\documentclass{rae}
\usepackage{graphicx}

\title{COMPOSITIONS OF DARBOUX-LIKE FUNCTIONS}
\author {Kenneth R. Kellum\\Department of Mathematics and Computer Science\\San 
Jos\'{e}
State University\\ San Jos\'{e}, California  95192-0103\\kellum@sjsumcs.sjsu.edu}


\date{}

\keywords{almost continuous function, Darboux function, connectivity function}

\MathReviews{Primary 26A15; Secondary 54C30}

\markboth{Kenneth R. Kellum}{Compositions of Darboux-Like Functions}

\newtheorem{example}{Example}
\newtheorem{theorem}{Theorem}
\newtheorem{lemma}{Lemma}
\def\proof{\noindent{\sc Proof.} }


\begin{document}\maketitle
\begin{abstract} 
An example is given of a real connectivity function which is not the composition of any finite
collection of almost continuous functions.  We also investigate conditions under which the
composition of two real  Darboux functions can be continuous. 
\end{abstract} 

In \cite{K} I asked if every Darboux function from $\Re$ to $\Re$ is the composition of two
almost continuous functions.  In the present note it is shown that the answer is ``no''.  In fact,
there exits a connectivity function which cannot be written as the composition of finitely many
almost continuous functions.  This example is not particularly difficult.  I think we have
overlooked this example until now because no one expected it to exist.  The composition of
almost continuous functions can be very nasty.  Our function which is not such a composition is
as well behaved as a Darboux function can be and not be almost continuous.

Natkaniec \cite{N}  has shown that if  $f$ is Darboux and nasty--that is,  $f^{-1}(x)$  is c-dense
for every $x$, then  $f$  is the composition of two almost continuous functions.  To look at the
other extreme, we need a Darboux function which is not almost continuous but as nice as
possible.  Since a Darboux function of Baire class 1 is almost continuous \cite{B}, the function
we want must be totally discontinuous on some perfect set.  Jones and Thomas \cite{JT}  give an
example of a function which is connectivity, continuous on the complement of the Cantor set, but
not almost continuous.  Our example is a simple modification of the Jones and Thomas example.

A function $f:\Re \to \Re$ is {\it almost continuous} if, given a closed set $K \subset \Re^2$ such
that $graph(f) \cap K = \emptyset$, there exists a continuous function  $g:\Re \to \Re$ such that
$graph(g) \cap K = \emptyset$.  $f$ is {\it Darboux} if $f(C)$ is connected whenever $C$ is
connected.  $f$ is a {\it connectivity function} if $graph(f|_C)$ is connected whenever $C$ is
connected.  In this setting, an almost continuous function is a connectivity function and a
connectivity function is Darboux.  The reader interested in learning about these classes of
functions should see the excellent survey articles \cite{GN}, \cite{N} and \cite{BC}.

We begin with two technical lemmas.  The letter $I$ denotes the closed unit interval 
$[0,1]$.

\begin{lemma}  Suppose $f:I \to I$, $U \subset I$ is an open interval, $f^{-1}(U)$ is open and
not connected, $f$ is continuous on each component of $f^{-1}(U)$ and  
$f$ is either increasing on every component of $f^{-1}(U)$ or is decreasing on all of them.   
Then $f$ cannot be almost continuous.
\end{lemma}

\proof\footnote{I thank the referee and Professors Jack Brown and 
Krzysztof Ciesielski  for pointing out errors in earlier versions of this proof.}
If $f$ is not Darboux, we are done, so we may assume that it is.

Assume $f$ is increasing on components of $f^{-1}(U)$. Choose a closed interval
$[p,q] \subset U$.  Pick two components $[a,b]$ and $[c,d]$ of $f^{-1}([p,q])$.  
The closed set 
$K_1 = (I \times [p,q]) \cup (\{a,c\} \times [q,1]) \cup (\{b,d\} \times
[0,p])$ intersects the graph of every continuous function from $I$ into itself. 

