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A remark on sets having the Steinhaus property

Mihai Ciucu
1996 Combinatorica  
A point set satisfies 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.  ...  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.  ... 
doi:10.1007/bf01261317 fatcat:ifyq2x2qqfentaohys5l75o2ui

A short proof of the converse to a theorem of Steinhaus [article]

Dang Anh Tuan
2015 arXiv   pre-print
A result of H. Steinhaus states that any positive Lebesgue measurable set has a property that its difference set contains an open interval around the origin. Y. V.  ...  Mospan proved that this result is the characterization of absolutely continuous measure. In this note we give a short proof of it.  ...  Definition 1 . 1 A Borel measure µ on R is said to have Steinhaus property (abbreviated SP) if for every µ−measurable set A with µ(A) > 0 the difference AA contains an open interval around zero.  ... 
arXiv:1511.05657v1 fatcat:5uypnnyarvbc5obhfort2ulewq

Nonmeasurability in Banach spaces [article]

Robert Ralowski
2010 arXiv   pre-print
We show that for a σ -ideal with a Steinhaus property defined on Banach space, if two non-homeomorphic Banach with the same cardinality of the Hamel basis then there is a nonmeasurable subset as image  ...  A subset of Z not in I will be called a I-positive set; sets in I will also be called I-null. Also, the σ-algebra generated by B(Z) ∪ I will be denoted by B(Z), called the I-completion of B(Z).  ...  Acknowledgement Author is very indebted to Professor Jacek Cichoń for help and critical remarks.  ... 
arXiv:1001.0073v1 fatcat:onxq5eaczje67c3c7bo3ihqp2q

Measurable Steinhaus sets do not exist for finite sets or the integers in the plane

Mihail N. Kolountzakis, Michael Papadimitrakis
2017 Bulletin of the London Mathematical Society  
A Steinhaus set S ⊆^d for a set A ⊆^d is a set such that S has exactly one point in common with τ A, for every rigid motion τ of ^d.  ...  An old result of Komjáth says that there exists a Steinhaus set for A = ×0 in ^2. We also show here that such a set cannot be Lebesgue measurable.  ...  Sierpinski [21] showed that a bounded set A which is either closed or open cannot have the lattice Steinhaus property (that is, intersect all rigid motions of Z 2 at exactly one point).  ... 
doi:10.1112/blms.12069 fatcat:ml4cne3w4vh3ndpnj5zd6yalxi

On Steinhaus sets

S.M. Srivastava, R. Thangadurai
2005 Expositiones mathematicae  
We give a common proof of several results on Steinhaus sets in R d for d 2 including the fact that a Steinhaus set in R 2 must be disconnected.  ...  Steinhaus sets have been the subject of several recent papers. In this note, we give a common proof of some results (which are known in R 2 ) and some new results on Steinhaus sets in any dimension.  ...  Let S ⊂ R d be a Steinhaus set, x ∈ R d and > 0. Then S ∩ A n (x, ) = ∅ for some n 1. Remark.  ... 
doi:10.1016/j.exmath.2005.01.011 fatcat:3p4r3ge7tjhtxfslti47njgcxm

On the Steinhaus property in topological groups

Hans Weber, Enrico Zoli
2006 Topology and its Applications  
Let G be a locally compact Abelian group and μ a Haar measure on G.  ...  We prove: (a) If G is connected, then the complement of a union of finitely many translates of subgroups of G with infinite index is μ-thick and everywhere of second category.  ...  Then the σ -ideal N of locally null sets with respect to a Haar measure on G has the Steinhaus property.  ... 
doi:10.1016/j.topol.2005.07.010 fatcat:o6groxauivdn5cg7oc623cc5rm

The Steinhaus tiling problem and the range of certain quadratic forms [article]

Mihail N. Kolountzakis, Michael Papadimitrakis
2000 arXiv   pre-print
In dimension d=2 (the original Steinhaus problem) the question remains open.  ...  We give a short proof of the fact that there are no measurable subsets of Euclidean space (in dimension d > 2), which, no matter how translated and rotated, always contain exactly one integer lattice point  ...  Assume from now on that the set E is a Steinhaus set in dimension d. Suppose now that we can find a lattice Λ * ⊂ B with det Λ * not an integer.  ... 
arXiv:math/0009207v1 fatcat:tylwb4xn3jfbljh7kyferolk2a

