# N-sphere math problem

$$\sum_{i=1}^{n+1} (x_i-c_i)^2{\leq}r^2$$ is not the set of all points at a certain distance from the center, but it is a "solid ball", so is it still an n-sphere?

Staff Emeritus
Gold Member
No, the n-sphere is the surface defined by the equality.

who what is the inequality?

Staff Emeritus
Gold Member
All points on $\sum_{i=1}^{n+1} (x_i-c_i)^2 = r^2$ belong in an n-sphere (not points inside).

i know that, i was asking what the points inside(the inequality) is called, since it isn't a sphere.

Staff Emeritus
Gold Member
I guess you'd simply call them "points inside the n-sphere" (?), though I'm not sure if 'inside' is well defined. In this case, though, you could define 'inside' as being on the same side of the surface as the center.

Last edited:
the volume which said n-sphere encapsulates?

Homework Helper
Gold Member
How about "interior of the (n+1)-ball"?

The convention in topology as I learned it is

$$\sum_{i=1}^{n} (x_i-c_i)^2{\leq}r^2$$
is a closed n-ball or radius r (or just "closed ball" when you're in R^n Euclidean space)

$$\sum_{i=1}^{n} (x_i-c_i)^2< r^2$$
is an open n-ball or radius r or "open ball" in R^n.

Note that in R^n, a ball is n-dimensional and a sphere is (n-1)-dimensional; i.e., a topologist's n-sphere is the boundary of an (n+1)-ball.