%\begin{figure}[htb]
%    \centering
%    \includegraphics[width=\textwidth]{blocking.eps}
%    \caption{The closed set K}
%    \label{fig:blocking}
%  \end{figure}


For each component $(u,v)$ of  $f^{-1}(U)$, we will remove a ``tube''  from $K_1$.
%(See Figure 1.)
Since $f(u)$ and $f(v)$ cannot be in $U$ and $f([u,v])$ is connected, it 
follows that $f((u,v)) = U$.
Let $s$
and $t$ denote the unique points in $(u,v)$ such that  $f(s) = p$ and $f(t) =q$.  
Choose
$\epsilon >0$ so that $u < s-\epsilon < s+\epsilon < t-\epsilon < t+ \epsilon < v$.  Let 
$V_{u,v}$ be the
union of all open line segments $(x-\epsilon, x+\epsilon) \times \{f(x)\}$ centered at points
$(x,f(x))$, where $s \leq x \leq t$.  Finally, let  $K = K_1 \setminus \bigcup V_{u,v}$, where the
union is taken over all components of $f^{-1}(U)$.  $K$ is a closed 
set, $K \cap graph(f) =
\emptyset$ and $K$ intersects the graph of every continuous function from $I$ into itself.  Thus,
$f$ is not almost continuous.  
\medskip


\begin{lemma} Suppose $f:I \to I$ and that there exist nondegenerate closed  intervals, $[a,b]$,
$[c,d]$, $[p,q]$ and $[u,v]$ such that $f^{-1}([p,q]) = [a,b]$, $f^{-1}([u,v]) = [c,d]$, 
$[a,b]\cap [c,d] = \emptyset$,  $f$ is continuous $[a,b] \cup [c,d]$, $f$ is increasing on $[a,b]$
and is decreasing on $[c,d]$.  Then $f$ is not almost continuous.
\end{lemma}

\proof  Choose  $x$ and $y$ such that $a < x < b$ and $c<y<d$.  Then
$(I \times [p,q]) \cup (\{x\} \times [q,1]) \cup (\{x\}\times [0,p]) \cup  (I \times [u,v]) \cup
(\{y\} \times [v,1]) \cup (\{y\}\times [0,u])$ is a closed set which intersects the graph of every
continuous function from $I$ into $I$.  We can now remove a tube from $I \times [p,q]$ above
$[a,b]$ and a tube from   $I \times [u,v]$ above $[c,d]$ and obtain a closed set with misses
$graph(f)$ but which  intersects the graph of every continuous function from $I$ into $I$.  This
completes the proof. 
\medskip


Let $\mathcal {C}$ denote the Cantor middle-thirds set.  By a {\it complementary interval} we 
mean the closure of a component of $I \setminus
\mathcal{C}$.  We define a function  $j:I \to I$ as follows.  For each complementary interval
$[a,b]$,  choose  $j|_{[a,b]}$ to be continuous and increasing  with $j(a)=0$ and $j(b)=1$ .  Let
$\mathcal {C^\circ}$ denote the points of $\mathcal{C}$ which are not endpoints of complementary
intervals.  If  $x \in \mathcal {C^\circ}$, let $j(x) = 0$.  It follows from Lemma 1 that $j$ is not almost
continuous.  Note that  $j$ is a connectivity function.

\begin{example}:  The function $j$  is not the composition of any finite
 collection of  almost continuous
functions from $I$ to $I$.
\end{example}

{\sc proof}  Assume that it is.  Let  $j = f_n \circ \dots \circ f_1$, where  $f_i$ is almost
continuous. 

Consider some complementary interval,  $[a,b]$.  Since  $j$ is one-to-one on $[a,b]$, $f_1$ must
also be one-to-one there.  Since a one-to-one Darboux function on an interval is continuous, it
follows that $f_1$  is monotone increasing or decreasing on $[a,b]$.  Thus,  $f_1([a,b])$  is a
closed interval,
and, by the same reasoning,  $f_2$  is one-to-one and continuous on  $f_1([a,b])$, 
$f_3$  is one-to-one and continuous on  $f_2 \circ f_1([a,b])$, and so on. 

Let $\mathcal {P}_0$ denote the set of complementary intervals.  For $i=1,\dots,n$, let $\mathcal
{P}_i = \{f_i(P) : P \in \mathcal{P}_{i-1}\}$.   Each member of  $\mathcal {P}_{i-1}$ is mapped
homeomorphically by $f_i$ onto a member of $\mathcal {P}_i$. 

Suppose  $P,Q \in \mathcal {P}_1$.  Then $f_n \circ \dots \circ f_2$ must map both $P$ and $Q$
homeomorphically onto $I$.  The endpoints of these intervals, and only the endpoints, are mapped
to $0$ and $1$. Thus $P \cap Q$ is either empty or is a common endpoint.  Similarly, for each
$i$, the interiors of  $\mathcal {P}_i$ are pairwise disjoint.

