\documentstyle{article}
\newcommand{\mathbb}{\bf}
\newcommand{\real}{{\mathbb R}}
\newcommand{\nat}{{\mathbb N}}
\newcommand{\inte}{{\mathbb Z}}
\newcommand{\co}{{\mathbb c}}
\newcommand{\skok}{\hfill\newline\hskip20pt} 

\title{Points of weak symmetric continuity}
\author{Marcin Szyszkowski 
\thanks{ This paper was written under supervision of 
K.~Ciesielski.
The author wishes to thank him for many helpful 
conversations.\skok
AMS classification numbers: Primary 26A15; Secondary 26A03.
\skok Key words: symmetric continuity.}\\
{\footnotesize
Department of Mathematics, West Virginia University,} \\
{\footnotesize  Morgantown, WV 26506-6310} \\
{\footnotesize  Marcin@math.wvu.edu}}
\date{}
 
\newtheorem{theorem}{Theorem}%[section]
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{problem}{Problem}%[section]
\newtheorem{example}{Example}%[section]
\newtheorem{factD}[theorem]{Fact}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{conjecture}{Conjecture}
 
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\newcommand{\la}{\langle}
\newcommand{\ra}{\rangle}
\def\qed{\hfill$\Box$}

\begin{document}

\maketitle

\begin{abstract} 
We show that every set of reals is a set of points of
weak symmetrical continuity for some function and compare
it with other generalized continuities. We also make some remarks on
points of weak symmetric continuity when function is finite valued.
\end{abstract}

\bigskip

Our terminology is standard and follows \cite{Ci:book}.
Let us recall a known fact about points of continuity.
(See e.g. \cite{K}.)

\fact{fact1}{The set of points of continuity of any function
(from $\real$ to $\real$) is a $G_{\delta}$ set and any 
$G_{\delta}$ set is the set of points of continuity
for some function.} 

Many authors have investigated what happens if we replace the ordinary continuity
by other types of continuity, i.e.,
which sets may be obtained as sets of points 
of different types of 
generalized continuity. First let us look at the symmetric continuity.

\defi{def1}{
A function $f\colon \real\to\real$ is symmetrically continuous at a point $x$ if
$$\forall_{\varepsilon>0}~ \exists_{\delta>0}~
\forall_{h<\delta}~~|f(x+h)-f(x-h)|~<\varepsilon. $$ }

Symmetrical continuity is obviously a weaker notion than 
ordinary continuity and 
Fact~\ref{fact1} is not  
valid any more for points of symmetric continuity. 
In fact it is an old problem of Marcus to characterize these points.
(For partial 
results see Thomson~\cite{t} and Jasku\l a, Szkopi\' nska~\cite{j}.)

Another notion we consider is weak continuity. 

\defi{def2}{A function $f\colon\real\to\real$ 
is weakly continuous at
a point $x$ if
there exist the sequences 
$\la x_n\ra$ and $\la y_n\ra$ 
such that $x_n\nearrow x$, $y_n\searrow x$,
and 
$$ 
\lim_{n\to\infty} f(x_n)= \lim_{n\to\infty} f(y_n)~=~f(x).$$
In particular $f(x_n)-f(y_n) \to 0$.}

The next, rather unexpected, theorem says that weak continuity is indeed a very
weak notion.

\thm{thm4}{{\rm \cite[Chap.~2]{Th2}}
Every function $f\colon\real\to\real$ is weakly continuous everywhere
except a countable set and every countable subset of $\real$ is the
set of points of weak discontinuity for some function.}
{\bf Proof.} For the first part see for example 
\cite[Chap.~2]{Th2}. 

