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\title[Decomposing with Smooth Sets]{Decomposing Euclidean Space with
a Small Number of Smooth Sets}
\author[J. Stepr\={a}ns]{Juris Stepr\={a}ns}
\address{Department of Mathematics, York University\\
4700 Keele Street,
North York, Ontario\\ Canada \ \ \  M3J 1P3}
\curraddr{}
\email{steprans@mathstat.yorku.ca}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%% SEPARATE \author COMMAND AND \address %%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%% ETC. FOR EACH AUTHOR %%%%%%%%%%%%%%%%%%%%%%%%%%
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%\dedicatory{}
%\date{}
\thanks{Research for this paper was partially supported by NSERC of
Canada. The author would also like to acknowledge that 
this paper has significantly benefitted from several remarks of A. Miller}
%\translator{}
\keywords{cardinal invariant, Sacks real, tangent
plane, covering number}
\subjclass{Primary 04A30; Secondary 28A15}
\begin{document}
\begin{abstract}
Let the cardinal invariant ${\mathfrak s}_{n}$ denote the least number
of continuously smooth 
\hyphenation{dim-en-sion-al}
$n$-dimensional 
surfaces  into which 
$(n+1)$-dimensional Euclidean space can be decomposed. It will be shown
to be consistent that ${\mathfrak s}_{n}$ is 
greater ${\mathfrak s}_{n+1}$. These cardinals will be shown to be
closely related to the invariants associated with the problem of
decomposing continuous functions into differentiable ones.
\end{abstract}
\maketitle
\tableofcontents
\bibliographystyle{plain}
\section{Introduction}
This paper is concerned with problems arising from partitioning Euclidean space
into subsets with certain smoothness properties. A geometric cardinal
invariant will be introduced which is determined by the least number of
smooth surfaces into which Euclidean space can be partitioned and it will be
shown that this invariant depends on the dimension of the Euclidean
space being considered. Results obtained about these cardinal
invariants will then be applied to the problem of decomposing a given
continuous function into differentiable functions. The relationship
established will answer a question of Morayne and Cichon who wondered whether
there is a continuous function from the reals to the reals which 
cannot be partitioned into fewer than $2^{\aleph_0}$ differentiable
functions. Finally, a question of Todorcevic will be answered by
showing that some of the cardinal invariants introduced distinguish between
the iteration and product of $\omega_2$ Sacks forcing partial orders.

The key interest of Theorem~\ref{main}, which will show that it is
possible to decompose Euclidean space into fewer than $2^{\aleph_0}$
smooth surfaces, is that smoothness is more difficult to handle than
other properties of analysis. For example, it is easy to use
$\sigma$-linked\footnote{A partial order is $\sigma$-linked if it is
the union of countably many subsets, each of which has the property
that any two of its elements have a lower bound.}
 forcing to decompose any Euclidean space into 
$\aleph_1$ continuous, one-to-one and even rectifiable\footnote{A
curve is rectifiable if it has a length --- in other words, the
supremum of the length of piecewise linear curves obtained by
connecting finite subsets on the curve exists.}
curves. To see that this is not so easily extended to differentiable
curves, consider the natural forcing for decomposing the plane into
rectifiable curves. It consists of triples $(c,x,\epsilon)$ where $c$
is a continuous curve of length less than 1, $\epsilon > 0$ and $x$ is
a finite subset of the $c$. Extension is defined by $(c,x,\epsilon)
\leq (c',x',\epsilon')$ if and only if the uniform norm of $c - c'$ is
less than  $\epsilon$, $x'
\supseteq x$ and $\epsilon' \leq \epsilon$. It is easy to see that
forcing with the product of  countably many copies of this partial
order will cover the ground model plane with countably many
rectifiable curves\footnote{In particular, if $G$ is generic for the
countable product let $\gamma_i$ be the
uniform limit of the curves $c$ such that there is some $p\in G$ such
that $p(i) = (c,x, \epsilon)$ for some $x$ and $\epsilon$.  Given
$w \in \Reals^2$ in the ground model and a condition $p$ choose a
coordinate $j$ such that $p(j) = \emptyset$ and extend $p$ to $p'$
such that $p'(j) = (c,\{w\},1)$ where $c$ is an arbitrary curve
containing $w$. It is the routine to check that $p'$ forces $\gamma_j$
to contain $w$.}.
  To see that the partial order
is $\sigma$-linked let $p$ be a curve of length $\delta < 1$ defined
by polynomials with rational coefficients. Let $\{A_i\}_{i\in k}$ be
open balls centred at points of $p$ such that the sum of the diameters
of the balls $A_i$ is less than $(1-\delta)/3$. If $(c^0,\{a^0_i\}_{i\in
k},\epsilon^0)$ and 
$(c^1,\{a_i^1\}_{i\in k},\epsilon^1)$ are such that $a_i^j\in A_i$,
$\epsilon^j > 4\epsilon $ and $\|c^j - p\| < \epsilon$ then 
$(c^0,\{a_i^0\}_{i\in k},\epsilon^0)$ and $(c^1,\{a_i^1\}_{i\in
k},\epsilon^1)$  are compatible. 

Trying to mimic this to obtain a smooth decomposition would entail
adding the derivative to the condition $(c,x,\epsilon)$ and
insisting that extension now means that derivatives stay within $\epsilon$ of
each other. However, this completely destroys the $\sigma$-linkedness
argument because, although it is possible to insist that $a^0_i$ is
close to $a^1_i$, it is not so easy to control the angle formed by
these two points. This paper provides an alternate approach which
allows this difficulty to be avoided.  

Section~2 contains the precise definitions of the cardinal invariants
discussed as well as definitions of intermediate invariants which will
play a role in the proof of the main theorem.  Section~3 contains lower
bounds and relations between the cardinal invariants which can be
proved without extra set theoretic assumptions. Section~4 contains the
definition and key properties of the generalization of Sacks forcing
which is used to establish later consistency results.  The key lemmas
that establish preservation properties of this 
partial order under countable support iteration are established in
Section~5.  From these, the main
theorem follows easily. Section~6 relates the previous results to the
problem of Cichon and Morayne concerning decomposing continuous
functions into differentiable ones.  Section~7 contains remarks about
the significance of the cardinal invariants discussed here to the
problem of iterating with countable support beyond $\omega_2$. A
question of Todorcevic is also answered and some open questions are posed.

\section{Basic Concepts}
Before introducing the main concepts some notation will be established.
Let $P_{k,n}$ denote the space of all subspaces of $\Reals^n$ of
dimension less than or equal to $k$. Define a metric $\rho$ on
$P_{k,n}$ by letting $\rho(V,W)$ measure the angle between $V$ and
$W$. To be precise, $\rho(V,W)$ is essentially  defined to be the Hausdorff
distance between $V$ and $W$ intersected with the unit sphere, or, in
other words, $\rho(V,W)$ is defined to be the maximum of $$\left\{\max_{x\in V^+}\min_{y\in W^+}
\arccos\left(\frac{x\cdot y}{\|x\|\|y\|}\right),
\max_{x\in W^+}\min_{y\in V^+}
\arccos\left(\frac{x\cdot y}{\|x\|\|y\|}\right)\right\}$$ 
where $V^+$ and $W^+$ are the
non-zero elements of the corresponding subspaces. If $v\in \Reals^n$
and $\langle v\rangle$  denotes the space
spanned by $v$,
then $\rho(\langle v\rangle,V)$ will be abbreviated to $\rho(v,V)$
and if $v'\in
\Reals^n$ then $\rho(\langle v\rangle,\langle v'\rangle)$ will be
abbreviated to $\rho(v,v')$. 

The following definition  plays a central
role in this paper. It generalizes the notion of a tangent plane to
the context of a set rather than a function. The case of one
dimensional  tangents was introduced by Besicovitch in \cite{besi.34} and
higher dimensional analogues have been considered in \cite{av.la.pr}.

\begin{defin}
If $A \subseteq \Reals^n$, $V \in P_{k,n}$ \label{TangentPlane}and $
b\in \Reals^n$ then 
$V$  will be said to 
be a $k$-dimensional tangent plane to $A$ at $b$ if and only if for
every $\epsilon > 0$ there is 
some $\delta > 0$ such that $\rho(V,b-a) < \epsilon$ for every $a
\in  A$ such that $0 < \|a-b\| < \delta$.
A subset $A \subseteq \Reals^n$ will be said to be a $k$-smooth
set if for every $b \in A$ there is $V\in P_{k,n}$ which is a
$k$-dimensional tangent plane\footnote{Observe 
that if $A\subseteq \Reals^n$ and $V$ is a $k$-dimensional
tangent plane to $A$ at $b$ and if there is no $m$-dimensional tangent
plane to $A$ at $b$ then $V$ is the unique $k$-dimensional tangent
plane to $A$ at $b$. However it is possible that a $k$-dimensional
smooth set in $\Reals^n$ may have $m$-dimensional tangent planes
at some --- or even all --- points, where $m < k$. In this respect,
the generalization of a tangent line to a tangent plane in
Definition~\ref{TangentPlane} 
differs from that found on page 263 of \cite{Saks}.}
 to $A$ at $b$.
\end{defin}

In the following discussion, there will also be occasion to refer to a
{\em continuously smooth set}, by which is meant a set to each point of
which it is possible to assign, in a continuous fashion, a tangent
plane. A subset $A \subseteq \Reals^n$ will be said to be
continuously-$k$-smooth
 if there is a continuous function $D:A \to P_{n,k}$ such that
each $D(a)$ is a $k$-dimensional tangent plane to $A$ at $a$.
By a $k$-smooth surface in $\Reals^m$ will be meant a subset of $\Reals^m$
which is isometric to (the graph of) some differentiable function
from $\Reals^k$ to $\Reals^{m-k}$. It is easy to see that a $k$-smooth
set need not be a $k$-smooth surface, nor even the  union of
finitely many $k$-smooth surfaces\footnote{Consider a sequence of
circles $\{C_n\}_{n\in\omega}$ each of which is tangent to the
horizontal axis at the origin and $C_n$ having radius $n$.}.  
However, it will be shown that
$k$-smooth sets can be covered by a sufficiently small number of
$k$-smooth surfaces so that some interesting results will follow. The next two
lemmas will, along with results from Section 3,  establish this fact.

\begin{lemma}
Let $X\subseteq \Reals^n$ be a closed subset, $\hat{X} \subseteq X$ be a
dense $G_\delta$ and  $F:X \to \Reals^k$ be 
a continuous 
function. If $D:\hat{X}\to P_{n+k,n}$ is a continuous function
 then the set of
points\label{DiifD} $x$ at which $D(x)$ is a tangent to the graph of $F$
 can be expressed as
the union of 
$\mathfrak d$ closed sets.
\end{lemma}
\begin{proof} Using continuity,  let $U\subseteq X$ be a countable dense
subset
and  notice that for every $\epsilon > 0$, $\delta > 0$ and $u \in U$
$$\{x\in \hat{X} : \|x-u\|\geq \delta \OR \rho((x,F(x)) - (u,F(u)),
D(x))
 \leq \epsilon\}$$
is closed in $\hat{X}$.
Hence, a direct calculation using the definition of a tangent yields
that the set of points $x$ 
at which $D(x)$ is tangent to the graph of $F$
 is a relative $G_{\delta\sigma}$ in $\hat{X}$ and, hence, is itself
a $G_{\delta\sigma}$.
Therefore, if this set of points 
is
$\bigcap_{n\in\omega}\bigcup_{m\in\omega}  C^n_m$ where each $C^n_m$
is closed  let $B_\xi =
\bigcap_{n\in\omega}\bigcup_{m\in\xi(n)}  C^n_m$ 
for $\xi \in \fomom$ and note 
 that each
$B_\xi$ is closed. If $\mathcal D$ is any
dominating family in $\fomom$ then $\bigcup_{\xi\in{\mathcal D}} {B_\xi}$ is
equal to the set of points at which the derivative of $F$ is given by
$D$.
\end{proof}

%It will be shown in Chapter 3 that the use of the invariant $\mathfrak d$
%in Lemma~\ref{DiifD} is optimal.

The proof of the next lemma is simplified by using the following definition.
\begin{defin} A set $E\subseteq \Reals^n$ is said to be $k$-Lipschitz
\label{lipschitz}if there is a $k$-dimensional hyperplane $H\subseteq \Reals^n$ and an
angle $\theta < \pi/2$ such
that $\rho(x-y,H) < \theta$ for any two points $x$ and $y$ from $E$.
\end{defin}

Observe that any $k$-Lipschitz set can be viewed as the graph  of a
function from $\Reals^k$ to $\Reals^{n-k}$ in a suitable coordinate
system. If   $E\subseteq \Reals^n$ is  $k$-Lipschitz
and this is witnessed by the hyperplane $H\subseteq \Reals^n$ then,
for each $z\in \Reals^k$, let $V_z$ be the $(n-k)$-dimensional
hyperplane orthogonal to $H$ and containing $z$. It follows that $E$
can intersect $V_z$ on at most one point --- if the intersection is not
empty then let $E^*(z)$ be this unique point. It is easy to see that
for each $\epsilon > 0$ the set of points at which the 
oscillation\footnote{The oscillation of a function $f:\Reals \to
\Reals$ at a point $x$ is $\lim_{\epsilon \to 0}\sup\{|f(a) - f(b)|
: |a - x|, |b - x| < \epsilon\}$.}
 of
$E^*$ is greater than $\epsilon$ must be a closed discrete set. Hence
$E^*$ is of Baire class 1 on its domain. Indeed, since the closure of
$E$ is still 
$k$-Lipschitz, it follows that $E^*$ is the restriction of a Baire
class 1 function defined on all of $\Reals^k$. 

The following theorem of Aversa, Laczkovich and Preiss \cite{av.la.pr}
 will be used in
the next lemma. 

\begin{theor}
If \label{avlapr} $H$ is a closed subset of $\Reals^n$, $F:H \to \Reals$ and 
$L:H \to P_{n, n+1}$ is a derivative of $F$ on $H$ then $F$ can be
extended to a function $\hat{F}$  everywhere differentiable on
$\Reals^n$
if and only if $L$ is Baire class 1.
\end{theor}

\begin{lemma} If $X\subseteq \Reals^m$ is a continuously  $k$-smooth set
then there are $\mathfrak d$ $k$-smooth surfaces \label{CoverByCurves} 
in $\Reals^m$ whose 
union contains $X$. 
\end{lemma}
\begin{proof}
 Let $X\subseteq \Reals^m$ be an arbitrary continuously $k$-smooth set.
The proof of the first theorem of \cite{besi.34} can be extended  to
 show that $X$
 is contained in a countable union of sets, each of which 
is $k$-Lipschitz. To see this, simply let $0 < \theta < \pi/4$, let
$\{H_n\}_{n\in\omega}$ enumerate a dense set of $k$-dimensional
hyperplanes in $\Reals^m$ and define $L(n,j)$ to be the set of all
$x\in X$ such that if $x'\in X\setminus \{x\}$ and $\|x - x'\|< 1/j$
then $\rho(x-x',H_n) < \theta$.

From the remarks following Definition~\ref{lipschitz} it follows that
$X$ can be covered by countably many
functions  of Baire class 1 and with closed domain, each with respect
to some coordinate 
system. It is shown in \cite{ci.mo} that any such
function can be covered using no more than $\mathfrak d$ continuous
functions. 

Let $f: \Reals^k \to \Reals^{m-k}$ be one of these continuous
functions where it is being supposed that, without loss
of generality,  its coordinate system is the standard one. Let
$f^* $ be the closure of $f\cap X$ and let $X^*$ be the domain of
$f^*$. 
Let $D:X \to P_{m,k}$ witness that $X$ is continuously $k$-smooth
and  define $\hat{D}(x)=D(x,f(x))$ for any $x$ such that $(x,f(x) )
\in X$. Let $\hat{X}$ be the $G_\delta$ consisting of all points to
which $\hat{D}$ can be extended continuously and denote this extension
also by $\hat{D}$. A direct computation
shows that the set of all $x\in\hat{X}$ such that $\hat{D}(x)$ is a
derivative to $f^*(x)$ is a relative $G_{\delta\sigma}$ and, since
$\hat{X}$ is itself a $G_\delta$, this set is a real $G_{\delta\sigma}$.
From Lemma~\ref{DiifD}, there are
$\mathfrak d$  closed sets, $\{B_\alpha\}_{\alpha\in {\mathfrak d}}$
such that 
$$\{x : (x,f(x)) \in X\}\subseteq \{x : 
 \hat{D}(x) \mbox{ is a
derivative to }f^*(x)\} =
 \bigcup_{\alpha\in {\mathfrak d}} B_\alpha$$ and
so $\hat{D}(x)$ is a derivative of
 $f^*\restriction B_\alpha$ for each $x\in B_\alpha$.
The continuity of $\hat{D}$ together with Theorem~\ref{avlapr} ensure that   
$f^*\restriction B_\alpha$ can be extended to a function which is
differentiable on all of $\Reals^k$. Its graph is therefore a
$k$-smooth surface.
\end{proof}   

Notice that if it is only desired to extend a smooth set to
a Borel set, rather than a function whose domain is all of Euclidean
space, then this is easy, as the following lemma demonstrates.

\begin{lemma}
If $S\subseteq \Reals^m$\label{??} is an $n$-smooth set then there is a
Borel set $S' \supseteq S$ which is also an $n$-smooth set. Similarly
if $S\subseteq \Reals^m$ is a continuously $n$-smooth set then there is a
Borel set $S' \supseteq S$ which is also a continuously $n$-smooth set.
\end{lemma}
\begin{proof} Choose $D$, a countable dense subset $S$ and let $S'$ be
the set of all those $x$ such that for
every $ \epsilon) > 0 $ there exists $\delta > 0$ if $ d_1$ and $ d_2$
belong to  $D$ and $\|d_1 - x\|$ and $ \|d_2 - x\|$ are both less than
$\delta$ then $\rho(x-d_1 ,
x -d_2) < \epsilon\}$.
 In the continuous case extend the derivative
continuously to a $G_{\delta}$.\end{proof}

The main object of study in this paper is decompositions of Euclidean
space into smooth surfaces and the next definition introduces
convenient cardinal invariants with which to discuss this.
\begin{defin}
Let ${\mathcal S}_{m,n}$ denote the $\sigma$-ideal generated by the
 $n$-smooth surfaces in $\Reals^m$. As well, define
 ${\mathcal D}_{m,n}$ to be the $\sigma$-ideal generated by the
$n$-smooth subsets of $\Reals^m$ and  ${\mathcal C}^1_{m,n}$  the
$\sigma$-ideal generated by the 
continuously $n$-smooth subsets of $\Reals^m$.
\end{defin}
The covering number of an ideal ${\mathcal I}$ will be denoted by
$\cov({\mathcal I})$ while $\unif({\mathcal I})$, $\cof({\mathcal I})$ and
$\add({\mathcal I})$ 
will represent the uniformity, cofinality and additivity invariants. 
The present paper is primarily concerned with
$\cov({\mathcal S}_{m,n})$ however some remarks about the other invariants
are in order. A. Miller has noticed the following identities.
\begin{propo} If $m > n$ then
$\cof({{\mathcal D}_{m,n}}) = \cof({{\mathcal S}_{m,n}}) = \cof({{\mathcal
C}^1_{m,n}})
 =  2^{\aleph_0}$ and
$\add({{\mathcal
D}_{m,n}})= \add({{\mathcal C}^1_{m,n}}) = \add({{\mathcal S}_{m,n}}) = \omega_1$.
\end{propo}
\begin{proof}
Only the case of ${\mathcal D}_{m,n}$ will be discussed since the others are
similar.  Both identities follow from the fact
that if $\mathcal H$ is an uncountable 
family of $n$-dimensional, parallel hyperplanes in $\Reals^m$ then 
$\cup{\mathcal H}\notin {\mathcal
D}_{m,n}$. To see that this is so, let $X \in {\mathcal
D}_{m,n}$ and let $X= \cup_{i\in\omega}X_i$ where each $X_i$ is
$n$-smooth and suppose that $\cup{\mathcal H} \subseteq X$.
Using Lemma~\ref{??}, it may be assumed that each $X_i$ has the
property of Baire. For $H\in \mathcal H$ let $i(H)\in \omega$ and $R(H)$
be a rational, $n$-dimensional rectangle such that $X_{i(H)}\cap H$ is
comeagre in $R(H)$. If $j$ is such that there is   a rational,
$n$-dimensional rectangle $R$ and there are uncountably many $H\in \mathcal
H$ such that $i(H) = j$ and $R(H) = R$ then $X_j$ can not be $n$-smooth.
\end{proof}

While $\cof$ and $\add$ are not monotone invariants with respect to
inclusion, the invariant $\unif$ is, and hence, it follows from the
fact that ${\mathcal S}_{m,n}\cup {\mathcal C}^1_{m,n}\subseteq {\mathcal
D}_{m,n}$ that 
$\unif({\mathcal S}_{m,n}) + \unif({\mathcal C}^1_{m,n})\leq \unif({\mathcal
D}_{m,n})$.
However the following questions remain open. 
\begin{quest} What is the value of 
$\unif({{\mathcal S}_{m,n}})$? What is the value of 
$\unif({{\mathcal C}^1_{m,n}})$? What is the value of 
$\unif({{\mathcal D}_{m,n}})$? In particular, can any of these be 
greater than $\omega_1$?
 \end{quest}

Finally, some simple inequalities will be established.
\begin{propo} If $m\geq m' > n$ then\label{si}
\begin{itemize}
\item $\cov({{\mathcal D}_{m,n}}) \geq\cov({{\mathcal D}_{m+1,n+1}}) $
\item $\cov({{\mathcal S}_{m,n}}) \geq\cov({{\mathcal S}_{m+1,n+1}}) $ \item
$\cov({{\mathcal C}^1_{m,n}}) \geq\cov({{\mathcal C}^1_{m+1,n+1}}) $
 \item 
$\cov({{\mathcal D}_{m,n}}) \geq\cov({{\mathcal D}_{m',n}}) $ \item
$\cov({{\mathcal S}_{m,n}}) \geq\cov({{\mathcal S}_{m',n}}) $ \item
$\cov({{\mathcal C}^1_{m,n}}) \geq\cov({{\mathcal C}^1_{m',n}}) $
\end{itemize}
and, moreover,
$\cov({{\mathcal C}^1_{m,n}})
\geq\cov({{\mathcal
D}_{m,n}})\geq \omega_1 $
and
$\cov({{\mathcal S}_{m,n}}) 
\geq\cov({{\mathcal
D}_{m,n}})\geq \omega_1 $. 
\end{propo}
\begin{proof} The proofs for each of the ideals are the same so only the
smooth case will be considered.
For the first inequality let $\{S_\xi\}_{\xi\in\kappa}$ be $n$-smooth
sets in $\Reals^m$ and note that $S_\xi\times \Reals$ is
$(n+1)$-smooth in $\Reals^{m+1}$. For the second inequality, take
any cover of $\Reals^m$ by $n$-smooth sets and intersect each member
with a fixed subspace of dimension $m'$. The first two inequalities of
the last clause are obvious and the last inequality follows from
observing that ${{\mathcal
D}_{m,n}}$ consists of meagre sets because $m>n$.
\end{proof}

\section{Absolute Lower Bounds}
This section will provide lower bounds for $\cov({\mathcal D}_{m,n})$
which, in light of Proposition~\ref{si}, provide lower bounds for
$\cov({\mathcal S}_{m,n})$ and $\cov({\mathcal C}^1_{m,n})$ as well. 
For the sake of the next theorem, let ${\mathcal D}_{m,0}$ denote the
$\sigma$-ideal generated by the
discrete subsets of $\Reals^m$ and note that $\cov({\mathcal
D}_{m,0}) = 2^{\aleph_0}$. Also, ${\mathcal D}_{m,n}^\lambda$ will denote
the ideal generated by unions of subsets of ${\mathcal D}_{m,n}$
of cardinality less than $\lambda$.

\begin{theor}
If $n' + k' = m$, $k \leq k'$,\label{MainLower} $n \leq n'$ and $ k +
n = j+1 $ 
$$\unif({\mathcal D}_{k',k-1}^\lambda)\cdot {\cov({\mathcal D}_{m,j})} <
\cov({\mathcal D}_{n',n-1})$$ 
  then $\cov({\mathcal
D}_{m,j}) \geq \lambda$.
\end{theor}
\begin{proof} Let $\cov({\mathcal D}_{m,j}) =  \kappa$ and let  $\{S_\alpha\}_{\alpha\in
\kappa}$ be $j$-smooth sets such that 
$\bigcup_{\alpha\in
\kappa}S_\alpha = \Reals^m$.
Since $m= n' + k'$ it is possible to let $V\subseteq \Reals^m$ be an
$n'$-dimensional subspace and let $W$ 
be a $k'$-dimensional subspace orthogonal to $V$. Letting
$\mu =  \unif({\mathcal D}_{k',k-1}^\lambda)$, let $B = \{w_\xi\}_{\xi\in
\mu}$ be a subset  of $W$ which can not be decomposed into less than
$\lambda$ elements of ${\mathcal D}_{k',k-1}$. For each $\xi \in \kappa  $ let
$S_{\xi,\alpha}$ consist of all those $s\in S_\alpha\cap (V+w_\xi)$
such that if $T$ is a tangent 
plane\footnote{Recall that tangent planes have been defined as subspaces so it does
not make sense to say, as would be natural, that  $V+ w_\xi$ does not
contain  a tangent 
plane to $S_\alpha$ at $s$.} to $ S_\alpha\cap (V+w_\xi)$ at
$s$ then $T/T\cap V$ has dimension at least $k$.
 Since $n+ k = j+1$ and $S_\alpha$ is
$j$-smooth, it follows that for each $s\in 
S_{\xi,\alpha}$ the  plane tangent to $S_\alpha$ at $s$ intersects $V$
on a subspace of dimension smaller than $n$ and so 
$S_{\xi,\alpha}$,  considered as subset of $V$ under the
obvious isomorphism, is a $(n-1)$-smooth set. Since $\kappa\cdot \mu <
 {\cov({\mathcal D}_{n',n-1})}$ it follows that there is some $v\in V$
such that $v+w_\xi\notin S_{\xi,\alpha}$ for all $\xi \in \mu$ and
$\alpha \in \kappa$.  

