% version of March 12, 2001

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\date{}

\title{On nowhere weakly symmetric functions 
and functions with two element range}

\author{
Krzysztof Ciesielski\\
{\footnotesize Department of Mathematics,}
{\footnotesize West Virginia University,}\\
{\footnotesize Morgantown, WV 26506-6310, USA}\\
{\footnotesize e-mail: K\_Cies@math.wvu.edu}\\
{\footnotesize web page: {\tt http://www.math.wvu.edu/\~{}kcies}}
\and
Kandasamy Muthuvel\\
{\footnotesize Department of Mathematics,}
{\footnotesize University of Wisconsin-Oshkosh}\\
{\footnotesize Oshkosh, Wisconsin 54901-8601}\\
{\footnotesize e-mail: muthuvel@vaxa.cis.uwosh.edu}
\and
Andrzej Nowik\\
{\footnotesize Current address: }
{\footnotesize Institute of Mathematics}
{\footnotesize Polish Academy of Sciences}\\
{\footnotesize Abrahama 18, Sopot, Poland}\\
%
{\footnotesize Permanent address:}
{\footnotesize Department of Mathematics,}
{\footnotesize Gda{\'n}sk University}\\
{\footnotesize ul. Wita Stwosza 57, 80-952 Gda{\'n}sk, Poland}\\
{\footnotesize e-mail: matan@paula.univ.gda.pl}\\
%
%{\footnotesize Department of Mathematics,}
%{\footnotesize Gda{\'n}sk University}\\
%{\footnotesize ul. Wita Stwosza 57, 80-952 Gda{\'n}sk, Poland}\\
%{\footnotesize e-mail: matan@paula.univ.gda.pl}\\
{\footnotesize web page: {\tt http://math.univ.gda.pl/\~{}tm/a.nowik}}
}

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\begin{document}
\maketitle

{\def\thefootnote{}%
\footnote[1]{$\!\!\!\!\!$AMS classification numbers: Primary 26A15;
Secondary  03E50, 03E35. \endgraf
Key words and phrases:
continuum hypothesis, second category sets, two-element range functions,
symmetric functions, symmetrically continuous functions,
anti-Schwartz functions. 
}
}


\begin{abstract}
A function $f\colon\real\to\{0,1\}$ is 
weakly symmetric (weakly symmetrically continuous)
at $x\in\real$
provided there is a sequence $h_n\to 0$ such that
$f(x+h_n)=f(x-h_n)=f(x)$
($f(x+h_n)=f(x-h_n)$, respectively) for every $n$. 
We will characterize the sets $S(f)$ of all points 
at which $f$ fails to be weakly symmetrically continuous
and show that $f$ 
must be weakly symmetric at some $x\in\real\setminus S(f)$. 
In particular, there is no 
$f\colon\real\to\{0,1\}$ which is 
nowhere weakly symmetric. 

It is also shown that if at each point $x$ we ignore 
some countable set from which we can choose the sequence $h_n$, then 
there exists a function $f\colon\real\to\{0,1\}$
which is nowhere weakly symmetric
in this weaker sense if and only if the continuum hypothesis holds. 
\end{abstract}

\section{Introduction}

The terminology of this note is standard and follows~\cite{CiBook}.

Recall that a function $f\colon\real\to\real$ is {\em symmetrically
continuous}\/ provided
\begin{equation}\label{con1}
\lim_{h\to 0} [f(x+h)-f(x-h)]=0
\end{equation}
for every $x\in\real$. A ``total negation'' of the condition (\ref{con1})
leads to a notion of 
{\em nowhere weakly symmetrically continuous}\/ function,%
\footnote{This terminology comes from 
a monograph Thomson~\cite{Thomson}.
Ciesielski and Larson~\cite{CL} 
and most other papers on this subject 
call such a function {\em uniformly antisymmetric}.}
that is,
\begin{equation}\label{con1a}
\lim_{n\to\infty}[f(x+h_n)-f(x-h_n)]\neq 0
\end{equation}
for every sequence $h_n \searrow 0$. It is easy to see that 
if the range of $f$
is discrete then (\ref{con1a}) 
is equivalent to: ``$f(x+h_n)\neq f(x-h_n)$ for 
all but finitely many $n$.'' 
In particular, if $f\colon\real\to\N$ then $f$ is 
not weakly symmetrically continuous at $x\in\real$
provided $f(x+h)\neq f(x-h)$ for all $h>0$ small enough,
that is, when there exists an $\e_x>0$ such that
the interval $(0,\e_x)$ is disjoint with the set
\[
S_x=\{h>0\colon f(x-h)=f(x+h)\}.
\]

In~\cite{CL} Ciesielski and Larson constructed a 
nowhere weakly symmetrically continuous
function $f\colon\real\to\N$ for which each set $S_x$ is finite.
(See also \cite[cor.~7.4.2]{CiBook}.)
They show also there that there is no 
nowhere weakly symmetrically continuous
function with two-element range.
The latter fact was proved earlier by Kostyrko~\cite{Kostyrko}.

