Is an N-Sphere Just a Solid Ball or Something More?

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An n-sphere is defined as the surface where the equation \sum_{i=1}^{n+1} (x_i-c_i)^2 = r^2 holds true, while the inequality \sum_{i=1}^{n+1} (x_i-c_i)^2 \leq r^2 describes a solid ball, referred to as the closed n-ball. The points inside this closed n-ball are not considered part of the n-sphere itself. In topology, the interior of the n-sphere is commonly called the interior of the (n+1)-ball, distinguishing it from the n-sphere's surface. The terminology varies between geometry and topology, with the closed ball encompassing all points within a certain radius from the center. Understanding these distinctions is crucial for accurate mathematical communication.
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\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?
 
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No, the n-sphere is the surface defined by the equality.
 
who what is the inequality?
 
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.
 
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?
 
How about "interior of the (n+1)-ball"?
 
  • #10
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
The interiot is commonly called a ball or a disc.
 

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