For the second part, let $A=\{a_n:n\in\omega\}$ be a 
countable set. The function $f(x)=\sum_{a_n<x} 2^{-n}$ is the 
one we are looking for. Indeed, if $x>a_n>y$ then 
$f(x)-f(y)\geq 2^{-n}$, so $f$ is not weakly continuous at 
any $a_n$. If $x\not\in A$ and if $y$ is closer to $x$ than 
any of the points $a_1,\ldots ,a_k$ then 
$|f(y)-f(x)|\leq\sum_{n=k+1}^\infty 2^{-n}=2^{-k}$ 
since we do not have any $a_n$'s in 
$(x-|x-y|,x+|x-y|)$ with $n\leq k$. Therefore $f$ is continuous 
(so weakly continuous) at $x\notin A$.\qed

Now if we replace weak continuity by its symmetric counterpart, we obtain
the following notion.

\defi{def3}{A function $f\colon \real\to\real$ is weakly 
symmetrically continuous at a point $x$ if
there exists a sequence $\la h_n\ra$ converging to $0$ such that 
$$\lim_{n\to \infty}~f(x+h_n)-f(x-h_n)=0. $$}
(Paradoxically, weak symmetric continuity is a stronger notion than weak 
continuity.)
We will  investigate the problem of characterizing the sets of points where a 
function can be weakly symmetrically continuous. The first result in this 
direction was obtained by Ciesielski and Larson, and it looks surprising in 
light of Theorem~\ref{thm4}. 

\thm{thmCL}{{\rm(Ciesielski, Larson \cite{c})} There is a function
$g\colon \real \to\nat$ that is weakly symmetrically discontinuous
everywhere.}

The next theorem, generalizing Theorem~\ref{thmCL}, answers the problem
completely.

\thm{pwsc}{Any set of reals is the set of points of weak symmetrical
continuity for some function $f\colon\real\to\nat$.}

Note that since the existence of weakly symmetrically discontinuous function
from $\real$ to $\real$ cannot be proved without 
the Axiom of Choice
\cite[Cor. 1.5]{c}, the same can be sad about Theorem~\ref{pwsc}.

The main part of the proof is in the following lemma. 
%We will use Axiom of Choice here, in fact it can not be avoided 
%as Lemma \ref{lem1} (and Theorem \ref{thmCL}) imply AC. 


\lem{lem1}{For any set $A\subset \real$ there exists a set $X$ such that
\begin{description}
\item[(a)]
for every $a \in A$ there is a sequence $\la h_n\ra$ with 
$h_n\nearrow 0$ and $a \pm h_n \in X$;
\item[(b)]
for every $b\in\real\setminus A$ there is no such a sequence.
\end{description}}

%***** Included part *************
Before we start the proof let us remark that 
the set $X$ may be always chosen to be meager and of 
measure zero. This partially answers the question stated 
at the Miniconference in Auburn in 1997 after presentation 
of Lemma \ref{lem1}. 

The set $X$ can be found as a subset of some special meager
 and measure zero set $C$. To make the below 
proof work, we need only that for all $x\in\real$ and  
for all $n\in\omega$ there are continuum many pairs 
$y,z\in C\cap \left(x-{1\over n},x+{1\over n}\right)$ with 
${{y+z}\over 2}=x$. Corollary \ref{mz} guarantees the 
existence of such set $C$. In the proof below however, 
for the sake of simplicity (and since we do not need that 
strengthened version to prove Theorem \ref{pwsc}), 
we take $C=\real$. The reader may notice that  
with no changes the proof works for other sets $C$.
%********** End of the included part **************
\smallskip

\noindent{\bf Proof of lemma \ref{lem1}.} 
We will construct the set $X$ inductively, using transfinite induction.
First note that (a) is equivalent to the following condition.
\begin{description}
\item{(a$^\prime$)}
For every $a \in A$ there are the sequences $\la x_n\ra$ and $\la y_n\ra$ 
from $X$ such that $x_n\nearrow a$, $y_n\searrow a$, and
${{x_n+y_n}\over 2}=a$ for every $n$.
\end{description}
Condition (b) than says that there are no such sequences 
$\la x_n\ra,\la y_n\ra$ for $b \notin A$.