Nowhere weakly symmetrically continuous
functions have been studied by many
authors. Komj\'{a}th and Shelah~\cite{KomSh} investigated %a question
under what set theoretical assumptions the
sizes of all sets $S_x$ can be bounded by
a fixed number. (See also~\cite{C:SumsAndDiff}.)
In~\cite{C0}  Ciesielski
proved that there is no nowhere weakly symmetrically continuous
function with three-element range and
in~\cite{C4}  he showed that the technique from the paper~\cite{C0}
cannot be used to prove that there is
no nowhere weakly symmetrically continuous
function with four-element range.
It is still an open question
(see \cite[Problem~1(a)]{CL}, \cite[Problem~2(a)]{56:surv}, or \cite{Thomson})
whether there exists
a nowhere weakly symmetrically continuous
function with finite range, though Ciesielski and Shelah~\cite{CSh}
constructed a nowhere weakly symmetrically continuous
function with bounded countable range.


For $f\colon\real\to\real$ let $S(f)$
denote
the set of all
points at which $f$ is not weakly symmetrically continuous, that is,
\[
S(f)=\{x\in\real\colon
\mbox{$f$ is not weakly symmetrically continuous at $x$}\}.
\]
We say that a set $A\subset\real$ is locally symmetric  provided
for every $x\in A$ there exists an $\e_x>0$ such that
\[
x-h\in A\ \Equi\ x+h\in A \ \ \ \ \mbox{for every $h\in (0,\e_x)$}.
\]
Our first theorem characterizes the family of all sets
$S(f)$ with $f\colon\real\to\{0,1\}$.


\thm{thA}{\begin{itemize}
\item[{\rm (a)}]
If $f\colon\real\to\{0,1\}$ then the set $S(f)$ is
countable and locally symmetric.
\item[{\rm (b)}] For every countable and locally symmetric subset $A$ of $\real$
there exists an $f\colon\real\to\{0,1\}$ such that $S(f)=A$.
\end{itemize}
}







A notion dual to weakly symmetrically continuous function
is that of a symmetric function:
a function $f\colon\real\to\real$ is
a {\em symmetric function}\/ provided
\begin{equation}\label{con2}
\lim_{h\to 0} [f(x+h)+f(x-h)-2 f(x)]=0
\end{equation}
for every $x\in\real$.
The ``total negation'' of (\ref{con2}) parallel to that done
for the case of  symmetrically continuous function leads to
a notion of {\em nowhere weakly symmetric}\/ function
and was introduced by Ciesielski in \cite{C4}.%
\footnote{Ciesielski~\cite{C4} calls such function {\em anti-Schwartz},
while current terminology follows Thomson~\cite{Thomson}.}
We will be interested in this notion only for
functions
$f\colon\real\to S$ when $S\subset\N$ contains no arithmetic progression
of length three. In this particular case
it is more convenient to adopt a simpler equivalent definition:
a function $f\colon\real\to S$ is nowhere weakly symmetric
provided for every $x\in\real$ there exists an $\e_x>0$ such that
the interval $(0,\e_x)$ is disjoint with the set
\[
T_x=\{h>0\colon f(x-h)=f(x)=f(x+h)\}.
\]
The paper \cite{C4} contains also a construction of
a nowhere weakly symmetric function $f\colon\real\to S\subset\N$ for which
$T_x=\emptyset$ for every $x\in\real$.
It also contains a question whether there exists a 
nowhere weakly symmetric function
with two-element range.
One of the goals of this paper is to answer this question in 
the negative:

\thm{thNoASc}{There is no nowhere weakly symmetric function
$f\colon\real\to\{0,1\}$.
}

In the second part of the paper we will consider 
the variations of weak symmetric continuity 
and weak symmetry associated with 
any ideal $\I$ on $\real$ in the following way:
for every $x\in\real$ we choose an exceptional set $E_x\in\I$,
and for this $x$ we consider only the sequences $h_n$ disjoint 
with $E_x$. 
The investigation of these notions for 
the ideal of countable subsets of $\real$
leads to the following theorem. 

\thm{th:almost}{
The following conditions are equivalent.
\begin{description}
\item[{\rm (i)}]   The Continuum Hypothesis.
\item[{\rm (ii)}]  There exists a function $f\colon\real\to\{0,1\}$
                   such that $|S_x|\leq\omega$
                   for every $x\in\real$.
\item[{\rm (iii)}] There exists a function $f\colon\real\to\{0,1\}$ such
                   that for every
                   $x\in\real$ there exists an $\e_x>0$ such that
                   $|S_x\cap(0,\e_x)|\leq\omega$.
\item[{\rm (iv)}]  There exists a function $f\colon\real\to\{0,1\}$
                   such that $|T_x|\leq\omega$ for every $x\in\real$.
\item[{\rm (v)}] There exists a function $f\colon\real\to\{0,1\}$ such
                   that for every
                   $x\in\real$ there exists an $\e_x>0$ such that
                   $|T_x\cap(0,\e_x)|\leq\omega$.
\end{description}
}
It should be pointed  here that the equivalences
(i)$\Leftrightarrow$(ii)$\Leftrightarrow$(iii)
are known and were proved in ~\cite{CL}.  
(See also \cite[thm. 7.4.4]{CiBook}.)

