\documentclass{rae}
\usepackage{amsmath}
\usepackage{amssymb}


%\documentstyle[amstex]{production}
\firstpagenumber{226}
\received{October 15, 1993}

\Volume{19}
\IssueNumber{1}
\Year{1993/94}


\def\integer{\Bbb Z}
\def\natural{\Bbb N}
\def\rational{\Bbb Q}
\def\real{\Bbb R}
% Gotic letters
     \newfont{\goticnine}{eufm9}
% continuum c
     \newcommand{\continuumnine}{\mbox{{\goticnine c}}} 
%End of proof marker
\def\qed{\hfill$\Box$}
%Proof
\def\pf{\noindent{\sc Proof.} }

\newcommand{\supp}{\operatorname{supp}}

%% Theorems, etc.

\newtheorem{theorem}{Theorem}[section]
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{example}[theorem]{Example}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{problem}{Problem}

\newcommand{\thm}[2]{\begin{theorem}\label{#1}#2\end{theorem}}
\newcommand{\cor}[2]{\begin{corollary}\label{#1}#2\end{corollary}}
\newcommand{\examp}[2]{\begin{example}\label{#1}#2\end{example}}
\newcommand{\lem}[2]{\begin{lemma}\label{#1}#2\end{lemma}}
\newcommand{\pr}[2]{\begin{problem}\label{#1}#2\end{problem}}
 
%%%%%%%%%%%%%%%%%%%%%   MISCELLANEOUS
 
\renewcommand{\P}{\mbox{$\cal P$}}
\newcommand{\B}{\mbox{$\cal B$}}
\newcommand{\e}{\mbox{$\varepsilon$}}

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

\title{UNIFORMLY ANTISYMMETRIC FUNCTIONS}

\author{Krzysztof Ciesielski,
Department of Mathematics,
West Virginia University,
Morgantown, WV 26506-6310,
(kcies@wvnvms.wvnet.edu)
\and
Lee Larson,
Department of Mathematics,
University of Louisville, Louisville, KY 40292,
(lmlars01@ulkyvx.louisville.edu)}

\date{}

\markboth{K.~Ciesielski and L.~Larson}{Uniformly Antisymmetric Functions}

\keywords{symmetric continuity, Hamel basis}
\MathReviews{26A15, 26A21}

\begin{document}\maketitle

\begin{abstract}
An example is given of a function $f:\real\to\real$ which satisfies the
property that for each $x$ there is an $\e(x)>0$ such that
$|f(x+h)-f(x-h)|\ge1$ whenever $0<h<\e(x)$. Further properties of such
functions are also discussed.
\end{abstract}

\section{An unbounded example}

A function $f:\real\to\real$ is {\em weakly continuous} at $x_0$ if
there are sequences $a_n\nearrow0$ and $b_n\searrow0$ such that
\[
\lim_{n\to\infty}f(x_0+a_n)=f(x_0)=\lim_{n\to\infty}f(x_0+b_n).
\]
A surprising theorem from cluster set theory 
\cite[p. 82]{CollingwoodLohwater:ThClSets}
states that an {\em arbitrary} function is weakly continuous everywhere on the
complement of a countable set. 

A function $f\colon\real\to\real$ is symmetrically continuous at 
$x_0\in\real$ if 
\[
\lim_{h\to 0^+} f(x_0+h)-f(x_0-h)=0;
\]
i.e., if for every $\e>0$
\begin{equation}\label{quant1}
(\exists d>0)(\forall 0<h<d)\ |f(x_0+h)-f(x_0-h)|<\e.
\end{equation}
Motivated by the definitions given above, we define a function to be {\em
weakly symmetrically continuous} at $x_0$ if there exists a sequence
$h_n\searrow0$ such that
\[
\lim_{n\to\infty}f(x_0+h_n)-f(x_0-h_n)=0.
\]

It is easy to construct functions which are nowhere symmetrically continuous.
For example, the characteristic function of any dense Hamel basis is such a
function.%
\footnote{In fact, a Hamel basis \B\ can be chosen to be both first category and
measure zero. For example, \B\ can be constructed such that $\B\setminus C$
is dense and countable, where $C$ is the Cantor ternary set. Thus
$\charf{\B}$ can
be measurable and have the Baire property. 
(Compare this with Theorem \ref{th2}.)}\label{footA}
Any nowhere symmetrically continuous function $f$ must satisfy the
property that for every $x$ there exists $\e>0$ such that
\begin{equation}\label{NegSym}
(\forall d>0)(\exists 0<h<d)\ |f(x-h)-f(x+h)|\geq\e.
\end{equation}
In asking how 
badly the symmetric continuity of $f$ can break, several people%
\footnote{Evans and Larson, 7th Real Analysis Exchange Conference, Santa
Barbara, 1984. P. Kostyrko, 15th Summer Symposium on Real Analysis, Smolenice,
1991.}
have asked whether a function can be nowhere weakly symmetrically continuous.
This is the same as asking whether the quantifiers in
(\ref{NegSym}) can be exchanged.  That is, whether
there exists a function $f\colon\real\to\real$ such that for every $x\in\real$ 
there exists $\e>0$ such that
\[
(\exists d>0)(\forall 0<h<d)\ |f(x-h)-f(x+h)|\geq\e.
\]
It is clear that this last question is equivalent to asking whether
the following statement holds:
\begin{description}
\item[($\star$)] There exists a function $f\colon\real\to\real$ such that
for every 
$x\in\real$ there is a $d>0$ (depending on $x$) so that whenever
$0<h<d$,
\[
|f(x-h)-f(x+h)|\geq d.
\]
\end{description}
A function $f$ satisfying ($\star$) will be called a 
{\em uniformly antisymmetric function.} In view of the theorem cited above for
weak continuity, the existence of such a function is by no means obvious.

The purpose of this note is to show that uniformly antisymmetric functions
exist.
The existence of such functions can be inferred from the following theorem.

\thm{th1}{ There exists a partition $\P=\{P_n\colon n\in\natural\}$ of
$\real$ such that for every $x\in\real$ the set
\begin{equation}\label{ConTh1}
S_x=\bigcup_{n\in\natural}\{h>0\colon x-h,x+h\in P_n\}
\end{equation}
is finite.
}

Before proving this theorem, we first show how it implies the existence of a
uniformly antisymmetric function.

\cor{cor1}{ Let $K\subset\natural$. If $\P=\{P_n\colon n\in K\}$ is a partition
of $\real$ satisfying (\ref{ConTh1}), then there exists a uniformly
antisymmetric 
function $f\colon\real\to K$. In particular, there 
exists a uniformly antisymmetric 
function $f\colon\real\to\natural$. }

\pf Let $\P=\{P_n\colon n\in K\}$ be a partition 
satisfying (\ref{ConTh1}) and for $x\in\real$ define
\[
f(x)=n\ \mbox{ if and only if }\ x\in P_n.
\]
Then $f$ is uniformly antisymmetric, since for every $x\in\real$
the set 
\[
S_x=\bigcup_{n\in\natural}\{h>0\colon x-h,x+h\in P_n\}
=
\{h>0\colon f(x-h)=f(x+h)\}
\]
is finite and
$|f(x-h)-f(x+h)|\geq 1$ for every $h>0$ such that $h\not\in S_x$.
Thus ($\star$) is satisfied by $d=\min S_x\cup\{1\}$. \qed
%
\medskip
%\newline
%
%{\sc Proof of Theorem \ref{th1}.}

{\noindent{\sc Proof of Theorem \ref{th1}.} }
Let \B\ be a Hamel basis for $\real$; i.e., a linear basis of $\real$ over
$\rational$. Consider $\rational$ with the discrete topology and 
$\rational^{\cal B}$ with the product topology. Then, 
$\rational^{\cal B}$ is separable, 
being a product of continuum many separable spaces.%
\footnote{ This is the Hewitt-Marczewski-Pondiczery Theorem 
\cite{Engelking:GenTop}.
In fact, if $\cal U$ is a countable basis for ${\cal B}\subset\real$ then 
the set 
$D\subset\rational^{\cal B}$ of all functions of the form
$\Sigma_{i<n} q_i \charf{U_i}$
(where $n\in\natural$,  $q_i\in\rational$  and $U_i\in{\cal U}$)
is countable and dense in $\rational^{\cal B}$.}
Let $D=\{d_n\colon n\in\natural\}$ be a dense subset of $\rational^{\cal B}$.

