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\title{Sums of Darboux-like functions from $\real^{n}$ to $\real^m$}

%\MathReviews{Primary:  26A15; Secondary: 03E75, 54A25.}
%\keywords{cardinal invariants; extendable, Darboux, almost continuous
%and peripherally continuous functions; functions with perfect road. }


\author{{\small Francis Jordan}%
\thanks{
AMS classification numbers: Primary 26A15
Secondary 54A25 \endgraf
Key words and phrases: cardinal invariants, extendable functions, connectivity
functions, peripherally continuous functions, almost continuous functions,
Darboux functions, Sierpi\'{n}ski-Zygmund functions.
\endgraf This paper was written under supervision of K.~Ciesielski.
The author wishes to thank him for many helpful conversations.},
\small
Department of Mathematics, West Virginia University,\\
Morgantown, WV 26506-6310\\
(fejord01@.athena.louisville.edu)}

\date{}



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\newcommand{\acon}{{\operatorname {AC}}}
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\begin{document}\maketitle

\begin{abstract}
The additivity $\add(\F)$ of a family $\F\subseteq\real^{\real}$ is the minimum
cardinality of a $G\subseteq\real^\real$ with the property that $f+G\subseteq\F$
for no $f\in\real^{\real}$.  The values of $\add$ have
been calculated for many families of Darboux-like functions in $\real^\real$.
We extend these results to include some families of Darboux-like functions in
$\real^{\real^n}$.  To do this we must define $(n,k)$-additivity
which is much more flexible than additivity.
\end{abstract}


\section{Preliminaries and $(n,k)$-additivity}
We use standard notation as in \cite{CiBook}.  In
particular, $\real$ will stand for the real numbers.  For any $r\in\real$ let
$\floor{r}$ denote the greatest integer less than or equal to $r$.  For sets $X$
and $Y$ we denote the set of all functions from $X$ into $Y$ by $Y^X$.  If
$f$ and
$g$ are functions with domain $X$ we let $[f=g]=\{x\in X\colon f(x)=g(x)\}$.
For a family of functions $\F\subseteq\ Y^X$, where $Y$ has an appropriate
algebraic structure, we let $-\F=\{-f\colon f\in \F\}$ and
$n\F=\{f_1+\cdots+f_n\colon f_i\in\F\}$. The cardinality of a set
$X$ will be denoted by $|X|$.  We let
$\cuum$ stand for the cardinality of $\real$ and $\omega$ will denote the
cardinality of the natural numbers.  We say a subset $A$ of $\real$ is
$\cuum$-dense provided that $|A\cap(a,b)|=\cuum$ for all $a,b\in\real$ such
that $a<b$.  For a cardinal
$\kappa$ we let
$\kappa^+$ denote its cardinal successor.  The symbol $\oplus$ will stand for
cardinal addition.   Recall that $\kappa\oplus\lambda=\kappa+\lambda$ if
$\kappa$ and
$\lambda$ are finite cardinals, and that
$\kappa\oplus\lambda=\max(\kappa,\lambda)$ if either one of
$\kappa$ or $\lambda$ is an infinite cardinal.  Given a topological space
$X$ and a natural  number $n$ we let $X^{n}$ stand for the product of the space
$X$ with itself $n$-times with the usual topology.

We now discuss the notion of $(n,k)$-additivity.
In what follows we will assume that $X$ is a non-empty set and $Y$ is an
Abelian group (i.e., $(\forall x,y\in X)(x+y=y+x)$) with identity $0$.  We
denote by
$\theta\in Y^X$ the function with range $\{0\}$. We will assume throughout
our
discussion that $Y$ contains more than one member; so $|Y^X|>1$.  The cardinal
function
$\add$ (the additivity function) has been defined in \cite{nat} for families
$\F\subseteq Y^X$:
\begin{equation*}
\add(\F)=\min(\{|F|\colon F\subseteq Y^{X} \& (\forall
g\in Y^{X}) (\exists f\in F) (f+g\notin \F)\}\cup \{(|Y^X|)^{+}\}).
\end{equation*}
We refer to the cardinal $\add(\F)$ as the additivity of
$\F$.

%\[
%\add(\F)=\min%\left
%(\{|F|\colon F\subseteq \real^{\real}\ \&\
%(\forall g\in\real^{\real})(\exists f\in F)(f+g\in\comp \F)\}
%\cup \{(2^{\cuum})^{+}\}%\right
%).
%\]

We define the repeatability of $\F$ to be
\begin{equation*}
\rep(\F)=\min(\{n\in\omega\colon n\F=Y^X\}\cup\{\omega\}).
\end{equation*}

The reader interested in these and other cardinal functions
in real analysis is referred to the survey article \cite{CiSurvey}.
Below we list some basic facts about the additivity function which can be found
in \cite{jord}.
\prop{prop:one}{Let $\G,\F\subseteq Y^X$.  Then,
\begin{description}
\item[(i)] $\add(\F) = 1 \text{ if and only if } \F=\emptyset$;
\item[(ii)] $\add(\F)\leq |Y^X| \text{ if and only if } \F\neq Y^X$;
\item[(iii)] if $\F\subset\G$ then $\add(\F)\leq\add(\G)$;
\item[(iv)]$2<\add(\F) \text{ if and only if } \F - \F = Y^X$.
\end{description}}

When $\F\subseteq Y^X$ has the property that $\F=-\F$
the statements (ii) and
(iv) of Proposition~\ref{prop:one} imply, using the notation of repeatability,
\[(*)\ \ \ \ \ 2<\add(\F)\leq |Y^X|\text{ if and only if } \rep(\F)=2.\]
One implication of ($*$) is that for reasonable families $\F$ of functions the
additivity is a generalization of repeatability when $\rep(\F)=2$.
One might also notice that under certain conditions we have a nice restatement
of the the definition of $\add$ in terms of coding functions as sums of
functions in $\F$.
\prop{prop:will}{Let $\F=-\F$ and $\add(\F)\geq\omega$.  Then
$\add(\F)$ is the largest cardinal $\kappa$ such that for any $F\subseteq Y^X$
of cardinality less than $\kappa$ there exists a $g\in\F$ such that
\begin{equation}\label{equation:last}
(\forall f\in F)(\exists g_f\in\F)(g+g_f=f).
\end{equation}}
\proof
Let $F\subseteq Y^X$ and $|F|<\add(\F)$.  We show that there is a $g\in\F$
which
satisfies (\ref{equation:last}).  Since $\add(\F)\geq\omega$, we may assume the
zero function $\theta$ is an element of $F$.  There is a $g'\in Y^X$ such that
$g'+F\subseteq\F$, notice that since $\theta\in F$ we actually have $g'\in\F$.
For each $f\in F$ let $g_f=g'+f\in\F$.  Since $g'\in\F$ and $-\F=\F$ we have
$-g'\in\F$.  Now $g=-g'$ satisfies (\ref{equation:last}).

Now assume $\kappa$ is
a cardinal that is larger than $\add(\F)$.  We will find a family $F\subseteq
Y^X$ such that $|F|<\kappa$ and there is no $g\in\F$ which satisfies
(\ref{equation:last}).  Let $F\subseteq Y^X$ witness the definition of
$\add(\F)$, i.e. $|F|=\add(\F)<\kappa$ and
\[(\forall g\in Y^{X}) (\exists f\in F) (f+g\notin \F).\]
By way of contradiction assume that there is some $g\in\F$ satisfying
(\ref{equation:last}).
Then $f+(-g)=g_f\in\F$ for each $f\in F$ which contradicts our choice of
$F$.\qed

When
the repeatability of the family of functions is greater than 2 the additivity
function tells us at most that the repeatability is not 2.
Since we will be considering
families of functions with repeatabilities that may be larger than 2
we want to define a cardinal function which
will be of use for these families.  We would like for this cardinal function
(actually it will be a family of cardinal functions) to satisfy some statements
similar to ($*$) and Proposition~\ref{prop:will}.  It would also be appealing if
we could have a statement like ($*$) that held for all $\F\subseteq Y^X$ not
just those  where $\F=-\F$.  We might also want to remove some of the
hypothesizes of Proposition~\ref{prop:will} as well.  Removal of the need
for the
hypothesises $\F=-\F$ and
$\add(\F)\geq\omega$ is fairly simple.  We define
\begin{equation*}
\add^*(\F)=\min(\{|F|\colon F\subseteq Y^{X} \& (\forall
g\in \F) (\exists f\in F) (f-g\notin \F)\}\cup \{(|Y^X|)^{+}\}).
\end{equation*}

We must now ask how closely $\add^*$ corresponds to $\add$ as a function.
The following proposition shows that the correspondence is very close for
reasonable families of functions.

\prop{prop:3}{Let $\F\subseteq Y^X$ be such that $\F=-\F$.
Then $\add(\F)=1\oplus\add^*(\F)$.  In particular, if
$\add(\F)\geq\omega$, then
$\add^*(\F)=\add(\F)$.}
\proof
We first note that the inequality
\begin{equation}\label{equation:fa}
\add^*(\F)\leq\add(\F)
\end{equation}
is immediate from the definitions.
We show that $\add(\F)\leq 1\oplus\add^*(\F)$.
Let $F\subseteq Y^X$ be such that $|F|=\add^*(\F)$ and
\begin{equation*}
(\forall g\in \F) (\exists f\in F) (f+g\notin \F).
\end{equation*}
Put $F_1=F\cup\{\theta\}$ and notice that
\begin{equation*}
(\forall g\in Y^X) (\exists f\in F_1) (f+g\notin \F).
\end{equation*}
Thus,
\begin{equation}\label{equation:fb}
\add(\F)\leq 1\oplus\add^*(\F).
\end{equation}

For the other inequality we consider two cases.
First assume $\add(\F)\geq\omega$.  Then, by (\ref{equation:fb}),
$\add^*(\F)$ is infinite so $1\oplus\add^*(\F)=\add^*(\F)$.
Now by
(\ref{equation:fa}) we have $1\oplus\add^*(\F)=\add^*(\F)\leq\add(\F)$.

Next assume $\add(\F)<\omega$.  Note that (\ref{equation:fa})
implies that $\add^*(\F)$ is finite.
We will show that $1\oplus\add^*(\F)\leq\add(\F)$.  First assume that
$\add^*(\F)=0$.  Since the empty set $\emptyset$ is the only subset of
$Y^X$ with
cardinality zero we have, by the definition of $\add^*(\F)$,
\begin{equation*}
(\forall g\in \F) (\exists f\in\emptyset) (f-g\notin \F).
\end{equation*}
The only way the above statement may be true is if
$\F=\emptyset$.  By Proposition~\ref{prop:one},
$\add(\emptyset)=1$.  So in this case $1\oplus\add^*(\F)=\add(\F)$.  So we may
assume that $\add^*(\F)>0$.  Take
$F\subseteq Y^X$ with $|F|<1\oplus\add^*(\F)$.  Fix $f\in F$ and put
$F_1=(-f+F)\setminus\{\theta\}$.  We have
$|F_1|<\add^*(\F)$. So there is a $g\in\F$ such that
$-g+F_1\subseteq\F$.   Since $\F=-\F$, we have
$-g+(F_1\cup\{\theta\})\subseteq\F$.  Now
$(-g-f)+F=-g+(F_1\cup\{\theta\})\subseteq\F$.
Thus,
$1\oplus\add^*(\F)\leq\add(\F)$.
\qed

Now the $\add^*$-analog of ($*$) is that for any $\F\subseteq Y^X$
\[(*_1)\ \ \ \ \  1<\add^*(\F)\leq |Y^X|\text{ if and only if } \rep(\F)=2.\]
The version of Proposition~\ref{prop:will} for $\add^*$ is more simple to state
and prove.
\prop{prop:was}{$\add^*(\F)$ is the largest cardinal $\kappa$ such that for
any $F\subseteq Y^X$ of cardinality less than $\kappa$ there
exists a $g\in\F$ such that \begin{equation}\label{equation:tour}
(\forall f\in F)(\exists g_f\in\F)(g+g_f=f).
\end{equation}}
\proof
Let $F\subseteq Y^X$ and $|F|<\add^*(\F)$.  We show that there is a $g\in\F$
which satisfies (\ref{equation:tour}).  Clearly there is a $g\in\F$ such that
$-g+F\subseteq\F$.  For each $f\in\F$ let $g_f=-g+f\in\F$.  Then $g$
satisfies (\ref{equation:tour}).

