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\title{$S_4$ continua in the sense of Michael are dendrites}


\author{{\small Francis Jordan}%
\thanks{AMS classification numbers: Primary 54C65, 54E40;
Secondary 54F15.
\endgraf  Key words and phrases: connectivity functions,
continuous selections, monotone maps, $S_4$ spaces, irreducible continua.  The results in this paper were found while the author was visiting Loyola University, New Orleans, LA between 2000 and 2001.}
\endgraf
\small  Department of Mathematics, \\
University of Mississippi,\\
University, MS 38677}

\date{}


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\begin{document}\maketitle
\begin{abstract}
We show that $S_4$ spaces in the sense of Michael are dendrites.  The proof involves functions whose graphs are connected and monotone maps onto the arc.
\end{abstract}
\section{Introduction}
We let $2^X$ denote the space of all non-empty, compact subsets of
a metric space $X$; $2^{X}$ has the Vietoris topology or,
equivalently, the Hausdorff metric topology \cite{HYPER1}.

Let  $\F\subseteq 2^X$.  A continuous selection for $\F$ is
a continuous function $\sigma\colon\F\to X$ such that
$\sigma(F)\in F$ for each $F\in\F$.

Michael \cite[p.178]{MICH1} defined an $S_4$ space to be a
space $X$ such that there is a continuous selection for every
$\F\subseteq 2^X$ such that the members of $\F$ are mutually
disjoint and $\bigcup\F=X$.  The question of which spaces are
$S_4$ spaces is asked in \cite[p.155]{MICH1}, and some partial
answers are given in \cite[pp.178--179]{MICH1}.  In particular, it is 
mentioned that no $S_4$ space can contain a 
simple closed curve.  A dendrite 
is a continuum that is locally connected and contains no simple closed curve.
The question of whether $S_4$ continua are dendrites is due
to Gail S. Young \cite{YNG}.

In \cite{jornad} $X$ is defined to be a weak $S_4$ space
provided that there is a continuous selection for every
$\F\subseteq F_2(X)$, where $F_2(X)$ denotes the $2$-fold
symmetric product of $X$ \cite{HYPER1}, such that the members of
$\F$ are mutually disjoint and $\bigcup\F=X$. Clearly, every $S_4$
space is a weak $S_4$ space.  Also, no simple closed curve is a
weak $S_4$ space.  Both the $S_4$ and weak $S_4$ property are
hereditary with respect to subsets \cite{jornad}.


We use the ideas developed in \cite{jornad}, namely maps with
connected graphs.  To prove the following theorem:
 
\medskip

\noindent Theorem : {\it If $X$ is a weak $S_4$ continuum, then $X$ is an dendrite.}
\medskip

\noindent It is still unknown if the converse of this theorem is true.

I would like to thank the referee for many helpful comments regarding the presentation of this work, particularly, his or her comments on Lemma~\ref{lem:21} and Corollary~\ref{cor:tip}.

\section{Terminology}
We denote the cardinality of $\real$ by $\cuum$ and the first
uncountable cardinal by $\omega_1$.  By $\omega$ we denote
simultaneously the non-negative integers and the cardinality of
the non-negative integers. For an arbitrary set $S$ we denote its
cardinality by $|S|$.  The closed unit interval $[0,1]$ will be
denoted by $I$.  

A continuum is a non-empty compact connected metric space.  We use
$H(A,B)$ to denote the Hausdorff distance between two sets $A$ and
$B$. Given sets $A,B\subseteq X$ and a metric $\dist$ on $X$ we
let $\dist(A,B)=\inf\{\dist(x,y)\colon x\in A\ \text{and}\ y\in B\}$ and
$\diam(X)=\sup\{\dist(x,y)\colon x,y\in X\}$.  Given a function $f$
we denote the set of its discontinuity points by $\dis(f)$.  We use 
$\cl$ to denote closure.  


If $X$ is a topological space and $A\subseteq X$ is such that
$X\setminus A$ is separated (i.e., not connected), we say $A$ is a
separator of $X$.  For a disconnected space $X$ we write $X=A|B$ to 
denote a partition of $X$ into disjoint closed subsets.

Given a functon $f\colon X\to Y$ we denote the graph of $f$ by $\Gamma(f)$.   
A function $f\colon X\to Y$ is called a connectivity function
provided that the $\Gamma(f|C)$ is connected for every connected
subset $C$ of $X$.  Connectivity functions have been studied
extensively (e.g., \cite{GNsurvey}, \cite{HAG} and, \cite{HAM}).


