% for latex 2.09 (Old) % needs epsf.tex to proceed picture file kcgraph.eps \documentstyle{article} \input{epsf.tex} %\headheight=0in %\headsep = 0in %\topmargin=0in %\textheight=9.12in %\textwidth=6.8in %\oddsidemargin=0in %\evensidemargin=0in %\parindent=0.2in \title{Uniformly approachable maps% \thanks{Work partially supported by the NATO Collaborative Research Grant CRG~950347 %research projects 60\% and 40\% of the Italian Ministero dell' %Universit\`{a} e della Ricerca Scientifica e Tecnologica. %\\ {\it Key words:} \endgraf AMS classification numbers: Primary 54C30, 41A30; \endgraf Secondary 54E35, 54B30, 26A15. \endgraf Key words and phrases: {\it metric spaces, uniformly continuous maps, uniform approachability, complex analytic functions.} }} \author{ Krzysztof Ciesielski\\ {\footnotesize Department of Mathematics, West Virginia University,} \\ {\footnotesize Morgantown, WV 26506-6310} \\ {\footnotesize KCies@wvnvms.wvnet.edu} \and Dikran Dikranjan \\ {\footnotesize Dipartimento di Matematica e Informatica, Universit\`{a} di Udine} \\ {\footnotesize Via Zanon 6, 33100 Udine, Italy} \\ {\footnotesize dikranja@dimi.uniud.it} } \date{} %*****************theorems and such \newtheorem{theorem}{Theorem}[section] \newtheorem{corollary}[theorem]{Corollary} \newtheorem{conjecture}[theorem]{Conjecture} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{problem}[theorem]{Problem} \newtheorem{em-example}[theorem]{Example} % example (auxiliary) \newtheorem{em-def}[theorem]{Definition} % definition (auxiliary) \newtheorem{em-remark}[theorem]{Remark} % remark (auxiliary) \newtheorem{em-question}[theorem]{Question} \newenvironment{example}{\begin{em-example} \em }{ \end{em-example}} \newenvironment{definition}{\begin{em-def} \em }{ \end{em-def}} \newenvironment{remark}{\begin{em-remark} \em }{\end{em-remark}} \newenvironment{question}{\begin{em-question} \em }{\end{em-question}} \newenvironment{proof}{Proof.}{QED \bigskip} \newenvironment{claim}{\underline{Claim}: \em}{\medskip} %commonly used newcommands %*********boolean connectives******** \newcommand{\club}{{\bf N.B.}} \newcommand{\limplies}{\mathrel{\rightarrow}} % Logical Implication \newcommand{\liff}{\mathrel{\leftrightarrow}} % Logical IFF Symbol \newcommand{\bigor}{\bigvee} %big or \newcommand{\bigand}{\bigwedge} %big and \newcommand{\false}{\bot} % false propositional constant \newcommand{\true}{\top} % true propositional constant \newcommand{\metaimplies}{\Rightarrow} %implication in the metatheory \newcommand{\metaimplied}{\Leftarrow} %reverse impl. in metatheory \newcommand{\metaiff}{\Longleftrightarrow} %iff in metatheory \newcommand{\metaand}{\mbox{and \ }} %can be used in math. mode \newcommand{\metaor}{\mbox{or \ }} \newcommand{\bi}{{\bf b}} \newcommand{\metanot}{\mbox{not }} %***********set operations and relations******************* \newcommand{\union}{\cup} %set union \newcommand{\diam}{\mbox{diam\,}} \newcommand{\included}{\subseteq} \newcommand{\includes}{\supseteq} \newcommand{\contains}{\supseteq} %set inclusions \newcommand{\intersect}{\cap} \newcommand{\nin}{\notin} %negated membership \newcommand{\iu}{{\bf i}} %*******other commands in alphabetical order************** \newcommand{\composed}{\circ} %function composition \newcommand{\gn}[1]{\lceil #1 \rceil} %Godel number \newcommand{\Image}{\mbox{Image}} \newcommand{\map}{\mapsto} %arrow for functions \newcommand{\proves}{\vdash} %provability sign \newcommand{\st}{\; | \;} \newcommand{\summation}[2]{{\sum_{#1}#2}} %summation \newcommand{\ov}[1]{\overline{#1}} \newcommand{\Un}{\mbox{\bf Unif}} \newcommand{\R}{{\bf R}} \newcommand{\N}{{\bf N}} \newcommand{\C}{{\bf C}} \newcommand{\Q}{{\bf