aa r X i v : . [ m a t h . GN ] J u l Key words:outer measure, rational distances,Bernstein set, Vitali set, nonmeasurableset
Author: Marcin Michalski A note on sets avoiding rational distances A BSTRACT
In this paper we shall give a short proof of the result originally obtained by AshutoshKumar that for each A ⊂ R there exists B ⊂ A full in A such that no distance between two distinct pointsfrom B is rational. We will construct a Bernstein subset of R which also avoids rational distances. Wewill show some cases in which the former result may be extended to subsets of R , i. e. it remains truefor measurable subsets of the plane and if non ( N ) = co f ( N ) then for a given set of positive outermeasure we may find its full subset which is a partial bijection and avoids rational distances.
1. INTRODUCTION AND PRELIMINARIES
We will use a standard set-theoretic notation based on [3]. We denote the real lineby R and the set of rationals by Q . By α , β , γ , ... we denote ordinal numbers, with anexception of λ ( λ ∗ respectively) which will stand for Lebesgue measure (outer Lebesguemeasure respectively). By ω we denote the set of natural numbers and by c = ω = | R | we denote the continuum - the cardinality of R . We will say that a set A is countable if | A | ≤ ω .We call a family of sets I an ideal of sets if I is closed under finite unions and takingsubsets. We call it a σ -ideal of sets if it is an ideal closed under countable unions. For σ -ideal I and the space X we define following cardinal coefficients from the Cicho´n’sdiagram: non ( I ) = min {| A | : A ⊂ X ∧ A / ∈ I } co f ( I ) = min {| A | : A ⊂ I ∧ ( ∀ I ∈ I )( ∃ A ∈ A )( I ⊂ A ) } The author was supported by grant S40012/K1102 from the Faculty of Fundamental Problems of Technol-ogy of Wrocław University of Technology. This work is a conference paper accepted to 13 th Students’ Science Conference (2015), Polanica Zdrój,Poland. We shall consider the classical examples of σ -ideals: M - the family of meager sets and N - the family of null sets. We denote Borel subsets of the considered space by B .D EFINITION
We say thay a set A is • I − positive Borel set if it is Borel and does not belong to I , • I -nonmeasurable if A / ∈ σ ( B ∪ I ) , • completely I -nonmeasurable, if it intersects each I − positive Borel set, butalso does not contain any such set. The classic example of completely nonmeasurable set for any reasonable σ -ideal(nontrivial, containing points, possessing Borel base) is a Bernstein set - a set that inter-sects every perfect set, but does not contain any such set. For I = N sets which are I -measurable are Lebesgue measurable and for I = M sets which are I -measurableposses the Baire property.For A ⊂ R and x , y ∈ R we define a horizontal slice of A : A y = { x ∈ R : ( x , y ) ∈ A } anda vertical slice of A : A x = { y ∈ R : ( x , y ) ∈ A } .Let us consider a well known equivalence relation ∼ over R : x ∼ y ↔ x − y ∈ Q . Let V be a selector of the partition of R induced by ∼ , that is | V ∩ [ x ] ∼ | = x ∈ R . Aset with such a property we call a Vitali set. Let us proceed to main definitions of thepaper.D EFINITION
We say that a set A avoids rational distances (shortly: A is an
ARD set) if for each x , y ∈ A , x = y, the distance between x and y is irrational. D EFINITION
We say that a set A ⊂ B is full (in B) if for each X of positivemeasure we have λ ∗ ( X ∩ B ) = λ ∗ ( X ∩ A ) . It is clear that every Vitali set is an ARD set. Observe also that for a set B of finiteouter measure A ⊂ B is full in B iff A has the same outer measure as B .Let us recall the following result of Gitik and Shelah (see [2]).T HEOREM (Gitik, Shelah) Let ( A n ) n ∈ ω be a sequence of subsets of R n . Thenthere exists a sequence ( B n ) n ∈ ω such that for every n ∈ ω we have B n ⊂ A n and λ ∗ ( B n ) = λ ∗ ( A n )
2. ON THE REAL LINE
In 2012 Ashutosh Kumar (see [5]) proved that for each A ⊂ R there exists an ARDset B ⊂ A full in A . We shall give a short proof of this result.T HEOREM (Kumar) Let A ⊂ R be a set of positive outer measure. Then thereexists an ARD set B ⊂ A full in A. P ROOF . Let V be a Vitali set and let us enumerate rationals Q = { q n : n ∈ ω } . Foreach n ∈ ω let: A n = { v ∈ V : v + q n ∈ A } By Theorem 1.1 let us take a sequence ( B n ) n ∈ ω such that for each n B n ⊂ A n and λ ∗ ( B n ) = λ ∗ ( A n ) . Since S n ∈ ω ( A n + q n ) = A it is easy to verify that B = S n ∈ ω ( B n + q n ) is the set. (cid:3) It is a nontrivial exercise to prove that there exists a Vitali set of full outer measure.We will construct a Vitali set with a bit stronger property.T