Now assume $\kappa$ is a cardinal that is
larger than $\add^*(\F)$.  We find a family $F\subseteq Y^X$ such that
$|F|<\kappa$ and there is no $g\in\F$ which satisfies (\ref{equation:tour}).
Let
$F\subseteq Y^X$ witness the definition of $\add^*(\F)$, i.e.,
$|F|=\add^*(\F)<\kappa$ and
\[(\forall g\in\F) (\exists f\in F) (f-g\notin \F).\]
By way of contradiction assume that there is some $g\in\F$ satisfying
\begin{equation*}
(\forall f\in F)(\exists g_f\in\F)(g+g_f=f)
\end{equation*}
Then $f-g=g_f\in\F$ for each $f\in F$ which contradicts our choice of
$F$.\qed

Now our goal will be be to construct a family of cardinal functions that will
allow us to generalize $(*_1)$ and Proposition~\ref{prop:was} to include
families with repeatabilities greater than $2$.   Let $n,k\in\omega$ be
such that
$k<n$ and let $\F\subseteq Y^X$.  We define the $(n,k)$-additivity of $\F$ to be
\begin{equation*}
\add_{n,k}(\F)=
\min(\{|F|\colon F\subseteq Y^X \&\ \Psi_{n,k}(F,\F) \text{
holds}\}\cup\{|Y^X|^+\})
\end{equation*}
where $\Psi_{n,k}(F,\F)$ denotes the statement
\begin{equation*}
(\forall g\in (n-k)\F)
(\exists f\in F)(f-g\notin (k+1)\F).
\end{equation*}
Notice that $\add^*=\add_{1,0}$.  We can now restate
Proposition~\ref{prop:3}  in this language.
\prop{prop:again}{Let $\F\subseteq Y^X$ be such that $\F=\{-f\colon f\in\F\}$.
Then $\add(\F)=1\oplus\add_{1,0}(\F)$.  In particular, if
$\add(\F)\geq\omega$, then
$\add_{1,0}(\F)=\add(\F)$.}

We see
that Proposition~\ref{prop:was} is generalized as follows. We leave the proof to
the reader since it is of the same form as the proof of
Proposition~\ref{prop:was}.
\prop{prop:was1}{$\add_{n,k}(\F)$ is the largest cardinal
$\kappa$ such that for any $F\subseteq Y^X$ of cardinality less than $\kappa$
there exist $g^1,\ldots,g^{n-k}\in\F$ such that
\[(\forall f\in F)(\exists g^1_f,\ldots,g^{k+1}_f\in\F)(g^1+\ldots+g^{n-k}
+g^1_f+\ldots+g^{k+1}_f=f).\]}

We state an expanded version of Proposition~\ref{prop:one} for
$(n,k)$-additivity and include some other facts.

\prop{prop:two}{Let $\F,\G\subseteq Y^X$ and $n\in\omega$.  Then
\begin{description}
\item[(i)] if $k<n$, then $\add_{n,k}(\F)=0 \text{ if and only if
}\F=\emptyset$;
\item[(ii)] if $k<n$, and $\F\subseteq\G$ then
$\add_{n,k}(\F)\leq\add_{n,k}(\G)$;
\item[(iii)] if $i<k<n$, then $\add_{n,i}(\F)\leq\add_{n,k}(\F)$;
\item[(iv)] if $k<n$, then $\add_{n,k}(\F)\leq |Y^X| \text{ if and only if }
\rep(\F)> k+1$;
\item[(v)] if $k<n$, then $1<\add_{n,k}(\F)\text{ if and only if }
\rep(\F)\leq n+1$;
\item[(vi)] $1<\add_{n,k}(\F)\leq |Y^X| \text{ for all }k<n\text{ if and only
if }
\rep(\F)=n+1$;
\item[(vii)] if $\rep(\F)\geq n+1$ and $\F=-\F$, then $\add_{n,k}(\F)\leq 2$
for all $k<\floor{n/2}$;
\item[(viii)] if $k<n$, then $\add_{n,k}(\F)\leq\add((k+1)\F)$.
\end{description}}
\proof
Items (i) and (ii) are straight forward and will be left without proof.

We prove (iii).  Let $F\subseteq Y^X$ and $|F|<\add_{n,i}(\F)$.  By
Proposition~\ref{prop:was1} there exist $g^1,\ldots,g^{n-i}\in\F$ such that
\begin{equation}\label{equation:slap}
(\forall f\in F)(\exists g^1_f,\cdots,g^{i+1}_f\in\F)
(g^1+\cdots+g^{n-i}+g^1_f+\cdots+g^{i+1}_f=f).
\end{equation}
Let $m=k-i$.  By
(\ref{equation:slap}) we have for each $f\in F$,
\begin{equation}\label{equation:smick}
g^1_f+\cdots+g^{i+1}_f+g^1+\cdots+g^m=f-(g^{m+1}+\cdots+g^{n-i}).
\end{equation}
Note that $n-i-(m+1)+1=n-k$ so $g=g^{m+1}+\cdots+g^{n-i}\in (n-k)\F$.  Notice
also that $i+1+m=k+1$, so $g^1_f+\cdots+g^{i+1}_f+g^1+\cdots+g^m\in (k+1)\F$.
So (\ref{equation:smick}) implies that there is a $g\in (n-k)\F$ such that
$-g+F\subseteq (k+1)\F$.  Thus $\add_{n,i}(\F)\leq\add_{n,k}(\F)$.

We prove (iv).  Suppose $\add_{n,k}(\F)=|Y^X|^+$.  Then there is a
$g\in(n-k)\F$ such that $(-g)+(Y^X)\subseteq (k+1)\F\subseteq Y^X$.  Thus
$(k+1)\F=Y^X$.  Hence $\rep(\F)\leq k+1$.  To see the other implication assume
that $\rep(\F)\leq k+1$ which is to say $(k+1)\F=Y^X$.
Then $g+(k+1)\F=Y^X$ for any $g\in Y^X$.  In particular, we may pick $g$ to be
in $(n-k)\F$.  Thus, $(-g)+(Y^X)\subseteq (k+1)\F$ for some $g\in (n-k)\F$.  So
$\add_{n,k}(\F)=|Y^X|^+$.

We prove (v). Suppose $\add_{n,k}(\F)\geq 2$.  Then for any $f\in Y^X$
there is a $g\in (n-k)\F$ such that $f-g\in (k+1)\F$, which is to say
$f\in(n+1)\F$.  Thus $(n+1)\F=Y^X$ and so $\rep(\F)\leq n+1$.
Now, assume $\add_{n,k}(\F)\leq 1$ and let
$G=\{g\}$ witness the definition of $\add_{n,k}(\F)$ (i.e.,
$f-g\notin (k+1)\F$ for every $g\in(n-k)\F$ ).  Clearly, $f\notin (n+1)\F$
so $(n+1)\F\neq Y^X$ which is to say $\rep(\F)>n+1$.

Item (vi) is a direct consequence of (iv) and (v).

We now prove (vii).
Let $k<\floor{n/2}$.
Since $\rep(\F)\geq n+1$ there is a function $h\colon X\to Y$
that is not an element of $(2k+2)\F$.  By way of contradiction assume
$\add_{n,k}(\F)>2$.  Pick $f_1,f_2\in Y^X$ such that $f_1-f_2=h$.  Since
$\add_{n,k}(\F)>2$, there is a $g\in (n-k)\F$ such that
$\{f_1-g,f_2-g\}\subseteq(k+1)\F$.  Note that $-(f_2-g)\in (k+1)\F$ since
$\F=-\F$.  Thus,
\[h=f_1-f_2=(f_1-g)-(f_2-g)\in(2k+2)\F\] contradicting our choice
of $h$.

We prove (viii).  Let $F\subseteq Y^X$ and $|F|<\add_{n,k}(\F)$.  There
is a $g\in (n-k)\F$ such that $-g+F\subseteq (k+1)\F$.  Thus,
$\add_{n,k}(\F)\leq\add((k+1)\F)$. \qed

Notice that (vi) of
Proposition~\ref{prop:two} is a good generalization of $(*_1)$.

\section{Results}
We calculate the generalized additivities of the following
families of functions.  Descriptions of these functions for
general topological spaces are given.
\begin{description}
\item[$\dar$:] $f\in Y^X$ is a Darboux function if and only if $f[C]$ is
connected
in $Y$ for any connected set $C$ of $X$.
\item[$\conn$: ] $f\in Y^X$ is a connectivity function if and only if the
graph of
$f$ restricted to $C$ is connected in $X\times Y$ for every connected
set $C$ of $X$.
\item[$\acon$: ] $f\in Y^X$ is an almost continuous function
if and only if every open set in $X\times Y$ containing $f$ also
contains some continuous function $g\in Y^X$.
\item[$\ext$:] $f\in Y^X$ is an extendable function if and only if
there is a connectivity function $g:X\times [0,1]\to Y$ such that
$f(x)=g(0,x)$ for every $x\in X$.
\item[$\phc$: ] $f\in Y^X$ is a peripherally continuous function if and only if
for every $x\in X$ and pair of open sets $U\subset X$ and $V\subset Y$
such that $x\in U$ and $f(x)\in V$ there is an
open neighborhood $W$ of $x$ with $\cl(W)\subset U$ and
$f[\bd(W)]\subset V$, where $\cl(W)$ and $\bd(W)$ denote the boundary and
closure  of $W$, respectively.
\item[$\sz$: ] $f\in Y^X$ is a Sierpi\'{n}ski-Zygmund function if and only if
$f$  is continuous on no set of cardinality $|X|$.
\end{description}
We note that in \cite{HAG} it is shown that
$\conn(\real^n,\real^m)=\phc(\real^n,\real^m)$ when
$n>1$.  The equality $\ext(\real^n,\real)=\conn(\real^n,\real)$ is shown in
\cite{CNW} for
$n>1$.  The relationships of containment between the above families for the
cases $n=1$ and
$n>1$ are shown in the diagrams below.  Each arrow denotes proper containment
(see
\cite{BHL} for $n=1$ and for $n>1$ see \cite{st} and
\cite[Examples~1.6 and 1.7]{nat}).

\begin{picture}(300,50)

\thicklines
\put(0,20){\makebox(0,0){$n=1$:}}
\put(40,20){\makebox(0,0){$\ext$}}
\put(110,20){\makebox(0,0){$\acon$}}
%\put(160,15){\makebox(0,0){$\pr$} }
\put(180,20){\makebox(0,0){$\conn$}}
\put(250,20){\makebox(0,0){$\dar$} }
\put(320,20){\makebox(0,0){$\phc$}}

\put(60,20){\vector(1,0){30}}
\put(130,20){\vector(1,0){30}}
\put(200,20){\vector(1,0){30}}
\put(270,20){\vector(1,0){30}}
%\put(30,35){\vector(4,-1){100}}
%\put(190,10){\vector(4,1){100}}

\end{picture}

\begin{picture}(300,100)

\thicklines
\put(20,40){\makebox(0,0){$n>1$:}}

\put(170,40){\makebox(0,0){$\ext=\phc=\conn$}}
%\put(260,15){\makebox(0,0){$\pr$} }
\put(260,70){\makebox(0,0){$\dar$}}
\put(260,10){\makebox(0,0){$\acon$} }
%\put(320,40){\makebox(0,0){$\phc$}}


\put(220,40){\vector(1,1){30}}
\put(220,40){\vector(1,-1){30}}
%\put(260,80){\vector(3,-2){40}}
%\put(30,35){\vector(4,-1){100}}
%\put(190,10){\vector(4,1){100}}

\end{picture}



To state some of the theorems we will need to define the following
cardinal number.
\[\diff_{\cuum}=\min\{|F|\colon F\subseteq\cuum^{\cuum}\ \&\ (\forall
g\in\cuum^{\cuum})(\exists f\in F)(|[f=g]|<\cuum)\}.\]
It is easy to prove that $\cuum<\diff_{\cuum}\leq 2^{\cuum}$ and in
\cite{CM} it is shown that this
is about all that can be said about its value in ZFC.
We intend to prove the following six theorems.  The first three deal with the
generalized additivities of some the families of functions listed above.  The
last three theorems are concerned with some containments between between the
above families of functions.