Define a one-to-one function $E\colon \real\to\rational^{\cal B}$
by putting $E(x)(b)=q_b$, where
$x=\Sigma_{b\in{\cal B}} q_b b$  is the unique representation of
$x$ in the Hamel basis $\B$. Notice that for every $x\in\real$ the set
$\supp(x)=\{b\in\B: E(x)(b)\neq 0\}$ is finite.

For $x\in\real$ let 
\[
[x]=\{g\in\rational^{\cal B}\colon g(b)=E(x)(b)\ 
\mbox{ for all }\ b\in\supp(x)\}.
\] 
Clearly, $[x]$ is a basic open set in $\rational^{\cal B}$.
Define 
\[
f(x)=\min\{n\colon d_n\in[x]\}
\]
and let $P_n=f^{-1}(n)$ for every $n\in\natural$.
We will show that the 
partition $\P=\{P_n\colon n\in\natural\}$ of
$\real$ satisfies (\ref{ConTh1}).

First notice that the following condition implies (\ref{ConTh1}):
\begin{description}
\item[(i)]  for every $x\in\real$ there exists only a finite number of
$h>0$ such that
$[x-h]\cap[x+h]\neq\emptyset$.
\end{description} 

To show (i), choose $x,h\in\real$, $h>0$, such that 
$[x-h]\cap[x+h]\neq\emptyset$.

Notice that 
\[
E(x-h)(b)+E(x+h)(b)=2\,E(x)(b)\ \text{ for all }\ b\in\B
\]
and, since $[x-h]\cap[x+h]\neq\emptyset$, 
\[
E(x-h)(b)=E(x+h)(b)=E(x)(b)\ \text{ for all } b\in\supp(x-h)\cap\supp(x+h).
\]
Hence, 
\begin{description}
\item[(A)]  $\supp(x-h)\cup\supp(x+h)=\supp(x)$
\end{description}
and for all $b\in\supp(x)$ exactly one of the following holds:
\begin{description}
\item[(B)] $b\in\supp(x-h)\cap\supp(x+h)$ and $E(x-h)(b)=E(x+h)(b)=E(x)(b)$,
\item[(C)] $b\in\supp(x-h)\setminus\supp(x+h)$ and $E(x-h)(b)=2\,E(x)(b)$,
$E(x+h)(b)=0$, or
\item[(D)] $b\in\supp(x+h)\setminus\supp(x-h)$ and $E(x+h)(b)=2\,E(x)(b)$,
$E(x-h)(b)=0$.
\end{description}

It is clear that there are at most $3^n$ numbers $h>0$ satisfying
(A)--(D), where $n$ is the cardinality of $\supp(x)$. 
Thus, Theorem \ref{th1} has been proved. \qed

The next theorem shows that unlike nowhere symmetrically continuous functions,
the uniformly antisymmetric functions can be neither measurable nor have
the Baire property. (See footnote \ref{footA})

\thm{th2}{ If $f\colon\real\to\real$ is uniformly antisymmetric, then
$f$ is neither measurable nor does it have the Baire property.}

\pf Our proof uses the following lemma, in which a point $x$ is said
to be a {\em qualitative point} of a set $S\subset\real$, if there is 
a neighborhood $U$ of $x$ such that $U\setminus S$ is of the first
category. \qed

\lem{lem1}{ If $d>0$ and $x$ is either a density point of $S$ or a qualitative
point of $S$, 
then there is $0<h<d$ such that $-h,h\in S$. }

\pf If $x$ is a density point of $S$, choose $0<\e<d$ such that
the Lebesgue measure of $S\cap(x-\e,x+\e)$ is greater that \e. 
If $x$ is a qualitative point of $S$, choose $0<\e<d$ such that
$(x-\e,x+\e)\setminus S$ is of first category. In both cases
it is easy to find  $0<h<\e$ such that $-h,h\in S$. \qed

It is well known \cite{Kur1}
that for a Baire function $f\colon\real\to\real$ there exists a first
category set $A$ such that every point $x\in\real\setminus A$
is a qualitative point of $f^{-1}((f(x)-d/2,f(x)+d/2))$ for every $d>0$.
A point $h$, chosen as in Lemma~\ref{lem1}, contradicts ($\star$).

The argument when $f$ is measurable follows the same path,
if we recall that a measurable function is approximately continuous
almost everywhere; i.e., that almost every point $x$ 
is a density point of $f^{-1}((f(x)-d/2,f(x)+d/2))$ for every $d>0$.

\cor{cor2}{ The existence of a uniformly antisymmetric function cannot
be proved without using of the Axiom of Choice.} 

\pf Shelah \cite{Shelah1} proved that there exists
a model of set theory ZF (without the Axiom of Choice) with 
the Dependent Choice Axiom DC (i.e., standard induction) 
in which every subset of $\real$ has the Baire property.
In this model every real function is a Baire function. So, by
Theorem \ref{th2},%
\footnote{More precisely, we should notice that the Theorem \ref{th2}
has been proved in ZF+DC.}
it is not uniformly antisymmetric. \qed

\section{ Bounded antisymmetric functions }

For the results in this section, we need the following generalization
of ($\star$).

Let $K\subset\real$ be an additive subgroup of $\real$. A function
$f\colon K\to\real$ is said to be {\em uniformly antisymmetric} 
provided for every 
$x\in K$, there exists $d>0$ such that for every
$0<h<d$, $h\in K$,
\begin{equation}\label{conGen1}
|f(x-h)-f(x+h)|\geq d.
\end{equation}

The next theorem has been also proved, by a different method, in
\cite{Kostyrko} for the case $K=\real$.

\thm{th21}{ If $K\subset\real$ is an uncountable additive subgroup of $\real$ 
then there is no uniformly antisymmetric function $f\colon K\to\{0,1\}$.%
\footnote{ This theorem is also true in case  $f\colon K\to\{0,1,2\}$.
This proof will appear in future work.}
In particular, a uniformly antisymmetric function $f\colon\real\to\real$
cannot be a characteristic function. }

\pf By way of contradiction assume that there exists
a uniformly antisymmetric function $f\colon K\to\{0,1\}$
and for every $x\in K$ let $n_x\in\natural$ be such that for every
$0<h<1/n_x$, $h\in K$,
\[
|f(x-h)-f(x+h)|\geq 1/n_x.
\]
Fix $n\in\natural$ such that $L=\{x\in K\colon n_x=n\}$ is uncountable. 
Then, 
\begin{equation}\label{conth21a}
f(x-h)\neq f(x+h)\ \mbox{ for every }\ x\in L\ \mbox{ and }\ h\in(0,1/n)\cap K. 
\end{equation}
Choose $x,y,z\in L$ such that $x<y<z<x+1/n$ and let
\[
a=x-y+z,\ \ \ b=x+y-z,\ \ \ c=-x+y+z.
\]
Then, for $h=z-y\in(0,1/n)\cap K$ we have $a=x+h$ and $b=x-h$. 
Hence, by (\ref{conth21a}),
$f(a)\neq f(b)$. Similarly,
$f(b)\neq f(c)$ and $f(c)\neq f(a)$. But this is impossible, since the
points $a$, $b$ and $c$ are distinct and $f$ attains only two values.
This contradiction finishes the proof of Theorem \ref{th21}. \qed

The next theorem is the main step in the proof
that the assumption that $K$ is uncountable in Theorem \ref{th21}
is essential.

\thm{th22}{ There is a partition $\P=\{A_0,A_1\}$ of the set $D_+$ of positive
dyadic numbers, $k/2^n$, $k,n\in\natural$, such that
for every $x\in D_+$ the set
\begin{equation}\label{ConTh22}
D_x=\bigcup_{i\in\{0,1\}}\{h\in D_+\cap(0,x)\colon x-h,x+h\in A_i\}
\end{equation}
is finite.
}

