one can further prove that the sphere S**n−1 can be partitioned into as many pieces as there are real numbers (that is,
pieces)
Would the answer to OP be some argument along the lines of defining the surface area of the ball as the sum of the partitioned balls surface areas then?
IDK what OP is even going on about. This just seemed relevant.
I would assume if and only if the radius is infinite
U’d think, right?!
I read it as surface area, thus being the amount of space on the sphere itself.
A=4πr2 is the formula if I remember correctly, so I just figure only radius can be altered to match infinity.
Maybe someone will tell me I’m missing something
I believe the surface area of an n-dimensional hypersphere is (n - 1) pi r^{n - 1}. In that case (I may have some factors wrong here, just going off memory), an infinite-dimensional hypersphere has infinite surface area as long as it has non-zero radius.
It does indeed scale with r^(n-1), but your factors are not close at all. It involves the gamma function, which in this case can be expanded into various factorials and also a factor of sqrt(pi) when n is odd. According to Wikipedia, the expression is 2pi(n/2)r(n-1)/Gamma(n/2).
Edit: ignore the below, I forgot my pi-factor in the gamma function for half-integers…
Edit 2: Since you’re right, my missing gamma-factor completely changes this. An infinite-dimensional hypersphere will have zero surface area for any (finite?) radius.
Original dum-dum:
While I’m completely open that my factor is likely wrong here, the expression you provided is definitely wrong in the 3D case (I’m assuming the r superscript on the pi was a typo), since it doesn’t give n = 3 => A = 4 pi r^2.