\documentclass{rae} \usepackage{amssymb} \usepackage{amstex} \usepackage{epic} %\input{StandardMacros} %\bysection \received{October 26, 1994} \MathReviews{Primary: 26A15; Secondary: 03E75, 54A25.} \keywords{cardinal invariants, extendable functions, functions with prefect road, peripherially continuous functions. } \firstpagenumber{459} \markboth{K. Ciesielski and I. Rec\l aw}% {Sum of Extendable and Peripherally Continuous Functions} \author{Krzysztof Ciesielski, Department of Mathematics, West Virginia University, Morgantown, WV 26506-6310, email: kcies@@wvnvms.wvnet.edu \and Ireneusz Rec\l aw\footnotemark,\addtocounter{footnote}{-1} Department of Mathematics, University of Scranton, Scranton, PA 18510-4666, email: ReclawI1@@jaguar.uofs.edu\\ Permanent address: Institute of Mathematics, Gda\'{n}sk University, Wita Stwosza~57, 80-952 Gda\'{n}sk, Poland, email: matir@@halina.univ.gda.pl} \title{CARDINAL INVARIANTS CONCERNING EXTENDABLE AND PERIPHERALLY CONTINUOUS FUNCTIONS} %\title{Cardinal invariants concerning extendable %and peripherally continuous functions.} % IF YOUR VERSION OF LATEX DOES NOT RECOGNIZE COMMANDS \frak, \mathbb, OR % \operatorname UNCOMMENT NEXT 3 LINES %\def\frak{\bf} %\def\operatorname{\rm} %\def\mathbb{\Bbb} %\pagestyle{myheadings} \renewcommand{\thefootnote}{\fnsymbol{footnote}} \def\real{{\mathbb R}} \def\rational{{\mathbb Q}} \def\integer{{\mathbb Z}} \def\continuum{{\frak c}} \def\co{\continuum} \def\add{{\operatorname{A}}} \def\mul{{\operatorname M}} \def\ext{{\operatorname{Ext}}} \def\pr{{\operatorname{PR}}} \def\pc{{\operatorname{PC}}} \def\ac{{\operatorname{AC}}} \def\conn{{\operatorname{Conn}}} \def\darb{{\operatorname D}} \def\cl{{\operatorname{cl}}} \def\bd{{\operatorname{bd}}} \def\cf{{\operatorname{cf}}} \def\diam{{\operatorname{diam}}} \def\rl{\real} \def\reals{\real} \def\ds{\bigcup} \def\di{\bigcap} \def\sb{\subseteq} \def\su{\subseteq} \newcommand{\propsub}{\subsetneqq} \def\sm{\setminus} \def\es{\emptyset} \def\C{{\cal C}} \def\F{{\cal F}} \def\G{{\cal G}} \def\H{{\cal H}} \def\J{{\cal J}} \def\T{{\cal T}} \def\e{\varepsilon} \def\la{\langle} \def\ra{\rangle} % characteristic function \newcommand{\charf}[1]{\mbox{\raise.48ex\hbox{$\chi$}$_{#1}$}} %% Theorems, etc. \newtheorem{theorem}{Theorem}[section] \newtheorem{corollary}[theorem]{Corollary} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{problem}{Problem}[section] \newtheorem{example}{Example}[section] \newtheorem{definition}{Definition} \newtheorem{remark}{Fact}[section] \newcommand{\thm}[2]{\begin{theorem}\label{#1}#2\end{theorem}} \newcommand{\cor}[2]{\begin{corollary}\label{#1}#2\end{corollary}} \newcommand{\prop}[2]{\begin{proposition}\label{#1}#2\end{proposition}} \newcommand{\lem}[2]{\begin{lemma}\label{#1}#2\end{lemma}} \newcommand{\prob}[2]{\begin{problem}\label{#1}#2\end{problem}} \newcommand{\ex}[2]{\begin{example}\label{#1}#2\end{example}} \newcommand{\defi}[2]{\begin{definition}\label{#1}#2\end{definition}} \newcommand{\rem}[2]{\begin{remark}\label{#1}#2\end{remark}} %End of proof marker \def\qed{\hfill$\Box$} %Proof \def\pf{\noindent{\sc Proof.} } \begin{document} \maketitle \addtocounter{footnote}{1} \footnotetext{The article was written when the second author was working at Institut fur Mathematik~II, FU~Berlin, Arnimallee 3, 14195~Berlin, Germany, under the support of Alexander von Humboldt Foundation.} \begin{abstract} Let $\F$ be a family of real functions, $\F\su\real^\real$. In the paper we will examine the following question: for which families $F\su\real^\real$ does there exist $g\colon\real\to\real$ such that $f+g\in\F$ for all $f\in F$? More precisely, we will study a cardinal function $\add(\F)$ defined as the smallest cardinality of a family $F\su\real^\real$ for which there is no such $g$. We will prove that $\add(\ext)=\add(\pr)=\co^+$ and $\add(\pc)=2^{\co}$, where $\ext$, $\pr$ and $\pc$ stand for the classes of extendable functions, functions with perfect road and peripherally continuous functions from $\real$ into $\real$, respectively. In particular, equation $\add(\ext)=\co^+$ implies immediately that every real function is a sum of two extendable functions. This solves a problem of Gibson~\cite{Gib1}. We will also study the multiplicative analogue $\mul(\F)$ of the function $\add(\F)$ and we prove that $\mul(\ext)=\mul(\pr)=2$ and $\add(\pc)=\co$. This article is a continuation of papers \cite{N2,CM,NR} in which functions $\add(\F)$ and $\mul(\F)$ has been studied for the classes of almost continuous, connectivity and Darboux functions. \end{abstract} \section{ Introduction.