\section{Preliminaries}

\textbf{Metric Spaces}. Let (X,d) be a metric space. The
\textit{diameter} of a set B is denoted by diam(B).
 A \textit{path} in X is a map $\gamma : I \rightarrow X $ where I is an interval
in $\mathbb R$. A path $\gamma $ joins two points x and y in X if I [a,b] and $\gamma$(a) = x , $\gamma$(b) = y . The path $\gamma$ is called an
infinite ray starting from $x_0$ if I=[0,$\infty$) and $\gamma(0)=x_0$
. A geodesic, a geodesic ray or a geodesic segment in X is an
isometry $\gamma : I \rightarrow X$ where I is $\mathbb R$ or $[0,\infty)$ or
a closed segment in $\mathbb R$. We use the term geodesic,
geodesic ray etc for the images of $\gamma$ without discrimination. On
a path connected space X given two points x,y we define the path
metric to be $\rho(x,y) = inf \{length(p)\}$ where the infimum is
taken over all paths $p$ that connect $x$ and $y$ (of course
$\rho(x,y)$ might be infinite). It is easy to see that inside a
ball B(x,n) of the hyperbolic plane or the euclidian plane the
path metric and the usual metric coincide. A metric space $(X,d)$
is called \textit{geodesic metric space} if $d=\rho $ (the path
metric is equal to the metric).


 \textbf{Hyperbolic Spaces}. Let (X,d) be a metric space .
Given three points x,y,z in X we define the \textit{Gromov
Product} of x and y with respect to the basepoint w to be
:$$(x|y)_w = \frac{1}{2} (d(x,w)+d(y,w)-d(x,y))$$  A space is said
to be \textit{$\da$- hyperbolic} if for all x,y,z,w in X we have:
$$(x|z)_w \geq min\{(x|y)_w,(y|z)_w\} - \da$$ A sequence of
points $\{x_i\}$ in X is said to converge at infinity if:
$$\lim_{i,j\rightarrow \infty} (x_i|x_j)_w = \infty $$ Two sequences
$\{x_i\}$ and $\{y_i\}$ are equivalent if: $$\lim_{i,j\rightarrow \infty}
(x_i|y_j) = \infty$$ This is an equivalence relation which does
not depend on the choice of w (easy to see). The boundary $\bd X$
of X is defined as the set of equivalence classes of sequences
converging at infinity. Two sequences are 'close' if
 $\liminf_{i,j\rightarrow \infty}(x_i|y_j) $ is big. This defines a
 topology on the boundary.

 The boundary of every proper hyperbolic space is a
compact metric space.

 If $X$ is a geodesic hyperbolic metric space
and $x_0\in X$ then $\bd X$ can be defined as the set of geodesic
rays from $x_0$ where we define to rays to be equivalent if they are
contained in a finite Hausdorf neighborhood of each other. We equip
this with the compact open topology.


A metric $d$ on the boundary $\bd X$ of X is said to be
\textit{visual} if there are $x_0 \in X , a>1 $ and $ c_1,c_2 > 0$
such that $$c_1 a^{-(z,w)_{x_0}}\leq d(z,w) \leq
c_2a^{-(z,w)_{x_0}}$$ for every z,w in $\bd X$. The boundary of a
hyperbolic space always admits a visual metric (see \cite{GRO1}).

A hyperbolic space X is called \textit{visual} if for some $x_0
\in X$ there exists a $D>0$ such that for every $x \in X$ there
exists a geodesic ray $r$ from $x_0$ in $\bd X$ such that $d(x,r)
\leq D$ (see more on \cite {Bo-S}). It is easy to see that if
$X$ is visual with respect to a base point $x_0$ then it is visual
with respect to any other base point.
\newline

\textbf{Topological Dimension}. A covering $\{B_i\}$ has
\textit{multiplicity} $n$ if no more than $n + 1$ sets of the
covering have a non empty intersection. The \textit{mesh} of the
covering is the largest of the diameters of the $B_i$.

We will use in this paper the following definition of topological
dimension for compact metric spaces which is equivalent to the
other known definitions : A compact metric space has
\textit{dimension $\leq n$} if and only if it has coverings
of arbitrarily small mesh and order $\leq n$ . (see \cite
{H-W})\newline


\textbf{Asymptotic Dimension}.
 A metric space Y is said to be d - disconnected or that
it has dimension 0 on the d - scale if  $$ Y \,\, = \bigcup_{i\in I}
 B_i $$ such that: $sup \{ diam B_i , i \in I \}= D < \infty
 $, dist($B_i,B_j$) $\geq d$ $\forall  i \neq j$ where
dist($B_i,B_j$) = $\inf $ \{dist(a,b) a$\in B_i$ , b $\in B_j
\}$\newline
 \textit{(Asymptotic Dimension 1)}. We say that a space X has asymptotic
dimension n if n is the minimal number such that for every $d > 0$
 we have : $X =\bigcup X_k$ for k = 1,2, ... n and all $X_k$ are d-disconnected. We then write asdim = n

