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%TCIDATA{Created=Fri Oct 11 19:19:45 1996}
%TCIDATA{LastRevised=Thu Oct 08 13:09:15 1998}
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\begin{document}
\title{Extending connectivity functions on $\Bbb{R}^{n}$}
\author{Krzysztof Ciesielski}
\address[K. Ciesielski and J. Wojciechowski]{Department of Mathematics\\
West Virginia University\\
PO Box 6310\\
Morgantown, WV 26506-6310, USA}
\email[K. Ciesielski]{kcies@wvnvms.wvnet.edu}
\author{Tomasz Natkaniec}
\address[T. Natkaniec]{Department of Mathematics\\
Gda\'{n}sk University\\
Wita Stwosza 57, 80-952 Gda\'{n}sk, Poland}
\email[T. Natkaniec]{mattn@ksinet.univ.gda.pl}
\author{Jerzy Wojciechowski}
\email[J. Wojciechowski]{jerzy@math.wvu.edu}
\subjclass{Primary 26B05; Secondary 54C30}
\keywords{connectivity functions, extendable functions, Darboux-like
functions\smallskip .}
\thanks{This work was partially supported by NSF Cooperative Research Grant
INT-9600548 with its Polish part being financed by Polish Academy of Science
PAN}
\maketitle

\begin{abstract}
A function $f:\Bbb{R}^{n}\rightarrow \Bbb{R}$ is a \emph{connectivity
function} if for every connected subset $C$ of $\Bbb{R}^{n}$ the graph of
the restriction $f|C$ is a connected subset of $\Bbb{R}^{n+1}$, and $f$ is
an \emph{extendable connectivity function} if $f$ can be extended to a
connectivity function $g:\Bbb{R}^{n+1}\to \Bbb{R}$ with $\Bbb{R}^{n}$
embedded into $\Bbb{R}^{n+1}$ as $\Bbb{R}^{n}\times \left\{ 0\right\} $.
There exists a connectivity function $f:\Bbb{R}\rightarrow \Bbb{R}$ that is
not extendable. We prove that for $n\ge 2$ every connectivity function $f:%
\Bbb{R}^{n}\rightarrow \Bbb{R}$ is extendable.
\end{abstract}

\section{Introduction}

Given functions $f:\Bbb{R}^{n}\rightarrow \Bbb{R}$ and $g:\Bbb{R}^{n+1}\to 
\Bbb{R}$, we say that $g$ \emph{extends} $f$ if $g$ extends the composition $%
f\circ \tau :\Bbb{R}^{n}\times \left\{ 0\right\} \rightarrow \Bbb{R}$, where 
$\tau :\Bbb{R}^{n}\times \left\{ 0\right\} \rightarrow \Bbb{R}^{n}$ and 
\begin{equation}
\tau (\left\langle x_{1},x_{2},\dots ,x_{n},0\right\rangle )=\left\langle
x_{1},x_{2},\dots ,x_{n}\right\rangle ,  \label{b}
\end{equation}
for every $\left\langle x_{1},x_{2},\dots ,x_{n}\right\rangle \in \Bbb{R}%
^{n} $. A function $f:\Bbb{R}^{n}\rightarrow \Bbb{R}$ is a \emph{%
connectivity function} if for every connected subset $C$ of $\Bbb{R}^{n}$
the graph of the restriction $f|C$ is a connected subset of $\Bbb{R}^{n+1}$,
and $f$ is an \emph{extendable connectivity function} if there exists a
connectivity function $g:\Bbb{R}^{n+1}\to \Bbb{R}$ extending $f$.

It follows immediately from the definition that every extendable
connectivity function is a connectivity function. Cornette \cite{Cornette:
connectivity and peano} and Roberts \cite{Roberts: zero-dim blocking
connectivity} proved that there exists a connectivity function $f:\Bbb{R}%
\rightarrow \Bbb{R}$ that is not extendable. This result was surprising and
sparked the interest in the family of extendable connectivity functions.
Ciesielski and Wojciechowski \cite{CieWoj: sums of connectivity} asked
whether there exists a connectivity function $f:\Bbb{R}^{n}\rightarrow \Bbb{R%
}$, with $n\ge 2$, that is not extendable. In this paper we will show that
the answer to that question is negative.

\begin{theorem}
\label{thm: connectivity are extendable}If $n\ge 2$ then every connectivity
function $f:\Bbb{R}^{n}\rightarrow \Bbb{R}$ is extendable.
\end{theorem}

To prove Theorem \ref{thm: connectivity are extendable} we will use ideas
from Gibson and Roush \cite{GibRoush: characterization of extendable} where
is formulated a necessary and sufficient condition for a connectivity
function $f:\left[ 0,1\right] \rightarrow \left[ 0,1\right] $ to be
extendable to a connectivity function $f:\left[ 0,1\right] ^{2}\rightarrow
\left[ 0,1\right] $ (if one considers $\left[ 0,1\right] $ to be embedded in 
$\left[ 0,1\right] ^{2}$ as $\left[ 0,1\right] \times \left\{ 0\right\} $).

Our basic terminology and notation is standard. (See \cite{Ciesielski: set
theory for working} or \cite{Engelking: general topology}.) In particular,
if $A$ is a subset of a metric space $X$, then $\func{bd}A$, $\func{cl}A$
and $\func{diam}A$ will denote the boundary, closure, and diameter of $A$ in 
$X$ respectively, and if $f$ is a function and $A$ is a subset of its
domain, then $f[A]$ is the image of $A$ under $f$.

The following additional terminology will be useful in our proof. Given a
function $f:\Bbb{R}^{n}\rightarrow \Bbb{R}$, a \emph{peripheral pair} (\emph{%
for }$f$) is an ordered pair $\left\langle A,I\right\rangle $ with $I$ being
a closed interval in $\Bbb{R}$ and $A$ being an open bounded subset of $\Bbb{%
R}^{n}$ with $f[\func{bd}A]\subseteq I$. Given $\varepsilon >0$, an $%
\varepsilon $\emph{-peripheral pair} is a peripheral pair $\left\langle
A,I\right\rangle $ with $\func{diam}A<\varepsilon $ and $\func{diam}%
I<\varepsilon $. Given a point $x\in \Bbb{R}^{n}$, a peripheral pair for $f$ 
\emph{at }$x$ is a peripheral pair $\left\langle A,I\right\rangle $ for $f$
with $x\in A$ and $f(x)\in I$. A function $f:\Bbb{R}^{n}\rightarrow \Bbb{R}$
is said to be \emph{peripherally continuous} if for every $x\in \Bbb{R}^{n}$
and $\varepsilon >0$ there is an $\varepsilon $-peripheral pair for $f$ at $%
x $.

