= Solution
The original spherical coordinates have $t\in\mathbb R$, $0\leq r<\infty$, $0\leq\theta\leq\pi$ and $0\leq\phi<2\pi$, with $\phi$ periodic. The angular coordinates degenerate at the polar axis and all angles identify at $r=0$; a regular spherical chart uses $r>0$, $0<\theta<\pi$ and suitable overlapping angular patches.
The <retarded time> $u=t-r$ and advanced coordinate $v=t+r$ both range over $\mathbb R$, with the joint restriction $u\leq v$. Their inverse relations are $t=(u+v)/2$ and $r=(v-u)/2$, so
$$
d\widehat s^2=du\,dv-\frac14(v-u)^2d\Sigma^2.
$$
Set $p=\arctan u$, $q=\arctan v$. Then
$$
-\frac\pi2<p\leq q<\frac\pi2,\qquad du=\sec^2p\,dp,\qquad dv=\sec^2q\,dq.
$$
The <conformal factor> is $\Omega=2\cos p\cos q>0$. Since $(\tan q-\tan p)\cos p\cos q=\sin(q-p)$, multiplication by $\Omega^2$ gives
$$
ds^2=4dp\,dq-\sin^2(q-p)d\Sigma^2.
$$
Finally $T=p+q$ and $R=q-p$ give $4dp\,dq=dT^2-dR^2$. Hence
$$
\boxed{ds^2=dT^2-dR^2-\sin^2R\,d\Sigma^2.}
$$
This is a region of the <Einstein static universe>, with positive time and negative spatial signature. The exact joint coordinate domain is
$$
\boxed{-\pi<T<\pi,\qquad 0\leq R<\pi-|T|,}
$$
or equivalently $R\geq0$ and $|T|+R<\pi$. The regular radial chart has $R>0$; $R=0$ is the ordinary centre. The angular ranges are unchanged. It is important not to treat the stated individual ranges of $T,R$ as independent. The inverse map and <conformal factor> provide a useful check:
$$
t=\frac{\sin T}{\cos T+\cos R},\qquad
r=\frac{\sin R}{\cos T+\cos R},\qquad
\Omega=\cos T+\cos R.
$$
The resulting <Minkowski conformal compactification> has a triangular radial <Penrose diagram>. The two sloping boundary edges are
$$
\mathcal I^+:\ T+R=\pi,\quad 0<R<\pi,\qquad
\mathcal I^-:\ T-R=-\pi,\quad 0<R<\pi.
$$
These are <future null infinity> and <past null infinity>. The three vertices are <future timelike infinity> $i^+=(\pi,0)$, <past timelike infinity> $i^-=(-\pi,0)$ and <spacelike infinity> $i^0=(0,\pi)$. They are ideal boundary points, not events of physical <Minkowski spacetime>.
For a complete inertial <timelike geodesic>, $\mathbf x=\mathbf b+\mathbf vt$ with $|\mathbf v|<1$. As $t\to+\infty$, both $t-r$ and $t+r$ tend to $+\infty$, so $p,q\to\pi/2$ and the curve ends at $i^+$. As $t\to-\infty$, both tend to $-\infty$, giving $i^-$. These endpoints occur at infinite physical <proper time>.
For a complete <null geodesic>, write $\mathbf x=\mathbf b+\mathbf nt$, $|\mathbf n|=1$. To the future, $r=t+\mathbf n\cdot\mathbf b+O(1/t)$, so $u\to-\mathbf n\cdot\mathbf b$ is finite and $v\to+\infty$. Thus $q\to\pi/2$ with $p$ finite: the endpoint lies on $\mathcal I^+$. To the past, $v\to-\mathbf n\cdot\mathbf b$ remains finite while $u\to-\infty$, giving $\mathcal I^-$. Physical <affine parameter> is infinite at both ends. Along a radial null path, $dT=\pm dR$, so the diagram represents its null directions at $45$ degrees.
For a complete <spacelike geodesic>, take $t=t_0+a\lambda$, $\mathbf x=\mathbf b+\mathbf w\lambda$ with $|\mathbf w|>|a|$. At either end $r>|t|$ asymptotically, so $u\to-\infty$, $v\to+\infty$ and $(T,R)\to(0,\pi)$. Both ends therefore reach $i^0$, again at infinite physical <affine parameter>. These are the <geodesic endpoints in Minkowski conformal compactification>.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2007/iii/paper-61-conformal-geodesics.png]
{title=Radial conformal compactification of Minkowski spacetime with images of complete timelike, null and spacelike geodesics}
Only <conformal preservation of null geodesic paths> is guaranteed. The pictured timelike and spacelike curves are images of physical geodesics, not generally geodesics of the unphysical metric. For a null path, an unphysical affine parameter satisfies $d\widetilde\lambda/d\lambda\propto\Omega^2$. Since $\Omega=O(1/|\lambda|)$ near a generic null-infinity endpoint, the unphysical parameter can converge while the physical parameter diverges. At $i^\pm$ and $i^0$, the usual smooth-boundary condition $d\Omega\ne0$ degenerates; these vertices should not be treated as regular points of null infinity.
\b[Complete Minkowski spacetime has no black-hole event horizon.] Every physical event can send an outgoing null signal to $\mathcal I^+$, so the black-hole region $\mathcal M\setminus J^-(\mathcal I^+)$ is empty. There is likewise no particle or future event horizon for an eternal inertial observer: any finite event can eventually signal that observer. Null infinity is a conformal boundary, not a physical horizon. Uniformly accelerated observers can have <Rindler horizons>, but that is an observer-dependent restriction and is not a horizon of the full inertial <Minkowski spacetime>.
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