We also show 
(Theorems~\ref{thAntiSym} and~\ref{thAntiSch}) 
that for the ideal of meager subsets of $\real$
a similar theorem cannot be proved. 






\section{Every function with two-element range is somewhere weakly symmetric
%There is no nowhere weakly symmetric function with two-element range
}

\noindent{\sc Proof of Theorem~\ref{thA}}(a).
We will first show that $S(f)$ is countable.
For $0<n<\omega$ let
\[
S_n(f) = \{x \in \real\colon (0,1/n)\cap S_x=\emptyset\}.
\]
Then clearly
\[
S(f)=\bigcup_{0<n<\omega}S_n(f).
\]
We will show that each $S_n(f)$ is discrete, so countable.

Fix an $n<\omega$, $n>0$.
We will show that if $a,b\in S_n(f)$ are such that
$0<b-a<1/n$ then the points $a$ and $b$ are isolated in $S_n(f)$.
This clearly implies that $S_n(f)$ is discrete.

So take $a,b\in S_n(f)$ such that
$0<b-a<1/n$ and consider the set
\[
Z = \left(b-\frac{1}{n},a+\frac{1}{n}\right) \setminus \{a, b\}.
\]
We will show that $Z\cap S_n(f)=\emptyset$, so that
$a$ and $b$ are isolated in $S_n(f)$.

Thus, take any $z \in Z$. Since the
diameter of $\{a,b,z\}$
is less than $1/n$
we have
\[
f(2a - z) = 1-f(z)=f(2b - z).
\]
But the mid point $(a+b)-z$ of $2a - z$ and $2b - z$ is of distance less
than
$1/n$ from $\{2a - z,2b - z\}$. So $(a+b) - z \notin S_n(f)$.

This implies that the set $Z=(a+b)-Z$ is disjoint with $S_n(f)$, as
required.

It is also easy to see that if $x\in S(f)$ and $\e_x>0$ is such that
$(0,\e_x)\cap S_x=\emptyset$
then
\[
x-h\in S(f)\ \Equi\ x+h\in S(f) \ \ \ \ \mbox{for every $h\in (0,\e_x)$}.
\]
Thus, $S(f)$ is locally symmetric. \qed

%\medskip

\noindent{\sc Proof of Theorem~\ref{thA}}(b).
The proof presented here is an elaboration of that
for \cite[thm.~1]{KomSh}, in which the authors prove that
for any countable subset $A$ of $\real$ there exists 
an $f\colon\real\to \{0,1\}$ with $A\subset S(f)$. 

Let $\{a_n\colon 0<n<\omega\}$ be an enumeration of $A$
and $\{d_n\colon 0<n<\omega\}$ be an enumeration of $G(A)\setminus A$,
where $G(A)$ is an additive group generated by $A$. 
Following \cite{KomSh} for every $n<\omega$ we define inductively
a finite collection $\I_n = \{I_\gamma\colon\gamma \in \Gamma_n\}$
of open intervals such that
$\emptyset = \Gamma_0 \subseteq \Gamma_1 \subseteq \cdots$,
each $I_{\gamma}$ is of the form
$(b_\gamma - h_\gamma, b_\gamma + h_\gamma)$,
and the following properties hold for any $\gamma,\gamma'\in\Gamma_n$. 
\begin{enumerate}
\item[(1)]
  If $\gamma \neq \gamma^\prime$, then either 
  $I_\gamma \cap I_{\gamma^\prime} = \emptyset$ or one of the intervals 
  $I_\gamma$ and $I_{\gamma^\prime}$ is a subset of another. 

\item[(2)]
  If $I_{\gamma^\prime}\subseteq I_\gamma$, then
  either $I_{\gamma'} \subseteq (b_\gamma - h_\gamma, b_\gamma)$
  or $I_{\gamma'} \subseteq (b_\gamma, b_\gamma + h_\gamma)$.

\item[(3)] $\{a_1,\ldots,a_n\}\subseteq B_n \subseteq A$,
   where $B_n = \{b_{\gamma}\colon \gamma\in\Gamma_n\}$.

\item[(4)] $b_\gamma \pm h_\gamma \notin A$. 

\item[(5)]
  If $I_{\gamma^\prime} \subseteq I_\gamma$, then
  $\p_\gamma[I_{\gamma^\prime}]\in {\cal I}_n$,
  where $\p_\gamma\colon I_\gamma\to I_\gamma$ 
  is a symmetry function with respect to
  $b_\gamma$, that is, $\p_\gamma(x) = 2 b_\gamma - x$. 

\item[(6)] If $x\in I_\gamma$ then 
           $x \in A$ if and only if $\p_{\gamma}(x) \in A$. 

\item[(7)] If $I_\gamma\notin\I_{n-1}$ then  $d_n\notin I_\gamma$.
\end{enumerate}
Conditions (1)--(5) are present in the proof of \cite[thm.~1]{KomSh},
while (6) and (7) are specific for our construction. 

