\documentclass[a4paper,12pt]{article}
\usepackage{epsfig}
\newcommand{\ds}{\displaystyle}
\newcommand{\un}{\underline}
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\begin{document}

{\bf Question}

Given two sets $X$ and $Y$ in ${\bf H}$, define
\[ {\rm d}_{\bf H}(X,Y) = \inf\{ {\rm d}_{\bf H}(x,y)\: |\: x\in X,
y\in Y\}. \]

\noindent Now, let $\ell_1$ be the hyperbolic line contained in
the Euclidean line $\{ {\rm Re}(z) = 4 \}$, let $\ell_2$ be the
hyperbolic line contained in the Euclidean line $\{ {\rm Re}(z) =
-14 \}$, and let $\ell_3$ be the hyperbolic line contained in the
Euclidean circle with Euclidean center $0$ and Euclidean radius
$1$.

\medskip
\noindent Calculate the three numbers ${\rm d}_{\bf H}(\ell_1,
\ell_2)$, ${\rm d}_{\bf H}(\ell_2, \ell_3)$, and ${\rm d}_{\bf
H}(\ell_1, \ell_3)$.

\medskip
\noindent Use this to prove that ${\rm d}_{\bf H}(\cdot, \cdot)$
is {\em not} a metric on the set of subsets of ${\bf H}$.

\medskip

{\bf Answer}

\un{$d_{\bf{H}}(\ell_1\ell_2)=0$}

Consider (for $\lambda>0$) $4+\lambda \in \ell_1$ and $-14+\lambda
\in \ell_2$. The horizontal euclidean line segment from
$4+\lambda$ has hyperbolic length $\ds\frac{18}{\lambda}$ and so
$d_{\bf{H}}(4+\lambda,-14+\lambda)<\ds\frac{18}{\lambda}$ (since
the hyperbolic distance is the length of the hyperbolic line
segment, which is less than the hyperbolic length of the euclidean
line segment).

Hence $d_{\bf{H}}(\ell_1\ell_2) \leq
{\rm{inf}}\left\{\left.\ds\frac{18}{\lambda}\right|\lambda>0\right\}=0$

\bigskip

\un{For $d_{\bf{H}}(\ell_1\ell_3)$ and
$d_{\bf{H}}(\ell_2\ell_3)$}, these two use the same method:

\un{$d_{\bf{H}}(\ell_1\ell_3)$} first draw the picture.

\begin{center}
\epsfig{file=407-9-1.eps, width=80mm}
\end{center}

The distance from $\ell_1$ to $\ell_3$ is equal to the hyperbolic
length of the \un{common\ perpendicular}.

\newpage
Any hyperbolic line perpendicular to $\ell_1$ is contained in a
euclidean circle with center 4: it intersects $\ell_3$ at
$e^{i\theta}$ if (by the euclidean pythagorean theorem)

\hspace{1in} $\begin{array} {l}
16=1+|4-e^{i\theta}|^2\\16=1+16-4e^{i\theta}-4e^{-i\theta}+1\\ 8
\cos(\theta)=2\\
\cos(\theta)=\ds\frac{1}{4}\\\sin(\theta)=\ds\frac{\sqrt 15}{4}\ \
\ \theta \sim 1.3181... \end{array}$

Parametrize the common perpendicular by $f(t)=4+re^{it}$ where
\begin{eqnarray*}
r=|4-e^{i\theta}| & = & |4-\cos(\theta)-i\sin(\theta)|\\ & = &
\sqrt{(4-\frac{1}{4})^2+\frac{15}{16}}\\ & = & \ds\frac{4\sqrt
15}{4} = \sqrt 15 \end{eqnarray*}

and where $\frac{\pi}{2} \leq t \leq \pi-\alpha$, where $\alpha$
is as in the picture.

$\cos(\alpha)=\ds\frac{1}{\sqrt 15}(4-\cos(\theta))=\ds\frac{\sqrt
15}{4}$

$\sin(\alpha)=\ds\frac{1}{\sqrt 15}\sin(\theta)=\ds\frac{1}{4}$

So, \begin{eqnarray*} d_{\bf{H}}(\ell_1\ell_3) & = &
\ds\int_{\frac{\pi}{2}}^{\pi-\alpha} \ds\frac{1}{\sin(t)} \,dt\\ &
= & \ln|\csc(\pi-\alpha)-\cot(\pi-\alpha)|-\ln(\alpha)|\\ & = &
\ln|\csc(\alpha)+\cot(\alpha)|\\ & = & \ln
\left|\ds\frac{1+\cos(\alpha)}{\sin(\alpha)}\right|\\ & = &
\un{\ln(4+\sqrt 15)} \end{eqnarray*}

\newpage
\un{$d_{\bf{H}}(\ell_2\ell_3)$}

\begin{center}
\epsfig{file=407-9-2.eps, width=80mm}
\end{center}

$\phi$ determined by $|-14-e^{i\phi}|+1=14^2$

$$14^2+14e^{i\phi}+14e{-i\phi}+2=0$$

\begin{eqnarray*} 28\cos(\phi) & = & -2\\ \cos(\phi) & = &
\ds\frac{-1}{14}\\ \sin(\phi) = \ds\frac{\sqrt {195}}{14}
\end{eqnarray*} (so $\phi$ is a bit more than $\frac{\pi}{2}$, as
indicated by the picture)

$r=|-14-e^{i\phi}=\sqrt{(14+\cos(\phi))^2+\sin^2(\phi)} =
\sqrt{195}$

\begin{eqnarray*} d_{\bf{H}}(\ell_2\ell_3) & = &
\ds\int_{\beta}^{\frac{\pi}{2}} \ds\frac{1}{\sin(t)} \,dt\\ & = &
\ln|\csc(\frac{\pi}{2})-\cot(\frac{\pi}{2})|-\ln|\csc{\beta}-\cot{\beta}|\\
& = & -\ln\left|\ds\frac{1-\cos(\beta)}{\sin(\beta)}\right|\\ & =
& \ln\left|\ds\frac{\sin(\beta)}{1-\cos(\beta)}\right|\ \
\cos(\beta)=\ds\frac{\sqrt {195}}{14}\ \
\sin(\beta)=\ds\frac{1}{14}\\ & = &
\ln\left|\ds\frac{1}{14-\sqrt{195}}\right| = \ln(14+\sqrt{195})
\end{eqnarray*}

Note that since $d_{\bf{H}}(\ell_1\ell_2)=0$ but $\ell_1 \ne
\ell_2$, this cannot be a metric on the set of subsets of
${\bf{H}}$.

\end{document}