\pf Put 
\[
A_0=\left\{\frac{4k+1}{2^n}\colon k,n\in\natural\right\}, \ \ \  
A_1=\left\{\frac{4k+3}{2^n}\colon k,n\in\natural\right\}
\]
and define 
\[
D_n=\left\{\frac{2k+1}{2^n}\colon k\in\natural\right\}
\]
for every $n\in\natural$. It is easy to see that the sets $D_n$
form a partition of $D_+$.
Notice also that for $m<n$
\[
\frac{4k+1}{2^n}+\frac{4l+1}{2^n}=\frac{2(k+l)+1}{2^{n-1}}\in D_{n-1}
\ \mbox{ and }\ 
\frac{4k+1}{2^m}+\frac{4l+1}{2^n}\in D_n.
\]
Hence, for $x\in D_n$ we have $2x\in D_{n-1}$ and if $a,b\in A_0$ are such 
that $a+b=2x$, then $a,b\in\bigcup_{m\leq n}D_m$.
Similarly we can show that for 
$x\in D_n$, if $a,b\in A_1$ are such 
that $a+b=2x$, then $a,b\in\bigcup_{m\leq n}D_m$. In particular,
if $x\in D_n$, $a=x-h$, $b=x+h$ 
and $a,b\in A_i$ for $i\in\{0,1\}$, then $a,b\in\bigcup_{m\leq n}D_m$.
But, since $a,b>0$
it is easy to see that there are only finitely many
pairs $\langle a,b\rangle $ such that 
$a+b=2x$ and $a,b\in\bigcup_{m\leq n}D_m$. So, the set $D_x$ is finite. 
\qed

\cor{cor23}{ If $D$ is the set of all dyadic 
numbers $k/2^n$, $n\in\natural$, $k\in\integer$, then there exists 
a uniformly antisymmetric function $f\colon D\to\{0,1\}$. In particular,
the uncountability of $K$ in Theorem \ref{th21}
is essential.
}

\pf As in Corollary \ref{cor1}, we can use the partition of
Theorem \ref{th22} to show the existence of a uniformly antisymmetric 
function $f\colon D_+\to\{0,1\}$ by putting $f(x)=i$ for $x\in A_i$. 
Extend this function to $D$ by 
defining $f(0)$ arbitrarily and $f(-d)=1-f(d)$ for $d\in D_+$.
It is easy to see that such a function is uniformly antisymmetric. \qed

The following problems remain open.%
\footnote{ The preliminary versions of this paper have been circulated 
since the Fall of 1992. Since then, several problems have been solved, 
as mentioned in the footnotes.}

\pr{pr1}{ Is there a uniformly antisymmetric function
          $f\colon\real\to\real$ such that its range $f[\real]$ is
          (a) finite? (b) bounded? }

\pr{pr2}{ Is there a finite
partition $\P$ of $\real$ such that for every $x\in\real$ the set
\[
\bigcup_{P\in{\cal P}}\{h>0\colon x-h,x+h\in P\}
\]
is finite?%
\,\footnote{ A negative answer has been given by 
P.~Komj\'ath and S.~Shelah \cite{KomShel}. J.~Baumgartner \cite{Baumgartner}
also
showed independently the somewhat stronger result that
for every finite
partition $\P$ of $\real$ and a cardinal number
$\kappa<\continuumnine$
there exists $x\in\real$ such that
the set 
$\bigcup_{P\in{\cal P}}\{h>0\colon x-h,x+h\in P\}$ 
has cardinality $\geq\kappa$.}
}

Notice that Corollary \ref{cor1} shows that a positive answer for 
Problem \ref{pr2}
implies a positive answer for Problem \ref{pr1} and that,
by Theorem \ref{th21} and Corollary \ref{cor1},
a partition \P\ from Problem \ref{pr2} must have at least $3$ elements.

\pr{pr3}{ Is there a countable subfield $K$ of $\real$ for which
          we can find a uniformly antisymmetric function
          $f\colon K\to\{0,1\}$? What about $K=\rational$?%
\,\footnote{  
P.~Komj\'ath and S.~Shelah \cite{KomShel} showed that 
a uniformly antisymmetric function $f\colon K\to\{0,1\}$
exists for every countable subfield $K$ of $\real$. They also 
proved that 
there exists a function
$f\colon\rational\to\{0,1,2,3\}$ such that the set 
$S_x$ is finite for every $x\in\rational$. Such a function is uniformly 
antisymmetric.}
}


\section{ Generalizations }

The following theorem generalizes some of the results given above. 
Part (ii) was inspired by discussions with Paul Erd\"{o}s. 
The implication (i)$\Rightarrow$(ii) is due to Sierpi\'{n}ski
\cite{Sierpinski}. The equivalence of (i) and (ii) is also implicitly
contained in Freiling \cite{Freiling2}.


