Allan's argument applies to the sphere, but not to the plane, which has infinite measure. Jim On Fri, Aug 7, 2020 at 4:25 PM Allan Wechsler <acwacw@gmail.com> wrote:
Maybe I'm not understanding, but I think we can prove right away that the measurability requirement is not attainable.
Two congruent measurable sets, after all, must have the same measure. And I think that if A is congruent to the union of the disjoint sets B and C, its measure must be at least |B| + |C|. But maybe I am forgetting some horrible pathological case.
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
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun