A Remark on Sets Having the Steinhaus Property
نویسنده
چکیده
A point set satisses the Steinhaus property if no matter how it is placed on a plane, it covers exactly one integer lattice point. Whether or not such a set exists, is an open problem. Beck has proved 1] that any bounded set satisfying the Steinhaus property is not Lebesgue measurable. We show that any such set (bounded or not) must have empty interior. As a corollary, we deduce that closed sets do not have the Steinhaus property, fact noted by Sierpinski 3] under the additional assumption of boundedness. The purpose of this paper is to prove the following Theorem. Any set S having the Steinhaus property has empty interior. Our proof requires a number of preliminary lemmas. For m; n 2 Z, denote by A m;n the unit square of the lattice Z 2 having its upper-left corner at (m; n). For any subset M A m;n we denote by M(mod 1) the set M + (?m; ?n). For 2 0; 2), S() is the set obtained from S by a rotation of angle around the origin. The unit squares A m;n are considered to contain their north and west sides, and not their south and east sides, so that they form a partition of R 2. Lemma 1. A set S satisses the Steinhaus property if and only if fS()\A m;n (mod 1)g m;n is a partition of A 0;0 , for all 2 0; 2).
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ورودعنوان ژورنال:
- Combinatorica
دوره 16 شماره
صفحات -
تاریخ انتشار 1996