\thm{thm:two}{Let $n>1$. Then,
\begin{equation*}
\add_{n,n-1}(\ext(\real^n,\real))=\add(n\ext(\real^n,\real))=\cuum^+.
\end{equation*}}

\thm{thm:one}{Let $n,m\geq 1$.  Then,
\[\add_{n,n-1}(\dar(\real^n,\real^m))=\add(n\dar(\real^n,\real^m))=
\diff_{\cuum}.\]}

\thm{thm:onea}{Let $n,m\geq 1$.  Then,
$\add(\acon(\real^n,\real^m))=\diff_{\cuum}$.}

\thm{thm:three}{Let $n,m\geq 1$. Then
\begin{equation}\notag
\add_{n,j}(\dar(\real^n,\real^m))=\add((j+1)\dar(\real^n,\real^m))=\cuum^+
\end{equation}
for $n-1>j\geq \floor{n/2}$.}

Before stating the other three main Theorems we consider some
implications of the two Theorems above.  Using Propositions \ref{prop:3} and
\ref{prop:two}(vi) we see that Theorems \ref{thm:two},
\ref{thm:one} and \ref{thm:onea} generalize the following results.
\prop{prop:foura}{
\begin{description}
\item[ ]
\item[(i)] \rm{\cite{CR}} $\add(\ext(\real,\real))=\cuum^{+}$.
\item[(ii)] \rm{\cite{CW}} $\rep(\F(\real^n,\real))=n+1$ for
$\F\in\{\ext,\conn\}$.
\item[(iii)] \rm{\cite{CM}}
$\add(\dar(\real,\real))=\add(\acon(\real,\real))=\diff_{\cuum}$.
\end{description}}

Theorem~\ref{thm:one} has as a corollary, using Proposition~\ref{prop:two}(v),
the answer to a question of Ciesielski and Wojciechowski \cite{CW}.
\cor{cor:9}{Let $n,m\geq 1$.  Then $\rep(\dar(\real^n,\real^m))=n+1$.}
Theorems \ref{thm:two} and \ref{thm:one} also have the following two
corollaries:
\cor{cor:20}{If $k<\floor{n/2}$ then $\add_{n,k}(\ext(\real^n,\real))=2$, and
if $m>k\geq n$ then
$\add_{m,k}(\ext(\real^n,\real))=(2^{\cuum})^+$.}
\proof
By Theorem~\ref{thm:two} and Proposition~\ref{prop:two}(v)
$\rep(\ext(\real^n,\real))=n+1$.  Since
$\ext(\real^n,\real)=-\ext(\real^n,\real)$, Proposition~\ref{prop:two}(vii)
implies that if
$k<\floor{n/2}$, then $\add_{n,k}(\ext(\real^n,\real))=2$.  Since
$\rep(\ext(\real^n,\real)=n+1$, it follows from Proposition~\ref{prop:two}(iv)
that if $m>k\geq n$ then $\add_{m,k}(\ext(\real^n,\real))=(2^{\cuum})^+$.\qed

\cor{cor:21}{If $k<\floor{n/2}$ then $\add_{n,k}(\dar(\real^n,\real^m))=2$,
and if $m>k\geq n$, then
$\add_{m,k}(\dar(\real^n,\real^m))=(2^{\cuum})^+$.}
\proof
Repeat the proof of Corollary~\ref{cor:20} replacing $\ext(\real^n,\real)$ with
$\dar(\real^n,\real^m)$ and using Theorem~\ref{thm:one} instead of
Theorem~\ref{thm:two}.\qed

The above results for $n>1$ are summarized in the
tables below.

%\begin{figure}[h]\label{fig:4}
\begin{center}
\begin{tabular}{|l|l|l|l|}\hline
    & $k<p<n$  & $\ \ \ \ k<n=p$ &$n\leq k< p$ \\ \cline{1-4}
$\F=\ext(\real^n,\real)$   &\ \ \ \ \ \ $1$ &see table below &\ \ \ \
$\left(2^{\cuum}\right)^+$ \\ \cline{1-4}
$\F=\dar(\real^n,\real^m)$   &\ \ \ \ \ \ $1$ &see table below &\ \ \ \
$\left(2^{\cuum}\right)^+$  \\ \cline{1-4}
\end{tabular}
%\caption{Values of $\add_{p,k}(\F)$.}
\end{center}
%\end{figure}
\begin{center}
Figure 3.1
\end{center}

\medskip

%\begin{figure}[h]\label{fig:5}
\begin{center}
\begin{tabular}{|l|l|l|l|l|}\hline
    & $k < \floor{n/2}$   & $\floor{n/2}\leq k\leq n-2$  & $k=n-1$ \\
\cline{1-4}
$\F=\ext(\real^n,\real)$   &\ \ \ \ \ \ $2$   &\ \ \ \ \ \ \ \ \ \ \ ?
&\ \ \
\ \ $\cuum^+$
\\ \cline{1-4}
$\F=\dar(\real^n,\real^m)$   &\ \ \ \ \ \ $2$   &\ \ \ \ \ \ \ \ \ \ \
$\cuum^+$     &\ \ \ \ \ $\diff_{\cuum}$
  \\ \cline{1-4}
\end{tabular}
%\caption{Values of $\add_{n,k}(\F)$.}
\end{center}
%\end{figure}
\begin{center}
Figure 3.2
\end{center}

The first table gives the values of $\add_{p,k}(\F)$; the second table
gives the values of $\add_{n,k}$.  We now state the remaining theorems.
\noindent\thm{thm:four}{$\acon(\real^n,\real^m)\subsetneq
n\dar(\real^n,\real^m)$.
Moreover, there is an almost continuous function $f$ such that $f\notin
 (n-1)\dar(\real^n,\real^m)$.}

\thm{thm:five}{$n\dar(\real^n,\real)\cap\sz(\real^n,\real)=\emptyset$ for
$n>1$.}

An immediate Corollary of Theorems \ref{thm:four} and \ref{thm:five} is

\cor{cor:10}{If $n>1$
then $\acon(\real^n,\real)\cap\sz(\real^n,\real)=\emptyset$.}

In \cite{BCN} it is shown that the conclusions of
Theorem~\ref{thm:five} and Corollary~\ref{cor:10} are independent of ZFC when
$n=1$.

Corollary~\ref{cor:9} and Theorem~\ref{thm:four} might lead one to conjecture
that every function from $\real^n$ into $\real$ is the sum of an almost
continuous function and a Darboux function.
The following example which is constructed in Section \ref{sec:ex1} shows
that this is very much not the case when $n>1$.

\ex{ex:1}{There exist a Baire class 1 function $f\colon\real^n\to\real$
such that
$f\notin n\dar(\real^n,\real)$.  Moreover, for $n>1$ we can also assume
$f$ is not the sum of an
almost continuous function and $n-1$ Darboux functions.}


\section{Proofs of Theorems \ref{thm:one} and \ref{thm:five}.}

\noindent\lem{lem:a}{Let $n,m\geq 1$, $f\colon\real^n\to\real^m$ be a
function and
$A,B\subseteq\real$ be a partition of $\real$ into two disjoint $\cuum$-dense
sets.  If $\pi:\real^n\to\real$ denotes the projection of $\real^n$ onto a fixed
coordinate and
\begin{description}
\item[(i)] $f|_{\pi^{-1}(y)}\in\dar(\pi^{-1}(y),\real^m)$ for every $y\in B$,
\item[(ii)] $f|_{\pi^{-1}(y)}$ is constant for every $y\in A$, and
\item[(iii)] $\{y\in A\colon f[
\pi^{-1}(y)]=\{r\}\}$ is dense in $\real$
for every $r\in\real^m$,
\end{description}
then $f\in\dar(\real^n,\real^m)$.}
\proof
Suppose $C\subseteq\real^n$ is connected.  We show that $f[C]$ is a connected
subset of $\real^m$.  There are two possible cases that may occur.
The first case is when there is a $y\in\real$ such that
$C\subseteq\pi^{-1}(y)$.
In this case, one of (i) or (ii) applies to show that $f[C]$ is a connected
set.
The other case is when there exist distinct $y_1<y_2\in\real$ such that
$C\cap\pi^{-1}(y_1)\neq\emptyset\neq C\cap\pi^{-1}(y_1)$.  Since $C$ is
connected,
we have $C\cap\pi^{-1}(y)\neq\emptyset$ for all $y\in [y_1,y_2]$.
So, by (iii), $f[C]=\real^m$, which is connected.\qed



We now prove one of the main inequalities of Theorem~\ref{thm:one}.
\noindent\lem{lem:ins}{$\add_{n,n-1}(\dar(\real^n,\real^m))\geq\diff_{\cuum}$.}
\proof
We proceed by induction on $n$. The inequality is proven in \cite{CM} for
the case $\dar(\real,\real)$, and the methods used in \cite{CM} can clearly
be used to establish the inequality for $\dar(\real,\real^m)$ when $m>1$.  So
we may assume that $n>1$ and
$\add_{n-1,n-2}(\dar(\real^{n-1},\real^m))\geq\diff_{\cuum}$.
Let $F\subseteq(\real^m)^{\real^n}$ be an arbitrary collection of functions
such
that $|F|<\diff_{\cuum}$.  We must find a $g\in\dar(\real^n,\real^m)$ such that
$-g+F\subseteq n\dar(\real^n,\real^m)$. Let $\{A_k\}_{k\in n+1}$ be a partition
of $\real$ into $n+1$ $\cuum$-dense sets.  Define
$h\colon\real\to\real^m$ so that for each $p\in\real^m$ and $k\in n+1$
\begin{equation*}
\{y\in A_k\colon h(y)=p\} \text{ is dense in $\real$}.
\end{equation*}
Let $\pi\colon\real^n\to\real$ denote the projection of $\real^n$ onto a fixed
coordinate.  Note that $\pi^{-1}(y)$ is homeomorphic to
$\real^{n-1}$ for every
$y\in\real$.  For each $y\in\real$ let $F^y=\{f|_{\pi^{-1}(y)}-h(y)\colon f\in
F\}$.  Since $|F^y|\leq |F|<\diff_{\cuum}$, we may apply the inductive
hypothesis
to find for each $l\in n$ and $y\in A_l$ a Darboux function
$g^{y}\colon\pi^{-1}(y)\to\real^m$ such that
$-g^y+F^y\subseteq(n-1)\dar(\pi^{-1}(y),\real^m)$.  So, for every $f\in F$ and
$y\in A_l$ there exist
$\{g^{f,y}_{k}\in\dar(\pi^{-1}(y),\real^m)\colon k\in n\setminus\{l\}\}$ such
that
\begin{equation}\label{eq:gtrav}
g^{y}+\sum_{\{k\in n\colon k\neq l\}} g^{f,y}_{k}=f|_{\pi^{-1}(y)}-h(y).
\end{equation}
Define $g\colon\real^n\to\real^m$ by
\begin{equation*}
g(p)=
\begin{cases}
h(y)& \text{ if $\pi(p)=y\in A_{n}$, }\\
g^{y}(p)& \text{ if $\pi(p)=y\notin A_{n}$.}
\end{cases}
\end{equation*}
We claim $g$ is as desired.
To see that $g$ is Darboux let $A=A_{n}$,
$B=\real\setminus A_n$,
and apply Lemma \ref{lem:a}.  By inductive hypothesis,
$\add_{n-1,n-2}(\dar(\pi^{-1}(y),\real^m)
\geq\diff_{\cuum}$ for each $y\in A_{n}$.  By
Proposition~\ref{prop:two}(v), $\rep(\dar(\pi^{-1}(y),\real^m))\leq n$.
Thus, for each $y\in A_{n}$ and $f\in F$, we may find
$\{g^{f,y}_k\in\dar(\pi^{-1}(y),\real^m)\colon k\in n\}$ such that
\begin{equation}\label{eq:trav1}
\sum_{k=0}^{n-1} g^{f,y}_k=f|_{\pi^{-1}(y)}-h(y).
\end{equation}
For each $f\in F$ and $k\in n$ define $g^f_k\colon\real^n\to\real^m$ so that
\begin{equation}\label{howl}
g_k^f(p)=
\begin{cases}
h(y)& \text{ if $\pi(p)=y\in A_{k}$,}\\
g^{f,y}_k& \text{ if $\pi(p)=y\notin A_{k}$.}
\end{cases}
\end{equation}
Note that $g^f_k\in\dar(\real^n,\real^m)$ for each $f\in F$ and $k\in n$.
To see it, take $A=A_{k}$, $B=\real\setminus A_{k}$ and apply Lemma
\ref{lem:a}.   We now show that $-g+F\subseteq n\dar(\real^n,\real^m)$.  More
precisely we show that for each $f\in F$
\begin{equation}\label{eq:40b}
g+\sum_{k=0}^{n-1} g^f_k=f.
\end{equation}
Let $p\in\real^n$.  We must consider two cases.
First, assume there is an $l\in n$ and a $y\in A_l$ such that
$p\in\pi^{-1}(y)$.
Then,
\begin{eqnarray}
g(p)+\sum_{k=0}^{n-1} g^f_k(p)
& = & g^{y}(p)+\left(\sum_{\{k\in n\colon k\neq l\}}g^{f,y}_k(p)
+g^f_l(p)\right)\notag
\\ & = & \left(g^y(p)+\sum_{\{k\in n\colon
k\neq l\}}g^{f,y}_k(p)\right)+g^f_l(p)\notag\\
& = & (f(p)-h(y))+h(y)\label{xues}\\
& = & f(p), \notag
\end{eqnarray}
where (\ref{xues}) follows from (\ref{eq:gtrav}) and
(\ref{howl}).   We now consider the case in which there exists a $y\in
A_{n}$ such that
$p\in\pi^{-1}(y)$.  Then,
\begin{eqnarray}
g(p)+\sum_{k=0}^{n-1} g_k^f(p)
& = & h(y)+\sum_{k=0}^{n-1} g^{f,y}_k(p)\notag \\
& = & h(y)+(f(p)-h(y)) \label{gor} \\
& = & f(p), \notag
\end{eqnarray}
where (\ref{gor}) follows from (\ref{eq:trav1}).
So, (\ref{eq:40b}) holds completing the inductive step.\qed



Our next goal is to show that $\add(n\dar(\real^n,\real^m))\leq\diff_{\cuum}$.
Since Proposition \ref{prop:two}(viii) implies that
$\add_{n,n-1}(\dar(\real^n,\real^m))\leq\add(n\dar(\real^n,\real^m))$, this
will complete the proof.  We will need some results and definitions from
dimension theory, all of which  may be found in \cite{dim}.


