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\markright{Almost Continuous ...
\ \ \today }


\title{Almost continuous images of $\real$ and $\infty$-ods}

%\MathReviews{Primary: 54C08, 54F15.}
%\keywords{cardinal functions; extendable, Darboux, almost continuous
%and peripherially continuous functions; functions with perfect road. }


\author{{\small Francis Jordan,\ }%
\thanks{AMS classification numbers: Primary: 54C08, 54F15 
\endgraf  Key words and phrases: almost continuous functions,
almost Peano, continuum, real line, infinite-od.}
\endgraf
\small  Department of Mathematics, \\ 
Loyola of University,\\ 
New Orleans, LA 70118\\ (fejord@hotmail.com) }

\date{}


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%\title{The minimum number of Darboux functions needed to cover Baire
%class 1}
%\author{Francis Jordan\\
%\small Dept. of Mathematics, West Virginia Univ., Morgantown, WV
%26506-6310}

\date{}


\pagestyle{myheadings}

\markright{Almost continuous images ...
\ \ \today }

\begin{document}\maketitle
\begin{abstract}
We investigate conditions under which almost Peano continua contain dense 
arc components.  These conditions involve the existence of almost continuous functions
from the real line onto continua and excluding $\infty$-ods from continua.  It is also
shown that a topological space of cardinality  less than or equal to continuum which
has a dense path component is the  almost continuous image of the real line.
\end{abstract}
 


\section{Introduction}
All terminology can be found in the Definitions and 
Terminology section of this paper.  
We consider conditions on an almost continuous image $X$ of
$[0,1]$ or
$\real$ which guarantee the existence of a dense arcwise
connected subset in $X$.  These results will 
involve excluding $\infty$-ods from a continuum $X$.  In
particular, we prove:

\thm{thm:1}{Let $X$ be an almost continuous image of $[0,1]$.  
If $X$ is a decomposable continuum which contains no
$\infty-od$, then $X$ has a dense arc-component.}

\thm{thm:2}{Let $X$ be an almost continuous image of $\real$.  
If $X$ is a continuum which contains no $\infty-od$, then $X$ has a dense
arc-component.}  
Theorem~\ref{thm:1} is no longer true if we do not assume
that $X$ is decomposable since there is an indecomposable arc-like continuum
which is the almost continuous image of $[0,1]$ but has no dense
arc-component
\cite[Example 2]{HAIG1}.  Moreover, the assumption that $X$ contain no $\infty$-od is also essential, \cite[Example 7]{GHV}.  By Theorem~\ref{thm:2}, \cite[Example
2]{HAIG1} gives an example of a continuum which is the
almost continuous image of $[0,1]$ but not $\real$.   

It is not known which spaces, or even continua, are the
almost continuous image of $\real$.  We have a partial
result in this direction.    
\thm{thm:3}{If a space $X$ has a dense path-component and 
$|X|\leq\cuum$, then $X$ is the almost continuous image of
$\real$.} 

An immediate corollary for continua is:
\cor{cor:1}{If a continuum $X$ has a dense arc-component, then $X$ is the almost continuous image of $\real$.}
 I do not
know if the converse of Theorem~\ref{thm:3} or Corollary~\ref{cor:1} is true or not.

We will use the
following theorem due to Kellum \cite{KEL1} many times:
\thm{thm:0}{If $X$ is a second countable $T_1$ space then 
$X$ is the almost continuous image of a Peano continuum if and
only if $X$ is almost Peano.}

\section{Definitions and Terminology}
The symbol $\cuum$ will denote the cardinality of the real
line $\real$.  The unit interval will be denoted by $[0,1]$. 
Let $X$ and $Y$ be sets and $A\subseteq X\times Y$.  We denote
the natural projection of $A$ into $X$ or $Y$ by $\pi_{X}(A)$
or $\pi_{Y}(A)$, respectively. 

