Why can't a neighbourhood contain the entire plane? Context:

Let $S$ be the spherical surface of radius 1 with center at the origin of $R^{3}$, let $p=(0,0,1)$ be the north pole of $S,$ and let $f: S-p \rightarrow R^{2}$ be the stereographic projection from $p$ of $S-p$ onto the equatorial plane. Construct a diagram which shows that $f$ is continuous. Over what part of $S-p$ is $f$ a contracting function? Show that $f$ is $1-1,$ and that $f^{-1}: R^{2} \rightarrow S-p$ is also continuous. What is the image under $f$ of the deleted neighborhood $N(p, r, S)-p$ for $r$ -values less than $1 ?$ Why is it impossible to define $f p$ so that the extended function is continuous?


Solution given:

Contracts distances for any pair of points in the lower hemisphere. $f^{-1}$ sends $y \in R^{2}$ into the intersection of the ray py with S. See Fig. S3.
The same picture shows that the inverse is continuous. The image of $N(p, r, S)-p$ is the exterior of a circle with center at the origin. If $f p$ were defined and $f$ were continuous at $p$, then any $N(fp,\epsilon)$ would have to contain the entire exterior of some circle; but this is impossible.

My question:

Why is it impossible for a neighbourhood to contain the entire exterior or the entire plane, excluding the circle itself?

Is it because :

*

*A neighbourhood cannot have infinite radius?

*A neighbourhood is an open set, thus in the plane where it exists, there will always be points on the plane that is not in the neighbourhood, in this case any point $q$ such that $d(O,q)=\epsilon=\infty$, assuming there exists the center of this neighbourhood, $O$? (Though the authors have not introduced the concept of open sets)

*A neighbourhood is an open disk, while the plane has no definite shape, thus it cannot extend over the whole plane?

Thanks in advance.
Source

Section 1.3, Exercise 3, Chinn, Steenrod: First Concepts of Topology

 A: There's a simple argument: $S^2$ is compact and so every continuous image of $S^2$ is compact. And thus $\mathbb{R}^2$ cannot be an image of $S^2$ because it is not compact. In particular no continuous function mapping $S^2\backslash\{p\}$ onto $\mathbb{R}^2$ can be extended to full $S^2$.

If $f p$ were defined and $f$ were continuous at $p$, then any $N(fp,\epsilon)$ would have to contain the entire exterior of some circle; but this is impossible.

Introducing interior and exterior of neighbourhoods is already not an easy task. Also it seems to strongly depend on the fact that $f p$ takes circles to circles. I think that the argument here is that there is a neighbourhood $U$ of $f(p)$ that is bounded (every point in $\mathbb{R}^2$ has such a neighbourhood), but in $S^2$ every neighbourhood of $p$ is mapped onto an unbounded subset of $\mathbb{R}^2$ via the stereographic projection.
The compactness argument is straightforward and easy.

I'll try to address your confusing questions now.


*

*A neighbourhood cannot have infinite radius?


Depends on how you define "neighbourhood" and "radius". Are you asking whether every neighbourhood is bounded? No, it does not have to be. In fact every open subset of $S^2$ or $S^2\backslash\{p\}$ is bounded, unlike in $\mathbb{R}^2$.



*A neighbourhood is an open set, thus in the plane where it exists, there will always be points on the plane that is not in the neighbourhood, in this case any point $q$ such that $d(O,q)=\epsilon=\infty$, assuming there exists the center of this neighbourhood?


I really don't understand what you are trying to say here. What is "the center of this neighbourhood"? What do you mean by $d(0,q)=\epsilon=\infty$? What do you mean by "$d(0,q)$"?



*A neighbourhood is an open disk, while the plane has no definite shape, thus it cannot extend over the whole plane?


I don't understand this question as well. What do you mean by "extending a neighbourhood over the whole plane"? Typically we extend functions, not subsets. Also what do you mean by "no definite shape" or "shape" alone? That doesn't make much sense.
Mathematics is a science of precision. Most of confusion can be avoided if we focus on mastering definitions.
A: The keyword here is "any". It is definitely possible for some neighborhood of $f(p)$ to contain the entire plane. But not all of them do.
A neighborhood of a point $p$ is any open set that contains $p$. Since the entire plane is an open set in itself, it is a neighborhood of every point.
