= Solution
{c}
The <inverse tangent> $T(x)=\arctan x$ is a <homeomorphism> from the <real line> with its <Euclidean metric> onto $I=(-\pi/2,\pi/2)$ with its <subspace topology>: $T$ and its inverse $\tan$ are <continuous>. It is also an <isometry> from $(\mathbb R,\rho)$ onto $I$, because $\rho(x,y)=|T(x)-T(y)|$. Consequently, for $U\subseteq\mathbb R$,
$$
\boxed{U\text{ is }d\text{-open}\iff T(U)\text{ is open in }I\iff U\text{ is }\rho\text{-open}.}
$$
For a direct <open ball> interpretation, $B_\rho(x,r)=T^{-1}((T(x)-r,T(x)+r)\cap I)$ is $d$-open by <continuity> of $T$. Conversely, <continuity> of $\tan$ at $T(x)$ supplies a sufficiently small $\rho$-ball inside any prescribed $d$-ball about $x$. Thus both <metrics> generate exactly the same <open sets>.
Back to article page
