# N-sphere math problem

1. Jul 24, 2005

$$\sum_{i=1}^{n+1} (x_i-c_i)^2{\leq}r^2$$ 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?

2. Jul 24, 2005

### Gokul43201

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

3. Jul 24, 2005

who what is the inequality?

4. Jul 24, 2005

### Gokul43201

Staff Emeritus
5. Jul 24, 2005

### Gokul43201

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

6. Jul 24, 2005

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

7. Jul 24, 2005

### Gokul43201

Staff Emeritus
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: Jul 24, 2005
8. Jul 24, 2005

the volume which said n-sphere encapsulates?

9. Jul 25, 2005

### robphy

How about "interior of the (n+1)-ball"?

10. Jul 25, 2005

### 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.

11. Jul 25, 2005

### matt grime

The interiot is commonly called a ball or a disc.