The class of peripherally continuous functions $f:\Bbb{R}\rightarrow \Bbb{R}$
is strictly larger than the class of connectivity functions. However, the
following result holds.

\begin{theorem}
\label{thm: connectivity and peripherally cont}If $n\ge 2$ then a function $%
f:\Bbb{R}^{n}\rightarrow \Bbb{R}$ is peripherally continuous if and only if
it is a connectivity function.
\end{theorem}

The implication that a connectivity function is peripherally continuous in
Theorem~\ref{thm: connectivity and peripherally cont} was proved by Hamilton 
\cite{Hamilton: fixed points for noncontinuous} and Stallings \cite
{Stallings: fixed point for connectivity}, and the opposite implication was
proved by Hagan \cite{Hagan: equivalence connectivity peripherally}.

Let $f:\Bbb{R}^{n}\rightarrow \Bbb{R}$ be a function and $\mathcal{P}$ be a
family of peripheral pairs for $f$. We say that $\mathcal{P}$ \emph{locally
converges to }$0$ if for every $\varepsilon >0$ and every bounded set $%
X\subseteq \Bbb{R}^{n}$ the set 
\begin{equation*}
\left\{ \left\langle A,I\right\rangle \in \mathcal{P}:A\cap X\neq \emptyset 
\text{ and }\func{diam}A\ge \varepsilon \right\}
\end{equation*}
is finite, and that $\mathcal{P}$ has the \emph{intersection property}
provided $I\cap I^{\prime }\neq \emptyset $ for any $\left\langle
A,I\right\rangle ,\left\langle A^{\prime },I^{\prime }\right\rangle \in 
\mathcal{P}$ such that each of the sets $A\cap A^{\prime }$, $A\setminus
A^{\prime }$, and $A^{\prime }\setminus A$ is nonempty. Given $X\subseteq 
\Bbb{R}^{n}$, we say that $\mathcal{P}$ is an $f$\emph{-base for} $X$ if for
every $\varepsilon >0$ and $x\in X$ there exists an $\varepsilon $%
-peripheral pair for $f$ at $x$ that belongs to $\mathcal{P}$. Note that a
function $f:\Bbb{R}^{n}\rightarrow \Bbb{R}$ is peripherally continuous if
and only if there exists an $f$-base for some set $X\subseteq \Bbb{R}^{n}$
that contains all points of discontinuity of $f$. A \emph{peripheral family
for} $f:\Bbb{R}^{n}\rightarrow \Bbb{R}$ is a countable family of peripheral
pairs for $f$ that locally converges to $0$, has the intersection property,
and is an $f$-base for $\Bbb{R}^{n}$.

Theorem \ref{thm: connectivity are extendable} follows from Theorem \ref
{thm: connectivity and peripherally cont} and the following two results.

\begin{theorem}
\label{thm: exist peripheral families}If $n\ge 2$ and $f:\Bbb{R}%
^{n}\rightarrow \Bbb{R}$ is a peripherally continuous function, then there
exists a peripheral family for $f$.
\end{theorem}

If $\left\langle A,I\right\rangle $ is a peripheral pair (for some $f:\Bbb{R}%
^{n}\rightarrow \Bbb{R}$), then the \emph{cylindrical extension} of $%
\left\langle A,I\right\rangle $ is a pair $\left\langle A^{\prime
},I\right\rangle $, where 
\begin{equation*}
A^{\prime }=A\times \left( -\func{diam}A,\func{diam}A\right) \subseteq \Bbb{R%
}^{n+1}.
\end{equation*}
If $\mathcal{P}$ is a set of peripheral pairs, then the \emph{cylindrical
extension} of $\mathcal{P}$ is the set of cylindrical extensions of all the
elements of $\mathcal{P}$.

The case $n=1$ of the following theorem is a modification of a result of
Gibson and Roush \cite{GibRoush: characterization of extendable}.

\begin{theorem}
\label{thm: use peripheral families}If $n\ge 1$ and $\mathcal{P}$ is a
peripheral family for $f:\Bbb{R}^{n}\rightarrow \Bbb{R}$, then there exists
a continuous function 
\begin{equation*}
h:\Bbb{R}^{n+1}\setminus \left( \Bbb{R}^{n}\times \left\{ 0\right\} \right)
\rightarrow \Bbb{R}
\end{equation*}
such that every element of the cylindrical extension of $\mathcal{P}$ is a
peripheral pair for the function 
\begin{equation*}
g=h\cup \left( f\circ \tau \right) :\Bbb{R}^{n+1}\rightarrow \Bbb{R},
\end{equation*}
where $\tau :\Bbb{R}^{n}\times \left\{ 0\right\} \rightarrow \Bbb{R}^{n}$ is
the bijection as in (\ref{b}).
\end{theorem}

The proof of Theorem \ref{thm: exist peripheral families} is given in
section \ref{sect: proof exist peripheral}, and the proof of Theorem \ref
{thm: use peripheral families} can be found in section \ref{sect: proof
extendable}. Now we shall give the proof of Theorem \ref{thm: connectivity
are extendable}.\medskip

\noindent \textbf{Proof of Theorem \ref{thm: connectivity are extendable}. }%
Let $n\ge 2$ and $f:\Bbb{R}^{n}\rightarrow \Bbb{R}$ be a connectivity
function. Since $f$ is peripherally continuous, it follows from Theorem \ref
{thm: exist peripheral families} that there exists a peripheral family $%
\mathcal{P}$ for $f$. Let $\mathcal{Q}$ be the cylindrical extension of $%
\mathcal{P}$. By Theorem \ref{thm: use peripheral families} there exists a
function $g:\Bbb{R}^{n+1}\rightarrow \Bbb{R}$ such that $g$ extends $f$, the
restriction of $g$ to $\Bbb{R}^{n+1}\setminus \left( \Bbb{R}^{n}\times
\left\{ 0\right\} \right) $ is continuous, and every element of $\mathcal{Q}$
is a peripheral pair for $g$. The proof will be complete when we show that $%
\mathcal{Q}$ is a $g$-base for $\Bbb{R}^{n}\times \left\{ 0\right\} $ since
then it will follow that $g$ is peripherally continuous and hence a
connectivity function.

