Why does $\mathbb{R}$ have two infinities instead of one?

Question

I know as a matter of fact, that $\mathbb{R}$ compactifies to a circle $S^1$. So there should, in my visualization, exist a single infinity. If I want to go from $S^1$ back to $\mathbb{R}$ I have to pick any point on $S^1$ and perform a cut. This cut can be done in two possible orientations. And the infinity in $\mathbb{R}$ is two fold separated.

Does this picture generalize to something obvious? I am eager to see a connection between this, and the existence of an orientation to a Riemannian manifold.

Answer 1

There are two natural compactifications of the reals:

• you add to to one point at infinity and you get $S^1$;
• you add to it two points at infinity and you get $[0,1]$.

In distinct contexts, you may use one of them or the other one.

Answer 2

You are right, we can compactify $\mathbb R$ by adding one point to create a space homeomorphic to the circle $S^1$. In fact, one-point compactifications exist in general.

However, in many cases we would want more specific information about the behavior of some topological space "at infinity" than a one-point compactification can afford us. A useful concept related to this is the idea of the ends of a topological space. The idea is to choose a nested family of compact sets whose union is the whole space, and then say something about the connected components of the complement at each stage. In the case of $\mathbb R$, there are two ends, "$+\infty$" and "$-\infty$". The study of ends of various kinds of topological spaces, including manifolds, is a rich area of study!

Answer 3

The correct concept here is that of a compactification, which is a compact space containing the original space as a dense subspace. You list two different compactifications of $\mathbb R$, the one point, and the "two point."

These can be characterized by which real valued functions can be extended to the compactification. For the one point compactification, the limits on the left and right have to be the same. For the two point, the two limits simply have to exist. But it doesn't end there. Considering the homeomorphic space $(0,1)$, the topologist's sine curve is also a compactification, adding an entire segment at the left hand side. This allows you to extend, in particular, $\sin(1/x)$.

The biggest compactification is the Stone-Cech compactification, which allows all bounded real valued functions to be extended. This is a seemingly nice property, but the resulting space is horrifyingly complicated and huge in cardinality.

Answer 4

In essence: the complement of a compact set in $\mathbb{R}$ has $2$ unbounded connected components, while in $\mathbb{R}^n$ ($n>1$) there is but one....

Answer 5

The distinguishing factor here is order. $\mathrm R$ is the unique complete, ordered field (well, provided it's also archimedean).

The complex plane has no such order. Thus, the two point compactification can give us additional information in a non-ambiguous way than a one-point compactification would, when we need such information -- this is always linked to the order on the real line.

In sum, it depends on what one is doing and which of the extensions is more convenient for what you're doing. However, sometimes we want to know not just that some quantity becomes unbounded, but whether it goes to infinity through negative or positive values -- direction of blowup tells us more than just mere blowup.