A {\em continuum} is a nonempty compact connected metric space.  For a set $A$ 
we let $\cl(A)$ and $\interior(A)$ denote the closure and interior of 
$A$, respectively.  Given
two subcontinua $A$ and $B$ of a continuum $X$, we let $H(A,B)$
denote the standard Hausdorff distance, as defined in
\cite[4.1]{NAD1}, between $A$ and
$B$.  A space is a {\em Peano continuum} provided that it is a
locally connected, continuum.  Two sets
$A,B\subseteq X$ are said to be {\em mutally separated} provided
that $\cl(A)\cap B=\emptyset$ and $\cl(B)\cap A=\emptyset$.  A
{\em $\infty$-od} is a continuum 
$X$ such that there exist a continuum $S\subseteq X$ such that
$X\setminus S$ has infinitely many components.  
We say a continuum $X$ is {\em decomposable}
provided that $X=A\cup B$ where $A$ and $B$ are proper
subcontinua of $X$.  If a continuum is not decomposable we say
that it is {\em indecomposable}.  The generalized open ball of radius 
$\epsilon>0$ about a set $A$ is denoted by $\ball_{\epsilon}(A)$ and consists 
of all points which are less than distance $\epsilon$ from some point in $A$.  
When $A$ has the form $\{x\}$ we write 
$\ball_{\epsilon}(x)$ for $\ball_{\epsilon}(\{x\})$.
 
A space is
called a {\em path} if it is the continuous image of $[0,1]$.  A
space $X$ is pathwise connected provided that any two points
of $X$ lie in some path contained in $X$.  A {\em path
component} of a space
$X$ is a maximal pathwise connected subset of $X$.  A space is 
{\em almost Peano} provided that for any finite
collection of nonempty open sets
$\{U_i\}_{i=1}^{n}$ there is a path $S$ such that $S\cap
U_i\neq\emptyset$ for each $1\leq i\leq n$.  The Hahn-Mazurkiewicz 
Theorem \cite[8.14]{NAD1}, implies that when $X$ is Hausdorff we may 
replace the word path with the word Peano continuum in the defintion of almost Peano.  
 
A function $f\colon X\to Y$ is an {\em almost continuous}
function provided that every open set in $X\times Y$
containing $f$ contains some continuous function
$g\colon X\to Y$.  We will denote the collection of almost
continuous functions from a space $X$ into a space $Y$ by
$\acon(X,Y)$.  We say a subset $B$ of $X\times Y$ is a {\em
blocking set} for
$\acon(X,Y)$ provided that $B$ is closed, $B$ has nonempty
intersection with each continuous function, and $B$ contains
no set of the form $\{x\}\times Y$ where $x\in X$.  It is
easily checked that a function $f\colon X\to Y$ is almost
continuous if and only if $f$ has nonempty intersection with
every blocking set for $\acon(X,Y)$.    


\section{Proof of Theorem~\ref{thm:1}}

We first prove a lemma which will help us identify
$\infty$-ods.  

\lem{lem:0}{Let $X$ be a continuum and let $Z$ and
$\{F_n\}_{n=1}^{\infty}$ be subcontinua of $X$ such that
for every $n$ we have:
\begin{description}
\item[(i)] $F_i\cap F_j\subseteq Z$, for all
$1\leq i<j\leq n$ and
\item[(ii)] $Z\subsetneq F_n$ and $H(F_n,Z)<1/n$. 
\end{description}
If $X=\bigcup_{n=1}^{\infty}F_n$, then $X$ is an $\infty-od$.}
\proof
It is enough for us to show that $F_n\setminus Z$ is open in
$X\setminus Z$ for every $1\leq n<\infty$.  Notice that
$F_n\setminus Z$ is nonempty by (ii) and $F_n\setminus Z$ is closed 
in $X\setminus Z$ by definition of the subspace topology.

Fix $n$.  By way of contradiction assume that $F_n\setminus Z$ is 
not open $X\setminus Z$.  Then, there is an $x\in F_n\setminus Z$ and 
a sequence of points $\{x_k\}_{k=1}^{\infty}$ in $X\setminus (F_n\cup Z)$ such that 
$\lim_{k\to\infty} x_k=x$.  Notice that, by (i), $F_k\cap\{x_k\colon 1\leq k<\infty\}$ 
is finite for every $1\leq k<\infty$.  So, by (ii), $\lim_{k\to\infty} x_k$ must be 
in $Z$, but $x\notin Z$.  Thus, $F_n\setminus Z$ is open in
$X\setminus Z$.\qed  

{\sc Proof of Theorem~\ref{thm:1}}
Assume that $X$ is a decomposable continuum which is the almost
continuous image of $[0,1]$ and has no dense arc-component.  We
show that $X$ must contain an $\infty$-od.  By
Theorem~\ref{thm:0}, $X$ is almost Peano.

Let $X=A\cup B$ where $A$ and $B$ are proper subcontinua;
note that both $A$ and $B$ have non-empty interior.   

