\documentclass{rae} %\usepackage{amssymb} \usepackage{amstex} %\input{StandardMacros} % The contents of StandardMacros is explained in StandardMacros.info %\bysection %\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} \newtheorem{problem}[theorem]{Problem} \newtheorem{claim}[theorem]{Claim} \newtheorem{example}[theorem]{Example} \newtheorem{definition}[theorem]{Definition} \newtheorem{remark}[theorem]{Remark} \newtheorem{question}[theorem]{Question} \newtheorem{Fact}[theorem]{Fact} \newcommand{\thm}[2]{\begin{theorem}\label{#1}{\sl #2}\end{theorem}} \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}} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \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}