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%\coverauthor{K. Ciesielski and M. Szyszkowski}
%\covertitle{A symmetrically  continuous function which 
%is not countably continuous}

\received{April 26, 1996}

\MathReviews{Primary 26A15; Secondary 26A03.}

\keywords{symmetric continuity, countable continuity,
Sierpi\'{n}ski-Zygmund functions}

\firstpagenumber{1}

\markboth{K. Ciesielski and M. Szyszkowski}{A Symmetrically  Continuous 
Function }


\author{Krzysztof Ciesielski\thanks{Partially
supported by the NSF grant INT-9600548.} and 
Marcin Szyszkowski, 
Department of Mathematics, West Virginia University,
Morgantown, WV 26506-6310, USA, email: kcies@@wvnvms.wvnet.edu and 
marcin@@math.wvu.edu
}

\title{A SYMMETRICALLY  CONTINUOUS FUNCTION WHICH IS NOT COUNTABLY CONTINUOUS}


%%%%%%Put Author's Definitions Below Here %%%%%%%%%%%

\def\Bbb{\mathbb}
\def\B{{\cal B}}
\def\D{{\cal D}}


\newcommand{\real}{{\Bbb R}}
\newcommand{\nats}{{\Bbb N}}

\def\goth{\frak}
\def\continuum{{\goth c}}

% characteristic function
     \newcommand{\charf}[1]{\mbox{\raise.48ex\hbox{$\chi$}$_{#1}$}}


\def\pc{\limfunc{PC}}
\def\bd{\limfunc{bd}}
\def\cl{\limfunc{cl}}
\def\darb{\limfunc{D}}
\def\ac{\limfunc{AC}}
\def\ext{\limfunc{Ext}}
\def\dim{\operatorname{ind}}
\def\la{\langle}
\def\ra{\rangle}
\def\A{{\cal A}}
\def\B{{\cal B}}
\def\qed{\hfill$\Box$}


\newtheorem{theorem}{Theorem}[section]
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{lemma}[theorem]{Lemma}
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\newtheorem{remark}[theorem]{Remark}
\newtheorem{question}[theorem]{Question}
\newtheorem{Fact}[theorem]{Fact}

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\newcommand{\cor}[2]{\begin{corollary}\label{#1}{\sl #2}\end{corollary}}
\newcommand{\prop}[2]{\begin{proposition}\label{#1}{\sl #2}\end{proposition}}
\newcommand{\lem}[2]{\begin{lemma}\label{#1}{\sl #2}\end{lemma}}
\newcommand{\pr}[2]{\begin{problem}\label{#1}{\rm #2}\end{problem}}
\newcommand{\ex}[2]{\begin{example}\label{#1}{\sl #2}\end{example}}
\newcommand{\defi}[2]{\begin{definition}\label{#1}{\rm #2}\end{definition}}
\newcommand{\rem}[2]{\begin{remark}\label{#1}{\rm #2}\end{remark}} 
\newcommand{\fact}[2]{\begin{Fact}\label{#1}{\rm #2}\end{Fact}} 


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\begin{document}

\maketitle


%\begin{center}
%{\small Department of Mathematics, West Virginia University, 
%PO Box 6310, Morgantown, WV 26506-6310, USA}
%\end{center}

\begin{abstract}
We construct a symmetrically continuous function
$f\colon\real\to\real$ such that for some $X\subset\real$
of cardinality continuum
$f|X$ is of Sierpi\'{n}ski-Zygmund type. 
In particular such an
$f$ is not countably continuous. This
gives an answer to a question of Lee Larson. 

\end{abstract}

\section{Preliminaries}

This paper concerns the relation between the following
two notions of generalized continuity.
(See \cite[Ch 3, 70--84]{BT} or \cite{T}.)

We say that a function $f\colon\real\to\real$ is 
{\em symmetrically continuous at point $x\in\real$} if 
\[
\lim_{k\to 0} f(x+k)-f(x-k)=0.
\]
Function $f\colon\real\to\real$ is 
{\em symmetrically continuous}
if it is symmetrically continuous at every point $x\in\real$.

For $X\subset \real$ a function $f\colon X\to\real$ is 
{\em countably continuous} 
if there is a countable cover $\{X_n\colon n\in\nats\}$ of
$X$ (by arbitrary sets) such that each restriction
$f|X_n$ is continuous. 

