3 ms·
Don't forget the 1D case, which has a 100% overlap.
by dTal 9y ago
Don't forget the 1D case, which has a 100% overlap.
- pbhjpbhj 9y agoDoesn't a sphere require all points to be separated from the centre point by the same distance r, where r>0? A point isn't a sphere.
- detaro 9y agoReally depends on who you are asking for a definition. E.g. there is the term "point sphere" for the special case of a sphere that is a point.
- pbhjpbhj 9y agoIs there any interesting result that comes from casting a point as being any shape (presumably the point-sphere is also a point-dodecahedron, etc.)?
- chestervonwinch 9y agoDefine a topology on any set you'd like and a sphere is just the set of limit points of any open ball.
- gus_massa 9y agoAre you taking about the orange peel, the orange juicy part or the whole fruit? If you consider only the "orange peel", if the center is 0 and r=1, then the 1D version is just two points {-1} and {1}, not only one point. If you consider only the "whole orange" then the 1D case is the segment [-1, 1], if the center is 0 and r=1 In both case it is equal to the 1D "square" equivalent.
- dTal 9y agoThe 1D case is a line segment, not a point. A point is the 0D case and indeed a degenerate case.