\documentclass[11pt,fleqn]{amsart} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{corollary}[theorem]{Corollary} \newcommand{\R}{\mbox{\bbb R}} \newcommand{\rR}{\mbox{\bbb Q}} \newcommand{\N}{\mbox{\bbb N}} \newcommand{\iN}{\mbox{\bbc N}} \newcommand{\sbfd}{\mbox{\bba d}} \newcommand{\diam}{\hbox{\rm diam}} \newcommand{\conv}{\hbox{\rm conv}} \newcommand{\tcap}{\textstyle\bigcap} \newcommand{\tcup}{\textstyle\bigcup} \newcommand{\Int}{\hbox{\rm Int}} \begin{document} \setlength{\unitlength}{0.01in} \linethickness{0.01in} \begin{center} \begin{picture}(474,66)(0,0) \multiput(0,66)(1,0){40}{\line(0,-1){24}} \multiput(43,65)(1,-1){24}{\line(0,-1){40}} \multiput(1,39)(1,-1){40}{\line(1,0){24}} \multiput(70,2)(1,1){24}{\line(0,1){40}} \multiput(72,0)(1,1){24}{\line(1,0){40}} \multiput(97,66)(1,0){40}{\line(0,-1){40}} \put(143,66){\makebox(0,0)[tl]{\footnotesize Proceedings of the Ninth Prague Topological Symposium}} \put(143,50){\makebox(0,0)[tl]{\footnotesize Contributed papers from the symposium held in}} \put(143,34){\makebox(0,0)[tl]{\footnotesize Prague, Czech Republic, August 19--25, 2001}} \end{picture} \end{center} \vspace{0.25in} \setcounter{page}{147} \title[Fell-continuous selections]{Fell-continuous selections and topologically well-orderable spaces II} \author{Valentin Gutev} \address{School of Mathematical and Statistical Sciences, Faculty of Science, University of Natal, King George V Avenue, Durban 4041, South Africa} \email{gutev@nu.ac.za} \thanks{Valentin Gutev, {\em Fell-continuous selections and topologically well-orderable spaces II}, Proceedings of the Ninth Prague Topological Symposium, (Prague, 2001), pp.~147--153, Topology Atlas, Toronto, 2002; {\tt arXiv:math.GN/0204129}} \begin{abstract} The present paper improves a result of \cite{gutev:00d} by showing that a space $X$ is topologically well-orderable if and only if there exists a selection for $\mathcal{F}_2(X)$ which is continuous with respect to the Fell topology on $\mathcal{F}_2(X)$. In particular, this implies that $\mathcal{F}(X)$ has a Fell-continuous selection if and only if $\mathcal{F}_2(X)$ has a Fell-continuous selection. \end{abstract} \subjclass[2000]{Primary 54B20, 54C65; Secondary 54D45, 54F05} \keywords{Hyperspace topology, selection, ordered space, local compactness} \maketitle \section{Introduction} Let $X$ be a topological space, and let $\mathcal{F}(X)$ be the family of all non-empty closed subsets of $X$. Also, let $\tau$ be a topology on $\mathcal{F}(X)$ and $\mathcal{D}\subset \mathcal{F}(X)$. A map $f:\mathcal{D}\to X$ is a \emph{selection} for $\mathcal{D}$ if $f(S)\in S$ for every $S\in\mathcal{D}$. A map $f:\mathcal{D}\to X$ is a \emph{$\tau$-continuous} selection for $\mathcal{D}$ if it is a selection for $\mathcal{D}$ which is continuous with respect to the relative topology $\tau$ on $\mathcal{D}$ as a subspace of $\mathcal{F}(X)$. Two topologies on $\mathcal{F}(X)$ will play the most important role in this paper. The first one is the \emph{Vietoris topology} $\tau_V$ which is generated by all collections of the form \[ \langle\mathcal{V}\rangle = \left\{S\in \mathcal{F}(X) : S\cap V\ne \emptyset,\ V\in \mathcal{V},\ \mbox{and}\ S\subset \bigcup \mathcal{V}\right\}, \] where $\mathcal{V}$ runs over the finite families of open subsets of $X$. The other one is the \emph{Fell topology} $\tau_F$ which is defined by all basic Vietoris neighbourhood $\langle \mathcal{V}\rangle$ with the property that $X\setminus \bigcup\mathcal{V}$ is compact. Finally, let us recall that a space $X$ is \emph{topologically well-orderable} (see Engelking, Heath and Michael \cite{engelking-heath-michael:68}) if there exists a linear order ``$\prec$'' on $X$ such that $X$ is a linear ordered topological space with respect to $\prec$, and every non-empty closed subset of $X$ has a $\prec$-minimal element. Recently, the topologically well-orderable spaces were characterized in \cite[Theorem 1.3]{gutev:00d} by means of Fell-continuous selections for their hyperspaces of non-empty closed subsets. \begin{theorem}[\cite{gutev:00d}]\label{th:Fell-1} A Hausdorff space $X$ is topologically well-orderable if and only if $\mathcal{F}(X)$ has a $\tau_F$-continuous selection. \end{theorem} In the present paper, we improve Theorem \ref{th:Fell-1} by showing that one may use $\tau_F$-continuous selections only for the subset $\mathcal{F}_2(X)=\{S\in \mathcal{F}(X): |S|\le 2\}$ of $\mathcal{F}(X)$. Namely, the following theorem will be proven. \begin{theorem}\label{th:Fell-2} A Hausdorff space $X$ is topologically well-orderable if and only if $\mathcal{F}_2(X)$ has a $\tau_F$-continuous selection. \end{theorem} About related results for Vietoris-continuous selections, the interested reader is referred to van Mill and Wattel \cite{mill-wattel:81}. Theorem \ref{th:Fell-2} is interesting also from another point of view. According to Theorem \ref{th:Fell-1}, it implies the following result which may have an independent interest. \begin{corollary}\label{cr:Fell-3} If $X$ is a Hausdorff space, then $\mathcal{F}(X)$ has a $\tau_F$-contin\-uous selection if and only if $\mathcal{F}_2(X)$ has a $\tau_F$-continuous selection. \end{corollary} A word should be said also about the proof of Theorem \ref{th:Fell-2}. In general, it is based on the proof of Theorem \ref{th:Fell-1} stated in \cite{gutev:00d}, and is separated in a few different steps which are natural generalizations of the corresponding ones given in \cite{gutev:00d}. In fact, the paper demonstrates that all statements of \cite{gutev:00d} remain true if $\mathcal{F}(X)$ is replaced by $\mathcal{F}_{2}(X)$. Related to this, the interested reader may consult an alternative proof of Theorem \ref{th:Fell-2} given in \cite{artico-marconi:00} and based again on the scheme in \cite{gutev:00d}. \section{A reduction to locally compact spaces} In the sequel, all spaces are assumed to be at least Hausdorff. In this section, we prove the following generalization of \cite[Theorem 2.1]{gutev:00d}. \begin{theorem}\label{th:step1} Let $X$ be a space such that $\mathcal{F}_{2}(X)$ has a $\tau_F$-continuous selection. Then $X$ is locally compact. \end{theorem} \begin{proof} We follow the proof of \cite[Theorem 2.1]{gutev:00d}. Namely, let $f$ be a $\tau_F$-continuous selection for $\mathcal{F}_2(X)$ and suppose, if possible, that $X$ is not locally compact. Hence, there exists a point $p\in X$ such that $\overline{V}$ is not compact for every neighbourhood $V$ of $p$ in $X$. Claim that there exists a point $q\in X$ such that \begin{equation}\label{eq:point-q} q\ne p\ \mbox{and}\ f(\{p,q\})=p. \end{equation} To this purpose, note that there exists $F\in\mathcal{F}(X)$ such that $F$ is not compact and $p\notin F$. Then, $f^{-1}(X\setminus F)$ is a $\tau_F$-neighbourhood of $\{p\}$ in $\mathcal{F}_2(X)$, so there exists a finite family $\mathcal{W}$ of open subsets of $X$ such that $X\setminus \bigcup \mathcal{W}$ is compact and \[ \{p\}\in\langle \mathcal{W}\rangle\cap\mathcal{F}_2(X) \subset f^{-1}(X\setminus F). \] Then, $F\cap W\neq\emptyset$ for some $W\in \mathcal{W}$ because $F$ is not compact. Therefore, there exists a point $q\in F\cap \left(\bigcup \mathcal{W}\right)$. This $q$ is as required. Let $q$ be as in (\ref{eq:point-q}). Since $X$ is Hausdorff, $f(\{q\})\neq f(\{p,q\})$, and $f$ is $\tau_F$-continuous, there now exist two finite families $\mathcal{U}$ and $\mathcal{V}$ of open subsets of $X$ such that $X\setminus \bigcup \mathcal{U}$ is compact, $\{q\}\in \langle \mathcal{U}\rangle$, $\{p,q\}\in \langle \mathcal{V}\rangle$, and $\langle \mathcal{U}\rangle\cap \langle \mathcal{V}\rangle=\emptyset$. Then, \begin{equation}\label{eq:point} p\in V_p = \bigcap\{V\in \mathcal{V}: p\in V\} \subset X\setminus \bigcup \mathcal{U}. \end{equation} Indeed, suppose there is a point $\ell\in V_p\cap \left(\bigcup \mathcal{U} \right)$. Then, $\{\ell,q\}\in\langle \mathcal{U}\rangle$ because $\{q\} \in \langle \mathcal{U}\rangle$. However, we also get that $\{\ell,q\}\in\langle \mathcal{V}\rangle$ because $q\notin V$ for some $V\in \mathcal{V}$ implies $p\in V$, hence $\ell\in V_p\subset V$. Thus, we finally get that $\{\ell,q\}\in\langle \mathcal{U}\rangle\cap\langle \mathcal{V}\rangle$ which is impossible. So, (\ref{eq:point}) holds as well. To finish the proof, it remains to observe that this contradicts the choice of $p$. Namely $V_p$ becomes a neighbourhood of $p$ which, by (\ref{eq:point}), has a compact closure because $X\setminus \bigcup \mathcal{U}$ is compact. \end{proof} \section{A reduction to compact spaces} For a locally compact space $X$ we will use $\alpha X$ to denote the one point compactification of $X$. For a non-compact locally compact $X$ let us agree to denote by $\alpha$ the point of the singleton $\alpha X\setminus X$. In what follows, to every family $\mathcal{D}\subset \mathcal{F}(X)$ we associate a family $\alpha(\mathcal{D})\subset \mathcal{F}(\alpha X)$ defined by \[ \alpha(\mathcal{D}) = \{S\in \mathcal{F}(\alpha X): S\cap X\in \mathcal{D}\cup \{\emptyset\}\}. \] The following extension theorem was actually proven in \cite[Theorem 3.1]{gutev:00d}. \begin{theorem}\label{th:step2} Let $X$ be a locally compact non-compact space $X$, and $\mathcal{D}\subset \mathcal{F}(X)$. Then, $\mathcal{D}$ has a $\tau_F$-continuous selection if and only if $\alpha(\mathcal{D})$ has a $\tau_V$-continuous selection $g$ such that $g^{-1}(\alpha)=\{\{\alpha\}\}$. \end{theorem} \begin{proof} Just the same proof as in \cite[Theorem 3.1]{gutev:00d} works. Namely, if $f$ is a $\tau_{F}$-continuous selection for $\mathcal{D}$, we may define a selection $g$ for $\alpha(\mathcal{D})$ by $g(S)=f(S\cap X)$ if $S\cap X\neq\emptyset$ and $g(S)=\alpha$ otherwise, where $S\in \alpha(\mathcal{D})$. Clearly $g^{-1}(\alpha)=\{\{\alpha\}\}$ and, as shown in \cite[Theorem 3.1]{gutev:00d}, $g$ is $\tau_V$-continuous. If now $g$ is a $\tau_{V}$-continuous selection for $\alpha(\mathcal{D})$, with $g^{-1}(\alpha)=\{\{\alpha\}\}$, then $g(S\cup\{\alpha\})\in S$ for every $S\in \mathcal{D}$, so we may define a selection $f$ for $\mathcal{D}$ by $f(S)=g\left(S\cup\{\alpha\}\right)$, $S\in\mathcal{D}$. The verification that $f$ is $\tau_F$-continuous was done in \cite[Theorem 3.1]{gutev:00d}. \end{proof} \section{Special selections and connected sets} In what follows, to every selection $f:\mathcal{F}_2(X)\to X$ we associate an order-like relation ``$\prec_f$'' on $X$ (see Michael \cite{michael:51}) defined for $x\ne y$ by \[ x_1\prec_f x_2\ \mbox{iff}\ f(\{x_1,x_2\})=x_1. \] Further, we will need also the following $\prec_{f}$-intervals: \[ (x,+\infty)_{\prec_f}=\{z\in X:x\prec_f z\} \] and \[ [x,+\infty)_{\prec_f}=\{z\in X:x\preceq_f z\}. \] Now, we provide the generalization of \cite[Theorem 4.1]{gutev:00d} for the case of $\mathcal{F}_{2}(X)$. \begin{theorem}\label{th:connected} Let $X$ be a space, $a\in X$, and let $A\in \mathcal{F}(X)$ be a connected set such that $|A|>1$ and $a\in A\cap \overline{X\setminus A}$. Also, let $f:\mathcal{F}_{2}(X)\to X$ be a $\tau_V$-continuous selection for $\mathcal{F}_2(X)$. Then, $f^{-1}(a)\neq\{\{a\}\}$. \end{theorem} \begin{proof} Suppose, on the contrary, that $f^{-1}(a)=\{\{a\}\}$. By hypothesis, there exists a point $b\in A$, with $b\ne a$. Since $f$ is $\tau_{V}$-continuous, $f(\{a,b\})=b$ and $a\in \overline{X\setminus A}$, we can find a point $c\in X\setminus A$ such that $f(\{b,c\})=b$. Then, $B=A\cap (c,+\infty)_{\prec_{f}}$ is a clopen subset of $A$ because $B=A\cap [c,+\infty)_{\prec_{f}}$, see \cite{michael:51}. However, this is impossible because $b\in A\setminus B$, while $a\in B$. \end{proof} \section{A further result about special selections} Following \cite{gutev:00d}, we shall say that a point $a\in X$ is a \emph{partition} of $X$ if there are open subset $L,R\subset X\setminus \{a\}$ such that $\overline{L}\cap\overline{R}=\{a\}$ and $L\cap R=\emptyset$. We finalize the preparation for the proof of Theorem \ref{th:Fell-2} with the following result about special Vietoris continuous selections and partitions which generalizes \cite[Theorem 5.1]{gutev:00d}. \begin{theorem}\label{th:partit} Let $X$ be a compact space, $f$ a $\tau_V$-continuous selection for $\mathcal{F}_2(X)$, and let $a\in X$ be a partition of $X$ such that $f^{-1}(a)=\{\{a\}\}$. Then, $X$ is first countable at $a$. \end{theorem} \begin{proof} By definition, there are open sets $L,R\subset X\setminus\{a\}$ such that $\overline{L}\cap\overline{R}=\{a\}$ and $L\cap R=\emptyset$. Hence, both $L$ and $R$ are non-empty. Take a point $\ell_0\in L$. Then, by hypothesis, $f(\{\ell_0,a\})=\ell_0$. Since $f$ is $\tau_V$-continuous, this implies the existence of a neighbourhood $L_0\subset L$ of $\ell_0$ and a neighbourhood $V_0$ of $a$ such that \[ L_0\cap V_0 = \emptyset \mbox{ and } f(\langle \{L_0,V_0\}\rangle\cap\mathcal{F}_2(X))\subset L_0. \] Since $a\in\overline{R}$, there exists a point $r_0\in V_0\cap R$. Observe that $f(\{a,r_0\})=r_0\in V_0$. Hence, just like before, we may find a neighbourhood $R_0\subset R\cap V_0$ of $r_0$ and a neighbourhood $W_0\subset V_0$ of $a$ such that \[ R_0\cap W_0 = \emptyset \mbox{ and } f(\langle \{R_0,W_0\}\rangle\cap\mathcal{F}_2(X))\subset R_0. \] Thus, by induction, we may construct a sequence $\{\ell_n:n<\omega\}$ of points of $L$, a sequence $\{r_n:n<\omega\}$ of points of $R$, and open sets $L_n,V_n,R_n,W_n\subset X$ such that \begin{equation}\label{eq:ln} \begin{array}{l} \ell_n\in L_n,\\ a\in V_n,\\ L_n\cap V_n=\emptyset \mbox{ and}\\ f(\langle \{L_n,V_n\}\rangle \cap\mathcal{F}_2(X))\subset L_n, \end{array} \end{equation} \begin{equation}\label{eq:rn} \begin{array}{l} r_n\in R_n,\\ a\in W_n,\\ R_n\cap W_n=\emptyset \mbox{ and}\\ f(\langle \{R_n,W_n\} \rangle \cap\mathcal{F}_2(X))\subset R_n, \end{array} \end{equation} and \begin{equation}\label{eq:lr} \begin{array}{l} {V_{n+1}}\subset W_n\subset V_n,\\ L_{n+1}\subset