Then $B = \bigcup_{\alpha\in \kappa}C_\alpha$ where $C_\alpha =
\{w\in B : v+w\in S_\alpha\}$.
If $\kappa < \lambda$ it follows that there is $\alpha \in \kappa$
such that $C = \{v+w\}_{w \in C_\alpha}$ does not belong to ${\mathcal
D}_{k',k-1}$. This means that that there is a point $v + w_\zeta\in C$
such that $C$ has no $(k-1)$-dimensional tangent plane at $v + w_\zeta$. However,
$S_\alpha$ does have a $j$-dimensional tangent plane, $T$, at $v + w_\zeta$. The
definition of $S_{\zeta,\alpha}$ implies that the dimension of $T/V\cap T$ 
is less than $k$. This contradicts the fact that $T/V\cap T$ is a
tangent to $C$ at $v + w_\zeta$ and, hence, shows that 
$\cov({\mathcal
D}_{m,j}) = \kappa \geq \lambda$.
\end{proof}

\begin{corol} If $m > n\geq 1$ then \label{LB2}
$\cov({\mathcal D}_{m,n})^+ \geq \cov({\mathcal D}_{m-1,n-1})$. 
\end{corol}
\begin{proof}  If $\cov({\mathcal D}_{m,n}) \geq \cov({\mathcal D}_{m-1,n-1})$ then
there is nothing to prove. Otherwise, 
$\cov({\mathcal D}_{m,n}) < \cov({\mathcal D}_{m-1,n-1})$ and so, if $\lambda <
\cov({\mathcal D}_{m-1,n-1})$ then 
$$\unif({\mathcal D}^\lambda_{1,0})\cdot\cov({\mathcal D}_{m,n}) < \cov({\mathcal
D}_{m-1,n-1})$$ and so it follows  from
Theorem~\ref{MainLower} that $\cov({\mathcal D}_{m,n}) \geq \lambda$.
Since $\lambda <
\cov({\mathcal D}_{m-1,n-1})$ was arbitrary, it follows that $$\cov({\mathcal
D}_{m,n})^+ 
\geq \cov({\mathcal D}_{m-1,n-1})$$.\end{proof}

Note that from Proposition~\ref{si} and Corollary~\ref{LB2} it follows that
$$\cov({{\mathcal D}_{m,n}})^+ \geq\cov({{\mathcal D}_{m-1,n-1}}) \geq
\cov({{\mathcal D}_{m,n}})$$ provided that $m>n$. 

\begin{corol} If $m \geq 2$ then 
$\cov({\mathcal D}_{m,1})^+ \geq 2^{\aleph_0} $ \label{LB1}. 
\end{corol}
\begin{proof}  Let $n=1$ in Corollary~\ref{LB2} and note that 
$\cov({\mathcal D}_{m-1,0}) = 2^{\aleph_0}$ provided that $m\geq 2$.
\end{proof}

It is worth noting that if $k < n$ and $S\subseteq \Reals^n$ is
$k$-smooth then $S$ is nowhere dense. Hence $\cov({\mathcal D}_{m,n}) \geq
\cov({\mathcal M})$ where $\mathcal M$ denotes the meagre ideal. Moreover, it
will be shown in Theorem~\ref{mora} that
$\cov({\mathcal D}_{2,1}) \geq \cov({\mathcal N})$ where $\mathcal N$ is the ideal
of null sets with respect to Lebesgue measure, but, this is also 
 a consequence of the next  proposition. The proof of this
proposition will require some
preliminary definitions.

\begin{defin}
Given an integer $n\geq 1$, $1\leq j \leq n$ and $q\in\Rationals$ let
$H^n_{q,j}\subseteq \Reals^n$ be the $(n-1)$-dimensional hyperplane
which is orthogonal to the $j^{\mbox{th}}$ coordinate axis and
intersects this axis at $q$.
 A subset $X\subseteq  \Reals^{n}
\setminus \Rationals^{n}$ will be said to 
be very nowhere dense if for each $q\in \Rationals$ and each
$1\leq j \leq n$ no point $x\in H^n_{q,j}$ is in the closure of
 $X$ from both sides of the hyperplane $H^n_{q,j}$.
The cardinal invariant ${\mathfrak d}_n$ is defined to be the covering
invariant\footnote{For more details on this
see
Definition 2.3 and Lemma 3.1 of \cite{step.30} where ``very nowhere
dense'' is misleadingly called ``almost nowhere dense''.}
 of the ideal
 generated by
the very nowhere dense subsets of $\Reals^n$.
\end{defin} 

Since closed subsets of the irrationals are clearly very nowhere
dense and it is well known that $\mathfrak d$ is the equal to least number
of closed sets of irrationals required to cover the irrationals, it
follows that ${\mathfrak d}_1\leq \mathfrak d$. Moreover, if $A_1$ and
$ A_2$
 are  very nowhere dense subsets of $\Reals^{n_1}$ and $\Reals^{n_2}$ then
$A_1 \times A_2$ is a very nowhere dense subset of $\Reals^{n_1}\times \Reals^{n_2}$.
Notice as well that the Baire Category Theorem implies that
${\mathfrak d}_n$ can not be countable.
 This establishes the following.

\begin{propo}
If\label{DUpperBound} $n\geq 1$ then $\omega_1\leq {\mathfrak d}_n\leq
{\mathfrak d}_1 \leq \mathfrak d$. 
\end{propo}

%The next result due to S. Solecki\cite{sole}, will be 
%used\footnote{In fact, Corollary~3.7 of \cite{sole} is
%stronger than that which is quoted here.} 
%to show, as promised that Lemma~\ref{DiifD}
%provides the best possible lower bound.
%\begin{theor}
%If $F:\Reals\to \Reals$ is a Baire class 1 function\label{cor3.7} and there do not
%exist countably many closed sets $\{C_n\}_{n\in\omega}$ such that
%$F\restriction C_n$ is continuous for each $n\in \omega$ then
%there do not
%exist  less than $\mathfrak d$ closed sets $\{C_\xi\}_{\xi\in\lambda<
%{\mathfrak d}}$ such that
%$F\restriction C_\xi$ is continuous for each $\xi$.
%\end{theor}
%
%Let $L:[0,1]\to [0,1]$ be a function --- due to Lebesgue ---
%which is monotone and continuous on precisely the irrationals and let
%$F(x) = \int_0^xL(t)dt$. Then $L$ is of Baire class 1 and
% $F$ is monotone, continuous and
%differentiable at the irrationals. Moreover, its upper left Dini
%derivative is less than the lower right Dini 
%derivative at  each rational. Therefore, if $F\restriction Z$ is
%differentiable it follows that $Z$ must be very nowhere dense and
%hence $L\restriction \overline{Z}$ must be continuous where
%$\overline{Z}$ denotes the closure of $Z$.
%
%
%it is easy to see that there do not exist countably many closed sets
%$\{Z_n\}_{n\in\omega}$ such that $[0,1] = \bigcup_{n\in\omega}Z_n$ and
%$L\restriction Z_n$ is continuous for each $n\in\omega$ because each
%of these closed sets would have to be nowhere dense. It follows
%from Theorem~\ref{cor3.7} that 
%there does not exist $\lambda < \mathfrak d$ and closed sets
%$\{Z_\xi\}_{\xi\in\lambda}$ such that $[0,1] =
%\bigcup_{\xi\in\lambda}Z_\xi$ and 
%$L\restriction Z_\xi$ is continuous for each $\xi$. It follows
%that the set of points at which $F$ is differentiable
% is not the union of less than $\mathfrak d$ closed sets.

 \begin{propo}
If $ n\geq 1$ then $\cov({\mathcal S}_{n+1,n}) \geq {\mathfrak d}_n$\label{llbb}.
\end{propo}
\begin{proof}  Let $\{p_n\}_{n\in\omega}$ enumerate
$\Rationals^n$. Let
$$L(x_1, x_2, \ldots x_n) = \sum\{\frac{1}{2^i} : p_i =(q_1,q_2,\ldots
q_n) \AND (\forall j\leq n)(q_j < x_j)\}$$
and define
$$H(x_1, x_2, \ldots x_n) =
\int_{-\infty}^{x_1}\int_{-\infty}^{x_2}\ldots
\int_{-\infty}^{x_n} L(t_1,t_2,\ldots t_n)dt_1dt_2\ldots dt_n$$

Now,  if
$\{X_\alpha\}_{\alpha\in\kappa}$ are $n$-smooth surfaces in $\Reals^{n+1}$
then the projection onto $\Reals^n$ of each set $X_\alpha\cap
H$ must be a very nowhere dense set in $\Reals^n$. 
To see this, fix $\alpha \in \kappa$ and suppose that  $1\leq j \leq n$, $q\in
\Rationals$ and $\bar{x} = (x_1,x_2,\ldots
x_{j-1}, q\ldots  x_n)$ belongs to the hyperplane which intersects
the $j^{\mbox{th}}$ axis at $q$ and is parallel to the other axes.
Suppose also that
$$\lim_{k\to\infty}(w^k_1,w^k_2,\ldots w^k_n) =
\lim_{k\to\infty}(z^k_1,z^k_2,\ldots z^k_n) = \bar{x}$$ are sequences such that $w^k_j < q <
z^k_j$ for each $k\in\omega$ and that
$(w^k_1,w^k_2,\ldots w^k_n) $ and 
$(z^k_1,z^k_2,\ldots z^k_n)$ belong to the projection of $ X_\alpha$ for each $k\in \omega$.

