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Physicist here, although math inclined but I certainly won't brag about it on a math forum, I am encountering a confusion with the compactification of the complex plane. I am learning conformal field theory in 2d and, as we use conformal map mapping some points to infinity and vice versa, we work on the Riemann sphere defined as $$\hat{\mathbb{C}} = \mathbb{C} \cup \{\infty\}$$ I have a doubt on having only "one infinity" and not seeing it as a full circle whose radius tends to $\infty$. For example compactifying $\mathbb{R}$ means adding the two infinities $\pm \infty$. Is it because all those infinities could "topologically" connected when one needs to cross all the real axis to go from $+\infty$ to $-\infty$?

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Compactifying $\mathbb R$ usually means adding one infinity, making $\mathbb R$ into a circle. The Riemann sphere is just the 2d version of this. See – lhf Mar 28 '13 at 1:23
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The Riemann sphere is not just $\mathbb{C} \cup \{ \infty\}$. It this space endowed with a particular topology. You can think of that topology as arising from adding infinities at the end of some "infinitely large" circle, and then collapsing all those infinities to a point. This perspective is highlighted when considering the stereographic projection of $S^2$ onto $\mathbb{R}^2 $ or $ \mathbb{C}$.

enter image description here

Horizontal circles wrapping around the sphere are mapped to circles in the plane. As we push the circles higher on the Riemann sphere, the corresponding circles in the plane grow. At the north pole, we should have an "infinitely large" circle in the plane, but since the north pole is just a point, that "infinitely large" circle is actually just a point, which we call the point at infinity.

From a technical, point-set topology perspective, this is an instance of the Alexandroff one-point compactification. Another way to view the Riemann sphere is as the complex projective line.

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It might help to think first about compactifying $\mathbb R$ by adding one point rather than two. Here is one way: identify $(-\infty,\infty)$ with $(-\pi,\pi)$ in the usual way, say using $x \mapsto 2\arctan x$, and then instead of compactifying $(-\pi,\pi)$ to the closed interval $[-\pi,\pi],$ instead compactify it to a circle by wrapping around the unit circle (in terms of formulas, you can use $\theta \mapsto e^{i \theta}$): this adds just one point instead of two.

(If you don't want to make all these changes of variables, you just think about taking $[-\infty,\infty]$ and then identifying the endpoints to get a circle, just as identifying the endpoints of any closed interval gives a circle.)

So you can form a one-point compactification of $\mathbb R$ to get a space that is topologically a circle.

Note that one nice thing about this is that the one-point compactification is topologically homogeneous (no point looks different to any other; it is a one-dimensional manifold); this is different to the two-point compactification, where the endpoints have different topology to all the other points.

Now we can form the one-point compaticification of $\mathbb C$ to get the Riemann sphere. Note that if we close up $\mathbb R$ inside, we get the one-point (circle) compatification of $\mathbb R$ discussed above.

The Riemann sphere is also topoogically homogeneous.

We could also compactify $\mathbb C$ by adding a circle at infinity instead of just a point, but this isn't as useful when studying complex analysis and related questions (for example because holomorphicity involves considering derivatives along multiple directions --- think of the Cauchy--Riemann equations --- and so adding points at infinity for each direction from the origin isn't so helpful when studying complex analysis). And it wouldn't be topologically homogeneous.

There are situations where you can see a "circle at infinity" in the context of the Riemman sphere. For example, the one-point compactification of $\mathbb R$ cuts the Riemann sphere into two halves (the upper and lower halfplanes), and we can think of the one-point compatification of $\mathbb R$ as being a "circle at infinity" for each of these half-planes.

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Thank you for the example of single point compactification of $\mathbb{R}$, it was very insightful. – Learning is a mess Mar 28 '13 at 10:30

A compactification of a topological space $X$ is a topological space $Y$ such that

  • $X\subseteq Y$, and
  • $Y$ is compact, and
  • $Y$ is only as big as necessary for the points above to be true, in the sense that every point in $Y$ is a limit point of $X$.

Every locally compact Hausdorf space has a "one-point compactification", but may also have many other compactifications. The real line with a single $\infty$ added, that can be approached by going in either direction, is the one-point compactification. The line also has a two-point compactification, with $\pm\infty$ added. It has no three-point compactification (I posted a question here some time in 2012, asking how to prove that) but it has some interesting compactifications with infinitely many points.

Only one $\infty$ is normally considered in complex analysis because it makes it possible to apply results to that point that apply to other points, i.e. it makes sense to speak of a function being holomorphic at $\infty$.

The one-point compactification of the real line makes sense as the codomain of the trigonometric functions on the line because it makes them continuous. The one-point compactification of both the domain and the codomain of rational functions makes sense regardless of whether those are on $\mathbb R$ or $\mathbb C$ or some yet more exotic spaces.

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Do you have a reference handy for compactifications of $\mathbf{R}$ with infinitely many points? My curiosity is piqued – Daenerys Naharis Mar 28 '13 at 5:54

You can define the Riemann sphere, $\hat{\Bbb C}$ as the one point compactification of $\Bbb C$. As a set we can think of this as $\hat{\Bbb C}= \Bbb C \cup \{\infty\}$, where if $\infty \not\in U$ then $U \subset \hat{\Bbb C}$ is open if $U$ is open in the standard topology on $\Bbb C$. If $\infty \in U$, then $U$ is open if $U \cap \Bbb C=K^{c}$, where $K \subset \Bbb C$ is compact. In short, the open sets of $\hat{\Bbb C}$ are the standard open sets of $\Bbb C$, and open sets containing $\infty$ are the complements of compact sets in $\mathbb C$.

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