Let $\varepsilon >0$ and $x=\left\langle x_{1},\dots ,x_{n}\right\rangle \in 
\Bbb{R}^{n}$. Since $\mathcal{P}$ is an $f$-base for $\Bbb{R}^{n}$, there is 
$\left\langle A,I\right\rangle \in \mathcal{P}$ such that $\func{diam}%
A<\varepsilon /\sqrt{5}$, $\func{diam}I<\varepsilon $, $x\in A$, and $%
f(x)\in I$. Then the cylindrical extension $\left\langle A^{\prime
},I\right\rangle \in \mathcal{Q}$ of $\left\langle A,I\right\rangle $ is an $%
\varepsilon $-peripheral pair for $g$ at $\bar{x}=\left\langle x_{1},\dots
,x_{n},0\right\rangle $ implying that $\mathcal{Q}$ is a $g$-base for $\Bbb{R%
}^{n}\times \left\{ 0\right\} $.

\section{Peripheral families for connectivity functions \label%
{sect: proof exist peripheral}}

In this section we are going to prove Theorem \ref{thm: exist peripheral
families}. First, let us introduce some more terminology. Throughout this
section we will assume that $n$ is a fixed integer and that $n\ge 2$.

Given $X,Y\subseteq \Bbb{R}^{n}$, the \emph{boundary} of $X\cap Y$ in $X$
will be denoted by $\limfunc{bd}_{X}Y$. The \emph{inductive dimension} $%
\func{ind}X$ of a subset $X\subseteq \Bbb{R}^{n}$ is defined inductively as
follows. (See for example Engelking \cite{Engelking: general topology}.)

\begin{enumerate}
\item[(i)]  $\func{ind}X=-1$ if and only if $X=\emptyset $.

\item[(ii)]  $\func{ind}X\le m$ if for any $p\in X$ and any open
neighborhood $W$ of $p$ there exists an open neighborhood $U\subseteq W$ of $%
p$ such that $\func{ind}\limfunc{bd}_{X}U\le m-1$.

\item[(iii)]  $\func{ind}X=m$ if $\func{ind}X\le m$ and it is not true that $%
\func{ind}X\le m-1$.
\end{enumerate}

\noindent A fundamental result of dimension theory states that $\func{ind}%
\Bbb{R}^{n}=n$.

Given a set $A\subseteq \Bbb{R}^{n}$ and an integer $m\ge 1$, we say that $A$
is an \emph{$m$-dimensional Cantor manifold} if $A$ is compact, $\func{ind}%
A=m$, and for every $X\subseteq A$ with $\func{ind}X\le m-2$, the set $%
A\setminus X$ is connected. (See \cite{HurWall: dimension theory}.) Given a
subset $A$ of $\QTR{userA}{\Bbb{R}^{n}}$, we say that $A$ is a \emph{%
quasiball} if $A$ is a bounded and connected open set, and $\limfunc{bd}A$
is an $\left( n-1\right) $-dimensional Cantor manifold. (See \cite{CieWoj:
sums of connectivity}.) A peripheral pair $\left\langle A,I\right\rangle $
with $A$ being a quasiball will be called a \emph{nice} peripheral pair.
Given $\varepsilon ,\delta >0$, an $\left\langle \varepsilon ,\delta
\right\rangle $\emph{-peripheral pair} is a peripheral pair $\left\langle
A,I\right\rangle $ with $\func{diam}A<\varepsilon $ and $\func{diam}I<\delta 
$. The following theorem follows immediately from Corollary 5.5 in \cite
{CieWoj: sums of connectivity}.

\begin{theorem}
\label{thm: periheral pairs for functions}If $f:\Bbb{R}^{n}\rightarrow \Bbb{R%
}$ is a peripherally continuous function, then for any $\varepsilon ,\delta
>0$ and $x\in \Bbb{R}^{n}$ there exists a nice $\left\langle \varepsilon
,\delta \right\rangle $-peripheral pair for $f$ at $x$.
\end{theorem}

We say that quasiballs $A$ and $A^{\prime }$ are \emph{independent} if each
of the sets $A\cap A^{\prime }$, $A\setminus A^{\prime }$, and $A^{\prime
}\setminus A$ is nonempty. The following lemma is a restatement of Lemma 5.6
in \cite{CieWoj: sums of connectivity}.

\begin{lemma}
\label{lem: independent quasiballs}If $A$ and $A^{\prime }$ are independent
quasiballs in $\Bbb{R}^{n}$, then $\func{bd}A\cap \func{bd}A^{\prime }\neq
\emptyset $.
\end{lemma}

The following lemma follows immediately from Lemma \ref{lem: independent
quasiballs}.

\begin{lemma}
\label{lem: intersection property}If $\mathcal{P}$ is a family of nice
peripheral pairs, then $\mathcal{P}$ has the intersection property.
\end{lemma}

For every positive integer $i\in \Bbb{N}$, let 
\begin{equation*}
\QTR{userA}{D}_{i}=\left\{ \frac{-4i^{2}}{4i},\frac{-4i^{2}+1}{4i},\dots ,%
\dfrac{4i^{2}}{4i}\right\}
\end{equation*}
and 
\begin{equation*}
\mathcal{J}_{i}=\left\{ J_{i,q}:q\in D_{i}\right\} ,
\end{equation*}
where $J_{i,q}$ is the open interval 
\begin{equation*}
J_{i,q}=\left( q-\frac{1}{4i},q+\frac{1}{4i}\right) ,
\end{equation*}
for each $q\in D_{i}$.