 Then 
$$\lim_{k\to\infty}\frac{H(w^k_1,w^k_2,\ldots w^k_n) -
H(x_1,x_2,\ldots x_n)}{w^k - q} <$$ $$ 
\lim_{k\to\infty}\frac{H(z^k_1,z^k_2,\ldots z^k_n) -
H(x_1,x_2,\ldots x_n)}{z^k - q} $$
contradicting that $\bar{x}\in X_\alpha$ and $X_\alpha$ has a tangent plane at $\bar{x}$.
\end{proof}

Before continuing, it will be noted that the cardinal invariant ${\mathfrak
d}_n$ is even more closely related to the function $H$ of the proof of
Proposition~\ref{llbb} than is suggested in that proof. In fact, the
least number of smooth functions required to cover $H$ is precisely
${\mathfrak d}_n$. The proof of Proposition~\ref{llbb} already shows that
at least ${\mathfrak d}_n$ smooth functions are required, but appealing to
Theorem~\ref{avlapr} shows that  this
number also suffices.

\begin{propo} If\label{lkj} $m \geq n+1$ then $\cov({\mathcal S}_{m,n}) = \cov({\mathcal
S}_{n+1,n})$.
\end{propo}
\begin{proof} From Proposition~\ref{si} it follows that it suffices to show
that $\cov({\mathcal S}_{m,n})$ is not greater  than $ \cov({\mathcal
S}_{n+1,n})$. Proceed by 
induction on $m$ and assume that $m > n+1$. Let $
\{f_\alpha:\Reals^n\to \Reals^m\}_{\alpha\in
\kappa}$ be smooth parametrizations of smooth surfaces whose union
covers $\Reals^m$.  Let $f_\alpha(x) =
(f_\alpha^1(x),f_\alpha^2(x),\ldots f_\alpha^n(x))$ and
define
$$H_{\alpha,\beta}(x)= (f_\alpha(f^1_\beta(x),f^2_\beta(x),\ldots
f^n_\beta(x)), f^{n+1}_\beta(x) )$$ and note
that each $H_{\alpha,\beta}$ is smooth\footnote{However, it may not be the
parametrization of a function with respect to any coordinate axes; but
 this is easily dealt with by decomposing it into a finite
number of pieces using the implicit function theorem.}.
  If $(z,w)\in
\Reals^m\times\Reals$ then there is some $z'\in \Reals^n$ and $\alpha\in
\kappa$ such that $f_\alpha(z') = z$. Choose $\beta\in\kappa$ and $x$
such that $f_\beta(x) = (z',w,t)$ for some $t$. Then
$H_{\alpha,\beta}(x) = (z,w)$
\end{proof}
  
Proposition~\ref{lkj} provides the justification for defining ${\mathfrak
s}_{n} =  \cov({\mathcal S}_{n+1,n})$. 
From Lemma~\ref{CoverByCurves} it follows that $\cov({\mathcal
S}_{m,n}) \leq \cov({\mathcal C}^1_{m,n})\cdot\mathfrak d$ and this
immediately implies the following.

\begin{corol}
If\/ ${\mathfrak d}_n = \mathfrak d$ \label{eqeqe} then $
{\mathfrak s}_n \leq \cov({\mathcal C}^1_{m,n})\cdot{\mathfrak d}$. 
\end{corol}

It may also be noted that 
from the same hypothesis as that of Corollary~\ref{eqeqe} it can
almost be concluded that
${\mathfrak s}_n \in \{\cov({\mathcal C}^1_{m,n}) , {\mathfrak d}\}$. 
It has already been observed that
${\mathfrak s}_n \leq \cov({\mathcal C}^1_{m,n})\cdot{\mathfrak d}$. 
On the other hand, from
Proposition~\ref{llbb} and the hypothesis  
 it follows that 
${\mathfrak s}_n \geq  {\mathfrak d}_n = \mathfrak d$. 
Hence,
${\mathfrak d} \leq {\mathfrak s}_n \leq \cov({\mathcal C}^1_{m,n})\cdot{\mathfrak d}$.
 There are two cases to consider. If $\cov({\mathcal
C}^1_{m,n}) \leq {\mathfrak d} $ then
$${\mathfrak d} \leq {\mathfrak s}_n \leq \cov({\mathcal
C}^1_{m,n})\cdot{\mathfrak d} = {\mathfrak d}$$ and 
so ${\mathfrak s}_n = \mathfrak d$. Otherwise, 
$\cov({\mathcal C}^1_{m,n}) > {\mathfrak d} $ and so
$${\mathfrak s}_n \leq \cov({\mathcal C}^1_{m,n})\cdot{\mathfrak d} =
\cov({\mathcal C}^1_{m,n}) $$ and so,
if $\cov({\mathcal C}^1_{m,n})\leq {\mathfrak s}_n$, it follows that ${\mathfrak s}_n = \cov({\mathcal
C}^1_{m,n})$. However, the gap in this argument is still an open
question.
\begin{quest} Is the inequality
$\cov({\mathcal C}^1_{m,n})\leq {\mathfrak s}_n$ true?\end{quest}

\section{Forcing a Lower Bound}

Sacks forcing will be denoted by $\Sacks$. The following definition
generalizes Sacks forcing. 
\begin{defin}
By $\Sacks(n)$ will be denoted the partial order, under inclusion,
 of all subtrees $T\subseteq
\wpresup{(n+1)}$ such that
\begin{itemize}
\item  for every $t \in T$ the cardinality of the set of immediate successors of $t$ in
$T$ does not exceed $n+1$ 
\item  for every $t \in T$ there is some
$t'\supseteq t$ such that $t'$ has  $n+1$ immediate successors
in $T$ --- such a $t'$ is known as a branching node.
\end{itemize}
 \end{defin}

For each $n\in \omega$ choose $\epsilon(n) > 0$, $\delta(n) > 0$ and
unit vectors $\{\vec{x}_i\}_{i=0}^n \subseteq \Reals^n$
such that, if $B_{i}$ is the ball of radius $\delta(n)$ and centre
$\vec{x}_i$ then

\begin{multline}\label{eq:k}
\IF V \subseteq \Reals^n \text{ is a } (n-1) \text{ -dimensional
hyperplane and } y\in B_j\cap V\\ \text{ then there exists }
 i\leq n \text{ such that } 
\rho(V,y-z) > \epsilon(n) \text{ for all } z \in B_i
\end{multline}

Let $\{\nu_n\}$ be a
sequence of positive reals such that if
$ \sum_{i=k+1}^\infty\nu_i < \nu_k\delta(n)$.
Next,
 define a mapping $\Phi_n : (n+1)^\omega \to
\Reals^n$ by $$\Phi_n(f) = \sum_{i\in\omega} \nu_i\vec{x}_{f(i)}$$
and observe that $\Phi_n$ is  one-to-one and continuous. It
follows that $\Phi_n(\overline{T})$ is a compact subset of $\Reals^n$
homeomorphic to $\overline{T}$ for any
$T\in \Sacks(n)$.

\begin{lemma} If\label{takw}
$T\in \Sacks(n)$ and $S$ is a
$(n-1)$-dimensional smooth set in $\Reals^m$ for $m \geq n$ then $S\cap \Phi_n(\overline{T})$ is meagre
relative to $\Phi_n(\overline{T})$.
\end{lemma}
\begin{proof} Let $\Reals^n$ be embedded in $\Reals^m$ in the natural way.
By Lemma~\ref{??} it may as well be assumed that $S$ is Borel and,
hence, satisfies the Property of Baire. If $S$ is not meagre relative
to $\Phi_n(\overline{T})$ then let $U$ be an open set in $\Phi_n(\overline{T})$ such that
$\Phi_n(\overline{T})\cap U\cap S$ is comeagre in
$\Phi_n(\overline{T})\cap U$. 
Choose $b\in \overline{T}$ such that $\Phi_n(b) \in U\cap S$ and
 let $V$ be a $(n-1)$-dimensional
tangent plane to $S$ at $\Phi_n(b)$ and let $\delta$ be such that
$\rho(V,a - \Phi_n(b)) < \epsilon(n)$ for every $a\in A$ such that $0 < |a-\Phi_n(b)|<\delta$.
   
Let $j\in\omega$ be such that $\Phi_n(\{b'\in \overline{T} :
b'\restriction j = b\restriction j\})$ is contained in the
intersection of $U$ with the ball of radius $\delta$
 around $\Phi_n(b)$. Let $k > j$ be such that $b\restriction k$ has
$n+1$ immediate successors $s_0,s_1, s_2, \ldots s_n$. Let $s$ be the common
root of  $s_0,s_1, s_2, \ldots s_n$ and let $W =  \sum_{i=0}^k
\nu_i\vec{x}_{f(i)}$.
 Then the image under $\Phi_n$ of $\{b' \in \overline{T}
: s_i\subseteq b'\}$ is contained in $\nu_kB_{i} + W$ for $i\leq n$.
 Let $i$ be such that $b\supseteq s_i$ and use the choice
of $\epsilon(n)$ and the $\{\nu_m\}_{m\in\omega}$ to find $j$ such
that $\rho(V,\Phi_n(b) - z) > \epsilon(n)$ for any $z\in \nu_kB_{j} + W$. Choosing
$z\in (\nu_k B_{j} + W)\cap S$ is possible because $S$ is comeagre in $U$ and
$B_{j,k} \subseteq U$.
This is a contradiction.
\end{proof}    

\begin{corol}
If $g\in (n+1)^\omega$ is the \label{GoUp} generic real added by forcing with
$\Sacks(n)$ then $\Phi_n(g)$ does not belong to any
$n-1$-smooth set with a Borel code in the ground model.
\end{corol}
\begin{proof} Suppose that $S$ is a $n-1$-smooth set and $T\in
\Sacks(n+1)$ is such that $$T\forces{\Sacks(n+1)}{\Phi_n(g)\in S}$$ From
Corollary~\ref{takw} it follows that $\overline{T}\setminus \Phi_n^{-1}(S)$
is a dense $G_\delta$ relative to $\overline{T}$.
It is then standard to find $T' \subseteq T$ such that $T' \in
\Sacks(n+1)$ and $T'\cap \Phi_n^{-1}(S) = \emptyset$.
 This immediately yields a contradiction since
$T'\forces{\Sacks(n+1)}{\Phi_n(g)\notin S}$. 
\end{proof}

\section{The Iteration and Upper Bounds}
The countable
support iteration of length $\alpha$ of forcings 
$\Sacks(n)$ will be denoted by $\Poset_\alpha^{n}$. As a
convenience, by countable support will be meant that if $p \in  
\Poset_\alpha^{n}$ then $p(\gamma) = \wpresup{(n+1)}$ for all but countably
many $\gamma \in \alpha$.  From Corollary~\ref{GoUp} 
of the previous section it follows that if $V$ is a model
of the Generalized Continuum Hypothesis and
$G$ is  $\Poset_{\omega_2}^{n}$ generic over $V$ then ${\cov({\mathcal D}_{m,n-1})}
= \omega_2$  in $V[G]$. It will be shown in this
section that $\cov({\mathcal C}^1_{m,n})
= \omega_1$ in $V[G]$. 

To this end, suppose that $V$ is a model
of the Continuum Hypothesis, $p^* \in \Poset_{\omega_2}^{n}$ and that 
$p^*\forces{\Poset_{\omega_2}^{n}}{x\in\Reals^m}$. It follows that
there must be some $\alpha \in \omega_2$ such that 
$p^*\forces{\Poset_{\alpha}^{n}}{x\in V[G\cap
\Poset_{\alpha}^{n}]}$ and
 $p^*\forces{\Poset_{\alpha}^{n}}{x\notin V[G\cap
\Poset_{\beta}^{n}]}$ for all $\beta \in \alpha$. It must be shown
that there is, in the ground model, a continuously $n$-smooth set $S
\subseteq\Reals^m$ as well as a condition $q
\leq p^*$ such that  $q\forces{\Poset_{\omega_2}^{n}}{x\in S}$.

Fix $x$ and let $\Model \prec H(\aleph_3)$ be a countable elementary
submodel containing $p^*$, $x$ and $\alpha$. Notice that if $p$ and
$q$ belong to $\Poset_{\alpha}^n\cap \Model$ then $p\wedge q $
also belongs to $\Model$. The exact definition of $p\wedge q$ is not
important here, only that it is definable in $\Model$. Observe that
$p\wedge q$ exists in $\Poset_{\alpha}^n$ because the meet of each factor of
the iteration is simply intersection.

