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\begin{document}
\title{Topological Dimension and Sums of Connectivity Functions}
\author{Krzysztof Ciesielski and Jerzy Wojciechowski}
\address{Department of Mathematics\\
West Virginia University\\
PO Box 6310\\
Morgantown, WV 26506-6310, USA}
\email[K. Ciesielski]{K\_Cies@math.wvu.edu \endgraf
{\it Web page address,} K. Ciesielski: {\tt http://www.math.wvu.edu/homepages/kcies}}
\email[J. Wojciechowski]{jerzy@math.wvu.edu}
\subjclass{Primary 54F45; Secondary 54C30, 26B40}
\keywords{inductive dimension, connectivity functions, Darboux functions}
\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}
The main goal of this paper is to show that the inductive dimension of a $%
\sigma $-compact metric space $X$ can be characterized in terms of
algebraical sums of connectivity (or Darboux) functions $X\rightarrow \Bbb{R}
$. As an intermediate step we show, using a result of Hayashi~\cite{Hayashi:
axiomatic dimension euclidean}, that for any dense $G_{\delta }$ set $G\in 
\Bbb{R}^{2k+1}$ the union of $G$ and some $k$ homeomorphic images of $G$ is
universal for $k$-dimensional separable metric spaces. We will also discuss
how our definition works with respect to other classes of Darboux-like
functions. In particular, we show that 
for the class of peripherally continuous functions on an
arbitrary separable metric space $X$  our parameter is equal to
either $\limfunc{ind}X$ or $\limfunc{ind}X-1$. Whether the later is at
all possible, is an open probem.
\end{abstract}

\section{Introduction}

Our terminology and notation is standard and follows~\cite{Ciesielski: set
theory for working}. Let $X$ be a non-empty set and $\mathcal{F}$ be a family
of functions from $X$ into $\Bbb{R}$. If $m$ is a nonnegative integer, then
let 
\begin{equation*}
m\mathcal{F}=\left\{ f_{1}+\dots +f_{m}:f_{1},\dots ,f_{m}\in \mathcal{F}%
\right\} ,
\end{equation*}
and let $\Bbb{R}^{X}$ be the family consisting of all functions from $X$
into $\Bbb{R}$. Let $\limfunc{DIM}\nolimits_{\mathcal{F}}X$ be defined by 
\begin{equation*}
\limfunc{DIM}\nolimits_{\mathcal{F}}X=\min \left( \left\{ m\in \Bbb{Z}:m\ge 0%
\text{ and }\left( m+1\right) \mathcal{F}=\Bbb{R}^{X}\right\} \cup \left\{
\infty \right\} \right) .
\end{equation*}

Given a metric space $X$, a function $f:X\rightarrow \Bbb{R}$ is a \emph{%
connectivity function} (\emph{Darboux function}) if for every connected
subset $C$ of $X$ the graph of the restriction $f|C$ is a connected subset
of $X\times \Bbb{R}$ (the image $f[C]$ is connected in $\Bbb{R}$). The
following theorem holds.

\begin{theorem}
\label{thm: connectivity on Rn}If $n$ is a positive integer and $\mathcal{F}$
is the family of connectivity functions or the family of Darboux functions
on $\Bbb{R}^{n}$, then 
\begin{equation*}
\limfunc{DIM}\nolimits_{\mathcal{F}}\Bbb{R}^{n}=n.
\end{equation*}
\end{theorem}

The proof of Theorem \ref{thm: connectivity on Rn} is given by Ciesielski
and Wojciechowski \cite{CieWoj: sums of connectivity}, except for the case $%
n=1$ that has been proved by Ciesielski and Rec\l aw \cite{CieRec:
invariants concerning extendable and}, and the inequality $\ge $ in the case
of Darboux functions that has been demonstrated by Jordan \cite{Jordan:
PhD,Jordan:darb}.

Theorem \ref{thm: connectivity on Rn} motivates the notation $\limfunc{DIM}%
\nolimits_{\mathcal{F}}X$ and shows that (with suitably chosen family $%
\mathcal{F}$) $\limfunc{DIM}\nolimits_{\mathcal{F}}X$ can be considered as a
sort of dimension of $X$ (\emph{dimension relative to} $\mathcal{F}$). In
this paper we are going to show that the dimension relative to the family of
connectivity (Darboux) functions coincides with the inductive dimension $%
\func{ind}$ on every $\sigma $-compact metric space.

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

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

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

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

\noindent Let $\mathcal{C}$ be the family of connectivity functions on $X$
and $\mathcal{D}$ be the family of Darboux functions on $X$. Our main result
is the following theorem.

\begin{theorem}
\label{thm: dimension}If $X$ is a $\sigma $-compact metric space, then 
\begin{equation*}
\limfunc{DIM}\nolimits_{\mathcal{C}}X=\limfunc{DIM}\nolimits_{\mathcal{D}}X=%
\func{ind}X.
\end{equation*}
\end{theorem}

Clearly 
\begin{equation}
\limfunc{DIM}\nolimits_{\mathcal{F}}X\ge \limfunc{DIM}\nolimits_{\mathcal{G}%
}X\quad \text{for any }\mathcal{F}\subseteq \mathcal{G}\subseteq \Bbb{R}^{X}
\label{eq1}
\end{equation}
Since $\mathcal{C}\subseteq \mathcal{D}$ for any space $X$, we have $%
\limfunc{DIM}\nolimits_{\mathcal{C}}X\ge \limfunc{DIM}\nolimits_{\mathcal{D}%
}X$, and so Theorem \ref{thm: dimension} follows immediately from the
following two results.

