# 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?

Gokul43201
Staff Emeritus
Gold Member
No, the n-sphere is the surface defined by the equality.

who what is the inequality?

Gokul43201
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.

Gokul43201
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.

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the volume which said n-sphere encapsulates?

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

rachmaninoff
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.

matt grime