It's well know that the Riemann sphere is constructed by adding a single point at infinity to the complex plane. The distinguishing feature of the point at infinity is that it's the inverse of the point $z=0$
$$0^{-1}=\infty$$
It's impossible to define what a point at infinity means unless you have previously defined what doing an inverse operation means. In other words, the point at infinity comes with an implicit definition of an inverse mapping, which other points don't possess.
If I want to remove any other point besides zero or infinity from the Riemann sphere, it's required to remove 2 points, since otherwise the inverse would not be well defined for the image of that point under the inverse operation, and the inverse operation is required for the definition of the point at infinity.
It's not possible to delete the definition of inverse while our space contains a point at infinity, and it's not possible to remove a point at infinity without without also removing the point 0 from the Riemann sphere, and then deleting the definition of inverse, and then adding the point $0$ back to the complex plane.
Is this wrong?
Best Answer
Your text is far away from the way we think about these things.
Adding a single point $\infty$ to ${\mathbb C}$ means that we define at free will new set $\bar{\mathbb C}:={\mathbb C}\cup\{\infty\}$. A priori this is a "purely geometric" operation, and there are no algebraic rules associated with it. In particular, $\bar{\mathbb C}$ is not a field, and we cannot expect that the special "number" $\infty$ has a multiplicative inverse in $\bar{\mathbb C}$, or that expressions like $0\cdot\infty$ give a nice result in $\bar{\mathbb C}$.
Nevertheless there are certain "exception handling rules". This point $\infty$ also has obtained its "neighborhood filter". In this way formulas like $$\lim_{z\to0} {1\over z}=\infty, \qquad\lim_{z\to\infty}{1\over z}=0\ ,$$ and similar, make sense. When we deal with Moebius transformations $T$ we, e.g., find that $$\lim_{z\to\infty} T(z)=\lim_{z\to\infty}{3z-4\over2z+25}={3\over2}\ ,$$ and therefore put $T(\infty):={3\over2}$.
Similarly, the function $$f(z):={z-3\over (z+1)(z-2)}$$ is taking the value $\infty$ at $z:=-1$ and $z:=2$.