The Steinhaus tiling problem and the range of certain quadratic forms

Mihail N. Kolountzakis, Michael Papadimitrakis
2002 Illinois Journal of Mathematics  
In dimension d = 2 (the original Steinhaus problem) the question remains open.  ...  We give a short proof of the fact that there are no measurable subsets of Euclidean space (in dimension d ≥ 3) which, no matter how translated and rotated, always contain exactly one integer lattice point  ...  Assume from now on that the set E is a Steinhaus set in dimension d. Suppose now that we can find a lattice Λ * ⊂ B with det Λ * not an integer.  ... 
doi:10.1215/ijm/1258130994 fatcat:ghg7oosqrbahjoqpx3incrx46m

Page 6473 of Mathematical Reviews Vol. , Issue 2001I [page]

2001 Mathematical Reviews  
The second remark is another way of phrasing the Steinhaus problem: A set E is a Steinhaus set if and only if any rotation of the set E translated at the locations Z” forms a tiling of the plane.  ...  of a plane set with a certain property.  ... 

ON THE STEINHAUS TILING PROBLEM FOR Z3

D. Goldstein, R. D. Mauldin
2013 Quarterly Journal of Mathematics  
Steinhaus asked in the 1950's whether there exists a set in R 2 meeting every isometric copy of Z 2 in precisely one point. Such a "Steinhaus set" was constructed by Jackson and Mauldin.  ...  Is there a subset S of R 3 meeting every isometric copy of Z 3 in exactly one point? We offer heuristic evidence that the answer is "no".  ...  However, it is known that no Steinhaus set in R 2 is a Borel set or even has the Baire property [7] .  ... 
doi:10.1093/qmath/hat025 fatcat:vhlpl2utu5dzncdnicted2q37u

Beyond Lebesgue and Baire IV: Density topologies and a converse Steinhaus–Weil theorem

N.H. Bingham, A.J. Ostaszewski
2018 Topology and its Applications  
The second was our reliance [BinO8] on a localized version of the Steinhaus-Weil property: in S the relative open neighbourhoods of all points were to have the Steinhaus-Weil property.  ...  (Boundedness of a subadditive function on A and B yields its boundedness on AB and hence on an open set, provided AB has the interior-point property -see §6.9.)  ...  We thank the Referee for a careful and scholarly reading of the paper, and for some very useful presentational suggestions.  ... 
doi:10.1016/j.topol.2017.12.029 fatcat:tugxj6movjcd7em2jjjs6x3ik4

On the Steinhaus tiling problem in three dimensions [article]

Daniel Goldstein, R. Daniel Mauldin
2013 arXiv   pre-print
Steinhaus asked in the 1950's whether there exists a set in the plane R^2 meeting every isometric copy of Z^2 in precisely one point. Such a "Steinhaus set" was constructed by Jackson and Mauldin.  ...  Is there a subset of R^3 meeting every isometric copy of Z^3 in exactly one point? We offer heuristic evidence that the answer is "no".  ...  However, it is known that no Steinhaus set in R 2 is a Borel set or even has the Baire property [7] .  ... 
arXiv:1304.8047v1 fatcat:whdzegvvgfeotcwgilxgr7v5ke

Beyond Lebesgue and Baire IV: Density topologies and a converse Steinhaus-Weil Theorem [article]

N. H. Bingham, A. J. Ostaszewski
2017 arXiv   pre-print
The Steinhaus-Weil interior-points theorem ('on AA^-1') plays a crucial role here; so too does its converse, the Simmons-Mospan theorem.  ...  One can often handle the (Baire) category case and the (Lebesgue, or Haar) measure cases together, by working bi-topologically: switching between the original topology and a suitable refinement (a density  ...  We thank the Referee for a careful and scholarly reading of the paper, and for some very useful presentational suggestions.  ... 
arXiv:1607.00031v2 fatcat:igizyst4erf2pgtvywmgxynl2q

Page 8683 of Mathematical Reviews Vol. , Issue 2004k [page]

2004 Mathematical Reviews  
A set with this property is called a Steinhaus set. The relation of Steinhaus sets to tiling is clear. A set S is a Steinhaus set if and only if every rotation of S tiles the plane.  ...  Recently the authors of the present paper have answered this long-standing question of Steinhaus by showing that there is a Steinhaus set. This solution was published in [J. Amer. Math.  ... 

On the difference property of families of measurable functions

Rafał Filipów
2003 Colloquium Mathematicum  
We show that, generally, families of measurable functions do not have the difference property under some assumption.  ...  We also show that there are natural classes of functions which do not have the difference property in ZFC. This extends the result of Erdős concerning the family of Lebesgue measurable functions.  ...  First, we have the well known theorems which explain the names of the Ostrowski property and Steinhaus property. They concern Lebesgue measure on R.  ... 
doi:10.4064/cm97-2-4 fatcat:fjuufahwrzhpnnmtahcp5zih74
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