Let $\kappa$ be the cardinality of $A$ and let 
$A=\{a_\alpha\colon\alpha<\kappa\}$ 
be an enumeration of $A$. By induction on $\alpha<\kappa$ 
we construct the sequences
$\la x_n^\alpha\colon n<\omega\ra$ and $\la y_n^\alpha\colon n<\omega\ra$
witnessing (a$^\prime$) for $a=a_\alpha$, aiming for
$X=\bigcup_{\alpha<\kappa}\{x_n^\alpha,y_n^\alpha\colon n<\omega\}$. 
The challenge will be in maintaining condition (b).
We will use the following auxiliary sets:
$$
X_\alpha=\bigcup_{\beta<\alpha}
\left(\left\{x_n^\beta\colon n<\omega\right\}\cup
\left\{y_n^\beta\colon n<\omega\right\}\right),
$$
$$
B_\alpha=\left\{{{x+y}\over 2}\colon x,y\in X_\alpha\right\}\setminus A,
$$
and the functions $f_\alpha\colon B_\alpha\to\real$ defined by
$$ 
f_\alpha (b)=\inf\left\{ |b-x|\colon x\in X_\alpha %\ \&\ % 
\mbox{ and }
\exists_{y\in X_\alpha}\frac{x+y}{2}=b \right\}.
$$
(Here $|b-x|$ denotes the distance between the numbers $b$ and $x$.)

In the construction we will maintain 
the following two inductive conditions:
\begin{itemize}
\item[(c)] $f_\alpha (x) > 0$ for each $x\in B_\alpha$, and
\item[(d)] $f_\beta \subset f_\alpha$ for $\beta \leq \alpha$.
\end{itemize}
Conditions (c) and (d) imply that (b) holds for every point in 
$\bigcup_{\alpha<\kappa} B_\alpha$, since 
$f=\bigcup_{\alpha<\kappa}f_\alpha$ is
positive  on $\bigcup_{\alpha<\kappa}B_\alpha$. 
Hence, maintaining (c) and (d) during the construction will 
imply that (b) holds. 

For every point outside $\bigcup_{\alpha<\kappa} B_\alpha $, condition
(b) is automatically satisfied and the set $X$ will have the desired property.

Take the first sequence $\la x_n^0\ra$ to be $x_n^0=a_0+{1\over n}$. Then
$y_n^0=a_0-{1\over n}$, $X_0=\{x_n^0,y_n^0\colon n<\omega\}$, and 
$B_0=\{{{x+y}\over 2}\colon x,y\in X_0\}\setminus A$.
Since the only accumulation point of $X_0$ is 
$a_0\in A$ then $f_0(x) >0$ for all 
$x\not= a_0$ and (c) is satisfied.

Assume that for some $\alpha<\kappa$ 
we have already defined the sequences $\la x_n^\beta\ra$ and 
$\la y_n^\beta\ra$ for all $\beta<\alpha$. 
We choose sequences $\la x_n^\alpha\ra$ and $\la y_n^\alpha\ra$ 
satisfying (a$^\prime$) so that
$$\forall_{b\in B_\beta} \forall_n \ 2b-x_n^\alpha \not
\in \bigcup_{\beta<\alpha}X_\beta \cup \{x_k^\alpha,y_k^\alpha
\colon k<n\} $$
and
$$\forall_{b\in B_\beta} \forall_n \ 2b-y_n^\alpha \notin
\bigcup_{\beta<\alpha}X_\beta\cup\{x_k^\alpha,y_k^\alpha\colon k<n\}.
$$
Such sequences exist since we may choose points $x_n^\alpha$ 
and $y_n^\alpha$ inductively so that 
$$x_n^\alpha,y_n^\alpha\notin\left\{2b-x\colon b\in 
\bigcup_{\beta<\alpha}B_\beta \ \&\ 
x\in \left(\bigcup_{\beta<\alpha}X_\beta\cup \{x_k^\alpha,
y_k^\alpha\colon k<n\}\right)\right\}
$$
(that set has cardinality 
$|\bigcup_{\beta<\alpha}B_\beta|\otimes|\bigcup_{\beta<\alpha}X_\beta|\leq
|\alpha|\otimes\omega<\co$)
and ${x_n+y_n \over 2} = a_\alpha$.
It is clear that both inductive conditions are preserved.

