5 ms·

It doesn't actually work in 2D for infinity, just for arbitrarily large integers (and there is an infinite amount of them). If it actually were infinite, three

by maweki 9y ago

It doesn't actually work in 2D for infinity, just for arbitrarily large integers (and there is an infinite amount of them). If it actually were infinite, three "neighbouring" points wouldn't actually describe a face but just a point, since they all fall onto each other. You could look at the Banach–Tarski paradox for a better understanding. "Half" the set of points is still enough to describe a sphere since there are infinitely many.

pbhjpbhj 9y ago
Surely then there's arbitrarily large numbers of polygons but not infinite, because if they're infinite then the spacing between points will be zero? In Banach-Tarski In thought that the assemblage of points relied on creating, eg, a new pair of spheres that worked under different non-Euclidean measures. But I'll need to review it, it's outside my domain (heh!).
- maweki 9y ago
  I was trying to convey that a sphere-sphere is an infinite amount of points with some unintuitive properties, like half the points are still an infinite amount. For a set of faces defined by an arbitrarily large but finite subset of points of the sphere, the important properties no longer hold.
- kuschku 9y ago
  > Surely then there's arbitrarily large numbers of polygons but not infinite, because if they're infinite then the spacing between points will be zero? If we describe the set of these polgons as P, then |P| = |N|, for N referring to the Natural Numbers. This is usually called "abzählbar unendlich" (countable infinite).
  - pbhjpbhj 9y ago
    That's kinda my point, if there are aleph_0 polygons, then surely the length of a side has to tend to 0. As you increase the number of vertices then the difference between the points occupied by the vertices and those of a coincidental circle reduces, ultimately the vertices must describe a circle if they form a set of size aleph_0? If you want the sides of the polygons to have a length then the polygons, to me, seem necessarily to be unable to have an infinite number of vertices, they could only have a very large number but one necessarily less than infinity. Perhaps it's my misunderstanding how a curve works? If you form a second set from an infinite number of equally spaced points between two arbitrary points on a continuous curve (the first set) then it seems, to me [ie intuitively, always a risk in maths], that the second set must also be a continuous curve? The corollary of that would seem to be that if you have an infinite number of vertices for your polygon it also forms a circle, to not form a circle it simply has to have a number of sides less than infinite. My contention would then be that either the set of polytopes in 2D is < aleph_0 or the set in 3D is 6 (ie applying the same rules makes circles and spheres both parts of the set of polytopes in the particular n-dimensional space).
    - kuschku 9y ago
      The problem seems to be your definition of infinity – the definiton as |N| is exactly defined the way you said it. An arbitrarily large number.
- bo1024 9y ago
  For each integer n >= 3, we can create a perfect n-sided polygon in two dimensions. That polygon will have some nonzero spacing between the points, roughly a distance like 2pi/n. Okay, now that is true for every n >= 3. This means that there are an infinite number of n for which it is true, i.e. an infinite number of polygons. However, each particular polygon has a nonzero length between points. However again, by picking n as large as you want, you can find polygons where this length is as small as you want, as long as it's strictly greater than 0. None of them are zero.

by maweki 9y ago

pbhjpbhj 9y ago

Surely then there's arbitrarily large numbers of polygons but not infinite, because if they're infinite then the spacing between points will be zero? In Banach-Tarski In thought that the assemblage of points relied on creating, eg, a new pair of spheres that worked under different non-Euclidean measures. But I'll need to review it, it's outside my domain (heh!).

maweki 9y ago
I was trying to convey that a sphere-sphere is an infinite amount of points with some unintuitive properties, like half the points are still an infinite amount. For a set of faces defined by an arbitrarily large but finite subset of points of the sphere, the important properties no longer hold.
kuschku 9y ago
> Surely then there's arbitrarily large numbers of polygons but not infinite, because if they're infinite then the spacing between points will be zero? If we describe the set of these polgons as P, then |P| = |N|, for N referring to the Natural Numbers. This is usually called "abzählbar unendlich" (countable infinite).
- pbhjpbhj 9y ago
  That's kinda my point, if there are aleph_0 polygons, then surely the length of a side has to tend to 0. As you increase the number of vertices then the difference between the points occupied by the vertices and those of a coincidental circle reduces, ultimately the vertices must describe a circle if they form a set of size aleph_0? If you want the sides of the polygons to have a length then the polygons, to me, seem necessarily to be unable to have an infinite number of vertices, they could only have a very large number but one necessarily less than infinity. Perhaps it's my misunderstanding how a curve works? If you form a second set from an infinite number of equally spaced points between two arbitrary points on a continuous curve (the first set) then it seems, to me [ie intuitively, always a risk in maths], that the second set must also be a continuous curve? The corollary of that would seem to be that if you have an infinite number of vertices for your polygon it also forms a circle, to not form a circle it simply has to have a number of sides less than infinite. My contention would then be that either the set of polytopes in 2D is < aleph_0 or the set in 3D is 6 (ie applying the same rules makes circles and spheres both parts of the set of polytopes in the particular n-dimensional space).
  - kuschku 9y ago
    The problem seems to be your definition of infinity – the definiton as |N| is exactly defined the way you said it. An arbitrarily large number.
bo1024 9y ago
For each integer n >= 3, we can create a perfect n-sided polygon in two dimensions. That polygon will have some nonzero spacing between the points, roughly a distance like 2pi/n. Okay, now that is true for every n >= 3. This means that there are an infinite number of n for which it is true, i.e. an infinite number of polygons. However, each particular polygon has a nonzero length between points. However again, by picking n as large as you want, you can find polygons where this length is as small as you want, as long as it's strictly greater than 0. None of them are zero.