\documentclass{rae} %\usepackage{amstex} \usepackage{amsmath}\usepackage{amssymb} \usepackage{graphicx} \Volume{24}\IssueNumber{2}\Year{1998/9} \firstpagenumber{599} \received{April 27, 1998} \author{Krzysztof Ciesielski% \thanks{The first author was partially supported by NSF Cooperative Research Grant INT-9600548 with its Polish part financed by KBN.\endgraf Papers authored or co-authored by a Contributing Editor are managed by a Managing Editor or one of the other Contributing Editors.}, Department of Mathematics, West Virginia University, Morgantown, WV 26506-6310, USA e-mail: {\tt KCies@wvnvms.wvnet.edu}\\ Kenneth R. Kellum, Department of Mathematics and Computer Science, San Jos\'e State University, San Jos\'e, California 95192-0103 e-mail: {\tt kellum@sjsumcs.sjsu.edu} } \title{Compositions of Darboux and connectivity functions} \date{} \MathReviews{Primary 26A15, 26A30; Secondary 54C30, 54C08} \keywords{%Key words and phrases: connectivity function, Darboux function, com\-po\-sition.} \newtheorem{theorem}{Theorem} \newtheorem{proposition}{Proposition} \newtheorem{corollary}{Corollary} \def\qed{\hfill$\Box$} %\def\qed{\hfill\vrule height6pt width6pt depth1pt%\medskip %} %Proof \def\proof{\noindent{\sc Proof.} } \newcommand{\op}{\operatorname} \newcommand{\cl}{{\op{cl}}} \newcommand{\hatbbP}{\widehat {\mathbb{P}}} \newcommand{\bbP} {\mathbb{P}} \newcommand{\bbR} {\mathbb{R}} \newcommand{\bbL} {\mathbb{L}} \renewcommand{\P}{\mathcal{P}} \newcommand{\calI} {\mathcal{I}} \newcommand{\I}{\mathcal{I}} \def\la{\langle} \def\ra{\rangle} \begin{document} \maketitle \markboth{Krzysztof Ciesielski and Kenneth R. Kellum}{Compositions of Darboux and Connectivity Functions} \begin{abstract} We describe here an example of a Darboux function $k$ from the unit interval $\I=[0,1]$ onto itself such that $k$ is not the composition of any finite collection of connectivity functions from $\I$ into $\I$. This answers a question of Ceder \cite{C}. \end{abstract} In \cite{K} the second author proved that there exists a connectivity function from $\I$ into $\I$ which cannot be written as the composition of finitely many almost continuous functions. In the present paper we show that the techniques developed in the earlier paper can be extended to answer a question of Ceder~\cite{C}. We prove that there exits a Darboux function which cannot be factored into a finite composition of connectivity functions. This stands in contrast with the following fact. \begin{proposition}$\!\!\!${\bf .} {\rm (Natkaniec~\cite{TN})} Assume that $\I$ is not a union of less than continuum many of its meager subsets. Then every function from $I$ into $I$ with dense level sets can be expressed as the composition of two almost continuous functions. \end{proposition} \noindent Since every almost continuous function $f\colon\I\to\I$ is connectivity, this theorem shows that our example must be relatively nice. (Cannot have dense level sets.) Our terminology is standard and follows \cite{Ci:book}. In particular, functions will be identified with their graphs. Recall also the following definitions. (See \cite{GN} for more on these.) A function $f\colon X\to Y$ from a topological space $X$ into a topological space $Y$ is {\it Darboux\/} if $f[C]$ is connected in $Y$ whenever $C\subset X$ is connected in $X$; $f$ is a {\it connectivity function\/} if the restriction $f \restriction C$ of $f$ to $C$ is a connected subset of $X\times Y$ for every connected subset $C$ of $X$. We will be interested in these functions only when $X=Y=\I$. Then $f\colon\I\to\I$ is Darboux if an $f$-image of any interval is an interval. Clearly every connectivity function $f\colon\I\to\I$ is Darboux. %\chKC For $U\subset\bbR$ we will write ${\rm int}(U)$ to denote the interior of $U$. Recall also that connectivity can be characterized in terms of continua, where a {\it continuum\/} is a compact connected set. We say that a continuum $M\subset X\times Y$ {\it cuts\/} the function $f\colon X\to Y$ if $M \cap f= \emptyset$ and there exist points $\la x_1,y_1\ra$ and $\la x_2,y_2\ra$ in $M$ such that $f(x_1) < y_1$ and $f(x_2) > y_2$. A function $f\colon\calI \to \calI$ is a connectivity function if and only if no continuum in $\I^2$ cuts $f$. (See e.g.