%REMARKS  ABOUT  CONNECTED  SPACES

\documentclass[12pt]{article}
\usepackage{epsf}

\newtheorem{theorem}{Theorem}
\newtheorem{lemma}{Lemma}
\newcommand{\thm}[2]{\begin{theorem}\label{#1}#2\end{theorem}}
\newcommand{\lem}[2]{\begin{lemma}\label{#1}#2\end{lemma}}

\def\proof{\noindent {\sc Proof. }}

\def\cl{{\rm cl}}
\def\real{{\bf R}}
\def\natural{{\bf N}}

 
\begin{document}

\setcounter{page}{181}

%\vspace*{-.3in}
\noindent{\large Q \&\ A in General Topology, Vol. 10 (1992) }

\bigskip

\begin{center}
{\Large REMARKS  ABOUT  CONNECTED  SPACES}

\bigskip\bigskip


by

\bigskip\bigskip

{\large KRZYSZTOF  CIESIELSKI}

\bigskip\bigskip

Department of Mathematics, West Virginia Univ., Morgantown, WV 26506.

\end{center}

\bigskip\bigskip

  
\begin{abstract}
The main result of the paper generalize Kuratowski's Theorem  
``if $X$ is a connected space,  $K\subset X$  is connected and
$S$ is a component of $X\setminus K$ then 
$X\setminus S$ is connected''  [Ku, Ch.V,46, III Theorem 5]  to the case when 
$K$ has finite number of components and $S$ consists of finite number of either
components or quasi components of $X\setminus K$.  Some examples are also
discussed.
\end{abstract}

\bigskip

In what follows we will use standard topological notation. For a topological
space $X$ and $A\subset X$ the symbol $A^c$ denotes the complement of $A$ with
respect to $X$;  i.e., $A^c=X\setminus A$.  A quasi component is said to be 
{\em proper}, if it is not a component.

\bigskip

\thm{th1}{Let $X$be a connected topological space, $n$ be a natural number,
and let $K\subset X$ have precisely $n$ components. If $C$ is a union of
finite number of components of $K^c$ and $Q$ is a union of finite number of
proper quasi components of $K^c$, then  $(C\cup Q)^c$ has at most $n$
components.}

\bigskip

In the proof we will need the following easy facts.

\pagebreak

\lem{lem1}{Let $Y$ be any topological space and let $k > 0$ be
a natural number.
\begin{itemize}
\item[(a)]  If $C_1,C_2,\ldots,C_k,C_{k+1}$ are different
nonempty components of $Y$ then there exists a nonempty clopen
set $W$ in $Y$ such that $W\cup C_i=\emptyset$ for all $i\leq
k$.

\item[(b)]  Let $Q_1,Q_2,\ldots,Q_k$ be quasi components of $Y$
and let 
$y\in Y\setminus(Q_1\cup Q_2\cup\ldots\cup Q_k)$.  Then there
exists a clopen subset $V$ of $Y$ containing $y$ and disjoint
from every $Q_i$.
\end{itemize}
}







\proof  
(a)  By [Ku, V,46, III Theorem 6] we can find $k+1$-element open partition  $\cal R$  of
$Y$.  Then, by the Pigeon Hole Principle, there is $U\in \cal R$ such that $C_i\cap
U=\emptyset$ for every  $i\leq k$, as every component can intersect at most one element of 
$\cal R$.

(b)  By the definition of quasi component for every $i=1,2,\ldots ,k$ there exists a clopen
set $V_i\subset Y$ containing $y$ and disjoint with $Q_i$.  Then $V=V_1\cap V_2\cap \ldots
\cap V_k$ works.

\bigskip

\noindent{\bf Proof of Theorem~1.}  
By way of contradiction let us assume that  $(C\cup Q)^c$ has more
than $n$ components.  Then $K$ is contained in at most $n$ components of $(C\cup Q)^c$
and,  by Lemma 1(a), there is a nonempty clopen subset $U$ of $(C\cup Q)^c$ such that
$U\cap K=\emptyset$.

Let $x\in U$. By Lemma 1(b), there exists a clopen subset $V$ of $K^c$ containing $x$ and
disjoint from $Q$.  Let us consider the space $Z=(U\cap V)\cup C\subset K^c$.  Then,
components forming $C$ are also components of $Z$ and  $x\in Z\setminus C$. Hence, $Z$ has
more components that $C$ and, by Lemma 1(a), there exists a nonempty clopen subset $W$ of
$Z$ such that $W\cap C=\emptyset$.