\begin{lemma}
\label{lem: then is an f-base}Let $f:\Bbb{R}^{n}\rightarrow \Bbb{R}$ be a
function and, for every $i\in \Bbb{N}$ and $q\in D_{i}$, let 
\begin{equation*}
\mathcal{P}_{i,q}=\left\{ \left\langle A_{\gamma },I_{\gamma }\right\rangle
:\gamma \in \Gamma _{i,q}\right\} 
\end{equation*}
be a family of $\left( 1/i\right) $-peripheral pairs for $f$ such that 
\begin{equation*}
f^{-1}(J_{i,q})\subseteq \bigcup_{\gamma \in \Gamma _{i,q}}A_{\gamma }\text{%
\qquad and\qquad }J_{i,q}\subseteq \bigcap_{\gamma \in \Gamma
_{i,q}}I_{\gamma }.
\end{equation*}
Then 
\begin{equation*}
\mathcal{P}=\bigcup_{i\in \Bbb{N}}\bigcup_{q\in D_{i}}\mathcal{P}_{i,q}
\end{equation*}
is an $f$-base for $\Bbb{R}^{n}$.
\end{lemma}

\begin{proof}
Let $\varepsilon >0$ and $x\in \Bbb{R}^{n}$. Then there are $i\in \Bbb{N}$
and $q\in D_{i}$ with $1/i\le \varepsilon $ and $f(x)\in J_{i,q}$. Since 
\begin{equation*}
f^{-1}(J_{i,q})\subseteq \bigcup_{\gamma \in \Gamma _{i,q}}A_{\gamma },
\end{equation*}
there is $\delta \in \Gamma _{i,q}$ such that $x\in A_{\delta }$. Since 
\begin{equation*}
J_{i,q}\subseteq \bigcap_{\gamma \in \Gamma _{i,q}}I_{\gamma },
\end{equation*}
it follows that $\left\langle A_{\delta },I_{\delta }\right\rangle $ is an $%
\varepsilon $-peripheral pair for $f$ at $x$.
\end{proof}

Now we are ready to prove Theorem \ref{thm: exist peripheral families}%
.\medskip

\noindent \textbf{Proof of Theorem \ref{thm: exist peripheral families}.}%
\textit{\ }Let $n\ge 2$ and $f:\Bbb{R}^{n}\rightarrow \Bbb{R}$ be a
peripherally continuous function. Fix $i\in \Bbb{N}$ and $q\in D_{i}$. By
Theorem \ref{thm: periheral pairs for functions} for each $x\in
f^{-1}(J_{i,q})$ there exists a nice $\left\langle 1/i,1/4i\right\rangle $%
-peripheral pair $\left\langle A_{i,q,x},I_{i,q,x}\right\rangle $ for $f$ at 
$x$. Let 
\begin{equation*}
\mathcal{T}_{i,q}=\left\{ \left\langle A_{i,q,x},\limfunc{cl}(I_{i,q,x}\cup
J_{i,q})\right\rangle :x\in f^{-1}(J_{i,q})\right\} .
\end{equation*}
Note that since 
\begin{equation*}
f(x)\in I_{i,q,x}\cap J_{i,q}\neq \emptyset
\end{equation*}
for every $x\in f^{-1}(J_{i,q})$, the elements of $\mathcal{T}_{i,q}$ are $%
\left\langle 1/i,3/4i\right\rangle $-peripheral pairs for $f$.

Let $j,k\in \Bbb{N}$ be any positive integers with $j>i$. Set 
\begin{equation*}
\mathcal{T}_{i,q}^{k}=\left\{ \left\langle A,I\right\rangle \in \mathcal{T}%
_{i,q}:A\cap B_{k}\neq \emptyset \text{ and }A\cap B_{k^{\prime }}=\emptyset 
\text{ for every }k^{\prime }<k\right\} ,
\end{equation*}
where $B_{k}$ is the open ball of center $\left\langle 0,0,\dots
,0\right\rangle $ and radius $k$, and 
\begin{equation*}
\mathcal{T}_{i,q}^{k,j}=\left\{ \left\langle A,I\right\rangle \in \mathcal{T}%
_{i,q}^{k}:\frac{1}{j}\le \func{diam}A<\frac{1}{j-1}\right\} .
\end{equation*}
Moreover, let 
\begin{equation*}
C_{i,q}^{k,j}=\func{cl}\left( \bigcup_{\left\langle A,I\right\rangle \in 
\mathcal{T}_{i,q}^{k,j}}A\right) ,
\end{equation*}
and 
\begin{equation*}
E_{i,q}^{k,j}=C_{i,q}^{k,j}\setminus \bigcup_{\left\langle A,I\right\rangle
\in \mathcal{T}_{i,q}^{k,j}}A.
\end{equation*}

Fix $y\in E_{i,q}^{k,j}$. Let $\left\langle A_{y},I_{y}^{\prime
}\right\rangle $ be a nice $\left\langle 1/j,1/4i\right\rangle $-peripheral
pair for $f$ at $y$. Since 
\begin{equation*}
E_{i,q}^{k,j}\subseteq \func{cl}\left( \bigcup_{\left\langle
A,I\right\rangle \in \mathcal{T}_{i,q}^{k,j}}\func{bd}A\right) ,
\end{equation*}
there is $\left\langle A,I\right\rangle \in \mathcal{T}_{i,q}^{k,j}$ such
that 
\begin{equation*}
A_{y}\cap \func{bd}A\neq \emptyset .
\end{equation*}
Since $\func{diam}A_{y}<\func{diam}A$, it follows that the quasiballs $A$
and $A_{y}$ are independent and so Lemma \ref{lem: independent quasiballs}
implies that $I\cap I_{y}^{\prime }\neq \emptyset $. Let $I_{y}=I\cup
I_{y}^{\prime }$ for every $y\in E_{i,q}^{k,j}$ and 
\begin{equation*}
\mathcal{S}_{i,q}^{k,j}=\mathcal{T}_{i,q}^{k,j}\cup \left\{ \left\langle
A_{y},I_{y}\right\rangle :y\in E_{i,q}^{k,j}\right\} .
\end{equation*}
Note that $J_{i,q}\subseteq I$ for every $\left\langle A,I\right\rangle \in 
\mathcal{S}_{i,q}^{k,j}$. Since the set $C_{i,q}^{k,j}$ is compact and 
\begin{equation*}
C_{i,q}^{k,j}\subseteq \bigcup_{\left\langle A,I\right\rangle \in \mathcal{S}%
_{i,q}^{k,j}}A,
\end{equation*}
there is a finite subset $\mathcal{P}_{i,q}^{k,j}$ of $\mathcal{S}%
_{i,q}^{k,j}$ such that 
\begin{equation*}
C_{i,q}^{k,j}\subseteq \bigcup_{\left\langle A,I\right\rangle \in \mathcal{P}%
_{i,q}^{k,j}}A.
\end{equation*}