\thm{th31}{ The following conditions are equivalent.
\begin{description}
\item[{\em (i)}]   The Continuum Hypothesis.
\item[{\em (ii)}]  There exists a partition $\P=\{A_0,A_1\}$ of $\real$
such that
             for every $x\in\real$ the set 
                \begin{equation}\label{G1}
                    T_x=\bigcup_{i\in\{0,1\}}\{h>0\colon x-h,x+h\in A_i\}
                \end{equation}
             is at most countable.
\item[{\em (iii)}] There exists a function $f\colon\real\to\{0,1\}$ such
that for every 
             $x\in\real$ there is a $d>0$ with the property that
\end{description}
\begin{equation}\label{G2}
       |f(x-h)-f(x+h)|\geq d\ \mbox{ for all but countable many }\ h\in(0,d).
\end{equation}
}

\pf (i)$\Rightarrow$(ii). Let $\B=\{b_\zeta\colon \zeta<\omega_1\}$
be a Hamel basis. For $x\in\real\setminus\{0\}$ let $q(x)=q_n$ where
$x=q_1 b_{\zeta_1}+\ldots+q_n b_{\zeta_n}$
is the unique
representation of $x$ in the basis \B, with
$\zeta_1<\ldots<\zeta_n$ and $q_i\neq 0$.
Put
\[
x\in A_0\ \mbox{ if and only if }\ q(x)>0
\]
and $A_1=\real\setminus A_0$.
We will show that the partition $\P=\{A_0,A_1\}$ satisfies (\ref{G1}).

For $\xi<\omega_1$ let $K_\xi$ be the linear subspace of $\real$
generated by $\{b_\zeta\colon\zeta\leq\xi\}$. Notice that
$\real=\bigcup_{\xi<\omega_1}K_\xi$ and that, by the Continuum Hypothesis, 
every $K_\xi$ is countable.
We will show that if $x\in K_\xi$ then $T_x\subset K_\xi$.
This will finish the proof. 

But it is easy to see that for $h\in\real\setminus K_\xi$ and $x\in K_\xi$,
\[
x\in A_0\ \text{ if and only if }\ x+h\in A_0.
\]
Since $h$ and $-h$ cannot belong to the same $A_i$ for $h\neq 0$, we have
$T_x\subset K_\xi$.

(ii)$\Rightarrow$(iii). The proof of this implication is the same as the one for
Corollary \ref{cor1}, if we put $f(x)=i$ for $x\in A_i$.

(iii)$\Rightarrow$(i). Let $f$ be as in (iii) and
for every $x\in\real$ let $n_x\in\natural$ be such that for every
$0<h<1/n_x$, 
\begin{equation}\label{con3a}
|f(x-h)-f(x+h)|\geq 1/n_x\ \mbox{ for all but countable many }\ 0<h<d
\end{equation}
and let $C_x$ be a countable set of exceptional points for (\ref{con3a}); i.e.,
\[
C_x=\{h<1/n_x\colon f(x-h)\neq f(x+h)\}.
\]
By way of 
contradiction, assume that the Continuum Hypothesis fails and let $B$ be a
linearly
independent subset of $\real$ over $\rational$ of cardinality $\omega_2$.
Choose $K\subset B$ of cardinality $\omega_2$ such that for some 
$n\in\natural$ we have $n_x=n$ for all $x\in K$. 
Let $U$ be an open interval of length less than $1/n$ such that the set
$L=K\cap U$ has cardinality $\omega_2$. Then, in particular, 
\begin{equation}\label{con3b}
f(x-h)\neq f(x+h)\ \mbox{ for every }\ x\in L\ \mbox{ and }\ 
h\in(0,1/n)\setminus C_x,
\end{equation}
and
\begin{equation}\label{con3c}
|x-y|<1/n\ \mbox{ for every }\ x,y\in L.
\end{equation}
Define, by transfinite induction, a sequence
$\{t_\xi\in L\colon \xi<\omega_2\}$ such that
\begin{equation}\label{con3ca}
t_\xi\not\in T_\xi\ \mbox{ for every }\ \xi<\omega_2,
\end{equation}
where $T_\xi$ is the smallest linear subspace of $\real$
containing $\{t_\zeta\colon\zeta<\xi\}$ and such that 
\begin{equation}\label{con3cb}
C_x\subset T_\xi\ \mbox{ for every }\ x\in T_\xi.
\end{equation} 
Evidently every 
set $T_\xi$ has cardinality $\leq\omega_1$, so
the induction can be done easily.

