\documentclass{rae}
\usepackage{amsfonts}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{amsthm}

\title{EVERY ALMOST CONTINUOUS FUNCTION IS POLYGONALLY ALMOST CONTINUOUS}
\author{Piotr Szuca\thanks{The author is a student.
                           This paper will be a part of his Master Thesis,
                           written under the supervision
                           of Professor Tomasz Natkaniec.},
        Department of Mathematics,
        Gda{\'n}sk University, Wita Stwosza 57, 80-952 Gda{\'n}sk, Poland.}
\markboth{P.~Szuca}{Every AC function is PAC}
\date{}

\keywords{Darboux functions, almost continuous functions,
          polygonal functions, polygonally almost continuous functions.}
\MathReviews{26A15.}

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\newtheorem{theorem}{Theorem}
\newtheorem{corollary}{Corollary}

\newcommand{\cl}[1]{{\rm cl(}#1{\rm )}}

\newcommand{\reals}{{\mathbb R}}
\newcommand{\mathI}{{\mathbb I}}
\newcommand{\AC}{{\rm AC}}
\newcommand{\PAC}{{\rm PAC}}
\newcommand{\Darb}{{\rm D}}
\newcommand{\R}{{\cal R}}

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\begin{document}
\maketitle

\begin{abstract}
We show that every almost continuous function $f\colon\mathI\to\reals$
is also polygonally almost continuous. This solves a problem
posed by Agronski, Ceder and Pearson (see \cite{ACP}).
\end{abstract}

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\section{Preliminaries}

By $\reals$ we denote the set of all reals,
by $\mathI$ we denote the interval $[0,1]$.
For every set $A$, by $\cl{A}$ we will denote closure of $A$.

We will consider following classes of functions
from the interval $\mathI$ to $\reals$:
\begin{description}
\item[AC]      Function $f\colon\mathI\to\reals$ is almost continuous ($\AC$)
               if whenever $U\subset\mathI\times\reals$ is an open set
               containing the graph of $f$, then $U$ contains
               the graph of a continuous function $g\colon\mathI\to\reals$.
\item[PAC]     Function $f\colon\mathI\to\reals$ is
               polygonally almost continuous ($\PAC$)
               if whenever $U\subset\mathI\times\reals$ is an open set
               containing the graph of $f$, then $U$ contains
               the graph of a polygonal (piecewise linear continuous)
               function $h\colon\mathI\to\reals$ with all its vertices on $f$.
\item[D]       Function $f\colon\mathI\to\reals$ is Darboux ($\Darb$)
               if the image of $C\subset\mathI$ is connected in $\reals$
               whenever $C$ is connected in $\mathI$.
\end{description}

For properties of these and other Darboux-like classes
see e.~g.~the survey \cite{GN}.
In particular, it is known that $\AC\subset\Darb$.
Clearly every $\PAC$ function is $\AC$.
Recently Agronsky, Ceder and Pearson asked whether the opposite
implication holds (\cite{ACP}).
In this note we answer this question in the positive.
(We would like to thank Professor Kenneth Kellum
for drawing the autor's attention to this problem.
In particular, Kellum proved that every extendable function
as well as every $\AC$ function with dense graph
is $\PAC$ (private communication)).

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\section{The Result}

\begin{theorem}
    Every $\AC$ function $f\colon\mathI\to\reals$ is $\PAC$.
\end{theorem}

\proof

Let $f\colon\mathI\to\reals$ be $\AC$. Suppose, $f$ is not $\PAC$.

\medskip

For every $x\in\mathI$, define
$$H_x=\{h\colon [0,x]\to\reals\ |\ \hbox{\rm $h$ is polygonally continuous
                                         with all its vertices on $f$}\}.$$

Let $G\subset\mathI\times\reals$ be an open set such that $f\subset G$
and there does not exist a polygonal function
$h_1\in H_1$, $h_1\subset G$.

\medskip

Define:
\begin{itemize}
\item $E=\{(x,y)\in f\ |\ (\exists h_x\in H_x)\ h_x\subset G\}$;
\item $N=\{(x,y)\in f\ |\ (\neg\exists h_x\in H_x)\ h_x\subset G\}$.
\end{itemize}
Clearly $E\cup N=f$, $E\cap N=\emptyset$,
$(0,f(0))\in E$, and by the supposition $(1,f(1))\in N$.

