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First use the original null coordinates $u=t+m/\lambda$, $v=t-m/\lambda$. They obey $u-v=2m/\lambda$, and the radial metric tensor becomesApply the given substitutions $u=\tan (U/2)$ and $v=-\cot (V/2)$. The identitiesgive the analytic extension of a spacetimeThe poles of the individual tangent and cotangent substitutions are now ordinary coordinate lines wherever $\cos ((U-V)/2)\ne 0$. Set $T=(U+V)/2$ and $X=(U-V)/2$. Allowing these coordinate functions to continue across every such line givesThis is the universal cover of two-dimensional anti-de Sitter spacetime of radius $m$, times a two-sphere of radius $m$: the simply connected Bertotti-Robinson spacetime. In defining this maximal analytic extension, $T$ ranges over the whole real line without periodic identification. A periodic identification would instead create closed timelike curves and would be an additional global choice.

For clarity, the original patch is embedded by the branches $-\pi <U<\pi$ and $0<V<2\pi$. The inequality $u>v$ is equivalent on these branches to $|U-V|<\pi$, so the original patch occupiesThe first spacetime extension in part (a) occupies the larger band $0<T-X<2\pi$, with $|X|<\pi /2$. Its finite-$v$ Killing horizon is the interior line $T+X=\pi$. The edges $T-X=0,2\pi$ of that band are further regular Poincaré horizons; continuing through them produces the full conformal strip of two-dimensional anti-de Sitter spacetime.

The original coordinate $\lambda$ has a smooth continuation throughout the strip:Consequently the complete set of zero-level Killing horizons isThese are two alternating families of parallel null lines; different families meet only at the excluded conformal boundaries, never inside the spacetime. The continuation of the original Killing vector field isThe field never vanishes in the interior, including on the Killing horizons.

To obtain the Penrose diagram, suppress the two-sphere and multiply by the conformal factor ${\cos }^{2}X/m^2$. The resulting radial metric tensor is $-d{T}^2+d{X}^2$, so the diagram is an infinitely tall strip of width $\pi$, with null lines at 45 degrees and timelike edges $X=\pm \pi /2$. Generically,At the special edge points where the two families of Killing horizons end, the displayed formula is a $0/0$ limit and $\lambda$ has no unique limiting value. Thus the infinity labels on the diagram refer to the open edge segments, not those vertices. Each interior point of the diagram represents a two-sphere of fixed radius $m$; neither edge is a centre where that two-sphere shrinks.

Finally, these timelike edges are conformal boundaries, not omitted regular points of a further spacetime extension. One way to establish maximality is geodesic completeness. The two-dimensional anti-de Sitter spacetime factor embeds locally as $Z\cdot Z=-m^2$ in a flat space with metric signature $(-,-,+)$. Along an affinely parametrized geodesic with constant $ϵ=\dot{Z}\cdot \dot{Z}$, the ambient equation isIts oscillatory, linear or hyperbolic solutions for negative, zero or positive $ϵ$ exist for all real affine parameters and lift to the universal cover. The two-sphere is also geodesically complete, and geodesics of the product split into geodesics of its factors. Hence $\tilde{g}$ is geodesically complete. A proper regular spacetime extension would furnish a geodesic reaching its boundary in finite affine parameter, contradicting that completeness. This proves that the displayed simply connected analytic spacetime is maximal.