The dimension of a topological space is defined as follows.
\begin{description}
\item[(i)] $\dim X=-1$ if and only if $X=\emptyset$.
\item[(ii)] $\dim X\leq n$ 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
$\dim\bd_X(U)\leq n-1$.
\item[(iii)] $\dim X=n$ if $\dim X\leq n$ and it is not true that $\dim
X\leq n-1$.
\end {description}
An n-dimensional Cantor manifold ($n \geq 1$) is a compact n-dimensional space
that
cannot be disconnected by a subset of dimension less than $n-1$.  We will
need three facts about n-dimensional Cantor manifolds.  We say $X$ is a
{\em continuum} if
$X$ is a compact, connected, nonempty metric space.
We say a continuum $X$ is {\em degenerate} if $|X|=1$; otherwise, we say $X$ is
{\em non-degenerate}.

\noindent\prop{prop:three}{Every n-dimensional Cantor manifold is a
continuum.}

\noindent\prop{prop:four}{If $X$ is a compact n-dimensional space, then $X$
contains an
n-dimensional Cantor manifold.}

\noindent\prop{prop:five}{$[0,1]^n$ is an $n$-dimensional Cantor manifold
for all $n>0$.}


To establish the inequality $\add(n\dar(\real^n,\real^m))\leq\diff_{\cuum}$, we
must first prove some lemmas.

\noindent\lem{lem:c}{Suppose $n>1$ and M is an n-dimensional Cantor manifold.
If $B\subseteq M$ disconnects $M$, then
there is an $(n-1)$-dimensional Cantor manifold contained in $B$ .}
\proof
It is widely known that if a subset $S$ of a continuum $X$ disconnects $X$,
then there is a compact set $F\subseteq S$ that disconnects $X$.  In
particular, we may find a compact set $C\subseteq B$ such that $B$ disconnects
$M$.  By the definition of Cantor manifold, the dimension of
$C$ is at least
$n-1$.   So, by Proposition~\ref{prop:four}, there is an $(n-1)$-dimensional
Cantor manifold $N$  such that $N\subseteq C\subseteq B$.\qed

\noindent\lem{lem:d}{Assume that $n>1$, $M$ is an n-dimensional Cantor
manifold, and
$f\colon M\to\real$ is Darboux.  There is a collection
$\{N_{\alpha}\}_{\alpha\in\cuum}$
of pairwise disjoint $(n-1)$-dimensional Cantor manifolds such that
$f|_{N_{\alpha}}$ is constant for each $\alpha\in\cuum$.}
\proof
We first assume $f$ is not constant.  Let $r$ be in the interior of $f[M]$.
Since $f[M]\setminus\{r\}$ is  not connected and $f$ is Darboux, it follows that
$M\setminus f^{-1}(\{r\})$ is not  connected.  By Lemma~\ref{lem:c} there is a
$(n-1)$-dimensional Cantor manifold $N$ such that $N\subseteq f^{-1}(\{r\})$.
Thus, $f|_N$ is constant.  Since we may do this for each $r$ in the interior
of $f[M]$ and  preimages of distinct points  are disjoint,
we can find the desired collection of continua.

If $f$ is a constant function, the lemma reduces to the question of whether
there exist $\cuum$-many pairwise disjoint $(n-1)$-dimensional Cantor
manifolds contained in $M$.  By the first part of the proof of this lemma, it
is enough to show there is a non-constant Darboux function $h\colon
M\to\real$.  Fix $x_0\in M$ and let $h\colon M\to\real$ be the
continuous function defined by
$h(x)=\rm{dist}(x,x_0)$.  Clearly, $h$ is non-constant and Darboux.
\qed


\noindent\lem{lem:b}{Suppose that $n>1$ and $M$ is a $n$-dimensional Cantor
manifold.  If
$n\geq k\geq 1$ and $f\in k\dar(M,\real)$, then there is a
collection of pairwise disjoint non-degenerate continua
$\{C_{\alpha}\}_{\alpha\in\cuum}$
such that $f|_{C_{\alpha}}$ is Darboux for all $\alpha\in\cuum$.
Moreover, if $n>k$ then we may assume
$f|_{C_{\alpha}}$ is constant for all $\alpha\in\cuum$.}
\proof
Let $M$ be an $n$-dimensional Cantor manifold and
$f\in k\dar(M,\real)$.  We proceed by induction on $n$.  First, we establish
the lemma for $n=2$.
When $n=2$ and $k=1$, the result is immediate by Lemma \ref{lem:d}.
When $n=2$ and $k=2$, then $f=g_1+g_2$ where $g_1,g_2\in\dar(M,\real)$.  By
Lemma \ref{lem:d}, there is a collection of pairwise disjoint
$1$-dimensional Cantor manifolds
$\{C_{\alpha}\}_{\alpha\in\cuum}$ such that
$g_1|_{C_{\alpha}}$ is constant for every $\alpha\in\cuum$.
It follows that $f|_{C_{\alpha}}$ is Darboux for every $\alpha\in\cuum$.
This completes the proof
of the lemma for $n=2$ since each $C_{\alpha}$ is a continuum.

So, suppose the lemma holds for all $m$ less than $n$.  We show it also
holds for
$n$.  If $k=1$, the result is immediate by Lemma
\ref{lem:d}.  So, we  may assume $k>1$.
Let $f=g_1+\cdots+g_k$ where $g_1,\ldots,g_k\in\dar(M,\real)$.  By Lemma~
\ref{lem:d} there is an $(n-1)$-dimensional Cantor manifold $N$ such that
$g_1|_N$ is a constant function.  Thus, $f|_N\in (k-1)\dar(N,\real)$.  By
inductive hypothesis there  is a collection of pairwise disjoint continua,
$\{C_{\alpha}\}_{\alpha\in\cuum}$, such that
$C_{\alpha}\subseteq N\subseteq M$ and
$f|_{C_{\alpha}}$ is Darboux for every $\alpha\in\cuum$.  Finally, if
$n>k$ then $n-1>k-1$.  So we may assume by inductive hypothesis that
$f|_{C_{\alpha}}$ is constant for every $\alpha\in\cuum$
since $f|_{N}$ is constant.
So, the lemma holds for all $n\geq 2$.\qed

\noindent\lem{lem:e}{Let $n>1$,
$n\geq k\geq 1$, and $f\in k\dar(\real^n,\real)$.
Then, there is a collection of pairwise disjoint non-degenerate continua
$\{C_{\alpha}\}_{\alpha\in\cuum}$   such that $f|_{C_{\alpha}}$ is Darboux
for each
$\alpha\in\cuum$.   Moreover, if $n>k$, then we may assume
$f|_{C_{\alpha}}$ is constant for every $\alpha\in\cuum$.}
\proof
If $f$ is Darboux then its restriction to $[0,1]^n$ is also
Darboux.  By Proposition~\ref{prop:five}, $[0,1]^n$ is an $n$-dimensional
Cantor
manifold.  The lemma now follows from Lemma \ref{lem:b}.\qed

Lemma~\ref{lem:e} has an easy consequence.

\noindent\lem{lem:f}{Let $n>1$,
$n\geq k\geq 1$, and $f\in k\dar(\real^n,\real)$.  Then either $f$ is constant
on some non-degenerate continuum or there is a rational number $q$
such that $|f^{-1}(\{q\})|=\cuum$.}
\proof
Let $\{C_{\alpha}\}_{\alpha\in\cuum}$ be as in Lemma \ref{lem:e} with
$M=\real^n$.  If $f$ is
constant on no non-degenerate  continuum, $f[C_{\alpha}]$ must be a
non-degenerate interval for each $\alpha\in\cuum$ and must contain a rational
number.   It follows that $|f^{-1}(q)|=\cuum$ for some $q\in\rational$.\qed

We now prove Theorem~\ref{thm:five}.

\bigskip
\noindent{\sc{Proof of Theorem~\ref{thm:five}.}}
By Lemma \ref{lem:f}, if $f\in n\dar(\real^n,\real)$, then $f$ is constant on
a set of cardinality continuum.  In particular, $f$ is not a
Sierpi\'{n}ski-Zygmund function.\qed

To get the upper bound of Theorem~\ref{thm:one}, we will
need the following combinatorial lemma which concerns the
cardinal
$\diff_\cuum$.

\noindent\lem{lem:h}{$\diff_\cuum=\kappa$ where
\begin{equation}\notag
\kappa=\min\{|F|\colon F\subseteq \cuum^{\cuum} \&\ (\forall G\in
[\cuum^\cuum]^\omega) (\exists f\in F)(\forall g\in G)(|[f=g]|<\cuum)\}.
\end{equation}}
\proof
We show that $\kappa\leq\diff_\cuum$.
Let $V=\cuum\times\omega$ and $W=\cuum^\omega$.
Take $F\subseteq \cuum^V$, witnessing the definition of $\diff_{\cuum}$, i.e.,
$|F|=\diff_{\cuum}$ and
\begin{equation}\label{eq:31}
(\forall g\in \cuum^V)(\exists f\in F)(|[f=g]|<\cuum).
\end{equation}
It is enough to construct a family $F^{*}\subseteq W^{\cuum}$ such that
$|F^*|\leq |F|$ and
\begin{equation}\label{eq:32}
(\forall G\in [W^\cuum]^\omega)(\exists f\in F^*)(\forall g\in
G)(|[f=g]|<\cuum).
\end{equation}
For every $f\in F$ let $f^*\in W^{\cuum}$ be defined so that
$f^{*}(\alpha)(n)=f(\alpha,n)$
for every $\alpha\in\cuum$ and $n\in\omega$.
Let $F^*=\{f^*\colon f\in F\}$ and note that $|F^*|\leq |F|$.  We show that
$F^*$
satisfies
(\ref{eq:32}).  Let $G\in [W^{\cuum}]^{\omega}$ and enumerate $G$ by
$G=\{g_n \colon n\in\omega\}$.  Define $g'\in \cuum^V$ so that
$g'(\alpha,n)=g_n(\alpha)(n)$ for $\alpha\in\cuum$ and $n\in\omega$.
By (\ref{eq:31}), there is an $f\in F$ such that $|[f=g'|<\cuum$.
We show that $|[f^*=g_n]|<\cuum$ for every $n\in\omega$.  By way of
contradiction,
assume that $|[f^*=g_n]|=\cuum$ for some $n\in\omega$.  Then,
$f^*(\alpha)(k)=g_n(\alpha)(k)$ for every $k\in\omega$ and $\alpha\in
[f^*=g_n]$.
In particular, we have
$f^*(\alpha)(n)=g_n(\alpha)(n)$ for each $\alpha\in [f^*=g_n]$.
So, $f(\alpha,n)=g'(\alpha,n)$ for every $\alpha\in [f^*=g_n]$.
Since $|[f^*=g_n]|=\cuum$ we have $|[f=g']|=\cuum$ , contradicting
our choice of $f$.
So $F^*$ satisfies (\ref{eq:32}) and
$\kappa\leq\diff_\cuum$.   The other inequality is trivial.\qed

We may now prove the remaining inequality for $m=1$.

\noindent\lem{lem:i}{If $n>1$, then
$\add(n\dar(\real^n,\real))\leq\diff_{\cuum}$.}
\proof
By Lemma \ref{lem:h} there is a family $F\subseteq (\real)^{\real^n}$ such that
$|F|=\diff_{\cuum}$ and
\begin{equation}\label{eq:33}
\left(\forall H\in
\left[(\real)^{\real^n}\right]^\omega\right)\left(\exists f\in
F\right)\left(\forall h\in H\right)\left(|[f=h]|<\cuum\right).
\end{equation}
It is enough to show that $F$ satisfies
\begin{equation}\label{eq:34}
\left(\forall g\in \left(\real\right)^{\real^n}\right)\left(\exists f\in
F\right)\left(f+g\notin n\dar\left(\real^n,\real\right)\right).
\end{equation}
So let $g\in (\real)^{\real^n}$ be arbitrary.  To find the appropriate
element of
$F$ we must first define some other functions.

