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If I parameterise $\mathbb{R}^n$ with generalised polar coordinates $(r, \Theta)$ it is possible to partition $\mathbb{R}^n$ into three parts

$$A = \{x \in \mathbb{R}^n \mid r < 1\}$$ $$B = \{x \in \mathbb{R}^n \mid r = 1\}$$ $$C = \{x \in \mathbb{R}^n \mid r > 1\}$$

$A$ is the inside and $C$ is the outside, of a sphere given by $B$. I can then define a function the maps most of the outside to the inside, for example:

$$f: C\to A$$ $$f(r,\Theta) = (1/r, \Theta)$$

This function works almost everywhere except for where $x\in A$ is the origin. It would seem impossible to define a bijective function that completely maps $A$ to $C$ and that you'll always have a single point which is problematic. Basically, $C$ has a hole and $A$ does not. My question is, is the presence of this single point generally considered the basis on which inside and outside are different?

Similarly, if we add a point at infinity to $\mathbb{R}^n$, what does this do to the notion of inside and outside, is there still a difference?

Lucas
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    If you walk naked inside, it's fine; if you walk naked outside it's not fine. That's the difference! :-) – Asaf Karagila Jun 20 '13 at 11:18
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    @AsafKaragila My experience has been that neither of those are strictly true :P – Lucas Jun 20 '13 at 11:20
  • Sometimes I lol so much reading comments on Math.SE :-) – Avitus Jun 20 '13 at 11:21
  • @Lucas which properties should $f$ satisfy? Continuity, bijectivity, $C^{\infty}$...? – Avitus Jun 20 '13 at 11:23
  • @Avitus continuity yes, bijective yes, $C^\infty$ not so sure. I was interested in what fundamentally differentiates inside and outside, $f$ was supposed to be illustrative rather than a formal definition. – Lucas Jun 20 '13 at 11:29

1 Answers1

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Yes. Often when you look at the complement of an embedded sphere, not just the standard embedding $r=1$ in your question, the two components are referred to as "the bounded component" and "the unbounded component." The unbounded component can be distinguished topologically by the fact that it is not compact.

When you add a point at $\infty$, then the two components are homeomorphic, so there is no distinction. Indeed, for the $r=1$ sphere, the two components correspond to the upper and lower hemisphere of $S^3$ under stereographic projection.

Cheerful Parsnip
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