\begin{theorem}
\label{thm: upper bound}If $X$ is a separable metric space, then 
\begin{equation*}
\limfunc{DIM}\nolimits_{\mathcal{C}}X\le \func{ind}X.
\end{equation*}
\end{theorem}

\begin{theorem}
\label{thm: lower bound}If $X$ is a $\sigma $-compact metric space, then 
\begin{equation*}
\limfunc{DIM}\nolimits_{\mathcal{D}}X\ge \func{ind}X.
\end{equation*}
\end{theorem}

A natural question is whether Theorem \ref{thm: lower bound} can be extended
to all separable metric spaces or perhaps all that are complete. The answer
is `no' in both cases since Mazurkiewicz \cite{Mazurkiewicz: problemes
Urysohn} has shown that for each positive integer $n$ there exists a
complete separable metric space $X$ of inductive dimension $n$ which is
totally disconnected, that is, single points are its only connected
subspaces. (See also \cite[Example~II~16]{HurWall: dimension theory}.) Since
for every totally disconnected space $X$ we have 
\begin{equation*}
\limfunc{DIM}\nolimits_{\mathcal{C}}X=\limfunc{DIM}\nolimits_{\mathcal{D}}X=0
\end{equation*}
(any function $f:X\rightarrow \Bbb{R}$ is a connectivity and Darboux), we
get 
\begin{equation}  \label{eq2}
\limfunc{DIM}\nolimits_{\mathcal{C}}X=\limfunc{DIM}\nolimits_{\mathcal{D}}X
=0<n=\func{ind}X,
\end{equation}
for every space of Mazurkiewicz of inductive dimension $n>0$. It might be
interesting to answer the question whether the equation 
\begin{equation}  \label{eq3}
\limfunc{DIM}\nolimits_{\mathcal{C}}X=\limfunc{DIM}\nolimits_{\mathcal{D}}X
\end{equation}
holds for all separable metric spaces $X$ or at least all that are complete.

To prove Theorem \ref{thm: upper bound} we will prove the following result
which seems to be of independent interest. We say that a separable metric
space $X$ is $m$\emph{-dimensional} if $\func{ind}X=m$. If $Y$ is a metric
space such that for every $m$-dimensional separable metric space $X$ there
is a subspace of $Y$ homeomorphic to $X$, then we say that $Y$ is \emph{%
universal\/} for $m$-dimensional separable metric spaces.

\begin{theorem}
\label{thm: union is universal}If $G$ is a dense $G_{\delta }$ set in $\Bbb{R%
}^{2k+1}$, then there are homeomorphisms $h_{j}:\Bbb{R}^{2k+1}\rightarrow 
\Bbb{R}^{2k+1}$, for $j=1,\dots ,k$, such that $G\cup
\bigcup_{j=1}^{k}h_{j}[G]$ is universal for $k$-dimensional separable metric
spaces.
\end{theorem}

Theorem~\ref{thm: union is universal} will be used to prove the following
fact, that easily implies Theorem \ref{thm: upper bound}.

\begin{proposition}
\label{prop} For every positive integer $k$ there exists a dense $G_{\delta }
$-set $H$ in $\Bbb{R}^{2k+1}$ such that

\begin{itemize}
\item[(i)]  $H$ is universal for $k$-dimensional separable metric spaces, and

\item[(ii)]  for every $\varphi \colon\Bbb{R}^{2k+1}\to \Bbb{R}$ there are
connectivity functions $g_{0},\ldots ,g_{k}\colon\Bbb{R}^{2k+1}\to \Bbb{R}$
such that $(g_{0}+\cdots +g_{k})(x)=\varphi (x)$ for every $x\in H$.
\end{itemize}
\end{proposition}

The proof of Theorem \ref{thm: union is universal} will be based on Lemma~%
\ref{lem: implicit} and Theorem~\ref{thm: negligible G-delta}, that are
proved in \cite{CieWoj: sums of connectivity}, and on Theorem \ref{thm:
hayashi}, which is proved by Hayashi~\cite{Hayashi: axiomatic dimension
euclidean}. Theorem \ref{thm: union is universal} is proved in Section \ref
{sec: universal}, the proof of Theorem \ref{thm: upper bound} is presented
in Section \ref{sec: upper}, while Theorem \ref{thm: lower bound} is proved
in Section \ref{sec: lower}. The authors would like to thank Roman Pol for
directing their attention to the results of Hayashi~\cite{Hayashi: axiomatic
dimension euclidean} and Mazurkiewicz \cite{Mazurkiewicz: problemes Urysohn}.

\section{\label{sec: universal}A $k$-dimensional universal set}

In this section we are going to present a proof of Theorem \ref{thm: union
is universal}.

Let a \emph{countable dense grid in }$\Bbb{R}^{n}$ be a product $B_{1}\times
\dots \times B_{n}\subseteq \Bbb{R}^{n}$ where $B_{1},\dots ,B_{n}$ are
countable dense subsets of $\Bbb{R}$. If $B=B_{1}\times \dots \times B_{n}$
is a countable dense grid in $\Bbb{R}^{n}$ and $i\le n$, then let $B^{\left(
i\right) }$ consist of those points in $\Bbb{R}^{n}$ that differ from a
point in $B$ at at most $i$ coordinates, that is, 
\begin{equation*}
B^{\left( i\right) }=\left\{ \left\langle x_{1},\dots ,x_{n}\right\rangle
\in \Bbb{R}^{n}:\left| \left\{ j:x_{j}\notin B_{j}\right\} \right| \le
i\right\} .
\end{equation*}
Note that in particular $B^{\left( 0\right) }=B$. Let $\Bbb{Q}$ be the set
of rational numbers and $I$ be the closed interval $\left[ 0,1\right] $.

Our proof of Theorem \ref{thm: union is universal} uses the following result
of Hayashi \cite{Hayashi: axiomatic dimension euclidean}.
(See also \cite{GS} for similar results.) 

\begin{theorem}
\label{thm: hayashi}If $G$ is a $G_{\delta }$-set in $I^{2k+1}$ containing $%
\left( \Bbb{Q}^{2k+1}\right) ^{\left( k\right) }\cap I^{2k+1}$, then 
$G$ is universal 
for $k$-dimensional separable metric spaces.
\end{theorem}

First notice that Theorem \ref{thm: hayashi} implies immediately the
following corollary.

\begin{corollary}
\label{cor: of hayashi}If $B$ is a countable dense grid in $\Bbb{R}^{2k+1}$
and $G$ is a $G_{\delta }$-set in $\Bbb{R}^{2k+1}$ containing $B^{\left(
k\right) }$, then $G$ is universal 
for $k$-dimensional separable metric spaces.
\end{corollary}

\begin{proof}
Let $B=B_{1}\times \dots \times B_{2k+1}$. Let $g_{1},\dots ,g_{2k+1}:\Bbb{R}%
\rightarrow \Bbb{R}$ be increasing homeomorphisms such that $B_{i}=g_{i}[%
\Bbb{Q}]$ and 
\begin{equation*}
g=g_{1}\times \dots \times g_{2k+1}:\Bbb{R}^{2k+1}\rightarrow \Bbb{R}^{2k+1}.
\end{equation*}
Then 
\begin{equation*}
\left( \Bbb{Q}^{2k+1}\right) ^{\left( k\right) }\cap I^{2k+1}\subseteq
g^{-1}[G]\cap I^{2k+1}.
\end{equation*}
Let $X$ be a $k$-dimensional separable metric space. It follows from Theorem 
\ref{thm: hayashi} that there is a subspace $Y$ of $g^{-1}[G]\cap I^{2k+1}$
that is homeomorphic to $X$. Then $g[Y]$ is a subspace of $G$ that is
homeomorphic to $X$.
\end{proof}

To prove Theorem \ref{thm: union is universal} we will also need a result
proved implicitly in \cite{CieWoj: sums of connectivity}. We will first
introduce the notation used there. If $\langle B_{i}:i\in n\rangle $ is a
family of subsets of $\Bbb{R}$ and $f$ is a function from $\left\{ 1,\dots
,n\right\} $ into $\left\{ 0,1\right\} $, then let 
\begin{equation*}
\prod_{i=1}^{n}\left( B_{i}\vee _{f}\QTR{userA}{\Bbb{R}}\right)
=B_{1}^{\prime }\times \dots \times B_{n}^{\prime },
\end{equation*}
where 
\begin{equation*}
B_{i}^{\prime }=\left\{ 
\begin{array}{cl}
B_{i} & \text{if }f(i)=0, \\ 
\Bbb{R} & \text{if }f(i)=1.
\end{array}
\right.
\end{equation*}
The following lemma is stated implicitly and proved in \cite{CieWoj: sums
of connectivity} (the inductive condition (8) in the proof of
Proposition~2.4, page 419).

\begin{lemma}
\label{lem: implicit}If $G$ is a dense $G_{\delta }$-set in $\Bbb{R}^{n}$,
then there are countable dense sets $B_{i}\subseteq \Bbb{R}$ and
homeomorphisms $h_{i}:\Bbb{R}^{n}\rightarrow \Bbb{R}^{n}$, for $i=1,\dots ,n$%
, such that 
\begin{equation*}
\prod_{i=1}^{n}\left( B_{i}\vee _{f}\QTR{userA}{\Bbb{R}}\right) \subseteq
G\cup \bigcup_{i=1}^{k}h_{i}[G]
\end{equation*}
for every $k\in \left\{ 0,1,\dots ,n\right\} $ and every function $f:\left\{
1,\dots ,n\right\} \to \left\{ 0,1\right\} $ such that $\left|
f^{-1}(1)\right| =k$.
\end{lemma}

Lemma \ref{lem: implicit} implies immediately the following result.

\begin{theorem}
\label{thm: cover grid}If $G$ is a dense $G_{\delta }$-set in $\Bbb{R}^{n}$
and $k\le n$, then there is a countable dense grid $B$ in $\Bbb{R}^{n}$ and
homeomorphisms $h_{1},\dots ,h_{k}:\Bbb{R}^{n}\rightarrow \Bbb{R}^{n}$ such
that 
\begin{equation*}
B^{\left( k\right) }\subseteq G\cup \bigcup_{j=1}^{k}h_{j}[G].
\end{equation*}
\end{theorem}

\begin{proof}
Let $G$ be a dense $G_{\delta }$-set in $\Bbb{R}^{n}$. For $i=1,\dots ,n$,
let $B_{i}\subseteq \Bbb{R}$ be countable dense sets and $h_{i}:\Bbb{R}%
^{n}\rightarrow \Bbb{R}^{n}$ be homeomorphisms as in Lemma \ref{lem:
implicit}. Then 
\begin{equation*}
B=B_{1}\times \dots \times B_{n}
\end{equation*}
is a countable dense grid in $\Bbb{R}^{n}$ and 
\begin{equation*}
B^{(k)}=\bigcup \left\{ \prod_{i=1}^{n}\left( B_{i}\vee _{f}\QTR{userA}{\Bbb{%
R}}\right) :\left| f^{-1}(1)\right| =k\right\} .
\end{equation*}
It follows from Lemma \ref{lem: implicit} that 
\begin{equation*}
B^{\left( k\right) }\subseteq G\cup \bigcup_{j=1}^{k}h_{j}[G].
\end{equation*}
\end{proof}

