aa r X i v : . [ m a t h . L O ] A ug ON A RELATION BETWEEN THIN SETS ANDXOR-SETS
PAWEŁ PASTECZKA
Abstract.
We study properties of thin sets introduced recentlyby T. Banakh and E. Jabłońska. Using Banach-Mazur games weprove that all thin sets are Baire spaces and generic (in particularmaximal) thin sets are not Borel.We also show that the Cantor cube can be decomposed to twosets of this type. This result is related to so-called xor-sets definedby D. Niwiński and E. Kopczyński in 2014. Introduction
There are various definitions of sets which are “small” (is some sense).Few of them are for example Haar-null sets (Christensen [5]); openlyHaar-null sets (Solecki [12]); Haar-meager sets (Darji [6]) and Null-finite sets (Banakh–Jabłońska [3]). We study properties of thin setsintroduced in Banakh–Jabłońska [3] (there are also an alternative no-tion of thin sets which is outside the scope of this paper; Ger [7]).Denote the set of all nonnegative and positive integers by ω and N + ,respectively. A subset T of the Cantor cube Z ω is called thin if for everynumber n ∈ ω the restriction pr n | T of the projection pr n : Z ω → ω \{ n } given by pr n : x x | ω \{ n } is injective. Intuitively, each two distinctelements of T differs at at least two bits (see Lemma 1 for precisewording of this statement). This is a necessary and sufficient conditionfor the family of infinite streams of bits which allows to recognize thesimplest error (single bit differs). In this sense thin sets are the infinitecounterpart of a parity bit.Some properties of this family has been already studied by Banakh-Głąb-Jabłonska-Swaczyna [2]. In particular it is known [2, Proposi-tion 9.3] that each Borel thin subset of the Cantor cube is meager andhas Haar measure zero. We deliver some further properties of thin sets. Mathematics Subject Classification.
Key words and phrases. thin sets; xor sets; Banach-Mazur game; capturing strat-egy; decomposition of Cantor cube.Author is greatful to M. Bojańczyk, E. Jabłońska and D. Niwiński for theirvaluable remarks.
In turns out that they are deeply connected with Banach-Mazur games.The key results are obtained using the folk “capture-the-strategy” idea.We also study some special subtype of this family, so-called xor-setsintroduced by Niwiński–Kopczyński [10]. This allows us to prove thatCantor cube can be partitioned into two thin sets (see section 4 fordetails). 2.
Auxiliary results
Few properties of thin sets.
First, as a straightforward im-plication of the definition, we can prove that a subset of a thin setis also thin. Furthermore this family is closed under union of chains.These properties follows from analogous asserts of injective mappings(understood as a set of pairs with suitable assumptions).Next, we formulate a decomposition principle for thin sets. Used willbe so-called
Hamming distance [8], that is a function hd : Z ω × Z ω → ω ∪ { + ∞} given by hd( x, y ) := |{ n ∈ ω : x ( n ) = y ( n ) }| ( x, y ∈ Z ω ) . This notion is deeply connected with the errors recognition, correctionand checksums. One can easy show that hd is an extended metric on Z ω . Therefore we can define the equivalence relation ∼ on Z ω by x ∼ y : ⇐⇒ hd( x, y ) < + ∞ . Obviously hd is a metric on every element of Z ω / ∼ . In what follows weestablish the simple observation binding Hamming distance and thinsets. Lemma 1.
Let T ⊂ Z ω . The following statements are equivalent:(i) T is thin;(ii) every class of T / ∼ is thin;(iii) hd( x, y ) = 1 for all x, y ∈ T .Proof. Implication ( i ) ⇒ ( ii ) is obvious as every subset of a thin set isthin.To prove ( ii ) ⇒ ( iii ) assume to the contrary that hd( x, y ) = 1 forsome x, y ∈ T . Then x ∼ y and { n ∈ ω : x ( n ) = y ( n ) } = { n } for some n ∈ ω . Thus we have pr n ( x ) = pr n ( y ) . By the definition of thinset this implies x = y , and consequently hd( x, y ) = 0 contradicting theassumption.To show ( iii ) ⇒ ( i ) assume that T is not a thin set. Then there exists n ∈ ω and two distinct elements x, y ∈ T such that pr n ( x ) = pr n ( y ) .Then { n ∈ ω : x ( n ) = y ( n ) } = { n } , i.e. hd( x, y ) = 1 . (cid:3) N A RELATION BETWEEN THIN SETS AND XOR-SETS 3
By the above results and Zorn Lemma we obtain next proposition.
Proposition 1.