L\cap W_n \mbox{ and}\\ R_{n}\subset R\cap V_n. \end{array} \end{equation} Since $X$ is compact, $\{\ell_{n}:n<\omega\}$ has a cluster point $\ell$, and $\{r_{n}:n<\omega\}$ has a cluster point $r$. We claim that $\ell=r$. Indeed, suppose for instance that $\ell\prec_{f} r$ (the case $r\prec_{f} \ell$ is symmetric). Then, there are disjoint open sets $U_{\ell}$ and $U_{r}$ such that $\ell\in U_{\ell}$, $r\in U_{r}$, and $x\prec_{f} y$ for every $x\in U_{\ell}$ and $y\in U_{r}$, see \cite{gutev-nogura:01a}. Next, take $\ell_{n}\in U_{\ell}$ and $r_{m}\in U_{r}$ such that $n>m$. Then, we have $\ell_{n}\prec_{f} r_{m}$. However, by (\ref{eq:ln}), (\ref{eq:rn}) and (\ref{eq:lr}), we get that $\{r_{m},\ell_{n}\}\in \langle\{R_{m},W_{m}\}\rangle\cap\mathcal{F}_2(X)$, and therefore $f(\{r_{m},\ell_{n}\})=r_{m}$. This is clearly impossible, so $\ell=r$. Having already established this, let us observe that $b=\ell=r$ implies $b\in \overline{L}\cap \overline{R}$ because $\ell\in \overline{L}$ and $r\in\overline{R}$. However, $\overline{L}\cap \overline{R}=\{a\}$ which finally implies that $b=a$. We are now ready to prove that, for instance, $\{W_n:n<\omega\}$ is a local base at $a$. To this end, suppose if possible that this fails. Hence, there exists an open neighbourhood $U$ of $a$ such that $W_n\setminus U\ne\emptyset$ for every $n<\omega$. Next, whenever $n<\omega$, take a point $t_n\in W_n\setminus U$. Since $X$ is compact, $\{t_n:n<\omega\}$ has a cluster point $t\not\in U$. Then, $t\prec_{f} a$ and, as before, we may find disjoint open sets $U_{t}$ and $U_{a}$ such that $t\in U_{t}$, $a\in U_{a}$, and $x\prec_{f} y$ for every $x\in U_{t}$ and $y\in U_{a}$. Next, take $t_{n}\in U_{t}$ and $r_{m}\in U_{a}$ such that $n>m$. Then, $t_{n}\prec_{f} r_{m}$, while, by (\ref{eq:rn}) and (\ref{eq:lr}), $r_{m}\prec_{f} t_{n}$ because $\{r_{m},t_{n}\}\in \langle\{R_{m},W_{m}\}\rangle\cap\mathcal{F}_2(X)$. The contradiction so obtained completes the proof.~\end{proof} \section{Proof of Theorem \ref{th:Fell-2}} In case $X$ is a topologically well-orderable space, we may use Theorem \ref{th:Fell-1}. Suppose that $\mathcal{F}_2(X)$ has a $\tau_F$-continuous selection. If $X$ is compact, then Theorem~\ref{th:Fell-2} is, in fact, a result of van Mill and Wattel \cite{mill-wattel:81}. Let $X$ be non-compact. By Theorem \ref{th:step1}, $X$ is locally compact. Then, by Theorem \ref{th:step2}, $\mathcal{F}_2(\alpha X)$ has a $\tau_V$-continuous selection $f$ such that $f^{-1}(\alpha)=\{\{\alpha\}\}$. Relying once again on the result of \cite{mill-wattel:81}, $\alpha X$ is a linear ordered topological space with respect to some linear order ``$<$'' on $\alpha X$. It now suffices to show that there exists a compatible (with the topology of $\alpha X$) linear order ``$\prec$'' on $\alpha X$ such that $\alpha$ is either the first or the last element of $\alpha X$, see \cite[Lemma 4.1]{engelking-heath-michael:68}. We show this following precisely the proof of Theorem \ref{th:Fell-1} in \cite{gutev:00d}. Namely, let \[ L = \{x\in \alpha X:x<\alpha\}\ \mbox{and}\ R = \{x\in \alpha X:\alpha< x\}. \] Note that $L,R\subset \alpha X\setminus \{\alpha\}=X$ are open subsets of $\alpha X$. In case one of these sets is also closed, the desired linear order ``$\prec$'' on $\alpha X$ can be defined by exchanging the places of $L$ and $R$. Namely, by letting for $x,y\in \alpha X$ that $x\prec y$ if and only if \[ \begin{array}{l} x,y\in \overline{L}\ \mbox{and}\ x