Re: [math-fun] Paradoxical decomposition of the plane or sphere?
I'll have to think about Andy's post when it's not so close to my bedtime. But meanwhile I noticed that Hausdorff proved that there *does* exist a paradoxical decomposition of the sphere S^2 - D where D is countable, using the Axiom of Choice: S^2 - D = A + B + C such that A, B, C are all disjoint and isometric to each other, and also A is isometric to B + C. —Dan Andy Latto wrote: ----- No, this is not possible. The key fact here is that it's possible to assign a nonnegative real number, the "content" of the set, to every subset of the plane, such that this agrees with the Lebesgue measure on all Lebesgue-measurable sets, and is finitely additive. It cannot of course be countably additive, because there exist non-measureable sets. Any isometry of the plane can be decomposed into a translation, a rotation, and possibly a reflection. Consider this decomposition for the isometries that take A-> B, B->C, and A+B -> C. Take a disk containing the three points of rotation of these three isometries, that is very large compared to both the distance between the three points of rotation and the lengths of the three translations. Looking at the content of the intersections of this disk with A, B, and C will produce a contradiction. Andy Intersecting A, B, and C with a disk that is very large compared to the length of the On Fri, Aug 7, 2020 at 2:50 PM Dan Asimov <dasimov@earthlink.net> wrote:
Does there exist a decomposition of the plane into three disjoint sets A, B, C such that all three are isometric to each other but also A is isometric to the union B + C ???
Ideally each set would be connected. Also measurable, but I would be surprised if measurable is possible.
And ideally, without needing the Axiom of Choice.
How about the same thing but with some points omitted:
A + B + C = R^2 - Z
where Z is a set of measure 0 ???
And if not the plane, how about the sphere S^2 ???
—Dan
On Fri, Aug 7, 2020 at 10:48 PM Dan Asimov <dasimov@earthlink.net> wrote:
I'll have to think about Andy's post when it's not so close to my bedtime.
But meanwhile I noticed that Hausdorff proved that there *does* exist a paradoxical decomposition of the sphere S^2 - D where D is countable, using the Axiom of Choice:
S^2 - D = A + B + C
such that A, B, C are all disjoint and isometric to each other, and also A is isometric to B + C.
Yes, that's the bais for the Banach-Tarski paradox, once you apply it to the concentric spheres that make up the ball, and do a little fudging for the countable set D in each sphere and the center of the sphee (which takes a few more pieces, but still a small finite number). The reason you can do this with the sphere and not the plane is that the group of rigid motions of the sphere, O(3) is an amenable group, while the rigid motions of the plane is non-amenable. There's a wonderful book, The Banach-Tarski paradox, by Tomkowicz and Wagon, that explains all of this. Reading it transformed the Banach Tarski paradox for me from something that seems bizarre and impossible to something that makes perfect sense. Andy
—Dan
Andy Latto wrote: ----- No, this is not possible.
The key fact here is that it's possible to assign a nonnegative real number, the "content" of the set, to every subset of the plane, such that this agrees with the Lebesgue measure on all Lebesgue-measurable sets, and is finitely additive. It cannot of course be countably additive, because there exist non-measureable sets.
Any isometry of the plane can be decomposed into a translation, a rotation, and possibly a reflection. Consider this decomposition for the isometries that take A-> B, B->C, and A+B -> C. Take a disk containing the three points of rotation of these three isometries, that is very large compared to both the distance between the three points of rotation and the lengths of the three translations. Looking at the content of the intersections of this disk with A, B, and C will produce a contradiction.
Andy
Intersecting A, B, and C with a disk that is very large compared to the length of the
On Fri, Aug 7, 2020 at 2:50 PM Dan Asimov <dasimov@earthlink.net> wrote:
Does there exist a decomposition of the plane into three disjoint sets A, B, C such that all three are isometric to each other but also A is isometric to the union B + C ???
Ideally each set would be connected. Also measurable, but I would be surprised if measurable is possible.
And ideally, without needing the Axiom of Choice.
How about the same thing but with some points omitted:
A + B + C = R^2 - Z
where Z is a set of measure 0 ???
And if not the plane, how about the sphere S^2 ???
—Dan
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