\noindent \textbf{Proof of Theorem \ref{thm: union is universal}. }Let $G$
be a dense $G_{\delta}$-set in $\Bbb{R}^{2k+1}$. By Theorem \ref{thm: cover
grid}, there is a countable dense grid $B$ in $\Bbb{R}^{2k+1}$ and
homeomorphisms $h_{1},\dots ,h_{k}:\Bbb{R}^{2k+1}\rightarrow \Bbb{R}^{2k+1}$
such that $B^{\left( k\right) }\subseteq G\cup \bigcup_{j=1}^{k}h_{j}[G]$.
By Corollary \ref{cor: of hayashi}, 
$G\cup \bigcup_{j=1}^{k}h_{j}[G]$ is universal for 
$k$-dimensional separable
metric spaces. $\blacksquare $\medskip

\section{\label{sec: upper}Inductive dimension as the upper bound}

Now we shall prove Theorem \ref{thm: upper bound}. Beside Theorem \ref{thm:
union is universal} we will need the following result. (See 
\cite[Proposition 2.3]{CieWoj: sums of connectivity}).

\begin{theorem}
\label{thm: negligible G-delta}For every $n>1$, there exists a function $f:%
\QTR{userA}{\Bbb{R}}^{n}\rightarrow \QTR{userA}{\Bbb{R}}$ and a dense $%
G_{\delta }$ subset $G$ of $\QTR{userA}{\Bbb{R}}^{n}$ such that any function 
$g:\QTR{userA}{\Bbb{R}}^{n}\rightarrow \QTR{userA}{\Bbb{R}}$ with $g(x)=f(x)$
for $x\notin G$ is a connectivity function.
\end{theorem}

Let us now introduce some notation. If $f,g:\Bbb{R}^{n}\rightarrow \Bbb{R}$
and $A\subseteq \Bbb{R}^{n}$, then we will write $g\equiv _{A}f$ if and only
if $g(x)=f(x)$ for every $x\in \Bbb{R}^{n}\setminus A$. Notice that if $%
g\equiv _{A}f$ and $A\subseteq A^{\prime }$, then $g\equiv _{A^{\prime }}f$.
Also $g\equiv _{\emptyset
 }f$ if and only if $g=f$, and $g\equiv _{\Bbb{R}%
^{n}}f$ for any $f,g:\Bbb{R}^{n}\rightarrow \Bbb{R}$. The following two
lemmas are easy observations.

\begin{lemma}
\label{lem: simple property1}Let $f,g:\Bbb{R}^{n}\rightarrow \Bbb{R}$ and $%
A\subseteq \Bbb{R}^{n}$. If $h:\Bbb{R}^{n}\rightarrow \Bbb{R}^{n}$ is a
bijection, then $g\equiv _{h[A]}\left( f\circ h^{-1}\right) $ if and only if 
$\left( g\circ h\right) \equiv _{A}f$.
\end{lemma}

\begin{proof}
Assume $g\equiv _{h[A]}\left( f\circ h^{-1}\right) $. Then $%
g(x)=f(h^{-1}(x)) $ for every $x\in \Bbb{R}^{n}\setminus h[A]$. If $y\in 
\Bbb{R}^{n}\setminus A $, then $h(y)\in \Bbb{R}^{n}\setminus h[A]$ so 
\begin{equation*}
\left( g\circ h\right) (y)=g(h(y))=f(h^{-1}(h(y)))=f(y),
\end{equation*}
implying that $\left( g\circ h\right) \equiv _{A}f$.

The opposite implication is proved similarly.
\end{proof}

\begin{lemma}
\label{lem: simple property2}Let $g_{0}^{\prime },\dots ,g_{k}^{\prime }:%
\Bbb{R}^{n}\rightarrow \Bbb{R}$. If $A\subseteq \Bbb{R}^{n}$, and $\left\{
A_{0},\dots ,A_{k}\right\} $ is a partition of $A$, then for any $\varphi
:A\rightarrow \Bbb{R}$ there are $g_{0},\dots ,g_{k}:\Bbb{R}^{n}\rightarrow 
\Bbb{R}$ such that 
\begin{equation*}
g_{i}\equiv _{A_{i}}g_{i}^{\prime },\quad \quad i=0,\dots ,k,
\end{equation*}
and the restriction of $g_{0}+\dots +g_{k}$ to $A$ is equal to $\varphi $.
\end{lemma}

\begin{proof}
Define $g_{i}:\Bbb{R}^{n}\rightarrow \Bbb{R}$ by 
\begin{equation*}
g_{i}(x)=\left\{ 
\begin{array}{ll}
\varphi (x)-\sum_{j\neq i}g_{j}^{\prime }(x) & \text{if }x\in A_{i}, \\ 
g_{i}^{\prime }(x) & \text{if }x\notin A_{i}.
\end{array}
\right.
\end{equation*}
Then $\varphi (x)=g_{0}(x)+\dots +g_{k}(x)$ for every $x\in A$.
\end{proof}

