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\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} }
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 \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}}

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\newcommand{\R}{{\bf R}}
\newcommand{\N}{{\bf N}}
\newcommand{\Q}{{\bf Q}}
\newcommand{\C}{{\bf C}}
\newcommand{\restricted}[1]{_{| #1}}
\newcommand{\ov}{\overline}
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\newcommand{\eps}{\varepsilon}
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\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 $i<k$ there is $P \in {\cal P}$
containing $x_i$ and $x_{i+1}$. 

If $X$ is metric and ${\cal P}$ consists of all open balls of diameter
$\eps$, then a ${\cal P}$-chain is also called a {\bf $\eps$-chain}. 
\end{definition}

\begin{definition} An open cover ${\cal Q}$ is a {\bf refinement} of an open 
cover ${\cal P}$ if every $Q
\in {\cal Q}$ is contained in some $P \in {\cal P}$. 

A {\bf uniform cover} of a uniform space $X$ is an open cover ${\cal P}$
such that for some member $U \subseteq X
\times X$ of the uniformity of
$X$, the collection of all the open sets $U[x] = \{y \mid (x, y) \in U\}$ is
a refinement of ${\cal P}$.

If $X$ is metric this means that there is
$\eps > 0$ 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 $\alpha<{\bf b}$. Hence one cannot hope to get in 
this way an example of a separable space with the above properties. 

\begin{question}
Is it true that every (complete) metric thin separable space is $UA$?
\end{question}

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\end{document}
