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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