\noindent \textbf{Proof of Proposition \ref{prop}.} Let $n=2k+1$. By Theorem 
\ref{thm: negligible G-delta} there exists a function $f\colon \QTR{userA}{%
\Bbb{R}}^{n}\rightarrow \QTR{userA}{\Bbb{R}}$ and a dense $G_{\delta }$%
-subset $G$ of $\QTR{userA}{\Bbb{R}}^{n}$ such that any function $g\colon 
\QTR{userA}{\Bbb{R}}^{n}\rightarrow \QTR{userA}{\Bbb{R}}$ with $g\equiv_{G}f$
is a connectivity function. By Theorem \ref{thm: union is universal}, there
are homeomorphisms $h_{i}\colon \Bbb{R}^{n}\rightarrow \Bbb{R}^{n}$, for $%
i=1,\ldots ,k$, such that the $G_\delta$ set $H=G\cup
\bigcup_{j=1}^{k}h_{j}[G]$ is universal for $k$-dimensional separable metric
spaces. Let $\left\{ A_{0},\ldots ,A_{k}\right\} $ be the partition of $H$
defined inductively by 
\begin{equation*}
A_{0}=G,\quad A_{j}=h_{j}[G]\setminus \left( A_{0}\cup \cdots \cup
A_{j-1}\right) ,\quad j=1,\dots ,k.
\end{equation*}

Let $\varphi\colon\Bbb{R}^{n}\rightarrow \Bbb{R}$ be an arbitrary function,
and $h_{0}:\Bbb{R}^{n}\rightarrow \Bbb{R}^{n}$ be the identity function. It
follows from Lemma \ref{lem: simple property2}, that there are functions $%
g_{0},\dots ,g_{k}:\Bbb{R}^{n}\rightarrow \Bbb{R}$ such that 
\begin{equation*}
g_{i}\equiv _{A_{i}}\left( f\circ h_{i}^{-1}\right) ,\quad \quad i=0,\dots
,k,
\end{equation*}
and the restriction of $g_{0}+\dots +g_{k}$ to $H$ is equal to $%
\varphi\restriction H$. It remains to prove that $g_i$'s are connectivity
functions.

Let $i\in \left\{ 0,\dots ,k\right\} $. Since $A_{i}\subseteq h_{i}[G]$, we
have 
\begin{equation*}
g_{i}\equiv _{h_{i}[G]}\left( f\circ h_{i}^{-1}\right) ,
\end{equation*}
and so Lemma \ref{lem: simple property1} implies that 
\begin{equation*}
g_{i}\circ h_{i}\equiv _{G}f.
\end{equation*}
Thus $g_{i}\circ h_{i}$ (and hence $g_{i}$) is a connectivity function on $%
\Bbb{R}^{n}$. $\blacksquare $ \medskip 


\noindent \textbf{Proof of Theorem \ref{thm: upper bound}.} 
 Let $X$ be a $k$%
-dimensional separable metric space. If $k=0$, then any function $%
X\rightarrow \Bbb{R}$ is a connectivity function, so we can assume that $%
k\ge 1$. Let $H$ be a $G_{\delta }$-set from Proposition~\ref{prop}. Then
there is a subspace $A$ of $H$ homeomorphic to $X$. Take an arbitrary $%
\varphi _{0}\colon A\to \Bbb{R}$. We have to show that $\varphi _{0}$ is a
sum of $k+1$ connectivity functions on $A$.

Let $\varphi \colon\Bbb{R}^{n}\rightarrow \Bbb{R}$ be an arbitrary extension
of $\varphi _{0}$ and let $g_{0},\ldots ,g_{k}:\Bbb{R}^{n}\rightarrow \Bbb{R}
$ be connectivity functions such that $(g_{0}+\cdots +g_{k})(x)=\varphi (x)$
for all $x\in H$. Then the functions $g_{i}\upharpoonright A$ are
connectivity and $(g_{0}\upharpoonright A)+\cdots +(g_{k}\upharpoonright
A)=\varphi _{0}$. $\blacksquare $

\section{\label{sec: lower}Inductive dimension as the lower bound}

In this section we are going to prove Theorem \ref{thm: lower bound}. In the
proof that follows we will need some additional definitions and results from
dimension theory. (See for example~\cite{HurWall: dimension theory}).

\begin{lemma}
\label{lem: countable union}If $X$ is a separable metric space and 
\begin{equation*}
X=\bigcup_{i=1}^{\infty }X_{i},
\end{equation*}
where $X_{i}$ is closed in $X$ and $\limfunc{ind}X_{i}\le m$, for $%
i=1,2,\dots $, then $\limfunc{ind}X\le m$.
\end{lemma}

Given $X\subseteq \QTR{userA}{\Bbb{R}}^{n}$ and an integer $m\ge 1$, we say
that $X$ is an \emph{$m$-dimensional Cantor-manifold} if $X$ is compact, $%
\limfunc{ind}X=m$, and for every $Y\subseteq X$ with $\limfunc{ind}Y\le m-2$%
, the set $X\setminus Y$ is connected.

The following lemma is proved in~\cite{HurWall: dimension theory}.

\begin{lemma}
\label{lem: Cantor inside}For any compact $Y\subseteq \QTR{userA}{\Bbb{R}}%
^{n}$ with $\limfunc{ind}Y\ge m$ there exists an $m$-dimensional Cantor
manifold $X\subseteq Y$.
\end{lemma}

We will also need the following result of Francis Jordan. (See 
\cite[Lemma 3.3.8]{Jordan: PhD} or \cite[Lemma 3.8]{Jordan:darb}.) A \emph{%
perfect set} is a non-empty closed set without isolated points.

\begin{lemma}
\label{lem: Jordan}Let $n>1$ and $M$ be an $n$-dimensional Cantor manifold.
If $n\ge k\ge 1$ and $f\in k\mathcal{D}$, where $\mathcal{D}$ is the family
of Darboux functions $M\rightarrow \Bbb{R}$, then there is a connected perfect set 
$P\subseteq M$ such that the restriction of $f$ to $P$ is Darboux.
\end{lemma}

A \emph{Bernstein set}, is a set $B\subseteq \QTR{userA}{\Bbb{R}}^{n}$ such
that $B\cap P\neq \emptyset $ and $B\setminus P\neq \emptyset $ for every
perfect set $P\subseteq \QTR{userA}{\Bbb{R}}^{n}$. Note that the
characteristic function of a Bernstein set is not Darboux on any perfect set.