A continuum $X$ is said to be irreducible between two points $p$
and $q$ provided that no proper subcontinuum of $X$ contains both
$p$ and $q$. By 4.34 of \cite{NAD1}, it is easy to see that given
any two distinct points $p$ and $q$ in a continuum there is a
minimal continuum $K$ containing $p$ and $q$.  Clearly, $K$ must
be irreducible between $p$ and $q$.

Let $X$ be a continuum.  We say $f\colon X\to I$ is a  $T$ map
provided that $f$ is a monotone continuous surjection such that
$f^{-1}(w)$ is nowhere dense in $X$ for every $w\in I$.  We use
$T$ map to denote these maps because of the following proposition
due to E. S. Thomas \cite[Theorem 10]{thom}: \prop{prop:1}{If $X$
is hereditarily decomposable and irreducible between two points,
then there is a $T$ map from $X$ onto $I$.}

For $f\colon X\to I$ and $0<w<1$ define
$A^f_w=\cl(f^{-1}([0,w)))\cap\cl(f^{-1}((w,1])))$.  For $w\in\{0,1\}$ we define 
$A^f_w=f^{-1}(w)$.  Observe that
if $f$ is a $T$ map, then $A^f_w\neq\emptyset$ for all $w\in I$.

Given a compact metric space $X$ and $f\colon X\to I$ continuous, we define $f^*\colon I\to 2^X$ to be $f^*(w)=f^{-1}(w)$.  Recall that a function $f\colon X\to Y$ is said to be of Borel Class 1 provided that $f^{-1}(U)$ is a countable union of closed sets for every open set $U\subseteq Y$.  We say $A\subseteq X$ is meager or of the first category provided that $A$ is contained in the countable union of closed nowhere dense sets.


We state some well known and useful facts about $f^*$.
\prop{prop:use1}{The function $f^*$ has the following properties:
\begin{description}
\item[(i)] $f^*$ is in Borel Class 1,
\item[(ii)] $\dis(f^*)$ is a meager set, and
\item[(iii)]  if $K\subseteq X$ is closed, then
$\{w\in I\colon f^*(w)\cap K\neq\emptyset\}$ is closed.
\end{description}}
\proof We outline a proof.  First one may verify (iii) using the
fact that $f[K]$ is closed by continuity and that $f[K]= \{w\in
I\colon f^*(w)\cap K\neq\emptyset\}$.  Next (i) can be verified by
looking at the basic open sets of $2^X$ and using the fact that
every open set in $X$ is the countable union of closed sets and
(iii). Finally (ii) follows from (i) and the well known fact, see for example \cite{KUR}, that the points of discontinuity of a Borel
Class 1 function form a meager set.\qed

Now, we introduce a simple version of the Cantor-Bendixson Theorem which we state
below.  We will only use the result for $I$, but we state it more generally.  We will use this proposition in the proofs of Lemma~\ref{lem:00} and Lemma~\ref{lem:tip}.  Let $X$ be a topological space and $K\subseteq X$.  Inductively define
sequences $\{K_{\alpha}\}_{0<\alpha<\omega_1}$ and
$\{W_{\alpha}\}_{\alpha<\omega_1}$ so that
\begin{description}
\item[(a)] $K_0=K$,
\item[(b)] $W_{\beta}=\{x\in K_{\beta}\colon x\text{ is isolated in
}K_{\beta}\}$,
\item[(c)] if $\alpha$ has the form $\beta+1$, then
$K_{\alpha}=K_{\beta}\setminus W_{\beta}$, and
\item[(d)] if $\alpha$ is a limit cardinal, then
$K_{\alpha}=\bigcap_{\beta<\alpha}K_{\beta}$.
\end{description}

\prop{prop:use0}{If $X$ is a complete metric space, $K\subseteq X$ is closed, and we define
$\{K_{\alpha}\}_{0<\alpha<\omega_1}$ and
$\{W_{\alpha}\}_{\alpha<\omega_1}$ as above, then $K_{\alpha}$ is closed for all $\alpha\in\omega_1$ and there is an
$\zeta<\omega_1$ such that
\begin{description}
\item[(i)] $K_{\zeta}=K_{\beta}$ for $\beta\geq\zeta$ and 
\item[(ii)] $K_{\zeta}=\emptyset$ if and only if $K$ is
countable if and only if $K=\bigcup_{\alpha<\omega_1}W_{\alpha}$.
\end{description}}