By induction we construct a sequence
$\{F_n\}_{n=1}^{\infty}$ of mutally disjoint Peano continua such that 
$F_n\cap (X\setminus A)\neq\emptyset\neq F_n\cap A$ for every
$n$.   

Since $X$ is almost Peano, there is a Peano continuum $P_1$
with the property that
$P_{1}\cap(X\setminus A)\neq\emptyset\neq P_{1}\cap\interior(A)$. 
Let 
$F_1=P_1$.  

Suppose now that $n\geq 1$ and we have chosen 
$\{F_k\}_{k=1}^{n}$.  We show how to pick $F_{n+1}$. 
For each $1\leq k\leq n$ let $D_{k}$ be the closure of the
arc-component of $X$ which contains $F_{k}$.   By assumption,
$D_{k}\neq X$ for each $k$.  So there is an
$\epsilon>0$ such that for every $D_k$ there is a $x_k$ such
that $\ball_{\epsilon}(x_k)\cap D_k=\emptyset$.  
Since $X$ is almost Peano, there is a
Peano continuum $P_{n+1}$ such that 
$P_{n+1}\cap\ball_{\epsilon}(x_k)\neq\emptyset$ for every $k$ 
and $P_{n+1}\cap(X\setminus A)\neq\emptyset\neq
P_{n+1}\cap\interior(A)$.   Since
$P_{n+1}\cap\ball_{\epsilon}(x_k)\neq\emptyset$ for every
$k$, it follows that $P_{n+1}$ is not contained in any
$D_{k}$.  So, $P_{n+1}\cap F_{k}=\emptyset$ for every $k$. 
Also, $P_{n+1}\cap(X\setminus A)\neq\emptyset\neq P_{n+1}\cap
A$.  Thus, 
$F_{n+1}=P_{n+1}$ will have the desired properties. 

We now construct the $\infty$-od.  For every $1\leq n
<\infty$, 
$A$ is a proper subcontinuum of $A\cup F_n$.  There is, by
boundary bumping \cite{NAD1}, a continuum $C_n$ such that
$A\subsetneq C_n\subseteq A\cup F_n$ and $H(A,C_n)<1/n$.  
Notice that for any $1\leq i<j<\infty$, we have $C_i\cap
C_j\subseteq A$ since $F_i\cap F_j=\emptyset$.  By
Lemma~\ref{lem:0}, 
$C=\bigcup_{n=1}^{\infty}C_n$ is an $\infty$-od.\qed

\section{Proof of Theorem~\ref{thm:2}}

\lem{lem:1}{If a continuum $X$ is an almost continuous image of 
$\real$, then $X$ is almost Peano.  In particular, $X$ is the almost
continuous image of $[0,1]$.}
\proof
Let $f\colon\real\to X$ be an almost continuous map onto $X$.  
Let $\{U_k\}_{k=1}^{n}$ be a collection of nonempty open sets in $X$. 
Since $f$ is an onto function, there is for every $U_k$ an
$x_k\in\real$ such that $f(x_k)\in U_k$.  Let 
\[V=(\real\times X)\setminus\bigcup\{\{x_k\}\times(X\setminus U_k)\colon 1\leq k\leq
n\}.\] Notice that $V$ is open and $f\subseteq V$.  By almost continuity, there is 
a continuous function $g\colon\real\to X$ such that $g\subseteq V$.  Let
$J\subseteq\real$ be a compact interval such that $x_1,\ldots, x_n\in J$. 
Now $g[J]$ is a Peano continuum and, by our choice of $V$, we have $g[J]\cap
U_k\neq\emptyset$ for each $1\leq k\leq n$.  Therefore,
$X$ is almost Peano.\qed