The construction is pretty much repetition of that in \cite{KomSh}.
We start with $\Gamma_0=\I_0=\emptyset$. 
If $\I_{n-1}$
%  = \{I_\gamma\colon\gamma \in \Gamma_n\}$
is already constructed, we construct $\I_n$ as follows. 
If $a_n\in B_{n-1}$ we put $\I_n=\I_{n-1}$. So assume that 
$a_n\notin B_{n-1}$ and notice that the set $B$
of all (well defined) numbers $\p_{\gamma_1}\circ\cdots\circ\p_{\gamma_r}(a_n)$
has at most $2^{|\I_{n-1}|}$ elements and that $a_n\in B\subset A\setminus B_{n-1}$.
Find an $h>0$ small enough that for every $b\in B$ the interval $(b-h,b+h)$
is disjoint with the set 
$\{d_1,\cdots,d_n\}\cup B_n\cup
\{b_\gamma\pm h_\gamma\colon\gamma\in\Gamma_{n-1}\}$
and that $A\cap (b-h,b+h)$ is symmetric with respect to $b$. 
Moreover, decreasing $h$ if necessary, we also require that
$\{b\pm h\colon b\in B\}$ is disjoint with $A$.
We define $\I_n$ as $\I_{n-1}\cup\{(b-h,b+h)\colon b\in B\}$.
It is easy to see that $\I_n$ is as desired. 

In what follows we will write 
$y=\p_{\gamma_1}\circ\cdots\circ\p_{\gamma_r}(x)$
only when all $\gamma_i$'s are from $\Gamma=\bigcup_{n<\omega}\Gamma_n$ 
and the value of the composition is well defined.
Note also that if $y=\p_{\gamma_1}\circ\p_{\gamma_2}(x)$ and 
$I_{\gamma_1}\subsetneq I_{\gamma_2}$ then we have also
$y=\p_{\gamma_2}\circ\p_{\gamma_1^\prime}(x)$,
where $b_{\gamma_1^\prime}=\p_{\gamma_2}(b_{\gamma_1})$. 
From this and the fact that $\p_\gamma\circ\p_\gamma(z)=z$
for every $z$ we can conclude that
\begin{description}
\item{($*$)} If $y=\p_{\gamma^\prime_1}\circ\cdots\circ\p_{\gamma^\prime_s}(x)$
    there exists an $r\leq s$ of the same parity that $s$ and 
    $\gamma_1,\ldots,\gamma_r\in\Gamma$ such that
    $y=\p_{\gamma_1}\circ\cdots\circ\p_{\gamma_r}(x)$
    and $I_{\gamma_1}\supsetneq\cdots\supsetneq I_{\gamma_r}$.
\end{description}

Notice also the following two facts. 
\begin{description}
\item{(A)} If $y=\p_{\gamma_1}\circ\cdots\circ\p_{\gamma_r}(x)$
    and $r$ is odd then $y\neq x$. 
\end{description}

Indeed, by ($*$) we can assume that 
$I_{\gamma_1}\supsetneq\cdots\supsetneq I_{\gamma_r}$. 
But this means that $x$ and $y$ are on the different sides of 
$b_{\gamma_1}$, so they are not equal. 

\begin{description}
\item{(B)} For every $x_0\in\real\setminus A$ there exists
a $\delta=\delta_x>0$ such that for every $x\in(x_0-\delta,x_0+\delta)$
if $y=\p_{\gamma_1}\circ\cdots\circ\p_{\gamma_r}(x)$ 
and $r$ is odd then $y\neq 2x_0-x$. 
\end{description}

To see it, first notice that 
\begin{eqnarray}\label{eqThm1}
y=\p_{\gamma_1}\circ\cdots\circ\p_{\gamma_r}(x)
= -x - 2 \sum_{i=1}^r(-1)^{i}b_{q_i}.
\end{eqnarray}
Now, if $2x_0-x=y=\p_{\gamma_1}\circ\cdots\circ\p_{\gamma_r}(x)$ 
then $x_0=-\sum_{i=1}^r(-1)^{i}b_{q_i}\in G(A)$. 
In particular, if $x_0\notin G(A)$ then (B) holds for any $\delta>0$.
Thus, we can assume that $x_0\in G(A)\setminus A$,
that is, $x_0=d_n$ for some $n<\omega$. 
Note also that if $x_0$ does not belong to any $I_\gamma$ then 
$y$ stays on the same side of $x_0$ that $x$ does and, once again, 
any $\delta>0$ works. 
So we can assume that $x_0$ belongs to some $I_\gamma$ which, by (7),
must belong to $\I_{n-1}$. 

Let $I_{\gamma}$ be the unique 
shortest interval such that $x_0 \in I_{\gamma}$ and let $\delta>0$ be 
such that
$(x_0-\delta,x_0+\delta)
\subset(b_\gamma - h_\gamma,b_\gamma + h_\gamma)
\setminus\{b_\gamma\}$. We will show that this $\delta$ satisfies (B).

