sum_{t\in 2^k}|f_m(x_t)|\right)\\
& = &
\sum_{t\in 2^k}\sum_{m=j}^\infty|f_m(x_t)|
+\sum_{m=0}^\infty \e_m^{k-1}\\
& = &
\sum_{t\in 2^k}g_j(x_t)
+\sum_{m=0}^\infty \e_m^{k-1}\\
& \leq &
2^k \ 4^{-k}+\sum_{m=0}^\infty\e_m^{k-1}
=
2^{-k}+\sum_{m=0}^\infty\e_m^{k-1}.
\end{eqnarray*}
This finishes the proof of Proposition~\ref{prop1}. \qed

In Proposition~\ref{prop1}, if we require only that the sequence
$\la \e_n\colon n<\omega\ra$ converges to zero then,
by Egoroff's theorem,
we can additionally assume that the set $K$ has a positive measure. 
However, the set $K$ from Proposition~\ref{prop1} need not have positive measure.


In the proof that follows we will use the following simple fact.

\fact{fact2}{For every $X\subset\real$ and 
$\la f_n\in\C(X)\colon n<\omega\ra$
such that $S(\la f_n\ra)=X$ there exists a Borel set $B\subset\real$ containing $X$
and extensions $f_n^*\in\C(B)$ of the functions $f_n$ such that
$S(\la f_n^*\ra)=B$.
}

\proof For every $n<\omega$ there exists a $G_\delta$ set $G_n\subset\real$
containing $X$ and an extension $\hat f_n\in\C(G_n)$ of $f_n$. 
Let $G=\bigcap_{n<\omega}G_n\supset X$
and let us define 
$B$ as $\bigcup_{m<\omega}\bigcap_{k<\omega}
\left\{x\in G\colon \sum_{n\leq k}|\hat f_n(x)|\leq m\right\}$.
Then $B$ is a Borel set containing $X$ and 
$B=S(\la \hat f_n\restriction G\ra)$.
Thus functions $f_n^*=\hat f_n\restriction B$ are as required. 
\qed

\medskip

\noindent{\sc Proof of Theorem~\ref{th:bigSS}.} 
Let $\B$ be a fixed countable clopen basis for $\Cantor$.
For $G\subset\real$ let $\K(G)$ be the family of all
sequences $\la f_n\in\C(G)\colon n<\omega\ra$
with $S(\la f_n\ra)=G$ and let $\la \la f_n^\xi\ra_{n<\omega}\colon \xi<\omega_1\ra$ 
be an enumeration of the family 
\[
\K=\bigcup\{\K(G)\colon G\mbox{ is a Borel subset of }\Cantor\}
\]
such that every $f^0_n$ is the constant zero function on $\Cantor$. 
By induction on $\xi<\omega_1$ we will construct a sequence
$\la \la x_\xi,\la\e_n^\xi\ra_{n<\omega},\la P_n^\xi\ra_{n<\omega},P_\xi,\G_\xi\ra\colon
\xi<\omega_1\ra$ such that $x_0=0$, $\e_n^0=2^{-n}$, $P_0=P_n^0=\Cantor$, 
$\G_0=\{\Cantor\}$, and the following conditions 
are satisfied for every $0<\xi<\omega_1$
and $k<\omega$. 
\begin{itemize}
\item[(a)] $P_\xi=\bigcup_{n<\omega}P_n^\xi$, $0<\e_k^\xi<\infty$, and
           $\sum_{n<\omega}\e_n^\xi<\infty$.

\item[(b)] if $\bigcap_{\eta<\xi}P_\eta\not\subset S(\la f_n^\xi\ra_{n<\omega})$
           then 

   \begin{itemize}
   \item $x_\xi\in\bigcap_{\eta<\xi}P_\eta\setminus S(\la f_n^\xi\ra)$,
           $\e_k^\xi=2^{-k}$, and $P_k^\xi=\Cantor$; 
    \end{itemize}
otherwise, 
   \begin{itemize}
   \item $P_k^\xi=\bigcap_{k<n<\omega}\{x\in G_\xi\colon |f_n^\xi(x)|\leq\e_n^\xi\}$,
            where $G_\xi$ is the domain of the functions $f_n^\xi$, $n<\omega$;
            $x_\xi\in\bigcap_{\eta\leq\xi}P_\eta\setminus\{x_\eta\colon\eta<\xi\}$.

   \end{itemize}


\item[(c)] $\G_\xi$ is a non-empty 
           countable family of perfect subsets of $\bigcap_{\eta\leq\xi}P_\eta$ such that

           \begin{itemize}
           \item  if $T\in\G_\xi$, $U\in\B$, and $T\cap U\neq\emptyset$
                        then $T\cap U\in\G_\xi$;
           \item every $H\in\bigcup_{\eta<\xi}\G_\eta$ 
                        contains a subset $T$ from the family $\G_\xi$.
           \end{itemize}
\end{itemize}
It should be clear that if we put $X=\{x_\xi\colon\xi<\omega_1\}$ 
then $X$ is uncountable and belongs to $\Sigma\Sigma^*$.

