% version of March 12, 2001 \documentclass[12pt]{article}\usepackage{amssymb} \newcommand{\cont}{{\mathfrak c}} \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}} } \newcommand{\oo}{\omega} \newcommand{\sym}{\underline{S}} \newcommand{\p}{\varphi} \newcommand{\ol}{\overline} \newcommand{\forces}{\mathrel{\|}\joinrel\mathrel{-}} \newcommand{\suc}{{\rm succ}} \newcommand{\ccc}{{\rm con}} \newcommand{\supp}{{\rm supp}} \newcommand{\nor}{{\rm norm}} \newcommand{\norsup}{{\rm\underline{norm}}} \newcommand{\sq}{\subseteq} \newcommand{\real}{{\mathbb R}} \newcommand{\R}{{\real}} \newcommand{\rational}{{\mathbb Q}} \newcommand{\Q}{{\rational}} \newcommand{\integer}{{\mathbb Z}} \newcommand{\N}{{\mathbb N}} \newcommand{\magic}{{\mathbb M}} \newcommand{\Z}{{\mathbb Z}} \newcommand{\NN}{{\cal N}} \newcommand{\nnn}{{\rm N}} \newcommand{\Sg}{{\Sigma}} \newcommand{\s}{{\sigma}} \newcommand{\h}{{\aleph}} \newcommand{\la}{{\langle}} \newcommand{\ra}{{\rangle}} \newcommand{\restr}{{\hbox{$\,|\grave{}\,$}}} \newcommand{\srestr}{{\hbox{${\scriptstyle\,|\grave{}\,}$}}} \newcommand{\A}{{\cal A}} \newcommand{\M}{{\cal M}} \newcommand{\F}{{\cal F}} \newcommand{\G}{{\cal G}} \newcommand{\C}{{\rm C}} \newcommand{\B}{{\cal B}} \newcommand{\T}{{\cal T}} \newcommand{\D}{{\cal D}} \newcommand{\E}{{\cal E}} \newcommand{\I}{{\cal I}} \newcommand{\e}{{\varepsilon}} \newcommand{\g}{{\gamma}} \renewcommand{\d}{{\delta}} \newcommand{\const}{{\rm Const}} \newcommand{\bor}{\mbox{${\cal B}or$}} \newcommand{\mathF}{{\mathbb F}} \newcommand{\mathS}{{\mathbb S}} \newcommand{\cf}{{\rm cf}} \newcommand{\psm}{{\rm PSMA}} \newcommand{\cpa}{{\rm CPA}} %double equivalence \newcommand{\Equi}{\mbox{$\Longleftrightarrow$}} \newcommand{\cm}{{{\rm C}_n}} \newcommand{\cc}{{{\rm C}_c}} \newcommand{\nd}{{\cal ND}} \newcommand{\continuum}{{\cont}} \def\poset{{\mathbb P}} \def\range{\mathop{\rm range}} \def\card{\mathop{\rm Card}} \def\cl{\mathop{\rm cl}} \def\inter{\mathop{\rm int}} \def\cof{\mathop{\rm cof}} \def\k{{\kappa}} \def\dom{{\rm dom}} \def\proof{\noindent {\sc Proof. }} \def\P{{\cal P}} %\def\qed{\hfill\box\vrule height6pt width6pt depth1pt\medskip} \def\qed{\hfill$\Box$ \medskip} \def\endproof{{\qed}} \def\AA{{\cal A}} \newcommand{\tr}{{\rm tr}} \newcommand{\sru}{{\rm sru}} \newcommand{\cra}{{\rm CRA}} \def\implies{\longrightarrow} %% Theorems, etc. \newtheorem{theorem}{Theorem}%[section] \newtheorem{corollary}[theorem]{Corollary} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{problem}[theorem]{Problem} \newtheorem{example}[theorem]{Example} \newtheorem{definition}[theorem]{Definition} \newtheorem{remark}[theorem]{Remark} \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}{ #2}\end{Fact}} \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 $00$. We will show that if $a,b\in S_n(f)$ are such that $00$ 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 $\{a_n\colon 00$ 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 $I_{q_i}$, $10$ 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 00\ 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 $t\in S_{x-h}\setminus T_{x-h}$. 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