%%% version 6.01.00 %% submitted %% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \documentstyle[12pt]{article} %-----------------DATE --------------------------- \date{} %----------------- T I T L E ------------------------ \title{Uniform quasi components, thin spaces and compact separation \thanks{The first and the second author were partially supported by a Research Grant of MURST. The third author was partially supported by the grant GA \v CR 201/97/0216.\endgraf AMS classification numbers: Primary 54C30, 54F55; %\endgraf Secondary 41A30, 54E35. \endgraf} } %----------------- A U T H O R S ------------------------ \author{Alessandro Berarducci \\ {\footnotesize Dipartimento di Matematica, Universit\`{a} di Pisa} \\ {\footnotesize Via Buonarroti 2,56127 Pisa, Italy} \and Dikran Dikranjan \\ {\footnotesize Dipartimento di Matematica e Informatica, Universit\`{a} di Udine} \\ {\footnotesize Via Zanon 6, 33100 Udine, Italy} \and Jan Pelant \\ {\footnotesize Institute of Mathematics, Czech Academy of Sciences} \\ {\footnotesize \v Zitna 25 11567 Praha 1, Czech Republic}} \headheight=0in \headsep = 0in \topmargin=0in \textheight=8.85in \textwidth=6.6in \oddsidemargin=0in \evensidemargin=0in \parindent=0.2in %--------------------ENVIRONEMENTS-------------------------------- \newtheorem{theorem}{Theorem}[section] \newtheorem{corollary}[theorem]{Corollary} \newtheorem{conjecture}[theorem]{Conjecture} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{problem}[theorem]{Problem} \newtheorem{notation}[theorem]{Notation} \newtheorem{em-example}[theorem]{Example} \newtheorem{em-def}[theorem]{Definition} \newtheorem{em-remark}[theorem]{Remark} \newtheorem{em-question}[theorem]{Question} \newtheorem{fact}[theorem]{Fact} \newenvironment{example}{\begin{em-example} \em }{ \end{em-example}} \newenvironment{definition}{\begin{em-def} \em }{ \end{em-def}} \newenvironment{remark}{\begin{em-remark} \em }{\end{em-remark}} \newenvironment{question}{\begin{em-question} \em }{\end{em-question}} \newenvironment{proof}{Proof.}{QED \bigskip} \newcommand{\R}{{\bf R}} \newcommand{\N}{{\bf N}} \newcommand{\Q}{{\bf Q}} \newcommand{\C}{{\bf C}} \newcommand{\restricted}[1]{_{| #1}} \newcommand{\ov}{\overline} \newcommand{\nin}{\not\in} \newcommand{\eps}{\varepsilon} \newcommand{\nbd}{neighbourhood } \newcommand{\CSP}{\mbox{\em CSP}} \newcommand{\St}{\mbox{\em St}} \begin{document} \maketitle \begin{abstract} We prove that every complete metric space $X$ that is thin (i.e., every closed subspace has connected uniform quasi components) has the compact separation property (for any two disjoint closed connected subspaces $A$ and $B$ of $X$ there is a compact set $K$ disjoint from $A$ and $B$ such that every neighbourhood of $K$ disjoint from $A$ and $B$ separates $A$ and $B$). The real line and all compact spaces are obviously thin. We show that a space is thin if and only if it does not contain a certain forbidden configuration. Finally we prove that every metric $UA$-space (see [BD1]) is thin. The $UA$-spaces form a class properly including the Atsuji spaces. \end{abstract} \section{Introduction} The class of the topological spaces $X$ having connected quasi components is closed under homotopy type, it contains all compact Hausdorff spaces (see \cite[Theorem 6.1.23]{E}) and every subset of the real line. Some sufficient conditions are given in \cite{GN} (in terms of existence of Vietoris continuous selections) and in \cite{CMP} (in terms of the quotient space $\Delta X$ in which each quasi-component is identified to a point), but an easily-stated description of this class does not seem to be available (see \cite{CMP}). The situation is complicated even in the case when all connected components of $X$ are trivial, i.e., when $X$ is {\em hereditarily disconnected} \cite{E}. (The term {\em totally disconnected} is used for spaces having trivial quasi components \cite{E}.) In these terms the question is to distinguish between hereditarily disconnected and totally disconnected spaces (examples to this effect go back to Knaster and Kuratowski \cite{KK}). The connectedness of the quasi component (i.e., the coincidence of the quasi component and the connected component) in topological groups is also a rather hard question. Although a locally compact space does not need to have connected quasi components \cite[Example 6.1.24]{E}, all locally compact groups have this property (this is an easy consequence of the well known fact that the connected component of a locally compact group coincides with the intersection of all open subgroups of the group \cite[Theorem 7.8]{HR}). Recently all countably compact groups were shown to have this property too (\cite{D3}, see also \cite{D2,D4}). Many examples of pseudocompact group where this property strongly fails in different aspects, as well as further information on quasi components in topological groups, can be found in (\cite{D1,D2,D4}, see also \cite{U} for a plane group with non-connected quasi component). Given a uniform space $X$ and a point $x \in X$ we have the following inclusions $$C_x(X) \subseteq Q_x(X) \subseteq Q^u_x(X),$$ where $C_x(X)$ denotes the connected component of $x$, $Q_x(X)$ the quasi component of $x$, and $Q^u_x(X)$ the uniform quasi component of $x$. \begin{definition} A uniform space $X$ is {\bf thin} if for every closed subset $Y$ of $X$ and every $y \in Y$, the uniform quasi component of $y$ in $Y$ is connected. \end{definition} By the above mentioned classical result (\cite[Theorem 6.1.23]{E}) the compact uniform spaces are thin (since the uniform quasi components coincide with the quasi components in this case). In this paper we study the thin spaces and our main result is establishing a separation property for every complete metric thin space. The relevant separation property is defined as follows. \begin{definition} We say that a uniform space $X$ has the {\bf compact separation property} (briefly $\CSP$), if for any two disjoint closed connected subspaces $A$ and $B$ there is a compact set $K$ disjoint from $A$ and $B$ such that every neighbourhood of $K$ disjoint from $A$ and $B$ separates $A$ and $B$ (consequently $K$ intersects every closed connected set which meets both $A$ and $B$, see Definition \ref{separated}). \end{definition} Note that for a locally compact space $X$, the $CSP$ can be equivalently expressed in the following simpler form: for any two disjoint closed uniformly connected subspaces $A$ and $B$ there is a compact set $K$ disjoint from $A$ and $B$ which separates $A$ and $B$. The real line, every zero-dimensional space and every compact space have $\CSP$. On the other hand the real plane $\R^2$ does not have $\CSP$: take $A, B$ to be two parallel lines. The main result of this paper, which we prove in section 4, is the following: \proclaim Theorem A. Every complete thin metric space has $\CSP$. The proof is based on a criterion for thinness given by Proposition \ref{new_prop}. We show that a space is thin if and only if it does not contain a certain forbidden configuration (that we call {\bf garland}, cf. Definition \ref{garland-configuration}). Another result, given in section 3, is that every metric $UA$-space (see Definition \ref{Def_UA}) is thin, so every complete metric $UA$-space has $CSP$. The $UA$-spaces are those uniform spaces where every continuous real-valued function can be approximated in a suitable way by uniformly continuous ones. They obviously comprise the compact spaces. The relevant definitions are as follows. \begin{definition}(\cite[Definition 2.1]{BD}) A function $f \colon X\to Y$ between two uniform spaces is {\bf uniformly approachable}, briefly $UA$, if for every compact set $K \subseteq X$ and every subset $M \subseteq X$, there exists a uniformly continuous $g \colon X \to Y$, called a $(K,M)$-approximation of $f$, such that $f = g$ on $K$ and $g(M) \subseteq f(M)$. \end{definition} This a modification of the notion originally introduced in \cite[Definition 4.1]{DP} (with the same name) in connection with the study of closure operators in the sense of \cite{DT,DP}. (The difference is that in \cite{DP} the compact set $K$ was assumed to be a single point.) The $UA$-functions have also been studied in \cite{BD2,BDP,BDP2,B,CD1,CD2}. Every uniformly continuous function is obviously $UA$, and it is easy to see that a $UA$ function $f$ must necessarily be continuous \cite[Lemma 4.2]{DP}. Moreover the $UA$-functions are closed under composition, so they define a category which sits between the category of uniform spaces with all continuous functions and the category of uniform spaces with all uniformly continuous functions. A characterization of the $UA$-functions is still missing, although a complete answer exists in the special case of polynomial maps $f \colon \R^n \to \R$: such an $f$ is $UA$ if and only if any pair of distinct fibers of $f$ are at positive distance \cite{BDP}. \begin{definition}\label{Def_UA}(\cite[Definition 2.4]{BD}) A uniform space $X$ is a {\bf $UA$-space} if every continuous $f \colon X \to \R$ is $UA$. \end{definition} Besides the compact spaces, the $UA$-spaces include the {\bf Atsuji spaces}, which are those metric spaces $X$ such that every continuous $f \colon X \to \R$ is uniformly continuous \cite{A1,A2}, and also the real line \cite[Proposition 3.5]{BD}. Other examples and general results on $UA$-spaces are given