Now we are ready to prove Theorem \ref{thm: lower bound}.\medskip

\textbf{Proof of Theorem~\ref{thm: lower bound}. }Suppose, by way of
contradiction, that there exists a $k$-dimensional $\sigma $-compact metric
space $X$ such that 
\begin{equation*}
\limfunc{DIM}\nolimits_{\mathcal{D}}X<\func{ind}X=k,
\end{equation*}
where $\mathcal{D}$ is the family of Darboux functions on $X$. We can assume
that $X\subseteq \Bbb{R}^{m}$ for some positive integer $m$. Then 
\begin{equation*}
X=\bigcup_{i=1}^{\infty }X_{i},
\end{equation*}
with $X_{i}$ compact, $i=1,2\dots $ and it follows from Lemma \ref{lem:
countable union} that there is a positive integer $j$ with $\limfunc{ind}%
X_{j}\ge k$. By Lemma \ref{lem: Cantor inside} there is a $k$-dimensional
Cantor manifold $M\subseteq X_{j}$.

Let $B\in \Bbb{R}^{m}$ be a Bernstein set and $f:X\rightarrow \Bbb{R}$ be
the characteristic function of $B\cap X$. Since $\limfunc{DIM}\nolimits_{%
\mathcal{D}}X<k$, we have $f\in k\mathcal{D}$. Hence the restriction of $f$
to $M$ is in $k\mathcal{D}^{\prime }$ where $\mathcal{D}^{\prime }$ is the
family of Darboux functions on $M$. It follows from Lemma \ref{lem: Jordan}
that the restriction of $f$ to some perfect set in $\Bbb{R}^{m}$ is Darboux.
Since no restriction of the characteristic function of a Bernstein set to a
connected 
perfect set can be Darboux, we got a contradiction proving that $\limfunc{DIM%
}\nolimits_{\mathcal{D}}X\ge \func{ind}X$. $\blacksquare $

\section{Dimension relative to other classes of Darboux-like functions.}

In this section we will consider how our definition of dimension works with
some other classes of Darboux-like functions. (See \cite{GN} or \cite{CiJa}%
.) Given a topological space $X$, a function $f\colon X\to \Bbb{R}$ is:

\begin{itemize}
\item  \emph{almost continuous\/} (in sense of Stallings) if each open
subset of $X\times \Bbb{R}$ containing the graph of $f$ contains also the
graph of a continuous function from $X$ to $\Bbb{R}$;

\item  \emph{extendable}\textit{\/} provided there exists a
connectivity function $F\colon X\times [0,1]\to \Bbb{R}$ such that $%
f(x)=F(x,0)$ for every $x\in X$;

\item  \emph{peripherally continuous}\textit{\/} if for every $x\in X$ and
for all pairs of open sets $U$ and $V$ containing $x$ and $f(x)$,
respectively, there exists an open subset $W$ of $U$ such that $x\in W$ and $%
f[\limfunc{bd}(W)]\subset V$.
\end{itemize}

The classes that are defined above are denoted by $\limfunc{AC}(X)$, $%
\limfunc{Ext}(X)$, and $\limfunc{PC}(X)$, respectively. The following
inclusion relations hold when $X=\Bbb{R}^{n}$. (See \cite{GN} or~\cite{CiJa}%
.) 
%TCIMACRO{
%\TeXButton{def vvv}{\def\vvv{\mbox{ }\put(0,3){\vector(1,0){15}}\hskip18pt}}}
%BeginExpansion
\def\vvv{\mbox{ }\put(0,3){\vector(1,0){15}}\hskip18pt}%
%EndExpansion
\begin{equation*}
\limfunc{Ext}(\Bbb{R})\vvv\limfunc{AC}(\Bbb{R})\vvv\mathcal{C}(\Bbb{R})\vvv%
\mathcal{D}(\Bbb{R})\vvv\limfunc{PC}(\Bbb{R})
\end{equation*}
and, for $n>1$,

%TCIMACRO{
%\TeXButton{inclusions for n>1}{\begin{picture}(0,90)
% \put(130,55){\makebox(0,0){$\Ext(\real^n)=\Conn(\real^n)=\PC(\real^n)$}}
%   \put(205,55){\vector(1,0){15}}
%   \put(270,55){\makebox(0,0){$\AC(\real^n)\cap\darb(\real^n)$}}
% \put(320,60){\vector(2,1){18}}
% \put(362,70){\makebox(0,0){${\AC(\real^n)}$}}
% \put(360,40){\makebox(0,0){${\darb(\real^n)}$}}
% \put(320,52){\vector(2,-1){18}}
%\end{picture}
%\vskip-18pt
%}}
%BeginExpansion
\begin{picture}(0,90)
 \put(130,55){\makebox(0,0){$\Ext(\real^n)=\Conn(\real^n)=\PC(\real^n)$}}
   \put(205,55){\vector(1,0){15}}
   \put(270,55){\makebox(0,0){$\AC(\real^n)\cap\darb(\real^n)$}}
 \put(320,60){\vector(2,1){18}}
 \put(362,70){\makebox(0,0){${\AC(\real^n)}$}}
 \put(360,40){\makebox(0,0){${\darb(\real^n)}$}}
 \put(320,52){\vector(2,-1){18}}
\end{picture}
\vskip-18pt
%
%EndExpansion

\noindent where $\vvv$ denotes a strict inclusion.