} We will use the following terminology and notation. Functions will be identified with their graphs. The family of all functions from a set $X$ into $Y$ will be denoted by $Y^X$. Symbol $|X|$ will stand for the cardinality of a set $X$. The cardinality of the set $\real$ of real numbers is denoted by $\co$. For a cardinal number $\kappa$ we will write $\cf(\kappa)$ for the cofinality of $\kappa$. A cardinal number $\kappa$ is regular, if $\kappa=\cf(\kappa)$. For $A\su\real$ its characteristic function is denoted by $\charf{A}$. In particular, $\charf\emptyset$ stands for the zero constant function. In his study of the class $\darb$ of Darboux functions (see definition below) Fast \cite{fast} proved that for every family $F\su\reals^\reals$ of cardinality at most continuum there exists $g\colon\reals\to\reals$ such that $f+g$ is Darboux for every $f\in F$. Natkaniec \cite{N2} proved the similar result for the class $\ac$ of almost continuous functions and defined the following two cardinal invariants for every $\F\su\reals^\reals$. \begin{eqnarray*} \add(\F) & = & \min\{|F|\colon F\su\reals^\reals\ \&\ \neg\exists g\in\reals^\reals\ \forall f\in F\ f+g\in\F\}\cup\{(2^\co)^+\}\\ & = & \min\{|F|\colon F\su\reals^\reals\ \&\ \forall g\in\reals^\reals\ \exists f\in F\ f+g\not\in\F\}\cup\{(2^\co)^+\} \end{eqnarray*} and \begin{eqnarray*} \mul(\F)\!\! & = &\! \min\{|F|\colon F\su\reals^\reals\ \&\ \neg\exists g\in\reals^\reals\setminus\{\charf\emptyset\}\ \forall f\in F\ f\cdot g\in\F\}\cup\{(2^\co)^+\}\\ & = &\! \min\{|F|\colon F\su\reals^\reals\ \&\ \forall g\in\reals^\reals\setminus\{\charf\emptyset\}\ \exists f\in F\ f\cdot g\not\in\F\}\cup\{(2^\co)^+\}. \end{eqnarray*} Thus, Fast and Natkaniec effectively showed that $\add(\darb)>\co$ and $\add(\ac)>\co$. The extra assumption that $g\neq\charf\emptyset$ is added in the definition of $\mul$ since otherwise for every family $\F\su\real^\real$ containing constant zero function $\charf\emptyset$ we would have $\mul(\F)=(2^\co)^+$. Notice the following basic properties of functions $\add$ and $\mul$. \prop{prop1}{ Let $\F\su\G\su\reals^\reals$. \begin{description} \item[(1)] $\add(\F)\le\add(\G)$. \item[(2)] $\add(\F)\geq 2$ if $\F\neq\emptyset$. \item[(3)] $\add(\F)\leq 2^\co$ if $\F\neq\real^\real$. \end{description} } \pf (1) is obvious. To see (2) let $h\in\F$ and $F=\{f\}$ for some $f\in\real^\real$. Then $f+g\in\F$ for $g=h-f$. To see (3) notice that for $F=\real^\real$ and every $g\in\real^\real$ there is $f\in F$ with $f+g\not\in\F$, namely $f=h-g$, where $h\in\real^\real\setminus\F$. \qed %\pagebreak \prop{prop2}{ Let $\F\su\G\su\reals^\reals$. \begin{description} \item[(1)] $\mul(\F)\le\mul(\G)$. \item[(2)] $\mul(\F)\geq 2$ if $\charf\emptyset,\charf\real\in\F$. \item[(3)] $\mul(\F)\leq\co$ if $r\charf{\{x\}}\not\in\F$ for every $r,x\in\real$, $r\neq 0$. \end{description} } \pf (1) is obvious. To see (2) let $F=\{f\}$ for some $f\in\real^\real$. If there is $x\in\real$ such that $f(x)=0$ take $g=\charf{\{x\}}$. Otherwise, define $g(x)=1/f(x)$ for every $x\in\real$. Then $f\cdot g\in\{\charf\emptyset,\charf\real\}\su\F$. To see (3) notice that for $F=\{\charf{\{x\}}\colon x\in\real\}$ and every $g\in\real^\real\setminus\{\charf\emptyset\}$ there is $f\in F$ with $f\cdot g\not\in\F$, namely $f=\charf{\{x\}}$, where $x$ is such that $g(x)=r\neq 0$. \qed \prop{prop3}{ Let $\charf\emptyset\in\F\su\reals^\reals$. Then $\add(\F)=2$ if and only if $\F-\F=\{f_1-f_2\colon f_1,f_2\in\F\}\neq\real^\real$. } \pf ``$\Rightarrow$'' Assume that $\F-\F=\real^\real$. We will show that $\add(\F)>2$. So, pick arbitrary $f_1,f_2\in\real^\real$ and put $F=\{f_1,f_2\}$. It is enough to find $g\in\real^\real$ such that $f_1+g,f_2+g\in\F$. But $f_1-f_2\in\real^\real=\F-\F$. So, there exist $h_1,h_2\in\F$ such that $f_1-f_2=h_1-h_2$. Put $g=h_1-f_1=h_2-f_2$. Then $f_i+g=f_i+(h_i-f_i)=h_i\in\F$ for $i=1,2$. ``$\Leftarrow$'' By Proposition~\ref{prop1}(2) we have $\add(\F)\geq 2$. To see that $\add(\F)\leq 2$ let $h\in \real^\real\setminus (\F-\F)$, take $F=\{\charf\emptyset,h\}$ and choose an arbitrary $g\in\real^\real$. It is enough to show that $f+g\not\in\F$ for some $f\in F$. But if $g=\charf\emptyset+g\in\F$ and $h+g\in\F$ then $h\in\F-g\subset\F-\F$, contradicting the choice of $h$. \qed \medskip Now, let $X$ and $Y$ be topological spaces. In what follows we will consider the following classes of functions from $X$ into $Y$. (In fact, we will consider these classes mainly for $X=Y=\real$.) \begin{description} \item{$\darb(X,Y)$} of {\em Darboux functions} $f\colon X\to Y$, i.e., such that $f[C]$ is connected in $Y$ for every connected subset $C$ of $X$. \item{$\conn(X,Y)$} of {\em connectivity functions} $f\colon X\to Y$, i.e., such that the graph of $f$ restricted to $C$ (that is $f\cap[C\times Y]$) is connected in $X\times Y$ for every connected subset $C$ of $X$. \item{$\ac(X,Y)$} of {\em almost continuous functions} $f\colon X\to Y$, i.e., such that every open subset $U$ of $X\times Y$ containing the graph of $f$, there is a continuous function $g\colon X\to Y$ with $g\subset U$. \item{$\ext(X,Y)$} of {\em extendable functions} $f\colon X\to Y$, i.e., such that there exists a connectivity function $g\colon X\times[0,1]\to Y$ with $f(x)=g(x,0)$ for every $x\in X$. \item{$\pr$} of {\em functions $f\colon\real\to\real$ with perfect road} ($X=Y=\real$), i.e. such that for every $x\in\real$ there exists a perfect set $P\su\real$ having $x$ as a bilateral limit point for which restriction $f|_P$ of $f$ to $P$ is continuous at $x$. \item{$\pc(X,Y)$} of {\em peripherally continuous functions} $f\colon X\to Y$, i.e., such that for every $x\in X$ and any pair $U\su X$ and $V\in Y$ of open neighbourhoods of $x$ and $f(x)$, respectively, there exists an open neighbourhood $W$ of $x$ with $\cl(W)\su U$ and $f[\bd(W)]\su V$, where $\cl(W)$ and $\bd(W)$ stand for the closure and the boundary of $W$, respectively. \end{description} We will write %simply $\darb$, $\conn$, $\ac$, $\ext$, and $\pc$ in place of $\darb(X,Y)$, $\conn(X,Y)$, $\ac(X,Y)$, $\ext(X,Y)$, and $\pc(X,Y)$ if $X=Y=\real$. Notice also, that function $f\colon\rl\to\rl$ is peripherally continuous ($f\in\pc$) if and only if for every $x\in\rl$ there are sequences $a_{n}\nearrow x$ and $b_{n}\searrow x$ such that $\lim_{n\to\infty} f(a_{n})=\lim_{n\to\infty} f(b_{n})=f(x)$. In particular, if graph of $f$ is dense in $\real^2$ then $f$ is peripherally continuous. For the classes of functions (from $\real$ into $\real$) defined above we have the following proper inclusions $\subset$, marked by arrows $\longrightarrow$. (See \cite{bhl}.) \begin{picture}(300,100) \thicklines \put(10,40){\makebox(0,0){$\ext$}} \put(80,80){\makebox(0,0){$\ac$}} \put(160,15){\makebox(0,0){$\pr$} } \put(160,80){\makebox(0,0){$\conn$}} \put(240,80){\makebox(0,0){$\darb$} } \put(320,40){\makebox(0,0){$\pc$}} \put(30,50){\vector(1,1){30}} \put(100,80){\vector(1,0){30}} \put(190,80){\vector(1,0){30}} \put(260,80){\vector(3,-2){40}} \put(30,35){\vector(4,-1){100}} \put(190,10){\vector(4,1){100}} \end{picture} In what follows we will also use the following theorem due to Hagan \cite{Hagan}. \thm{thHagan}{ If $n\geq 2$ then $\conn(\real^n,\real)=\pc(\real^n,\real)$. \qed} Functions $\add$ and $\mul$ for classes $\ac$, $\conn$ and $\darb$ were studied in \cite{N2,CM,NR}. In particular, the following is known. \thm{thNR}{ {\rm \cite{NR}} $\mul(\ac)=\mul(\conn)=\mul(\darb)=\cf(\co)$. \qed} \thm{thCM}{ {\rm \cite{CM}} $\co^+\leq\add(\ac)=\add(\conn)=\add(\darb)\leq 2^\co$, $\cf(\add(\darb))>\continuum$ and it is pretty much all that can be shown in ZFC. More precisely, it is consistent with ZFC that $\add(\darb)$ can be equal to any regular cardinal between $\continuum^+$ and $2^{\continuum}$ and that it can be equal to $2^{\continuum}$ independently of the cofinality of $2^{\continuum}$. \qed} The goal of this paper is to prove the following theorem. \pagebreak \thm{main}{ \begin{description} \item[(1)] $\mul(\pc)=\co$. \item[(2)] $\mul(\ext)=\mul(\pr)=2$. \item[(3)] $\add(\pc)=2^\co$. \item[(4)] $\add(\ext)=\add(\pr)=\co^+$. \end{description} } This will be proved in the next sections. Notice only that the equation $\add(\ext)=\co^+$ and Proposition~\ref{prop3} imply immediately the following corollary, which gives a positive answer to a question of Gibson \cite{Gib1}. (Compare also \cite{Rosen1} and \cite{Rosen2}.) \cor{corGib}{ Every function $f\colon\real\to\real$ is a sum of two extendable functions. \qed} \section{Proof of Theorem~\ref{main}(1), (2) and (3).} %$\mul(\ext)=\mul(\pr)=2$, $\mul(\pc)=\co$, $\add(\pc)=2^{\frak c}$.} {\noindent{\sc Proof of $\mul(\ext)=\mul(\pr)=2$.}} The inequalities $2\leq\mul(\ext)\leq\mul(\pr)$ follow from Proposition~\ref{prop2}. To see that $\mul(\pr)\leq 2$ take $F=\{\charf{B},\charf{\real\setminus B}\}$ where $B\subset\real$ is a Bernstein set. Then for every $g\in\real^\real\setminus\{\charf\emptyset\}$ we have $f\cdot g\neq\pr$ for some $f\in F$. To see it, take $x\in\real$ such that $g(x)=r\neq 0$. If $x\in B$ then $\charf{B}\cdot g$ does not have a perfect road at $x$, since $(\charf{B}\cdot g)(x)=r\neq 0$ and $(\charf{B}\cdot g)^{-1}(0)\cap P\neq\emptyset$ for every perfect set $P\su\real$. Similarly, $\charf{\real\setminus B}\cdot g$ does not have a perfect road at $x$ if $x\in \real\setminus B$. \qed \medskip {\noindent{\sc Proof of $\mul(\pc)=\co$.