2

I'm aware that there's only one 'infinity' in the complex numbers, and I remember one of my professors demonstrating the fact that indeed all numbers tend to the same point as their modulus goes to infinity. I do not remember exactly how he did it, but I remember the outline:

  1. Set up a equivalence relationship over $\mathbb{C}$. I believe that equivalence was invariant under scalar multiplication
  2. Use that equivalence relationship in a smart way so that for any $z$ we have something like $$z \sim (a,1)$$ for some a. This step is by far the blurriest step in my memory.
  3. Since the equivalence is invariant under scalar multiplication, we can multiply by $\frac{1}{n}$ and make $n$ tend towards infinity. In this way all points with a modulus that tends to infinity are equivalent to $(0,0)$.

As I said, the memory is very blurry, so everything I wrote above is to take with a pinch of salt (I am aware some of it barely makes any sense). What I'm hoping for is that someone recognises what I'm talking and could hint at what the equivalence relationship should be for this to work.

Quantaliinuxite
  • 1,517
  • Maybe this: https://proofwiki.org/wiki/Definition:Neighborhood_of_Infinity_(Complex_Analysis) –  Jul 20 '17 at 14:42
  • In a sense there's no canonical way to assign an infinity. There could be just one, or you could make a whole circle's worth of them, with a different point with infinite modulus for every angle. Other options are possible as well. Having only one infinity gives the resulting space nice properties though geometrically. – Matt Samuel Jul 20 '17 at 14:45
  • 1
    While the one-point compactification/Riemann sphere is the most common way to understand limits at infinity in the complex plane, it is not the only one: you can add one infinite point for each direction, in the same way that the real line can be compactified to $[-\infty,\infty]$. This is helpful for functions like the exponential or the Airy function, where the limiting behaviour is different in different directions. – Chappers Jul 20 '17 at 14:47
  • I emailed the teacher, and he actually used projective spaces to show that the complex line plus a point is topologically the same as $S^2$ – Quantaliinuxite Jul 20 '17 at 22:47

0 Answers0