It is known that a symmetrically continuous
function $f\colon\real\to\real$
is relatively close to being continuous. For 
example the set of points of discontinuity
of $f$ is nowhere dense and of
measure zero. (See e.g.~\cite[Sec.~2.7]{T}.) 
Thus, Lee Larson (private communication) asked whether every 
symmetrically continuous function is countably
continuous. The main aim of this note is to give a negative answer 
to this question. 

We will also use the following notion. 
For $X\subset\real$ a function $f\colon X\to\real$ is 
said to be of {\em Sierpi\'{n}ski-Zygmund type} 
if $f|Y$ is discontinuous for every $Y\subset X$ of
cardinality $\continuum$, the cardinality of $\real$.  

The following fact describes 
the basic relation between these classes of functions.

\fact{fact1}{ Let $f\colon\real\to\real$. If there exists 
$X\subset\real$ of cardinality $\continuum$ such that
$f|X$ is of Sierpi\'{n}ski-Zygmund type then 
$f|X$ and $f$ are not countably continuous.
}

Proof. If $\{X_n\subset X\colon n\in\nats\}$ is 
a cover of $X$ then there exists $n\in\nats$ such that
$X_n$ has cardinality $\continuum$. (Since the cofinality
of $\continuum$ is uncountable.) Thus, $f|X_n$ is discontinuous, 
as $f|X$ is of Sierpi\'{n}ski-Zygmund type. \qed


\section{Technical Lemmas}

The construction presented below is, in a big part,
based on the technique developed in~\cite{Ch}. In particular,
the following lemmas are the modifications of 
their counterparts from~\cite{Ch}. 

In what follows we will use the following notation. 
For $A,B\subset\real$ we define
\[
2A=\{2x\colon x\in A\}\ \ \ \text{ and }\ \ \ 
A+B=\{x+y\colon x\in A\ \&\ y \in B\}.
\]
In the case when $B=\{b\}$, we write $A+b$ instead of $A+\{b\}$.
The symbol $\charf{A}$ will denote the characteristic 
function of a set $A\subset\real$. 

The set of centers of symmetry of a set $ A\subset\real$
will be denoted by $A^\star$, i.e., 
\[
A^\star=
\{x\in\real\colon(\forall k\in\real)(x+k\in A \iff x-k\in A)\}.
\]
For any function $f\colon X\to\real$ with $X\subset\real$  the symbol 
$C(f)$ will stand for the set of 
continuity points of $f$ and $D(f)$
for the set of points of discontinuity. Thus,  
$D(f)=\real\setminus C(f)$.

\lem{lem1}{ Let $h\colon\real\to\real$ be symmetrically 
continuous and $\{A_\alpha\}_{\alpha\in\A}$
be a family of disjoint subsets of $\real$ such that
for every $x\in\real$
\begin{equation}\label{eq1}
\lim_{y\to x}h(y)=0\ \ \ \text{ or }\ \ \ 
x\in\bigcap_{\alpha\in\A}A^\star_\alpha.
\end{equation}
If $r_\alpha\in[0,1]$ for every $\alpha\in\A$ then the function
\[
f=h\cdot\sum_{\alpha\in\A} r_\alpha \charf{A_\alpha}
\]
is symmetrically continuous.}

Proof. This is a modification of Lemma~2 from~\cite{Ch}. 
First note that 
$0\leq\sum_{\alpha\in\A} r_\alpha \charf{A_\alpha}\leq 1$
since the sets $A_\alpha$ are disjoint. 
Let $x\in\real$. 

If $\lim_{y\to x}h(y)=0$ then
\[
\lim_{y\to x}f(y)=
\lim_{y\to x}
\left(h\cdot\sum_{\alpha\in\A} r_\alpha \charf{A_\alpha}\right)(y)=0
\]
since $\sum_{\alpha\in\A} r_\alpha \charf{A_\alpha}$ is bounded. 
Hence $f$ is
symmetrically continuous at $x$.

If $x\in\bigcap_{\alpha\in\A}A^\star_\alpha$ then 
$\sum_{\alpha\in\A}r_\alpha \charf{A_\alpha}(x-k)
=\sum_{\alpha\in\A} r_\alpha \charf{A_\alpha}(x+k)$
for every $k\in\real$, and so
\[
\lim_{k\to 0}f(x+k)-f(x-k)=
\lim_{k\to 0}[h(x+k)-h(x-k)]
\sum_{\alpha\in\A} r_\alpha\charf{A_\alpha}(x+k)=0.
\]
Thus, once again, $f$ is symmetrically continuous at $x$.
\qed

\lem{lem2}{If $G\subset\real$ is an additive subgroup
then $G\subset(2G+x)^\star$ for every $x\in G$.}

Proof. 
Let $g\in G$, $a\in 2G+x$ and $b$ be a point symmetric to
$a$ with respect to $g$, i.e., such that $a+b=2g$. 
We have to prove that $b\in 2G+x$.

