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{\bf Question} %note ref to qu 1%

Prove that the identity map $g: {\bf H}\rightarrow {\bf H}$,
defined by $g(x) = x$, gives a homeomorphism between the metric
space $({\bf H}, {\rm d}_{\bf H})$ and the metric space $({\bf H},
f)$, where $f$ is as defined in Problem 1 of this sheet.
\medskip

{\bf Answer}

Let $\begin{array} {rcl} \cup_{\epsilon}(x) &amp; = &amp; \{y \in
{\bf{H}}|d_{\bf{H}}(x,y)&lt;\epsilon\}\ {\rm{and}}\\
\cup_{\epsilon}^f (x) &amp; = &amp; \{y \in {\bf{H}}|f(x,y)&lt;\epsilon\}
\end{array}$

\bigskip

First note that $g$ is a bijection, and so it remains only to show
that $g:({\bf{H}},d_{\bf{H}}) \rightarrow ({\bf{H}},f)$ and
$g:({\bf{H}},f) \rightarrow ({\bf{H}},d_{\bf{H}})$ are continuous
(as $g^{-1}=g$).

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$g:({\bf{H}},d_{\bf{H}}) \rightarrow ({\bf{H}},f)$ is continuous
at $a$ if for every $\epsilon &gt;0$ there is $\delta &gt;0$ so that
$g(\cup_{delta}(a)) \subseteq \cup_{\epsilon}^f(a)$; that is, that
$\cup_{delta}(a) \subseteq \cup_{\epsilon}^f(a)$.

\bigskip

\hspace{1in} $\bullet$\ If $d_{\bf{H}}(a,x)&lt;\delta$, then
$f(a,x)=\ds\frac{d_{\bf{H}}(a,x)}{1+d_{\bf{H}}(a,x)}$


\bigskip

$&lt;d_{\bf{H}}(a,x)&lt;\delta,$

\bigskip

so we may take $\delta=\epsilon$.

\bigskip

Going in the other direction, consider $g:({\bf{H}},f) \rightarrow
({\bf{H}},d_{\bf{H}})$.

\bigskip

Given $\epsilon&gt;0$, choose
$\delta=\ds\frac{\epsilon}{1+\epsilon}$.

\bigskip

Then, if
$f(a,x)=\ds\frac{d_{\bf{H}}(a,x)}{1+d_{\bf{H}}(a,x)}&lt;\delta=\ds\frac{\epsilon}{1+\epsilon}$
then $d_{\bf{H}}(a,x)&lt;\epsilon$ as desired (since
$g(t)=\ds\frac{t}{1+t}$ is increasing in $t$).
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