\documentclass{rae}
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\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.} }


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\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,
$f(v_1)=h(0)<h(v_1)$ and $f(u_2)=h(1)>h(u_2)$
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 $a<b$, $\P$, $K$, and $h$ be as in
($\star$).
Then case (i) holds. Also, by (\ref{eq3}),
there exists $P'\in \bbP_R$, $P'\subset(a,b)$, such that
$f[P']\cap K=\emptyset$. Notice that $f[P']$ is below $K$,
as $f\restriction(a,b)$ is below $h$, and that
\begin{equation}\label{eq7}
\!\!\!%\!
\mbox{there is $Q'\in\bbP\setminus\{P'\}$, 
%\chKC between $a$ and $b$ 
$Q'\subset(a,b)$,
such
that $f[{\rm int}(P')]\cap f[{\rm int}(Q')]\neq\emptyset$. }%\!\!\!\!\!
\end{equation}

To see (\ref{eq7}) assume, to the contrary, that such a $Q'$ does
not exist. By (\ref{eq2}) we can choose $y\in f[P']\setminus
f[C]$. Pick $c\in P'$ with $\la c,y\ra$ above $f\restriction P'$
and note that if $f\restriction P'$ is increasing then
$[a,c]\times\{y\}$ cuts $f$, and if $f\restriction P'$ is
decreasing, then $[c,b]\times\{y\}$ cuts $f$. (Recall that $\P$ is
dense in $(a,b)$ and $K$ is above $f[P']$.) This contradiction
proves (\ref{eq7}).

Next, assume that $P'$ is below $Q'$ let $a'<b'$, $\P'$, $K'$, and
$h'$ satisfy ($\star$). Notice that $(a',b')\subset(a,b)$ and that
$f\restriction(a',b')$ must be above $h'$. Only two possibilities
remain.

\medskip

\noindent {\sc Case 1.} There is $P''\in \bbP$ such that
$P''\subset(a',b')$ and $f[P'']\cap (K\cup K')=\emptyset$.

Then, as in (\ref{eq7}), we can find another $Q''\in
\bbP\setminus\{P''\}$ such that $Q''\subset(a',b')$ and
$f[P'']\cap f[Q'']\neq\emptyset$. Using ($\star$) again, we can
find appropriate $a''<b''$, $\P''$, $K''$, and $h''$. The range of
$h''$ is strictly between $K$ and $K'$, and, since the domain of
$h''$ contains either 0 or 1, $h''$ cuts $f$, a contradiction.


\medskip

\noindent {\sc Case 2.} $f[P'']\subset K\cup K'$ for every
$P''\in \bbP$ with $P''\subset(a',b')$.

Note that $K\cap K'=\emptyset$. So $J=(e_{P'},e_P)$ is non-empty.
Take $y\in J\setminus f[C]$, which exists by (\ref{eq2}). Then
$[a',b']\times\{y\}$ cuts $f$, a contradiction.

This last contradiction completes the proof. \qed


\medspace

\begin{corollary}$\!\!\!${\bf .}  The function $k$ cannot  
be written as a finite
composition of connectivity functions from $\calI$ onto itself.
\end{corollary}

\proof
Assume $k$ is such a composition and let $n$ be the smallest
integer such that $k = f_n \circ \dots \circ f_1$, where each of
$f_1 \dots f_n$ is a connectivity function from $\calI$ onto
$\calI$. Since $f_n \circ \dots \circ f_2$ is Darboux, by Theorem
1, $f_1$ is continuous.  By Theorem 4 of \cite{KR}, $f_2 \circ f_1$
is connectivity, contradicting the definition of $n$.
\qed

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\bibitem{C}  J.~G.~Ceder, {\it On composition with connected functions}, 
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\bibitem{Ci:book} K.~Ciesielski,
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\bibitem{GNK} B.~D.~Garrett, D.~Nelms, and K.~R.~Kellum, {\it
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\bibitem{GN}
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http://www.math.wvu.edu/homepages/kcies/STA/STA.html})




\bibitem{K} K.~R.~Kellum, {\it Compositions of Darboux-like functions}, 
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\bibitem{KR} K.~R.~Kellum and H.~Rosen {\it Compositions of continuous
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\bibitem{TN} T.~Natkaniec, {\em On compositions and products of
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\end{thebibliography}


\end{document}