}} Inequality $\mul(\pc)\leq\co$ follows from Proposition~\ref{prop2}. So, it is enough to show that $\mul(\pc)\geq\co$. Let $F\su\real^\real$ be a family of cardinality less then or equal to $\kappa$ with $\omega\leq\kappa<\co$. We will find $g\in\real^\real\setminus{\charf\emptyset}$ such that $f\cdot g\in\pc$ for every $f\in F$. For $f\colon\real\to\real$ let $[f\ne 0]$ stand for the set $\{x\in\real\colon f(x)\neq 0\}$ and define \[ A_f=\{x\in\real\colon f(x)\ne 0\ \&\ [f\ne 0]\ \text{ is not bilaterally $\kappa^+$-dense at } x \}, \] where set $S\su\real$ is said to be bilaterally $\kappa^+$-dense at $x$ if for every $\e>0$ each of the sets $S\cap[x-\e,x]$ and $S\cap[x,x+\e]$ has cardinality at least $\kappa^+$. Notice that $|A_f|\leq\kappa$ for every $f\colon\real\to\real$. This is the case, since for every $x\in A_f$ there exists a closed interval $J$ with non-empty interior such that $x\in J$ and $|[f\ne 0]\cap J|\leq\kappa$. Now, if $\J$ is the family of all maximal intervals $J$ with non-empty interior such that $|[f\ne 0]\cap J|\leq\kappa$ then $|\J|\leq\omega$, $A_f\su\bigcup_{J\in\J}([f\ne 0]\cap J)$ and $|A_f|\leq|\bigcup_{J\in\J}([f\ne 0]\cap J)|\leq\kappa$. Let $A=\ds_{f\in F}A_f$. Then $|A|\leq\kappa$. Notice that the set $[f\ne 0]\sm A$ is bilaterally $\kappa^+$-dense at $x$ for every $f\in F$ and $x$ from $[f\ne 0]\sm A$. To define $g$ let $\la\la f_\alpha,q_\alpha,I_\alpha\ra\colon \alpha<\kappa\ra$ be the sequence of all triples with $f_\alpha\in F$, $q_\alpha\in\rational$ and $I_\alpha$ is an open interval with rational end points. Define by induction on $\alpha<\kappa$ a one-to-one sequence $\la x_\alpha\colon\alpha<\kappa\ra$ by choosing \begin{description} \item[(i)] $x_\alpha\in[f_{\alpha}\ne 0]\cap(I_\alpha\sm A)\sm\{x_\beta\colon\beta<\alpha\}$ if the choice can be made, and \item[(ii)] $x_\alpha\in(I_\alpha\sm A)\sm\{x_\beta\colon\beta<\alpha\}$ otherwise. \end{description} Now, for $x\in\real$ we put \[ g(x)= \begin{cases} \frac{q_\alpha}{f_\alpha(x)} & \text{ if there is $\alpha<\kappa$ with $x=x_\alpha$ and $f_\alpha(x)\neq 0$}\\ 1 & \text{ if there is $\alpha<\kappa$ with $x=x_\alpha$ and $f_\alpha(x)=0$}\\ 0 & \text{ otherwise.} \end{cases} \] Then $g\neq\charf\emptyset$ and for each $q\in\rational$ and $f\in F$ the set $(g\cdot f)^{-1}(q)$ is bilaterally dense at every $x$ from $[f\ne 0]\sm A$. Moreover, $(g\cdot f)(x)=0$ outside of $[f\ne 0]\sm A$ and $(g\cdot f)^{-1}(0)$ is bilaterally dense at every $x\in\rl$. So, $g\cdot f\in\pc$ for every $f\in F$. \qed \medskip To prove $\add(\pc)=2^\co$ we will use the following result. \thm{thComfNeg}{ Let $A$ and $B$ be such that $|A|=\omega$ and $|B|=\continuum$. Then there exists a family $\C\sb A^B$ of size $2^\co$ such that for every one-to-one sequence $\la g_a\in\C\colon a\in A\ra$ there is $b\in B$ with $g_a(b)=a$ for every $a\in A$.} \pf The theorem is proved in \cite[Corollary 3.17, p. 77]{CN} for $A=\omega$ and $B=\continuum$. The generalization is obvious. \qed \medskip From this we will conclude the following lemma. \lem{lem1}{ If $B\su\real$ has cardinality $\co$ and $H\su\rational^B$ is such that $|H|<2^{\co}$ then there is $g\in\rational^B$ such that $h\cap g\ne\es$ for every $h\in H$.} \pf Let $\C$ be as in Theorem~\ref{thComfNeg} with $A=\rational$. For each $h\in H$ there only finitely many $g\in\C$ for which $h\cap g =\es$, since any countable infinite subset of $\C$ can be enumerated as $\{g_a\in\C\colon a\in\rational\}$. So there is $g\in\C$ such that $h\cap g\ne\es$ for every $h\in H$. \qed \medskip {\noindent{\sc Proof of $\add(\pc)=2^\co$.}} By Proposition~\ref{prop1} to prove $\add(\pc)=2^\co$ it is enough to show that $\add(\pc)\geq 2^\co$. So, let $F\su\real^\real$ be such that $|F|<2^{\co}$. We will find $g\colon\real\to\real$ such that $f+g\in\pc$ for every $f\in F$. Let $\G$ be the family of all triples $\la I,p,m\ra$ where $I$ is a non-empty open interval with rational end points, $p\in\rational$ and $m<\omega$. For each $\la I,p,m\ra\in\G$ define a set $B_{\la I,p,m\ra}\su I$ of size $\co$ such that $B_{\la I,p,m\ra}\cap B_{\la J,q,n\ra}=\es$ for any distinct $\la I,p,m\ra$ and $\la J,q,n\ra$ from $\G$. Next, fix $\la I,p,m\ra\in\G$ and for each $f\in F$ choose $h^f_{\la I,p,m\ra}\colon B_{\la I,p,m\ra}\to\rational$ such that \[ |p-(f(x)+h^f_{\la I,p,m\ra}(x))|< \frac{1}{m}\ \text{ for every } x\in B_{\la I,p,m\ra}. \] Then, by Lemma~\ref{lem1} used with a set $H_{\la I,p,m\ra}=\{h^f_{\la I,p,m\ra}\colon f\in F\}$, there exists $g_{\la I,p,m\ra}\colon B_{\la I,p,m\ra}\to\rational$ such that \[ \forall f\in F\ \exists x\in B_{\la I,p,m\ra}\ h^f_{\la I,p,m\ra}(x)=g_{\la I,p,m\ra}(x). \] In particular, if $g\colon\real\to\rational$ is a common extension of all functions $g_{\la I,p,m\ra}$ then for every $\la I,p,m\ra\in\G$ and every $f\in F$ there exists $x\in B_{\la I,p,m\ra}\su I$ such that \[ |p-(f(x)+g(x))|< \frac{1}{m}. \] So, for every $f\in F$ the graph of $f+g$ is dense in $\real^2$. Thus, $f+g\in\pc$. \qed \section{ Proof of Theorem~\ref{main}(4): $\add(\ext)=\add(\pr)=\co^+$.