Let 
\begin{equation*}
\mathcal{P}_{i,q}=\bigcup_{k\in \Bbb{N}}\bigcup_{j>i}\mathcal{P}%
_{i,q}^{k,j}=\left\{ \left\langle A_{\gamma },I_{\gamma }\right\rangle
:\gamma \in \Gamma _{i,q}\right\} .
\end{equation*}
It is clear that the elements of $\mathcal{P}_{i,q}$ are $\left( 1/i\right) $%
-peripheral pairs and 
\begin{equation*}
J_{i,q}\subseteq \bigcap_{\gamma \in \Gamma _{i,q}}I_{\gamma }\text{.}
\end{equation*}
Moreover, 
\begin{equation*}
f^{-1}(J_{i,q})\subseteq \bigcup_{\left\langle A,I\right\rangle \in \mathcal{%
T}_{i,q}}A\subseteq \bigcup_{\left\langle A,I\right\rangle \in \mathcal{T}%
_{i,q}}\func{cl}A\subseteq \bigcup_{k\in \Bbb{N}}\bigcup_{j>i}C_{i,q}^{k,j}%
\subseteq \bigcup_{\gamma \in \Gamma _{i,q}}A_{\gamma },
\end{equation*}
implying, by Lemma \ref{lem: then is an f-base}, that 
\begin{equation*}
\mathcal{P}=\bigcup_{i\in \Bbb{N}}\bigcup_{q\in D_{i}}\mathcal{P}_{i,q}
\end{equation*}
is an $f$-base for $\Bbb{R}^{n}$. Of course $\mathcal{P}$ is countable and
since all peripheral pairs in $\mathcal{P}$ are nice, it follows from Lemma 
\ref{lem: intersection property} that $\mathcal{P}$ has the intersection
property. It remains to prove the following claim.\medskip

\noindent \textbf{Claim. }The family $\mathcal{P}$ locally converges to $0$%
.\medskip

We are going now to prove the claim. First note that if $\left\langle
A,I\right\rangle \in \mathcal{T}_{i,q}^{k,j}$ and $k^{\prime }<k$, then $%
A\cap B_{k^{\prime }}=\emptyset $, implying that $y\notin B_{k^{\prime }}$
(and hence $A_{y}\nsubseteq B_{k^{\prime }}$) for any $y\in E_{i,q}^{k,j}$.
Therefore 
\begin{equation}
A\nsubseteq B_{k^{\prime }}\text{ for any }\left\langle A,I\right\rangle \in 
\mathcal{S}_{i,q}^{k,j}\text{ and }k^{\prime }<k\text{.}  \label{prepare1}
\end{equation}
Also note that 
\begin{equation}
\func{diam}A<\frac{1}{j^{\prime }}\text{ for any }\left\langle
A,I\right\rangle \in \mathcal{S}_{i,q}^{k,j}\text{ and }j^{\prime }<j\text{.}
\label{prepare2}
\end{equation}
Now let $\varepsilon >0$ and $X\subseteq \Bbb{R}^{n}$ be a bounded set. Then
there are $j^{\prime },k^{\prime }\in \Bbb{N}$ such that $1/j^{\prime
}<\varepsilon $ and $X$ is a subset of the ball $B_{k^{\prime }-1}$. Let $%
\left\langle A,I\right\rangle \in \mathcal{P}$ be such that $A\cap X\neq
\emptyset $ and $\func{diam}A\ge \varepsilon $. Since $A\cap B_{k^{\prime
}-1}\neq \emptyset $ and $\func{diam}A<1$, it follows that $A\subseteq
B_{k^{\prime }}$. Therefore, since $\func{diam}A\ge 1/j^{\prime }$, it
follows from (\ref{prepare1}) and (\ref{prepare2}) that if $\left\langle
A,I\right\rangle \in \mathcal{P}_{i,q}^{k,j}\subseteq \mathcal{S}%
_{i,q}^{k,j} $, then $k\le k^{\prime }$ and $j\le j^{\prime }$. Thus 
\begin{equation*}
\left\langle A,I\right\rangle \in \mathcal{P}^{k^{\prime },j^{\prime
}}=\bigcup_{k\le k^{\prime }}\bigcup_{j\le j^{\prime
}}\bigcup_{i<j}\bigcup_{q\in D_{i}}\mathcal{P}_{i,q}^{k,j}\text{.}
\end{equation*}
Since the set $\mathcal{P}^{k^{\prime },j^{\prime }}$ is finite, the proof
of the claim, and hence of the theorem is complete.

\section{Connectivity functions are extendable \label{sect: proof extendable}
}

In this section we are going to prove Theorem \ref{thm: use peripheral
families}.

A \emph{partial order} on a set $T$ is a binary relation $\preccurlyeq $ on $%
T$ that is reflexive, transitive and antisymmetric (that is, $t\preccurlyeq
s $ and $s\preccurlyeq t$ imply $t=s$ for every $s,t\in T$). We say that $%
\preccurlyeq $ has the \emph{finite predecessor property} if for every $t\in
T$ the set $\left\{ s\in T:s\preccurlyeq t\right\} $ of $\preccurlyeq $%
-predecessors of $t$ is finite. A partial order $\preccurlyeq ^{*}$ on a set 
$T$ is an $\omega $\emph{-order} if there is a bijection $f:\omega
\rightarrow T$ (where $\omega =\left\{ 0,1,\dots \right\} $) such that $%
f(t)\preccurlyeq ^{*}f(s)$ if and only if $t\le s$. Given partial orders $%
\preccurlyeq $ and $\preccurlyeq ^{*}$ on $T$, we say that $\preccurlyeq
^{*} $ \emph{extends} $\preccurlyeq $ if and only if $t\preccurlyeq s$
implies $t\preccurlyeq ^{*}s$ for every $s,t\in T$.

\begin{lemma}
\label{lem: partial order}If $\preccurlyeq $ is a partial order on an
infinite countable set $T$ with the finite predecessor property, then there
is an $\omega $-order $\preccurlyeq ^{*}$ on $T$ that extends $\preccurlyeq $%
.
\end{lemma}

\begin{proof}
It is enough to show that there is a bijection $f:\omega \rightarrow T$ such
that $f(i)\preccurlyeq f(j)$ implies $i\le j$. Let $\lesssim $ be any fixed $%
\omega $-order on $T$. We shall define the value $f(i)$ by induction on $i$.
Let $i\in \omega $ and assume that $f(j)$ has been defined for every $j<i$.
Let 
\begin{equation*}
T_{i}=T\setminus \left\{ f(j):j<i\right\} ,
\end{equation*}
and let $T_{i}^{\prime }$ consist of all $\preccurlyeq $-minimal elements in 
$T_{i}$. For every $t\in T_{i}$ the set of $\preccurlyeq $-predecessors of $%
t $ is finite so there is $s\in T_{i}^{\prime }$ with $s\preccurlyeq t$. In
particular, $T_{i}^{\prime }$ is nonempty. Let $f(i)$ be the $\lesssim $%
-minimal element of $T_{i}^{\prime }$.

