Infinite temporal distances in Anti-deSitter space The following is a figure from Christodoulou. 

I am trying to understand the following claim. 

In Anti-deSitter space there are points $p$ and $q$ as shown for which (the temporal distance) $\tau(p,q) = \infty$. For, the length of the timelike segment of the casual curve joining $p$ and $q$ in the figure can be made arbitrarily large by making the segment approach the hyperboloid. 

Why is this true? In particular, why are the regions beyond the hyperboloids inaccessible in finite proper time?
 A: What you see is AdS in the coordinates where
$$d s^2 = \frac{1}{1 + \bar{x}^2/R^2 - t^2/R^2} (- dt^2 + d \bar{x}^2)$$
I like to keep things dimensional, so I introduced the AdS radius $R$. Consider a massive particle (thus, on a time-like trajectory) approaching the coordinate position which I will call the "edge of the universe"
$t^2-\bar{x}^2 = R^2$
The four-velocity normalization $g_{\alpha \beta} u^\alpha u^\beta = -1$, $u^\mu = dx^\mu/d\tau$, will force it to elapse a diverging amount of proper time $\tau$ for any small amount of the coordinate time $t$ to pass. 
Now we know that the temporal distance $\tau(p,q)$ is defined as the supremum over proper times of any time-like trajectories connecting the two points $p,q$. In Minkowski and de-Sitter space-time without holes this means that you simply choose the connecting geodesic and its proper time. This is easy to see in Minkowski and by recalling from special relativity that proper time passes slower when you move faster with respect to the lab frame. Now choose your lab frame to be the one in which both $p,q$ happen at the same spatial coordinate point just at a different time (such a frame always exists), and you see that the space-time trajectory with longest elapsed proper time is simply the one waiting at that coordinate point from event $p$ to $q$.
However, in Anti-de-Sitter, a weird thing happens. One can have two points at such a coordinate separation such that it pays off to travel faster to the edge of the universe, elapse a lot of proper time, and come back. Consider points $p,q$ which are simply at $\bar{x} = 0$ and at some $t$ separation. How long does the $t$ separation need to be to allow for this kind of travel there and back again? For sure the edge of the universe must be in the causal future of $p$ and the causal past of $q$. If you draw a diagram, the edges of the causal future and the causal past will be diagonals at $45^\circ$ in the right directions (the coordinates are conformally flat). Then we see on the image you posted that $p$ and $q$ are exactly such points (if they were at $\bar{x} = 0$ and inside the diamond on the picture, they would not have this property). One can thus send the time-like trajectory on an almost light-like (null) trajectory to reach the edge of the universe, spend an arbitrary large amount of proper time at arbitrarily small cost of coordinate time and return to $q$ on another almost light-like trajectory. It is obvious that once this is possible, there is no upper bound on the elapsed proper time and $\tau(p,q) = \infty$
