%11/14/2002

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\markboth{A. Bartoszewicz and K. Ciesielski}{Errata: MB-representations and topological algebras
\ \ \ \UpdateDate}

\title{Errata: MB-representations and topological algebras}


\author{
Artur Bartoszewicz, Institute of Mathematics, {\L}\'od\'z Technical University,
Al. Politechniki 11, 90-924 {\L}\'od\'z, Poland  (arturbar@ck-sg.p.lodz.pl)
\and
Krzysztof Ciesielski%
%\thanks{The second author was partially supported by NATO Grant PST.CLG.977652.
%\endgraf
%Papers authored or
%co-authored by a Contributing Editor are managed
%by a Managing Editor or one of the other Contributing Editors.
%}%
,
Department of Mathematics, West Virginia
University, Morgantown, WV 26506-6310, USA
(K\_Cies@math.wvu.edu)
{\tt http://www.math.wvu.edu/\~{}kcies}
}


\date{}

\MathReviews{Primary 54E52; Secondary 06E25, 03E35}
\keywords{Generalized Marczewski's sets, topology.}

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%\title%
%[MB-representations and topological algebras]
%{MB-representations and topological algebras}
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%\author[A.Bartoszewicz, K.Ciesielski]{Artur Bartoszewicz, Krzysztof Ciesielski}
%\address{Technical University of Lodz,
%         Institute of Mathemathics,
%         Al. Politechniki 11,
%         90-924 Lodz,
%         Poland}


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\begin{document}

\vspace*{-1.0in}

\maketitle

Theorem~5(b) from \cite{BC} says the following:

\begin{quote}
\textit{Let $|X|=\kappa\geq\omega$, % be an infinite set of cardinality $\kappa$, 
$\F_0\subset[X]^\kappa$ be an almost disjoint family, and 
$\F=\{F\triangle A\colon F\in\F_0\ \&\ A\in[X]^{<\kappa}\}$.
If $|\F_0|> \kappa$ then the algebra 
$S(\F)=\{A \subset X\colon (\forall P \in\F) (\exists Q \in \F)
(Q \subset A \cap P \mbox{ or } Q \subset P \setminus A) \}$ is not topological,
that is, $\A\neq S(\tau\setminus\{\emptyset\})$ for any topology $\tau$ on $X$. 
}
\end{quote}
Unfortunately, the printed proof has a gap. (It shows only that the pair 
$\la S(\F),S_0(\F)\ra$ is not topological.) A correct proof 
for this result follows. 

\medskip 

\proof  By way of contradiction suppose that there exists a topology $\tau$ on $X$
such that $S(\F)=S(\tau_0)$, where 
$\tau_0=\tau\setminus\{\emptyset\}$.

Notice that for every $F\in\F$ we have $F\in S(\F)=S(\tau_0)$ and
\begin{equation}\label{eq1}
\mbox{$U\notin S(\F)$ for every $U\in[F]^\kappa$ with $|F\setminus U|=\kappa$.
}
\end{equation}
In particular, $\P(F)\not\subset S(\tau_0)$, so 
$F$ does not belong to $S_0(\tau_0)$,
which is defined as $\{A \subset X\colon (\forall P \in\F) (\exists Q \in \F)
(Q \subset P \setminus A) \}$. 
Thus, 
\begin{equation}\label{eq2}
\inter_\tau(F)\neq\emptyset \ \mbox{ for every } F\in\F.
\end{equation}
For every $F\in\F_0$ let $\V_F$ be a maximal pairwise disjoint
subfamily of $\tau\cap[F]^{<\kappa}$ and notice that 
$\left|\bigcup\V_F\right|<\kappa$.
Indeed, otherwise we could find a subfamily $\V$ of $\V_F$ with
$\left|\bigcup\V\right|=\left|F\setminus\bigcup\V\right|=\kappa$.
But then $U=\bigcup\V\in\tau\subset S(\tau_0)=S(\F)$
would contradict (\ref{eq1}). 
So, $F\setminus\bigcup\V_F\in\F$ and $V_F=\inter_\tau\left(F\setminus\bigcup\V_F\right)$
is nonempty by (\ref{eq2}).
To finish the proof it is enough to notice
that $\{V_F\colon F\in\F_0\}$ is a family of nonempty pairwise disjoint 
subsets of $X$, contradicting the fact that $|\F_0|>\kappa=|X|$. 
\qed

\vspace*{-.1in}

\begin{thebibliography}{abc}

\bibitem[BC]{BC} A.~Bartoszewicz, K.~Ciesielski, 
{\em MB-representations and topological algebras }, 
Real Anal. Exchange {\bf 27}(2) (2001--2002), 749--755. 

\end{thebibliography}


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