} By Proposition~\ref{prop1} we have $\add(\ext)\leq\add(\pr)$. Thus, it is enough to prove two inequalities: $\add(\pr)\leq\co^+$ and $\add(\ext)\geq\co^+$. First we will prove $\add(\pr)\leq\co^+$. For this we need the following lemma. \lem{lemaddpr}{ There is a family $F\sb\rl^{\rl}$ of size $\co^{+}$ such that for every distinct $f,h\in F$, every perfect set $P$ and every $n<\omega$ there exists an $x\in P$ with $|f(x)-h(x)|\ge n$. } \pf The family $F=\{f_\xi\colon \xi<\co^+\}$ is constructed by induction using standard diagonal argument. If for some $\xi<\co^+$ the functions $\{f_\zeta\colon \zeta<\xi\}$ are already constructed, we construct $f_\xi$ as follows. Let $\la\la P_{\alpha},h_{\alpha},n_{\alpha}\ra\colon\alpha<\co\ra$ be an enumeration of all triples $\la P,h,n\ra$ where $P\su\real$ is perfect, $h=f_\zeta$ for some $\zeta<\xi$ and $n<\omega$. By induction on $\alpha<\co$ choose $x_{\alpha}\in P_{\alpha} \sm\{x_{\beta}\colon\beta < \alpha\}$ and define $f_\xi(x_{\alpha})=h_{\alpha}(x_{\alpha})+n_\alpha$. Then any extension of such defined $f_\xi$ onto $\real$ will have the desired properties. \qed \medskip {\noindent{\sc Proof of $\add(\pr)\leq\co^+$.}} Now let $F$ be a family from Lemma~\ref{lemaddpr}. We will show that for every $g\colon\real\to\real$ there exists $f\in F$ such that $f+g\not\in\pr$. By way of contradiction assume that there exists a function $g\colon\real\to\real$ such that $f+g\in\pr$ for every $f\in F$. Then, for every $f\in F$ there exists a perfect set $P_f$ such that $0$ is a bilateral limit point of $P_f$ and $(f+g)|_{P_f}$ is continuous at $0$. Since there are $\co^{+}$-many functions in $F$ and only $\co$-many perfect sets there are distinct $f,h\in F$ with $P_f=P_h$. Then the function $((f+g)-(h+g))|_{P_f}=(f-h)|_{P_f}$ is continuous at $0$ contradicting the choice of family $F$. \qed \medskip The proof of $\add(\ext)\geq\co^+$ is based on the following facts. \lem{lem11}{ For every meager subset $M$ of the real line $\rl$ there exists a family $\{h_\xi\in\real^\real\colon \xi<\co\}$ of increasing homeomorphisms such that $h_\zeta[M] \cap h_\xi[M]=\es$ for every $\zeta<\xi<\co$.} \pf Let $\{D_\zeta:\zeta <\co\}$ be a family of pairwise disjoint $\co$-dense, meager $F_\sigma$-sets. Then by \cite[Lemma 4]{gor} there are homeomorphisms $\{h_\zeta:\reals \to \reals: \zeta < \co\}$ such that $h_\zeta[M] \subset D_\zeta$. \qed \medskip For $f\in\F\subset\real^\real$ we say that a set $G\su\real$ is {\em $f$-negligible for the class $\F$} provided $g\in\F$ for every $g\colon\real\to\real$ such that $g|_{\rl\sm G}=f|_{\rl\sm G}$. Thus, $G\su\real$ is $f$-negligible for $\F$ if we can modify $f$ arbitrarily on $G$ remaining in the class~$\F$. \thm{thConnExamp}{ There exists a connectivity function $f\colon\rl^2 \to \rl$ with graph dense in $\real^3$ such that some dense $G_{\delta}$ subset $G$ of $\real^2$ is $f$-negligible for the class $\conn(\real^2,\real)$. } The proof of this theorem will be postponed to the end of the section. \cor{corExtExamp}{ There exists an extendable function $\hat{f}\colon\rl \to \rl$ with graph dense in $\real^2$ such that some dense $G_{\delta}$ subset $\hat{G}$ of $\real$ is $\hat{f}$-negligible for the class $\ext$.} \pf Let $f\colon\rl^{2}\to\rl$ and $G$ be as in Theorem~\ref{thConnExamp}. Then there exists $y\in\rl$ with $G^y=\{x\colon\la x,y\ra\in G\}$ being a dense $G_{\delta}$ subset of $\rl$. Clearly the set $\hat{G}=G^y$ and the function $\hat{f}\colon\rl\to\rl$ defined by $\hat{f}(x)=f(x,y)$ for every $x\in\real$ satisfy the requirements. \qed \medskip The existence of a function as in Corollary~\ref{corExtExamp} was first announced by H.~Rosen at the 10th Auburn Miniconference in Real Analysis, April 1995. However, the construction presented at that time had a gap. This gap had been removed later, as described in \cite{Rosen2}. The construction presented in this paper is an independently discovered fix-up of the original Rosen's gap. It is also more general than that of \cite{Rosen2}, since \cite{Rosen2} does not contain any example similar to that of Theorem~\ref{thConnExamp}. Next, we will show how Lemma~\ref{lem11} and Corollary~\ref{corExtExamp} imply $\add(\ext)\geq\co^+$. The argument is a modification of the proof of Corollary~\ref{corGib}. (Compare also \cite{N11} and \cite{Rosen2}.) \medskip {\noindent{\sc Proof of $\add(\ext)\geq\co^+$.