Q}} \newcommand{\restricted}[1]{_{| #1}} \newcommand{\sq}{\subseteq} \newcommand{\Reals}{{\bf R}} \newcommand{\im}{\mbox{Im}} \newcommand{\re}{\mbox{Re}} \newcommand{\pr}{{\rm pr}} %\maketitle \begin{document} \maketitle % \begin{sloppypar} \section{Preliminaries}\label{Preliminaries} Throughout the paper we will use the standard definitions and notation (\cite{ABR}, \cite{E}). Let $X$ and $Y$ be metric spaces. The goal is to study some intermediate classes of functions between the class $C_{uc}(X,Y)$ of all uniformly continuous mappings (briefly, $UC$) from $X$ into $Y$ and the class $C(X,Y)$ of all continuous functions $f\colon X\to Y$. These classes, defined below, have been intensively studied in \cite{BD} and \cite{BDP} mainly in the case when $Y=\R$. (Compare also \cite{B,DP}.) In this paper we will study them for general $Y$. In particular, we will consider the case when $X=Y$ is the complex plane $\C$. Let $X$ and $Y$ be metric spaces and let $f\colon X\to Y$. \begin{enumerate} \item For $K,M \subseteq X$ we say that $g\colon X\to Y$ is a {\em $(K,M)$-approximation} of $f$ if $g$ is a $UC$ map such that $g[M]\subseteq f[M]$ and $g(x)=f(x)$ for each $x\in K$. \item The function $f$ is {\it uniformly approachable} (briefly, $UA$) if $f$ has a $(K,M)$-approximation for every compact $K \subseteq X$ and every $M\subseteq X$. \item The function $f$ is {\it weakly uniformly approachable} (briefly, $WUA$) if $f$ has a $(\{x\},M)$-approximation for every $x\in X$ and every $M\subseteq X$. \end{enumerate} The class of all uniformly approachable (weekly uniformly approachable, respectively) functions $f\colon X\to Y$ will be denoted by $C_{ua}(X,Y)$ ($C_{wua}(X,Y)$, respectively.) Notice that \begin{equation}\label{eq1} C_{uc}(X,Y)\subseteq C_{ua}(X,Y)\subseteq C_{wua}(X,Y)\subseteq C(X,Y). \end{equation} (See \cite{DP}. Compare also \cite[Section 12]{BD}.) Indeed, every $UC$ function $f$ is $UA$, since $g=f$ is a $(K,M)$-approximation of $f$ for every $M,K\subseteq X$. Every $UA$ function is $WUA$, since $\{x\}$ is compact for every $x\in X$. To see that every $WUA$ map $f$ is continuous, it is enough to show that $f[\ov{M}]\subseteq\ov{f[M]}$ for every $M\subseteq X$, where $\ov{M}$ stands for the closure of $M$. So, take $x\in\ov{M}$ and an $(\{x\},M)$-approximation $g$ of $f$. Then, $f(x)=g(x)\in g[\ov{M}]\subseteq\ov{g[M]}\subseteq\ov{f[M]}$. We will discuss possible equations in (\ref{eq1}) for different spaces $X$ and $Y$ in Section \ref{sec:Eq}. For a topological space $X$ we say that a function $g\in C(X,\R)$ is a {\em truncation} of $f\in C(X,\R)$ if $g$ is constant on every connected component of $\{x\in X\colon f(x)\neq g(x)\}$. In what follows we will use also the following fact which can be proved by straightforward transfinite induction diagonal argument. \begin{proposition}\label{prop:magic} {\rm \cite[Thm 8.1]{BD}} Let $X$ be a separable topological space. Then there is a set $M \subseteq X$ such that for every $f,g\in C(X,\R)$, if $g[M]\subseteq f[M]$ and $f^{-1}(y)$ is at most countable for every $y\in\R$ then $g$ is a truncation of $f$. \end{proposition} The set $M$ from Proposition \ref{prop:magic} will be called a {\em magic set (for $X$).