\lem{lem:2}{Suppose that $f\colon\real\to X$ is an almost
continuous function onto a continuum $X$.  If $f[[M,\infty)]$ is dense in
$X$ for every
$M>0$, then $X$ has a dense arc-component.}
\proof
Let ${\cal B}=\{\ball_n\colon 1\leq n\in\infty\}$ be a countable
base for $X$.  Let $M_0=0$.  Since $f[[1,\infty)]$ is dense in
$X$, there exists an $M_1\in\real$ with the property that
$f[[1,M_1]]\cap\ball_1\neq\emptyset$.  Since
$f[[M_1+1,\infty)]$ is dense in $X$, there is an $M_2\in\real$
such that $f[[M_1+1,M_2]]\cap\ball_2\neq\emptyset$.  Continuing
inductively we may find an increasing sequence
$\{M_n\}_{n=0}^{\infty}$ of real numbers such that
$f[[M_{n-1}+1,M_n]]\cap B_n\neq\emptyset$.  For each $n$ pick 
$t_n\in [M_{n-1}+1,M_n]]$ such that $f(t_n)\in\ball_n$.  Since 
$\{t_n\colon 1\leq n <\infty\}$ is a discrete set of points
all of which are distinct, the set $U\subseteq\real\times X$
defined by
\[U=(\real\times
X)\setminus\left(\bigcup_{n=1}^{\infty}\{t_n\}\times
(X\setminus\ball_n)\right)\] is open.  Since $f\subseteq U$ and
$f$ is almost continuous,  there is a continuous function
$g\colon\real\to X$ such that $g\subseteq U$.  Since
$g\subseteq U$, we have $g(t_n)\in\ball_n$ for every $n$.  So
$g[\real]$ is dense in $X$.  Obviously, $g[\real]$ is arcwise
connected so
$X$ has a dense arc-component.\qed 

{\sc Proof of Theorem~\ref{thm:2}}
Let $f\colon\real\to X$ be an almost continuous onto function.  
Assume that $X$ contains no $\infty$-od.  We show that $X$ has
a dense arc-component.

If $X$ is decomposable, then Lemma~\ref{lem:1} and 
Theorem~\ref{thm:1} imply that $X$ must have a dense
arc-component.  

We may now assume that $X$ is indecomposable and
nondegenerate.  By way of contradiction, assume that $X$ has
no dense arc-component.  By Lemma~\ref{lem:2}, it follows
that for some $M>0$, $f[[M,\infty)]$ is not dense in
$X$.

Let 
$Z=\cl(f[[M,\infty)])$.  Since $f[[M,\infty)]$ is connected
\cite[Theorem 1.7]{NATsurvey}, $Z$ is a continuum.  Since
$f[[M,\infty)]$ is not dense, $Z$ is a proper subcontinuum of $X$
and is nowhere dense in $X$ \cite[5.20a \& 11.17]{NAD1}.   

By induction we construct a
sequence $\{W_n\}_{n=1}^{\infty}$ of 
subcontinua of $X$ such that for every $n$: 
\begin{description}
\item[(C1)] $W_i\cap W_j\subseteq Z$ for every $1\leq i<j\leq
n$, 
\item[(C2)] $Z\subsetneq W_n$, and
\item[(C3)] there is a $t_n\in\real$ and a continuous function
$g_n\colon[t_n,\infty)\to X$ such that
$W_n=g_n[[t_n,\infty)]\cup Z$.
\end{description}
Define $U\subseteq\real\times X$ to be the open set 
\[U=((-\infty,M)\times X)\cup
\left(\bigcup_{n=0}^{\infty}[M+n,M+(n+1))\times\ball_{1/n+1}(Z)\right).\]
Notice that $f\subseteq U$.  To construct $W_1$ pick a point
$p_1\in X\setminus Z$.  Since $p_1\notin Z$, there is a
$t_1\in (-\infty,M)$ such that $f(t_1)=p_1$.  There is an
$\epsilon_1>0$ such
$\ball_{\epsilon_1}(p_1)\cap Z=\emptyset$.  Let $U_1=U\setminus 
\left[\{t_1\}\times
(X\setminus\ball_{\epsilon_1}(p_1))\right]$.  Notice that
$U_1$ is open and $f\subseteq U_1$.  Since $f$ is almost
continuous, there is a continuous $g_1\colon\real\to X$ such
that $g_1\subseteq U_1$.  Let $W_1=g_1[[t_1,\infty)]\cup Z$. 
Using the fact that $g_1\subseteq U$ it is easy to verify that 
$W_1$ is a continuum.  Clearly, $Z\subsetneq W_1$ 
since $f(t_1)\notin Z$.  Thus,
$W_1$ satisfies (C1), (C2), and (C3).

