%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}