Let $\{r_{\alpha}\}_{\alpha\in\cuum}$ be an enumeration of $\real$ and
$\{B_{\alpha}\}_{\alpha\in\cuum}$ be a partition of $\real^n$ into
Bernstein sets.  Recall that a Bernstein set has the property that both it and
its complement have nonempty intersection with every perfect set.  Notice
also, that $|C\cap B_{\alpha}|=\cuum$ for any non-degenerate continuum
$C\subseteq\real^n$.  Define
$k\colon\real^n\to\real$ so that
$k^{-1}(\{r_{\alpha}\})=B_{\alpha}$ for  each $\alpha\in\cuum$.  Also, for each
$q\in\rational$ let
$k_q\colon\real^n\to\real$ be such that $k_q[\real^n]=\{q\}$.

Now, let $H=\{k_q-g\colon q\in\rational\}\cup\{k-g\}$.  By (\ref{eq:33})
there is an
$f\in F$ such that $|[f=h]|<\cuum$ for all $h\in H$.  We now show that
$f+g\notin n\dar(\real^n,\real)$ using Lemma \ref{lem:f}.  We first claim
that $f+g$ is constant on
no non-degenerate continuum in $C\in\real^n$.  By way of contradiction,
assume there is a
non-degenerate continuum $C\in\real^n$ such that $(f+g)|_{C}$ is constant.
Then, there is
an $\alpha\in\cuum$ such that $(f+g)[C]=\{r_{\alpha}\}$.  Since $B_{\alpha}$
is a Bernstein set, $|C\cap B_{\alpha}|=\cuum$.  But
$f(x)=r_{\alpha}-g(x)=(k-g)(x)$ for each $x\in C\cap B_\alpha$, which
contradicts our
choice of $f$.  So the claim is established.  Next, we claim that
$|(f+g)^{-1}(\{q\})|<\cuum$  for each $q\in\rational$.  This follows from our
choosing $f$ so that
$|[f=k_q-g]|<\cuum$
for each $q\in\rational$.   Thus, $(f+g)$ is constant on no
non-degenerate continuum and $|(f+g)^{-1}(q)|<\cuum$ for every $q\in\rational$.
So, by  Lemma~\ref{lem:f}, $f+g\notin n\dar(\real^n,\real)$.  Therefore, $F$
satisfies (\ref{eq:34}), which completes the proof.\qed




Finally, we generalize Lemma \ref{lem:i} to include more general range spaces.

\noindent\lem{lem:j}{If $n,m \geq 1$, then
$\add(n\dar(\real^n,\real^m))\leq\diff_{\cuum}$.}
\proof
By Lemma \ref{lem:i} there is an $F\subseteq\real^{\real^n}$ such that
$|F|=\diff_{\cuum}$ and $F$ satisfies the definition of
$\add(n\dar(\real^n,\real))$, i.e.,
\begin{equation}\label{eq:35}
\left(\forall g\in (\real)^{\real^n}\right)(\exists f\in F)(f+g\notin
n\dar(\real^n,\real)).
\end{equation}
Let $\pi\colon\real^m\to\real$ be the projection function of $\real^m$ onto
some fixed
coordinate.  Since $\pi$ is onto, for every $f\in F$ there
is an
$f^*\colon\real^n\to\real^m$ such that $f=\pi\circ f^*$.  Let $F^*=\{f^*\colon
f\in F\}$ and note that $|F^*|\leq|F|=\diff_{\cuum}$.  We will be done if we
show that $F^*$  satisfies
\begin{equation}\label{eq:vill}
\left(\forall g\in (\real^m)^{\real^n}\right)(\exists f^{*}\in
F^{*})(f^{*}+g\notin n\dar(\real^n,\real^m)).
\end{equation}
So let $g\in (\real^m)^{\real^n}$ be arbitrary and put $g_1=\pi\circ g$.  By
(\ref{eq:35}), there is an $f\in F$ such that $(f+g_1)\notin
n\dar(\real^n,\real)$.   We claim that $(f^*+g)\notin n\dar(\real^n,\real^m)$,
which will complete the proof.   Assume that $(f^*+g)\in
n\dar(\real^n,\real^m)$. Then there  exist
$\{d_j\in\dar(\real^n,\real^m)\}_{j=1}^{n}$ such that
\begin{equation}\notag
f^*+g=d_1+\ldots+d_{n}.
\end{equation}
Since the projection onto a coordinate is additive and continuous,
we now have
\begin{eqnarray*}
f+g_1
& = & (\pi\circ f^*)+(\pi\circ g) \notag \\
& = & \pi\circ (f^*+g) \notag \\
& = &(\pi\circ d_1)+\ldots+(\pi\circ d_{n})\in n\dar(\real^n,\real), \notag
\end{eqnarray*}
which contradicts our choice of $f$.  Thus, $F^*$ satisfies
(\ref{eq:vill}).\qed

\section{Example \ref{ex:1}}\label{sec:ex1}
In this section, we construct the
Baire class 1 function of Example \ref{ex:1} and use Lemmas \ref{lem:e} and
\ref{lem:f} to show it has the required properties.  We denote the
linear span of a collection of  real numbers $A$ over $\rational$ by
$\lin_{\rational}(A)$.  Recall that if
$A$ is countable, then so is $\lin_{\rational}(A)$.
\noindent\lem{lem:b1}{Let $n\geq1$.  There is a Baire class 1 function
$f\colon\real^n\to\real$
such that $|f[\real^n]|=\omega$ and $f[C]$ is an interval for no non-trivial
connected subset $C$ of
$\real^n$.}
\proof
Let $A=\{\alpha_{i}\colon i\in n\}$ be a collection of distinct real
numbers which are
linearly independent over the rationals.  Let $\{q_{k}\}_{k=1}^{\infty}$ be
an enumeration of
$\rational$.  For each $i\in n$, define $g_{i}\colon\real\to\real$ by
\begin{equation*}
g_i(x)=
\begin{cases}
\alpha_{i}/k& \text{if $x=q_{k}$}\\
0& \text{if $x\notin\rational$.}
\end{cases}
\end{equation*}
Notice that each $g_i$ is in Baire class 1.  Define $f\colon\real^n\to\real$ by
\[f(\langle x_1,\ldots ,x_n\rangle)=\sum_{i=1}^n g_{i}(x_i).\]
It is easy to check that $f$ is in Baire class 1.  We show that $f$ has the
desired
properties.  We first notice that $f[\real^n]\subseteq\lin_{\rational}(A)$ so
$|f[\real^n]|=\omega$. We now check the other property.
Let $C\subseteq\real^n$ be a non-trivial connected set.  Since $C$ is
non-trivial,
there is some $1\leq i\leq n$ such that the projection $\pi_i[C]$ of $C$
onto the
$i^{\text{th}}$ coordinate is an interval with non-empty
interior.  Let
$t\in \pi_{i}[C]\cap\rational$, $s\in\pi_{i}[C]\setminus\rational$, and
$p,q\in C$  be such that $\pi_i(q)=t$ and $\pi_i(p)=s$.  By definition of $f$ we
have
$f(q)\notin\lin_{\rational}(A\setminus\{\alpha_i\})$ and
$f(p)\in\lin_{\rational}(A\setminus\{\alpha_i\})$.  In particular, $f(q)\neq
f(p)$ so $f|_C$ is not constant.  Since $f|_C$ is not constant and
$|f[\real^n]|=\omega$, it is clear that
$f[C]$ is not an interval.\qed

It is immediate by Lemma \ref{lem:e} that $f\notin n\dar(\real^n,\real)$ for all
$n\geq 1$.  We now show that $f$ is not the sum of an almost continuous
function, $h$, and $n-1$ Darboux functions when $n>1$.  By Lemma
\ref{lem:e}, for
any
$g\in (n-1)\dar(\real^n,\real)$ there is a non-degenerate continuum $C$
upon which
$g$ is constant.  Thus, $(h+g)|_{C}$ would be
almost continuous (Restrictions of almost continuous functions to closed sets
are almost continuous \cite{st}.); in which case $(h+g)[C]$ is an interval
\cite[Theorem~1.7]{nat}.  Therefore, $h+g\neq f$.


\section{Proof of Theorem~\ref{thm:four}}
We first show that the containment $\acon(\real^n,\real^m)\subseteq
n\dar(\real^n,\real^m)$ holds.
\noindent\lem{lem:hap}{$\acon(\real^n,\real^m)\subseteq
n\dar(\real^n,\real^m)$.}
\proof
The lemma is known for the case $n=1$; see \cite{BHL}.  We proceed by
induction on $n$.  Assume $n-1\geq 1$ and
$\acon(\real^{n-1},\real^m)\subseteq(n-1)\dar(\real^{n-1},\real^m)$.
Let $g\in\acon(\real^n,\real^m)$.  We show $g\in n\dar(\real^n,\real^m)$.

Let $\{A_k\}_{k\in n}$ be a partition of $\real$ into $n$ $\cuum$-dense sets.
Define
$h\colon\real\to\real^m$ so that for each $p\in\real^m$ and $k\in n$
\begin{equation*}
\{y\in A_k\colon h(y)=p\} \text{ is dense in $\real$}.
\end{equation*}
Let $\pi\colon\real^n\to\real$ denote the projection of $\real^n$
onto a fixed coordinate. Since restrictions of almost
continuous functions to closed sets are almost continuous
\cite{st},
$g|_{\pi^{-1}(y)}$ is almost continuous for every $y\in\real$.  Note that
$\pi^{-1}(r)$ is homeomorphic to $\real^{n-1}$ for every $r\in\real$.
By inductive hypothesis, for each $l\in n$ and $y\in A_l$ we may find Darboux
functions
$\{g^{y}_k\colon\pi^{-1}(y)\to\real^m\colon k\in n\setminus\{l\}\}$ such that
\begin{equation*}
\sum_{\{k\in n\colon k\neq l\}}  g^y_k=g|_{\pi^{-1}(y)}-h(y).
\end{equation*}
Now for each $k\in n$, define $g_{k}\colon\real^n\to\real^m$ by
\begin{equation}\label{eq:39}
g_k(p)=
\begin{cases}
h(y)& \text{ if $\pi(p)=y\in A_{k}$, }\\
g^y_k(p)& \text{ if $\pi(p)=y\notin A_{k}$.}
\end{cases}
\end{equation}
We claim that the functions of (\ref{eq:39}) are as desired.  Let $k\in
n$.  To see that $g_k$ is Darboux put $A=A_k$ and
$B=\cup\{A_i\colon i\in n \text{ and } i\neq k\}$,
then apply Lemma~\ref{lem:a}.  We now show that
\begin{equation}\label{eq:40}
\sum_{k=0}^{n-1} g_k=g.
\end{equation}
If $p\in\real^n$, then there is an $l\in n$ such that $y\in A_l$,
$p\in\pi^{-1}(y)$, and
\begin{eqnarray*}
\sum_{k=0}^{n-1} g_k(p)
& = & h(y)+\sum_{\{k\in n\colon k\neq l\}}g_{k}^y(p) \notag \\
& = & h(y)+(g(p)-h(y)) \notag \\
& = & g(p). \notag
\end{eqnarray*}
So (\ref{eq:40}) holds, completing the inductive step.
\qed

We now show that the containment of Lemma \ref{lem:hap} is proper. In fact, we
show more.  We note that the Lemma below has been shown for the case $n=2$ in
\cite[Example 1.6]{nat} and for $n=1$ in \cite{BHL}.
\noindent\lem{lem:hap1}{
$\dar(\real^n,\real^m)\setminus\acon(\real^n,\real^m)\neq\emptyset$.}
\proof
Since $\real$
can be embedded in $\real^m$ for any $m\geq 1$, it is enough to show
that
\begin{equation}\label{eq:rrr}
\dar(\real^n,\real)\setminus\acon(\real^n,\real)\neq\emptyset.
\end{equation}
Take $f\in\dar(\real,\real)\setminus\acon(\real,\real)$, $n>1$ and let
$\pi\colon\real^n\to\real$ be the projection of $\real^n$ onto the first
coordinate, i.e., $\pi\langle r_0,\ldots ,r_{n-1}
\rangle=r_0$.  Define
$g\colon\real^n\to\real$ by $g=f\circ\pi$.  Since
$f$ is Darboux and $\pi$ is continuous, $g\in\dar(\real^n,\real)$.  We show
that $g\notin\acon(\real^n,\real)$.  Consider the set
$S=\{p\in\real^n\colon p=\langle r,0,\ldots,0\rangle\ \&\, r\in\real\}$.  Since
$g|_S$ is an exact copy of $f$, it follows that
$g|_S\notin\acon(S,\real)$.  However, restrictions of almost continuous
functions to closed sets are almost continuous \cite{st}, so we must conclude
that
$g$ is not almost continuous.\qed

We now prove the last part of Theorem~\ref{thm:four}.  We
will need the following fact which may be found in
\cite{nat}.