Indeed, 
$X$ is uncountable, since 
$S(\la f_n^\xi\ra_{n<\omega})=\Cantor\supset\bigcap_{\eta<\xi}P_\eta$ 
for uncountably many $\xi<\omega_1$. 
To see that $X\in\Sigma\Sigma^*$
take a sequence $\la f_n\in\C(X)\colon n<\omega\ra$
for which $S(\la f_n\ra)=X$. By Fact~\ref{fact2} 
we can find a Borel set $B\subset\real$ containing $X$
and extensions $f_n^*\in\C(B)$ of the functions $f_n$ such that
$S(\la f_n^*\ra)=B$. 
Then, there exists a $\xi<\omega_1$ such that 
$\la f^\xi_n\ra_{n<\omega}=\la f^*_n\ra_{n<\omega}$.
Notice that $\bigcap_{\eta<\xi}P_\eta\subset S(\la f_n^\xi\ra_{n<\omega})$,
since otherwise $x_\xi\in X\setminus S(\la f_n^\xi\ra_{n<\omega})$ contradicting 
$X=S(\la f_n\ra)\subset S(\la f_n^\xi\ra_{n<\omega})$.
Therefore, 
$\{x_\zeta\colon \xi\leq\zeta<\omega_1\}\subset P_\xi=
\bigcup_{k<\omega}\bigcap_{k<n<\omega}\{x\in G_\xi\colon |f_n^\xi(x)|\leq\e_n^\xi\}$.
Since the sequence $\la \e_n^\xi\ra_{n<\omega}$ works for all but countably many
points of $X$, the existence of appropriate sequence 
$\la \e_n\ra_{n<\omega}$ follows from an obvious fact that the class $\Sigma\Sigma^*$
is countably additive. 

Thus, it is enough to show that the inductive construction is
possible. The key point here is to prove
that the intersection $\bigcap_{\eta<\xi}P_\eta$ is uncountable,
so that we could have a chance to choose the next $x_\xi$.
This will be done with the help of families $\G_\eta$. 

First notice that 
\begin{equation}
\mbox{for every $H\in\bigcup_{\eta<\xi}\G_\eta$ there is a perfect set 
                      $H^*\subset H\cap\bigcap_{\eta<\xi}P_\eta$.}
\end{equation}
If $\xi$ is a successor ordinal, say $\xi=\alpha+1$, 
this is obvious: by the inductive assumption there is
an $H^*\in\G_\alpha$ which is a subset of 
$H\cap\bigcap_{\eta\leq\alpha}P_\eta=H\cap\bigcap_{\eta<\xi}P_\eta$. 
So, assume that $\xi$ is a limit ordinal
and let $H\in\G_\eta$ for some $\eta<\xi$.
Let $\eta=\eta_0<\eta_1<\cdots$ be such that $\xi=\bigcup_{n<\omega}\eta_n$.
Define a family $\{H_s\colon s\in 2^{<\omega}\}$
by induction on the length of $s$. We put $H_\emptyset=H$
and, if for some $s\in 2^n$ the set 
$H_s\in\G_{\eta_n}$ 
is already defined, choose disjoint 
perfect subsets $H_{s\hat{\ }0}$ and $H_{s\hat{\ }1}$
from $\G_{\eta_{n+1}}$. 
The choice can be made by the inductive assumption (c) 
applied to the family $\G_{\eta_{n+1}}$ 
and disjoint clopen portions of $H_s$.
Then $H^*=\bigcap_{n<\omega}\bigcup_{s\in 2^n} H_s$ 
is a perfect subset of $\bigcap_{\eta<\xi}P_\eta$.

Now, let $\H=\{H^*\colon H\in\bigcup_{\eta<\xi}\G_\eta\}=\{H_k\colon k<\omega\}$.
We need to find a sequence $\la\e_n^\xi\ra_{n<\omega}$
for which 
\begin{equation}
\mbox{$H^*\cap P_\xi$ contains a perfect subset for every $H^*\in\H$.}
\end{equation}

If $\bigcap_{\eta<\xi}P_\eta\not\subset S(\la f_n^\xi\ra_{n<\omega})$
then $P_\xi=\Cantor$, and this is obvious.
So, assume that $\bigcap_{\eta<\xi}P_\eta\subset S(\la f_n^\xi\ra_{n<\omega})$.
Then, by Proposition~\ref{prop1}, 
for every $k<\omega$ 
there exists a sequence 
$\la \hat\e_n^k>0\colon n<\omega\ra$
such that $\sum_{n<\omega}\hat\e_n^k<\infty$ and there is
a perfect subset 
$T_k$ of $\bigcap_{n<\omega}\{x\in H_k\colon |f_n(x)|\leq\hat\e_n^k\}$.
Let $\e_n^\xi=\max\{\hat\e_n^k\colon k\leq k(n)\}$,
where the sequence $k(0)\leq k(1)\leq\cdots$ 
converges to $\infty$ slowly enough that 
$\sum_{n<\omega}\e_n^\xi<\infty$.
Then $T_n\subset P_\xi$ for every $n<\omega$
and the family
\[
\G_\xi=\{T_n\cap B\neq\emptyset\colon n<\omega\ \&\ B\in\B\}
\]
satisfies (c). \qed 

We like to thank an anonymous referee for several suggestions
that lead to an improvement of the paper. 

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\end{document}

