Past exam of the metric spaces and complex analysis course of the University of Oxford / 2021 / 1 / b

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 $U\subseteq \mathbb{R}$,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.

The arctangent pullback metric is not complete. Take $x_n=n$. Since $\arctan n\to \pi /2$, its transformed sequence is Cauchy, and hence $(x_n)$ is a Cauchy sequence for $\rho$. If it converged in $\rho$ to $x\in \mathbb{R}$, thenwould force $\arctan x=\pi /2$, which is impossible. The missing endpoint of $I$ explains the failure of completeness, even though part (i) shows that the metric topologies agree.