\noindent\prop{prop:a1}{Let $n,m\in\omega\setminus\{0\}$.  There exists a
family $\mathcal{B}$ of closed sets in
$\real^n\times\real^m$, a blocking family, with the following properties:
\begin{description}
\item[(i)] $f\in\acon(\real^n,\real^m)$ if and only if $f\cap
B\neq\emptyset$ for each
$B\in\mathcal{B}$ and
\item[(ii)] for every $B\in\mathcal{B}$ the projection of $B$ onto
$\real^n$ is a non-degenerate connected set.
\end{description}}

\noindent\lem{lem:jaba}{If $n\geq 2$ and $m \geq 1$, then there is an
almost continuous
function $f$ such that $f\notin (n-1)\dar(\real^n,\real^m)$.}
\proof
Let $\pi\colon\real^m\to\real$ be the projection function of $\real^m$ onto
some fixed
coordinate.  Since $\pi$ is additive and continuous, if
$f\in (n-1)\dar(\real^n,\real^m)$, then $(\pi\circ f)\in
(n-1)\dar(\real^n,\real)$.   It then follows from Lemma \ref{lem:e} that
$(\pi\circ f)$ is  constant on some non-degenerate continuum in $\real^n$.  So,
if we show there is a $g\in\acon(\real^n,\real^m)$ such that $g_1=\pi\circ g$
is constant on no non-degenerate continuum in $\real^n$, we will be done.  We
now construct $g$.  Let
$\{P_{\langle\alpha,i\rangle}\}_{\langle\alpha,i\rangle\in\cuum\times 2}$ be a
partition of $\real^n$ into Bernstein sets.  Let
$\{B_{\alpha}\}_{\alpha\in\cuum}$ be an enumeration of the elements of the
blocking family of
Proposition~\ref{prop:a1}, and for each $\alpha\in\cuum$ let
$B_{\alpha}^*$ denote the
projection of $B_{\alpha}$ onto $\real^n$.  Take
an enumeration $\{r_{\alpha}\}_{\alpha\in\cuum}$ of $\real^n$.  By
Proposition~\ref{prop:a1},
$|B_{\alpha}^*|=\cuum$ and $B^*_{\alpha}$ is an $F_{\sigma}$-set. It
follows that
$|P_{\langle\alpha,i\rangle}\cap B_{\alpha}^*|=\cuum$ for every
$\langle\alpha,i\rangle\in\cuum\times 2$.
For each $\alpha\in\cuum$, define $h^{\alpha}\colon\real^n\to\real^m$ so that
$h^{\alpha}|_{P_{\langle\alpha,0\rangle}\cap B^*_{\alpha}}\subseteq
B_{\alpha}$, $h^{\alpha}[P_{\langle\alpha,1\rangle}\cap
B^*_{\alpha}]=\{r_{\alpha}\}$, and let
$h^{\alpha}$ be arbitrary elsewhere.  Let $g\colon\real^n\to\real^m$ be defined
by
\begin{equation*}
g(x)=
\begin{cases}
h^{\alpha}(x)& \text{if $x\in B^*_{\alpha}\cap(P_{\alpha,0}\cup
P_{\alpha,1})$}\\ 0& \text{otherwise.}
\end{cases}
\end{equation*}
We claim $g$ is as desired.   Since
$g|_{P_{\alpha,0}\cap B^*_{\alpha}}=h^{\alpha}|_{P_{\alpha,0}\cap
B^*_{\alpha}}\subseteq B_{\alpha}$ for every $\alpha\in\cuum$,
$g\in\acon(\real^n,\real^m)$.  We now show that
$g_1=\pi\circ g$ is constant on no continuum.  Since Bernstein sets intersect
all perfect sets, they intersect all non-degenerate continua.  It follows from
the way we defined $g$ that if $C\subseteq\real^n$ is a non-degenerate
continuum,
then $(\pi\circ g)[C]=\pi[\real^m]=\real$.  Thus, $g_1$ is constant on no
non-degenerate subcontinuum of $\real^n$.\qed


\section{Proof of Theorem~\ref{thm:onea}}
By Theorem~\ref{thm:four} $\acon(\real^n,\real^m)\subseteq
n\dar(\real^n,\real^m)$.  By Proposition~\ref{prop:one}(iii), we have, using
Theorem~\ref{thm:one},
$\add(\acon(\real^n,\real^m))\leq\add(n\dar(\real^n,\real^m))=\diff_{\cuum}$.
So we only need to prove that $\add(\acon(\real^n,\real^m))\geq\diff_{\cuum}$.
To see this,
let $F\subseteq (\real^m)^{\real^n}$ and $|F|<\diff_{\cuum}$.
We must find a $g\colon\real^n\to\real^m$ such that
$g+F\subseteq\acon(\real^n,\real^m)$.  Let
$\{P_{\alpha}\}_{\alpha\in\cuum}$ be a partition of $\real^n$ into
Bernstein sets,
$\{B_{\alpha}\}_{\alpha\in\cuum}$ be an enumeration of the elements
of the blocking family of
Proposition~\ref{prop:a1}; and for each $\alpha\in\cuum$ let $B_{\alpha}^*$
denote the
projection of $B_{\alpha}$ onto $\real^n$.  By Proposition~\ref{prop:a1}
$|B_{\alpha}^*|=\cuum$ and $B_{\alpha}^*$ is an $F_{\sigma}$-set.
It follows that $|P_{\alpha}\cap B_{\alpha}^*|=\cuum$ for each
$\alpha\in\cuum$.  For each $\alpha\in\cuum$, define
$h^{\alpha}\colon\real^n\to\real^m$ so that $h^{\alpha}|_{P_{\alpha}\cap
B^*_{\alpha}}\subseteq B_{\alpha}$ and let $h^{\alpha}$ be arbitrary elsewhere.
Put
\[F^*=\{h^{\alpha}-f\colon f\in F \text{ and } \alpha\in\cuum\}.\]  Since
$\cuum<\diff_{\cuum}$, it follows that $|F^*|<\diff_{\cuum}$.  So, for
every $\alpha\in\cuum$ there is a function
$g_{\alpha}\colon (P_{\alpha}\cap B^*_{\alpha})\to\real^m$ such that
$|\{x\in P_{\alpha}\cap B^*_{\alpha}\colon
g_{\alpha}(x)=(h^{\alpha}-f)(x)\}|=\cuum$.  Let $g\colon\real^n\to\real^m$
be defined by
\begin{equation*}
g(x)=
\begin{cases}
g_{\alpha}(x)& \text{if $x\in P_{\alpha}\cap B^*_{\alpha}$}\\
0& \text{otherwise.}
\end{cases}
\end{equation*}
We claim $g$ is as desired.  Let $f\in F$ and $B\in\mathcal{B}$.  There is an
$\alpha\in\cuum$ such that $B_{\alpha}=B$.  By the way we defined $g$,
we have
\begin{eqnarray*}
\left|(f+g)|_{P_{\alpha}\cap
B^*_{\alpha}}=h^{\alpha}|_{P_{\alpha}\cap B^*_{\alpha}}\right|
& = &\left|f|_{P_{\alpha}\cap
B^*_{\alpha}}+g_{\alpha}=h^{\alpha}|_{P_{\alpha}\cap B^*_{\alpha}}\right|\\  & =
& \left|g_{\alpha}=(h^{\alpha}-f)|_{P_{\alpha}\cap B^*_{\alpha}}\right|\\  & = &
\cuum.
\end{eqnarray*}  Thus,
$|(g+f)|_{P_{\alpha}\cap B^*_{\alpha}}=h^{\alpha}|=\cuum$.  In particular,
$(f+g)\cap B_{\alpha}\neq\emptyset$.  Since $B$ was arbitrary, we conclude that
$f+g\in\acon(\real^n,\real^m)$.\qed



\section{Proof of Theorem~\ref{thm:two} }
Our first goal is to show that $\add(n\ext(\real^n,\real))\leq\cuum^+$
for $n>1$.   The following proposition can be found in \cite[Lemma 3.1]{CR}
where it is stated for
$\real^{\real}$.  The proof is essentially the same for $(\real)^{\real^n}$.

\noindent\prop{prop:9}{There is a family $F\subseteq (\real)^{\real^n}$ of
cardinality $\cuum^+$ such that for  every distinct $f,h\in F$, every
perfect set
$P\subseteq\real^n$, and every
$k<\omega$ there exists an $x\in P$ such that $|f(x)-h(x)|\geq k$.}

The next proposition follows immediately from \cite[Proposition~2.8]{CW} and
the fact that $\conn(\real^n,\real)=\ext(\real^n,\real)$ for $n>1$.

\noindent\prop{prop:10}{If $n>1$ and $g\in n\ext(\real^n,\real)$, then there
exists a perfect set
$P$ such that $g|_{P}$ is continuous.}

\noindent\lem{lem:m}{If $n>1$, then
$\add(n\ext(\real^n,\real))\leq\cuum^+$.}
\proof
Let
$F\subseteq (\real)^{\real^n}$ be as in Proposition~\ref{prop:9}.  We claim
that $g+F\subseteq
n\ext(\real^n,\real)$ for no $g\colon\real^n\to\real$.   By way of
contradiction, assume that such a
$g$ exists.  By Proposition~\ref{prop:10}, for every $f\in F$ there is a
perfect set $P_{f}$ such that the restriction of $g+f$  to $P_{f}$ is
continuous.  Since
$|F|=\cuum^+$ and there are only
$\cuum$-many perfect sets, there are $f,h\in F$ such that $P_{f}=P_{h}$.  It
follows that $f-h=(g+f)-(g+h)$  is continuous on $P_{f}$, which contradicts our
choice of $F$.\qed

Since $\add_{n,n-1}(\ext(\real^n,\real))\leq\add(n\conn(\real^n,\real))$ by
Proposition~\ref{prop:two} (viii), the proof of Theorem~\ref{thm:two} will be
complete if we show that $\cuum^+\leq\add_{n,n-1}(\ext(\real^n,\real))$.

For what follows we will use the notation of \cite[Sec. 6]{CW}.
For sets $\{ A_i\colon i\in n\}$ and $\{B_i\colon i\in n\}$
and for $f\colon n\to 2$ we let
\begin{equation}\notag
A_i\vee_f B_i=
\begin{cases}
A_i& \text{if } f(i)=0 \\
B_i& \text{if } f(i)=1.
\end{cases}
\end{equation}
If $j\in n$ and $C$ is a set, we define
\begin{equation}\notag
A_i\vee_f B_i\vee_j C=
\begin{cases}
C& \text{if $i=j$}\\
A_i& \text{if $i\neq j$ and $f(i)=0$} \\
B_i& \text{if $i\neq j$ and  $f(i)=1$.}
\end{cases}
\end{equation}
We call $M\subseteq\real$ a thick meager set provided that $M$ is dense and
is a
countable  union of nowhere dense perfect sets.  Notice that any thick meager
set is
$\cuum$-dense in $\real$.
We will also need the following four propositions from \cite{CW}.

\noindent\prop{prop:29}{{\rm(\cite[Lemma 4.1]{CW})}If $G$ is a dense
$G_{\delta}$-set in
$\real^n$, then for each $i\in n$ there is a countable dense set
$B_i\subseteq\real$ and a thick meager set $Y_i\subseteq\real$ such that
$B_i\cap Y_i=\emptyset$ and
\[\prod_{i=0}^{n-1}(B_i\cup Y_i)\subseteq G.\]}

\noindent\prop{prop:30}{{\rm(\cite[Proposition 2.3]{CW})}Let $n\geq 1$.
There is a dense
$G_{\delta}$-set
$G\subseteq\real^n$ and  a function $f\colon\real^n\to\real$ such that for any
$g\colon\real^n\to\real$ if
$g(x)=f(x)$ for every $x\notin G$, then $g\in\conn(\real^n,\real)$.}
Since $\conn(\real^n,\real)=\ext(\real^n,\real)$ for $n>1$,
we have an obvious corollary of Proposition~\ref{prop:30}.