Natkaniec~\cite[prop. 1.7.1]{AC} proved that every function $f\colon \Bbb{R}%
^{n}\to \Bbb{R}$ is a sum of two almost continuous functions. This implies
that 

\begin{equation}
\limfunc{DIM}\nolimits_{\limfunc{AC}}\Bbb{R}^{n}=1\text{\quad for every }%
n=1,2,3,\ldots 
\end{equation}
making the class $\limfunc{AC}$ useless in our definition of dimension. The
situation is different for the remaining two classes.

Let $X$ be a separable metric space. Since
\begin{equation*}
\limfunc{Ext}(\Bbb{R}^{2k+1})=\mathcal{C}(\Bbb{R}^{2k+1})=\limfunc{PC}(\Bbb{R%
}^{2k+1})
\end{equation*}
for $k\geq 1$, and since any function $X\rightarrow \Bbb{R}$ is both
peripherally continuous and extendable when $\limfunc{ind}X=0$, the
inequalities 
\begin{equation}
\limfunc{DIM}\nolimits_{\limfunc{Ext}}X\leq \func{ind}X\quad \text{and}\quad 
\limfunc{DIM}\nolimits_{\limfunc{PC}}X\leq \func{ind}X  \label{eq11}
\end{equation}
follow from Proposition~\ref{prop} in precisely the same way as Theorem~\ref
{thm: upper bound} does. Moreover, it is immediate to see that the analog of
Theorem~\ref{thm: dimension} for the class $\limfunc{Ext}$ is also true.

\begin{theorem}
\label{thm:dimension2} If $X$ is a $\sigma $-compact metric space, then 
\begin{equation*}
\limfunc{DIM}\nolimits_{\limfunc{Ext}}X=\func{ind}X.
\end{equation*}
\end{theorem}

\begin{proof}
The inequality $\limfunc{DIM}\nolimits_{\limfunc{Ext}}X\leq \func{ind}X$ is
a restatement of (\ref{eq11}). The other inequality holds since for every $%
\sigma $-compact metric space~$X$ we have $\limfunc{DIM}\nolimits_{\mathcal{C%
}}X=\func{ind}X$ and the inequality $\limfunc{DIM}\nolimits_{\limfunc{Ext}%
}X\ge \limfunc{DIM}\nolimits_{\mathcal{C}}X$ is implied by $\limfunc{Ext}%
(X)\subseteq \mathcal{C}(X)$ and (\ref{eq1}).
\end{proof}

In the case of the class $\limfunc{PC}$ the situation is quite different.
Unlike for the classes $\mathcal{C}$, $\mathcal{D}$, and $\limfunc{Ext}$,
(see (\ref{eq2}) which holds also for $\limfunc{DIM}\nolimits_{\limfunc{Ext}%
}X$) the dimension relative to the class $\limfunc{PC}$ is very close to the
inductive dimension for every separable metric space. However, it is not
clear whether we have equality even for all compact metric spaces.

%On one side, there are no Polish spaces that
%make the number
%$\limfunc{DIM}\nolimits_{\PC}X$
%as useless as the example from (\ref{eq2})
%makes in case of numbers
%$\limfunc{DIM}\nolimits_{\mathcal{C}}X$ and
%$\limfunc{DIM}\nolimits_{\mathcal{D}}X$.
%On the other side, we do not know
%whether the equation
%$\limfunc{DIM}\nolimits_{\PC}X=\func{ind}X$
%holds, in general, even for the compact metric spaces.

\begin{theorem}
\label{thm:seq5} If $X$ is a separable metric space, then
\begin{equation}
\limfunc{ind}X-1\leq \limfunc{DIM}\nolimits_{\limfunc{PC}}X\leq \limfunc{ind}%
X.  \label{11}
\end{equation}
\end{theorem}

\begin{proof}
Let $k=\limfunc{ind}X$. The inequality $\limfunc{DIM}\nolimits_{\limfunc{PC}%
}X\leq k$ is a restatement of (\ref{eq11}). To prove the other inequality we
will show that

\begin{itemize}
\item[($*$)]  for every $g_{1},\ldots ,g_{k-1}\in \limfunc{PC}(X)$ and $%
\varepsilon >0$ there exist a closed subset $Y$ of $X$ of cardinality
continuum such that 
\begin{equation*}
\left| g_{i}(x)-g_{i}(y)\right| <\varepsilon 
\end{equation*}
for every $x,y\in Y$ and $i=1,2,\dots ,k-1$.
\end{itemize}

We prove ($*$) by induction on $k\ge 1$. If $k=1$, take $Y=X$. The
cardinality of $X$ cannot be smaller than continuum since for some $x\in X$
and $r>0$ the boundaries of the open balls in $X$ with center $x$ and radius
smaller than $r$ are non-empty and pairwise disjoint.

