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Let $x_n=(\pi /2+2\pi n{)}^{-1}$. The points $p_n=(x_n,1)$ lie on the specified graph and converge in ${\mathbb{R}}^2$ to $(0,1)$. They form a Cauchy sequence in the induced metric, but their only possible limit is absent from the space. Adding $(0,0)$ does not supply that missing limit. Hence the specified space is not complete.

For the unheaded request on $\mathbb{R}$, use the arctangent pullback metricPositivity, symmetry and the triangle inequality follow from the ordinary distance, and injectivity of $\arctan$ gives $d(x,y)=0$ only if $x=y$. The sequence $n$ is Cauchy for this metric, because $\arctan n\to \pi /2$, but cannot converge to any finite real $x$: that would require $\arctan x=\pi /2$. Thus this metric makes $\mathbb{R}$ incomplete, even though it induces its usual topology.

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.