2
$\begingroup$

If we take an n-dimensional Euclidean space and cut off a ball centered at origin, we get a set that has boundary equal to the surface area of the cut off ball.

I wonder whether there were any attempts or suggestions to consider such body a sphere with a kind of infinite or negative or otherwise non-real radius?

Of course, if we consider a geometry on a sphere, the circle divides the space into two disks, but what about Euclidean space? Does the Euclidean space show the part outside the sphere behaves like a ball?

What about other possible Riemannian or Minkowski spaces?

My own opinion, the resulting figure is not a sphere even if a sphere with infinite radius is permitted (particularly, they are not members of Lie sphere geometry).

Particularly, if we take any inequality defining a ball, it seems we cannot "invert" it by just going to any limit.

Improve this question
$\endgroup$
3
  • $\begingroup$ What if the one-point compactification $\widehat{\mathbb{R}^n} $ the Euclidean space $\mathbb{R}^n$; in this case the resulting space $\widehat{\mathbb{R}^n}\setminus \overline{B(x,r)}$, where $B(x,r)$ is any (open) ball centered at $x$ of radius $r$ becomes homeomorphic to ball itself. $\endgroup$
    – Jack L.
    Nov 1, 2020 at 21:48
  • $\begingroup$ @JackL. well, it would be another space where there are still no spheres of infinite radius, or am I wrong? $\endgroup$
    – Anixx
    Nov 1, 2020 at 21:54
  • $\begingroup$ No, you’re not wrong because $\widehat{\mathbb{R}^n}$ is obviously compact. $\endgroup$
    – Jack L.
    Nov 1, 2020 at 22:03

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.