\noindent\cor{cor:49}{Let $n\geq 1$.  There is a dense $G_{\delta}$-set
$G\subseteq\real^n$ and  a function $f\colon\real^n\to\real$ such that for any
$g\colon\real^n\to\real$ if
$g(x)=f(x)$ for every $x\notin G$, then $g\in\ext(\real^n,\real)$.}
\proof
When $n>1$ the proposition follows from Proposition~\ref{prop:30} and the
fact that
$\conn(\real^n,\real)=\ext(\real^n,\real)$.  When $n=1$ then the
proposition reduces to \cite[Corollary 3.4]{CR}.\qed

\noindent\prop{prop:32}{{\rm(\cite[Lemma 4.3]{CW})}
Let $n>0$ and $G\subseteq\real^n$ be a $G_{\delta}$-set.  If
$f\colon n\to 2$ is a function, $i\in n$, and
$\langle b_0,\ldots,b_{n-1}\rangle\in\real^n$, then the set
\[
\left\{x\in\real\colon\prod_{j=0}^{n-1}
(\{b_j\}\vee_f \real\vee_i \{x\})
\subseteq  G\right\}
\]
is a $G_{\delta}$-subset of $\real$.
}

The next proposition is a more detailed statement of
\cite[Proposition~2.4]{CW}.
\noindent\prop{prop:33}{Let $G$ be a dense $G_{\delta}$-subset of
$\real^n$.  Then there exist countable dense sets
$\{B_i\}_{i=0}^{n-1}$ of $\real$ and homeomorphisms
$h_1,\ldots,h_n\colon\real^n\to\real^n$ such that
\begin{equation*}
\prod_{i=0}^{n-1}(B_i\vee_{f}\real)\subseteq G\cup\left(\bigcup_{i=1}^{k}
h_i(G)\right)
\end{equation*}
for each $f\in 2^n$ with $|f^{-1}(1)|=k$.  In particular, if $k=n$ then,
\begin{equation*}
\real^n=G\cup\left(\bigcup_{i=1}^{n} h_i(G)\right).
\end{equation*}}
\proof
The statement of the proposition follows directly from consideration of the
inductive step of the proof of Proposition~2.4 in \cite{CW}.  \qed

\noindent\lem{lem:51}{Let $n\geq 1$ and $G\subseteq\real^n$ be a dense
$G_{\delta}$.
There exist homeomorphisms $h_1,\ldots, h_{n-1}\colon\real^n\to\real^n$ and
meager subsets $\{M_i\}_{i=0}^{n-1}$ of $\real$ such that
\begin{equation}\label{eq:a}
\real^n\setminus \left(G\cup \left(\bigcup_{j=1}^{n-1}
h_j[G]\right)\right)\subseteq\prod_{j=0}^{n-1}M_j.
\end{equation} }
\proof
Let $\{B_i\}_{i=0}^{n-1}$ be the countable dense subsets of $\real$ from
Proposition~\ref{prop:33} and
$h_{1},\ldots ,h_{n-1}\colon\real^n\to\real^n$ be the first $n-1$
homeomorphisms of Proposition~\ref{prop:33}.  Then
\begin{equation}\label{eq:b}
\prod_{j=0}^{n-1}(B_j\vee_{f}\real)\subseteq
G\cup\left(\bigcup_{j=1}^{n-1} h_j(G)\right)
\end{equation}
for each $f\in 2^n$ such that $|f^{-1}(1)|=n-1$.  Let $i\in n$ and
$f_i\colon n\to 2$ be such that $f_i^{-1}(0)=\{i\}$.
By Proposition~\ref{prop:32}, for each $b=\langle b_0,\ldots,
b_{n-1}\rangle\in B_0\times\cdots\times B_{n-1}$ there is a $G_{\delta}$-subset
$K_i^b$ of $\real$ such that
\[\prod_{j=0}^{n-1}(\{b_j\}\vee_{f_i}\real\vee_i K_i^b)\subseteq
G\cup\left(\bigcup_{j=1}^{n-1}h_j[G]\right).\]   By (\ref{eq:b}), we know
that
$B_i\subseteq K_i^b$; so, $K_i^b$ is a dense
$G_{\delta}$-subset of $\real$.  So the set
\[K_i=\bigcap\{K_i^b\colon b\in B_0\times\cdots\times B_{n-1}\}\]
is a dense $G_{\delta}$-subset of $\real$, satisfying
\[\prod_{j=0}^{n-1}(B_j\vee_{f_i}\real\vee_i K_i)\subseteq
G\cup\left(\bigcup_{j=1}^{n-1}h_j[G]\right).\]   Since
$f_{i}[n\setminus\{i\}]=\{1\}$, we have
\[\prod_{j=0}^{n-1}(K_j\vee_{f_i}\real)=\prod_{j=0}^{n-1}(B_j\vee_{f_i}\real
\vee_i
K_i)\subseteq G\cup\left(\bigcup_{j=1}^{n-1}h_j[G]\right).\]   Letting
$M_i=\real\setminus K_i$ for each $i\in n$, we have
\begin{equation*}
\real^n\setminus \left(G\cup \left(\bigcup_{j=1}^{n-1}h_j[G]\right)\right)
\subseteq\real^n\setminus\bigcup_{i=0}^{n-1}\left(\prod_{j=0}^{n-1}(K_j\vee_
{f_i}\real)\right)
=\prod_{i=0}^{n-1}M_i.
\end{equation*}
Thus, (\ref{eq:a}) holds. \qed

\noindent\lem{lem:53}{Let $\{M_i\}_{i=0}^{n-1}$ be meager subsets of
$\real$ and
$M=\prod_{i=0}^{n-1}M_i\subseteq\real^n$.  For any
dense $G_{\delta}$-set $G\subseteq\real^n$ there exist homeomorphisms
$\{h_{\xi}\colon\real^n\to\real^n\}_{\xi\in\cuum}$ such that
\begin{description}
\item[(i)] $h_{\xi}[M]\cap h_{\zeta}[M]=\emptyset$ for all $\zeta<\xi<\cuum$
and
\item[(ii)] $h_{\xi}[M]\subseteq G$ for all $\xi<\cuum$.
\end{description}}
\proof
By Proposition~\ref{prop:29} there exist thick meager subsets
$\{N_i\}_{i=0}^{n-1}$ of $\real$ such that
\begin{equation}\label{eq:68}
\prod_{i=0}^{n-1} N_i\subseteq G.
\end{equation}
Let $\{N_{i,\xi}\}_{\xi\in\cuum}$ be a partition of
$N_i$ into thick meager sets.  By \cite[Lemma 3.2]{WG}, for each $i\in n$ and
$\xi\in\cuum$ there exists a
homeomorphism $h_{i,\xi}\colon\real\to\real$ such that
\begin{equation}\label{eq:69}
h_{i,\xi}[M_i]\subseteq N_{i,\xi}.
\end{equation}
For each $\xi\in\cuum$ let
$h_{\xi}=h_{0,\xi}\times\cdots\times h_{n-1,\xi}\colon\real^n\to\real^n$.  The
homeomorphisms $\{h_{\xi}\}_{\xi\in\cuum}$ satisfy (i) and (ii).   Indeed,
$\{N_{i,\xi}\}_{\xi\in\cuum}$ partitions $N_i$ for each $i\in n$ so (i) follows
from (\ref{eq:69}).  Using (\ref{eq:68}) and (\ref{eq:69}), we
conclude (ii).\qed


\noindent\lem{lem:54}{Let $n\geq 1$.  There exist a meager subset $M$ of
$\real^n$,
meager subsets $\{M_i\colon i\in n\}$ of $\real$, and a function
$f\colon\real^n\to\real$ such that
\[M=\prod_{i=0}^{n-1} M_i\]
and for any $g\colon\real^n\to\real$, if $g|_M=f|_M$, then
$g\in n\ext(\real^n,\real)$.}
\proof
By Corollary~\ref{cor:49} there is an extendable function
$l\colon\real^n\to\real$  and a dense $G_{\delta}$-set $G\subseteq\real^n$
such that for any function $k\colon\real^n\to\real$ if
$k(x)=l(x)$ for every $x\notin G$, then $k\in\ext(\real^n,\real)$.
Let $h_{0}\colon\real^n\to\real^n$
be the identity homeomorphism. Pick $h_1,\ldots,
h_{n-1}\colon\real^n\to\real^n$
and $\{M_i\}_{i=0}^{k-1}$, as in Lemma~\ref{lem:51} for $G$.  Put
\[M=\prod_{i=0}^{n-1} M_i\]
and let $f\colon\real^n\to\real$ be defined by
\begin{equation*}
f(x)=
\begin{cases}
\sum_{i=0}^{n-1} (l\circ h_{i}^{-1})(x)& \text{ if $x\in M$}\\
0& \text{ otherwise.}
\end{cases}
\end{equation*}
We show $f$ is as desired.  Let $g\colon\real^n\to\real$ be such that
$g|_M=f|_M$.
We show $g\in n\ext(\real^n,\real)$.  Let $G_{0}=G$, and for each $0<i\leq
n$ let
\[G_i=h_i[G]\setminus\left( \bigcup_{j=0}^{i-1} h_j[G]\right).\]
Note that $\{G_i\colon i\in n\}$ is a collection of pairwise disjoint sets
such that \[\real^n\setminus M\subseteq \bigcup_{i=0}^{n-1} G_i.\]
For each $i\in n$ we define $g_i\colon\real^n\to\real$ by
\begin{equation*}
g_i(x)=
\begin{cases}
g(x)-\sum_{\{j\in n\colon j\neq i\}}(l\circ h^{-1}_j)(x)& \text{ if $x\in
G_i$}\\
(l\circ h_{i}^{-1})(x)& \text{ if $x\notin G_i$.}
\end{cases}
\end{equation*}
Now $g=g_0+\cdots +g_{n-1}$.  We show that each $g_i$ is in
$\ext(\real^n,\real)$.  Fix an $i\in n$.  If
$x\in\real^n\setminus G$, we have
$g_i(h_i(x))=(l\circ h_{i}^{-1})(h_{i}(x))=l(x)$ since $h_i(x)\notin G_i$.
Thus,
$(g_i\circ h_i)\in\ext(\real^n,\real)$ by our choice of $l$.   Since the
composition of a extendable function and a homeomorphism is extendable
\cite[Lemma 1]{what}, it then follows that
$g_i=((g_i\circ h_i)\circ h^{-1}_i)\in\ext(\real^n,\real)$.\qed


\noindent\lem{lem:55}{$\add_{n,n-1}(\ext(\real^n,\real))\geq\cuum^+$.}
\proof
Let $\{g_{\xi}\}_{\xi\in\cuum}$ be a collection of functions from $\real^n$
into
$\real$.  Suppose $M$ and $f\colon\real^n\to\real$ are as in
Lemma \ref{lem:54}.  Take
an extendable function $f^*\colon\real^n\to\real$ and
a dense $G_{\delta}$-set $G\subseteq\real^n$, as in Corollary~\ref{cor:49}.
By Lemma \ref{lem:53} there exist homeomorphisms
$\{h_{\xi}\colon\real^n\to\real^n\}_{\xi\in\cuum}$ such that
\begin{description}
\item[(i)] $h_{\xi}[M]\cap h_{\zeta}[M]=\emptyset$ for all $\zeta<\xi<\cuum$.
\item[(ii)] $h_{\xi}[M]\subseteq G$ for all $\xi\in\cuum$.
\end{description}
Define $l\colon\real^n\to\real$ so that
\begin{equation*}
l(x)=
\begin{cases}
g_{\xi}(x)-(f\circ h^{-1}_{\xi})(x)& \text{if $x\in h_{\xi}[M]$ for some
$\xi\in\cuum$}\\ f^*(x)& \text{ otherwise.}
\end{cases}
\end{equation*}
Note that $l\in\ext(\real^n,\real)$ by our choice of $f^*$ and $G$.  We
will be done
if we show that $g_{\xi}-l\in n\ext(\real^n,\real)$ for each $\xi\in\cuum$.
So fix
$\xi\in\cuum$.  If $x\in M$, then
$(g_{\xi}-l)(h_{\xi}(x))=g_{\xi}(h_{\xi}(x))-(g_{\xi}(h_{\xi}(x))-(f\circ
h^{-1}_{\xi})(h_{\xi}(x)))=(f\circ h^{-1}_{\xi})(h_{\xi}(x))=f(x)$.
So for each $x\in M$ we have
$(g_{\xi}-l)(h_{\xi}(x))=f(x)$.  By our choice  of $M$ and $f$ it follows
that there exist
$f_{0,\xi},\ldots,f_{n-1,\xi}\in\ext(\real^n,\real)$ such that
\begin{equation*}
(g_{\xi}-l)\circ h_{\xi}=\sum_{i=0}^{n-1}f_{i,\xi}.
\end{equation*}
So,
\begin{equation*}
(g_{\xi}-l)=\left(\sum_{i=0}^{n-1}f_{i,\xi}\right)\circ
h^{-1}_{\xi}=\sum_{i=0}^{n-1}(f_{i,\xi}\circ h^{-1}_{\xi}).
\end{equation*}
Since the composition of an extendable function and a homeomorphism
is extendable, $(f_{i,\xi}\circ h^{-1}_{\xi})\in\ext(\real^n,\real)$ for each
$i\in n$.  Thus, $(g_{\xi}-l)\in n\ext(\real^n,\real)$. The proof is
complete.\qed

\section{Proof of Theorem~\ref{thm:three}}
In this section we will use the notation of the section above.
Additionally for $k\leq n$, we make the definition $F_{k}^n=\{f\in
2^n\colon |f^{-1}(0)|=k\}$.
We first show that
$\add_{n,\floor {n/2}}(\dar(\real^n,\real^m))\geq\cuum^+$.