\section{Proof of Theorem}\label{cats}


\lem{lem:00}{Let $X$ be a continuum and $f\colon X\to I$ be a $T$
map. If $C$ is a closed separator of $X$, then $|f[C]|=\cuum$ or
there is a $w\in(0,1)$ such that $A^f_w\subseteq C$.} \proof Let
$K=f[C]$.  By way of contradiction, assume that $|K|\neq\cuum$ and $A^f_w\setminus C\neq\emptyset$ for all $w\in I$.  Since $K$ is closed,
$|K|\leq\omega$. Letting $K_0=K$, define sequences of sets
$\{K_{\alpha}\}_{\alpha\in\omega_1}$ as we did in the discussion prior to the statement of Proposition~\ref{prop:use0}.  By Proposition~\ref{prop:use0}(ii), there is an $\zeta\in\omega_1$
such that $K_{\zeta}=\emptyset$.  Notice that $C$ must be nowhere dense in $X$, since $f$ is a $T$ map and $f[C]$ is countable. 

We claim that for every $\beta\leq\zeta$ if $J$ is a
complementary interval of $K_{\beta}$, then $f^{-1}(J)\setminus C$
is connected.  For $\beta=0$ the claim follows immediately from
the monotonicity of $f$.  So assume we have shown the claim for all
$\xi<\beta$.  We now establish the claim for $\beta$.  Let $J$ be
a complementary interval of $K_{\beta}$.  Consider the following
collection of open intervals:
\[{\cal C}=\{L\subseteq J\colon f^{-1}(L)\setminus C
\text{ is connected}\}.\] Notice that ${\cal C}\neq\emptyset$
since $K_0$ is countable and closed.  It is easily checked that
any $\subseteq$-increasing chain in ${\cal C}$ has an upper bound
in ${\cal C}$, so by the Hausdorff Maximal Principle, there is a
$\subseteq$-maximal element $M$ of ${\cal C}$.  We show that $M=J$
which will prove the claim for $\beta$. Let $M=(a,b$ and
$J=(c,d)$.  By way of contradiction, assume that $b\neq d$.  Since
$b\notin K_{\beta}$, there is a maximal $\xi<\beta$ such that
$b\in K_{\xi}$.  Since $b\notin K_{\xi+1}$, we have that $b$ is
isolated in $K_{\xi}$. So, there is an $e\in J\cup\{d\}$ such that
$(b,e)$ is a complementary interval of $K_{\xi}$. Now
$f^{-1}((a,b))\setminus C$ is connected and, by inductive
hypothesis, $f^{-1}((b,e))\setminus C$ is connected.  Since $C$ is
nowhere dense $\cl(f^{-1}((b,e))\setminus C)=\cl(f^{-1}((b,e)))$.
So, $(f^{-1}((b,e))\setminus C)\cup H$ is connected for any
$H\subseteq A^f_b$. Similarly, $(f^{-1}((a,b))\setminus C)\cup H$
is connected for any $H\subseteq A^f_b$. By assumption, $A^f_b
\setminus C\neq\emptyset$. Thus, letting $H=A^f_b\setminus C$, we
have $f^{-1}((a,e))\setminus C=((f^{-1}((b,e))\setminus C)\cup
H)\cup ((f^{-1}((a,b))\setminus C)\cup H)$ connected which
contradicts maximality of $M$.  Thus, $b=d$.  Similarly, we can show that
$a=c$.  Hence, $J=M$.  So, the claim holds for $\beta$. Therefore,
the claim holds for all $\beta\leq\zeta$.