Let ${\mathcal P}(\Poset_{\alpha}^n\cap  \Model)$ be given the
Tychonoff product topology --- in other words, it is homeomorphic to the
Cantor set. Let ${\mathcal F}$ be the set of all prime filters on
$\Poset_{\alpha}^n\cap  \Model$ --- in other words, $F\in {\mathcal
P}(\Poset_{\alpha}^n\cap  \Model)$ such that
\begin{itemize}
\item if $p \in F$ and $q \geq p$ then $ q\in F$
\item if $ p\in F$ and $q \in F$ then $p\wedge q \in F$ and $p\wedge q
\neq \emptyset$
\item if $p \in F$, $\{q_i\}_{i\in k}\subseteq \Poset_{\alpha}^n$
and there does not exist any $ r \in  \Poset_{\alpha}^n$ such that
$r \leq p$ and $r \wedge q_i = \emptyset$ for each $ i\in k$ then
$F\cap \{q_i\}_{i\in k} \neq \emptyset$.
\end{itemize}
Notice that $\mathcal F$ is a closed set in ${\mathcal
P}(\Poset_{\alpha}^n\cap \Model)$. Next, let $\mathcal G$ be the set
of all $F\in \mathcal F$ such that $F$ is $\Poset_{\alpha}^n$ generic
over $ \Model$ and $p^*\in F$. To see that $\mathcal G$ is a dense
$G_\delta$ relative to $\mathcal F$, it suffices to show that if $D$ is a
dense subset of $\Poset_{\alpha}^n\cap \Model$ then $\{F\in {\mathcal F} : F\cap
D\neq
\emptyset\}$ is a dense open set in $\mathcal F$. In order to verify this,
let ${\mathcal V}$  be an open set in $\mathcal F$. It may be assumed that
$${\mathcal V} = \{F \in {\mathcal F} : p \in F  \AND \{q_i\}_{i\in k}\cap F =
\emptyset\}$$ for some $p\in \Poset_{\alpha}^n\cap \Model$ and $\{q_i\}_{i\in k}
\subseteq  \Poset_{\alpha}^n\cap \Model$. Because ${\mathcal V}$ is a non-empty
open set in $\mathcal F$ it follows from elementarity that there is some $r \in
\Poset_{\alpha}^n\cap \Model$ such that
$r \leq p$ and $r \wedge q_i = \emptyset$ for each $ i\in k$. Let $r'
\in D\cap \Model$ be such that $r' \leq r$ and observe  that $\{F\in {\mathcal F}
: r'\in F\} \subseteq {\mathcal V}$. Hence ${\mathcal G}$ is a dense $G_\delta$
in a closed subspace of ${\mathcal P}(\Poset_{\alpha}^n)$.
It is worth noting that the third
condition in the definition of $\mathcal F$ implies that $\mathcal G$ has a
base consisting of sets of the form ${\mathcal U}(p) = \{F\in
{\mathcal G} : p \in F\}$. 

Let $\Psi :{\mathcal G}\mapsto \Reals^m$ be the mapping defined by
$\Model[G]\text{``}\models x^G =
\Psi(G)\text{''}$ where $x^G$ is the interpretation of the name $x$ in
$\Model[G]$.  Notice that $\Psi$ is continuous.

Before continuing, some notation concerning fusion arguments will be
established. Let $\{f_i\}_{i\in\omega}$ be a sequence of finite functions
satisfying the following properties:
\begin{itemize}
\item $f_i : \alpha \cap \Model \to \omega$
\item for each $i\in \omega$ there exists a unique ordinal $\beta(i)
\in \alpha \cap \Model$ such that $f_{i+1}(\gamma) = f_i(\gamma)$
unless $\gamma = \beta(i)$ and $f_{i+1}(\beta(i)) = f_i(\beta(i))+1$
where $f_i(\beta(i))$ is defined to be 0 if $\beta(i)$ is not in the
domain of $f_i$
\item $\sup_{i\in\omega}f_i(\gamma) = \omega$ for each $\gamma\in
\alpha\cap \Model$.
\end{itemize}

%For each condition $p \in \Poset_{\alpha}^n$ the definition of
%$\Sacks(m,n)$ guarantees that, for each $\gamma \in \alpha \cap
%\Model$, there is a name $\{C^\gamma_\xi : \xi 
%\in {}^{\stackrel{\omega}{\smile}}(n+1)\}$ such that
%$p\restriction\gamma$ forces that $\{C^\gamma_\xi : \xi 
%\in {}^{\stackrel{\omega}{\smile}}g(n+1)\}$ is a witness to the fact
%that  $p(\gamma) \in \Sacks(m,n)$ and each $C^\gamma_\xi$ is an open
%ball whose radius and centre are rational. It will also be assumed that
%$ C^\gamma_\emptyset = \Reals^m$ for each $\gamma$.  

Let $\Sacks^*(n)$ be the set of all $T \in \Sacks(n)$  with the
additional property that for every $t\in T$ the cardinality of the set
of immediate 
successors of $t$ in $T$ is either $1$ or $n+1$.
Applying a recursion argument to thin down an arbitrary condition in 
$\Sacks(n)$  shows that
$\Sacks^*(n)$ is dense in $ \Sacks(n)$. 
For $p \in \Sacks^*(n)$ let $I_p:\wpresup{(n+1)} \to p$ be the
unique lexicographic order preserving mapping from
$\wpresup{(n+1)}$ onto the branching nodes of $p$. For $p\in
\Sacks^*(n)$ and $t \in \wpresup{(n+1)}$ let $p\langle t \rangle =
\{s \in T : s \subseteq I_p(t) \OR I_p(t) \subseteq s\}$.

Suppose that  $r \in \Poset_{\alpha}^{n}$, $\Gamma\in
[\alpha\cap \Model]^{<\aleph_0}$ and $h:\Gamma \to \wpresup{(n+1)}$.
 Then
$r[h]$ denotes the condition defined by 
$$r[h](\gamma) = \left\{\begin{array}{ll}
r(\gamma) &\IF \gamma\notin \Gamma\\
r(\gamma)\langle h(\gamma)\rangle &\IF \gamma\in \Gamma\end{array}\right.$$  
For $p \in \Poset_{\alpha}^n$ the notion of what
it means for $p$ to 
be determined is similar to that
in \cite{ba.la.sack}. In particular, let $\Gamma$ be the
support of $f$ 
and define a condition $p \in
\Poset_{\alpha}^n$ to be $f$-determined if
\begin{itemize}
\item $p\restriction\gamma\forces{}{p(\gamma) \in \Sacks^*(n)}$ for
every $\gamma \in \Gamma$  
\item for every $ h \in  \prod_{\gamma\in \Gamma}{}^{f(\gamma )}(n+1)$ the condition 
$p[h\restriction \gamma]$ decides the value of
$I_{p(\gamma)}(h(\gamma))$
 for each $\gamma\in \Gamma$.\end{itemize} 

By a fusion sequence will be meant a sequence of conditions
$\{p_i\}_{i\in\omega}\subseteq \Model$ such that
 each $p_i$ is $f_i$-determined and such that for each $j \leq i$ and
$h \in \prod_{\gamma\in\Gamma_j}{}^{f_j(\gamma)}(n+1)$ and $\gamma\in
\Gamma_j$ the condition $p_{i+1}[h\restriction \gamma]$ forces
$I_{p_i(\gamma)}(h(\gamma))$ and $I_{p_j(\gamma)}(h(\gamma))$ to have
the same value.
If $G\in \mathcal G$ then define $H_{i}(G)$ to be the unique member of
$\prod_{\gamma\in \Gamma_i}{}^{f_i(\gamma )}(n+1)$  
such that $p_i[H_i(G)]\in G$ provided that such a
maximal member exists at all.  Of course, 
$H_i(G)$ is only defined in the context of a given fusion sequence
but, since this will always be clear,
it will not be added to the notation.

Two facts are worth noting. First,  it is easy to verify that the fusion of
such a sequence is in $\Poset_{\alpha}^n$ and its domain is $\Model
\cap \omega_2$. Second,  if $G$ and $G'$
are in $\mathcal G$ and $H_i(G)$ and $H_i(G')$ are defined and equal for
all $i\in \omega$ then $G = G'$. 


Two cases will now be considered depending on whether or not $\alpha$
is a limit ordinal. 

\begin{lemma}
If $\alpha $ is limit \label{limit}ordinal then there is a continuously 
1-smooth set in
$\Reals^m$, $L$,  and a condition $ r \leq p^*$ such that $r
\forces{\Poset_{\alpha}^n}{x\in L}$. 
\end{lemma}
\begin{proof}
For each $G\in \mathcal G$ let $\tau(G)$ be the 
set\footnote{This definition is somewhat similar to Besicovitch's definition of
a contingent which can be found, for example,
on page 262 of \cite{Saks}.}  
 of all $V \in P_{1,m}$
such that for all $\epsilon > 0$, $q\in G$ and $\beta \in \alpha$ there is
$G'\in \mathcal G$ such that $G'\cap \Poset_{\beta}^n = G \cap
\Poset_{\beta}^n$, $q\in G'$, $\Psi(G) \neq \Psi(G')$ and
$\rho(V,\Psi(G) -\Psi(G')) < \epsilon$.
The first thing to observe is that if $G \in {\mathcal G}$ then $\tau(G)
\neq \emptyset$. To see this, first observe that
$\tau(G) = \bigcap_{\beta\in\alpha\cap \Model}\bigcap_{q\in
G}\overline{\tau(G,\beta,q)}$ where $\tau(G,\beta,q)$ is defined to be
$$\{\langle \Psi(G) - \Psi(G')\rangle : G'\cap \Poset_{\beta}^n = G \cap
\Poset_{\beta}^n, q\in G' \AND \Psi(G) \neq \Psi(G')\}$$
and, since $P_{1,m}$ is compact and the family
$\{\tau(G,\beta,q)\}_{\beta\in\alpha, q \in G}$ is centred, it
suffices to show that 
if $\beta \in \alpha \cap \Model$ and $q\in G$ then
$\tau(G,\beta,q)\neq \emptyset$.  Since
$p^*\forces{\Poset_{\alpha}^n}{x\notin V[G\cap \Poset_{\beta}^n]}$ for
every $\beta \in
\alpha$ it follows that there must be $G'$ containing both $p^*$ and $q$ 
which is $\Poset_{\alpha}^n/(G\cap \Poset_{\beta}^n)$ generic
over $\Model[G\cap\Poset_{\beta}^n]$ and such that the
interpretation of $x$ in $\Model[G\cap \Poset_{\beta}^n*G']$ is
different from the interpretation of $x$ in $\Model [G]$.

Since $\mathcal G$ and $P_{1,m}$ are both Polish spaces it follows
directly  from an analysis of the definition of the relation $\tau$ that it is an
analytic subset of ${\mathcal G}\times P_{1,m}$. It is then possible to
appeal to the von Neumann Selection Theorem to find a Baire measurable function
${\Delta} :{\mathcal G}\to P_{1,m}$ such that
${\Delta}(G) \in 
\tau(G)$ for 
each $G\in \mathcal G$. 
Let ${W}\subseteq {\mathcal G}$ be a dense $G_\delta$ such that
${\Delta}$ is continuous on $W$ and let  $W_i$ be dense open sets
such that $W = \bigcap_{i\in\omega}W_i$. 

The next step will be to construct a fusion sequence $\{p_i\}$ 
 so that the following conditions are
satisfied
\begin{itemize}
\item if $h \in \prod_{\gamma\in \Gamma_i}\presup{f(\gamma)}{n+1}$
(recall that $\Gamma_i$ is the domain of $f_i$)
  then ${\mathcal U}(p_i[h]) \subseteq
W_i$ 
\item if $h_0$ and $h_1$ are in  $\prod_{\gamma\in
\Gamma_i}\presup{f(\gamma)}{n+1}$ then 
the images
$\Psi({\mathcal U}(p_i [h_0]))$ and $ \Psi({\mathcal
U}(p_i [ h_1]))$ have disjoint 
closures
\item if $G$  belongs to $W$, $H_i(G) = H_i(G')$ and
$H_{i+1}(G) \neq H_{i+1}(G')$ then
$\rho( \Psi(G) -\Psi(G'), \Delta(G)) < 1/i$
\end{itemize}
First, it will be shown that if such a fusion sequence can be
constructed, then the proof is complete. In particular, let $L$ be the
image under $\Psi$ of the set of all $G\in \mathcal G$ such that $H_i(G)$
is defined for all $i \in \omega$. Let $r$ be the fusion of the
sequence $\{p_i\}_{i\in \omega}$ and note that
$r\forces{\Poset_{\alpha}^n}{x\in L}$. 

To see that $L$ is continuously 1-smooth, suppose that $z = \Psi(G)\in
L$ and let 
$\epsilon > 0$. It must first be shown
 that there is $\delta > 0 $ such 
that if $z' \in L$ and $\|z - z'\| < \delta$ then
$\rho({\Delta}(G),z-z') < \epsilon$. To this end, let $i >
1/\epsilon$ and let $\delta$ be so small that if $h$ and $h'$ are
distinct functions in  $\prod_{\gamma\in
\Gamma_i}\presup{f(\gamma)}{n+1}$
 then the distance between
$\Psi({\mathcal U}((p_i [ h ])))$ and $\Psi({\mathcal U}((p_i
[ h' ])))$ is greater than
$\delta$. Therefore, if $\|z- z'\| < \delta$  it follows that
$H_i(G') = H_i(G)$ where $G'$ is any member of $W$ such that $\Psi(G')
= z'$. Let $j\in 
\omega$ be the greatest integer such that $H_j(G) = H_j(G')$.  Since 
both $G$ and $G'$ belong to $W$ and,
since $H_{j+1}(G)
\neq H_{j+1}(G')$ it follows that $\rho({\Delta}(G),z-z') = 
\rho(\Delta(G),
\Psi(G)-\Psi(G')) < 1/j < 1/i < \epsilon$.
Finally,  that the preceding argument actually established that
$\Psi^{-1}$ is continuous on $L$. Hence so is $\Delta\circ\Psi^{-1}$
and this implies that $L$ is not only 1-smooth but also
continuously 1-smooth.