in \cite{BD,BD2}. We know for instance that every linear chain of compact sets, each attached to the next by a single point, is $UA$ (provided each subchain is closed, see \cite[Theorem 11.4]{BD}). This can be generalized to certain tree-like unions of compact sets \cite{BDP2}. A useful observation is the following: \begin{fact} \label{nonUAsubset} If a function is $UA$, its restriction to a subspace is $UA$. So if a normal uniform space $X$ contains a non-$UA$ closed subspace, then $X$ is not $UA$. \end{fact} This can be used to prove that $\R^2$ is not $UA$ \cite{BD} (see also \cite{B}). In this paper we prove: \proclaim Theorem B. Every metric $UA$-space is thin. This gives as a corollary: \proclaim Theorem C. Let $X$ be a complete metric $UA$-space. Then $X$ has $CSP$. In particular, $\CSP$ holds for every metric Atsuji space, since every Atsuji space is $UA$ and complete (one can also reason directly from the following characterization of an Atsuji space $X$: the set $X'$ of accumulation points of $X$ is compact, and for every $\eps >0$ there exists $\delta >0$ such that for all $x, y\in X$ with $d(x,X')>\eps$ and $d(y,X')>\eps$ one has $d(x,y)\geq\delta$ \cite{Hu}). In \S 5 we give examples showing that the implications in Theorems A and B cannot be inverted: \begin{itemize} \item[(A)] a complete separable metric space satisfying CSP need not be thin; \item[(B)] a complete (connected) metric thin space need not be $UA$. \end{itemize} In section 2 we give a self contained exposition of all the needed properties of the quasi components, the uniform quasi components, and the uniformly connected sets. \proclaim Notations. $C(X, Y)$ is the set of all continuous functions from $X$ to $Y$. We also write $C(X)$ for $C(X, \R)$. The abbreviation ``u.c.'' stands for ``uniformly continuous''. %%%%%%%%%%%%%%%%%%% Q U A S I C O M P O N E N T S %%%%%%%%%%%%%% \section{Quasi components} \begin{definition} \label{separated} Two subsets $A, B$ of a topological space $X$ are {\bf separated} if the closure of each of them does not meet the other (this is equivalent to say that $A$ and $B$ are clopen in $A \cup B$). So $X$ is connected if and only if it cannot be partitioned in two separated sets. A subset $S$ of $X$ {\bf separates} the nonempty sets $A$ and $B$ if the complement of $S$ can be partitioned in two separated sets, one of which containing $A$, the other containing $B$ (see \cite[\S 16, VI]{K}). We say that $S$ {\bf cuts} between $A$ and $B$ if $S$ intersects every connected set which meets both $A$ and $B$. (If $S$ is empty this means that there is no connected set which meets both $A$ and $B$.) \end{definition} If $S \subseteq X$ separates $A$ and $B$, then $S$ cuts between $A$ and $B$. The converse holds in a connected locally connected regular topological space [HY, Theorem 3-6], but it is false in general. \begin{example} Consider a subset $X$ of the plane consisting of two points $a$ and $b$ together with countably many parallel lines so that both $a$ and $b$ are at distance $1/n$ from the $n$-th line. The empty set cuts between $a$ and $b$ in $X$, but it does not separate $a$ and $b$. \end{example} Note that $X$ is connected if every $f \in C(X, \{0, 1\})$ is constant. \begin{definition} A uniform space $X$ is {\bf uniformly connected} if every u.c. function $f\in C(X, \{0, 1\})$ is constant. A subspace $A \subseteq X$ is uniformly connected if it is uniformly connected with respect to the induced uniformity, i.e. if every u.c. function $f \in C(A, \{0, 1\})$ is constant. \end{definition} \begin{definition} Given $x \in X$ the {\bf quasi component} of $x$ in $X$ is the set of all points $y \in X$ such that every function $f\in C(X, \{0, 1\})$ gives the same value to $x$ and $y$. Equivalently $x$ is in the same quasi component of $y$ in $X$ if $x$ cannot be separated from $y$, i.e. for every partition $X = A \cup B$ with $A, B$ open, $x$ and $y$ lie both in $A$ or both in $B$. So the quasi component of $x$ is the intersection of all clopen sets containing $x$. The {\bf uniform quasi component} of $x$ in $X$ is defined in the same way but requiring $f$ to be uniformly continuous. \end{definition} \begin{definition} Let ${\cal P}$ be an open cover of $X$. Given $x, y \in X$, a {\bf ${\cal P}$-chain from $x$ to $y$} is a finite sequence $x_0, x_1, \ldots, x_k$ of points in $X$ with $x_0 = x, x_k = y$ and for all $i 0$ such that each open set of diameter $< \eps$ is contained in some element $P$ of ${\cal P}$. \end{definition} We characterize the quasi component and the uniform quasi component in terms of ${\cal P}$-chains. \begin{lemma} \label{same} \begin{enumerate} \item Two points $x, y$ of a topological space $X$ belong to the same quasi component if and only if for every open cover ${\cal P}$ of $X$ there is a ${\cal P}$-chain from $x$ to $y$. \item Two points $x, y$ of a uniform space $X$ belong to the same uniform quasi component iff for every uniform open cover ${\cal P}$ of $X$ there is ${\cal P}$-chain from $x$ to $y$. \item In particular if $X$ is metric we obtain: $x, y$ belong to the same uniform quasi component iff for every $\eps > 0$, there is $\eps$-chain from $x$ to $y$. \end{enumerate} \end{lemma} \begin{proof} Suppose $x, y$ belong to the same (uniform) quasi component and let ${\cal P}$ be a (uniform) open cover of $X$. Given $x \in X$, define $O_x \subseteq X$ as the set of points reachable from $x$ by a ${\cal P}$-chain. Suppose for a contradiction that $y \nin O_x$. Clearly there is no $P\in {\cal P}$ which intersects both $O_x$ and $X \setminus O_x$. So the function with value $0$ on $O_x$ and $1$ on $X \setminus O_x$ is (uniformly) continuous and takes different values on $x$ and $y$, a contradiction. Conversely suppose that $x, y$ belong to two different (uniform) quasi components. Then there is a (uniformly) continuous function $f \in C(X, \{0, 1\})$ with $f(x) = 0$ and $f(y) = 1$. Let ${\cal P}$ be the (uniform) open cover consisting of the (uniformly) clopen sets $f^{-1}(0)$ and $f^{-1}(1)$. Then there is no ${\cal P}$-chain from $x$ to $y$. \end{proof} \begin{definition} Given an open cover ${\cal P}$ of a topological space $X$ and $A \subseteq X$ we say that $A$ is {\bf ${\cal P}$-connected} if every pair of points of $A$ can be joined by a ${\cal P}$-chain contained in $A$. \end{definition} For instance every ${\cal P}$-chain is ${\cal P}$-connected. We characterize connected and uniformly connected sets in terms of ${\cal P}$-chains. \begin{corollary} \label{connected} \begin{enumerate} \item A topological space $X$ is connected iff for every open cover ${\cal P}$ of $X$, $X$ is ${\cal P}$-connected. \item A uniform space $X$ is uniformly connected iff for every uniform open cover ${\cal P}$ of $X$, $X$ is ${\cal P}$-connected. \item In particular if $X$ is metric, then $X$ is uniformly connected iff for every $\eps > 0$ any two points of $X$ can be joined by a $\eps$-chain. \end{enumerate} \end{corollary} \begin{proof} $X$ is (uniformly) connected iff every pair of points $x, y \in X$ lie in the same (uniform) quasi component. Now apply Lemma \ref{same}. \end{proof} \begin{definition} Let ${\cal P}$ be an open cover of a topological space $X$. We say that a subset $S$ of $X\;$ {\bf ${\cal P}$-cuts} between two subsets $A$ and $B$ of $X$ if $S$ is disjoint from $A \cup B$ and intersects every ${\cal P}$-connected set (or equivalently every ${\cal P}$-chain) which meets both $A$ and $B$. \end{definition} If $S$ ${\cal P}$-cuts between $A$ and $B$, then $S$ cuts between $A$ and $B$. Actually more is true: \begin{lemma} \label{cuts} If $S$ ${\cal P}$-cuts between $A$ and $B$ in a topological space $X$, then $S$ separates $A$ and $B$ in $X$. \end{lemma} \begin{proof} Let $O_A$ be the set of points $x \in X \setminus S$ such that there is a ${\cal P}$-chain contained in $X \setminus S$ from some point of $A$ to $x$. Then $O_A$ contains $A$ and is disjoint from $S$ and from $B$. It suffices to show that $O_A$ is clopen in $X \setminus S$. It is open because if $x \in O_A$, then for every $P \in {\cal P}$ containing $x$, $P \cap (X \setminus S) \subseteq O_A$. It is closed because if $x \in X \setminus S$ and $x \nin O_A$, then for every $P \in {\cal P}$ containing $x$, $P \cap (X \setminus S)$ is disjoint from $O_A$. \end{proof} So ``${\cal P}$-cuts'' entails ``separates'' which implies ``cuts''. For open separators we have the following partial converse. \begin{lemma} \label{P-cuts} If an open set $S$ separates $A$ and $B$ in a topological space $X$, then for some open covering ${\cal P}$ of $X$, $S$ ${\cal P}$-cuts between $A$ and $B$ \end{lemma} \begin{proof} The complement of $S$ can be partitioned in two separated sets $H$ and $K$. Let ${\cal P}$ be the covering of $X$ consisting of $S$ together with all the open sets which do not intersect both $H$ and $K$. \end{proof} %%%%%%%%%%%%%% U A i m p l i e s u n i f. T H I N %%%%%%%%%%%%%%%%%% \section{$UA$ spaces have connected uniform quasi components} Given a uniform space $X$ we give a necessary condition for a function $f\in C(X)$ to be $UA$ in terms of uniform quasi components. \begin{lemma} \label{nonUAfunction} Let $f \in C(X)$. Suppose there are $y \neq z$ in $\R$ and two points $a \in f^{-1}(y)$ and $b \in f^{-1}(z)$ such that the uniform quasi component of $a$ in $f^{-1}(y) \cup f^{-1}(z)$ contains $b$. Then $f$ is not $UA$. \end{lemma} \begin{proof} Let $M = f^{-1}(y) \cup f^{-1}(z)$. If $f$ is $UA$, then its restriction $f_{|M}$ is $UA$. Let $g \in C(M)$ be a $(\{a,b\}, M)$-approximation of $f_{|M}$. Then $g(M) \subseteq \{y,z\}$, $g(a) = x$ and $g(b) = y$. Since $g$ is uniformly continuous, $g$ witnesses the fact that $b$ does not belong to the uniform quasi component of $a$ in $M$, a contradiction. \end{proof} \begin{lemma} \label{nonUAspace} Suppose that a metric space $X$ contains two disjoint closed sets $H$ and $K$ and a point $a \in H$ such that the uniform quasi component of $a$ in $H \cup K$ intersects $K$. Then $X$ is neither thin nor $UA$. \end{lemma} \begin{proof} Clearly, $X$ cannot be thin since the closed subspace $Y=H \cup K$ has a uniform quasi component (that of $a$) which is not connected since it hits both $H$ and $K$, which are closed and disjoint. The closed subspace $H \cup K$ is not $UA$ since we can apply Lemma \ref{nonUAfunction} to the characteristic function of $H$. Hence $X$ is not $UA$ by Fact \ref{nonUAsubset}. \end{proof} \begin{corollary} \label{positive} Two disjoint closed uniformly connected subsets $A, B$ of a thin metric space $X$ are at positive distance. \end{corollary} \begin{proof} Apply Lemma \ref{nonUAspace} with $H=A, K=B$. \end{proof} Clearly, one can prove the above corollary for $UA$ spaces as well, but we do not put it here explicitly since $UA$ spaces will be proved to be thin in the sequel. Obviously, a closed uniformly connected subspace of a thin space is connected. The counterpart of this fact for $UA$ was proved in \cite{BD}. Now we obtain a new proof of this result: \begin{corollary}\label{u.c.} A closed uniformly connected subspace $C$ of a $UA$ metric space $X$ is connected. \end{corollary} \begin{proof} If $S$ is not connected it can be partitioned in two non-empty closed sets $H$ and $K$. Apply Lemma \ref{nonUAspace} with this choice of $H$ and $K$. \end{proof} \begin{definition} \label{garland-configuration} Given two distinct points $a,b$ of a metric space $X$ such that the uniformly connected component of $a$ contains $b$, there exists for each $n$ a finite set $L_n \subset X$ whose points form a $1/n$-chain from $a$ to $b$. We say that the sets $L_n$, together with $a$ and $b$, form a ({\bf discrete}) {\bf garland}, if there is an open subset $V$ of $X$ which separates $a$ and $b$ and such that $V\cap\bigcup_n L_n$ is closed (and discrete). \end{definition} Let $X$ be a metric space and let $A$, $B$ be two disjoint closed uniformly connected subsets of $X$. A garland from a point $a$ of $A$ to a point $b$ of $B$ may appear in the following situation: \begin{itemize} \item[(1)] $d(A,B)=0$. Now fix $a_n\in A$ and $b_n\in B$ with $d(a_n,b_n)<1/n$ and take $L_n$ to be the union of two $1/n$-paths, one connecting $a$ to $a_n$ in $A$, the other connecting $b_n$ to $b$ in $B$. Now $V=X\setminus (A\cup B)$ witnesses that $a$, $b$, $\langle L_n \mid n \in \N \rangle$ is a discrete garland in which $V \cap \bigcup_n L_n$ is empty. \item[(2)] $b\in Q_a^u(X)$ and there exists an open set $V$ of $X$ disjoint from $Q_a^u(X)$ that separates $A$ and $B$ (this entails that $ Q_a^u(X)$ is not connected, we show in the proof of Proposition \ref{new_prop} how to produce a discrete garland from $a$ to $b$ under this assumption). Let us mention that while the garland in (1) has a very particular nature, the one produced here is a generic one (see Proposition \ref{new_prop} for more details). \end{itemize} In the following lemma we show that one can shrink an open separator. \begin{lemma} \label{small-separator} Let $X$ be a metric space and $V \subseteq X$ an open set which separates two subsets $A$ and $B$ of $X$. Then there is a open set $U \subseteq X$ which separates $A$ and $B$ and such that $\overline{U} \subset V$. \end{lemma} \begin{proof} The complement of $V$ can be partitioned in two closed sets $H \supseteq A$ and $K \supseteq B$. By normality $H$ and $K$ are included in two disjoint open sets with disjoint closures $H'$ and $K'$. The complement $U$ of $H' \cup K'$ works. \end{proof} Now we are in position to give a criterion for thinness for metric spaces in terms of existence of garlands in the space. \begin{proposition}\label{new_prop} For a metric space $X$ the following conditions are equivalent: \begin{itemize} \item[(a)] $X$ is thin; \item[(b)] $X$ contains no garlands; \item[(c)] $X$ contains no discrete garlands. \end{itemize} \end{proposition} \begin{proof} (a) $\to$ (b) The existence of a garland $a, b, \langle L_n \mid n\in\N\rangle$ in $X$ leads to the existence of an open set $V$ separating $a$ and $b$ such that $D=V \cap \bigcup_n L_n$ is closed. The complement of $V$ can be