Assume, by way of contradiction, that
there is an $x\in(x_0 - \delta, x_0 + \delta)$
such that 
$2x_0-x=y=\p_{\gamma_1}\circ\cdots\circ\p_{\gamma_r}(x)$ and $r$ is odd.
By ($*$) we can also assume that 
$I_{\gamma_1}\supsetneq\cdots\supsetneq I_{\gamma_r}$.

Since $b_{q_1}$ stays on one side of each $I_{q_i}$, $1<i\leq r$, 
the points $x$ and $\p_{\gamma_2}\circ\cdots\circ\p_{\gamma_r}(x)$ 
are on the same side of $b_{q_1}$. 
So $x$ and 
$2x_0-x=\p_{\gamma_1}\circ\cdots\circ\p_{\gamma_r}(x)$
are on the opposite sides of $b_{q_1}$. 
But $x,2x_0-x\in(x_0-\delta,x_0+\delta)\subset I_\gamma$,
implying that $b_{q_1}\in(x_0-\delta,x_0+\delta)\subset I_\gamma$. 
So, $I_{\gamma_1}\subsetneq I_\gamma$. 
However, $x$ and $2 x_0-x=y$ belong also to $I_{\gamma_1}$, and so does
$x_0$, which is between $x$ and $2 x_0-x$. 
This gives $x_0\in I_{\gamma_1}\subsetneq I_\gamma$, 
contradicting the choice of $I_\gamma$. This completes the proof of (B). 

With (A) and (B) in hand, we can construct the desired function $f$.
For this, consider $\real$ as vertices of an infinite graph
and declare numbers $x$ and $y$ as connected by an edge
provided $y=\p_\gamma(x)$ for some $\gamma\in\Gamma$. 
Also, for each $x\in\real$ let $C_x$ stands for all $y\in\real$ 
such that there exists a path from $x$ to $y$, that is, such that
$y=\p_{\gamma_1}\circ\cdots\circ\p_{\gamma_r}(x)$ 
for some $\gamma_1,\ldots,\gamma_r\in\Gamma$. 

Note that condition (A) says that our graph has no odd cycles, so 
we can color it by two colors, $0$ and $1$, such that 
no connected vertices have the same color. 
In fact, if we color each component $C_x$ that way, then the resulted 
coloring, a function $f\colon\real\to\{0,1\}$,
will have this property. It is easy to see that for any
such $f$ we will have $A\subset S(f)$.
To see that we can find such an $f$ with $A=S(f)$
first note that, by (\ref{eqThm1}), if $y\in C_x$ then 
$y$ belongs to the group generated by $A\cup\{x\}$.
Thus each component $C_x$ is at most countable. 
We color components $C_x$ by transfinite induction.
We enumerate $(\real\setminus A)\times\omega$ as 
$\la \la x_\xi,n_\xi\ra\colon \xi<\continuum\ra$.
Next, at each stage $\xi<\continuum$ we choose 
and color two new components $C_\xi$ and $C'_\xi$
in the following way. 
First we choose $x\in(x_\xi,x_\xi+\delta_{x_\xi}/2^{n_\xi})$
such that neither $x$ nor $2x_\xi-x$ is in previously chosen components.
Then we choose components $C_\xi$ and $C'_\xi$ containing 
$x$ and $2x_\xi-x$, respectively, and color each in such a way, that 
both $x$ and $2x_\xi-x$ get the same color.
Such a coloring can be obviously chosen if 
components $C_\xi$ and $C'_\xi$ are not equal. However, if 
$C_\xi=C'_\xi$ then, by (B), $x$ and $2x_\xi-x$ must be colored 
the same color automatically.

It is easy to see that a coloring function $f$ defined that way
is as desired. \qed

The characterization given in Theorem~\ref{thA}
does not hold for
functions with a 3-element range.
This follows from the fact that,
by \cite[thm.~2]{C4}, for every Hamel
basis
$H\subset\real$ there exists
an $f\colon\real\to\{0,1,2\}$ such that $H\subset S(f)$.
Since a Hamel
basis
can be a Bernstein set (see e.g. \cite[cor.~7.3.7]{CiBook})
the set $S(f)$ for such an $f$ need to be neither meager nor of measure zero.






With Theorem~\ref{thA} in hand we are ready to prove 
our second theorem. \medskip

\noindent{\sc Proof of Theorem~\ref{thNoASc}}.
By way of contradiction assume that $f$ is 
nowhere weakly symmetric.

 From Theorem~\ref{thA} we know that $|S(f)|\leq \oo$.
Translating $f$, if necessary, we can assume that $0 \notin S(f)$.
Let $V$ be an uncountable linear subspace of $\real$ over $\rational$
disjoint with $S(f)$.
We will obtain a contradiction with \cite[thm. 2.1]{CL} by showing that
the restriction function
\[
f\restriction V\colon V\to\{0,1\}
\]
is nowhere weakly symmetrically continuous. 

So, fix an $x\in V$. Since $f$ is not weakly symmetric
at $x$ there exists an $\e>0$
such that $(0,\e)\cap T_x=\emptyset$.
We will show that this implies that
we have also $(0,\e)\cap S_x\cap V=\emptyset$.
We will assume that
\[
f(x)=0,
\]
the argument for $f(x)=1$ being essentially the same.