Let T ⊂ Z ω be a thin set. Then there exists a maximalthin set T ⊂ Z ω such that T ⊆ T . Moreover for all Q ∈ Z ω / ∼ wehave that T ∩ Q is a maximal thin subset of Q .Proof. As thin sets are closed under union of chains, the first part isan immediate implication of Zorn lemma. To show the moreover partassume that there exists Q ∈ Z ω / ∼ such that T ∩ Q is not a maximalthin subset of Q . Then there exists q ∈ Q \ T such that ( T ∩ Q ) ∪ { q } is thin. Therefore applying the implication ( ii ) ⇒ ( i ) in Lemma 1 weobtain that T ∪ { q } is also thin contradicting the maximality. (cid:3) Banach-Mazur game.
Following Berwanger-Grädel-Kreutzer [4]consider a special type of Banach-Mazur game parameterized by a set F ⊂ Z ω (with the product, i.e. Tychonoff topology). Let G ( F ) be aninfinite two-player game with a complete information, where moves ofplayers consist of selecting and extending finite path through a com-plete binary tree Z ω by an element in Z +2 := S ∞ n =1 Z n . The playerswill be called Ego and Alter. The two players alternate turns, andeach player is aware of all moves before making the next one; Ego be-gins. All plays are infinite and the result outcome of each play is anelement of x ∈ Z ω . Ego wins if x ∈ F , otherwise Alter wins. Fordetailed history of this games we refer the reader to Oxtoby [11] andTelgársky [13].Using some unraveling techniques it is possible to embed G ( F ) to theclassical Banach-Mazur game on a tree Z ω (see [4] for details). Thus wecan reformulate the original Banach-Mazur theorem [1] in the flavour ofBerwanger-Grädel-Kreutzer. Prior to this we need to recall the notionof strategy.Ego’s and Alter’s strategy are the functions e : ∞ [ n =0 ( Z +2 ) n → Z +2 and a : ∞ [ n =1 ( Z +2 ) n → Z +2 , respectively. Denote sets of all Alter’s and Ego’s strategies by A and E . For a ∈ A and e ∈ E one can consider a sequence of moves ǫ := e ( ∅ ) ,α i := a ( ǫ , . . . , ǫ i − ) , i ∈ N + ǫ i := e ( α , . . . , α i ) , i ∈ N + and define a play (heareafer we use the classical abbreviation to con-tatenation of sequences) Play :
E × A ∋ ( e, a ) ( ǫ α ǫ α ǫ α . . . ) ∈ Z ω . PAWEŁ PASTECZKA
This is very usual notion in game theory – instead of sequence of movesplayers show whole strategy at the beginning. We also use the time-lapse approach to a strategy. We treat it as a sequence of replies for theopponent’s moves and write it in terms of pseudocode – it is a classicalapproach in game theory which is equivalent to the one above.We say that a ∈ A is an Alter’s winning strategy (in G ( F ) ) if Play( e, a ) / ∈ F for all e ∈ E . Analogously e ∈ E is an Ego’s winningstrategy (in G ( F ) ) if Play( a, e ) ∈ F for all a ∈ A . If one of playershas a winning strategy then the game G ( F ) is determined . Now we canrecall celebrated Banach-Mazur theorem. Theorem 1 (Banach-Mazur) . Let F ⊂ Z ω .(1) Alter has a winning strategy for the game G ( F ) if and only if F is meager.(2) Ego has a winning strategy for the game G ( F ) if and only ifthere exists finite word x ∈ Z +2 such that ( x · Z ω ) \ F is meager. As a result we have the following corollary.
Corollary 1.
Games G ( F ) are determined for all Borel sets F . In the following two propositions we present necessary conditions forEgo and Alter to have a winning strategy. The first result essentiallyfollows the idea of Niwiński and Kopczyński from [10].
Proposition 2.