partitioned in two separated sets $H \ni a$ and $T \ni b$. Then the closed subspace $Y=H\cup T\cup D$ of $X$ and $a\in Y$ witness the non-thinness of $X$. Indeed, the uniform quasi component $Q$ of $a$ in $Y$ is not connected since the sets $H$ and $T \cup D$ are closed and disjoint, and they both intersect $Q$. (b) $\to$ (c) is trivial. (c) $\to$ (a) Now assume that $X$ is not thin. In order to produce a discrete garland take a closed subspace $Y$ of $X$ and $a\in Y$ witnessing non-thinness of $X$. Then there exists a pair of closed disjoint non-empty subsets $H$ and $K$ of $Y$ such that $Q=Q_a^u(Y) = H\cup K$ with $a\in H$. Then $U=X\setminus (H\cup K)$ is an open set of $X$ that separates $H$ and $K$. By Lemma \ref{small-separator} there exists an open subset $V\subseteq U$ separating $H$ and $K$ such that $\overline{V} \subset U$. Fix a point $b\in K$. By Lemma \ref{same}, for each $n$ there is a finite set $L_n$ whose points form a $1/n$-chain from $a$ to $b$. We will show that $a, b, \langle L_n \mid n \in \N \rangle$ form a discrete garland. So it suffices to show that $D= V \cap \bigcup_n L_n$ is closed and discrete, i.e., the set $D$ has no accumulation points in $X$. In fact if $x$ is such an accumulation point, then any neighbourhood of $x$ meets infinitely many $L_n$, and therefore for every $n$ there is a $1/n$-chain from $a$ to $x$, showing that $x$ belongs to $Q^u_a = H \cup K$. But clearly $x$ belongs also to the closure $\overline{V}\subseteq U$, which is disjoint from $H \cup K$, and we have a contradiction.\end{proof} A more careful analysis of the above proof shows that if $X$ admits a garland $a, b, \langle L_n \mid n \in \N \rangle$, then the closed set and the uniform quasi component witnessing the fact that $X$ is not thin can be obtained by taking simply the closure $Z$ of the subset $\{a, b\}\cup \bigcup_{n \in \N}L_n$ and the uniform quasi component of $a$ in $Z$. The following proposition is the last step in the proof of Theorem B. \begin{proposition}\label{thin-ua} If a metric space $X$ contains a garland, then $X$ is not $UA$. \end{proposition} \begin{proof} Given a garland $a, b, \langle L_n \mid n \in \N \rangle$ in $X$, let $V$ be as required in Definition \ref{garland-configuration}. So $D = V \cap \bigcup_n L_n$ is closed and the complement of $V$ can be partitioned in two separated sets $H \ni a$ and $T \ni b$. The sets $H$ and $K$ are closed in the complement of $V$, hence they are closed also in $X$ (as $V$ is open). It follows that the set $K = T \cup D$ is closed as well. Since $\bigcup_n L_n$ is contained in $H \cup K$, the uniform quasi component of the point $a$ in $H \cup K$ contains $b$. Hence by Lemma \ref{nonUAspace} $X$ is not $UA$. \end{proof} This proves that every metric $UA$ space is thin, i.e., Theorem B. %%%%%%%%%%%% P R O O F O F M A I N T H E O R E M %%%%%%%%%%% \section{The compact separation theorem} \begin{definition} Given an open cover ${\cal P}$ of $X$ and $x\in X$ the {\bf star} $\St(x, {\cal P})$ is the union of all $P \in {\cal P}$ containing $x$. If $S \subseteq X$, $St(S, {\cal P})$ is the union of all the sets $\St(x, {\cal P})$ with $x\in S$, i.e. the union of all the $P\in {\cal P}$ which intersect $S$. \end{definition} \begin{definition} Let ${\cal P}$ be an open cover of a topological space $X$ and let $V \subseteq X$. We say that two sets $L_1, L_2 \subseteq X$ are {\bf $({\cal P}, V)$-apart} if for each $P \in {\cal P}$ which intersects $V$, $P$ can intersect at most one of the two sets $L_1$ and $L_2$. \end{definition} If $V = X$, this intuitively means that the sets $L_1$ and $L_2$ are far away from each other (by an amount measured by ${\cal P}$). If $V \subseteq X$ we obtain a relative notion: the portions of $L_1$ and $L_2$ which are close to $V$ are far away from each other. \begin{lemma} \label{accumulation} Let ${\cal P}$ be an open cover of a topological space $X$, let $V \subseteq X$ and let $\{L_i \mid i \in I\}$ be a family of finite subsets of $X$ which are pairwise $({\cal P}, V)$-apart. Then $V \cap \bigcup_i L_i$ has no accumulation point in $X$. \end{lemma} \begin{proof} If $x \in X$ is an accumulation point of $V \cap \bigcup_i L_i$, then any $P \in {\cal P}$ containing $x$ intersects $V$ and contains infinitely many points of $\bigcup_i L_i$. Since each $L_i$ is a finite set, $P$ must intersect infinitely many $L_i$, contradicting the $({\cal P}, V)$-apartness. \end{proof} The next lemma is the main step in the proof of the compact separation theorem. It permits to shrink an open separtor to a smaller open one that is covered a finitely many balls of a given radious $\eps>0$. \begin{lemma} \label{small-separator2} Let $X$ be a