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It's been a while since I've been out of school (now a high school Math teacher) and I'm working on just some standard bijections between different objects with some of our more advanced students.

The other day we were playing around with the stereographic projection in 1-dimension ($S^1 \setminus \{\text{a point}\} \rightarrow \mathbb{R}$) which got me thinking about the standard one (i.e. Riemann Sphere to $\mathbb{C}$ or $\mathbb{R}^2$). This kind of led me down a rabbit hole circling back to a question that I never fully answered for myself during undergrad which was the following:

Given that the stereographic projection works as a bijection from the Riemann Sphere to both $\mathbb{C}$ and $\mathbb{R}^2$ , why is it that we treat infinity as a point only in $\mathbb{C}$ and not in $\mathbb{R}^2$? Looking at the bijection, it seems we could make the same argument for $\infty$ being a point in $\mathbb{R}^2$ as we do in $\mathbb{C}$ (i.e. the north pole of the Riemann Sphere), so I'm curious about whether it's a complex-analysis-based choice (I understand holomorphicity is a much nicer condition and might benefit from the choice to make infinity a point) or if there is something more fundamental about the field structure that allows us to deal with $\infty$ as a point in $\mathbb{C}$ but not in $\mathbb{R}^2$?

Thanks for the help!

Barto_Wynne12
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  • I'm not sure I follow. The stereographic projection gives a bijection (in fact a homeomorphism) from $S^2 \setminus {p}$ to either $\mathbb C$ or $\mathbb R^2$. It extends as a homeomorphism from $S^2$ to the Riemann sphere $\mathbb{C} \cup {\infty}$, which is the one-point compactification of $\mathbb{C}$ (or equivalently the one-point compactification of $\mathbb{R}^2$). – Albert Oct 11 '24 at 16:29
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    When you say "we", who exactly do you refer to? There are several common compactifications of the Euclidean plane: One point compactification (Riemann sphere), real projective plane compactification, etc. One uses whatever compactification is most suitable for one's purposes. – Moishe Kohan Oct 11 '24 at 16:31
  • I guess my question stems from the fact that the one-point compactification argument seems to be used a lot more when talking about $\mathbb{C}$ than in $\mathbb{R}^2$. I have never been talked to about the possibility of treating the north pole of the Riemann Sphere as a point in $\mathbb{R}^2$ while that feels like a point that is commonly brought up in $\mathbb{C}$. Is there a field of study that makes one-point compactification much more useful in $\mathbb{C}$ or is it just that I happened to run into a group of weird professors who liked it for $\mathbb{C}$ and not for $\mathbb{R}^2$? – Barto_Wynne12 Oct 11 '24 at 16:39
  • the map $z \to 1/z$ allows to us to treat infinity (almost) as a finite point in the complex plane; for example a holomorphic function around infinity is just an expansion $\sum_{ n \ge 0} a_nz^{-n}$ that converges at a finite point (hence on a neighborhood of infinity which we can take a disc exterior if you want); when we regard the plane as $\mathbb R^2$ this map is not really natural – Conrad Oct 11 '24 at 16:41
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    I have never been talked to about the possibility --- I'm pretty sure that if you look through some topology texts (several are mentioned, sometimes by author only, in this MSE answer) you'll find discussions that refer to the one-point compactification of ${\mathbb R}^2$ (without reference, other than perhaps tangentially, to the complex numbers). Of related interest is A question about some special compactifications of $\mathbb{R}$ -- answers/comments also deal with ${\mathbb R}^n.$ – Dave L. Renfro Oct 11 '24 at 16:49
  • I never actually took a point-set topology course (shh!) so I'm definitely deficient in some of those regards. I think the answer above about the convergence of $\frac{1}{z}$ to a point being natural in the complex plane but not in $\mathbb{R}^2$ mostly answers my question, though! – Barto_Wynne12 Oct 11 '24 at 16:58

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Really the $\infty$ point is use in $\mathbb{R}^2$ as well, for instance with the one-point compactification of $\mathbb{R}^2$. Also you implicitly use this point when you treat limits at infinity with functions defined in $\mathbb{R}^2$. I suppose that the last fact is the reason we rarely use the point at infinity in $\mathbb{R}^2$, because we always treat it as an infinite limit.

J. W. Tanner
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Carlos Licea
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