Suppose  $x \in \mathcal {C^\circ}$.  Since $j(x)=0$, and interiors of members of $\mathcal {P}_i$
map onto interiors of $\mathcal {P}_{i+1}$, $f_1(x)$ is not in the interior of any member of
$\mathcal {P}_1$, $f_2 \circ f_1(x)$ is not in the interior of any member of $\mathcal {P}_2$, 
and so on.  Thus, for each $P \in \mathcal{P}_{i+1}$, $f_i^{-1}(int(P))$ is open, in fact, is the
union of interiors of elements of  $\mathcal{P}_{i}$.

Now, we will prove by induction that for each $i$ with $0< i \leq n$, no two members of
$\mathcal{P}_{i-1}$ are mapped by $f_i$ onto the same member of $\mathcal{P}_{i}$ and that
$f_i$ is either increasing on every member of $\mathcal{P}_{i}$ or is decreasing on all of them.

  

First, if  $f_1([c,d]) = f_1([a,b])$, since $f_n \circ
\dots \circ f_1(d) = f_n \circ \dots \circ f_1(b) =  1$, we must have $f_1(b)=f_1(d)$. Thus, if two
or more elements of  $\mathcal{P}_{0}$ map to the same element of $\mathcal{P}_{1}$, $f_1$
is increasing on all or is decreasing on all of these.  By Lemma 1, this is impossible.
Now, by Lemma 2, $f_1$ cannot be increasing on one element of $\mathcal{P}_{0}$ and
decreasing on another.


Assume now that we have the desired conditions for all $i$ with $0<i \leq k-1$.  Then $f_{k-1} \circ
\dots \circ f_1$ increases on all members of $\mathcal{P}_{0}$ or decreases on all of them.  If
$f_k([c,d]) = f_k([a,b])$, we can reason as above to conclude that $f_k(c) = f_k(a)$.  Lemma 1
then rules this possibility out.  Apply Lemma 2 again to complete the induction.

By the above argument, $f_i$ induces a one-to-one
correspondence between   $\mathcal{P}_{i}$ and $\mathcal{P}_{i+1}$.  Since
$\mathcal{P}_{n}$ has $[0,1]$ as its only element, we have a
contradiction and the proof is complete.

\medskip
{\bf Remark.}  Another way we could obtain a contradiction in the proof of Example 1
is as follows.  First, choose $n$ to be the smallest integer such that $j$ is the
composition of $n$ almost continuous functions.  Since the composition of a
continuous function and an almost continuous function is almost continuous, none of 
$f_1, \dots ,f_n$ could be continuous.  Once we have that $f_1$ is increasing 
(decreasing) on each complementary interval, we can use an argument similar to the 
lemmas to show that $f_1$ is increasing (decreasing) on $I\setminus \mathcal{C}$.
The next step is a little delicate, but, using the fact that $f_1$ sends points of
$\mathcal{C}^\circ$ only to endpoints, we can then argue that $f_1$ is continuous.


\medskip
In the proof of Example 1, we made repeated use of the fact that if the composition of two real Darboux
functions is strictly monotone, then each factor is continuous and strictly
monotone. It is natural to ask: If $g \circ f$ is continuous, where $g$ and $f$ are Darboux, then
what can be said about $g$ and $f$?

\begin{example}  There exist $f$ Darboux and discontinuous and $g$ continuous such that 
$g \circ f$ is continuous.
\end{example}

Define $f:I \to I$ as follows:  Let $f(\mathcal {C^\circ}) = [0,1/2]$ so that for each $y \in [0,1/2]$, 
$f^{-1}(y)$ is c-dense in $\mathcal{C}$.  For each complementary interval $[a,b]$, define
$[a,b]$ to be continuous so that $f(a) = f(b) = 1/2$ and $f([a,b]) = [1/2,1]$.

Define $g$ by $g(x) = 0$ for $0 \leq x \leq 1/2$ and $g(x) = x - 1/2$ otherwise. 

\medskip

We will now show that all such examples will be similar to the above, in that the discontinuities of
one factor are crushed by intervals of constancy of the other.