Since $K_{\zeta}=\emptyset$, we have $X\setminus
C=f^{-1}(I)\setminus C$ which is connected by the claim.  This
contradicts that $C$ is a separator of $X$.\qed

\lem{lem:000}{If $X$ is compact $f\colon X\to I$ is a function, then
$f^{-1}(w)=A^f_w$ for all but countably many $w\in I$.} \proof
Suppose the lemma is false. Without loss of generality, we may
assume that there is an uncountable $W\subseteq I$ and numbers
$\epsilon,\delta>0$ such that
$\dist(x_w,f^{-1}((w,w+\delta)))>\epsilon$ where for each $w\in W$
we have $x_w\in f^{-1}(w)$.   Since $W$ is uncountable, there is a
decreasing convergent sequence $\{w_n\}_{n=1}^{\infty}$ such that $w_n\in W$
for all $n$.  There is an integer $k$ large enough that
$|w_n-w_m|<\delta$ for all $m,n\geq k$.  By our choice of $\delta$
we must have $\dist(x_{w_n},x_{w_m})>\epsilon$ for all $n,m\geq
k$, this however contradicts the compactness of $X$.  Thus, we
have the lemma.\qed

Given a function $f\colon X\to Y$ we say $g\colon Y\to X$ is an
inverse of $f$ provided that $f(g(y))=y$ for all $y\in f[X]$ and
$g(f(x))=x$ for all $x\in g[Y]$.

\lem{lem:0}{Let $X$  be a continuum, $f\colon X\to I$ be a $T$
map. There is a connectivity function $g\colon I\to X$ such that
$g$ is an inverse of $f$ and $g[I]$ is dense in $X$. Moreover, if
$p\in f^{-1}(0)$ and $q\in f^{-1}(1)$ we may assume $g(0)=p$ and
$g(1)=q$.} \proof Let $\{C_{\alpha}\}_{\alpha\in\cuum}$ be an
enumeration of the closed separators $C$ of $X$ with the property
that $|f[C]|=\cuum$. Let $R=\{w\colon f^{-1}(w)\neq
A^f_w\}\cup\{0,1\}$, by Lemma~\ref{lem:000}, $R$ is countable. By
transfinite induction we can construct a sequence
$\{x_{\alpha}\}_{\alpha\in\cuum}$ such that for every
$\alpha\in\cuum$ we have,

\[x_{\alpha}\in
f[C_{\alpha}]\setminus(\{x_{\xi}\colon\xi<\alpha\}\cup R).
\]
Notice such choices can be made since $|f[C_{\alpha}]|=\cuum$ and $|\{x_{\xi}\colon\xi<\alpha\}\cup R|<\cuum$ for every
$\alpha\in\cuum$. For each $\alpha\in\cuum$ define $g(x_{\alpha})$
so that $g(x_{\alpha})\in f^{-1}(x_{\alpha})\cap C_{\alpha}$. For
$x\in I\setminus\{x_{\alpha}\colon\alpha\in\cuum\}$ define $g(x)$
so that $g(x)\in A^f_x$, in particular we may define $g(0)=p$ and
$g(1)=q$. It is important to notice at this point that $g(x)\in
A^f_x$ for all $x\in I$.  In particular, $g$ is an inverse of $f$ since 
$g(x)\in A_x^f\subseteq f^{-1}(x)$.

We claim that $g[I]$ is connected.  By way of contradiction, assume
that $g[I]$ is not connected. Since $g[I]\subseteq X$ and $X$ is
connected, there is, by \cite[Theorem 3, p.
155]{KUR}, a closed set $C\subseteq X$ such that $C$
separates $X$ and $g[I]\cap C=\emptyset$.  We must now consider two cases. First suppose
$|f[C]|=\cuum$.  Pick $\alpha\in\cuum$ such that $C=C_{\alpha}$.
By the way we defined $g$, we have $g(x_{\alpha})\in C_{\alpha}$
which contradicts that $g[I]\cap C=\emptyset$.  We now consider
the case when $f[C]$ is countable.  Since $C\subseteq X$, and $f$
is a $T$ map, Lemma~\ref{lem:00} tells us that there is a $w\in
(0,1)$ such that $A^{f}_w\subseteq C$.  Now $g(w)\in
A^f_w\subseteq C$ which contradicts that $C\cap g[I]=\emptyset$.
Thus, $g[I]$ is connected.  

Let $f_1\colon X\to I\times X$ be the
continuous injection defined by $f_1(x)=\langle f(x),x \rangle$.
Clearly, $f_1[g[I]]$ is connected.  If $x\in g[I]$, then 
\[f_1(x)=\langle f(x),x\rangle=\langle f(x), g(f(x))\rangle\in \Gamma(g).\]
Thus, $f_1[g[I]]\subseteq\Gamma(g)$.  If $\langle w,y\rangle\in\Gamma(g)$, then since $g$ is an inverse of $f$ and 
$f$ is a surjection, we have 
\[\langle w,y \rangle=\langle w,g(w)\rangle=\langle f(g(w)), g(w)\rangle=f_1(g(w)).\]
Thus, $\Gamma(g)\subseteq f_1[g[I]]$.  Therefore, $\Gamma(g)=f_1[g[I]]$ is
connected. Since the domain of $g$ is $I$, the fact that $\Gamma(g)$ is connected implies that $g$ is connectivity.