HEOREM
There exists a Vitali set that is also a Bernstein set. P ROOF . Let { P α : α < c } be an enumeration of all perfect subsets of R and { C α ; α < c } be an enumeration of all equivalance classes of relation ∼ . We will construct desiredset by transfinite induction.At the first step we choose arbitrarily p ∈ P then c ∈ C if p / ∈ C (otherwise we set c = p ) and e ∈ P , e = p , c .At the step ξ < c let us assume that we have transfinite sequences ( p α : α < ξ ) , ( c α : α < ξ ) , ( e α : α < ξ ) such that:(1) p α ∈ P α , c α ∈ C α , e α ∈ P α for all α < ξ ,(2) { p α : α < ξ } ∩ { e α : α < ξ } = /0,(3) { c α : α < ξ } ∩ { e α : α < ξ } = /0,(4) | ( { p α : α < ξ } ∪ { c α : α < ξ } ) ∩ C β | ≤ β < c ,(5) |{ e α : α < ξ } ∩ C β | ≤ β < c .We will extend our sequences in such a way that above properties will be preserved. Letus consider the following set: E ξ = { x ∈ C β : β < c , ( { p α : α < ξ } ∪ { c α : α < ξ } ) ∩ C β = /0 } ∪ { e α : α < ξ } . Its cardinality is at most | ξ | < c since each class C β is countable and our sequences havea length ξ . Every perfect set has the cardinality of c so the set P ξ \ E ξ is nonempty. Letus choose then p ξ ∈ P ξ \ E ξ . If p α ∈ C ξ for some α ≤ ξ , then set c ξ = p α , otherwise wechoose arbitrarily c ξ ∈ C ξ \{ e α : α < ξ } (the latter is nonempty since intersection of C ξ and { e α : α < ξ } is at most one point). Eventually let us consider the following set: E ′ ξ = { p α , c α : α ≤ ξ } ∪ { e α : α < ξ } ∪ { x ∈ C β : β < c , { e α : α < ξ } ∩ C β = /0 } . Similarly to the previous reasoning it has the cardinality of | ξ | , so we may pick e ξ ∈ P ξ \ E ′ ξ .This finishes the construction and V = { p α : α < c } ∪ { c α : α < c } is the set. (cid:3)
3. ON THE REAL PLANE
Since being in rational distance on the plane is not an equivalence relation, the ideabehind the Theorem 2.1 cannot be directly utilized to prove a similar result for the planeand higher dimensions. However, we are not totally helpless and there are some cases in which we are able to give some answers. First, we shall consider the case of measurablesets. Before we proceed let us denote by d a standard Euclidean metric on R , by B ( x , r ) an open ball with origin x and radius r and let S ( x , r ) = δ B ( x , r ) (the frame of ball B - in R it is a circle).T HEOREM
Let A be a measurable subset of R of positive measure. Then thereexists an ARD set B ⊂ A full in A. P ROOF . Since A is measurable, there exists F σ set F ⊂ A such that λ ( A ) = λ ( F ) .Let { B α : α < c } be an enumeration of all Borel subsets of F of positive measure. Weshall construct the desired set inductively.Let us choose point b ∈ B . Now, let assume that we have already chosen a transfinitesequence ( b α : α < ξ ) for ξ < c such that for each α < ξ we have b α ∈ B α and for all α , β < ξ we have d ( b α , b β ) ∈ Q . Let us pick b ξ ∈ B ξ \ (cid:0) [ α < ξ [ q ∈ Q + S ( b α , q ) (cid:1) . Such a choice can be made since by the Fubini theorem for each set of positive measurethere is a positive set A α ⊂ R such that for every x ∈ A α the set B α x has the cardinalityof c and its intersection with S α < ξ S q ∈ Q + S ( b α , q ) has a cardinality of | ξ | < c .This finishes the construction and B = { b α : α < c } is the set. (cid:3) It is quite easy to verify that the above theorem holds also in the case of R n for anynatural n and the proof remains almost intact.Peter Komjath (see [4]) proved that the real plane can be colored by countably manycolors such a way that each color is an ARD set. It implies that for each set of positiveouter measure there exists its ARD subset of positive measure too. With some additionalassumptions we may prove a little more surprising result. Let us recall that f ⊂ A ⊂ R is a partial bijection, if there exists a set A ⊂ R such that f : A → R is injection and { ( x , f ( x )) : x ∈ A } ⊂ A .T HEOREM
Assume that non ( N ) = co f ( N ) . Let A ⊂ R be a set of positiveouter measure. Then there exists a partial bijection f that is an ARD full subset of A. P ROOF . Let us assume that non ( N ) = co f ( N ) = κ . To construct a full subset B of A we only have to make sure that B intersects X ∩ A for each X of positive measuresuch that λ ∗ ( X ∩ A ) >