} The following fact is an easy but useful corollary of Proposition \ref{prop:magic}. It is a version of \cite[Cor 8.3]{BD}. \begin{corollary}\label{cor:magic} Let $X$ be a separable metric space. If $f\in C(X,\R)$ is a one-to-one function without non-constant $UC$ truncation then $f$ is not $WUA$. \end{corollary} \section{Basic facts}\label{BasicFacts} In this section we collect some basic properties of the classes of $UA$- and $WUA$-maps. \begin{theorem}\label{thm:composition} Composition of $UA$ ($WUA$, respectively) maps is a $UA$ ($WUA$, respectively) map. \end{theorem} \begin{proof} Let $f\colon X\to Y$ and $f'\colon Y\to Z$ be $UA$ maps. To show that $f'\circ f\colon X\to Z$ is $UA$ take a compact subset $K$ of $X$ and a set $M\sq X$. Then $f[K]$ is a compact subset of $Y$. Choose a $(K,M)$-approximation $g\colon X\to Y$ of $f$ and a $(f[K],f[M])$-approximation $g'\colon Y\to Z$ of $f'$. It is enough to show that $g'\circ g$ is a $(K,M)$-approximation of $f'\circ f$. Obviously $g'\circ g$ coincides with $f'\circ f$ on $K$. Also $g[M]\sq f[M]$ and $g'[f[M]]\sq f'[f[M]]$ implies that $(g'\circ g)[M]\sq (f'\circ f)[M]$. The proof is finished when we notice that $g'\circ g$ is $UC$ as a composition of two $UC$ functions. The argument for $WUA$ functions is identical, if we replace $K$ with $\{x\}$. \end{proof} \begin{proposition}\label{prop:restr} Let $F$ be a subset of a metric space $X$. If $f\colon X\to\R$ is $UA$ ($WUA$, respectively) then so is its restriction $f|_F\colon F\to\R$. \end{proposition} \begin{proof} Let $f\in C_{ua}$ and $M,K\sq F$, $K$ being compact. So, there exists $(K,M)$-approximation $g\colon X\to\R$ of $f$. It is easy to see that $g|_F$ is an $(K,M)$-approximation of $f|_F$. \end{proof} In \cite{BD} it was shown that every continuous function $f\colon\R\to\R$ is $UA$. In fact, one can show the following stronger property. \begin{theorem}\label{thm:R-R} If $X$ is a uniform space then every continuous function $f\colon\R\to X$ is $UA$. \end{theorem} \begin{proof} Let $K$ be a compact subset of $\R$ and $M\sq \R$. Choose an interval $[a,b]$ containing $K$ such that either $a\in M$ or $(-\infty,a]\cap M=\emptyset$ and that either $b\in M$ or $[b,\infty)\cap M=\emptyset$. Let $r\colon\R\to[a,b]$ be the retraction, i.e., $r(x)=x$ for $x\in[a,b]$, $r(x)=a$ for $xb$. Clearly $r$ is $UC$. Define $g$ by $g=f|_{[a,b]}\circ r$. Then $g$ is $UC$ as a composition of two $UC$ functions, and $g$ coincides with $f$ on $K\sq [a,b]$. It is also easy to see that $g[M]\sq f[M]$. \end{proof} In what follows we will often use the following criteria which is similar to that of \cite[Thm. 6.3]{BD}. \begin{proposition}\label{prop:crit} Let $(X,\rho)$ be a separable metric space and let $f\colon X\to \R$. If there exists an uncountable $Y\sq\R$ such that $f^{-1}(y)$ is non-empty and connected for every $y\in Y$ and that $\rho(f^{-1}(x),f^{-1}(y))=0$ for every $x,y\in Y$ then $f$ is not $WUA$. \end{proposition} \begin{proof} Replacing $X$ with $f^{-1}(Y)$, if necessary, we can assume that $X=f^{-1}(Y)$. Now, let $D$ be a countable dense subset of $X$ and put $M=f^{-1}(f[D])$. Then $M$ is dense in $X$, since $D\sq M$. Moreover, $M$ is a proper subset of $X$, since $Y=f[X]$ is uncountable unlike $f[M]=f[f^{-1}(f[D])]=f[D]$. Pick some $x\in X\setminus M$. Then, $x\not\in M=f^{-1}(f[D])=f^{-1}(f[f^{-1}(f[D])])=f^{-1}(f[M])$, i.e., $f(x)\not\in f[M]$. Our aim is to show that $f$ has no $(\{x\},M)$-approximation. By way of contradiction assume that there exists an $UC$ function $g\colon X\to \R$ such that $g[M]\sq f[M]$. Then $g[M]\sq f[M]=f[D]$ is countable, hence zero-dimensional. (See \cite[Chapter 7, \S2, Corollary]{E}.) Since $C_d=f^{-1}(f(d))\sq M$ is connected for every $d\in D$ and $g[C_d]\sq g[M]\sq f[M]=f[D]$, we conclude that $g$ is constant on each $C_d$. Since our hypothesis yields that $\rho(C_d, C_{d'})=0$ for every $d,d'\in D$, the uniform continuity of $g$ implies that all these constants coincide. Therefore $g|_M$ is constantly equal to some $b\in f[M]$. But $f(x)\not\in f[M]$. So, $g(x)=b\ne f(x)$. Therefore $g$ is not an $(\{x\},M)$-approximation of $f$. \end{proof} As an immediate corollary we obtain the following. (See also \cite[Lemma 5.4]{BD}.) \begin{corollary}\label{cor:crit} The function $h\colon\R^2\to\R$ given by formula $h(x,y)=x^2-y^2$ is not $WUA$. \end{corollary} \begin{proof} By Proposition \ref{prop:crit} function $h|_{[0,\infty)\times[0,\infty)}$ is not $WUA$. So, $h$ is not $WUA$ by Proposition \ref{prop:restr}. \end{proof} One of the most interesting properties of the class of $UA$ functions is the fact that every perfect map% \footnote{ A map $f\colon X\to\R^n$ is perfect if $f^{-1}(r)$ is compact for every $r\in\R^n$.} from $\R^n$ into $\R$ is $UA$ \cite[Thm 5.2]{BD}. However, this theorem does not generalize for all functions from $\R^2$ into $\R^2$, as shown by the next example. (Compare with Example \ref{Ex:zSq}.) \begin{em-example}\label{ex:perfect} The functions $f,g\colon\R^2\to\R$ given by the formulas $f(x,y)=x^2$ and $g(x,y)=y^2$ are $UA$, while the perfect map $H=(f,g)\colon\R^2\to\R^2$ is not $WUA$. \end{em-example} \begin{proof} The function $f$ is $UA$ by Theorem \ref{thm:composition} since it is a composite of a $UC$ function $f_1\colon\R^2\to\R$, $f_1(x,y)=x$, and a $UA$ function $f_2\colon\R\to\R$, $f_2(x)=x^2$. (See Theorem \ref{thm:R-R}.) Similarly we show that $g$ is $UA$. To see that $H$ is is not $WUA$ notice that the function $h\colon\R^2\to\R$ from Corollary \ref{cor:crit} is a composite of $H$ and a $UC$ function $h\colon\R^2\to\R$, $h(x,y)=x-y$. Thus, by Theorem \ref{thm:composition}, $H$ cannot be $WUA$. \end{proof} Example \ref{ex:perfect} shows also that a map $H=(f,g)\colon X\to Y\times Z$ does not have to be $WUA$ even if $f\colon X\to Y$ and $g\colon X\to Z$ are both $UA$. \medskip Now, if $X$ is a metric space then $C(X,\R)$ is a linear topological space and $C_{uc}(X,\R)$ is its linear subspace. However, the classes $C_{ua}(X,\R)$ and $C_{wua}(X,\R)$ are not closed under addition, as shown by Example \ref{ex:perfect} and Corollary \ref{cor:crit}. In fact, the next example shows that the sum of $WUA$ function and $UC$ function does not have to be $WUA$. To formulate it easier, we need the following notation and lemma. We will use the symbol $C$ to denote the unit circle on the complex plane: \[ C=\{z\in\C\colon |z|=1\} \] with the standard distance and we will write $\R_c$ to denote the real line with the metric of $C\setminus\{-1\}$, i.e., given by formula \[ \rho(x,y)=\left|e^{2 {\bf i} \arctan x}-e^{2 {\bf i} \arctan y}\right|. \] \begin{lemma}\label{lem:SinFunc} \begin{description} \item[(i)] Let $Y$ be a metric space. A function $f\in C(\R_c,Y)$ is $UC$ if and only if both limits $\lim_{x\to -\infty}f(x)$ and $\lim_{x\to\infty}f(x)$ exist and are equal. \item[(ii)] A function $f\in C(\R_c,\R)$ is $WUA$ provided\ ${\rm int}(f[(-\infty,x)]\cap f[(x,\infty)])\neq\emptyset$ for every $x\in\R_c$. \end{description} \end{lemma} \begin{proof} Since the spaces $\R_c$ and $C\setminus\{-1\}$ are isometric, we can assume that $f\colon C\setminus\{-1\}\to Y$. (i) Now, if both limits $\lim_{x\to -\infty}f(x)$ and $\lim_{x\to\infty}f(x)$ exist and are equal to $b\in\R$, then we can extend $f$ to a continuous function $F$ on $C$ by putting $F(-1)=b$. Clearly $F$ is $UC$, as a continuous function on the compact set $C$. So, $f$ is $UC$. Conversely, if $f\colon C\setminus\{-1\}\to Y$ is $UC$, then it can be extended uniquely to $UC$ function $F\colon C\to Y$ and then clearly both limits exist and are equal to $F(-1)$. (ii) Choose $x\in\R_c$ and $M\sq \R_c$. Clearly we can assume that $M\not\sq\{x\}$. If $M$ is not dense in $\R_c$ then we can find $a\arctan(4k-1)>-\frac{\pi}{2} \] and \[ f(4k+5)= %\arctan(4k+5) - \frac{12 \sin \frac{\pi}{2} (4k+5)}{(4k+5)^2+1}= \arctan(4k+5) - \frac{12}{(4k+5)^2+1}<\arctan(4k+5)<\frac{\pi}{2}. \] Thus, it is enough to show the inequality $f(4k-1)>f(4k+5)$, since then we have $(-\pi/2,f(4k-1))\sq f[(-\infty,x)]$, $(f(4k+5),\pi/2)\sq f[(x,\infty)]$, and $(-\pi/2,f(4k-1))\cap (f(4k+5),\pi/2)\neq\emptyset$. So, we can reduce our task to the proof of \begin{equation}\label{eq1:sin} \arctan(4k+5) - \arctan(4k-1)< \frac{12}{(4k+5)^2+1} + \frac{12}{(4k-1)^2+1}. \end{equation} Now, if $4k+5\leq 0$, then $\arctan$ is concave on the interval $[4k-1,4k+5]$ and using differential approximation at the point $4k+5$ we obtain \[ \arctan(4k+5) - \arctan(4k-1)\leq 6 \arctan^\prime(4k+5)= \frac{6}{(4k+5)^2+1}. \] So, (\ref{eq1:sin}) holds. The case when $4k-1\geq 0$ is handled by the fact that $f$ is an odd function. Now, if $4k-1<0<4k+5$ then $k=0$ or $k=-1$, i.e., $\{4k-1,4k+5\}\cap\{-1,1\}\neq\emptyset$, So, using the fact that the derivative of $\arctan$ is at most $1$ we obtain \[ \arctan(4k+5) - \arctan(4k-1)<6= \frac{12}{(\pm 1)^2+1}<\frac{12}{(4k+5)^2+1} + \frac{12}{(4k-1)^2+1}. \]\end{proof} \begin{problem}\label{problem:UcUa} If $X$ is a metric space, $f\in C_{ua}(X,\R)$ and $g\in C_{uc}(X,\R)$ is $f+g$ $UA$? \end{problem} \begin{problem}\label{problem:UcWua} If $f\in C_{wua}(\R^2,\R)$ and $g\in C_{uc}(\R^2,\R)$ is $f+g$ $WUA$? \end{problem} \section{Equations between $C_{uc}$, $C_{ua}$, $C_{wua}$ and $C$. }\label{sec:Eq} In this section we will discuss all the possible spectra of equations and sharp inclusions in the formula \begin{equation}\label{eq2} %C_{uc}(X,Y)\buildrel (a)\over \subseteq C_{ua}(X,Y)\buildrel (b)\over %\subseteq C_{wua}(X,Y)\buildrel (c)\over \subseteq C(X,Y). C_{uc}(X,Y)\subseteq C_{ua}(X,Y) \subseteq C_{wua}(X,Y)\subseteq C(X,Y). \end{equation} An important contribution for this discussion comes from the following fact. \begin{proposition}\label{prop:ats} {\rm \cite[Thm. 12.1]{BD}} For any metric space $X$ the equation $C_{uc}(X,\R)=C_{ua}(X,\R)$ implies that $C_{uc}(X,\R)=C(X,\R)$. \end{proposition} In particular, the space $Y$ in the following cases (ii)-(iv) cannot be equal to $\R$. On the other hand, we have $Y=\R$ in all remaining cases. \begin{description} \item[(i)] $C_{uc}(X,Y)=C_{ua}(X,Y)=C_{wua}(X,Y)=C(X,Y)$. Any compact space $X$ will do the job. For example $X=[0,1]$, $Y=\R$. \end{description} \begin{description} \item[(ii)] $C_{uc}(X,Y)=C_{ua}(X,Y)=C_{wua}(X,Y)\neq C(X,Y)$. Let $M$ be a magic set for $\R$ from Proposition \ref{prop:magic}. Then $X=Y=M$ considered with the usual metric have the above property. (See \cite[Example 13]{B}.) Spaces $X$ and $Y$ can be also chosen to be complete metric spaces \cite[Cor. 18]{B}. (Notice that in \cite{B} term $UA$ is used for what we call here $WUA$.) \end{description} \begin{description} \item[(iii)] $C_{uc}(X,Y)=C_{ua}(X,Y)\neq C_{wua}(X,Y)=C(X,Y)$. \end{description} We do not know such an example. So, the following problem remains open. \begin{problem}\label{problem:Equations} Do there exist metric spaces $X$ and $Y$ with the property that $C_{uc}(X,Y)=C_{ua}(X,Y)\neq C_{wua}(X,Y)=C(X,Y)$? \end{problem} \begin{description} \item[(iv)] $C_{uc}(X,Y)=C_{ua}(X,Y)\neq C_{wua}(X,Y)\neq C(X,Y)$. %Notice first, that as in (2) $Y$ cannot be equal to $\R$. Now, let $f %h=f_{c_0} \colon\R_c\to\R$ be as in Example \ref{ex:SinFunc} and let $M\sq\R_c$ be a magic set. We claim, that the above chain of equations and inequalities holds for $X=M$ and $Y=f[M]$. First notice that both $M$ and the complement of $M$ are dense in $\R_c$. Otherwise, if $(a,b)$ is a non-empty interval contained in either $M$ or $\R_c\setminus M$ and if $g\in C(\R_c,\R)$ is such that $g|_{\R_c\setminus(a,b)}=\arctan |_{\R_c\setminus(a,b)}$, $g[(a,b)]=\arctan [(a,b)]$, and $g\ne \arctan$, then $g$ is an $(\{a\},M)$-approximation of $\arctan$ which is not a truncation of $\arctan$. Next we will show that \begin{equation}\label{eq4:u_ext} \mbox{ any }h\in C_{wua}(M,f[M]) \mbox{ can be uniquely extended to }C(\R_c,\R). \end{equation} This immediately implies $C_{wua}(M,f[M])\ne C(M,f[M])$ since there exists a function $h\in C(M,f[M])$ with $h[M]$ having precisely two elements and, by (\ref{eq4:u_ext}), such $h$ is not $WUA$. To see (\ref{eq4:u_ext}) let $h\colon M\to \R$ be a function without continuous extension into $\R_c$. Then there exist $x\in \R_c\setminus M$, $\varepsilon>0$ and two sequences $\{x_n\}_{n\in\N}$ and $\{y_n\}_{n\in\N}$ in $M$ converging to $x$ such that $|h(x_n)-h(y_n)|\geq \varepsilon$ for all $n\in \N$. Let $a0$ such that $|f(x)-f(y)|<\varepsilon$ for every $x,y\in [a,b]$ with $|x-y|<\delta$. It is easy to see that this implies essentially the same property for any truncation $g$ of $f$, i.e., that \begin{equation}\label{eq5:**} |g(x)-g(y)|<\varepsilon \end{equation} for every truncation $g$ of $f$ and any $x,y\in [a,b]$ with $|x-y|<\delta$. Now, choose $k\in \N$ such that $|x_k-y_k|<\delta$ and put $ M_0=\{y_k\}$. We claim that if $g\colon M\to h[M]$ is an $(\{x_k\},M_0)$-approximation of $h$ then $g$ is not $UC$. Indeed, if $g$ is an $(\{x_k\},M_0)$-approximation of $h$ then $g(x_k)=h(x_k)$ and $\{g(y_k)\}=g[M]\sq h[M]=\{h(y_k)\}$. So \begin{equation}\label{eq6:***} |g(x_k)-g(y_k)|=|h(x_k)-h(y_k)|\geq\varepsilon. \end{equation} On the other hand, if $g$ were $UC$, then it could be extended to a $UC$ function $\hat g:\R_c\to \R$ which must be a truncation of $f$, since $\hat g[M]=g[M]\subseteq f[M]$. But this and (\ref{eq6:***}) contradict (\ref{eq5:**}). This finishes the proof of (\ref{eq4:u_ext}). Next we will show that for any $h\in C_{wua}(M,f[M])$ \begin{eqnarray} h\in C_{ua}(M,f[M]) & \mbox{ if and only if } & \hat{h} \mbox{ is a truncation of $f$ with} \\ \label{eq:o} & & \lim_{x\to\infty}\hat h(x)=\lim_{x\to-\infty}\hat h(x), \nonumber \end{eqnarray} %\begin{equation}\label{eq:o} %h\in C_{ua}(M,f[M]) \mbox{ if and only if } \hat{h} %\mbox{ is a truncation of $f$ with } %\lim_{x\to\infty}\hat h(x)=\lim_{x\to-\infty}\hat h(x), %\end{equation} where $\hat h\colon \R_c\to \R$ is the unique extension of $h$, existing by (\ref{eq4:u_ext}). Notice that by Lemma \ref{lem:SinFunc}(i) and (\ref{eq4:u_ext}) this will immediately imply $C_{uc}(M,f[M])=C_{ua}(M,f[M])$. The implication ``$\Leftarrow$" follows immediately from Lemma \ref{lem:SinFunc}(ii). To prove the other implication take $h\in C_{ua}(M,f[M])$. Then $\hat{h}$ is a truncation of $f$ by the definition of $M$. It is also easy to see that the limits under question exist. %\neq C(X,Y)$ choose $y_0,y_1\in h[M]$ such that $y_04\varepsilon. \end{equation} Moreover, if $b_0=\lim_{x\to-\infty}\hat g(x)$ then either $b_0=h(a_0)$, if $h(a_0)\not \in (-\pi/2-\varepsilon,-\pi/2+\varepsilon)$, or otherwise $b_0\in (-\pi/2-\varepsilon,-\pi/2+\varepsilon)$. In any case, $|b_0-h(a_0)|\leq 2\varepsilon$. Similarly, we argue that $|b_1-h(a_1)|\leq 2\varepsilon$, where $b_1=\lim_{x\to\infty}\hat g(x)$. Combining this with (\ref{eq:oo}) we obtain that $b_0\ne b_1$. So, by Lemma \ref{lem:SinFunc}(i), $\hat g$ is not $UC$. This contradiction finishes the proof of (\ref{eq:o}). To finish the proof it is enough to show that \begin{equation}\label{eq:ooo} f|_M\in C_{wua}(M,f[M])\setminus C_{ua}(M,f[M]). \end{equation} But in Example 2.9 we proved that $f$ is $WUA$, so