Let F ⊆ Z ω be a thin set. Then Ego has no winningstrategy in a game G ( F ) .Proof. Assume to the contrary that Ego has a winning strategy in G ( F ) .We play this game two times simultaneously – we call them “initial”and “mirror” play. Denote the Ego’s moves in the initial and mirrorplays as ( α i ) ∞ i =0 and ( β i ) ∞ i =0 , respectively. Obviously α = β as thefirst Ego’s move is fixed.First Alter’s reply in the initial play is . In the mirror play it is (1 α ) . From now on Alter capture the Ego’s strategy in the followingway: • each time Ego plays α k ( k ≥ ) in the initial play it is the Alterplays α k in a mirror play; • each time Ego plays β k ( k ≥ ) is the mirror play it is the Alterplays β k in the initial play.These two plays can be illustrated in the following table-like form: N A RELATION BETWEEN THIN SETS AND XOR-SETS 5
Initial playEgo α α α . . . α k . . . Alter β β . . . β k . . . Mirror playEgo α β β . . . β k . . . Alter α α . . . α k . . . The final outcome of the initial and mirror plays are a := ( α α β α β · · · ) and b := ( α α β α β · · · ) , respectively. As Ego has a winning strategy in G ( F ) then we obtain a, b ∈ F . However in this case we have pr | α | +1 ( a ) = pr | α | +1 ( b ) , i.e. pr | α | +1 is not injective. This implies that F is not a thin set, contra-dicting the assumptions. (cid:3) Krom [9] proved that Ego has no winning strategy in G ( F ) if andonly if F ⊂ Z ω is the Baire space. Thus applying Proposition 2 weimmediately obtain that thin sets are Baire spaces.Now we are heading toward the necessary condition to Alter’s win-ning strategy, however we need to introduce few notions first. For k ∈ ω define the function bit k : ω → { , } such that bit k ( x ) is the k -th bitfrom the right in the binary notation of x (counting from zero). Moreprecisely, for k, n ∈ ω we have bit k ( n ) = ( if ( n mod 2 k +1 ) ∈ { , . . . , k − } , if ( n mod 2 k +1 ) ∈ { k , . . . , k +1 − } . For n ∈ N , x ∈ { , } n and m ∈ { , , . . . , n − } define Θ( x, m ) ∈{ , } n as follows ( ⊕ stands for a binary xor) Θ( x, m ) := ( x k ⊕ bit k ( m )) k ∈{ ,...,n − } . For an infinite sequence x ∈ { , } ω and m ∈ ω define Θ( x, m ) ∈{ , } ω by Θ( x, m ) := ( x k ⊕ bit k ( m )) k ∈ ω . We can now proceed to formulate and proof the most technical propo-sition of this paper.
Proposition 3.
Let F ⊆ Z ω be a set such that Alter has a winningstrategy in a game G ( F ) . Then there exists an element X ∈ Z ω / ∼ suchthat X ∩ F = ∅ .Proof. Fix an Alter strategy. Now we consider infinitely many plays of G ( F ) and show that their output covers whole class of abstraction of ∼ . PAWEŁ PASTECZKA
Let ( v i ) ∞ n =1 of elements in S ∞ n =1 n be a sequence of Alter replies(in all plays), enumerated by the order of moves. Ego spreads Altersreplies among all plays in a way which are described by the algorithmbelow. There are two types of Ego’s moves: Start i ( α ) and Move i ( α ) for i ∈ ω and α ∈ Z +2 :1. Start i ( α ) – Ego starts Play i with the initial move α ;2. Move i ( α ) – Ego makes a subsequent move α in Play i .We now present the algorithmic description of the Ego’s strategy(in infinitely many plays). It depends on the Alter’s strategy which isemphasized as an argument (this is a sort of an input stream to thisprocedure). procedure Capture (Alter Strategy σ ) Start (0) Alter replies: v for i = 1 to + ∞ do Start i (cid:0) Θ(0 v . . . v i ( i +3)2 − , i ) (cid:1) Alter replies: v i ( i +3)2 for j = 0 to i do Move j (cid:0) v i ( i +1)2 +1+ j . . . v i ( i +1)2 + i + j (cid:1) Alter replies: v i ( i +3)2 + j +1 end forend forend procedure Similarly to the previous proof let us illustrate several first moves in atabular form. Play 0Ego v v v v v v Alter v v v v Play 1Ego v v v v v v v Alter v v v . . .Play 2Ego Θ(0 v v v v , v v . . .Alter v v Play 3Ego
Θ(0 v v v v v v v v , . . .Alter v . . .Obviously both players make infinitely many moves in each of plays.Furthermore, in order to show that this algorithm is correct, we need N A RELATION BETWEEN THIN SETS AND XOR-SETS 7 to show that Alter’s replies are properly enumerated (i.e. id of the ele-ment coincide with the replies number). This proof is a straightforwardapplication of the “loop invariant” method.Now let r i ∈ Z ω be the output of Play i ( i ∈ ω ). As Ego rewrites allAlter answers except the initial move we obtain r i = Θ( r , i ) for all i ∈ ω. Therefore if Alter has a winning strategy we get { r i : i ∈ ω } ∩ F = ∅ ,and thus [ r ] ∼ ∩ F = ∅ . (cid:3) Main result
In this brief section we present three results. First of them is pre-sented in the game setting approach, while second and third one aretopological properties of thin sets.
Proposition 4.
Let T ⊂ Z ω be a thin set such that T ∩ X = ∅ for all X ∈ Z ω / ∼ . Then the game G ( T ) is undetermined.Proof. Since T is thin, Proposition 2 implies that Ego has no winningstrategy in G ( T ) . On the other hand as T ∩ X = ∅ for all X ∈ Z ω / ∼ by Proposition 3 we get that Alter has no winning strategy in G ( T ) ,too. (cid:3) Now, applying Corollary 1, we can formulate the main result of thispaper.