To see that $g[I]$ is dense in $X$ it is enough, by \cite[Lemma
4.1]{jornad} to notice that $g[I]$ intersects each closed
separator of $X$.\qed


I have been informed by the referee that that the construction in the following lemma is what is known as a pullback.  In particular, the construction we use here is very similar to the one used in \cite[Theorem 3.9]{GNT}.

\lem{lem:1}{Let $X$ and $Y$ be continua. If $f\colon X\to I$ and
$g\colon Y\to I$ are $T$ maps and
$\dis(f^*)\cap\dis(g^*)=\emptyset$ then
\[R=\{\langle x,y\rangle\colon f(x)=g(y)\}\] is a continuum and
$h\colon R\to I$ defined by $h(\langle x,y\rangle)=f(x)=g(y)$ is a
$T$ map.} \proof First note that the function $(f,g)\colon X\times
Y\to I^2$ defined by $(f,g)(\langle x,y \rangle)=\langle
f(x),g(y)\rangle$ is continuous. It is easily checked that
$R=(f,g)^{-1}(\diag(I\times I))$ so $R$ is compact.

We now show that $h\colon R\to I$ is monotone.  Let $w \in I$. By
way of contradiction, assume that $h^{-1}(w)$ is not connected.
Let $h^{-1}(w)=A|B$. Since $g^{-1}(w)$ is connected, we have that
$\{x\}\times g^{-1}(w)\subseteq A$ or $\{x\}\times
g^{-1}(w)\subseteq B$ for all $x\in f^{-1}(w)$.  We now have
\[f^{-1}(w)=\{x\in f^{-1}(w)\colon \{x\}\times
g^{-1}(w)\subseteq A\}|\{x\in f^{-1}(w)\colon \{x\}\times
g^{-1}(w)\subseteq B\},\] which contradicts that $f^{-1}(w)$ is
connected.  Thus, $h$ is monotone.  Since $I$ is connected, we
also have that $R$ is a continuum.

We now show that $h^{-1}(w)$ is nowhere dense for all $w\in I$.
Let $\langle x,y\rangle\in h^{-1}(w)$.  Since $\dis(f^*)\cap\dis(g^*)=\emptyset$, we may  assume that $g^*$ is continuous at $w$.  Since $f$ is a $T$ map, there exists a sequence $\{x_n\in X\setminus f^{-1}(w)\}_{n\in\omega}$ converging
to $x$ such that $\{f(x_n)\}_{n\in\omega}$ converges to $f(x)=w$.
Since $g^*$ is continuous at $w$ and $f(x_n)\to w$, we have $g^*(f(x_n))\to g^*(w)$.  
So, we may pick a
sequence $\{y_n\in Y\}_{n\in\omega}$ converging to $y$ such that
$g(y_n)=f(x_n)$.  Now $\{\langle x_n,y_n\rangle\}_{n\in\omega}$
converges to $\langle x,y\rangle$ and  $\{\langle
h(x_n,y_n)\rangle\}_{n\in\omega}$ converges to $w$ and 
$\langle x_n,y_n\rangle\notin h^{-1}(w)$ for all $n\in\omega$. Thus,
$h^{-1}(w)$ is nowhere dense. \qed