\medskip

For $S_{(x,y)}$ being an open square with center $(x,y)$
let $3\cdot S_{(x,y)}$ denote the open square
with the center $(x,y)$ and with the diagonal 3 times
that of $S_{(x,y)}$.
For every $(x,y)\in f$ let $S_{(x,y)}$ be an open square
with the center $(x,y)$ such that:
\begin{itemize}
\item $3\cdot S_{(x,y)} \subset G$ for $x\in (0,1)$,\newline
      $3\cdot S_{(0,f(0))} \cap [0,+\infty)\times\reals \subset G$,\newline
      $3\cdot S_{(1,f(1))} \cap (-\infty,1]\times\reals \subset G$;
\item either $S_{(x,y)} \cap ([0,x)\times\reals) \cap f \subset E$
      or
      $S_{(x,y)}\cap ([0,x)\times\reals) \cap f \subset N$,
      for $x>0$;
\item either $S_{(x,y)} \cap ((x,1]\times\reals) \cap f \subset E$
      or
      $S_{(x,y)} \cap ((x,1]\times\reals) \cap f \subset N$,
      for $x<1$.
\end{itemize}
Such a $S_{(x,y)}$ exists for every $(x,y)\in f$.
Indeed, suppose for example, there exists $(x,y)\in f$ such that
for every $S_{(x,y)}\subset G$
there exist $x_1 \in [0,x)$ and $x_2 \in [0,x))$ such that
$$(x_1,f(x_1))\in E\cap S_{(x,y)}
  \hbox{\rm\ and }
  (x_2,f(x_2))\in N \cap S_{(x,y)}.$$
Now we can find $(x_1,f(x_1))\in E\cap S_{(x,y)}$
and $(x_2,f(x_2))\in N \cap S_{(x,y)}$, $x_1<x_2$.
But $(x_1,f(x_1))\rightarrow(x_2,f(x_2)) \subset S_{(x,y)} \subset G$,
with $\alpha\rightarrow\beta$ denoting the line segment linking
$\alpha$ and $\beta$ for every $\alpha\in\mathI\times\reals$,
$\beta\in\mathI\times\reals$, $\alpha<\beta$.
So, we can extend the polygonal function $h_{x_1}\in H_{x_1}$,
$h_{x_1}\subset G$
to a polygonal function $h_{x_2}\in H_{x_2}$, $h_{x_2}\subset G$,
contrary to $x_2\in N$.

\medskip

For every $(x,y)\in f$ let $R_{(x,y)}\subset S_{(x,y)}$ be an open
rectangular neighbourhood of $(x,y)$ fulfilling the following conditions
(with $x_l$ denoting $\inf\{a\ |\ (a,b)\in R_{(x,y)}\}$,
$x_r$ denoting $\sup\{a\ |\ (a,b)\in R_{(x,y)}\}$,
$y_l$ denoting $\inf\{b\ |\ (a,b)\in R_{(x,y)}\}$,
$y_u$ denoting $\sup\{b\ |\ (a,b)\in R_{(x,y)}\}$):
\begin{itemize}
\item[(1)] If $x>0$ and $S_{(x,y)} \cap ([0,x)\times\reals) \cap f \subset E$,
           then $f(x_l) \in (y_l,y_u)$.
\item[(2)] If $x<1$ and $S_{(x,y)} \cap ((x,1]\times\reals) \cap f \subset N$,
           then $f(x_r) \in (y_l,y_u)$.
\end{itemize}
Such a $R_{(x,y)}$ always exists, because $(x,y)$ is 
a left side limit point of $f$ for every $x\in (0,1]$
and  $(x,y)$ is a right side limit point of $f$ for every $x\in [0,1)$
($f$ is Darboux, so it satisfies the Young's condition, see e.~g.~\cite{GN}).

\medskip

Note also that $(x_l,f(x_l))$ is a right side limit point of $f$,
so if $R_{(x,y)} \cap ([0,x)\times\reals) \cap f \subset E$ 
then (from (1)) for every $a\in (x_l,x) \cap \mathI$ there exists
$b<a$ such that $(b,f(b)) \in E \cap R_{(x,y)}$.

Because $(x_r,f(x_r))$ is a left side limit point of $f$,
if $R_{(x,y)} \cap ((x,1]\times\reals) \cap f \subset N$ 
then (from (2)) for every $a\in (x_l,x_r) \cap \mathI$
there exists $c>a$ such that $(c,f(c)) \in N \cap R_{(x,y)}$,
so $R_{(x,y)} \cap ((a,x_r)\times\reals) \cap f \not\subset E$.

Now for every $R_{(x,y)}$ we have:
\begin{itemize}
\item[(A)] If $R_{(x,y)} \cap ((a,x)\times\reals) \cap f \subset E$
           for some $a\in (x_l,x) \cap \mathI$
           then $R_{(x,y)} \cap ([0,x)\times\reals) \cap f \subset E$,
           and there exists $b<a$ such that $(b,f(b)) \in E \cap R_{(x,y)}$.
\item[(B)] If $R_{(x,y)} \cap ((a,x_r)\times\reals) \cap f \subset E$
           for some $a\in (x_l,x_r) \cap \mathI$
           then $R_{(x,y)} \cap ((x,1]\times\reals) \cap f \subset E$,
           and there exists $b<a$ such that $(b,f(b)) \in E \cap R_{(x,y)}$.
\end{itemize}

\medskip

Let $H=\bigcup_{(x,y) \in f} R_{(x,y)}$. $H$ is open, $H \subset G$
and $f \subset H$.
So, there exists a continuous function $g\colon\mathI\to\reals$,
$g\subset H$.
Because the graph of $g$ is compact, there exists a finite family
of sets $\R\subset\{R_{(x,y)}\ |\ (x,y)\in f\}$
such that $g\subset\bigcup\R$.