 We say that a covering  $\{B_i\}$  has \textit{d - multiplicity} ,
k if and only if every d - ball in X meets no more than k sets $B_i$
of the covering.A covering has \textit{multiplicity} n if no more
than n + 1 sets of the covering have one a non empty intersection. A
covering $\{ B_i \} \,i \in I$ is \textit{D - bounded} if diam
$(B_i) \leq D \,\forall \,i\in I$
\newline \textit{(Asymptotic Dimension 2)}. We say that a space $X$ has asdim
= n if n is the minimal number such that $\forall \,d > 0$ there
exists a covering of X of uniformly  D - bounded sets $B_i$ such
that d - multiplicity of the covering $\leq n+1$. The two
definitions are equivalent. (see \cite {GRO1})\newline



 \textbf{The Hyperbolic Plane}. The hyperbolic plane
$\mathbb H$$^2$ is a visual hyperbolic space of bounded geometry.
It is easy to see that $asdim\mathbb H$$^2 = 2$ (see \cite
{GRO2}).
We will use  the
standard model of the hyperbolic plane given by the interior of a
disk in $\mathbb R
$$^2$.


\section{Constructing The "COMB" Space}
Let $\mathbb {H}$$^2$ be the hyperbolic plane and let
$a_1,a_2,....$ be  geodesic rays starting from a point $x_0$ and
extending to infinity such that the angle between $a_n,a_{n+1}$ is
$\frac{\pi}{2^n}$.

Let $S(a_n,a_{n+1})$ be the sector defined by the rays
$a_n,a_{n+1}$. In other words $S(a_n,a_{n+1})$ is the convex
closure of $a_n,a_{n+1}$.

 Since geodesics diverge in $\mathbb {H}$$^2$ there
is an $x\in S(a_n,a_{n+1})$ such that the ball of radius $n$ and
center $x$, $B(x,n)$ is contained in $S(a_n,a_{n+1})$. Let $N_n$
be such that $B(x,n)\subset B(x_0,N_n)$. Let
$$S(a_n,a_{n+1},N_n)=S(a_n,a_{n+1})\cap B(x_0,N_n)$$

Let's call $K_n$ the upper arc of $S(a_n,a_{n+1},N_n)$, i.e.
$$K_n=S(a_n,a_{n+1},N_n)\cap \partial B(x_0,N_n)$$

 We subdivide
$K_n$ into small pieces of length between $1/2$ and 1 marking the
vertices.
Then we consider the geodesic rays starting from $x_0$
to every vertex we defined and we extend them to infinity.

So we arrive at the "COMB" space which is the union of all the
$S(a_n,a_{n+1},N_n)$ together with these rays and looks like this:


\begin{figure}[htb]
\epsfxsize 10.0cm \epsfysize 8.0cm \epsfbox{comb1.eps} \caption{Comb
Space}

\end{figure}

\section{The Properties Of "COMB" Space}
\begin{itemize}
\item[a)] $dim(\partial X)=0$. For every $n$ we have that $K_n$ is
bounded. That means that we define a finite number of vertices on
every $K_n$ so we add a finite number of geodesic rays. So, all
the infinite geodesic rays are countable. So $\partial X $ is
countable. Now a countable metric space has dimension 0 (see \cite
{H-W} page 18). So dim($\partial X $)=0
\newline
\item[b)] X is a hyperbolic space with the "path" metric. That is
true since every pair of points of $X$ can be joined by a path of
finite length. Also let $l$ be a closed curve of X then $l$ is a
closed curve in $\mathbb H$$^2$ and $length(l)_X \geq
length(l)_{{\mathbb H} ^2}$. But since $\mathbb H$$^2$ is
hyperbolic we have the isoperimetric inequality $Area(l) \leq
c*length(l)_{{\mathbb H} ^2}$ so $Area(l) \leq c*length(l)_X$
which means that X is hyperbolic.(see \cite{GRO1}, \cite{Bow})
\newline
\item[c)] $asdim(X) = 2$. That is because $X$ contains arbitrarily
large balls $B(x,n)\subset \mathbb{H}$$^2$ for every $n \in
\mathbb N$.
\newline
\item[d)] $X$ is a visual hyperbolic space with $D = 1$ since for
every $x$ in $X$ there exists a geodesic from $x_0$ to $x$. Let's
call that $g_1$. If $g_1$ can be extended to infinity then we have
nothing to prove. Let $g_1$ be finite ,then $x$ must belong to a
sector $S(a_n,a_{n+1},N_n)$. We extend $g_1$ until it meets $K_n$
at a point $v_1$. Then by the construction of $X$ there exists an
infinite geodesic $r$ corresponding to the vertex on $K_n$ $v$
such that $d(v_1,v)$ is less than 1. Then obviously $d(x,r)$ is
less than 1.\newline


\end{itemize}

So $X$ is a visual hyperbolic metric space such that that $asdim X
> dim \bd X + 1$.

We remark that it is not very hard to see that $X$ is
quasi-isometrically embedded in $\mathbb{H}$$^2$.






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