\lem{lem:21}{Let $X$ and $Y$ be continua, $f\colon X\to I$ and
$g\colon Y\to I$ be $T$ maps such that $f^*$ is continuous at $0$
and $g^*$ is continuous at $0$.  There exist connected dense
subsets $N\subseteq X$ and $M\subseteq Y$ and a connectivty
injection $h\colon N \to M$ such that for $a\in f^{-1}(0)$, $b\in
f^{-1}(1)$, $c\in g^{-1}(0)$, and $d\in g^{-1}(1)$, we have
$h(a)=d$ and $h(b)=c$.} \proof Since $f$ and $g$ are continuous, 
we have, (using Proposition~\ref{prop:use1}(ii)) that $f^*$ and $g^*$,  
are continuous outside of meager sets $M_f$ and $M_g$, respectively.
By \cite{cset} and the fact that $0\notin M_f$ and $0\notin M_g$
there is a decreasing homeomorphism $l\colon I\to I$ such that
$l[M_f]\cap M_g=\emptyset$.  Let $f_1=l\circ f$. Notice that
$\dis(f^*_1)=l[M_f]$.  Since $f_1$, and $g$ are $T$ maps and
$\dis(f_1^*)\cap\dis(g^*)=\emptyset$, Lemma \ref{lem:1} yields 
that \[R=\{\langle x,y\rangle\in X\times Y\colon f_1(x)=g(y)\}\] is a continuum
and $k\colon R\to I$ defined by $k(\langle
x,y\rangle)=f_1(x)=g(y)$ is a $T$ map.  Notice that $\langle
a,d\rangle\in k^{-1}(1)$ and $\langle b,c\rangle\in k^{-1}(0)$,
since $l$ is decreasing. By Lemma \ref{lem:0}, There is a
connectivity inverse $j\colon I\to R$ of $k$ with the property
that $j(1)=\langle a,d \rangle$ and $j(0)=\langle b,c\rangle$ and
$j[I]$ is dense in $R$. 

Let $N=\{x\in X\colon\text{ there is a } y_x\in Y \text{ such that }\langle x,y_x\rangle\in j[I]\}$.  Since $j[I]$ is dense and connected in $R$, it follows that $N$ is dense and connected in $X$. 

Define $h\colon N\to Y$ by $h(x)=y_x$.  To see that $h$ is well defined  take $x\in N$ and suppose that $\langle x,y_x\rangle,\langle x,y_x^*\rangle\in j[I]$.  Since $j[I]\subseteq R$, we have $k(\langle x,y_x \rangle)=f_1(x)=k(\langle x, y_x^*\rangle)$.  Since $j$ is an inverse of $k$, $\langle x,y_x\rangle=j(k(\langle x,y_x\rangle))=j(k(\langle x,y_x^*\rangle))=\langle x,y_x^*\rangle$.  Hence, $y_x=y_x^*$.  Therefore, $h$ is well defined. 

The same argument that was applied to the first coordinate can be used to show that $h$ is injective.

We show that $h$ is connectivity.  Let
$P$ be a connected subset of the domain $N$ of $h$.  We must show that $\Gamma(h|_P)$ is connected.   Since $P$ is connected, $f_1$ is continuous and $j$ is connectivity; $j[f_1[P]]$ is connected.  So, it is enough for us to show that $\Gamma(h|_P)=j[f_1[P]]$.

If $\langle x,y\rangle\in\Gamma(h|_P)$, then $x\in P$ and there is a $w\in I$ such that $j(w)=\langle x,y \rangle$.  It follows that $w=k(\langle x,y\rangle)=f_1(x)$.  So, $\langle x,y\rangle\in j[f_1[P]]$.

Suppose $\langle x,y \rangle\in j[f_1[P]]$.  Let $x^*\in P$ be such that $j(f_1(x^*))=\langle x,y\rangle$.  Since $x^*\in P\subseteq N$, we have $\langle x^*,h(x^*)\rangle\in j[I]\subseteq R$.  So, $k(\langle x^*,h(x^*)\rangle)=f_1(x^*)$.  It follows that, $\langle x,y \rangle=j(f_1(x^*))=j(k(\langle x^*,h(x^*)\rangle))=\langle x^*,h(x^*)\rangle$.  Hence, $\langle x,y\rangle\in\Gamma(h)$.

Let $M=h[N]$.  Since $\Gamma(h|_N)=j[I]$ is dense and connected in $R$, it follows that $M$ is dense and connected in $Y$.  We also have that
$h(a)=d$ and $h(b)=c$.

\qed

\lem{lem:tip}{Suppose $X$ is a continuum and $f\colon X\to I$ is a
$T$ map.  If $\diam(X)>\epsilon$, then there exist $p,q\in X$ such
that $\dist(p,q)>\epsilon/2$, $f(p)\neq f(q)$, $f^{-1}(f(q))= A^f_{f(q)}$,
and $f^*$ is continuous at $f(p)$.} \proof Let $p\in X$ be such
that $f^*$ is continuous at $f(p)$.  Since $\diam(X)>\epsilon$,
there is a $q_1\in X$ such that $\dist(p,q_1)>\epsilon/2$.