thin metric space. Let $A, B$ be disjoint closed connected subsets of $X$, and let $V$ be an open set which separates $A$ and $B$. Then for every $\eps > 0$ there is an open set $S$, with $\overline{S} \subset V$, which separates $A$ and $B$ and is contained in the union of finitely many balls of radious $\eps$. \end{lemma} \begin{proof} Fix two points $a \in A$ and $b \in B$. By Lemma \ref{small-separator} there is an open set $U$, with closure included in $V$, which separates $A$ and $B$. Since $U$ is an open separator, by Lemma \ref{P-cuts} there is an open covering ${\cal P}$ of $X$ such that $U$ ${\cal P}$-cuts between $A$ and $B$. By refining ${\cal P}$ if necessary, we can assume that $\overline{\St(U, {\cal P})} \subset V$ and each member of ${\cal P}$ is contained in a ball of radious $\eps$. Fix a sequence $({\cal P}_n \mid n \in \N)$ of open coverings of $X$, each refining ${\cal P}$, and such that every member of ${\cal P}_n$ is contained in a ball of radious $1/n$. Consider a finite sequence $L_0, L_1, L_2, \ldots, L_k$ such that each $L_i$ is a finite set whose points form a ${\cal P}_i$-chain from some point of $A$ to some point of $B$ and the sets $L_n$ are pairwise $({\cal P}, U)$-apart. We claim that there is a maximal such sequence (possibly empty) $L_0, L_1, \ldots, L_k$. If this is not so, we would obtain a countable sequence $(L_n \mid n \in \N)$. Now $U \cap \bigcup_n L_n$ is closed by Lemma \ref{accumulation}. By Lemma \ref{connected} we can adjoin to each $L_n$ a portion of $A$ and $B$ to obtain a $1/n$-chain from $a$ to $b$, thus obtaining a garland which contradicts (by Proposition \ref{new_prop}) the fact that $X$ is thin. Thus we can fix a maximal sequence $L_0,L_1,\ldots, L_k$ as above and define $S$ as the union of all $P \in {\cal P}_{k+1}$ which intersect both $U$ and $L_0 \cup \ldots \cup L_k$. Then $S$ misses $A\cup B$ since $\St(U, {\cal P}_{k+1}) \subset V$. Moreover, $S$ ${\cal P}_{k+1}$-cuts between $A$ and $B$ because otherwise a chain $C$ witnessing the opposite would contradict the maximality of $L_0, \ldots, L_k$ (for we could set $L_{k+1} = C$). By construction $S\subseteq \St(U, {\cal P}_{k+1})$ and since ${\cal P}_{k+1}$ refines ${\cal P}$ we have have $\overline{S} \subseteq V$. The rest is clear. \end{proof} This lemma gives the following immediate corollary. \begin{corollary} \label{ostacolo} Let $X$ be a thin metric space and let $A, B$ be disjoint closed uniformly connected subsets of $X$. Then there is a collection $\{H_n \mid n \in \N\}$ of nonempty closed subsets of $X$ such that for every $n$, \begin{enumerate} \item $H_{n+1} \subseteq H_n$, \item $H_n$ separates $A$ and $B$, \item $H_n$ is contained in a finite union of balls of diameter $< 1/n$. \end{enumerate} \end{corollary} Now Theorem A follows from the following more general result: \begin{theorem} \label{main} Let $X$ be a complete thin metric space and let $A, B$ be disjoint closed uniformly connected subsets of $X$. Then: \begin{enumerate} \item there is a compact set $K$ such that each \nbd of $K$ disjoint from $A \cup B$ separates $A$ and $B$; \item hence $K$ intersects every closed connected set which meets $A$ and $B$; \item if $X$ is also locally compact, there is a compact set $K'$ which separates $A$ and $B$. \end{enumerate} \end{theorem} \begin{proof} To prove (1) let $H_n$ be as in Lemma \ref{ostacolo} and let $K = \bigcap_n H_n$. It suffices to prove that every open \nbd $U$ of $K$ contains some $H_n$. (This shows in particular that $K$ is non-empty, and the compactness of $K$ follows from the fact that $K$ is closed and totally bounded.) We reason by contradiction. So assume that for each $n$ there is $x_n \in H_n \setminus U$. We can then easily extract from $\{x_n\}$ a subsequence $\{y_i\}$ such that for each $n$ all but finitely many of the points $y_i$ lie in only one of the finitely many open balls of the fixed cover of $H_{n}$. It follows then that $\{y_i\}$ is a Cauchy sequence, hence it converges to a point $y$ which must lie in $K$ and also outside of $U$, a contradiction. To prove (2) consider a closed set $C$ avoiding $K$ and intersecting both $A$ and $B$. Then the complement of $C \cup A \cup B$ in $X$ is an open \nbd of $K$ disjoint from $A \cup B$. Hence $C$ cannot be connected by the separation property established in part (1). To prove (3) assume that $X$ is locally compact. Then we can find a compact \nbd $K'$ of $K$ disjoint from $A \cup B$ and we apply (1). \end{proof} %%%%%%%%%%% L a S T E L L A E U N I F. T H I N %%%%%%%%%%%%%%%%% \section{Examples and questions} We have seen that a complete thin metric space has $CSP$. This suggests the following \begin{question} Is it true