Translative covering of space with slabs

A slab (or plank) of width w is a part of the d-dimensional space that lies between two parallel hyperplanes at distance w from each other. It is conjectured that any slabs S1, S2, . whose total width is divergent have suitable translates that altogether cover Rd. We show that this statement is true if the widths of the slabs, w1, w2, . , satisfy the slightly stronger condition lim supn→∞ w1+w2+. +wn log(1/wn) > 0. This can be regarded as a converse of Bang’s theorem, better known as Tarski’s plank problem. We apply our results to a problem on simultaneous approximation of polynomials. Given a positive integer d, we say that a sequence of positive numbers x1 ≤ x2 ≤ . controls all polynomials of degree at most d if there exist y1, y2, . ∈ R such that for every polynomial p of degree at most d, there exists an index i with |p(xi) − yi| ≤ 1. We prove that a sequence has this property if and only if ∑∞ i=1 1 xi is divergent. This settles an old conjecture of Makai and Pach. 1 Tarski’s plank problem and its affine version The closed set of points S lying between two parallel hyperplanes in R at distance w from each other is called a slab (or plank) of width w. Given a convex body C ⊂ R, its width w(C) is the smallest number w such that there is a slab of width w that covers C. In 1932, Alfred Tarski [18] made the following attractive conjecture. Tarski’s plank problem. If a sequence of slabs covers a convex body C, then the total width of the slabs is at least w(C). Tarski found a beautiful proof of his conjecture in the special case where C is a closed disk. His argument was based on the following fact discovered by Archimedes more than two thousand years ago. Let D be a disk of radius 1 2 in the (x, y)-plane R, and let H denote the hemisphere of radius 1 2 , concentric with D, that lies in the closed half-space above R. *EPFL, Lausanne and MIPT, Moscow. Supported in part by the grant N 15-01-03530 of the Russian Foundation for Basic Research. E-mail: kupavskii@ya.ru. †EPFL, Lausanne and Renyi Institute, Budapest. Supported by Hungarian Science Foundation EuroGIGA Grant OTKA NN 102029, by Swiss National Science Foundation Grants 200020-144531 and 200021137574. E-mail: pach@cims.nyu. edu. 1 ar X iv :1 51 1. 08 11 1v 1 [ m at h. M G ] 2 5 N ov 2 01 5 No matter how we place a 2-dimensional slab S of width w in the plane such that both of its boundary lines intersect D, the surface area of the vertical projection of S ∩ D to the hemisphere H is always equal to 1 2 πw; see Fig. 1. Suppose now that a system of slabs S1, S2, . completely covers D. We can assume without loss of generality that the boundary lines of every slab intersect D, otherwise, we could replace one of the slabs with a narrower one. The vertical projections of the slabs cover H, therefore the sum of the areas of these projections, 1 2 πw(S1) + 1 2 πw(S2) + . , is at least Area(H) = π 2 . Thus, we have that w(S1) + w(S2) + . ≥ 1 = w(D), as required. Figure 1: Tarski’s proof for a disk of unit diameter. A slight modification of the above argument also settles the analogous problem in the case where C is a 3-dimensional ball B of unit diameter. Now let H denote the whole sphere bounding B, so that Area(H) = π. Suppose that S1, S2, . is a system of 3-dimensional slabs that altogether cover B, and they have the property that the boundary planes of each Si intersect B . By the Archemidean observation, for every i, the surface area of Si ∩H is equal to πw(Si). Since the slabs must also cover H, we have ∑