Now putting $X=\bigcup_{\alpha<\kappa}X_\alpha=
\bigcup_{\alpha<\kappa}\{x_n^\alpha,y_n^\alpha\colon n<\omega\}$, 
we see that (a) is obviously satisfied and (b) is satisfied since 
$f=\bigcup_{\alpha<\kappa}f_\alpha$ is a positive function and for 
every point $b$ not in $A$ either we do not have a pair of sequences 
contained in $X$ and symmetric about that point or 
these sequences cannot have elements closer to $b$ than $f(b)$.
\qed

\bigskip

\noindent{\bf{Proof of Theorem~\ref{pwsc}}.}
Let $ g\colon\real\to\nat$ be an everywhere weakly symmetrically 
discontinuous function and $A\subset\real$. We'll construct a function $f$
that has points of weak symmetric continuity exactly in the set A.
Let X be the set from Lemma~\ref{lem1} for the set A.
We define
$$
f(x)=\left\{
\matrix{ g(x)+1 & \hbox{ for }  x \notin X \cr
           0    & \hbox{ for } x \in X.   \cr }
\right. 
$$
Obviously every point of A is a point of weak symmetric continuity of $f$.
(There is a symmetric sequence in X converging to this point.) 
If $x\not \in A $ then we do not have a sequence symmetrically convergent to $x$ 
contained in $X$. We also do not have such a sequence contained in 
$\real \setminus X$  satisfying 
the definition of weak symmetric continuity for $x$, because $g$ is 
weakly symmetrically discontinuous everywhere. Finally any ``mixed'' 
sequence will not satisfy that definition either since
$f$ differ on points from $X$ and outside of $X$ by at least 1. \qed

\medskip

%****** added part **********************************
\parindent0pt
\quad Now we provide the proof of the existence of measure 
zero and meager set $C$. Since the main theorems of this 
paper do not require that set, the reader may skip that 
part. The author includes this since it seems to him that 
it is worth adding that ``small" sets can have big 
``weak symmetric closure."
Note also, that the property %of a set $\hat C$ 
stated in Theorem~\ref{cant} is closely related to the Steinhaus property
that every number in $[0,1]$ is the sum of two 
numbers from the Cantor ternary set. 
\smallskip

\quad We will use the base 7 expansion, i.e., for 
$x\in [0.1]$ we write $x=0.x_1x_2\ldots $ meaning that 
$x=\sum_{i=1}^\infty {{x_i}\over{7^i}}~,x_i\in
\{ 0,1,2,3,4,5,6\}$. 

\quad For every finite sequence $x_1 x_2 \ldots x_n$, where  
$x_i\in \{0,1,2,3,4,5,6\}$, let $$C_{x_1 x_2 \ldots x_n}=
\{x\in [0,1]: x=0.x_1x_2\ldots x_n x_{n+1}\ldots ~~x_i\not =3~
{\rm for}~i>n\}.$$
$C_{x_1x_2\ldots x_n}$ is a ``Cantor ${1\over 7}$" set. It is 
a perfect set of measure zero (thus, nowhere dense).

