4
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ Because there is an intrinsic order ($1$ is clearly distinguishable from $-1$ in a way $i$ is not from $-i$ since $1$ is a multiplicative unit) and hence plus and minus $\infty$ on $\mathbb{R}$; orientation is more about if you preserve or switch the order, rather than its existence $\endgroup$ Commented Apr 30, 2019 at 16:31
  • 1
    $\begingroup$ Related: math.stackexchange.com/questions/31148/… $\endgroup$ Commented Apr 30, 2019 at 16:46
  • 1
    $\begingroup$ The number of infinities of $\mathbb{R}$ will depend on the compactification you choose. If you identify (through homeomorphism) $\mathbb{R}$ with the interval $(0,1)$ you see that there are 2 possible compactifications. Either $(0,1)\subset[0,1]$ or $(0,1)=S^1\setminus\{p\}\subset S^1$ for any $p\in S^1$. The extra points will define the infinites. $\endgroup$ Commented May 1, 2019 at 13:21

5 Answers 5

Reset to default
7
$\begingroup$

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!

$\endgroup$
6
$\begingroup$

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.

$\endgroup$
7
  • $\begingroup$ Is Stone-Cech compactification's existence on the same shaky basis as the atlas is for a manifold? $\endgroup$ Commented Apr 30, 2019 at 16:39
  • $\begingroup$ @EEEB See en.wikipedia.org/wiki/Stone%E2%80%93%C4%8Cech_compactification for more info. $\endgroup$ Commented Apr 30, 2019 at 16:40
  • $\begingroup$ One other thing; can I say that $S^1$ is the resulting space under compactification of $\mathbb{R}$, under the function $\dfrac{1}{x}$? $\endgroup$ Commented Apr 30, 2019 at 16:44
  • 2
    $\begingroup$ @EEEB The boundedness can be defined for any real valued function from any space. It means the image is a bounded subset of $\mathbb R$. Every real valued function on a compact space is bounded. $\endgroup$ Commented Apr 30, 2019 at 16:55
  • 2
    $\begingroup$ @EEEB what do you mean by "shaky basis"? $\endgroup$ Commented May 1, 2019 at 3:03
2
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ Is there a general statement to relate how many points at infinity I add in $\mathbb{R}^n$ in general? This seems to be answerable by the cuts on $S^1$ $\endgroup$ Commented Apr 30, 2019 at 16:37
  • $\begingroup$ @EEEB for $n > 1$, $\mathbb R^n$ has one end. If you imagine removing larger and larger closed balls, the complements you get always have one connected component. $\endgroup$ Commented Apr 30, 2019 at 16:40
  • $\begingroup$ in $\mathbb{R}^n$ you can add a point and get $S^n$, but you can also add a hyperplane at infinity and get the projective space; however for $n \ge 2, \mathbb{R}^n$ has no order so there is no plus/minus, so no "two things/spaces of things to add" but only one in various ways - you can still do Stone-Cech too after all... $\endgroup$ Commented Apr 30, 2019 at 16:41
  • $\begingroup$ @EEEB if you think about the ways of "walking in a straight line" away from a point in $\mathbb R^n$, you could also imagine a sphere $S^{n-1}$ "at infinity", where each point on the sphere corresponds to a straight line (or ray, if you'd prefer). We define $S^0$ to be a space with two points, so this recovers the case of $\mathbb R^1$. $\endgroup$ Commented Apr 30, 2019 at 16:44
2
$\begingroup$

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....

$\endgroup$
0
2
$\begingroup$

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.

$\endgroup$
0

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.