By way of contradiction assume that there is an $h\in (0,\e)\cap S_x\cap V$.
Then $f(x-h)=f(x+h)$ is not equal to $f(x)=0$ since $h\notin T_x$.
In particular,
\[
f(x-h)=f(x+h)=1.
\]
Since $f$ is not weakly symmetric
at $x+h$ and at $x-h$ we can find a
$\delta\in(0,\e-h)$ such that
$(T_{x-h}\cup T_{x+h})\cap(0,\delta)=\emptyset$.

Since $x-h\in V\subset\real\setminus S(f)$
there exists a $t\in(0,\delta)\cap(0,h)\cap S_{x-h}$.
Then $f((x-h)-t)=f((x-h)+t)\neq f(x-h)=1$, since
$t\in S_{x-h}\setminus T_{x-h}$.
Thus,
\[
f((x-h)-t)=f((x-h)+t)=0.
\]
Since $h+t\in(0,\e)\subset\real\setminus T_x$ and $f((x-h)-t)=f(x)=0$
we conclude that
\[
f((x+h)+t)=1.
\]
Similarly $h-t\in(0,\e)\subset\real\setminus T_x$ and $f((x-h)+t)=f(x)=0$
imply that
\[
f((x+h)-t)=1.
\]
Thus
$f((x+h)+t)=f((x+h)-t)=f(x+h)=1$ contradicting the fact that $t\notin T_{x+h}$.
\qed

Now for $f\colon\real\to\real$ let $T(f)$ be the set of all
$x\in\real$ such that $f$ is not weakly symmetric at $x$.
In this notation Theorem~\ref{thNoASc} says that
$T(f)\neq\real$ for every function $f\colon\real\to\{0,1\}$.
One might suspect that
a characterization similar to that from Theorem~\ref{thA} may be true also
for $T(f)$.
This,
however, is not true.

\ex{ex1}{There exists an $f\colon\real\to\{0,1\}$ such that
$T(f)$ is not meager.
}

\proof Let $f_0\colon\real\to\N$ be a function from
\cite[cor.~1.2]{CL}. For this function none of the level sets
$f_0^{-1}(n)$ contains an arithmetic progression of length 3.
Let $n\in\N$ be such that
$f_0^{-1}(n)$ is not meager and let
$f$ be a characteristic function of the set $f_0^{-1}(n)$.
It is easy to see that $f_0^{-1}(n)\subset T(f)$. \qed

It is easy to see that the same argument gives also a function
$f\colon\real\to\{0,1\}$ for which the set
$T(f)$ is not of measure zero.

It is also worth to mention here the following fact.


\prop{pwsc}{{\rm (Szyszkowski~\cite{msz})}
For every $B\subset\real$ there exists an $f\colon\real\to\N$
for which $S(f)=B$.}

We do not know whether its weakly symmetric
analog
is also true. Therefore the following question remains open.

\pr{prob1}{Is it true that for
for every $B\subset\real$ there exists an $f\colon\real\to\N$
for which $T(f)=B$?}

\section{Nowhere almost weakly symmetric
function with two-element range}

We will start this section with the proof of Theorem~\ref{th:almost}. 
Since the equivalences
(i)$\Leftrightarrow$(ii)$\Leftrightarrow$(iii)
are known it remains to show 
(i)$\Leftrightarrow$(iv)$\Leftrightarrow$(v).
This follows immediately from the following theorem.

\thm{thm1}{Let $\kappa$ be an infinite cardinal number and let
$W$ be a linear subspace of $\real$ over $\rational$.
The following conditions are equivalent.
\begin{description}
\item[{\rm (i)}]   $|W|\leq\kappa^+$.
\item[{\rm (ii)}]  There exists a function $f\colon\real\to\{0,1\}$
                   such that $|T_x\cap W|\leq\kappa$ for every $x\in W$.
\item[{\rm (iii)}] There exists a function $f\colon\real\to\{0,1\}$ such
                   that for every
                   $x\in W$ there exists an $\e_x>0$ such that
                   $|T_x\cap W\cap(0,\e_x)|\leq\kappa$.
\end{description}
}

We will prove Theorem~\ref{thm1} only for $\kappa=\omega$
which is enough to get Theorem~\ref{th:almost}. 
The proof of the general
case is essentially the same. (Though it requires
also a simple generalization of 
the first part of Theorem~\ref{th:almost}.)
So, in the rest of the argument we will assume that $\kappa=\omega$.

Since $T_x\subset S_x$, the implication (i)$\Rightarrow$(ii)
follows from the first part of Theorem~\ref{th:almost}.
Clearly (ii) implies~(iii).
Thus, it is enough to prove only that (iii) implies~(i).

In the proof we will need the following lemma.