It is obvious from the construction that $f$ is injective and that $%
f(i)\preccurlyeq f(j)$ implies $i\le j$ for every $i,j\in \omega $. To see
that $f$ is surjective note that for any $i\in \omega $ and $t\in T_{i}$ the
set of $\preccurlyeq $-predecessors of $t$ is finite, so one of them is in $%
T_{i}^{\prime }$. This predecessor of $t$ will eventually become a value of $%
f$ since $\lesssim $ is an $\omega $-order. Then the number of unassigned $%
\preccurlyeq $-predecessors of $t$ becomes smaller and hence eventually $t$
itself must become a value of $f$.
\end{proof}

A family $\mathcal{A}$ of subsets of a metric space $X$ is \emph{locally
finite} if for every $x\in X$ some open neighborhood of $x$ intersects only
finitely many elements of $\mathcal{A}$. Let a \emph{Tietze family} for a
metric space $X$ be a countable family 
\begin{equation*}
\mathcal{F}=\left\{ \left\langle C_{\gamma },I_{\gamma }\right\rangle
:\gamma \in \Gamma \right\}
\end{equation*}
such that:

\begin{enumerate}
\item  $\mathcal{A}=\left\{ C_{\gamma }:\gamma \in \Gamma \right\} $ is a
locally finite closed cover of $X$ with any $C_{\gamma }$ intersecting only
finitely many elements of $\mathcal{A}$;

\item  for every $\gamma \in \Gamma $, $I_{\gamma }$ is either equal to $%
\Bbb{R}$ or is a closed interval in $\Bbb{R}$;

\item  for every $\Phi \subseteq \Gamma $%
\begin{equation*}
\text{if }\bigcap_{\gamma \in \Phi }C_{\gamma }\neq \emptyset \text{ then }%
\bigcap_{\gamma \in \Phi }I_{\gamma }\neq \emptyset .
\end{equation*}
\end{enumerate}

The following result will be the key step in our proof of Theorem \ref{thm:
use peripheral families}.

\begin{theorem}
\label{thm: inductive tietze}Let $X$ be a metric space and $\mathcal{F}%
=\left\{ \left\langle C_{\gamma },I_{\gamma }\right\rangle :\gamma \in
\Gamma \right\} $ be a Tietze family for $X$. Then there is a continuous
function $h:X\rightarrow \Bbb{R}$ such that $h[C_{\gamma }]\subseteq
I_{\gamma }$ for every $\gamma \in \Gamma $.
\end{theorem}

\begin{proof}
Let $\mathcal{A}=\left\{ C_{\gamma }:\gamma \in \Gamma \right\} $, and 
\begin{equation*}
T_{\mathcal{A}}=\left\{ \Phi \subseteq \Gamma :\bigcap_{\gamma \in \Phi
}C_{\gamma }\neq \emptyset \right\} .
\end{equation*}
Let $\preccurlyeq _{\mathcal{A}}$ be the partial order of reversed inclusion
on $T_{\mathcal{A}}$, that is, let $\Phi _{1}\preccurlyeq _{\mathcal{A}}\Phi
_{2}$ if and only if $\Phi _{2}\subseteq \Phi _{1}$. Since every element of $%
\mathcal{A}$ intersects only finitely many elements of $\mathcal{A}$, it
follows that the elements of $T_{\mathcal{A}}$ are finite sets and that $%
\preccurlyeq _{\mathcal{A}}$ has the finite predecessor property.

Let $\preccurlyeq _{\mathcal{A}}^{*}$ be an $\omega $-order extending $%
\preccurlyeq _{\mathcal{A}}$ and for every $\Phi \in T_{\mathcal{A}}$ let 
\begin{equation*}
C_{\Phi }=\bigcap_{\gamma \in \Phi }C_{\gamma }\neq \emptyset .
\end{equation*}
Take the enumeration $\Phi _{1},\Phi _{2},\dots $ of $T_{\mathcal{A}}$ with 
\begin{equation*}
\Phi _{1}\preccurlyeq _{\mathcal{A}}^{*}\Phi _{2}\preccurlyeq _{\mathcal{A}%
}^{*}\cdots
\end{equation*}
and for every $i=1,2,\dots $ let 
\begin{equation*}
C_{i}=\bigcup_{j\le i}C_{\Phi _{j}},\qquad C_{i}^{\prime }=C_{i}\cap C_{\Phi
_{i+1}},
\end{equation*}
and 
\begin{equation*}
I_{i}=\bigcap_{\gamma \in \Phi _{i}}I_{\gamma }\neq \emptyset .
\end{equation*}
We are going to define a sequence $h_{1},h_{2},\dots $ of continuous
functions $h_{i}:C_{i}\rightarrow \Bbb{R}$ such that for every $i=1,2,\dots $
the function $h_{i+1}$ is an extension of $h_{i}$ and 
\begin{equation}
h_{i}[C_{\gamma }\cap C_{i}]\subseteq I_{\gamma }\text{.}  \label{a}
\end{equation}
for every $\gamma \in \Gamma $. Having defined such a sequence of functions
our proof will be complete since it is easy to see that the function 
\begin{equation*}
h=\bigcup_{i=1}^{\infty }h_{i}
\end{equation*}
satisfies the required conditions. Indeed, (\ref{a}) implies that $%
h(C_{\gamma })\subseteq I_{\gamma }$ for every $\gamma \in \Gamma $, and
since $\mathcal{F}$ is a locally finite closed cover of $X$ it follows that $%
h$ is a continuous function on $X$.

Let $h_{1}:C_{1}\rightarrow I_{1}$ be any continuous function. Suppose that $%
h_{i}$ has been defined in such a way that (\ref{a}) is satisfied. Let $%
h_{i}^{\prime }$ be the restriction of $h_{i}$ to $C_{i}^{\prime }$. It
follows from (\ref{a}) that $h_{i}^{\prime }:C_{i}^{\prime }\rightarrow
I_{i+1}$. Since $C_{i}^{\prime }$ is a closed subset of $C_{\Phi _{i+1}}$,
it follows from Tietze Extension Theorem that $h_{i}^{\prime }$ can be
extended to a continuous function $h_{i}^{\prime \prime }:C_{\Phi
_{i+1}}\rightarrow I_{i+1}$. Let $h_{i+1}=h_{i}\cup h_{i}^{\prime \prime }$.
Since $C_{i}$ and $C_{\Phi _{i+1}}$ are closed subsets of $C_{i+1}$, the
function $h_{i+1}:C_{i+1}\rightarrow \Bbb{R}$ is continuous. It remains to
show that (\ref{a}) is satisfied for $h_{i+1}$.

