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From: Miroslav Chlebik
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Subject: Re: 7 papers
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Date: Mon, 26 Oct 1998 14:44:01 +0000 (GMT)
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\documentclass{rae}
\usepackage{amsmath,amsthm,amssymb}
%\coverauthor{Stanis\l aw Wro\' nski}
%\covertitle{On A Family of Functions Defined by The Boundary Operator}
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\received{We need this date!}
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\firstpagenumber{1}
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\markboth{Stanis\l aw Wro\' nski}{...}
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\author{Stanis\l aw Wro\' nski, ...}
\title{ON A FAMILY OF FUNCTIONS DEFINED BY THE BOUNDARY OPERATOR}
%%%%%%Put Author's Definitions Below Here %%%%%%%%%%%
\def\pf{\noindent{\sc Proof.} }
\def\specpf#1.{\noindent{\sc #1.} }
\newcommand{\consecutive}
{\newtheorem{theorem}{Theorem}
\newtheorem {lemma} {Lemma}
}
\def\Fr{\mbox{\rm Fr\,}}
\def\card{\mbox{\rm card\,}}
\def\cl{\mbox{\rm cl\,}}
\def\Int{\mbox{\rm int\,}}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{document}
\maketitle
\begin{abstract}
For a topological space $X$, let $M(X,R)$ denote the family of all functions
$f\in R^{X}$ such that $f(\Fr(A))\subseteq \Fr(f(A)).$ Let $N(X,R)$ denote the
family of all continuous functions $f\in R^{X}$ such that $\card(f^{-1}(c))=1$
for each $c\in \left( \inf_{x\in X}f(x),\sup_{x\in
X}f(x)\right) $. We show that $M(X,R)=N(X,R)$ if $X$ is a connected and
locally connected Hausdorff space.
\end{abstract}
\bigskip
\consecutive
We adopt the following notation:
\begin{itemize}
\item $X$ -- a topological space with the family $O$ of open sets,
\item $R$ -- the set of real numbers with the natural topology,
\item $C(X,R)$ -- the family of continuous functions from $X$ into~$R$.
\end{itemize}
For $f\in R^{X}$ let $i_{f}$ and $s_{f}$ abbreviate $\inf\limits_{x\in X}f(x)
$ and $\sup\limits_{x\in X}f(x)$ respectively. By $\mbox{int\,}\left( A\right) $, $
\mbox{cl\,}\left( A\right) $ and $\mbox{Fr\,}\left( A\right) $ we denote the interior, the
closure and the boundary of the set $A\subseteq X$.
Let us define two classes of functions: $M(X,R)$ and $N(X,R)$ in the
following manner: \footnote{Professor Ryszard Pawlak observed that instead of the family of continuous
functions one can take the family of functions having the Darboux property.}
\begin{align*}
M(X,R)&=\left\{ f\in R^{X}:\forall _{A\subseteq X}f(\Fr(A))\subseteq
\Fr(f(A))\right\}\\
N(X,R)&=\left\{ f\in C(X,R):\forall _{c\in \left( i_{f,}s_{f}\right)
}\card(f^{-1}(c))=1\right\}
\end{align*}
Then we have the following:
\begin{theorem}
$N(X,R)\subseteq M(X,R)\subseteq C(X,R)$
\end{theorem}
\pf For an indirect proof of the first inclusion suppose that
$N(X,R)\setminus M(X,R)\neq \emptyset$. Thus, there exist
$f\in N(X,R)$ and $A\subseteq X$ such that $f(\Fr(A))\setminus \Fr(f(A))\neq
\emptyset$. Let $y\in f(\Fr(A))\setminus \Fr(f(A))$, then there exists an $
x\in \cl(A)\setminus \Int(A)$ such that $y=f(x)$ and therefore $y\in \cl(f(A))$
because $f$ is continuous. Since $y\notin \Fr(f(A))$ then necessarily
$y\in \Int(f(A))$. Since $x\notin \Int(A)$ and $f$ is continuous then
$y\in \cl(f(X\setminus A))$. This means that $\Int(f(A))$, being an open
neighbourhood of $y$, intersects the set $f(X\setminus A)$ i.e.
$\Int(f(A))\cap f(X\setminus A)\neq \emptyset$. Let $c\in
\Int(f(A))\cap f(X\setminus A)$. Then there exist distinct elements
$x^{\prime }\in A$ and $x^{\prime \prime }\in X\setminus A$ such
that $f(x^{\prime })=f(x^{\prime \prime })=c$. Since $c\in \Int(f(A))$
we have $c\in \left( i_f,s_f\right)$ which contradicts the assumption
that $f^{-1}(c)$ is a singleton.
To prove the second inclusion assume that $f\in M(X,R)$ and
$A\subseteq X$. Then $f(\cl(A))=f(A\cup \Fr(A))=f(A)\cup f(\Fr(A))\subseteq
f(A)\cup \Fr(f(A))=\cl(f(A))$ which means that $f\in C(X,R)$.
\qed
\medskip
It is worth noticing that the family $C(X,R)$ can be larger than $M(X,R)$.