So, let $z\in G$ be such that $a=2z+x$. Then 
\[
b=2g-a=2g-(2z+x)=2(g-z-x)+x\in 2G+x
\]
since $g,x,z\in G$. \qed

\lem{lem3}{If $G\subset\real$ is an additive subgroup
then the sets $\{2G+x\colon x\in G\}$ form a partition of $G$. }

Proof. This is just the usual coset
decomposition of $G$ using the subgroup $2G$. To see it explicitly, 
assume that for some $x,y\in G$ there exists
$z\in(2G+x)\cap(2G+y)$. We have to prove that $2G+x=2G+y$,
i.e., that $x-y\in 2G$.
So, let $a,b\in G$ be such that $z=2a+x=2b+y$. Then
$x-y=2(b-a)\in 2G$. \qed


\lem{lem4}{There exists a symmetrically continuous function 
$h\colon\real\to\real$ with the property that 
\begin{description}
\item[(i)]   $C(h)=h^{-1}(0)$,
\item[(ii)]  $D(h)$ is an additive subgroup of $\real$, and 
\item[(iii)] there exists a subset $X$ of $D(h)$ of cardinality 
             $\continuum$ such that 
             \[
             (2D(h)+x)\cap(2D(h)+y)=\emptyset\ \ 
             \text{ for every distinct $x,y\in X$.}
             \]
\end{description}
}

Proof. Chleb\'{\i}k in~\cite{Ch} constructed a symmetrically
continuous function $h\colon\real\to\real$ satisfying (i)%
\footnote{Chleb\'{\i}k remarks only that 
$h=\lim_{m\to\infty}
\left(1+\sum_{n=1}^m\left|\frac{1}{n}\sin 3^n x\right|\right)^{-1}$ 
is continuous at every point of $h^{-1}(0)$,
implying that 
$h^{-1}(0)\subset C(h)$. However $h^{-1}(0)$ 
is clearly dense in $\real$. 
(For example for all $x$'s in the form
$\pi\left(m+\sum_{i=1}^\infty\frac{a_i}{3^i}\right)$, where
$m\in{\Bbb Z}$ and the numbers $a_i\in\{0,1,2\}$ are eventually equal
$1$.)}
and (ii) for which there exists a set $X\subset D(h)$ of 
cardinality $\continuum$ with the property that
\[
2D(h)+H_1\neq 2D(h)+H_2\ \ \ \text{ for every distinct 
$H_1,H_2\subset X$}.
\]
In particular, if $x\in X$ and $y\in X$ are distinct 
and $H_1=\{x\}$, $H_2=\{y\}$, then 
\[
2D(h)+x=2D(h)+H_1\neq 2D(h)+H_2=2D(h)+y.
\]
So, by Lemma~\ref{lem3}, $(2D(h)+x)\cap(2D(h)+y)=\emptyset$. \qed

\section{Main result} 

\thm{th:main}{There exists a symmetrically continuous function 
$f\colon\real\to\real$ and a
subset $X$ of $\real$ of cardinality $\continuum$  
such that $f|X$ is of a Sierpi\'{n}ski-Zygmund type.
}

Proof. Let
\[
\D=\{\la g,G\ra\colon G
\text{ is a $G_\delta$ subset of $\real$ and $g\colon G\to\real$
is continuous}\}
\]
and let $\la\la G_\alpha,g_\alpha\ra\colon\alpha<\continuum\ra$
be an enumeration of $\D$. Also, let $h\colon\real\to\real$ and $X$
be from Lemma~\ref{lem4} and pick a one-to-one enumeration
$\la x_\alpha\colon\alpha<\continuum\ra$ of $X$.

By transfinite induction on $\alpha<\continuum$ define
a sequence $\la r_\alpha\colon\alpha<\continuum\ra$
such that the following inductive condition is satisfied
for every $\alpha<\continuum$:
\begin{equation}\label{eq:Ind}
r_\alpha\in[0,1]\setminus
\left\{\frac{g_\beta(x_\alpha)}{h(x_\alpha)}
\colon\beta\leq\alpha\ \&\  
x_\alpha\in G_\beta\right\}.
\end{equation}
(Note that $h(x_\alpha)\neq 0$ since 
$x_\alpha\in D(h)=\real\setminus h^{-1}(0)$.)