Suppose that $\gamma \in \Gamma $ and $x\in C_{\gamma }\cap C_{i+1}$. If $%
x\in C_{i}$, then $h_{i+1}(x)=h_{i}(x)\in I_{\gamma }$ by the inductive
hypothesis. Otherwise $x\in C_{\Phi _{i+1}}$ and so $h_{i+1}(x)=h_{i}^{%
\prime \prime }(x)\in I_{i+1}$. It suffices to show that $\gamma \in \Phi
_{i+1}$.

Indeed, since $C_{\gamma }\cap C_{\Phi _{i+1}}\neq \emptyset $, it follows
that $\Phi _{i+1}\cup \left\{ \gamma \right\} \in T_{\mathcal{A}}$. Since 
\begin{equation*}
\Phi _{i+1}\cup \left\{ \gamma \right\} \preccurlyeq _{\mathcal{A}}\Phi
_{i+1}
\end{equation*}
and since $\preccurlyeq _{\mathcal{A}}^{*}$ extends $\preccurlyeq _{\mathcal{%
A}}$, it follows that there is $j\le i+1$ with 
\begin{equation*}
\Phi _{i+1}\cup \left\{ \gamma \right\} =\Phi _{j}.
\end{equation*}
Since $x\in C_{\gamma }\cap C_{\Phi _{i+1}}=C_{\Phi _{j}}$ and $x\notin
C_{i} $, it follows that $j=i+1$. Thus $\gamma \in \Phi _{i+1}$ and so the
proof is complete.
\end{proof}

\begin{lemma}
\label{lem: extension of family}Let $n\ge 1$, $f:\Bbb{R}^{n}\rightarrow \Bbb{%
R}$, $\mathcal{P}$ be a peripheral family for $f$, and $\mathcal{Q}$ be the
cylindrical extension of $\mathcal{P}$. If $\left\{ \left\langle
A_{j},I_{j}\right\rangle :1\le j\le k\right\} \subseteq \mathcal{Q}$ and $%
\func{bd}A_{i}\cap \func{bd}A_{j}\neq \emptyset $ for every $i,j\le k$, then 
$\bigcap_{j=1}^{k}I_{j}\neq \emptyset $.
\end{lemma}

\begin{proof}
First we shall prove the lemma for $k=2$. Suppose, by way of contradiction,
that there exist $\left\langle A_{1},I_{1}\right\rangle ,\left\langle
A_{2},I_{2}\right\rangle \in \mathcal{Q}$ with $\func{bd}A_{1}\cap \func{bd}%
A_{2}\neq \emptyset $ and $I_{1}\cap I_{2}=\emptyset $. Let $\left\langle
A_{1}^{\prime },I_{1}\right\rangle ,\left\langle A_{2}^{\prime
},I_{2}\right\rangle \in \mathcal{P}$ be such that 
\begin{equation*}
A_{1}=A_{1}^{\prime }\times \left( -a_{1},a_{1}\right) \text{\quad and\quad }%
A_{2}=A_{2}^{\prime }\times \left( -a_{2},a_{2}\right) ,
\end{equation*}
where $a_{1}=\func{diam}A_{1}^{\prime }$ and $a_{2}=\func{diam}A_{2}^{\prime
}$.

Since $f[\func{bd}A_{1}^{\prime }]\subseteq I_{1}$ and $f[\func{bd}%
A_{2}^{\prime }]\subseteq I_{2}$, we have 
\begin{equation*}
\func{bd}A_{1}^{\prime }\cap \func{bd}A_{2}^{\prime }=\emptyset \text{.}
\end{equation*}
It follows that $A_{1}^{\prime }\cap A_{2}^{\prime }\neq \emptyset $ since
otherwise we would have $\func{cl}A_{1}^{\prime }\cap \func{cl}A_{2}^{\prime
}=\emptyset $ in contradiction with $\func{bd}A_{1}\cap \func{bd}A_{2}\neq
\emptyset $. Since $\mathcal{P}$ has the intersection property, one of $%
A_{1}^{\prime }$, $A_{2}^{\prime }$ is a subset of the other.

Assume that $A_{1}^{\prime }\subseteq A_{2}^{\prime }$. Since $\func{cl}%
A_{1}^{\prime }\subseteq \func{cl}A_{2}^{\prime }$ and $\func{bd}%
A_{1}^{\prime }\cap \func{bd}A_{2}^{\prime }=\emptyset $, it follows that $%
\func{cl}A_{1}^{\prime }\subseteq A_{2}^{\prime }$. Since the set $\func{cl}%
A_{1}^{\prime }$ is compact, there are $x_{1},x_{2}\in \func{cl}%
A_{1}^{\prime }$ with $\func{diam}A_{1}^{\prime }$ equal to the distance
from $x_{1}$ to $x_{2}$. Since $x_{1},x_{2}\in A_{2}^{\prime }$ and $%
A_{2}^{\prime }$ is open, it follows that 
\begin{equation*}
a_{1}=\func{diam}A_{1}^{\prime }<\func{diam}A_{2}^{\prime }=a_{2},
\end{equation*}
and so 
\begin{equation*}
\func{bd}A_{1}=\func{bd}A_{1}^{\prime }\times \left[ -a_{1},a_{1}\right]
\cup A_{1}^{\prime }\times \left\{ -a_{1},a_{1}\right\} \subseteq
A_{2}^{\prime }\times \left( -a_{2},a_{2}\right) =A_{2},
\end{equation*}
contradicting our assumption that $\func{bd}A_{1}\cap \func{bd}A_{2}\neq
\emptyset $.