Theorem 2. If T ⊂ Z ω is a thin set such that T ∩ X = ∅ for all X ∈ Z ω / ∼ then T is not Borel. As a singleton is a thin set, by Proposition 1 we easily obtain
Corollary 2.
Maximal thin sets are not Borel. Applications to Xor-sets
We show that xor-sets introduced by Niwiński–Kopczyński [10] aremaximal thin sets. Before we go into details, let us introduce some sortof conjugency. For x ∈ Z ω and n ∈ ω we define x n ∈ Z ω by x n ( k ) := ( x ( k ) k ∈ ω \ { n } , − x ( n ) k = n. For fixed n ∈ ω the operator ( · ) n is a symmetry, i.e. ( x n ) n = x forall x ∈ Z ω . We are now in the position to present the main definitionof this section. PAWEŁ PASTECZKA
Definition 3.
A set
X ⊂ Z ω is called a xor-set if for every n ∈ ω and x ∈ Z ω we have x ∈ X ⇐⇒ x n / ∈ X .As ( · ) n is a symmetry we can easily check that for every xor-set X the set Z ω \X is a xor-set, too. We prove that xor-sets are maximal thinsets. Before we go into details let us introduce few technical notions.First, let us define the relation ≈ on Z ω by x ≈ y : ⇐⇒ hd( x, y ) is finite and even.Observe that x ≈ y implies x ∼ y . Moreover each element of Z ω / ∼ split into two elements of Z ω / ≈ . Therefore, as Z ω / ∼ has a cardinalitycontinuum, one consider a partition of Z ω to a family of disjoint sets U := { U η,j : η ∈ R and j ∈ { , }} such that U η, , U η, ∈ Z ω / ≈ and U η, ∪ U η, ∈ Z ω / ∼ ( i ∈ R ). These sets play an essential role in thetheory of xor-sets. Lemma 2.
Set
X ⊂ Z ω is a xor-set if and only if there exists a selector S of (cid:8) { U η, , U η, } : η ∈ R (cid:9) such that X = S S .Proof. Let η ∈ R . Observe that for all x ∈ U η, ∪ U η, and n ∈ ω wehave x n ∼ x and x n x . Therefore x ∈ U η, ⇐⇒ x n ∈ U η, .Furthermore for all y ∈ Z ω such that y ≈ x there exists k ∈ N and aset { i , . . . , i k } such that y = ( · · · (( x i ) i ) · · · ) i k .Therefore x ∈ X implies [ x ] ≈ ⊂ X .Furthermore x ∈ U η, ⇐⇒ y ∈ U η, and x ∈ U η, ⇐⇒ y ∈ U η, .Thus [ x ] ≈ ∈ { U η, , U η, } . As [ x ] ∼ contains two classes of abstraction of ≈ we get (cid:8) U η, , U η, (cid:9) = (cid:8) [ x ] ≈ , ([ x ] ∼ \ [ x ] ≈ ) (cid:9) . By the definition every xor-set contains exactly one element of eachpair, which implies our assertion. (cid:3)
Proposition 5.
Every xor-set
X ⊂ Z ω is a maximal thin set. Inparticular, X is not Borel.Proof. As X a xor-set we have that for all x, y ∈ X the distance hd( x, y ) is either infinite of even. Then the implication ( iii ) ⇒ ( i ) in Lemma 1yields that X is a thin set.To show the maximality assume to the contrary that there exist axor-set X and an element x ∈ Z ω \ X such that X ∪ { x } is thin. Bythe definition of xor-set we have x ∈ X . However π ( x ) = π ( x ) which lead to a contradiction as X ∪ { x } was supposed to be thin.The remaining part is a straightforward implication of Corollary 2. (cid:3) N A RELATION BETWEEN THIN SETS AND XOR-SETS 9
As a complementary of a xor-set is a xor-set we obtain the followinginteresting property.
Corollary 3.
There exists two non-Borel, thin and disjoint sets T , T ⊂ Z ω such that T ∪ T = Z ω . In fact we can also prove a sort of the reverse statement
Proposition 6.
Let T , T be two thin sets such that T ∪ T = Z ω .Then T and T are disjoint xor-sets.Proof. Indeed, as T is thin and T ∪ T = Z ω we have(4.1) x ∈ T ⇒ x n / ∈ T ⇒ x n ∈ T for all x ∈ Z ω and n ∈ ω. Similarly x ∈ T ⇒ x n ∈ T , which yields x ∈ T ⇐⇒ x n ∈ T .If there existed x ∈ T ∩ T then by (4.1) we would obtain x n ∈ T ,contradicting the fact that T is a thin set. Therefore T ∩ T = ∅ .Then we have x ∈ T i ⇐⇒ x n / ∈ T i for all x ∈ Z ω , n ∈ ω and i ∈ { , } which shows that both T and T are xor-sets. (cid:3) Remark 1.
Applying above results we can easily show that
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