Assume that $n\geq 1$ and that $W_1,\ldots, W_n$ have been
constructed so that (C1), (C2), and (C3) are satisfied.  We
construct
$W_{n+1}$ so that (C1), (C2), and (C3) are satisfied.  For
each $1\leq k<n+1$ let
$A_k$ be the arc-component of $X$ that contains
$g_k[[t_k,\infty)]$.  
By assumption,
$\cl(A_k)\neq X$ for each $1\leq k<n+1$.  Since $\cl(A_k)$ is a
proper subcontinuum, it follows that $\cl(A_k)\cup Z$ is nowhere
dense.  So, $T=Z\cup\bigcup_{k=1}^{n}\cl(A_k)$ is closed and
nowhere dense in $X$.  Pick a point
$p_{n+1}\in X\setminus T$.  Since $p_{n+1}\notin Z$, there is a
$t_{n+1}\in (-\infty,M)$ such that $f(t_{n+1})=p_{n+1}$.  There
is an
$\epsilon_{n+1}>0$ such
$\ball_{\epsilon_{n+1}}(p_{n+1})\cap Z=\emptyset$.  Let
$U_{n+1}=U\setminus 
\left[\{t_{n+1}\}\times
(X\setminus\ball_{\epsilon_{n+1}}(p_{n+1}))\right]$.  Notice
that
$U_{n+1}$ is open and $f\subseteq U_{n+1}$.  Since $f$ is almost
continuous, there is a continuous $g_{n+1}\colon\real\to X$ such
that $g_{n+1}\subseteq U_{n+1}$.  Let
$W_{n+1}=g_{n+1}[[t_{n+1},\infty)]\cup Z$.  Using the fact that
$g_{n+1}\subseteq U$ it is easy to verify that 
$W_{n+1}$ is a continuum.  Clearly, $Z\subsetneq W_{n+1}$ 
since $f(t_{n+1})\notin Z$.  So (C2) and (C3) are 
satisfied.  We show that (C1) is satisfied.  By inductive
assumption, it is enough for us to check that $W_{n+1}\cap
W_{k}\subseteq Z$ for every $k<n+1$.  Fix $k<n+1$.  Since
$f(t_{n+1})\notin T\supseteq A_k$ and
$g_{n+1}[[t_{n+1},\infty)]$ is arcwise connected, it follows
that
$g_{n+1}[[t_{n+1},\infty)]\cap A_k=\emptyset$.  Since
$g_k[[t_k,\infty)]\subseteq A_k$, we have $W_k\cap
W_{n+1}\subseteq Z$.  Thus, (C1) is satisfied completing the
induction.

Since $Z$ is a proper subcontinuum of $W_n$ for each $1\leq
n<\infty$, we may find by boundary bumping \cite{NAD1} a
subcontinuum $F_n$ such that $Z\subsetneq F_n\subseteq W_n$ and
$H(F_n,Z)<1/n$.  By (C1) $F_n\cap F_m\subseteq Z$, for all
$m\neq n$.  By Lemma~\ref{lem:0}, 
$\bigcup_{n=1}^{\infty}F_n$ is an $\infty$-od which
contradicts our original assumption.\qed   


\section{Proof of Theorem~\ref{thm:3}}
Let $X$ be a space and $[a,b]$ be a compact interval in
$\real$.  Fix $p,q\in X$.  We say a function $f\colon [a,b]\to
X$ is {\em $(p,q)$-almost continuous} provided that for every
open set $U\subseteq [a,b]\times X$ such that $f\subseteq U$
there is a continuous function $g\colon [a,b]\to X$ such that
$g\subseteq U$ and $g(a)=p$ and $g(b)=q$.  We say $B\subseteq
[a,b]\times X$ is a {\em $(p,q)$-blocking set} provided that 
\begin{description}
\item[(a)] $B$ is closed,
\item[(b)] $B$ contains no set of the form $\{w\}\times X$
where $w\in [a,b]$,
\item[(c)] $B\cap\{\langle a,p\rangle,\langle b,q
\rangle\}=\emptyset$, and
\item[(d)] $B\cap g\neq\emptyset$ for every continuous $g\colon
[a,b]\to X$ such that $g(a)=p$ and $g(b)=q$.
\end{description}
We say a $(p,q)$-blocking set $B$ is {\em irreducible} if no
$(p,q)$-blocking set is properly contained in $B$.  
It is easily checked that if $f\colon [a,b]\to X$ is such that
$f(a)=p$ and $f(b)=q$, then $f$ is $(p,q)$-almost continuous if
and only if $f$ has nonempty intersection with every
$(p,q)$-blocking set.  Notice that by the definition of
$(p,q)$-blocking set, if
$X$ is a single point, then there are no $(p,q)$-blocking sets
and every function $f\colon [a,b]\to X$ is $(p,q)$-almost
continuous.  We will denote the collection of $(p,q)$-almost continuous functions 
from $[a,b]$ into $X$ by $\acon_{p,q}([a,b],X)$.