Let $0<\delta<(\dist(p,q_1)-\epsilon/2)/4$.  Let $K=\{w\in
I\colon\dist(q_1,f^*(w))\leq\delta\}$.  By
Proposition~\ref{prop:use1}, $K$ is closed.  
Starting with $K_0=K$, define 
sequences $\langle W_{\alpha}\rangle_{\alpha\in\omega_1}$ and
$\langle K_{\alpha}\rangle_{\alpha\in\omega_1}$ as we did before 
the statement of Proposition~\ref{prop:use0}.  
Let $W=\bigcup_{\alpha<\omega_1}W_{\alpha}$.

We claim that for every $w\in W$ we have
$\dist(f^*(w),q_1)=\delta$. First suppose $w\in W_0$. Since $w$ is
isolated in $K_0=K$, there is an open set $U$ such that $w\in U$ and
for all $x\in U\setminus\{w\}$ we have $\dist(f^*(x),q_1)>\delta$.
Since $f^{-1}(w)$ is nowhere dense in $X$, we must have that
$\dist(f^*(w),q_1)=\delta$.  Suppose now that $\alpha\in\omega_1$
and we have shown that $\dist(f^*(w),q_1)=\delta$ for all
$w\in\bigcup_{\beta<\alpha}W_{\beta}$. Let $w\in W_{\alpha}$.
Since $w$ is isolated in $K_{\alpha}$, there is an open set $U$
such that $w\in U$ and for every $x\in U\setminus\{w\}$ we have
$x\in\bigcup_{\beta<\alpha}W_{\beta}$ or $x\notin K$.  It
follows that $\dist(f^*(x),q_1)\geq\delta$ for all $x\in
U\setminus\{w\}$. Since $f^{-1}(w)$ is nowhere dense in $X$, we
have $\dist(f^*(w),q_1)=\delta$.  Thus, $\dist(f^*(w),q_1)=\delta$
for every $w\in W$.

If $K$ is countable then, by Proposition \ref{prop:use0}(ii),
$K=W$.  It follows, then, that $\dist(q_1,f^*(w))=\delta$ for all
$w\in K$, which contradicts that $\dist(q_1,f^*(f(q_1)))=0$.
Thus, $K$ is uncountable.

Since $K\setminus W$ is uncountable, there is by
Lemma~\ref{lem:00} a $w\in K\setminus (W\cup\{f(p)\})$ such that
$f^{-1}(w)= A^f_{w}$.  Pick $q\in f^{-1}(w)$ such that
$\dist(q,q_1)\leq\delta$. Clearly, $f^{-1}(f(q))\in A^f_{f(q)}$ and $f(p)\neq
f(q)$.  By our choice of $\delta$ we have
$\dist(p,q)>\epsilon/2$.\qed

An immediate Corollary to Lemma~\ref{lem:tip} is:

\cor{cor:tip}{If $X$ is a continuum $f\colon X\to I$ is a  $T$ map
and $\diam(X)>\epsilon$, then there is a subcontinuum $Y$ of $X$
and a  $T$ map $f_1\colon Y\to I$ such that there exist points
$p\in f_1^{-1}(0)$ and $q\in f^{-1}_1(1)$ with
$\dist(p,q)>\epsilon/2$ and $f_1^*$ is continuous at $0$.} \proof
To see it let $p,q\in X$ be as in Lemma~\ref{lem:tip}.  Without
loss of generality, we may assume that $f(p)<f(q)$.  Let
$Y=f^{-1}([f(p),f(q)]$.  Clearly, $p,q\in Y$.  We check that $f|_Y$ is
a $T$ map.  Since $(f|_Y)^{-1}(w)=f^{-1}(w)$ for all $w\in[p,q]$, we know that $f|_Y$ is monotone.  It remains to check that $f^{-1}(w)$ is nowhere dense in $Y$ for all $w\in[p,q]$.  This is easily seen to be the case for $w\in (p,q)$.  To see it for $p,q$ notice that $f^{-1}(p)=A^f_{f(p)}\subseteq \cl((p,q])$ and $f^{-1}(q)=A^f_{f(q)}\subseteq \cl([p,q))$.  So, $f|_Y$ is a $T$ map.