$R_{(0,f(0))}\in\R$ and $R_{(1,f(1))}\in\R$.
Indeed, for every $0<x<1$ we have $3\cdot S_{(x,y)}\subset G$, so only
the square $S_{(0,f(0))}$ contains points with abscissa $0$,
and only the square $S_{(1,f(1))}$ contains points with abscissa $1$.
Moreover, since $\R$ is finite,
$$\sup\{x\in\mathI\ |\ (x,y)\in\bigcup(\R\backslash\{R_{(1,f(1))}\})\}<1.$$

Let
$$C=\{x\in\mathI\ |\ (\exists R\in\R)\ ((x,g(x))\in R \hbox{\rm\ and }
                 (\exists x_1\leq x)\ (x_1,f(x_1))\in E\cap R)\},$$
let $s=\sup C$.
Because $R_{(0,f(0))} \cap f \subset E$ and $(0,g(0))\in R_{(0,f(0))}$,
$C\not=\emptyset$ and $s$ is well defined.
Since $R_{(1,f(1))} \cap f \subset N$, $0<s<1$.

\medskip

$\R$ is finite and $g$ is continuous, so there exists
$R_{(a,b)}\in\R$ such that $(s,g(s))\in\cl{R_{(a,b)}}$
and there exists $x_1\leq s$ such that $(x_1,f(x_1))\in E\cap R_{(a,b)}$.

There exists also an open set $R_{(c,d)}\in\R$
such that $(s,g(s))\in R_{(c,d)}$.

Then $(s,g(s))\in\cl{R_{(a,b)}} \cap R_{(c,d)}$, so
$R_{(a,b)} \cap R_{(c,d)} \not=\emptyset$.
Note that $R_{(a,b)}\cup R_{(c,d)}\subset 3\cdot S_{(p,q)}$,
for $S_{(p,q)}$ being this square from $S_{(a,b)}$ and $S_{(c,d)}$
which has greater diameter.
Therefore we can connect every two points $\alpha$, $\beta$
of $R_{(a,b)}\cup R_{(c,d)}$
by the line segment $\alpha\rightarrow\beta$ whole contained
in $3\cdot S_{(p,q)}\subset G$.

\medskip

In $R_{(c,d)}$ we can find $x_2>s$ such that $(x_2,f(x_2))\in N$.
Indeed, suppose
$$R_{(c,d)}\cap ((s,1]\times\reals)\cap f \subset E.\eqno (\star)$$
$R_{(c,d)}$ is open and $g$ is continuous, so there exists
$t>s$ such that $(t,g(t))\in R_{(c,d)}$.
We have two cases:
\begin{enumerate}
\item If $t\leq c$, then $s<c$, and from ($\star$) we have
                         $R_{(c,d)}\cap ((s,c)\times\reals)\cap f \subset E$.
                         From (A) we have
                         $R_{(c,d)}\cap ([0,c)\times\reals) \cap f \subset E$
                         and
                         $(\exists w\leq t)\ (w,f(w))\in E\cap R_{(c,d)}$.
                         But now $t\in C$,
                         $\sup C\geq t>s$, a contradiction.
\item If $t>c$, then     from ($\star$) and (B) we have
                         $R_{(c,d)}\cap ((c,1]\times\reals)\cap f \subset E$
                         and
                         $(\exists w\leq t)\ (w,f(w))\in E\cap R_{(c,d)}$.
                         Now $t\in C$, $\sup C\geq t>s$, a contradiction.
\end{enumerate}

Now we have $x_1<x_2$, $(x_1,f(x_1))\in E\cap R_{(a,b)}$,
$(x_2,f(x_2))\in N\cap R_{(c,d)}$,
so we can extend the polygonal function $h_{x_1}\in H_{x_1}$,
$h_{x_1}\subset G$
via the line segment
$(x_1,f(x_1))\rightarrow(x_2,f(x_2))$
contained in $3\cdot S_{(p,q)} \subset G$
to a polygonal function $h_{x_2}\in H_{x_2}$,
$h_{x_2}\subset G$.
Thus we have $(x_2,f(x_2))$ belongs to $E$ rather than $N$.
This is a contradiction.
\qed

\bigskip

The following corollary gives a full answer to the question
from \cite{ACP}.

\begin{corollary}
    Given any $\varepsilon>0$ and any open neighbourhood $G$
    of an almost continuous function $f$,
    there exists a polygonal function $h$
    with the length of the longest line segment less than $\varepsilon$,
    such that $h\subset G$ and all vertices of $h$ belong to $f$.
\end{corollary}

\proof

It is easy to modify previous proof,
such that the length of every line segment
of polygonal function $h\subset G$ will be less than $\varepsilon$.
\qed

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\begin{thebibliography}{ACP}
\bibitem[ACP]{ACP} S.~J.~Agronsky, J.~G.~Ceder, T.~L.~Pearson,
                   {\it Some characterizations of Darboux Baire~1
                   functions},
                   Real Analysis Exchange {\bf 23}(2) (1997--98), 421--430.
\bibitem[GN]{GN}   R.~Gibson, T.~Natkaniec,
                   {\it Darboux like functions},
                   Real Analysis Exchange {\bf 22}(2) (1996--97), 492--533.
\end{thebibliography}

\end{document}