By the {\it cluster set} of $f$ at $a$, denoted by $C(f,a)$, we mean the set of all $y$ such that
$(a,y)$
is a limit point of the set $\{(x,f(x)): x \neq a \}$.  The cluster sets of a Darboux function are
connected.

\begin{theorem} Suppose $f$ and $g$ are Darboux functions from $\Re$ into itself and that $g
\circ f$ is continuous at $a$.  If $f$ is discontinuous at $a$, then $g$ is constant on
$C(f,a)$.
\end{theorem}

\proof  Pick $y \in int(C(f,a))$, where $y \neq f(a)$.  Since $f$ is Darboux, there exists a
sequence
$\{x_n\}$ converging to $a$ such that $f(x_n) = y$.  By hypothesis, $g \circ f(x_n)$ converges to
$g \circ f(a)$.  Since $g \circ f(x_n) = g(y)$, we must have $g(y) = g \circ f(a)$.  Since $g$ is
constant on $int(C(f,a))$ and is Darboux, the result follows.


\begin{theorem}  Suppose $f$ and $g$ are Darboux functions from $\Re$ into itself and that 
$g \circ f$ is continuous at $a$.  If $g$ is discontinuous from the left (right) at $f(a)$, then $f$
has a
local minimum (maximum) at $a$.  Thus, if $g$ is discontinuous from both sides at $f(a)$, then
$f$ is constant in a neighborhood of $a$.
\end{theorem}

\proof  Assume $g$ is discontinuous from the left at $f(a)$.  If $f$ does not have a local minimum
at $a$, there exists a sequence $\{x_n\}$
converging to $a$, where $f(x_n) < f(a)$.  Because $f$ is
Darboux, we may choose $\{x_n\}$ so that $\{f(x_n)\}$ converges to $f(a)$.  Since $g$ is Darboux, there exists an
increasing sequence $\{y_n\}$ converging to $f(a)$ where $g(y_n) =y$ for each $n$ and $y \neq
g \circ f(a)$.  Passing to a subsequence, we may assume that $y_n > f(x_n)$.  Since $f$ is
Darboux, there exists $z_n$ between $x_n$ and $a$ such that $f(z_n) = y_n$.  But then, 
$g \circ f(z_n) = y$, contradicting the continuity of $g \circ f$ at $a$.

\medskip

\begin{theorem}  Suppose  $f$ and $g$ are Darboux functions from $\Re$ to $\Re$
where $f$ is onto.
 If $g \circ f$ is continuous, then $g$ is continuous.
\end{theorem}


\proof  Assume, on the contrary, that $g$ is discontinuous at $a$, from, say, the 
right.  For each $x$ such that $f(x)=a$, $f$ has a maximum at $x$.   

First, we claim that $f^{-1}((- \infty,a])$ is closed.  For, if $x$ was a limit 
point of  $f^{-1}((- \infty,a])$ and
$f(x) > a$, then $f$ would be discontinuous at $x$ and $[a,f(x)] \subset  C(f,x)$.  By
Theorem 1, $g$ would then be constant on $[a,f(x)]$, a contradiction.

Now, pick $z \in f^{-1}(a+ 1)$.  At least one of $f^{-1}((- \infty,a]) \cap 
(-\infty,z]$ 
and $f^{-1}((-\infty,a]) \cap [z,\infty)$ is nonempty, so assume   
$f^{-1}((-\infty,a]) \cap (-\infty,z] \neq \emptyset$.  Let 
$t = max (f^{-1}((-\infty,a]) \cap (-\infty,z])$.  Then $f(t) \leq a$.  If $f(t) < a$, we have a contradiction
because $f$ is Darboux.  If $f(t)=a$, we have a contradiction because $f$ must have 
a maximum
at $t$.  This completes the proof.

\medskip
The techniques of this paper do not seem to apply to the following question, due to Ceder \cite
{C}, which remains open.

Question.  Is every real Darboux function the composition of two (finitely many) connectivity
functions?

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\bibitem{GN} R.~G.~Gibson and T.~Natkaniec, {\it Darboux like functions}, Real Anal.
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\bibitem{JT} F.~B.~Jones and E.~S.~Thomas, Jr., {\it Connected $G_\delta$-graphs}, Duke
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\bibitem{N} T.~Natkaniec, {\it Almost continuity}, Real Anal. Exchange, {\bf 17} (1991-92),
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\end{thebibliography}


\enddocument