\lem{lem1}{Let $f\colon\real\to\{0,1\}$
and let $V$ be a linear subspace of $\real$ over $\rational$.
If $Z=\{x\in V\colon\exists \e>0\  (|S_x\cap V\cap(0,\e)|\leq\omega)\}$
then $|Z|\leq\omega_1$.
}

\proof The proof is essentially identical to that for 
(iii)$\Rightarrow$(i) from Theorem~\ref{th:almost}
as presented in \cite[thm. 7.4.4]{CiBook}.
Just assume, by way of contradiction, that $|Z|>\omega_1$
and choose $B\subset Z$ of cardinality $\omega_2$ which is
linearly independent over~$\rational$. For this choice of $B$
the intact proof presented in \cite[thm.~7.4.4]{CiBook}
gives the desired contradiction.
\qed

\noindent{\sc Proof of Theorem~\ref{thm1}.}
Let $f$ be as in (iii) and, by way of contradiction, assume that
$|W|\geq\omega_2$. Let $Z_0$ be as in Lemma~\ref{lem1} with $V=W$.
By shifting the domain of~$f$, if necessary, we can assume that
$0\notin Z_0$.
Let $V$ be a linear subspace of $W$ over $\rational$
such that $|V|\geq\omega_2$ and $V\cap Z_0=\emptyset$.
Apply again Lemma~\ref{lem1}
to get a set $Z$ working for $f$ and this $V$.
Then $|V\setminus Z|\geq\omega_2$.

Fix an $x\in V\setminus Z$. By (iii) we can choose an
$\e>0$ with the property that $|T_x\cap W\cap(0,2\e)|\leq\omega$.
Since $x\notin Z$ we can also find an $h\in(0,\e)\cap S_x\cap V\setminus T_x$.
In particular $f(x-h)=f(x+h)$.
Next, assume that
\[
f(x)=0,
\]
the argument for $f(x)=1$ being essentially the same.
Then
\[
f(x-h)=f(x+h)=1,
\]
since $h\notin T_x$. Let $\delta\in(0,h)$ be such that
the set $T=(T_{x-h}\cup T_{x+h})\cap W\cap(0,\delta)$ is countable.
Since $x-h\in V\subset W\setminus Z_0$,
the set $S_{x-h}\cap W\cap(0,\delta)$ is uncountable.
In particular, there exists a $t\in S_{x-h}\cap W\cap(0,\delta)\setminus T$
such that
\[
h+t,h-t\in(0,2\e)\setminus T_x.
\]
Then $f((x-h)-t)=f((x-h)+t)\neq f(x-h)=1$, since
$t\in S_{x-h}\setminus T_{x-h}$.
Thus,
\[
f((x-h)-t)=f((x-h)+t)=0.
\]
Since $h+t\notin T_x$ and $f((x-h)-t)=f(x)=0$
we conclude that
\[
f((x+h)+t)=1.
\]
Similarly $h-t\notin T_x$ and $f((x-h)+t)=f(x)=0$
imply that
\[
f((x+h)-t)=1.
\]
Thus
$f((x+h)+t)=f((x+h)-t)=f(x+h)=1$ contradicting the fact that $t\notin T_{x+h}$.
\qed

Let us finish with a remark that
Theorem~\ref{thm1} can be proved with $W$ being replaced by
any additive subgroup of $\real$.

\bigskip

The above theorems show that, under CH, for a function $f\colon\real\to\{0,1\}$
the sets $S_x$ and $T_x$ can be countable.
Can we make them any smaller? Can they all be, for example, nowhere dense?
The following two theorems show that this is not possible.
In what follows $\M_0$ and $\M$
will stand for the ideals of nowhere dense and meager subsets
of $\real$, respectively.

\thm{thAntiSym}{Let $f\colon\real\to\{0,1\}$, $M\in\M$,
and $V$ be a linear subspace of $\real$ over $\rational$.
If $S_x^M=\{h\in S_x\cap V\colon h\notin \pm(x-M)\}$ and
\[
Z=\{x\in V\colon \exists \e>0\ S_x^M\cap(0,\e)\in\M_0\}
\]
then $Z\in\M$.
}

\proof By way of contradiction assume that $Z\notin\M$.
So, we have also $V\notin\M$.
For every $x\in Z$ let $n_x\in\N$ be such that
the set
\[
C_x=S_x^M\cap(0,1/n_x)
\]
is nowhere dense. The sets $Z_n=\{x\in Z\setminus M\colon n_x=n\}$
form a countable partition of $Z\setminus M\notin\M$, so there exists an
$n\in\N$
such that $Z_n\notin\M$.
Let $U$ be an open interval of length less than $1/n$ such that the set
$L=Z_n\cap U$ is nowhere meager in $U$.
Note that
$n_x=n$ for all $x\in L$. Moreover, if
\[
C_x^*=C_x\cup\pm(x-M)\in\M
\]
then $S_x\cap V\cap (0,1/n)\subset C_x^*$ so that
\begin{equation}\label{con3b}%\!\!\!\!\!\!
f(x-h)\neq f(x+h)\mbox{ for }x\in L\mbox{ and }
h\in(0,1/n)\cap V\setminus C_x^*,%\!\!\!\!\!\!
\end{equation}
and
\[
|x-y|<1/n\ \mbox{ for every }\ x,y\in L.
\]