\noindent\lem{lem:n}{Let $n\geq k\geq 1$, $\{M_i\subseteq\real\colon i\in
n\}$ be a
collection  of thick meager sets and
\[M=\bigcup_{f\in F_{k}^n}\prod_{i=0}^{n-1}(M_i\vee_f \real).\]
Then there is a function $f\colon\real^n\to\real^m$ such that for any
$g\colon\real^n\to\real^m$ if $g|_M=f|_M$, then $g\in k\dar(\real^n,\real^m)$.}
\proof
Throughout this proof we assume that $\real^n$ is written in the form
$\real^n=\{\langle r_0,\ldots, r_{n-1}\rangle\colon r_i\in\real\text{ for each
}0\leq i\leq n-1\}$.  We do an induction on $n$.

Suppose n=k=1.  In this case $\{M_i\colon i\in 1\}=\{M_0\}$ and $M=M_0$.  Since
$M$ is a thick meager set, $|(a,b)\cap M|=\cuum$ for all $a<b\in\real$.  So,
using transfinite induction, we may easily construct a function $f$ which
maps each $M\cap (a,b)$ onto $\real^m$ for all $a<b\in\real$.  Clearly,
$f$ has the desired property.

So, let
$n\geq k\geq 1$ and assume the lemma holds for $n-1\geq 1$.  We now show that
the lemma holds for $n$.
Let $\pi\colon\real^n\to\real$ be the projection of
$\real^n$ onto the last coordinate; and note that
$\pi^{-1}(r)$  is homeomorphic to $\real^{n-1}$ for each
$r\in\real$.  For every $r\in\real$, put
$M^r=M\cap \pi^{-1}(r)$.  Let $\{B_i\}_{i=0}^{k-1}$ be a partition of $M_{n-1}$
into $\cuum$-dense sets and $d\colon\real\to\real$ be a function such
that for each $y\in\real^m$ and $i\in k$
\begin{equation*}
\{x\in B_i\colon d(x)=y\} \text{ is dense in $\real$}.
\end{equation*}

Notice that for each $r\in M_{n-1}$ we have
\begin{equation}\label{eq:bee}
M^r=\left(\bigcup_{f\in F_{k-1}^{n-1}}\prod_{i=0}^{n-2}(M_i\vee_f
\real)\right)\times\{r\}.
\end{equation}
So if
$k>1$, we may use the inductive hypothesis to find for each $r\in M_{n-1}$ a
function
$f^r\colon \pi^{-1}(r)\to\real^m$ such that for every $g\colon
\pi^{-1}(r)\to\real^m$ if $g|_{M^r}=f^r|_{M^r}$, then $g\in
(k-1)\dar(\pi^{-1}(r),\real^m)$.  When $k=1$, then $M^r=\pi^{-1}(r)$,
and we define
$f^r\colon\pi^{-1}(r)\to\real^m$ so that $f^r[\pi^{-1}(r)]=\{d(r)\}$.

When $r\in\real\setminus M_{n-1}$, we must again consider two cases.
If $k<n$, then
\begin{equation}\label{eq:bee1}
M^r=\left(\bigcup_{f\in F_{k}^{n-1}}\prod_{i=0}^{n-1}(M_i\vee_f
\real)\right)\times\{r\}.
\end{equation}
So, by the inductive hypothesis, there is a function
$f^r\colon \pi^{-1}(r)\to\real^m$ such that for every $g\colon
\pi^{-1}(r)\to\real^m$ if $g|_{M_r}=f^r|_{M_r}$, then $g\in
k\dar(\pi^{-1}(r),\real^m)$.  In the case when $n=k$, we have $M_r=\emptyset$.
It follows from Corollary~\ref{cor:9} that $g\in
k\dar(\pi^{-1}(r),\real^m)$ for every
$g\colon\pi^{-1}(r)\to\real^m$. So, in this case, we may define
$f^r\colon\pi^{-1}(r)\to\real^m$
as we please.

Let $f=\bigcup_{r\in\real}f^{r}\colon\real^n\to\real^m$.
We show $f$ is as desired.  Let $g\colon\real^n\to\real^m$ be a function
such that $g|_M=f|_M$.  We must show that $g\in k\dar(\real^n,\real^m)$.  For
each $r\in\real$, let $g^r=g|_{\pi^{-1}(r)}$.   For $i\in k$ and $r\in
B_i\subseteq M_{n-1}$, there exist
$\{g^r_{j}\in\dar(\pi^{-1}(r),\real^m)\colon j\in k\setminus\{i\}\}$  such that
\begin{equation}\notag
\sum_{j\in k\setminus\{i\}} g^r_{j}=g^r
\end{equation}
since $g^r|_{M^r}=f^r|_{M^r}$.
The above sum does not make sense when $k=1$.
In this case, (\ref{eq:bee}) implies that $M^r=\pi^{-1}(r)$, so
$g^r=f^r\in\dar(\pi^{-1}(r),\real^m)$.

If $r\in\real\setminus M_{n-1}$ there exist $g^r_{0},\ldots
,g^r_{k-1}\in\dar(\pi^{-1}(r),\real^m)$  such that
\begin{equation}\notag
g^r_0+\cdots +g^r_{k-1}=g^r
\end{equation}
since $g^r|_{M^r}=f^r|_{M^r}$ or $k=n$.

For each $i\in k$, let
$g_i\colon\real^n\to\real^m$ be defined by,
\begin{equation*}
g_i(x)=
\begin{cases}
d(r)& \text{if $x\in \pi^{-1}(r)$ and $r\in B_{i}$}\\
g^r_i(x)-(d(r)/(k-1))& \text{if $x\in \pi^{-1}(r)$ and $r\in M_{n-1}\setminus
B_i$}\\ g^r_i(x)& \text{if $x\in \pi^{-1}(r)$ and $r\notin M_{n-1}$}.
\end{cases}
\end{equation*}
Notice there is no problem with division by zero when $k=1$ since in this case
$M_{n-1}=B_0$.  We claim that $g_i\in\dar(\real^n,\real^m)$ for each $i\in
k$.  To see this let $A=B_i$ and $B=\real\setminus B_i$ and apply Lemma
\ref{lem:a}.   It is easily checked that,
\begin{equation}\notag
\sum_{i=0}^{k-1} g_i=g.
\end{equation}
So $f\colon\real^n\to\real^m$ is as desired,
completing the inductive step.\qed

We may now prove one of the inequalities of Theorem~\ref{thm:three}
\noindent\lem{lem:q}{If $n\geq 1$, then
$\add_{n,\floor{n/2}}(\dar(\real^n,\real^m))\geq\cuum^+$.}
\proof
Let $\{g_{\xi}\colon\xi\in\cuum\}$ be a collection of functions from $\real^n$
into $\real^m$.  We must find
$f\in(n-\floor{n/2})\dar(\real^n,\real^m)$ such that
$g_{\xi}-f\in (\floor{n/2}+1)\dar(\real^n,\real^m)$ for every
$\xi\in\cuum$.

Let $\{M_{i,\xi}\subseteq\real\colon i\in n \text{ and } \xi\in\cuum\}$ and
$\{K_i\subseteq\real\colon i\in n\}$ be thick meager sets such that for every
$i\in n$
\begin{description}\label{eq:76}
\item[(i)] $M_{i,\xi}\cap M_{i,\zeta}=\emptyset$ for $\xi<\zeta<\cuum$ and
\item[(ii)] $K_i\cap (\bigcup_{\xi\in\cuum}M_{i,\xi})=\emptyset$.
\end{description}
For every $\xi\in\cuum$, let
\begin{equation}\label{eq:77}
M_{\xi}=\bigcup_{f\in
F^n_{\floor{n/2}+1}}\prod_{i=0}^{n-1}(M_{i,\xi}\vee_f\real).
\end{equation}
We let
\begin{equation}\label{eq:78}
K=\bigcup_{f\in F^n_{n-\floor {n/2}}}\prod_{i=0}^{n-1}(K_i\vee_f\real).
\end{equation}
We now claim that
\begin{description}
\item[(a)] $M_{\xi}\cap M_{\zeta}=\emptyset$ for $\zeta<\xi<\cuum$ and
\item[(b)] $K\cap(\bigcup_{\xi\in\cuum}M_{\xi})=\emptyset$.
\end{description}
We show (b).  Fix $\xi\in\cuum$.  Let \[x=\langle x_0,\ldots ,x_{n-1}\rangle\in M_{\xi}
\text{ and }
y=\langle y_0,\ldots ,y_{n-1}\rangle\in K.\]  We claim
that $x\neq y$.  By (\ref{eq:77}) we have
\[|\{i\in n\colon x_i\in M_{i,\xi}\}|\geq\floor{n/2}+1.\]
By (\ref{eq:78}) we have
\[|\{i\in n\colon y_i\in K_{i}\}|\geq n-\floor{n/2}.\]
Since $n-\floor{n/2}+\floor{n/2}+1=n+1$,
by the Pigeonhole Principle there is an $i\in n$ such that $x_i\in M_{i,\xi}$
and
$y_i\in K_{i}$.  So, by (ii), $x_i\neq y_i$ which implies that
$x\neq y$.  A similar argument together with the fact that
$2(\floor{n/2}+1)\geq n+1$ shows that (a) holds.

Using Lemma \ref{lem:n}, we may find for each $\xi\in\cuum$ a function
$f_{\xi}\colon\real^n\to\real^m$ such that for any
$g\colon\real^n\to\real^m$ if
$g|_{M_{\xi}}=f_{\xi}|_{M_{\xi}}$, then $g\in
(\floor{n/2}+1)\dar(\real^n,\real^m)$.   Again applying Lemma \ref{lem:n}, we
may find a function
$f^*\colon\real^n\to\real$ such for any $g\colon\real^n\to\real^m$ if
$g|_{K}=f^*|_{K}$ then, $g\in (n-\floor{n/2})\dar(\real^n,\real^m)$.
Define $f\colon\real^n\to\real^m$ by
\begin{equation}\notag
f(x)=\begin{cases}
f^*(x)& \text{if $x\notin\bigcup_{\xi\in\cuum}M_{\xi}$}\\
g_{\xi}(x)-f_{\xi}(x)& \text{if $x\in M_{\xi}$}.
\end{cases}
\end{equation}
We claim that $f$ has the desired property.  Since $f|_{K}=f^*|_K$, it follows
from (b) that
$f\in (n-\floor{n/2})\dar(\real^n,\real^m)$.
Now, we only need to show that for each $\xi\in\cuum$ we have
$g_{\xi}-f\in(\floor{n/2}+1)\dar(\real^n,\real^m)$.  Fix
$\xi\in\cuum$.  Let $x\in M_{\xi}$.  Then,
$(g_{\xi}-f)(x)=g_{\xi}(x)-(g_{\xi}(x)-f_{\xi}(x))=f_{\xi}(x)$.  Since
$(g_{\xi}-f)|_{M_{\xi}}=f_{\xi}|_{M_{\xi}}$, we have
$g_{\xi}-f\in(\floor{n/2}+1)\dar(\real^n,\real^m)$.\qed

To complete the proof it is enough to show that
$\add((n-1)\dar(\real^n,\real^m))\leq\cuum^{+}$ for $n>2$.  To see this,
notice that by (iii) of Proposition~\ref{prop:two} we have
\[\add_{n,\floor{n/2}}(\dar(\real^n,\real^m))\leq
\add_{n,n-2}(\dar(\real^n,\real^m)).\]
By Proposition~\ref{prop:two}(viii), we have
\[\add_{n,n-2}(\dar(\real^n,\real^m))
\leq\add((n-1)\dar(\real^n,\real^m)).\]  Finally, notice that
$\floor{n/2}<n-1$ only if $n>2$.

\noindent\lem{lem:r}{If $n>2$, then
$\add((n-1)\dar(\real^n,\real))\leq\cuum^+$.}
\proof
Let $F\subseteq (\real)^{\real^n}$ be as in Proposition~\ref{prop:9}.  We claim
that  there is no $g\colon\real^n\to\real$ such that
$g+F\subseteq (n-1)\dar(\real^n,\real)$.
By way of contradiction assume that there is such a $g$.
Since $n-1<n$, Lemma~\ref{lem:f} implies that for every $f\in F$ there is
a perfect set $P_{f}$ such that the restriction of $g+f$
to $P_{f}$ is continuous.  Since $|F|=\cuum^+$ and there are only
$\cuum$-many perfect
sets, there exist $f,h\in F$ such that $P_{f}=P_{h}$.  It follows that
$f-h=(g+f)-(g+h)$
is continuous on $P_{f}$, which contradicts our choice of $F$.\qed

We now work to generalize the range space of Lemma~\ref{lem:r} to
$\real^m$.

\noindent\lem{lem:s}{If $n,m \geq 1$ and $n>2$, then
$\add((n-1)\dar(\real^n,\real^m))\leq\cuum^+$.}
\proof
The proof essentially follows that of Lemma~\ref{lem:j}.




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\end{thebibliography}
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