\quad Denote also 
$$\hat C = \bigcup \{C_{x_1x_2\ldots x_n}: n\in\omega\}.$$
The set $\hat C$ being a countable union of measure zero 
and nowhere dense sets has measure zero and is meager. 
The next theorem says it is the set we are looking for. 
We say that a set $A\subset\real$ is symmetric about 
a point $x\in\real$ if 
$x+h\in A \iff x-h\in A~\forall_h$ (the same as $A=2x-A$).
\thm{cant}{For every $x\in (0,1)$ and for every open 
neighbourhood $U$ of $x$ there is a perfect set $P$ 
contained in $\hat C \cap U$ and symmetric about $x$.}
{\bf Proof.} Let $x\in (0,1)$ and let $0.x_1 x_2 \ldots $ be 
the expansion of $x$. We assume that not almost all $x_i's$ 
are equal to $6$. Let also $U$ be an 
arbitrary neighbourhood of $x$. We fix number $n\in\omega$ 
such that for all $y=0.y_1y_2\ldots $ with $y_i =x_i$ for 
$i\leq n$ we have $y\in U$.

\quad We consider three cases.
\smallskip

Case 1. In the sequence $x_1x_2\ldots $ we have infinitely many 
1's or 2's or 3's or 4's or 5's.

Say at first that we have infinitely many 1's and that 
$x_{i_j}=1, j=1,2,3\ldots $ and $i_j >n$.
We define numbers $y$ and $z$.

$y=0.y_1y_2\ldots $ and $z=0.z_1z_2\ldots $ where

$y_i =x_i$ for $i\leq n$, $z_i =x_i$ for $i\leq n$,

$y_{i_j} =0$ or $y_{i_j} =1$, $z_{i_j} =2-y_{i_j}$, 

$y_i =2$ if $x_i =3,~i>n$, $z_i =4$ if $x_i =3,~i>n$,

$y_i= x_i$, $z_i=x_i$ for all other $i$.

Clearly all such $y$'s form a ``Cantor-like" perfect set and
 so do $z$'s, every $y$ and $z$ belongs to 
$U\cap C_{x_1x_2\ldots x_n}$. For every pair of corresponding 
$y$ and $z$ we have ${{y_i+z_i}\over 2} =x_i$ so 
${{y+z}\over 2} =x$.

\quad If we have infinitely many other digits instead then 
we repeat the above construction changing only $y_{i_j}$ 
and $z_{i_j}$ according to the following formulas.

$x_{i_j}=2~\rightarrow ~y_{i_j}=0~or~y_{i_j}=4,~
z_{i_j}=4-y_{i_j}$, 

$x_{i_j}=3~\rightarrow ~y_{i_j}=0~or~y_{i_j}=6,~
z_{i_j}=6-y_{i_j}$, 

$x_{i_j}=4~\rightarrow ~y_{i_j}=2~or~y_{i_j}=4,~
z_{i_j}=8-y_{i_j}$, 

$x_{i_j}=5~\rightarrow ~y_{i_j}=5~or~y_{i_j}=6,~
z_{i_j}=10-y_{i_j}$. 

Once again we see that ${{y_i+z_i}\over 2}=x_i$ so 
${{y+z}\over 2}=x$ and that $y$'s and $z$'s form a 
symmetric perfect set around $x$ that is contained in 
$\hat C\cap U$.
\smallskip

Case 2. In $x_1x_2\ldots $ we have infinitely many $0$'s and 
$6$'s and Case 1 does not hold. 

In that case we have infinitely many pairs 
$x_i=6,~ x_{i+1}=0$. Let $x_{i_j}=6, x_{i_j+1}=0$ for 
$j=1,2,\ldots $ and $i_j >n$.

We define numbers $y=0.y_1y_2\ldots $ and $z=0.z_1z_2\ldots $.

$y_i=x_i,~z_i=x_i$ for $i\leq n$, 

($y_{i_j}=6$ and $y_{i_j+1}=0$) or ($y_{i_j}=5$ and 
$y_{i_j+1}=1$), 

$z_{i_j}=6$, $z_{i_j+1}=0$ if $y_{i_j+1}=0$
 and $z_{i_j+1}=5$ if $y_{i_j+1}=1$,

$y_i =2$ if $x_i =3,~i>n$, $z_i =4$ if $x_i =3,~i>n$,

$y_i= x_i$, $z_i=x_i$ in other places.

