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Minkowski spacetime conformal infinity: why not allow the full range of $R$?

Let $(\mathbb{R}^4,\eta)$ be Minkowski spacetime. We want to describe infinity as a place in some bigger manifold containing Minkowski spacetime. The idea is to work with null incoming/outgoing geodesics and compactify along these, to preserve the causal structure.

So define the advanced/retarded null coordinates $u=t-r$ and $v=t+r$. These have ranges $(-\infty,\infty)$ both with restriction $u\leq v$.

Then we define $U = \arctan u$ and $V=\arctan v$ both with ranges $(-\pi/2,\pi/2)$ and $U\leq V$.

We finaly define $$T=V+U,\quad R=V-U.$$

These have ranges $0\leq R<\pi$ and $|T|+R<\pi$. These coordinates with these ranges still just describe Minkowski spacetime, albeit in a rather awkward coordinate system.

Then we want to add the missing points describing infinity. So the obvious idea would be to extend the ranges of the coordinates allowing them to describe a bigger manifold and picture Minkowski spacetime as the submanifold defined by the above constraints inide of it.

Now the canonical way to do it is to extend $T$ to cover $(-\infty,\infty)$ and to extend $R$ to cover $0\leq R\leq \pi$. In this way the bigger manifold is $\mathbb{R}\times S^3$.

Now, why is that? If we further allow $R$ to run from $(0,\infty)$ or even $(-\infty,\infty)$ we still get a bigger manifold, properly containing the previous one, which in turn properly contains Minkowski spacetime.

There is certainly a reason behind this choice, but I can't see it.