The next claim will be used in showing that the desired fusion
sequence can be constructed. The basic idea is that at stage $i$ of
the induction some finite subset
$A$ of
$\alpha$ on which some fusion has already taken place is given, as well as
some ordinal $\beta(i)$ --- which comes from the bookkeeping functions
$\{f_j\}_{j\in\omega}$ --- in this finite set on which some new branching must be
frozen. On $A$ finitely many finite subtrees corresponding
to the coordinates of the  condition $p_i$  have already been determined.
For each maximal branch through these possibilities a possible
branching at coordinate $\beta(i)$ will be frozen. Moreover, these
branchings will be chosen in a coherent way so that,  in the end, they
can all be pasted together without contradiction. 
Essentially the differences between conditions corresponding to
different branches will occur above the maximum of $A$ and this is
where the hypothesis that $\alpha$ is a limit ordinal is used.
The chief
difficulty is that these branchings will differ on the values they
assign to the real $x$ and a way must be found to control the angles
between them. This is the role of Claim~1. 
 The ordinal $\beta$ in this
claim plays the part of $\beta(i)$ and the conditions $q_i$ provide
the branching at $\beta$. Notice their coherence below $\alpha$ and the
fact that the third condition controls the angle between the values
determined for $x$. These conditions are then pasted together by
following conditions (a) to (g).

\begin{claim}
Suppose that $\epsilon > 0$, $q\in
\Poset_{\alpha}^n\cap \Model$, $\beta
\in \alpha$ and $J\in\omega$. Then there exist conditions
$\{q_i\}_{i=0}^{ n}\subseteq \Model$ 
such that 
\begin{enumerate}
\item $q_i \leq q$ for $i\in n+1$
\item $q_i \restriction \beta = q_{i'}\restriction\beta$ for $i$ and
$i'$ in $ n+1$  
\item  if $\{G_i\}_{i=0}^{n} \subseteq W$ and  $q_i \in G_i$ for each
$i \leq n$ then
$$\rho(\Psi(G_i)-\Psi(G_j),\Delta(G_i)) < \epsilon$$ so long as 
 $i\neq j$
\item $q_i\forces{\Poset_{\alpha}^n}{x\in E_i}$ for closed balls
$E_i$ such that $E_i\cap E_{i'}= \emptyset$
if $i \neq i'$
\item $q_i\restriction \beta$ decides all the values of
$I_{q_i(\beta)}\restriction \presup{J}{(n+1)}$
\end{enumerate}
\end{claim}

Assuming the claim, suppose that $p_i$  of the fusion
sequence have been constructed satisfying the induction
requirements. Let $\{h_s\}_{s\in k}$ enumerate all the 
elements of $\prod_{\gamma\in \Gamma_i}{}^{f_i(\gamma )}(n+1)$ and 
let $\mu\in \alpha\cap \Model$ contain the domain of
$f_{i+1}$. Let $q_s^{0,j} = p_i[h_s]$ for each $s\in k$ and $j \leq
n$.  Proceed 
by induction on $y\in k$ to construct $q_s^{y,j}$ such that
\begin{itemize}
\item[{\bf a.}] $q_s^{y,j} \leq q_s^{y-1,j}$ for each $y\leq k$
\item[{\bf b.}] if $\mu' \leq \mu$ and $h_s\restriction \mu' = h_{s'}\restriction \mu'$ 
then  $q_s^{y,j}\restriction \mu' = q_{s'}^{y,j}\restriction \mu'$
\item[{\bf c.}] ${\mathcal U}( q_y^{y+1,j})\subseteq W_i$ for $y\leq
k$ and $ j\in n+1$  
\item[{\bf d.}] if $q_y^{y+1,j}\in G\in W$, $ q_y^{y+1,j'}\in G'\in W$ and
$j\neq j'$ then 
$\rho(\Psi(G) - \Psi(G'),\Delta(G)) < 1/i$
\item[{\bf e.}]  for all $y\leq k$ and $ j\leq n$
the condition $q_y^{y+1,j}\restriction \beta(i)$
decides the value of $ I_{q_y^{y+1,j}(\beta(i))}(h_y(\beta(i))\wedge
j)$
\item[{\bf f.}] $q_y^{y+1,j}\forces{\Poset_{\alpha}^n}{x\in E_y^j}$ where
$\{E_y^j\}_{j=0}^{ n}$ is a pairwise disjoint collection of closed balls 
\item[{\bf g.}]\label{ihg} if $y\leq s$ then $q_s^{y,j} = q_s^{y,j'}$ for all $j$ and $j'$
in $n+1$
\end{itemize}
 To see that this induction can be carried out, suppose that
$\{q_s^{y,j}\}_{s\in k}^{j\leq n}$ have been defined. From induction
hypothesis~(g.) it follows that there is some $q''$ such that $q'' =
q_y^{y,j}$ for all $j\in n+1$. Since $W_i$ is dense open it is
possible to find $q' \leq q''$ such that ${\mathcal U}( q')
\subseteq W_i$. Extend $q'$ to $q$ such that
$q\restriction \gamma\forces{}{q(\gamma) \in \Sacks^*(n)}$ for every $\gamma$ 
in  the domain of $f_i$ and, furthermore,
$q\restriction
\beta(i)$ decides the value of
$I_{q(\beta(i))}(h_y(\beta(i))\wedge
j)$ for each $j\leq n$. Moreover,  
use the claim to find conditions
$\{q_y^{y+1,j}\}_{j\in n+1}$ such that
\begin{itemize}
\item $q_y^{y+1,j} \leq q$ for $j\in n+1$
\item $q_y^{y+1,j} \restriction \mu = q_y^{y+1,j'}\restriction\mu$
for $j$  and $j'$ in $ n+1$
\item  if $\{G_j\}_{j\in n+1} \subseteq W$ and  $q_y^{y+1,j} \in G_j$ for each
$j \in n+1$
then $$\rho(\Psi(G_j)-\Psi(G_{j'}),\Delta(G_j)) < 1/(i+1)$$ so long as 
 $j\neq j'$
\item $q_y^{y+1,j}\forces{\Poset_{\alpha}^n}{x\in E_y^j}$ for closed balls
$E_y^j$ such that $E_y^j\cap E_y^{j'}= \emptyset$
if $j \neq j'$
\item $q_y^{y+1,j}\restriction \beta(i)$ decides the value of
$I_{q_y^{y+1,j}(\beta(i))}(h_y(\beta(i))\wedge 
j)$ 
\end{itemize}
If $s\neq y$ let $b(s)$ be the least member of the domain  of $f_i$
such that $h_s(b(s))\neq h_y(b(s))$ and define $q_s^{y+1,j}$ to be the
largest lower bound  of $ q_y^{y+1,j}\restriction b(s)$ and $q_s^{y,j}$.
It is easily verified that all of the induction requirements are
satisfied.

Now, if $h$ is a member of $\prod_{\gamma\in
\Gamma_{i+1}}{}^{f_{i+1}(\gamma )}(n+1)$ then let $j(h)\leq n$ and 
$h_s$ be a member of $\prod_{\gamma\in \Gamma_i}{}^{f_i(\gamma )}(n+1)$ such that $h(\beta(i)) =
h_s(\beta(i))\wedge j(h)$ and, if $\gamma
\neq \beta(i)$ then $h(\gamma) = h_s(\gamma)$.  
Define $p_{i+1}$ to be the join of all the
conditions $\{q_{s}^{k,j}\}$
and observe that $p_{i+1}[h] \leq q_s^{k,j(h)}$ and, furthermore,
because $p_{i+1}$ is defined as a join, 
$$p_{i+1}\forces{}{I_{p_{i+1}(\gamma}(h(\gamma)\restriction m) =
I_{p_{i}(\gamma}(h(\gamma)\restriction m)}$$ for $m\leq f_i(\gamma)$.
Hence $p_{i+1}$ satisfies the requirements of a fusion sequence.
Note that if $y\neq y'$ then, by the induction hypothesis on $p_i$,
$\Psi({\mathcal U}(p_i[ h_y]))$ and $\Psi({\mathcal
U}(p_i [ h_{y'}]))$ have disjoint 
closures so it may be assumed that $E^j_y\cap E^{j'}_{y'}= \emptyset$.
Note that if $H_i(G) = H_i(G')$ and $H_{i+1}(G) \neq H_{i+1}(G')$
then there is some $y \leq k$ and $j\neq j'$ such that $q_y^{y+1,j}
\in G$ and $q_y^{y+1,j'}
\in G'$. Hence the third induction requirement follows directly from (d.). 
Thus the three requirements of the desired fusion sequence are satisfied.

All that remains to be done is to prove the claim. To this end, let
$q$, $\epsilon$ and $\beta$ be given. 
Using the fact that $W$ is
dense, let $G^*\in W$ be arbitrary such that $q\in G^*$. Using the
continuity of $\Delta$ on $W$, find $q'\in G^*$ such that $q' \leq q$ and,
if $G'\in W$ and $q'\in G'$ then $\rho(\Delta(G^*),\Delta(G')) <
\epsilon/2$.
 Let $p^{-1} = q'$. 
First notice that it suffices to construct by
induction on $i$ a sequence
 $\{(E_i,E^i,G_i,p_i,p^i) \}_{i\in n+1}$
such that 
$\{(E_i,E^i,p_i,p^i) \}_{i\in n+1}\in \Model$ and
 such that for each $ i\in n+1$
\begin{enumerate}
\item $p_i \leq p^{i-1}$
\item $p^i \leq p^{i-1}$
\item $G_i\cap \Poset_{\beta }^n = G^*\cap \Poset_{\beta }^n$ 
\item $p^i \in G^*$ for each $i$
\item $p_i \in G_i$ for each $i$
\item $E^i$ and $E_i$ are disjoint closed subsets of $\Reals^m$
\item $E^{i+1}\cup E_{i+1}\subseteq E^i$
\item  $\Psi({\mathcal U}(p^i))$ is contained in the
interior of $E^i$
\item  $\Psi({\mathcal U}(p_i))$ is contained in the
interior of $E_i$
\item $\rho(x-x',\Delta(G^*)) < \epsilon/2$ if  $x \in E^i$ and $x'\in E_i$
\end{enumerate}
The reason this suffices is that, having done so, using conditions (3)
and (5) it is possible to find a single $p \in G^*\cap
\Poset_{\beta }^n$  extending each
$p_i\restriction \beta$ and such that
$p\restriction \beta$ decides all the values of
$I_{p(\beta)}\restriction \presup{J}{n}$. 
Let $q_i$ be the greatest lower bound of both $p_i$ and $p$. It
follows that  $q_i \leq p_i \leq p^{i-1} \leq p^{-1} = q'\leq q$. Moreover,
$q_i\restriction \beta = p$ for each $i\in n+1$. 
Also,  it follows from conditions (6) and (7) that
the sets $\{E_i\}_{i\in n+1}$ are pairwise disjoint closed sets and
$p_i\forces{}{x\in E_i}$ by condition (9). Finally, suppose that
$i\neq j$, $\{G_i,G_j\} \subseteq W$,  $q_i \in G_i$ and   $q_j \in G_j$.
Then  $\rho(\Delta(G_i),\Delta(G^*)) < \epsilon/2$ because $q_i  \leq q'$. 
Hence 
$$\rho(\Psi(G_i)-\Psi(G_j),\Delta(G_i)) < \epsilon$$ by condition (10).