We show that $W$ is clopen in $X$.  Since $W$ and $W^c$ are nonempty, this will contradict
the assumption that $X$ is connected.

First notice that
\[
\cl(W)\subset \cl(V)\subset V\cup K\subset Q^c\ \mbox{ and }\ \cl(W)\subset C^c
\]
since $V\subset Q^c$ is clopen in $K^c$ and $W\subset C^c$ is closed in $Z=(U\cap V)\cup
C$.  Hence,

\[
\cl(W)\subset (V\cup K)\cap (C\cup Q)^c.
\]
Moreover, $U\subset K^c$  is clopen in $(C\cup Q)^c$.  So, $\cl(W)\subset \cl(U)\subset
U\cup C\cup Q\subset K^c$  and
\[
\cl(W)\subset (U\cup C\cup Q)\cap K^c.
\]
But the last two displays imply that $W\subset \cl(W)\subset U\cap V\subset Z$ and $W$ is
closed in $Z$.  Thus, $W=\cl(W)$.  So, $W$ is closed in $X$.

To prove that $W^c$ is closed in $X$,  let us first notice that

\begin{eqnarray*}
W^c &= & [Z\setminus W]\cup Z^c\\
    &= & [Z\setminus W]\cup [U\cap V]^c\\
    &= & [Z\setminus W]\cup U^c\cup V^c\\
    &= & [Z\setminus W]\cup [(C\cup Q)^c\setminus U]\cup [K^c\setminus V]
\end{eqnarray*}
But, by the fact that $W$, $U$, and $V$ are clopen in $Z$, $(C\cup Q)^c$, and $K^c$, 
respectively, we conclude that
\[
\cl[Z\setminus W]\cap W=\emptyset,
\]
\[
\cl[(C\cup Q)^c\setminus U]\cap W\subset \cl[(C\cup Q)^c\setminus U]\cap U=\emptyset,
\]
\[
\cl[K^c\setminus V]\cap W\subset \cl[K^c\setminus V]\cap V=\emptyset.
\]
Hence $\cl[W^c]\cap W=\emptyset$, i.e., $W^c$ is closed in $X$.
This finishes the proof of the theorem.

\bigskip



It is easy to see that in Theorem 1 the space $(C\cup Q)^c$ can have precisely $n$
components. For example, it is enough to take define $X\subset \real^2$  by $X =
([1,n]\times \{3\}) \cup  (\{1,2,\ldots ,n\}\times [0,3])$,  and choose  $K_i=\{i\}\times
(1,3)$ and $C\cup Q=\{S_1\}$,  where  $S_1=[1,n]\times \{3\}$.  For  $n=4$, 
 see Picture~1.  
It is also easy to modify this example to work in the case when  $C\cup Q$  has any
finite number $m$ of elements. For  $m=2$  the construction is suggested in Picture~2.

\bigskip

\begin{figure}[htb]
     \[
         \vbox to 1.46in{\hrule width 4.3in height 0pt depth 0pt \vfill
         \special{illustration pict1and2.eps scaled 700}}
     \]
%     \caption{The function $g(x)$}
\end{figure}





In order to obtain the conclusion of Theorem 1 we can remove from $K^c$ an entire proper
quasi component or finite number of its components. It seems to be natural to ask whether
the conclusion remains valid if we remove from a proper quasi component of $K^c$ an
infinite number of components. The answer for this question is negative. To see this
define  $X\subset \real^2$  by  $X=K\cup Q\cup\{\{1+(1/2)^n\}\times [0,1]\colon 
n\in\natural\}$,  where $K=[0,2]\times \{1\}$, $Q=P\cup \{(0,0)\}$, and  $P=\{[0,1]\times
\{(1/2)^n\}\colon   n>0,\ n\in\natural\}$.  
It is easy to see that $Q$ is a proper quasi
component of $K^c$ and that  $P^c$  is disconnected,  while $K$ is connected.  
(See Picture~3.)

\begin{figure}[htb]
     \[
         \vbox to 1.46in{\hrule width 4.3in height 0pt depth 0pt \vfill
         \special{illustration pict3.eps scaled 700}}
     \]
%     \caption{The function $g(x)$}
\end{figure}