The next two lemmas are
modifications of Lemma 1 and Lemma 2 of \cite{KEL1} for
$(p,q)$-almost continuity; the proofs are essentially the same
but are included for completeness.

\lem{lem:39}{If $p,q\in X$ and 
$B$ is a $(p,q)$-blocking set for $\acon_{p,q}([0,1],X)$, then
$B$ contains an irreducible $(p,q)$-blocking set for
$\acon_{p,q}([0,1],X)$.}
\proof
Let $\cball$ be the collection of all $(p,q)$-blocking sets
contained in $B$.  Let $\cball^*\subseteq\cball$ be linearly
ordered by inclusion.  Suppose that $g\colon[0,1]\to X$ is a
continuous function such that $g(0)=p$ and $g(1)=q$.  Notice
that the graph of $g$ is homeomorphic to $[0,1]$ and is
compact.  By assumption, $B^*\cap g\neq\emptyset$ for every
$B^*\in\cball^*$.  So, $\{B^*\cap g\colon
B^*\in\cball^*\}$ is a collection of nonempty compact
subsets of $g$ linearly ordered by inclusion.  Thus,
$\bigcap\{B^*\cap g\colon B^*\in\cball^*\}\neq\emptyset$.  It follows
that $C=\bigcap\cball^*$ has nonempty
intersection with $g$.  
Since $g$ is arbitrary and
$C\subseteq B$, it follows that $C$ is a $(p,q)$-blocking
set.  The result now follows from the Zorn Lemma.\qed   

\lem{lem:4}{If $X$ is a path, $p,q\in X$ and 
$B$ is an irreducible $(p,q)$-blocking set for
$\acon([0,1],X)$, then
$\pi_{[0,1]}(B)$ is a nondegenerate interval.}
\proof
First note that $\pi_{[0,1]}(B)$ is not of the form $\{w\}$
where $w\in [0,1]$.  Assume there is such a $w$.  Clearly,
$w\notin\{0,1\}$ since $B\cap\{\langle
0,p\rangle,\langle 1,q\rangle\}=\emptyset$.  So, we may assume
that $w\in(0,1)$.  By path connectedness, there is for every
$x\in X$ a continuous function $g\colon[0,1]\to X$ such that
$g(0)=p$, $g(w)=x$, and $g(1)=q$.  Thus, $\langle w,x\rangle\in
B$.  Since $x$ is arbitrary, it follows that $\{w\}\times
X\subseteq B$, a contradiction.  Thus, $\pi_{[0,1]}(B)$ is not
a point.

If we assume that $\pi_{[0,1]}(B)$ is not connected then, by
compactness, there exist $a<b\in\pi_{[0,1]}(B)$ such that
$(a,b)\cap\pi_{[0,1]}(B)=\emptyset$.  Since $B$ is
irreducible, there exist continuous functions
$g_a,g_b\colon[0,1]\to X$ such that $g_a(0)=g_b(0)=p$,
$g_a(1)=g_b(1)=q$, $g_a|_{[b,1]}\cap B=\emptyset$, and
$g_b|_{[0,a]}\cap B=\emptyset$.  Again using path connectedness
we may find a continuous $h\colon[0,1]\to X$ such that
$h|_{[0,a]}=g_b|_{[0,a]}$, $h|_{[b,1]}=g_a|_{[b,1]}$, and $h\cap
B=\emptyset$, a contradiction.\qed