Now for $k>2$ the assertion follows easily from the fact that if $%
\{I_{j}:1\le j\le k\}$ is a family of intervals in $\Bbb{R}$ and $I_{j}\cap
I_{m}\neq \emptyset $ for every $j,m\le k$, then $\bigcap_{j=1}^{k}I_{j}\neq
\emptyset $.
\end{proof}

Now we are ready to prove Theorem \ref{thm: use peripheral families}.\medskip

\noindent \textbf{Proof of Theorem \ref{thm: use peripheral families}.}%
\textit{\ }Let $f:\Bbb{R}^{n}\rightarrow \Bbb{R}$ be a function, $\mathcal{P}
$ be a peripheral family for $f$, $\mathcal{Q}$ be the cylindrical extension
of $\mathcal{P}$, and 
\begin{equation*}
X=\Bbb{R}^{n+1}\setminus \left( \Bbb{R}^{n}\times \left\{ 0\right\} \right) .
\end{equation*}
We need to construct a continuous function $h:X\rightarrow \Bbb{R}$ such
that $h[\func{bd}A]\subseteq I$ for every $\left\langle A,I\right\rangle \in 
\mathcal{Q}$. The existence of the function $h$ will follow from Theorem \ref
{thm: inductive tietze} after we have constructed a Tietze family 
\begin{equation*}
\mathcal{F}=\left\{ \left\langle C_{\gamma },I_{\gamma }\right\rangle
:\gamma \in \Gamma \right\}
\end{equation*}
for $X$ such that for every $\left\langle A,I\right\rangle \in \mathcal{Q}$
there is $\Phi \subseteq \Gamma $ with 
\begin{equation}
X\cap \func{bd}A\subseteq \bigcup_{\gamma \in \Phi }C_{\gamma }\text{ and }%
I_{\gamma }=I\text{ for every }\gamma \in \Phi \text{.}  \label{aa}
\end{equation}

Let $\mathcal{K}$ consist of all closed intervals of the 
following %KC
forms: $\left[
i,i+1\right] $, $\left[ -i-1,-i\right] $, $\left[ 1/\left( i+1\right)
,1/i\right] $, and $\left[ -1/i,-1/\left( i+1\right) \right] $ for every $%
i=1,2,\dots $. Set 
\begin{equation*}
\mathcal{A}_{1}=\left\{ \left( \func{cl}B_{k}^{n}\setminus
B_{k-1}^{n}\right) \times \left[ a,b\right] \subseteq \Bbb{R}^{n+1}:\left[
a,b\right] \in \mathcal{K}\text{ and }k=1,2,\dots \right\} ,
\end{equation*}
where $B_{k}^{n}\subseteq \Bbb{R}^{n}$ is the open ball with center $%
\left\langle 0,0,\dots ,0\right\rangle $ and radius $k$. Note that $\mathcal{%
A}_{1}$ is a locally finite closed cover of $X$.

Define 
\begin{equation*}
\mathcal{F}_{1}=\left\{ \left\langle C,\Bbb{R}\right\rangle :C\in \mathcal{A}%
_{1}\right\}
\end{equation*}
and 
\begin{equation*}
\mathcal{F}_{2}=\left\{ \left\langle \func{bd}A\cap L,I\right\rangle
:\left\langle A,I\right\rangle \in \mathcal{Q}\text{ and }L\in \mathcal{L}%
\right\} ,
\end{equation*}
where 
\begin{equation*}
\mathcal{L}=\left\{ \Bbb{R}^{n}\times \left[ a,b\right] :\left[ a,b\right]
\in \mathcal{K}\right\} .
\end{equation*}
Let $\Gamma _{1}$ and $\Gamma _{2}$ be disjoint sets of indices such that

\begin{equation*}
\mathcal{F}_{1}=\left\{ \left\langle C_{\gamma },I_{\gamma }\right\rangle
:\gamma \in \Gamma _{1}\right\} \text{ and }\mathcal{F}_{2}=\left\{
\left\langle C_{\gamma },I_{\gamma }\right\rangle :\gamma \in \Gamma
_{2}\right\} .
\end{equation*}
Obviously, for every $\left\langle A,I\right\rangle \in \mathcal{Q}$ there
is $\Phi \subseteq \Gamma _{2}$ such that (\ref{aa}) holds. Thus to complete
the proof it remains to prove the following claim.\medskip

\noindent \textbf{Claim.} The family $\mathcal{F}_{1}\cup \mathcal{F}_{2}$
is a Tietze family for $X$.\medskip

Let 
\begin{equation*}
\mathcal{A}_{2}=\left\{ C_{\gamma }:\gamma \in \Gamma _{2}\right\} .
\end{equation*}
Obviously, $\mathcal{A}_{1}\cup \mathcal{A}_{2}$ is a closed cover of $X$.
Since the family $\mathcal{P}$ is locally convergent to $0$, every bounded
subset of an element of $\mathcal{L}$ intersects only finitely many elements
of $\mathcal{A}_{2}$. Since each point $x\in X$ has an open neighborhood
contained in at most two elements of $\mathcal{L}$, it follows that $%
\mathcal{A}_{2}$ is locally finite, and hence $\mathcal{A}_{1}\cup \mathcal{A%
}_{2}$ is locally finite.

Since every element $C$ of $\mathcal{A}_{1}\cup \mathcal{A}_{2}$ is a
bounded subset of an element of $\mathcal{L}$, it follows that $C$
intersects only finitely many elements in $\mathcal{A}_{2}$, and it is clear
that $C$ intersects only finitely many elements of $\mathcal{A}_{1}$. Thus
every element of $\mathcal{A}_{1}\cup \mathcal{A}_{2}$ intersects only
finitely many elements in $\mathcal{A}_{1}\cup \mathcal{A}_{2}$.

Now suppose that 
\begin{equation*}
\bigcap_{\gamma \in \Phi _{1}\cup \Phi _{2}}C_{\gamma }\neq \emptyset
\end{equation*}
for some $\Phi _{1}\subseteq \Gamma _{1}$ and $\Phi _{2}\subseteq \Gamma
_{2} $. Since $\bigcap_{\gamma \in \Phi _{2}}C_{\gamma }\neq \emptyset $, it
follows from Lemma \ref{lem: extension of family} that $\bigcap_{\gamma \in
\Phi _{2}}I_{\gamma }\neq \emptyset $. Since $I_{\gamma }=\Bbb{R}$ for $%
\gamma \in \Phi _{2}$, we have 
\begin{equation*}
\bigcap_{\gamma \in \Phi _{1}\cup \Phi _{2}}I_{\gamma }=\bigcap_{\gamma \in
\Phi _{2}}I_{\gamma }\neq \emptyset .
\end{equation*}
Thus $\mathcal{F}_{1}\cup \mathcal{F}_{2}$ is a Tietze family for $X$, and
so the proof of the claim and hence of the theorem is complete.

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\end{document}