We may get the desired $f_1$ by
letting $h\colon[f(p),f(q)]\to I$ be an increasing homeomorphism
and setting $f_1(x)=h(f(x))$ for all $x\in Y$.\qed

We are now ready to prove our theorem.
\newline

{\sc Proof of Theorem} Let $X$ be a weak $S_4$ continuum.
Since $X$ contains no simple closed curve, it is enough for us to
show that $X$ is locally connected.  We also note that, by
Corollary 7.2 of \cite{jornad}, $X$ is hereditarily decomposable.

By way of contradiction, assume that $X$ is not locally connected.
By Exercise 5.22 and Theorem 5.12 of \cite{NAD1} there exist
mutually disjoint subcontinua $K$ and $\{K_n\}_{n\in\omega}$ in
$X$ such that $\lim_{n\to\infty}K_n=K$ and $K$ is nondegenerate.
Since $K$ is nondegenerate, we may also assume that there is an
$\epsilon>0$ such that $\diam(K_n)>\epsilon$ for every
$n\in\omega$.

For each $K_n$ pick points $a_n,b_n\in K_n$ such that
$\dist(a_n,b_n)>\epsilon$.  For each $n\in\omega$ let
$L_n\subseteq K_n$ be a continuum irreducible between $a_n$ and
$b_n$. By Proposition \ref{prop:1}, there is a  $T$ map $g_n\colon
L_n\to I$. By Corollary ~\ref{cor:tip}, we may find a continuum
$Y_n\subseteq L_n$, a  $T$ map $f_n\colon Y_n\to I$, and
$p_n,q_n\in Y_n$ such that $\dist(p_n,q_n)>\epsilon/2$,
$f_n(p_n)=0$, $f_n(q_n)=1$, and $f_n^*$ is continuous at $0$.
Picking convergent subsequences if necessary, we may assume that
there exist $p,q\in K$ such that the $\lim_{n\to\infty} p_n=p$, and $\lim_{n\to\infty}q_n=q$.
Notice that $\dist(p,q)\geq\epsilon/2$. By Lemma~\ref{lem:21}, for
every $n\in\omega$ there are connected dense sets $N_n\subseteq
Y_{2n}$ and $M_n\subseteq Y_{2n+1}$ and a connectivity injection
$h_n\colon N_{n}\to M_{n}$ such that $h_n(p_{2n})=q_{2n+1}$, and
$h_n(q_{2n})=p_{2n+1}$.

Consider the collection of mutually disjoint doubletons
\[S=(\bigcup_{n=0}^{\infty}\{\{x,h_n(x)\}\colon x\in N_n\})\cup\{\{p,q\}\}.\]
We show there is no continuous selector for $S$.
By way of contradiction, suppose $g\colon S\to X$ is a continuous
selector.

Fix $n\in\omega$.  Assume $g(\{p_{2n},h_n(p_{2n})\})=p_{2n}$.
Since $\{\{x,h_n(x)\}\colon x\in N_n\}$ is connected and $K_{2n}$ and $K_{2n+1}$ 
are disjoint, we have
$g(\{q_{2n},h_n(q_{2n})\})=q_{2n}$ by continuity. By a similar argument, if
$g(\{p_{2n},h_n(p_{2n})\})=h_n(p_{2n})=p_{2n+1}$, then
$g(\{q_{2n},h_n(q_{2n})\})=h_n(q_{2n})=q_{2n+1}$.

Now suppose that $g(\{p_{2n},h_n(p_{2n})\})=p_{2n}$ for every $n$ in some infinite subset $A$ of $\omega$.  Since
$\lim_{n\to\infty}\{p_{2n},h_n(p_{2n})\}=\{p,q\}$, we have
$g(\{p,q\})=p$.  However, it is also the case that
$g(\{q_{2n},h_n(q_{2n})\})=q_{2n}$ for all $n\in A$.  Since
$\lim_{n\to\infty}\{q_{2n},h_n(q_{2n})\}=\{p,q\}$, we must have
$g(\{p,q\})=q$.  Since $p\neq q$, we have a contradiction. A
similar argument shows that if
$g(\{p_{2n},h_n(p_{2n})\})=h_n(p_{2n})=q_{2n+1}$ for infinitely many
$n$ in $\omega$, then one also gets a contradiction.  Therefore,
$X$ is locally connected and thus a dendrite.\qed


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\end{document}