Indeed, the function $f:R\rightarrow R$ such that $f(x)=\left| x\right| $
belongs to the set $C(R,R)\setminus M(R,R)$. If we take as $X$ the set of
the real numbers with the discrete topology then the characteristic function
of the interval $\left[ 0,\infty \right) $ belongs to $M(X,R)\setminus N(X,R)
$. However, if $X$ is assumed to be a connected and locally connected
Hausdorff space then one gets:
\begin{theorem}
$N(X,R)=M(X,R)$.\footnote{The Referee has remarked that it would be interesting to know
whether Theorem~2 holds for functions $f\in R^{X}$ where $X$ is only
assumed to be arcwise connected.}
\end{theorem}
To prove the above theorem we shall need several lemmas.
\begin{lemma}
If the sets $M$ and $N$ are disjoint, nonempty and closed in a connected and
locally connected space then the complement of $M\cup N$ has a component
whose closure intersects each of $M$ and~$N$.
\end{lemma}
\pf See \cite[p. 183]{j}. \qed
\begin{lemma}
For every open and connected $V\subseteq X$ and for any $a,b\in V$, if $
a\neq b$ then there exists an open and connected set $I$ such that $
I\subseteq V$ and $a\in I$, $b\in \Fr(I)$.
\end{lemma}
\pf Given $V\subseteq X$ and $a,b\in V$ are such as required
in the above lemma. Since the subspace $V$ must be locally connected
(see~\cite{d}) and $\left\{ a\right\}$, $\left\{ b\right\}$ are nonempty,
disjoint and closed in $V$ then, by Lemma~1,
the set $V\setminus \left\{ a,b\right\}$ in the subspace $V$ must have a component $U$
such that $\left\{a,b\right\} \subseteq \cl(U)$. Since the component $U$
is open and connected in $V$, it is also open and connected in~$X$.
Since $U$ is open and $\left\{ a,b\right\} \subseteq \cl(U)$ then
$\left\{ a,b\right\} \subseteq \Fr(U)$. Applying again the assumption
that $X$ is a locally connected Hausdorff space we get that there exists
a connected open set $W\subseteq X$ such that $a\in W\subseteq V$
and $b\notin W$. Since $a\in \Fr(U)$ then $W\cap U\neq\emptyset$.
Now, putting $I=W\cup U$ we obtain a connected open set such that
$a\in I\subseteq V$. We will show that $b\in \Fr(I)$. Indeed, on one hand
$b\in \Fr(U)\subseteq \cl(U)\subseteq \cl(I)$. On the other hand, $b\notin U$
because $b\in \Fr(U)$. Since at the same time $b\notin W$, we have
$b\notin W\cup U=I$.
This together with the fact proved before allows us to conclude
that $b\in \Fr(I)$. \qed
\medskip
In the lemmas that follow we shall assume that $f\in M(X,R)$.
\begin{lemma}
For every real number $c$, the set $\Fr(f^{-1}(c))$ has at most one element.
\end{lemma}
\pf For an indirect argument, let us assume that for some real number
$c$ there exist two distinct $x_1$, $x_2\in X$ such that
$\left\{ x_1,x_2\right\} \subseteq \Fr(f^{-1}(c))$. By the continuity of $f$
(see Theorem~1) it follows that $f^{-1}(c)$ is a closed set and therefore
$f(x_1)=f(x_2)=c$. Let $K$, $L$ be open sets such that $x_1\in
K$, $x_2\in L$, and $K\cap L=\emptyset$. From assumptions about the space $X$
it follows that there exist connected and open sets $A$, $B$ such that
$x_1\in A\subseteq K$, $x_2\in B\subseteq L$. Note that there must exist an
element $x_1^{\prime }\in A$ such that $f(x_1^{\prime })\neq c$. Indeed, in
the opposite case one gets that $A\subseteq f^{-1}(c)$ and consequently
$x_1\notin \Fr(f^{-1}(c))$. Analogously one can show the existence of an
element $x_2^{\prime }\in B$ such that $f(x_2^{\prime })\neq c$. Now we have
to consider the following four cases:
\begin{enumerate}
\item $f(x_1^{\prime })c$
\item $f(x_1^{\prime })>c$ i $f(x_2^{\prime })c$ i $f(x_2^{\prime })>c$
\item $f(x_1^{\prime })c$ then by the continuity of $f$, it follows that the whole
$f$-image of some neighbourhood of $x_0$ lays strictly above $c$ which
contradicts the assumption that the space $X$ is connected. \qed
\begin{lemma}
For every $c\in \left( i_{f},s_{f}\right) $ the set $f^{-1}(c)$ is nonempty.
\end{lemma}
\pf It is an immediate consequence of the assumption that
the function $f$ is continuous and the space $X$ is connected. \qed
\bigskip
\specpf Proof of the Theorem 2.%
By Lemma~5 we need only to prove that for every $c\in \left(
i_{f},s_{f}\right) $, the set $f^{-1}(c)$ has at most one element. Suppose
the contrary, i.e.\ for some $c\in \left( i_{f},s_{f}\right) $ the set $
f^{-1}(c)$ has more than one element. From the assumptions it follows that
the set $f^{-1}(c)$ is a closed subset of $X$ distinct from the empty set
and from~$X$. Since the space $X$ is connected then $f^{-1}(c)$ can not be
open and therefore $\Fr(f^{-1}(c))\neq \emptyset $. Let $x^{\prime }\in
Fr(f^{-1}(c))\subseteq f^{-1}(c)$ and let $x^{\prime \prime }$ be an element
of $f^{-1}(c)$ which is distinct from $x^{\prime }$. Then, by Lemma~3, $
x^{\prime \prime }\in \Int(f^{-1}(c))$. Let $U$ and $V$ be connected sets
such that $x^{\prime }\in U,$ $x^{\prime \prime }\in V,$ $U\cap V=\emptyset $.
By Lemma~4, there exist $a,b\in U$ such that $f(a)