Let $x=t_0\in L$ and let
$\{t_m\colon 0<m<\omega\}$ be a dense subset of $L\setminus(x\pm C_x^*)$.
Moreover, let $K$ be an additive subgroup generated by $\{t_m\colon m<\omega\}$
and choose a
\begin{equation}\label{conZz}
\mbox{$z\in L\setminus \bigcup_{i<\omega}(K\pm C_{t_i}^*)$,}
\end{equation}
$z\neq x$. Moreover,
choose $y=t_m\notin\{x,z\}$ such that
\begin{equation}\label{conZy}
y\notin x\pm C_z.
\end{equation}
Such a choice is possible  since $C_z$ is nowhere dense and $t_m$'s are
dense in $U$.
Next define
\[
a=x-y+z,\ \ \ b=x+y-z,\ \ \ c=-x+y+z.
\]
Then $(a+b)/2=x$, $(b+c)/2=y$, and $(c+a)/2=z$.
We will show that
\begin{equation}\label{con3d}
f(a)\neq f(b),\ \ \ f(b)\neq f(c),\ \ \ f(c)\neq f(a),
\end{equation}
which will give
us the desired contradiction, since the
points $a$, $b$, and $c$ are distinct and $f$ admits only two values.
But (\ref{con3d}) follows immediately from (\ref{con3b})
and
\begin{equation}\label{con3f}
|a-x|\not\in C_x^*,\ \ \
|b-y|\not\in C_y^*,\ \ \
|c-z|\not\in C_z^*,
\end{equation}
since we obviously have $|a-x|,|b-y|,|c-z|\in(0,1/n)\cap V$.
But (\ref{con3f}) follows easily from (\ref{conZz}) and (\ref{conZy}):


$|a-x|\not\in C_x^*$, since otherwise we would have
$z-y=a-x\in \pm C_x^*$ and
$z\in y\pm C_x^*$, contradicting (\ref{conZz}).

$|b-y|\not\in C_y^*$, since otherwise we would have
$z-x=y-b\in \pm C_y^*$ and $z\in x\pm C_y^*$,
again contradicting (\ref{conZz}).

$|c-z|\not\in C_z^*$, since otherwise point
$y-x=c-z$ would belong to the set $\pm C_z^*=\pm[C_z\cup \pm(z-M)]$.
However, $y-x\in \pm C_z$ implies that $y\in x\pm C_z$,  contradicting
(\ref{conZy}).
On the other hand if
$y-x\in \pm(z-M)$, then $z\in K+M$, contradicting (\ref{conZz}).
\qed



\thm{thAntiSch}{Let $f\colon\real\to\{0,1\}$. If
\[
U=\{x\in\real\colon \exists \e>0\ T_x\cap(0,\e)\in\M_0\}
\]
then $\real\setminus U\notin\M$.
}

\proof The proof is quite similar to that for Theorem~\ref{thm1}.
By way of contradiction assume that $U^c=\real\setminus U\in\M$.
Let $Z_0\in\M$ be from Theorem~\ref{thAntiSym} used with $V=\real$ and
$M=\emptyset$.
Then apply again Theorem~\ref{thAntiSym} with $V=\real$ and
$M=Z_0\cup U^c$ to get a set $Z\in\M$.

Next fix an $x\in\real\setminus (Z\cup U^c)$. We assume that
\[
f(x)=0,
\]
the argument for $f(x)=1$ being essentially the same.
Since $x\in U$ we can choose an
$\e>0$ such that $T_x\cap(0,\e)\in\M_0$.
Since $x\notin Z$ we have $S_x^M\cap(0,\e)\notin\M_0$.
Thus, there exists a non-empty open interval $J_0\subset(0,\e)\setminus T_x$
intersecting $S_x^M=S_x\setminus\pm(x-M)$.
Let $h\in S_x^M\cap J_0$. Then
\[
f(x-h)=f(x+h)=1,
\]
since $h\in S_x\setminus T_x$.
Since $h\notin \pm(x-M)$ we have $x\pm h\in \real\setminus M=U\setminus Z_0$.
So there exists a $\delta>0$ such that
the set $T=(T_{x-h}\cup T_{x+h})\cap(0,\delta)$ is nowhere dense.
Decreasing $\delta$, if necessary, we can also assume that
$J=(h-\delta,h+\delta)\subset J_0$.
Next, since $x-h\notin Z_0$,
we can find a $t\in S_{x-h}\cap (0,\delta)\setminus T$.

Then $f((x-h)-t)=f((x-h)+t)\neq f(x-h)=1$, since
$t\in S_{x-h}\setminus T_{x-h}$.
Thus,
\[
f((x-h)-t)=f((x-h)+t)=0.
\]
But $h\pm t\in J\subset J_0\subset(0,\e)\setminus T_x$ and
$f(x-(h\pm t))=f(x)=0$ imply that
\[
f(x+(h\pm t))=1.
\]
Thus
$f((x+h)\pm t)=f(x+h)=1$ contradicting the fact that $t\notin T_{x+h}$.
\qed





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\end{document}