To carry out the induction, suppose that
$\{(E_i,E^i,G_i,p_i,p^i) \}_{i\in J}$ have been constructed.
  From the
definition of $\Delta(G^*)$  it follows that there is $G_J$ such that
\begin{itemize}
\item $G_J\cap \Poset_{\beta}^n = G^*\cap \Poset_{\beta}^n$
\item  $p^{J-1}\in G_J$ 
\item $\rho(\Psi(G^*)-\Psi(G_J),\Delta(G^*)) < \epsilon/4$
\item $\Psi(G^*) \neq \Psi(G_J)$
\end{itemize}
Let $E^J$ and $E_J$ be disjoint closed neighbourhoods of $\Psi(G^*)$
and $\Psi(G_J)$ 
respectively such that $\rho(x-x',\Delta(G^*)) < \epsilon/2$ for
any $x \in E^J$ and $x'\in E_J$.  
 Since induction hypothesis (8) implies that
 $\Psi({\mathcal U}(p^{J-1}))$ is contained in the
interior of $E^{J-1}$ and
$ p^{J-1}\in G_J$ it follows that it may be assumed that
 $E^J\cup E_{J}\subseteq E^{J-1}$.
From the continuity of $\Psi$ it is
possible to find $p_J\in G_J$ and $p^J\in G^*$ extending $p^{J-1}$
  such that $\Psi({\mathcal U}(p_J))$ is contained in the interior of $ E_J$ and
$\Psi({\mathcal U}(p^J))$ is contained in the interior of $ E^J$.
All of the induction hypotheses are now satisfied.
\end{proof}
It is worth noting that nothing very specific to $\Sacks(n)$ was
used in the proof of Lemma~\ref{limit}. A similar result can be
obtained for almost any partial order which is sufficiently similar to
Sacks forcing that a fusion argument, like the one of
Lemma~\ref{limit}, can be applied to it.

The possibility that $\alpha$ is a successor must now be considered.
The proof has the same structure as the limit case but the details are
different. In particular, the successor case requires the use  of higher
dimensional tangent planes and so, unlike the limit case, the use of
$\Sacks(n)$ here will 
be crucial.

\begin{lemma}
If $\alpha $ is a successor \label{successor} ordinal then there is
a continuously $n$-smooth set in $\Reals^m$, $L$, and a condition $ r \leq p^*$
such that $r
\forces{\Poset_{\alpha}^n}{x\in L}$. 
\end{lemma}
\begin{proof} Let $\alpha = \beta + 1$. 
For each $G\in \mathcal G$ let $\tau(G)$ be the set of all $V \in P_{n,m}$
such that for all $\epsilon > 0$ and $q\in G$ there 
is a family $a \in [\mathcal G]^{n+1}$ such that 
\begin{enumerate}
\item \label{2ndlc1} $G\in a = \{G_0, G_1,\ldots G_n\}$
\item\label{2ndlc2} $G_i\cap \Poset_{\beta}^n = G \cap
\Poset_{\beta}^n$ for each $i \leq n$
\item\label{2ndlc3}  $q\restriction \beta * q(\beta)\langle
\{(0,i)\}\rangle \in G_i$ for each $i \leq n$
\item\label{2ndlc4} $\Psi\restriction a$ is one-to-one
\item\label{2ndlc5}  $\rho(V,\Psi(G_i) - \Psi(G_j)) < \epsilon$ for  $\{i,j\}\in
[n+1]^2$
\end{enumerate}
  As in Lemma~\ref{limit}, it must be noted that if $G \in {\mathcal G}$
then $\tau(G)
\neq \emptyset$.  
Notice that $\tau(G) \supseteq \bigcap_{q\in G}\overline{\tau(G,q)}$ 
where $\tau(G,q)$ is
defined to be the set of all spaces generated by $$\{\Psi(G'') -
\Psi(G') : \{G',G''\}\in [a]^2\} $$ where $a$ satisfies conditions (1)
to (4) with respect to $q$. Notice also that its dimension does not
exceed $n$ and hence, it belongs to $P_{n,m}$.
Since $P_{n,m}$ is compact and $\{\overline{\tau(G,q)}\}_{q\in G}$
has the finite intersection property, it suffices to show that
if  $q\in G$ then
$\tau(G,q)\neq \emptyset$.  In $\Model[G\cap \Poset_{\beta}^n]$
the name $q(\beta) $ is interpreted as a condition in $\Sacks(n)$.
It follows that in $\Model[G\cap \Poset_{\beta}^n]$ 
 there must be $q'\leq q\restriction \beta$ which decides the root of
$q(\beta)$ to be $r$. Let $q_i =q' \restriction \beta *
q(\beta)\langle \{(0,i)\}\rangle$.
 Since $$p^*\forces{\Poset_{\alpha}^n}{x\notin
\Model[G\cap \Poset_{\beta}^n]}$$ it is an easy matter to  
choose $G'_i$ which is $\Sacks(n)$ generic over
 $\Model[G\cap\Poset_{\beta}^n]$, $q_i \in G'_i$ and such that the
interpretation of $x$ in $\Model[G\cap \Poset_{\beta}^n*G'_i]$ is
different from the interpretation of $x$ in 
$\Model[G\cap \Poset_{\beta}^n*G'_j]$ unless $i=j$. It follows
that, letting $G_i = G\cap \Poset_{\beta}^n* G'_i$,
the space generated by  $$\{\Psi(G) - \Psi(G_i) \}_{i=1}^n  $$
belongs to $\tau(G,q)$.

As in Lemma~\ref{limit}, it is possible to find a Baire measurable function
${\Delta} :{\mathcal G}\to P_{n,m}$ such that
${\Delta}(G) \in \tau(G)$ for 
each $G\in \mathcal G$. Let ${W}$ be a dense $G_\delta$ on which
${\Delta}$ is continuous.
Let $W_n$ be dense open sets 
such that $W = \bigcap_{i\in\omega}W_n$. 

As before, the next step will be to construct a fusion sequence $\{p_i\}_{i\in\omega}$ 
so that the following conditions are
satisfied
\begin{itemize}
\item if $h$ is in $ \prod_{\gamma\in
 \Gamma_i}\presup{f(\gamma)}{n+1}$
 then ${\mathcal U}(p_i[h]) \subseteq
W_i$
\item if $h_0$ and $h_1$ are distinct elements of $ \prod_{\gamma\in \Gamma_i}\presup{f(\gamma)}{n+1}$ then
the images
$\Psi({\mathcal U}(p_i[h_0]))$ and $ \Psi({\mathcal U}(p_i[h_1]))$ have disjoint
closures
\item if $G$ belongs to $W$, $H_i(G) = H_i(G')$ and
$H_{i+1}(G) \neq H_{i+1}(G')$ then
$\rho(\Psi(G)-\Psi(G'), \Delta(G)) < 1/i$
\end{itemize}
Observe that if such a fusion sequence can be
constructed, then the proof is complete. In particular, let $L$ be the
image under $\Psi$ of the set of all $G\in \mathcal G$ such that $H_i(G)$
is defined for all $i \in \omega$. Let $r$ be the fusion of the
sequence $\{p_i\}_{i\in\omega}$. Then $r\forces{}{x\in L}$. 
The proof  that $L$ is continuously $n$-smooth is exactly the same as in
Lemma~\ref{limit}.  

To see that the fusion sequence can be found, suppose that
$p_i$  has been constructed
satisfying the induction requirements. Let $\{h_s\}_{s\in k}$
enumerate  $$\prod_{\gamma\in \Gamma_i}{}^{f_i(\gamma )}(n+1)$$ and 
let $q_s^{0,j} = p_i[A(h_s)]$ for
each $s\in k$.  Proceed by induction on $y$ to construct $q_s^{y,j}$
such that
\begin{itemize}
\item $q_s^{y,j} \leq q_s^{y-1,j}$ for each $y\leq k$
\item if $\mu \leq \alpha$ and $h_s\restriction \mu = h_{s'}\restriction \mu$ 
then  $q_s^{y,j}\restriction \mu = q_{s'}^{y,j}\restriction \mu$
\item ${\mathcal U}( q_y^{y+1,j})\subseteq W_i$ for $y\leq
k$ and $ j\in n+1$  
\item if $q_y^{y+1,j}\in G\in W$, $ q_y^{y+1,j'}\in G'\in W$ and $j\neq j'$ then
$\rho(\Psi(G) - \Psi(G'),\Delta(G)) < 1/i$
\item  for all $y\leq k$ and $ j\leq n$ the condition
$q_y^{y+1,j}\restriction \beta(i)$ decides the value of
$I_{q_y^{y+1,j}(\beta(i))}(h_y(\beta(i))\wedge
j)$ 
\item $q_y^{y+1,j}\forces{\Poset_{\alpha}^n}{x\in E_y^j}$ where
$\{E_y^j\}_{j\in n+1}$ is a pairwise disjoint collection of closed balls 
\item if $y\leq s$ then $q_s^{y,j} = q_s^{y,j'}$ for all $j$ and $j'$
in $n+1$
\end{itemize}
 To see that this induction can be carried out, suppose that
$\{q_s^{y,j}\}_{s\in k}$ have been defined. From the last induction
hypothesis it follows that there is some $q''$ such that $q'' =
q_y^{y,j}$ for all $j\in n+1$. Since $W_i$ is dense open it is
possible to find $q' \leq q''$ such that ${\mathcal U}( q')
\subseteq W_i$. Extend $q'$ to $q$ such that 
$q\restriction \beta(i)$ decides the value of
$I_{q(\beta(i))}(h_y(\beta(i))\wedge
j)$  for each $j\leq n$. 

Using the fact that $ W$ is dense in $\mathcal G$, let $G\in W$ be such
that $q\in G$.  Using the definition of $\tau(G)$, let $a\in [{\mathcal
G}]^{n+1}$ be such that Conditions~\ref{2ndlc1} to \ref{2ndlc5} hold.
It follows from Condition~\ref{2ndlc3} 
that  $\hat{q}_j = {q}\restriction \beta * q(\beta)\langle
\{(0,j)\}\rangle \in G_j$ for each $j \leq n$.
Using Condition~\ref{2ndlc5}, the continuity of $\Delta$ and $\Psi$ as
well as the fact  that $p^*\forces{\Poset_{\alpha}^n}{x\notin
\Model[G\cap \Poset_{\beta}^n]}$ it is possible to extend
each $\hat{q}_j$ to $q_y^{y+1,j}$ such that
\begin{itemize}
\item $q_y^{y+1,j} \leq q$ for $j\in n+1$
\item 
$q_y^{y+1,j} \restriction \beta = q_y^{y+1,j'} \restriction \beta $
for $j$ and $j'$ in $n+1$ 
\item  if $\{G_j\}_{j\in n+1} \subseteq W$ and  $q_y^{y+1,j} \in G_j$ for each
$j \in n+1$
then $$\rho(\Psi(G_j)-\Psi(G_{j'}),\Delta(G_j)) < 1/(i+1)$$ so long as 
 $j\neq j'$
\item $q_y^{y+1,j}\forces{\Poset_{\alpha}^n}{x\in E_y^j}$ for closed balls
$E_y^j$ such that $E_y^j\cap E_y^{j'}= \emptyset$
if $j \neq j'$
\end{itemize}

Note that guaranteeing that the second conditionis satisfied is easy
since $\hat{q}_j \in G_j$ for each $j \leq n$ and, hence, are all compatible.
If $s\neq y$ let $b(s)$ be the least member of the domain  of $f_i$
such that $h_s(b(s))\neq h_y(b(s))$ and define $q_s^{y+1,j}$ to be the
least upper bound  of $ q_y^{y+1,j}\restriction b(s)$ and $q_s^{y,j}$.
It is easily verified that all of the induction requirements are
satisfied.
Define $p_{i+1}$ to be the join of all the
conditions $q_s^{k,j}$.
This satisfies the three requirements of the desired fusion sequence
regardless of whether $\beta = \beta(i)$.
\end{proof}

 Lemmas \ref{successor} and \ref{limit} together show that the ground
model continuously $n$-smooth sets are sufficient to cover all reals added by
iteratively adding $\omega_2$ reals with the partial
order $\Sacks(n)$. Combined with Corollary~\ref{GoUp}, this
immediately gives the following theorem.