Now, put $x=t_0$, $z=t_{\omega_1}$ and, for $0<\xi<\omega_1$ consider the 
numbers $|-x+t_\xi|$.
All these numbers are different, so there is $0<\eta<\omega_1$ such that
\begin{equation}\label{con3cba}
|-x+t_\eta|\not\in C_z.
\end{equation}
Put $y=t_\eta$ and proceed as in Theorem \ref{th21}.
Define
\begin{equation}\label{con3cc}
a=x-y+z,\ \ \ b=x+y-z,\ \ \ c=-x+y+z.
\end{equation}
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.

To prove (\ref{con3d}), notice first that it follows from (\ref{con3b}) 
as long as
\begin{equation}\label{con3e}
|a-x|<1/n,\ \ \
|b-y|<1/n,\ \ \
|c-z|<1/n
\end{equation}
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}
But (\ref{con3e}) follows easily from (\ref{con3cc}) and (\ref{con3c}).
Finally, (\ref{con3f}) can be proved as follows.

$|a-x|\not\in C_x$, since otherwise we would have $z-y=a-x\in T_{\omega_1}$ and
$z\in y+T_{\omega_1}\subset T_{\omega_1}$, contradicting 
$z=t_{\omega_1}\not\in T_{\omega_1}$.

$|b-y|\not\in C_y$, since otherwise we would have $z-x=y-b\in T_{\omega_1}$ and
$z\in x+T_{\omega_1}\subset T_{\omega_1}$, 
contradicting $z=t_{\omega_1}\not\in T_{\omega_1}$.

$|c-z|\not\in C_z$, since otherwise we would have 
$|-x+t_\eta|=|-x+y|=|c-z|\in C_z$, contradicting (\ref{con3cba}).

This finishes the proof of Theorem \ref{th31}. \qed

We finish this paper with the following open problem.

\pr{pr4}{ 
Does there exist a partition $\P=\{P_n\colon n\in\natural\}$ of
$\real$ such that 
the set
\[
S_x=\bigcup_{n\in\natural}\{h>0\colon x-h,x+h\in P_n\}
\]
has at most one element.%
\footnote{  
P.~Komj\'ath and S.~Shelah \cite{KomShel} proved that 
a positive answer for this question is equivalent to the Continuum
Hypothesis. They also proved that if $\continuumnine\geq\aleph_n$
then for every partition $\P=\{P_n\colon n\in\natural\}$ of
$\real$ there exists
$x\in\real$ such that the set $S_x$ has $\geq 2^n-1$ elements.
}
}

\begin{thebibliography}{22}
\bibitem{Baumgartner} James E.~Baumgartner, A letter to K. Ciesielski, May 1993.
\bibitem{CollingwoodLohwater:ThClSets} E.~F.~Collingwood, A.~J.~Lohwater, 
The Theory of Cluster Sets, Cambridge University Press, Cambridge, 1966.
\bibitem{Engelking:GenTop} R.~Engelking, General Topology, 
Polish Scientific Publishers, Warsaw, 1977.
\bibitem{Freiling2} Chris Freiling, {\sl A converse to a theorem of Sierpinski
on almost symmetric sets}, Real Analysis Exchange {\bf 15} (1989-90), 760--767.
\bibitem{KomShel} P.~Komj\'ath and S.~Shelah, 
{\sl On uniformly antisymmetric functions}, 
Real Analysis Exchange {\bf19(1)} (1993-94), 218--225.
\bibitem{Kostyrko} P.~Kostyrko, 
{\sl There is no strongly locally antisymmetric set},
Real Analysis Exchange {\bf 17} (1991-92), 423--425.
\bibitem{Kur1} K. Kuratowski, 
Topology, Vol. I, Academic Press, New York, NY, 1968.
\bibitem{Shelah1} S. Shelah, {\sl Can you take Solovay's innaccessible away?}, 
Israel J. Math. {\bf 48} (1984), 1-47.
\bibitem{Sierpinski} W. Sierpi\'{n}ski, 
{\sl Sur une fonction non mesurable partout presque sy\-met\-rique}, 
Acta Litt. Scient (Szeged) {\bf 8} (1936), 1-6. 
{\sl Oeuvres Choisies} t. {\bf III}, 277-281.


\end{thebibliography}



\end{document}