that a complete thin uniform space has $CSP$? What about a complete $UA$ uniform space? \end{question} Our next question is about how much we use the fact that {\bf uniform} quasi components are connected. \begin{question} Is it true that every complete metric space $X$ such that every closed subspace of $X$ has connected quasi components has necessarily CSP ? \end{question} Our next examples show that the implications in Theorems A and B cannot be inverted. \subsection{CSP does not imply thin for complete separable metric spaces} There exist many examples of separable metric space with CSP that are not thin: \begin{itemize} \item[(i)] the circle minus a point (has two closed connected subsets at distance zero, so it cannot be thin by Corollary \ref{positive}); \item[(ii)] the rationals $\Q$ (uniformly connected non-connected, hence not thin). \item[(iii)] $\R\times \Q$ (uniformly connected non-connected, hence not thin). \end{itemize} None of the above examples is complete. Here we offer an example of a complete separable metric space with CSP that is not thin. \begin{example} Let $H_1$ and $H_2$ be the branches of hyperbolas $\{(x,y)\in \R^2: xy=1\}$ and $\{(x,y)\in \R^2: xy=2\}$, respectively, contained in the first quadrant. Then the space $X=H_1\cup H_2$ with the metric induced from $\R^2$ is a complete separable space. Since $H_1$ and $H_2$ are connected and at distance zero, it follows from Corollary \ref{positive} that $X$ is not thin. On the other hand, the empty set separates the closed connected sets $H_1$ and $H_2$. So if $A$ and $B$ are closed connected disjoint sets in $X$, it remains to consider only the case when both $A$ and $B$ are contained in the same component $H_i$ ($i=1,2$). Now $A$ and $B$ can be separated by a point. \end{example} We do not know whether completeness is necessary in Theorem A, in other words: \begin{question}\label{non-complete} Are there examples of thin spaces that do not have CSP? What about $UA$ spaces? \end{question} As the following example shows, neither thinness nor $UA$-ness is preserved by passage to completions, thus an immediate application of Theorem A (via passage to completions) cannot help in trying to answer negatively Question \ref{non-complete}. \begin{example} There is a $UA$ metric space whose completion is not thin (hence not $UA$). Let $X = \bigcup_{n\in \N} \{1/n\} \times I$, where $I$ is the unit interval $[0,1]\subset {\bf R}$, let $a=(0,0)$, $b=(0,1)$ and $Y=X\cup \{a,b\}$. We put on $Y$ the following metric. The distance between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $|y_1-y_2|$ if $x_1=x_2$. Otherwise the distance is the minimum between $y_1+y_2 + |x_1 - x_2|$ and $(1-y_1) + (1-y_2) +|x_1 - x_2|$. With this metric $Y$ is the completion of $X$ and the two points $a, b$ are the limits for $n \to\infty$ of $(1/n, 0)$ and $(1/n, 1)$ respectively. The space $Y$ is not thin since there is a garland consisting of $a$, $b$ and $\langle L_n \mid n \in {\bf N}\rangle$ where $L_n$ is a $1/n$-chain between $a$ and $b$ in $\{1/n\} \times I$. The space $X$ is $UA$ since $X$ is a union of a chain of compact sets, each attached to the next by at most one point (see \cite[Theorem 11.4]{BD} and the introduction). \end{example} \subsection{Thin does not imply UA for complete metric spaces} We give an example of a complete connected thin metric space that is not $UA$. \begin{example} For a cardinal $\alpha$ denote by $J(\alpha)$ be the hedgehog of $\alpha$ spikes (see \cite[Example 4.1.5]{E}). Let us see that $J(\alpha)$ is thin. By Proposition \ref{new_prop} if $J(\alpha)$ is not thin, it contains a discrete garland $a,b, \langle L_n \mid n \in {\bf N} \rangle$. Let $V$ be an open set separating $a,b$ such that $V \cap \bigcup_n L_n$ is closed and discrete. The minimal connected set $C$ containing $a,b$ must non-trivially intersect $V$, so it contains an open interval $I$ on one of the spikes. Now whenever $1/n$ is less than the diameter of $I$, $L_n$ must intersect $I$, so $V \cap \bigcup_n L_n$ has an accumulation point, which is a contradiction. \end{example} This gives the following immediate corollary based on one of the main results of \cite{BDP2} (see \cite{vD} for the definition of the cardinal number ${\bf b}$): \begin{corollary} For every $\alpha \geq {\bf b}$ the hedgehog $J(\alpha)$ is thin (so has the property CSP), but not UA. \end{corollary} The space $J(\alpha)$ is not separable for $\alpha>\omega$ On the other hand, ${\bf b}>\omega$ (\cite{vD}), hence the above examples are not separable. It was proved in \cite{BDP2} that the hedgehogs $J(\alpha)$ are $UA$ for all $\alpha<{\bf b}$. 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