~\cite{GNK}.) \section{The Example} Let $C$ denote the Cantor middle two-fifths set: \[ C=\left\{\sum_{n=1}^\infty\frac{i_n}{5^n}\colon i_n\in\{0,2,4\}\ \mbox{ for every $n$}\right\}. \] Geometrically, $C$ is obtained from $\calI$ by first removing the pair of intervals (1/5,2/5) and (3/5,4/5) from $\I$, then by removing similar pairs of intervals (the middle two fifths) from each of [0,1/5], [2/5,3/5], and [4/5,1], etc. Let $\bbP_L$ denote the set of closures of removed intervals which are the left members of a removed pair and $\bbP_R$ the set of closures of right members of removed pairs. (For example $[1/5,2/5]\in \bbP_L$ and $[3/5,4/5]\in \bbP_R$.) Thus $\bbP =\bbP_L \cup \bbP_R$ is the family of all closures of components of $\calI\setminus C$. Also, let $\Delta$ denote the diagonal $\{\la x,x\ra\colon x\in\I\}$ in $\calI \times \calI$ and let $C^\circ$ denote the points of $C$ which are not endpoints of the removed intervals, that is, $C^\circ=\I\setminus\bigcup\bbP$. We define a function $k\colon\calI \to \calI$ in the following way. On $C^\circ$ we define $k$ so that $k[C^\circ] = C^\circ$, $k\restriction C^\circ$ is one-to-one, and $x \neq k(x)$ for each $x\in C^\circ$. Thus, $\Delta\cap k\restriction C^\circ=\emptyset$. Next, suppose that $P$ and $Q$ are the closures of adjacent removed middle fifths and suppose that $P \in \bbP_L$. We define $k$ on $P\cup Q$ such that $k[P \cup Q] = \calI$, $k \restriction P$ and $k\restriction Q$ are each continuous and strictly increasing and so that $k \restriction P$ lies above $\Delta$ while $k\restriction Q$ lies below $\Delta$. So, $\Delta\cap k=\emptyset$. \begin{figure}[htb] \centering \includegraphics[width=\textwidth]{k.eps} \caption{The function $k$ (the dotted line represents $\Delta$).} \label{fig:k} \end{figure} Clearly, $k$ is not a connectivity function since its graph is separated by $\Delta$. It is also easy to see that $k$ is Darboux. Indeed, if an interval $J\subset\I$ is a subset of a closure $P$ of some interval from $\bbP$ then $k[J]$ is also an interval since $k\restriction P$ is continuous. But otherwise $J$ contains a pair $P$ and $Q$ of closures of adjacent removed middle fifths from the construction of $C$ and so $k[J]=\I$ is also an interval. Now we are ready for the main result of the paper. \begin{theorem}$\!\!\!${\bf .} If $k = g\circ f$, where $f$ and $g$ are functions from $\calI$ onto itself, $f$ is connectivity, and $g$ is Darboux, then $f$ is a homeomorphism. \end{theorem} \proof First note that \begin{equation}\label{eq1} \mbox{if $U\in\bbP$ then $f\restriction U$ and $g \restriction f[U]$ are homeomorphisms.} \end{equation} Indeed, since $k\restriction U$ is one-to-one then so is $f\restriction U$. Thus $f\restriction U$ is a Darboux function with closed level sets and so (see e.g. \cite[thm~5.2]{BC}) it is continuous. So, $f \restriction U$ is a homeomorphism. By the same reasoning $g\restriction f[U]$ also is a homeomorphism. Next note that \begin{equation}\label{eq2} \mbox{$f[C]$ does not contain any non-trivial interval.} \end{equation} Indeed, suppose for a moment that $f[C]$ contains a non-trivial interval, say $V$. Since $g[V]$ is connected, this would imply that an uncountable subset of $C$ is mapped by $k$ onto a connected set, a contradiction. Notice also that \begin{equation}\label{eq3} \mbox{if $U\in\bbP_L$ and $V\in\bbP_R$ then $f[{\rm int} (U)]\cap f[{\rm int} (V)]=\emptyset$.} \end{equation} To see it assume, by way of contradiction, that there are $U=[u_1,u_2]\in\bbP_L$ and $V=[v_1,v_2]\in\bbP_R$ such that %\chKC $f[{\rm int} (U)]$ and $f[{\rm int} (V)]$ do overlap. Then, by (\ref{eq1}), $g \restriction (f[U]\cup f[V])$ is a homeomorphism, say increasing. Thus $f\restriction U$ and $f \restriction V$ are increasing as well. Moreover, from the choice of $U$ and $V$, $g$ maps $f[U]\cup f[V]$ onto $\calI$, $g\circ f(u_2)=k(u_2)=1$, and $g\circ f(v_1)=k(v_1)=0$. So $h=(g \restriction f[U]\cup f[V])^{-1}$ is an increasing homeomorphism on $\calI$. Note also that $f(x)\neq h(x)$ for every $x\in\I$, since otherwise $k(x)=g(f(x))=g(h(x))=x$ contradicting the fact that $k\cap\Delta=\emptyset$. %\chKC Thus Therefore $h\cap f=\emptyset$. Moreover, $f(v_1)=h(0)h(u_2)$ since $h$ is increasing. Thus, the graph of $h$, which is a continuum, cuts $f$, contradicting the connectivity of $f$. Condition (\ref{eq3}) has been proved. For $P\in\bbP$ let $c_P\in P$ be the end-point of $P$ for which $k(c_P)\in\{0,1\}$ and let $e_P=f(c_P)$. So, by (\ref{eq1}), (\ref{eq3}), and the construction of $k$ we see that for every $P\in\bbP$ and $\P\subseteq\{U\in\bbP\colon f[{\rm int}(U)]\cap f[{\rm int}(P)]\neq\emptyset\}$ \begin{equation}\label{eq4} \mbox{$e_P\in f[U]$ for every $U\in\P$}, \end{equation} \begin{equation}\label{eq5} \mbox{$f\left[\bigcup\P\right]$ is an interval with one end point equal to $e_P$,} \end{equation} and \begin{equation}\label{eq6} \mbox{$g$ is a homeomorphism on $f\left[\bigcup\P\right]$.} \end{equation} In what follows, we say that a family $\P\subset\bbP$ is {\it dense} in an open interval $(a,b)\subset\I$ if $\bigcup\P\subset(a,b)$, $a=\inf\bigcup\P$, $b=\sup\bigcup\P$, and between any two members of $\P$ there is another. The main technical fact in the proof is the following. \begin{description} \item{($\star$)} Assume that $f[{\rm int}(P)]\cap f[{\rm int}(Q)]\neq\emptyset$ for some different $P,Q\in\bbP$ and let $\P$ be the set of all $U\in\bbP$ between $P$ and $Q$ %\chKC such that with $f[{\rm int}(U)]\cap f[{\rm int}(P)]\neq\emptyset$. Then %\chKC the family $\P$ is non-empty and dense in $(a,b)$, where $a=\inf\bigcup\P$ and $b=\sup\bigcup\P$. Moreover, there %\chKC is exists a homeomorphism $h$ into $K=f\left[\bigcup\P\right]$ such that either \begin{description} \item{(i)} $e_P\in K\subset[e_P,1]$ and the graph of $f\restriction(a,b)$ is below $h$, or \item{(ii)} $e_P\in K\subset [0,e_P]$ and the graph of $f\restriction(a,b)$ is above $h$. \end{description} and either \begin{description} \item{(iii)} the domain of $h$ is $[0,b)$ or \item{(iv)} the domain of $h$ is $(a,1]$ \end{description} \end{description} Indeed, to see that $\P$ is non-empty and dense in $(a,b)$ assume, to the contrary, that there are $U$ and $V$ in $\P\cup\{P,Q\}$ with no member of $\P$ between them. Suppose $U$ is to the left of $V$ and let $[c,d]=f[U\cap V]$. By (\ref{eq2}) we can choose $y\in(c,d)\setminus f[C^\circ]$. Also, by (\ref{eq1}), we can also choose $p\in U$ and $q\in V$ such that the continuum $[p,q]\times \{y\}$ is disjoint with $f\restriction U$ and $f\restriction V$. Thus $[p,q]\times \{y\}$ cuts the function $f$ since, by (\ref{eq6}), either both $f\restriction U$ and $f\restriction V$ are increasing or both are decreasing. But this contradicts the connectivity of $f$. To see the existence of $h$ as in ($\star$) assume that $P\in\bbP_R$ and that $f\restriction P$ is increasing, the other three cases being similar. We will show that this implies (i) and (iii). Under this assumption we have $k\left[\bigcup\P\right]=[0,b)$, since there are elements of $\P$ arbitrarily close to $b$. Also $K=[e_P,d)$ for some $d>e_P$ and, by (\ref{eq6}), $g\restriction K$ is an increasing homeomorphism between $K$ and $[0,b)$. Let $h=(g\restriction K)^{-1}$ and notice that $f(x)\neq h(x)$ for every $x\in[a,b]$, since otherwise $k(x)=g(f(x))=g(h(x))=x$ contradicting the fact that $k\cap\Delta=\emptyset$. Thus $h\cap (f\restriction(a,b))=\emptyset$ and, by the connectivity of $f\restriction(a,b)$, its graph lies either below or above $h$. It is below, since $e_P\in f[\bigcup\P]$ is the minimal value of $h$. This proves ($\star)$. Now, if $f[{\rm int}(P)]\cap f[{\rm int}(Q)]=\emptyset$ for any different $P,Q\in\bbP$, then $f$ is one-to-one on $\bigcup \{ {\rm int}(P): P \in \bbP \}$. It is also one-to-one on $C^\circ$, since so is $k$. Therefore $f^{-1}(z)$ has at most two points for every $z$ and so it is closed. Thus, $f$ is a continuous (see e.g. \cite[thm~5.2]{BC}) and so, it must be a homeomorphism. So, by way of contradiction, assume that there are different $P,Q\in\bbP$ for which $f[{\rm int}(P)]\cap f[{\rm int}(Q)]\neq\emptyset$. It follows from (\ref{eq3}) that either $P,Q\in\bbP_L$ or $P,Q\in\bbP_R$. We will assume that $P,Q\in\bbP_L$ and that $f$ is increasing on $P$, the other cases being similar. Let $a