Now let $A_\alpha=2 D(h)+x_\alpha$ for every $\alpha<\continuum$
and notice that, by Lemma~\ref{lem4}, the sets 
$\{A_\alpha\colon\alpha<\continuum\}$ are disjoint.
Define
\[
f=h\cdot\sum_{\alpha\in\A} r_\alpha \charf{A_\alpha}.
\]
Thus, $f$ is a well defined real function. 
Moreover, by Lemma~\ref{lem2},
\[
D(h)\subset
\bigcap_{\alpha<\continuum}(2 D(h)+x_\alpha)^\star=
\bigcap_{\alpha<\continuum}A_\alpha^\star,
\]
since $D(h)$ is an additive subgroup of $\real$. 
So, by Lemma~\ref{lem1}, $f$ is symmetrically
continuous, since $\real\setminus D(h)=C(h)=h^{-1}(0)$,
implying (\ref{eq1}). 
It remains to show that is $f|X$ is of 
Sierpi\'{n}ski-Zygmund type.

For this, by way of contradiction, assume that there exists
$Y\subset X$ of cardinality $\continuum$ such that
$f|Y$ is continuous. 
Then there exists a $G_\delta$ set $G\subset\real$ containing $Y$
and a continuous function $g\colon G\to\real$ such that
$g|Y=f|Y$. In particular, $\la g,G\ra\in\D$, and there exists 
$\beta<\continuum$ such that 
$\la g,G\ra=\la g_\beta,G_\beta\ra$.
Also, since $Y$ has cardinality $\continuum$, there exists
$\alpha<\continuum$, $\alpha\geq\beta$, such that
$x_\alpha\in Y$. But $h(x_\alpha)\neq 0$, since
$x_\alpha\in X\subset D(h)=\real\setminus C(h)
=\real\setminus h^{-1}(0)$. So, 
by (\ref{eq:Ind}), and the fact that $x_\alpha\in A_\alpha$
\[
f(x_\alpha)= h(x_\alpha)\cdot r_\alpha\neq 
h(x_\alpha) \frac{g_\beta(x_\alpha)}{h(x_\alpha)}=
g_\beta(x_\alpha)=g(x_\alpha)
\]
contradicting $g|Y=f|Y$. \qed

\cor{cor1}{There is a symmetrically continuous function 
$f\colon\real\to\real$ which is not countably continuous.}

Proof. By Theorem~\ref{th:main} and Fact~\ref{fact1}. \qed

\medskip

Notice also that in Fact~\ref{fact1} and Corollary~\ref{cor1}
we can conclude also that $f$ is not 
$\kappa$-continuous 
(the graph of $f$ cannot be covered by the graphs of 
$\kappa$  many continuous functions) 
where $\kappa$ is less then the cofinality of $\continuum$.

It is also worth to mention that neither Chleb\'{\i}k's theorem
nor Corollary~\ref{cor1} can be generalized for 
the class of symmetrically differentiable functions, i.e.,
the functions $f\colon\real\to\real$ for which the 
limit
\begin{equation}\label{eq11}
\lim_{k\to\infty}\frac{f(x+k)-f(x-k)}{2k}
\end{equation}
exists and is finite for every $x\in\real$. This follows
from a theorem of Charzy\'{n}ski \cite[Thm~2.9]{T}, since the
set $D(f)$ for every such function is at most countable. 
On the other hand, we do not know whether the same is true if we allow
the limit (\ref{eq11}) to be infinite. 





\medskip

\begin{thebibliography}{22}
\bibitem{Ch} M.~Chleb\'{\i}k, 
{\it There are $2^\continuum$ symmetrically continuous functions},
Proc. Amer. Math. Soc. {\bf 113} (1991), 683--688.

%\bibitem{LL} Lee Larson, 
%{\it The generalized Zahorski class structure of symmetric derivatives},
%Trans. Amer. Math. Soc. {\bf 282} (1984), 45--58.

\bibitem{BT} B.~S.~Thomson, {\it Real Functions},
Springer-Verlag, Lecture Notes in Math. 1170, 1985.

\bibitem{T}
B.~S.~Thomson, {\it Symmetric properties of real functions}, 
Marcel Dekker, Inc. 1993.

\end{thebibliography}

\end{document}