\begin{theor}
If  $1 \leq  n < m \in \omega$,\label{main} then it is consistent, relative to the consistency
of set theory itself, that $\cov({\mathcal D}_{m,n-1}) = \aleph_2$ and
$ \cov({\mathcal
C}^1_{m,n}) = \aleph_1$.
\end{theor}

\begin{corol}
It is consistent, \label{mainCorola}relative to the consistency
of set theory itself, that ${\mathfrak s}_{n-1} = \aleph_2$ and ${\mathfrak s}_{n} =
\aleph_1$ for $n > 1$. 
\end{corol}
\begin{proof} The model is the same as that for Theorem~\ref{main}. Since
${\mathfrak s}_{n-1} \geq \cov({\mathcal D}_{m,n})$ it follows that ${\mathfrak s}_{n-1} =
\omega_2$. 
Since the standard Sacks real argument shows that each partial order
$\Sacks(n)$ is $\fomom$-bounding, it follows from the preservation
of this property by countable support iteration \cite{shel.pf} that
this model also satisfies that ${\mathfrak d} =
\aleph_1$. It is shown in Proposition~\ref{DUpperBound} that $
\omega_1 \leq {\mathfrak d}_n
\leq {\mathfrak d}$.  
 Hence, ${\mathfrak d}_n = \mathfrak d = \omega_1$ in this model. From
Corollary~\ref{eqeqe} and the fact that $\cov({\mathcal C}^1_{n+1,n})
= {\mathfrak d}$ in this model it 
follows that, in this model,
 the equality
$\cov({\mathcal C}^1_{n+1,n}) = \cov({\mathcal
S}_{n+1,n}) = {\mathfrak s}_{n}$ must hold. Hence ${\mathfrak s}_{n} <
{\mathfrak s}_{n-1}$ in this model.
\end{proof} 
\section{Decomposing Continuous Functions}

In \cite{cimopaso} the authors consider the following question: If $\mathcal{A}$
and $\mathcal{B}$ are two families of functions between Polish spaces,
what is the least cardinal $\kappa$ such that every member of
$\mathcal{A}$ can be decomposed into $\kappa$ members of $\mathcal{B}$ --- this
cardinal $\kappa$ the authors call $\mbox{dec}(\mathcal{A},\mathcal{B})$.
The most natural class to consider for $\mathcal{B}$ is the class of
continuous functions and the problem to which it gives rise  had been
posed by Lusin, who wondered whether every Borel function could be
decomposed in countably many continuous functions.
In more recent times, it has been shown  by Abraham, Rubin and Shelah to be
consistent  that every monotone
function of size less than $\mathfrak c$ can be decomposed into countably
many continuous  functions \cite{ab.ru.sh}.
 Various results concerning $\mbox{dec}(\mathcal{B}_1,\mathcal{C})$
where ${\mathcal B}_1$ is the class of pointwise limits of continuous
functions and $\mathcal C$ is the class of continuous functions can be
found in  \cite{step.34},\cite{step.30} and \cite{sole}.

The question of determining $\mbox{dec}(\mathcal{C},\mathcal{D})$ where $\mathcal
D$ is the class of functions which are differentiable on their domain
was raised by  M. Morayne and J. Cichon. It was not known whether it
is consistent that $\mbox{dec}(\mathcal{C},\mathcal{D}) < 2^{\aleph_0}$ and
the best lower bound for $\mbox{dec}(\mathcal{C},\mathcal{D})$ has been noted by
Morayne to be the additivity of the null ideal. The following result
shows that 
$\mbox{dec}(\mathcal{C},\mathcal{D}) < 2^{\aleph_0}$ is indeed consistent.

\begin{propo} 
$\mbox{dec}(\mathcal{C},\mathcal{D}) \leq \cov({\mathcal D}_{2,1})$\label{Equiv}.
\end{propo}
\begin{proof} To show that
$\mbox{dec}(\mathcal{C},\mathcal{D}) \leq \cov({\mathcal D}_{2,1})$ let $\kappa = 
\mbox{cov}(\mathcal{D}_{2,1})$ and let $\{S_\eta\}_{\eta\in \kappa}$ be
a decomposition of $\Reals^2$ into 1-smooth sets. Given a continuous
function $f:\Reals \to \Reals$, let $f_\eta =  f\cap S_\eta$.

To see that $f_\eta$ is differentiable on its domain, let $x$ be in
the domain of $f_\eta$. Let $m$ be the slope of the tangent line to
$S_\eta$ at the point $(x,f_\eta(x))$ allowing the possibility that $m
= \infty$. Then, using the fact that $f$ is continuous, it is  
 easily verified that $f'_\eta(x) = m$.
\end{proof}

It is not known whether the inequality in Proposition~\ref{Equiv} can
be replaced by an equality. If the answer is yes, then this would
provided a better lower bound for $\mbox{dec}(\mathcal{C},\mathcal{D})$,
namely
$\mbox{dec}(\mathcal{C},\mathcal{D})^+ \geq 2^{\aleph_0}$.
However, it can be shown that there is a perfect set $A
\subseteq \Reals$ such that $A\times A$ can be covered by
$\mbox{dec}(\mathcal{C},\mathcal{D})$ 1-smooth sets.
Simply let $A$ be a perfect set such that there is
 $\theta \in (0,\pi/4)$ be such that 
if $a$ and $a'$ are distinct points in $A\times A$ then the angle formed
by the horizontal axis and the line connecting $a$ and $a'$ is
different from $\theta$. Let $H_\theta$ be the function which projects
$\Reals^2$ to the vertical axis along the line at angle $\theta$ with
respect to the horizontal axis. Let $H_{\pi/2}: \Reals\to \Reals$ be
the orthogonal projection onto the horizontal axis. Since
$H_\theta\restriction (A\times A)$ is one-to-one and $A\times A$ is
compact, it follows that $H_\theta^{-1}$ is continuous and, hence, so
is $H = H_{\pi/2}\circ H_\theta^{-1}$. Since the domain of $H$ is
compact, it can be extended to a continuous function on the entire
real line. 

Now suppose that $\{X_\eta\}_{\eta \in \kappa}$ are subsets of
$\Reals$ such that $H\restriction X_\eta$ is differentiable for each
$\eta \in \kappa$. To see that $H_\theta^{-1}(X_\eta)$ is a 1-smooth
set, suppose that $(x,y)\in H_\theta^{-1}(X_\eta)$ is a point at
which $H_\theta^{-1}(X_\eta)$ does not have a 1-dimensional tangent.
In this case it is possible to find distinct slopes $m_0$ and $m_1$
and sequences $\{(x^i_n,y^i_n)\}_{n\in\omega}$ for each $i\in 2$ such
that $$\lim_{n\to\infty}\frac{y - y^i_n}{x- x^i_n} = m_i$$ allowing
the possibility of an infinite limit. Let $w^i_n = H^{-1}_\theta
(x^i_n,y^i_n)$ and $w = H_\theta^{-1}(x,y)$. It follows from the
linearity of $H_\theta$ that there are distinct $M_0$ and $M_1$ such
that $$\lim_{n\to\infty}\frac{x - x^i_n}{w - w^i_n} = M_i$$ and this
contradicts the differentiability of $H$ at $w$ since $H(w^i_n) =
x^i_n$ and $H(w) = x$.

It has already been remarked that $\cov({\mathcal D}_{2,1}) \geq
\cov({\mathcal N})$. This will follow immediately from
Proposition~\ref{Equiv} and the following theorem of M. Morayne.

\begin{theor} There is a continuous function from $\Reals$ to $\Reals$
which \label{mora} can 
not be decomposed into fewer than $\cov({\mathcal N})$ differentiable
functions. \end{theor}
\begin{proof} Let $F:\Reals\to  \Reals$ be a continuous function\footnote{To find such
a function 
 let $\pi:2^\omega\times 2^\omega$ be the projection function, let $C$
be a Cantor set of positive measure, let $f_1:C\to 2^\omega$ 
and  $f_2:C\to 2^\omega\times 2^\omega$ be
homeomorphisms. Let $f=  f_1^{-1}\circ\pi\circ f_2 $ and let
$F\supseteq f$ be an arbitrary continuous extension of $f$.}  
such that there is a set $A$ of positive measure such that
$\card{F^{-1}\{a\}} = 2^{\aleph_0}$ for every $a\in A$. 
If $\kappa < \cov({\mathcal N})$ and $\{Z_\zeta\}_{\zeta\in\kappa}$ are
subsets of $\Reals$ let $Z_{\zeta}' = \{z\in Z_\zeta :
\card{F^{-1}\{F(z)\}} \geq 
\aleph_0\}$ then $\bigcup_{\zeta\in\kappa}F(Z_\zeta')\supseteq A$. It
is therefore possible to choose $\zeta\in \kappa$ such that the
measure of $F(Z_\zeta')$ is positive.

The following theorem of Banach can be found on page 278 of
\cite{Saks}: If $X\subseteq \Reals$ and $H:X\to \Reals$ is continuous,
then the set of points in $\Reals$ whose preimage under $H$ is
infinite is of measure zero if and only if the set of values attained
by $H$ at points at which $H$ is not differentiable has measure zero.
This  immediately implies that $F\restriction Z_{\zeta}'$ is not
differentiable at some point.\end{proof}

\section{Remarks and Open Questions}
An interesting byproduct of the results established in this paper has
to do with the difficulties encountered when iterating
with countable support beyond $\omega_2$. In this regard, it is
interesting to note a consequence of Corollary~\ref{LB1} and 
Theorem~\ref{main}. Since $\Sacks(1)$ is simply ordinary Sacks
forcing, it follows from Lemma~\ref{successor} and Lemma~\ref{limit}
that in the model obtained by iteratively adding 
$\omega_2$ Sacks reals to a model of  \CH \/ one obtains a model where
$\cov({\mathcal D}_{2,1}) = \aleph_1$. However, by forcing with the
product of $\aleph_3$ Sacks reals over this same model one obtains a
model where $2^{\aleph_0} = \aleph_3$ and hence, by
Corollary~\ref{LB1}, it follows that $\cov({\mathcal D}_{2,1}) \geq
\aleph_2$. A standard reflection argument yields that in some
intermediate model, obtained by forcing with the product of only
$\aleph_2$ Sacks reals, it must be  that $\cov({\mathcal D}_{2,1}) =
\aleph_2$. Homogeneity yields that forcing with the product of
$\aleph_2$ Sacks reals increases $\cov({\mathcal D}_{2,1})$ whereas, the
iteration does not. This yields
an answer to a question of Todorcevic who asked for a simple cardinal
invariant to  distinguish between the iteration and product of
$\omega_2$ Sacks reals. It also shows that there is no way of
extending the iteration of Sacks reals to obtain a model where
$2^{\aleph_0} >\aleph_2$ and so that most of the reasonable properties
of the usual iteration still hold --- reasonable here means, of
course, that which was used in the preceding argument. The following
are some natural questions which require overcoming this difficulty.
\begin{quest} Does the inequality $\cov({\mathcal D}_{2,1}) > \aleph_1$
imply that
$\cov({\mathcal D}_{2,1}) = 2^{\aleph_0}$ ?
\end{quest}

Similar questions can be asked about the ideals ${\mathcal S}_{2,1}$ and
${\mathcal C}^1_{2,1}$. The same is true of the following question.

\begin{quest}
Is it consistent that $$\cov({\mathcal D}_{n,m}) > \cov({\mathcal D}_{n',m'}) >
\cov({\mathcal D}_{n'',m''})$$ for some choice of integers?
\end{quest}

Other types of decompositions of the plane may also
be of interest. By a curve, will always be meant a one-to-one
function. Let ${\mathcal R}$ be the $\sigma$-ideal generated by
rectifiable curves in the plane, $\mathcal L$ the $\sigma$-ideal generated by
Lipshitz curves in the plane and ${\mathcal C}$ be the $\sigma$-ideal generated by
continuous curves   in the plane. Since ${\mathcal S}_{2,1}\subseteq {\mathcal C}\cap
  {\mathcal L}\cap {\mathcal R}$, it follows that $\cov({\mathcal X}) \leq
\cov({\mathcal S}_{2,1})$ if ${\mathcal X}\in \{ {\mathcal R}, {\mathcal L},{\mathcal C}\}$.
What else can be said about these ideals and their associated
invariants?

It has already been remarked in Section 3 that under various
 assumptions 
 $\cov({\mathcal C}^1_{m,n}) = {\mathfrak s}_n$.
\begin{quest} Is it consistent that  $\cov({\mathcal
C}^1_{m,n}) \neq {\mathfrak s}_n$?\end{quest}
\begin{quest}
Is it consistent that  $\cov({\mathcal
D}_{m,n}) \neq \cov({\mathcal D}_{m',n})$ for some $m$, $m'$ and $n$?\end{quest}

Finally, one can define related cardinal invariants by defining a set
to be $(n,m)$-smooth if the number of $m$-dimensional tangent
planes  at each point is no more than  $n$. In effect, this allows corners because, for example, the graph
of the absolute value function in $\Reals^2$ is $(2,1)$-smooth. One can
then define the associated ideals and consider their cardinal
invariants as before. What
can be said about these? For example if the plane is covered by
$\lambda$ $(7,1)$-smooth sets is it also covered by $\lambda$
$(6,1)$-smooth sets?

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\end{document}