\lem{lem:7}{Let $X$ and $Y$ be spaces and $X$ be compact.  If
$B\subseteq X\times Y$ is closed and $B$ contains no set of the form
$\{x\}\times Y$ where $x\in X$, then there exist a finite
collection
$\{V_{i}\}_{i\in n}$ of nonempty open subsets of
$Y$ such that for any set $S\subseteq Y$, if $S\cap
V_{i}\neq\emptyset$ for each $i\in n$, then $\{w\}\times
S\not\subseteq B$ for every $w\in X$.}
\proof
Let $x\in X$.  Since $\{x\}\times Y\not\subseteq B$ by
assumption, there is a $y\in Y$ such that 
$\langle x,y\rangle\notin B$.  Since $B$ is closed, there 
exist open sets $U_x\subseteq X$ and $V_x\subseteq Y$ such
that $B\cap (U_x\times V_x)=\emptyset$ and $\langle
x,y\rangle\in (U_x\times V_x)$.  Clearly,
$V_x\neq\emptyset$.  By compactness, there is a finite
collection $\{U_{x_i}\}_{i\in n}$ such that $X=\bigcup_{i\in
n}U_{x_i}$.  Let $S$ be a set such that $S\cap
V_{x_i}\neq\emptyset$ for every $i\in n$.  Let $w\in X$.  There
is an $i\in n$ such that $w\in U_{x_i}$.  By the way $U_{x_i}$
and $V_{x_i}$ were selected, we have $(\{w\}\times V_{x_i})\cap
B=\emptyset$.  Since $S\cap V_{x_i}\neq\emptyset$, it follows
that $\{w\}\times S\not\subseteq B$.  Thus, $\{V_{x_i}\}_{i\in
n}$ is the desired finite collection of nonempty open sets.\qed 


\lem{lem:13}{If $X$ is a second countable space and $Y$ is a
space with no more than $\cuum$-many open sets, then $X\times Y$
contains no more than 
$\cuum$-many open sets.  In particular, if $Y$ is a path,
then there are at most $\cuum$ many open subsets of
$[0,1]\times Y$.}
\proof
Let $\cball$ be a countable base for $X$.  Let ${\cal T}$ denote
the collection of open sets of $Y$.  The collection
${\cal M}=\{B\times V\colon B\in\cball\ \&\ V\in{\cal T}\}$ is
a base for $X\times Y$.  Clearly, $|{\cal M}|\leq\cuum$.  Let
$U\subseteq X\times Y$ be open.  The collection $\cball_U=\{
B\in\cball\colon (\exists M\in {\cal M}) (B=\pi_{X}(M)\ \&\ 
M\subseteq U)\}$ is countable.  For each $B\in\cball_U$ let
$V_B=\bigcup\{\pi_{Y}(M)\colon M\in {\cal
M}\ \&\ B=\pi_{X}(M)\ \&\  M\subseteq U\}$.  Clearly, $B\times
V_B\in{\cal M}$ and it is easily checked that $B\times
V_B\subseteq U$.  It is also the case that
$U\subseteq\bigcup_{B\in\cball_U}B\times V_B$.  Thus, $U$ is
the countable union of elements from ${\cal M}$.  Since $U$ was
arbitrary and $|{\cal M}|\leq\cuum$, it follows that $X\times Y$
has at most $\cuum$-many open sets. 

To see the second statement it is enough to notice that $[0,1]$
is second countable and that any continuous image of $[0,1]$ has at
most $\cuum$-many open sets.\qed

\lem{lem:6}{If $X$ has a dense pathwise connected subset $D$ and
$|X|\leq\cuum$, then there is a collection ${\cal S}$ of compact
subsets of
$\real\times X$ such that
\begin{description}
\item[(i)] $|{\cal S}|\leq\cuum$, 
\item[(ii)] if $f\colon\real\to X$ is such that $f\cap
S\neq\emptyset$ for every $S\in{\cal S}$, then 
$f\in\acon(\real,X)$, and 
\item[(iii)]  $\pi_{\real}(S)$ contains a
nondegenerate interval for every $S\in{\cal S}$.  
\end{description}}
\proof
We define ${\cal S}$.  Let ${\cal F}$ denote the nonempty
finite subsets of $D$.  For each
$F\in{\cal F}$ let $P_F$ be a path in $D$ such that
$F\subseteq P_F$.   Let ${\cal P}=\{P_F\colon F\in{\cal F}\}$. 
For each
$F\in{\cal F}$ and $n\in\integer$ let $\cball^n_{F}$ denote the
subsets of
$\real\times X$ which are irreducible
$(p,q)$-blocking sets or irreducible $(q,p)$-blocking sets
for $[n,n+1]$ and $P_{F}$ where $n\in\integer$ and 
$\{p,q\}\subseteq F$.  Let
\[{\cal S}=\bigcup\{\cball^n_{F}\colon F\in{\cal F}\ \&\
n\in\integer\}.\] 

By Lemma~\ref{lem:4}, we have (iii).  
Lemma~\ref{lem:13} and 
$|\{\cball^n_{F}\colon F\in{\cal F}\ \&\
n\in\integer\}|\leq\cuum$ imply (i).  We now show that (ii)
holds.