Assume that $k\ge 2$. Let $g_{1},\ldots ,g_{k-1}\in \limfunc{PC}(X)$ and $%
\varepsilon >0$. There is $p\in X$ and an open neighborhood $W$ of $p$ such
that $\limfunc{ind}\limfunc{bd}(U)=k-1$ for any open $U$ with $p\in
U\subseteq W$. Since $g_{1}$ is peripherally continuous, there is an open
neighborhood $U$ of $p$ such that $U\subseteq W$ and the image $g_{1}[%
\limfunc{bd}(U)]$ is contained in the open interval $\left(
g_{1}(p)-\varepsilon /2,g_{1}(p)+\varepsilon /2\right) $. Since $\limfunc{ind%
}\limfunc{bd}(U)=k-1$, it follows from the inductive hypothesis that there
is a closed subset $Y$ of $\limfunc{bd}(U)$ of cardinality continuum such
that 
\begin{equation}
\left| g_{i}(x)-g_{i}(y)\right| <\varepsilon   \label{abc}
\end{equation}
for every $x,y\in Y$ and $i=2,3,\dots ,k-1$. Then $Y$ is closed in $X$ and
it follows from the choice of $U$, that (\ref{abc}) holds also for $i=1$
completing the proof of ($*$).

Now we show that ($*$) implies that
\begin{equation*}
\limfunc{DIM}\nolimits_{\limfunc{PC}}X\ge k-1.
\end{equation*}
Let $Z$ be a subset of $X$ such that $A\cap Z\neq \emptyset
 $ and $%
A\setminus Z\neq \emptyset
 $ for every closed $A\subseteq X$ of
cardinality continuum. The existence of such $Z$ can be proved by listing
all closed subsets of $X$ of cardinality continuum 
in a sequence $\langle A_{\alpha }\rangle _{\alpha <%
\frak{c}}$ of length continuum, defining two sequences 
$\langle a_{\alpha
}\rangle _{\alpha <\mathfrak{c}}$ and 
$\langle b_{\alpha }\rangle _{\alpha <\frak{%
c}}$ of points in $X$ by transfinite induction so that
\begin{equation*}
a_{\alpha }\in A_{\alpha }\setminus \left( \left\{ a_{\beta }:\beta <\alpha
\right\} \cup \left\{ b_{\beta }:\beta <\alpha \right\} \right) ,
\end{equation*}
and
\begin{equation*}
b_{\alpha }\in A_{\alpha }\setminus \left( \left\{ a_{\beta }:\beta \le
\alpha \right\} \cup \left\{ b_{\beta }:\beta <\alpha \right\} \right) ,
\end{equation*}
for every $\alpha <\mathfrak{c}$, and putting
\begin{equation*}
Z=\left\{ a_{\alpha }:\alpha <\frak{c}\right\} .
\end{equation*}

Let $f:X\rightarrow \Bbb{R}$ be the characteristic function of the set $Z$.
The proof will be complete when we show that 
\begin{equation*}
f\notin \left( k-1\right) \limfunc{PC}(X).
\end{equation*}
Suppose, by way of contradiction, that 
\begin{equation*}
f=g_{1}+\cdots +g_{k-1}
\end{equation*}
for some $g_{1},\ldots ,g_{k-1}\in \limfunc{PC}(X)$. By ($*$) there is a
closed subset $Y$ of $X$ of cardinality continuum such that
\begin{equation*}
\left| g_{i}(x)-g_{i}(y)\right| <\frac{1}{k-1}
\end{equation*}
for every $x,y\in Y$ and $i=1,2,\dots ,k-1$. Therefore
\begin{equation*}
\left| f(x)-f(y)\right| <1
\end{equation*}
for every $x,y\in Y$. Since $Y\cap Z$ and $Y\setminus Z$ are both non-empty,
there are $x,y\in Y$ with $f(x)=0$ and $f(y)=1$ and we get a contradiction.
Thus the proof is complete.
\end{proof}

\begin{corollary}
If $X$ is a space of Mazurkiewicz of dimension $k\ge 2$, then the class $%
\limfunc{PC}(X)$ is not equal to either $\mathcal{C}(X)$, $\mathcal{D}(X)$,
or $\limfunc{Ext}(X)$.
\end{corollary}

\begin{proof}
If $X$ is a space of Mazurkiewicz of dimension $k\ge 2$, then
\begin{equation*}
\limfunc{DIM}\nolimits_{\mathcal{C}}X=\limfunc{DIM}\nolimits_{\mathcal{D}}X=%
\limfunc{DIM}\nolimits_{\mathrm{\limfunc{Ext}}}X=0<k-1\le \limfunc{DIM}%
\nolimits_{\mathrm{\limfunc{PC}}}X.
\end{equation*}
\end{proof}

\begin{problem}
Does there exist a separable (complete separable, $\sigma $-compact,
compact) metric space $X$ such that
\begin{equation}
\limfunc{DIM}\nolimits_{\limfunc{PC}}X=\func{ind}X-1?  \label{bcd}
\end{equation}
\end{problem}

It is clear that if $X$ satisfies (\ref{bcd}), then $X$ cannot be a finite
dimensional manifold since $\limfunc{DIM}\nolimits_{\limfunc{PC}}\Bbb{R}^{n}=%
\func{ind}\Bbb{R}^{n}$. Moreover, such a space must be at least
two-dimensional. Indeed, if $\func{ind}X=0$, then $X\neq \emptyset
 $ so $%
\limfunc{DIM}\nolimits_{\mathcal{F}}X\ge 0$ for every $\mathcal{F}\subseteq 
\Bbb{R}^{X}$. If $\func{ind}X=1$, then there is an $x\in X$ and an open
neighborhood $W$ of $x$ such that $\limfunc{bd}U\neq \emptyset
 $ for
every open $U$ with $x\in U\subseteq W$. If $f:X\rightarrow \Bbb{R}$ is the
characteristic function of the singleton $\left\{ x\right\} $, then $f$ is
not peripherally continuous implying that $\limfunc{DIM}\nolimits_{\limfunc{%
PC}}X\ge 1$.

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\end{document}