}} Let $F=\{f_\xi\in\real^\real\colon\xi<\co\}$. We will find $g\colon\real\to\real$ such that $f_\xi+g\in\ext$ for every $\xi<\co$. So, let $\hat{f}\colon\real\to\real$ and $\hat{G}\su\real$ be as in Corollary~\ref{corExtExamp}. Put $M=\rl\sm\hat{G}$ and take $\{h_\xi\in\real^\real\colon \xi<\co\}$ as in Lemma~\ref{lem11}. For $\xi<\co$ define $g$ on $h_\xi[M]$ to be $(\hat{f}\circ h_\xi^{-1}-f_\xi)|_{h_\xi[M]}$ and extend it onto $\real$ arbitrarily. To see that $f_\xi+g\in\ext$ notice that $f_{\xi}+g=\hat{f}\circ h_{\xi}^{-1}$ on $h_{\xi}[M]$. But the set $\rl \sm h_{\xi}[M]=h_{\xi}[\hat{G}]$ is $(\hat{f}\circ h_{\xi}^{-1})$-negligible for the class $\ext$. (See \cite{N11} for an easy proof.) So, each $f_{\xi}+g$ is extendable. \qed \medskip \begin{figure} \begin{picture}(300,220)(-40,0) \thicklines \multiput(10,10)(60,0){4}{ \drawline[20](0,0)(60,0)(30,52)(0,0)} \multiput(40,62)(60,0){3}{ \drawline[20](0,0)(60,0)(30,52)(0,0)} \multiput(70,114)(60,0){2}{ \drawline[20](0,0)(60,0)(30,52)(0,0)} \put(100,166){ \drawline[20](0,0)(60,0)(30,52)(0,0)} \end{picture} \caption{grid $S_n$\label{fig1}} \end{figure} {\noindent{\sc Proof of Theorem~\ref{thConnExamp}.}} We will construct a peripherally continuous function $f\colon\rl^{2}\to\rl$ with dense $G_{\delta}$ subset $G$ of $\rl^{2}$ which is $f$-negligible for the class $\pc(\real^2,\real)$. It is enough since, by Theorem~\ref{thHagan}, $\conn(\real^2,\real)=\pc(\real^2,\real)$. The construction is a modification of that from \cite{GR}, where a similar example of function from $[0,1]\times[0,1]$ onto $[0,1]$ was constructed. (Compare also \cite{B}.) The additional difficulty in our construction is to make sure that some sequences of points in the range of $f$ (=$\real$) have cluster points, what is obvious for all sequences in $[0,1]$. Also, our basic construction step will be based on a triangle, while construction in \cite{GR} was based on a square. The triangles work better, since for arbitrary three non-collinear points in $\rl^{3} $ there is precisely one plane passing through them, while it is certainly false for four points. \pagebreak {\bf Basic Idea}: We will construct, by induction on $n<\omega$, a sequence $\la S_n\colon n<\omega\ra$ of triangular ``grids'' formed with equilateral triangles of side length $1/2^{k_n}$, as in Figure~\ref{fig1}. The grid $S_n$ will be identified with the points on the edges of triangles forming it and we will be assuming that $S_n\su S_{n+1}$ for all $n<\omega$. With each grid $S_n$ we will associate a continuous function $f_n\colon S_n\to\real$ which is linear on each side of a triangle from $S_n$. Moreover, each $f_{n+1}$ will be an extension of $f_n$. Function $f$ will be defined as an extension of $\bigcup_{n<\omega}f_n$. \begin{figure} \begin{picture}(300,130)(-100,0) \thicklines \put(10,10){ \begin{picture}(150,150) \drawline[20](0,0)(120,0)(60,104)(0,0) \drawline[20](60,0)(90,52)(30,52)(60,0) \drawline[20](45,26)(75,26)(60,52)(45,26) \end{picture}} \end{picture} \caption{basic partition\label{fig2}} \end{figure} {\bf Terminology}: In what follows a {\em triangle} will be identified with the set of points of its interior or its boundary. For a grid $S$ we say that a {\em triangle $T$ is from $S$} if the interior of $T$ is equal to a component of $\real^2\setminus S$. For an equilateral triangle $T$, its {\em basic partition} will be its split into seven equilateral triangles, as in Figure~\ref{fig2}. The central triangle $\hat{T}$ of Figure~\ref{fig2} will be referred as {\em the middle quarter of $T$}. Thus, $\hat{T}\cap\bd(T)=\emptyset$ and the length of each side of $\hat{T}$ is equal to $1/4$ of the length of a side of $T$. If a function $F$ is defined on the three vertices of a triangle $T$, its {\em basic extension} is defined as the unique function $\hat{F}\colon T\to\real$ extending $F$ whose graph is a subset of a plane. Notice, that $\hat{F}$ is linear on each side of the triangle $T$ and that $\hat{F}$ extends $F$ even if the function $F$ has already been defined on some side of $T$ as long as $F$ is linear on this side. {\bf Inductive construction}: We will define inductively three increasing sequences: $\la S_n\colon n<\omega\ra$ of triangular grids as in Figure~\ref{fig1}, $\la f_n\in\real^{S_n}\colon n<\omega\ra$ of continuous functions and $\la k_n<\omega\colon n<\omega\ra$ of natural numbers such that the following inductive conditions are satisfied for every $n<\omega$. \pagebreak \begin{description} \item[(i)] $f_n\colon S_n\to[-2^n,2^n]$ and it is linear on each side of a triangle $T$ from $S_n$. \item[(ii)] The side length of each triangle from $S_{n}$ is equal to $1/2^{k_n}$. \item[(iii)] The variation of $f_n$ on each triangle from $S_{n}$ is $\le 1/2^{n}$. \item[(iv)] If $n>0$ then for every