Let $f\colon\real\to X$ be a function which has nonempty
intersection with every member of ${\cal S}$.  By way of
contradiction, assume that $f$ is not almost continuous.  There
is a blocking set $B\subseteq\real\times X$ for
$\acon(\real,X)$ such that
$B\cap f=\emptyset$.  
We may
find a function $j\colon\integer\to D$ such that 
$j\cap B=\emptyset$ since $D$ is dense and such that $j(n)=f(n)$ whenever $f(n)\in
D$.  

Fix $n\in\integer$.  
Let $B_n=B\cap ([n,n+1]\times X)$.  Since $B_n$ is
closed and contains no set of the form $\{w\}\times X$, there
is, by Lemma~\ref{lem:7}, a finite collection $\{V_{i}\}_{i\in
n}$ of nonempty open subsets of
$X$ such that for any set $S\subseteq X$, if $S\cap
V_{i}\neq\emptyset$ for each $i\in n$, then $\{w\}\times
S\not\subseteq B$ for every $w\in [n,n+1]$.  Since $D$ is dense
in
$X$, there is a finite set $F_1\subseteq D$ such that
$F_1\cap V_i\neq\emptyset$ for every $i\in n$.  Let
$F=F_1\cup\{j(n),j(n+1)\}$.  By our choice of $\{V_{i}\}_{i\in
n}$, we have $\{t\}\times P_F$ is not contained in $B$ for every $t\in [n,n+1]$.  So, we may find a function $k\colon [n,n+1]\to
P_F$ such that $k(n)=j(n)$, $k(n+1)=j(n+1)$, and $k\cap
B_n=\emptyset$.  Consider the function $f^*_n\colon [n,n+1]\to P_F$ defined by 
\begin{equation}
f^{*}_n(x)=
\begin{cases} 
k(x) & \text{if $f(x)\notin P_F$;}\\
f(x) & \text{if $f(x)\in P_F$.}
\end{cases}
\end{equation}
Notice that $f^{*}\in\acon_{j(n),j(n+1)}([n,n+1],P_F)$ since the range of $f^*_n$ is
contained in $P_F$ and 
$f^*_n$ has nonempty intersection with every member of ${\cal S}$,
and thus, every member of ${\cal B}^n_{F}$.   Since $f^{*}_n\cap B_n=\emptyset$, there is
a continuous function $g_n\colon [n,n+1]\to P_{F}$
such that
$g_n(n)=j(n)$, $g_n(n+1)=j(n+1)$, and $g_n\cap B_n=\emptyset$. 

For each $n\in\integer$ let $g_n$ be as constructed above.  Now,
$g=\bigcup_{n\in\integer}g_n$ is a continuous function such that $g\cap  B$, a contradiction.  
Thus, $f$ is almost continuous and (ii) holds.\qed


{\sc Proof of Theorem~\ref{thm:3}}
Let $D$ be a dense pathwise connected subset of $X$.  

By Lemma~\ref{lem:6}, there is a collection ${\cal S}$ of
compact subsets of
$\real\times X$ such that
\begin{description}
\item[(i)] $|{\cal S}|\leq\cuum$, 
\item[(ii)] if $f\colon\real\to X$ is such that $f\cap
S\neq\emptyset$ for every $S\in{\cal S}$, then 
$f\in\acon(\real,X)$, and 
\item[(iii)]  $\pi_{\real}(S)$ contains a
nondegenerate interval for every $S\in{\cal S}$.  
\end{description}

Let $\{S_{\alpha}\}_{\alpha\in\cuum}$ be an enumeration of
${\cal S}$.  Let $H_1,H_2\subseteq\real$ be a partition of
$\real$ such that
$|H_1|=|H_2|=\cuum$ and $|H_1\cap I|=\cuum$ for every
nondegenerate interval $I\subseteq\real$.  Using a standard
diagonalization argument and (iii) we may by transfinite
induction construct a function $f^*\colon H_1\to X$ such that
$f^*\cap S_{\alpha}\neq\emptyset$ for every $\alpha\in\cuum$.  
By (ii), any extension of $f^*$ to all of $\real$ will be
almost continuous.  Since $|X|\leq |H_2|=\cuum$, we may extend
$f^*$ to an onto almost continuous function $f\colon\real\to
X$.\qed

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\end{document}