We may easily check that ${{y+z}\over 2}=x$ 
(since ${{y_i+z_i}\over 2}=x_i$), $y$'s and $z$'s are in 
$U$ and in $C_{x_1x_2\ldots x_n}$. It's also clear 
that $y$'s and $z$'s form a perfect set.
\smallskip

Case 3. In $x_1x_2\ldots $ all $x_i$ are eventually equal $0$. 

Now $x=0.x_1x_2\ldots x_k000\ldots $ and $x_k\not =0$.

If $k>n$ then the set 
$C_{x_1x_2\ldots (x_k-1)}\cup C_{x_1x_2\ldots x_k}$ is a desired set.

If $k\leq n$ then the set 
$C_{x_1x_2\ldots (x_k-1)66\ldots 6}~\cup C_{x_1x_2\ldots x_k00\ldots 0}$ 
($n-k$ $0$'s and $n-k$ $6$'s) is a desired set.
\qed

\medskip
\quad Now if we have a copy of $\hat C$ in every interval 
$[n,n+1]$ ($n\in\inte$) that is if 
$$C = \bigcup_{n\in\inte}~\hat C + n$$
then the theorem \ref{cant} holds for any $x\in\real$, and 
we have the corollary.
\cor{mz}{The set $C$ is of measure zero and first 
category and for every $x\in\real$ and 
every open neighbourhood $U$ of $x$ there is a perfect 
set $P$ contained in $C \cap U$ and symmetric about $x$.}

%***************End of added part**************

\quad In theorem \ref{pwsc} 
we built a function with infinitely many values. An 
interesting problem that Ciesielski and Larson ask in 
\cite{c} is whether this is necessary, that is, if there 
is a nowhere weakly symmetrically continuous 
function with finitely many values.

\quad In case of 2-value function we have the following 
result. (Compare also \cite{c}.)

\thm{ThN}{{\rm(Nowik~\cite{n})} 
Every 2-valued function has only countably many points of weak
symmetric discontinuity.}

\quad Ciesielski in \cite{C1}
showed also that every 3-valued function has a point of weak symmetric 
continuity. We do not know if the same is true for 4-valued functions.
Recently Ciesielski and Shelah \cite{CS}
showed that there is an everywhere weakly symmetrically
discontinuous function with bounded range. 
 
\begin{thebibliography}{DbKl}
\bibitem[C1]{C1} K.~Ciesielski,  {\it On range of uniformly antisymmetric
functions}, Real Anal. Exchange {\bf 19} (1993-94), 616--619.


\bibitem[C2]{Ci:book}  K.~Ciesielski, 
{\it Set Theory for the Working Mathematician}, 
London Math. Soc. Student Texts {\bf 39}, 
Cambridge Univ. Press 1997.

\bibitem[CL]{c}
K.~Ciesielski, L.~Larson, {\em Uniformly antisymmetric functions}, 
Real Anal. Exchange {\bf 19(1)} (1993-94), 226--235.

\bibitem[CS]{CS}
K.~Ciesielski, S. Shelah, {\em Uniformly antisymmetric 
function with bounded range}, 
Real Anal. Exchange, to appear.

\bibitem[JS]{j}
J. Jasku{\l}a, B. Szkopi\'nska, 
{\em On the set of points of symmetric continuity},
An. Univ. Bucuresti Mat. {\bf 37} (1988), 29--35.

\bibitem[K]{K} K. Kuratowski, {\em Topology},
Vol. I, Acad. Press, New York, N.Y., 1966.

\bibitem[N]{n}
A. Nowik, private communication.

\bibitem[T1]{t}
B. S. Thomson, {\em Symmetric properties of real functions}, 
Marcel Dekker, 1994.

\bibitem[T2]{Th2}
B. S. Thomson, {\em Real Functions}, Springer-Verlag, 1985. 

\end{thebibliography}
\end{document}