triangle $T$ from $S_{n-1}$ and every dyadic number $i/2^{n}\in[-2^n,2^n]$ with $i\in\integer$ ($-4^{n}\leq i\leq 4^{n}$) there is a triangle $T_i\su\hat{T}$ such that $\bd(T_i)\su S_{n}$ and $f_{n}(x)=i/2^{n}$ for every $x\in\bd(T_i)$. \item[(v)] If $n>0$, $T$ is a triangle from $S_{n-1}$ and $T^\prime$ is a triangle from $S_{n}$ such that $T^\prime\su T$ and $T^\prime\not\su\hat{T}$ then $f_{n}[\bd(T^\prime)]\su [-M,M]$, where $M=\max\{|f_{n-1}(x)|\colon x\in\bd(T)\}$. \end{description} \begin{figure} \begin{picture}(300,220)(-40,0) \thicklines \multiput(10,10)(60,0){4}{ \dashline{4}(0,0)(60,0)(30,52)(0,0)} \multiput(40,62)(60,0){3}{ \dashline{4}(0,0)(60,0)(30,52)(0,0)} \multiput(70,114)(60,0){2}{ \dashline{4}(0,0)(60,0)(30,52)(0,0)} \put(100,166){ \dashline{4}(0,0)(60,0)(30,52)(0,0)} \multiput(55,36)(60,0){3}{ \drawline[20](0,0)(30,0)(15,26)(0,0)} \multiput(85,88)(60,0){2}{ \drawline[20](0,0)(30,0)(15,26)(0,0)} \put(115,140){ \drawline[20](0,0)(30,0)(15,26)(0,0)} \end{picture} \caption{some triangles of the grid of $\hat{T}$\label{fig3}} %KC \end{figure} %KC To start the induction, take $k_0=0$, define grid $S_0$ as in Figure~\ref{fig1} with all sides of length $1=1/2^0$ and choose $f_0\colon S_0\to\real$ as constantly equal $0$. It is easy to see that the conditions (i)-(v) are satisfied with such a choice. Next, assume that for some $n>0$ we already have $S_{n-1}$, $f_{n-1}$ and $k_{n-1}$ satisfying (i)-(v). We will define $S_n$, find $k_n$ and extend $f_{n-1}$ to $f_n\colon S_n\to\real$ such that (i)-(v) will still hold. Put $F_n=f_{n-1}$. Step 1. Let $T$ be a triangle from $S_{n-1}$ and extend $F_n$ into each vertex of its middle quarter $\hat{T}$ by assigning it the value $0$. Notice, that $F_n$ is defined on all vertices of the basic partition of $T$. Partition $\hat{T}$ into a grid $S$ such that the size of each triangle from $S$ is equal $1/2^{\hat{k}_n}$. The number ${\hat{k}_n}<\omega$ is chosen as a minimal number such that there are $2\cdot 8^n + 1$ disjoint triangles $\{T_i\colon i\in\integer, -4^n\leq i\leq 4^n\}$ from $S$ none of which intersects the boundary of $\hat{T}$. (See Figure~\ref{fig3}.) Notice that the value of ${\hat{k}_n}$ does not depend on $T$. On the vertices of each triangle $T_i$ define $F_n$ to be equal $i/2^{n}$. On the remaining undefined vertices of $S$ define $F_n$ to be equal $0$. Notice that $F_n$ is defined on all vertices of each triangle defined so far. Step 2. Extend $F_n$ into $\rl^{2}$, by defining it on every triangle $T$ constructed so far as the basic extension of $F_n|_{\bd(T)}$. Notice that if we extend grid $S_{n-1}$ to the grid $\hat{S}_n$ with side length of each triangle from $\hat{S}_n$ equal $1/2^{\hat{k}_n}$ and put $\hat{f}_n=F_n|_{\hat{S}_n}$ then the triple $\la \hat{S}_n,\hat{f}_n,\hat{k}_n\ra$ satisfies conditions (i), (ii), (iv) and (v). Step 3. We have to modify $\hat{S}_n$, $\hat{f}_n$ and $\hat{k}_n$ to get also condition (iii), while keeping the other properties. First notice that for every triangle $T$ from $\hat{S}_n$ and any interval $J$ inside $T$ the slope of $F_n$ on $J$ does not exceed the number \[ \frac{\text{length of the range of $F_n$}}{\text{length of a side of $T$}} \leq\frac{2\ 2^n}{2^{-\hat{k}_n}}=2^{n \hat{k}_n +1}. \] So, let $k_n\geq n\hat{k}_n+n+1$, in which case \[ \frac{1}{2^{k_n}} 2^{n \hat{k}_n +1}\leq \frac{1}{2^n}, \] let $S_n$ be a refinement of the grid $\hat{S}_n$ with triangles with side size $1/2^{k_n}$ and put $f_n=F_n|_{S_n}$. It is easy to see that this gives us (iii) while preserving the other conditions. This finishes the inductive construction. \medskip Let $S=\ds_{n<\omega}S_{n}$ and define $f$ on $S$ by $f=\ds_{n<\omega}f_{n}$. To extend it to $\rl^{2}\sm S$ notice that \begin{description} \item[($\star$)] for every $x\in\rl^{2}\sm S$ there exists a number $f(x)\in\rl$ and a sequence $\la T_k\colon k<\omega\ra$ of triangles with $x$ being an interior point of each $T_k$ such that $\lim_{k\to\infty}\diam(T_k)\to 0$ and \[ f(x)=\lim_{k \to \infty}\min f[\bd(T_k)]=\lim_{k \to \infty}\max f[\bd(T_k)]. \] \end{description} The proof of ($\star$) will finish the construction of $f$. To see ($\star$) fix $x\in\rl^{2}\sm S$ and let $T_{n}^{0}$ be the triangle from $S_{n}$ such that $x$ belongs to the interior of $T_{n}^{0}$. Let $N=\{n<\omega\colon n>0\ \&\ T_n^{0}\su\hat{T}_{n-1}^{0}\}$. There are two cases to consider. Case 1. Set $N$ is infinite. Then, let $\la n_k\colon k<\omega\ra$ be a one-to-one enumeration of $N$ and define $T_k=\hat{T}_{n_k - 1}^0$. It is easy to see that this sequence satisfies ($\star$) with $f(x)=0$. Case 2. Set $N$ is finite. Let $m<\omega$ be such that $T_n^{0}\not\su\hat{T}_{n-1}^{0}$ for every $n\geq m$ and let $M=\max\{|f_{m-1}(x)|\colon x\in\bd(T_{m-1}^{0})\}$. Then, by condition (v), $f_{n}[\bd(T_n^{0})]\su [-M,M]$ for every $n\geq m$. So, there exists an increasing sequence $\la n_k\geq m\colon k<\omega\ra$ such that the limit $L=\lim_{k \to \infty}\max f[\bd(T_{n_k}^0)]$ exists. It is easy to see that the sequence $\la T_k\ra =\la T_{n_k}^0\ra$ satisfies ($\star$) with $f(x)=L$, since the variation of $f$ on $\bd(T_{n_k}^0)$ tends to $0$ as $k\to\infty$. This finishes the construction of function $f$. It remains to show that $f$ has the desired properties. Clearly ($\star$) implies that $f$ is peripherally continuous at every point $x\in\rl^{2} \sm S$. To see that $f$ is peripherally continuous on $S$ take $x\in S$. Then, there exists $k<\omega$ such that $x \in S_{n}$ for every $n\geq k$. For any such $n$ let $\T_n$ be the set of all triangles from $S_n$ to which $x$ belongs. Notice that $\T_n$ has at most six elements and that $x$ belongs to the interior of the polygon $P_n=\bigcup\T_n$. Hence, the variation on the boundary of $P_n$ is at most $6/2^n$ and the diameter of $P_n$ is at most $1/2^{n-1}$. So, the sequence $\la P_n\ra$ guarantees that $f$ is peripherally continuous at $x$. To finish the proof it is enough to find a dense $G_{\delta}$ set $G$ which is $f$-negligible for $\pc(\real^2,\real)$. For any dyadic number $d$ and any $k\in\omega$ let $\F_d^k$ denote %be the family of all triangles $T$ for which there exists $n\geq k$ such that $T$ is from $S_n$ and $f_n(x)=d$ for every $x\in\bd(T)$. Let $G_d^k$ be the union of the interiors of all triangles $T\in\F_d^k$. Then, by condition (iv), each set $G_d^k$ is open and dense. Therefore, %Thus, $G=\di\{G_d^k\colon k\in\omega\ \&\ d\text{ is dyadic}\}$ is a dense $G_\delta$ set. It is easy to see, that $f$ is peripherally continuous if we redefine it on the set $G$ in an arbitrary way. This finishes the proof of Theorem~\ref{thConnExamp}. \section{Compositions of Lebesgue measurable functions.} We can consider similar problems for compositions of functions. For example, we know that every function is a composition of Lebesgue measurable functions \cite{Ruz}. (See also {\sl Problem 6378}, American Mathematical Monthly {\bf 90}, 573.) It is easy to make every function in $\real^\real$ measurable (in a sense of definition of $\add$) using composition with just one function. We simply take a composition with a constant function. So we need to define cardinal invariants in different way. The next definition will represent one of the way the problem can be approached. Instead of ``making family $H$ to be in $\F$'' we will try to recover all elements of $H$ with one ``coding'' function $\hat f\in\F$ and the class $\F$ of all codes. This leads to the following definitions. \[ C_{r}(\F) = \min\{|H|\colon H\su\reals^\reals\ \&\ \neg\exists \hat f\in\F\; \forall h\in H \exists f\in \F\ f\circ \hat f = h\}\cup\{(2^\co)^+\} \] and \[ C_{l}(\F) = \min\{|H|\colon H\su\reals^\reals\ \&\ \neg\exists \hat f\in\F\; \forall h\in H \exists f\in \F\ \hat f\circ f = h\}\cup\{(2^\co)^+\}. \] Let $\cal L$ be the family of all Lebesgue measurable functions from $\rl$ into $\rl$. \thm{th:comp}{ $C_{r}({\cal L})=(2^{\co})^{+}$ and $C_{l}({\cal L})=\co^{+}$. } \pf To see $C_{r}({\cal L})\geq(2^{\co})^{+}$ we will show that the family $H=\real^\real$ of all functions can be ``coded'' by one function $\hat f\in{\cal L}$. Simply, let $\hat f$ be a Borel isomorphism from $\rl$ onto Cantor ternary set $C$. For any function $h\in\real^\real$ we define $f_h\in{\cal L}$ by putting $f_{h}\equiv 0$ on the complement of $C$ and $f_{h}=h\circ \hat f^{-1}$ on $C$. Then $h=f_{h}\circ \hat f$. To see that $C_{l}({\cal L})\ge \co^{+}$ let $H=\{h_{\xi}\colon \xi <\co\} \sb \rl^{\rl}$ and let $\{C_{\xi}\colon\xi < \co\}$ be a partition of the Cantor ternary set $C$ into perfect sets. Then for every $\xi<\co$ take a Borel isomorphism $f_{\xi}\colon \rl \to C_{\xi}$ and define $\hat f(x)=(h_\xi\circ f_{\xi}^{-1})(x)$ for $x \in C_{\xi}$ and $\hat f(x)=0$ otherwise. It is easy to see that $\hat f\in{\cal L}$ and $\hat f\circ f_\xi=h_\xi$ for every $\xi<\co$ To prove $C_{l}({\cal L})\le \co^{+}$ take $\{h_{\xi }\colon \xi< \co^{+}\}$ from Lemma~\ref{lemaddpr}. Assume that there exists a sequence $\{f_{\xi}\colon \xi<\co^{+}\}$ of measurable functions and measurable function $\hat f$ such that $h_{\xi}=\hat f\circ f_{\xi}$ for every $\xi<\co^+$. For each $\xi<\co^{+}$ let $P_{\xi}$ be a perfect set such that $f_{\xi}|P_{\xi}$ is continuous. Then, by a cardinality argument, there are $\zeta<\xi<\co^+$ such that $P_\zeta=P_\xi$ and $f_\zeta|P_\zeta=f_\xi|P_\xi$. 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