Let us also notice that in the above simple examples the conclusion depends on the choice
of $K$ and $C\cup Q$.  This changes, when define $X$ as an indecomposable continuum (for
definition and construction see e.g. [HY, p.~139].)  In this case, we can choose an
arbitrary proper subset $K\subset X$ having any finite number $n$ of components and an
arbitrary nonempty set $C\cup Q$.  Then, $(C\cup Q)^c$ will have precisely $n$ components. 
For  $n=1$  it follows at once from Theorem 1.  For  $n>1$  it is the case, since $K^c$ is
connected and so,  $C\cup Q=K^c$,  i.e., $(C\cup Q)^c=K$.  The fact that $K^c$ is connected
can be deduced from the properties of composants of indecomposable continuum that can be
found in [HY] sec. 3-8. An alternative, direct proof of it goes as follows. Let
$K_1,K_2,\ldots,K_n$ be nonempty different components of $K$.  Then,  $\cl(K)=\cl(K_1)\cup
\ldots \cup \cl(K_n)\neq X$,  since otherwise $X$ would be a finite union of proper
subcontinua $\cl(K_1),\ldots ,\cl(K_n)$.  If  $U=X\setminus \cl(K)$  is disconnected,
separated by nonempty open sets $V_1$ and $V_2$ then,  by [Ku, V, 46 Thm 7 p. 134],  sets
$\cl(K)\cup V_i=\cl(K)\cup \cl(V_i)$,  for  $i=1,2$,  would have at most $n$ components and
each of these components would be a proper subcontinuum of $X$.  This would contradict
indecomposability of $X$.  Thus,  $U=X\setminus \cl(K)$  is connected.  But $U$ must be
dense in $X$, since otherwise the sets $\cl(K_1),\ldots,\cl(K_n)$  and  $\cl(U)$  would
form a forbidden decomposition of $X$.  Hence,  $U\subset K^c\subset X=\cl(U)$  and, by
connectedness of  $U$, $K^c$ is connected.

\smallskip %\bigskip

Theorem 1 and previous examples shows that the number of components of $(C\cup Q)^c$ is
bounded by number of components of $K$ and this upper bound is independent of the number
$m$  of components of $C\cup Q$.  However, within the limit given by Theorem 1, the number
of components of $(C\cup Q)^c$ can be determined by number $m$  of components of  $C\cup
Q$.  This is the case for a circle as can be seen from the theorem below.
 

\thm{th2}{Let $X$be a continuum.  The following properties are equivalent:

\begin{itemize}
\item[{\rm (1)}]	for every $K\subset X$ having a finite number of components and for every  $C\subset
K^c$  containing $m$ components of $K^c$ the space $C^c$ has $m$ components.

\item[{\rm (2)}]		for every $K\subset X$ having a finite number of components and for every component
$C$ of $K^c$ the space  $C^c$  is connected.

\item[{\rm (3)}]		for every  $2$-element set $K\subset X$ and for every component $C$ of $K^c$ the
space $C^c$  is connected.

\item[{\rm (4)}]		for every  $2$-element set $K\subset X$ the space $K^c$ is disconnected.

\item[{\rm (5)}]		$X$ is homeomorphic to a circle.

\end{itemize}
}

\proof   The implications  (1)$\Rightarrow$(2)$\Rightarrow$(3)  are obvious.

To prove  (3)$\Rightarrow$(4)  let us assume that  (4)  is false. Then for some  $2$-element set
$K\subset X$ we must have  $C=K^c$,  since $K^c$ is connected. Hence  $C^c=K$  is disconnected.

(4)$\Rightarrow$(5)  follows from the well known Moore's characterization of a circle (see [Ku] V,47
p. 180).

(5)$\Rightarrow$(1)  follows from the property of a circle $X$ that for every  $C\subset X$  the sets
$C$ and 
$C^c$  have the same number of components.

\vspace*{-.15in}

\begin{thebibliography}{22}
\bibitem[HY]{H}	J.G. Hocking, G.S. Young,{\em Topology}, Addison-Wesley Pub. Co., 1961.
\bibitem[Ku]{K}	K. Kuratowski, {\em Topology vol II}, Academic Press 1968.
\end{thebibliography}

\vspace*{-.1in}

\begin{flushright}
Received March 29, 1992. 

Revised May 30, 1992\vspace*{-.3in}
\end{flushright}

\end{document}