Proposition \ref{prop:restr} implies that $f|_M\in C_{wua}(M,f[M])$. The fact that $f|_M\not \in C_{ua}(M,f[M])$ follows immediately from (\ref{eq:o}). \end{description} \begin{description} \item[(v)] $C_{uc}(X,Y)\neq C_{ua}(X,Y)=C_{wua}(X,Y)=C(X,Y)$ holds for $X=Y=\R$ by Theorem \ref{thm:R-R}. \end{description} \begin{description} \item[(vi)] $C_{uc}(X,Y)\neq C_{ua}(X,Y)=C_{wua}(X,Y)\neq C(X,Y)$ holds for $X$ being the Hedgehog with continuum many spikes \cite[Example 4.1.3]{E} and $Y=\R$. For the proof see \cite[Theorem 5.9]{BDP}. (Actually it suffices to take just $b$ spikes, where $b$ is the smallest cardinality of an unbounded family in $\omega^\omega$.) \end{description} \begin{problem}\label{problem:Separable} Does there exist a separable metric space $X$ with the property that $C_{uc}(X,\R)\neq C_{ua}(X,\R)=C_{wua}(X,\R)\neq C(X,\R)$? \end{problem} \begin{description} \item[(vii)] $C_{uc}(X,Y)\neq C_{ua}(X,Y)\neq C_{wua}(X,Y) = C(X,Y)$ holds for $X=\R\setminus\{0\}$ and $Y=\R$. (See \cite[Corollary 7.2 and Lemma 3.1]{BD}.) \end{description} \begin{problem}\label{problem:Connected} Does there exist a connected metric space $X$ with the property that $C_{uc}(X,\R)\neq C_{ua}(X,\R)\neq C_{wua}(X,\R)=C(X,\R)$? \end{problem} \begin{description} \item[(viii)] $C_{uc}(X,Y)\neq C_{ua}(X,Y)\neq C_{wua}(X,Y) \neq C(X,Y)$. Holds for $Y=\R$ and $X=\R_c$. The relation $C_{wua}(\R_c,\R) \neq C(\R_c,\R)$ is witnessed by the identity function, which is not $WUA$ by Corollary \ref{cor:magic} and Lemma \ref{lem:SinFunc}(i). This and Proposition \ref{prop:ats} imply $C_{uc}(\R_c,\R)\neq C_{ua}(\R_c,\R)$. The relation $C_{ua}(\R_c,\R)\neq C_{wua}(\R_c,\R)$ is justified by the function $f$ from Example \ref{ex:SinFunc}. It is shown there that $f\in C_{wua}(\R_c,\R)$. The fact that $f\not\in C_{ua}(\R_c,\R)$ follows from (\ref{eq:ooo}) and Proposition \ref{prop:restr}. \end{description} \section{Complex polynomials and harmonic functions}\label{harmonic} In this section we will show that most complex analytic functions $f\colon\C\to\C$ are not $WUA$. We start with the following easy example. \begin{em-example}\label{Ex:zSq} The function $f\colon\C\to\C$ given by $f(z)=z^2$ is not $WUA$. \end{em-example} \begin{proof} If $f$ were $WUA$ than the real part of $f$, $\re f=\pr_x\circ f$, would be $WUA$, since $\pr_x\colon\R^2\to\R$, $\pr_x(x,y)=x$, is $UC$. But $\re f(x+{\bf i}y)=x^2-y^2$ is not $WUA$. (See Corollary \ref{cor:crit}.) \end{proof} Notice that $f(z)=z^2$ is perfect. So, even the simplest analytic perfect mappings do not have to be $WUA$. Clearly all linear functions from $\R^2$ into $\R^2$ are $UC$. Hence all mappings $\C\to \C$, $z\mapsto a z + b$ $(a, b \in \C)$ are $UC$. In fact, it seems that these are the only complex analytic functions that are $WUA$. We do not have a proof of this general statement. However, the examples below should explain the reasons that stands behind this conjecture. In all the examples that follow we will prove that the complex-valued function $f(z)$ in question is not $WUA$ by showing that its imaginary part $h(x,y)=\im f(x+\iu y)$ is not $WUA$. This is enough by Theorem \ref{thm:composition}. To prove that $h$ is not $WUA$ we will apply Proposition \ref{prop:restr} to an appropriate restriction $h_A$ of the function $h$ to a subspace $A$ of $\R^2$. The verification that $h_A$ is not $WUA$ will be done by applying Proposition \ref{prop:crit}. \begin{example}\label{two_functions} \begin{description} \item[(1)] The for $n>1$ the $n$-th power function $\C\to \C$, $z\mapsto z^n$, is not $WUA$. \item[(2)] The exponential function $\C\to \C$, $z\mapsto e^z$, is not $WUA$. \end{description} \end{example} \begin{proof} In case (1) we have $h(x,y)=(x^2+y^2)^{n/2}\sin [n\arctan(y/x)]$. Take as $A$ those points of the upper half-plain which have the argument in the interval $(0, \pi/n)$. Then for every $c>0$ the level curve $L_c= h^{-1}(c)$ is connected and has two asymptotes: the rays $\mbox{Arg}(z)=0$ and $\mbox{Arg}(z)=\pi/n$. %(See Figure 2.) Thus the distance between $L_c$ and $L_{c'}$ is zero for $c, c'>0$. The application of Proposition \ref{prop:crit} finishes the proof. In case (2) we have $h(x,y)=e^x\sin y$. Take $A=\{z\in \C: \re z> 0\ \&\ 0< \im z< \pi/2\}$. Then for $c>1$ the level curve $L_c= h^{-1}(c)=\{(\ln[c/\sin y],y)\colon 01}$ have as common asymptote the line $\im z= 0$. Thus the distance between $L_c$ and $L_{c'}$ is zero for $c, c'>1$. Now Proposition \ref{prop:crit} can be applied. \end{proof} In both cases above we used the fact that the function $h(x,y)=\im f(x+\iu y)$ has many level curves whose connected components are of distance zero. (This was done by noticing that they have the same asymptote.) The same argument works also for other elementary entire functions like $\sin z$, $\cos z$ and complex polynomial functions of degree greater than $1$. In fact, by Liouville's theorem for positive harmonic functions \cite[Theorem 3.1]{ABR}, %the Great Theorem of Picard easily implies that for every entire analytic function $f$ the level curves of $h=\im f$ are unbounded and it seems conceivable that for any non-linear $f$ the function $h$ always has many level curves whose connected components are of distance zero. This leads us to the following conjecture. \begin{conjecture}\label{conj:analituc} If $f\colon\C\to\C$ is entire analytic function then $f$ is $WUA$ if and only if $f$ is a linear. \end{conjecture} We will finish this discussion with one more example. \begin{example}\label{third_function} The inverse function $f\colon\C\setminus\{0\}\to\C$, $f(z)= 1/z$, is not $WUA$. \end{example} \begin{proof} Consider $A$ to be the first quadrant $\{(x,y)\colon x>0\ \&\ y>0\}$. Since $h(x,y)=\im f(x+\iu y)=-y/\sqrt{x^2+y^2}$ it is easy to see that for $-10$. So, Proposition \ref{prop:crit} can be applied. \end{proof} It is not difficult to extend the argument of Example \ref{third_function} to all rational functions with essential poles. This suggests that Conjecture \ref{conj:analituc} can be true also for all meromorphic functions. \medskip We also do not know the answer for the following natural question. (Compare with Example \ref{ex:SinFunc}.) \begin{question} If $f\in C_{ua}(\C,\C)$ ($f\in C_{wua}(\C,\C)$, respectively) and $g:\C\to \C$ is linear, is it always true that $f+g$ is $UA$ ($WUA$, respectively)? \end{question} \begin{thebibliography}{Barendregt99} \bibitem[ABR]{ABR} S. Axler, P. Bourdon and W. Ramey, {\it Harmonic Function Theory}, Springer-Verlag, Berlin-New York 1992. \bibitem[BD]{BD} A. Berarducci and D. Dikranjan, {\it Uniformly approachable functions and $UA$ spaces}, Rend. Ist. Matematico Univ. di Trieste {\bf 25} (1993), 23--56. \bibitem[BDP]{BDP} A. Berarducci, D. Dikranjan and Jan Pelant, {\it Totally continuous functions}, Preprint. \bibitem[B]{B} M. Burke, {\it Characterizing uniform continuity with closure operators}, Topology Appl. {\bf 59} (1994), 245--259. \bibitem[DP]{DP} D. Dikranjan and J. Pelant, {\it The impact of closure operators on the structure of a concrete category}, Quaestiones Mathematicae {\bf 18} (1995), 381--396. \bibitem[E]{E} R. Engelking, {\it General Topology, Revised and Completed Edition}, Sigma Series in Pure Mathematics, vol. {\bf 6}, Heldermann Verlag, Berlin 1989. \end{thebibliography} \end{sloppypar} \end{document}