Packing index of subsets in Polish groups

P A CKING INDEX OF SUBSETS IN POLISH GROUPS T ARAS BANAKH, NAD Y A L Y ASK OVSKA, DU ˇ SAN REPO V ˇ S Abstract. F or a subset A of a Polish group G , w e study the (almost) packing index ind P ( A ) (resp. Ind P ( A )) of A , equal to the supremum of cardinalities | S | of subset s S ⊂ G suc h that the family of s hi fts { xA } x ∈ S is (almost) disjoi nt (in the s ense that | xA ∩ yA | < | A | f or any distinct p oints x, y ∈ S ). Sub sets A ⊂ G with small (almost) pac king index are small in a geomet ric sense . W e s how that ind P ( A ) ∈ N ∪ {ℵ 0 , c } for any σ -compact subset A of a Polish group. If A ⊂ G is Borel, then the packing indices ind P ( A ) and Ind P ( A ) cannot tak e v alues in the half-interv al [ s q (Π 1 1 ) , c ) where s q (Π 1 1 ) is a certain uncoun table cardinal that is small er than c i n some models of ZFC. In each non-discrete Po lish Ab elian group G we construct tw o closed subsets A, B ⊂ G with ind P ( A ) = ind P ( B ) = c and Ind P ( A ∪ B ) = 1 and then apply this result to sho w that G cont ains a nowhere dense Haar null subset C ⊂ G with ind P ( C ) = Ind P ( C ) = κ for any given cardinal num ber κ ∈ [4 , c ]. 1. Introduction Given a Polish gro up G and a non-empt y subset A ⊂ G with nice descriptive prop erties, we study all po ssible v alues of the packing index ind P ( A ) = sup  | S | : S ⊂ G and { xA } x ∈ S is disjo in t  of A , which indicates the smallnes s of the subset A in a geometric sense. The pap ers [BL1], [BL2], [L1] and [L2] are devoted to constructing subsets with a given packing index. In pa rticular, for every non-ze ro ca rdinal num b er κ ≤ c one ca n ea sily construct a subset A ⊂ R with ind P ( A ) = κ . After discussing thos e r esults o n the top ological seminar at W ro claw Universit y the second author w as a sked by Kr ysztof Omiljanowski ab out p ossible r estrictions on the packing index ind P ( A ) of subsets A ⊂ R ha ving go o d descriptive prop erties (lik e b eing compact σ -compact, Borel, measurable or meager). This question was pr obably motiv ated by the well-known fact that the Contin uum Hypo thesis (though inreso lv able in ZFC) has pos itive solution in the realm of B orel s ubsets of the rea l line: each uncountable Bor el subset A ⊂ R has cardinality c of contin uum. In this pa pe r w e shall give several partial answers to Omiljanowski’s question. On the one hand, we show that σ -co mpact subsets A in Polish groups cannot hav e an intermediate packing index ℵ 0 < ind P ( A ) < c . F or a Bo rel subset A of a Polish gr oup w e hav e a w eaker r esult: ind P ( A ) cannot take the v alue in the in terv al sq (Π 1 1 ) ≤ ind P ( A ) < c where sq (Π 1 1 ) stands for the smallest ca rdinality κ s uch that each coana lytic subset X ⊂ 2 ω × 2 ω contains a square S × S of size | S × S | = c provided X contains a square of size ≥ κ . The v alue of the small uncoun table 2000 Mathematics Subje ct Classific ation. 22A99, 05D99. Key wor ds and phr ases. P olish group, packing index, Borel subset. This reseac h w as supported the SRA grants P1-0292 00101-04, J1-9643-0101 and BI-UA/07- 08-001. 1 2 T ARAS BANAKH, NADY A L Y ASKO VSKA, DU ˇ SAN REPOV ˇ S cardinal sq (Π 1 1 ) is not completely determined b y ZF C Axioms: both the equality sq (Π 1 1 ) = c and the strict inequalit y s q (Π 1 1 ) < c are consistent w ith Martin Axiom, see [Sh]. On the other hand, for every infinite cardina l num ber κ ≤ c in each non-discr ete Polish Ab elian g roup G we shall co nstruct a nowhere dense Haar null subset A ⊂ G with ind P ( A ) = Ind P ( A ) = κ . Here Ind P ( A ) = sup {| S | : S ⊂ G a nd { xA } x ∈ S is almost disjoin t } is the almost p acking index of A . In the above definition, a family of shifts { xA } x ∈ S is defined to b e almost disjoint if | xA ∩ y A | < | A | for all distinct x, y ∈ S . T o co nstruct the nowhere dens e Haar n ull subset A ⊂ G with a giv en (almost) packing index, in each non- discrete Polish Abelian group G we first construct a closed nowhere dens e Haa r null subset C ⊂ G with Ind P ( C ) = 1. The set C , b eing nowhere dense and Haa r null, is small in the sense of categor y and measure, but is large in the geometric sense b ecause for an y t wo distinct po int s x, y ∈ G the shifts xC and y C hav e in tersectio n of car dinality | xC ∩ y C | = | G | . In pa rticular, C C − 1 = G and th us C is a closed nowhere dense Haar n ull subset that algebr aically generates the group G . This r esult can be seen as an extension of a res ult of S.Solecki [So] who prov ed that each non- lo cally compact Polish Abelia n group G is algebr aically gener ated by a nowhere dense subset. Also it extends some results of [BP, § 1 3]. In fact, the closed Haar null subset C ⊂ G with Ind P ( C ) = 1 is constructed as the union C = A ∪ B of tw o closed subsets A, B ⊂ G with ind P ( A ) = ind P ( B ) = c . This shows that the packing index is highly non-additive. Notation. By ω we denote the first infinite ordinal, N = ω \ { 0 } s tands for the set of positive integers. Ca rdinals are iden tified with the initial or dinals of g iven cardinality; c stands for the cardinalit y of con tin uum. All top olog ical gr oups G considered in this pap er will b e supplied with an in v aria nt metric ρ gener ating the top ology of G . By e we denote the iden tit y element of G . F or a p oint x ∈ G and a real num b er r by B ( x, r ) = { g ∈ G : ρ ( g , x ) < r } w e denote the o pe n r -ball centered at x . Also for x ∈ G w e put k x k = ρ ( x, e ). The in v a riantness of ρ implies k x − 1 k = k x k and k xy k ≤ k x k + k y k fo r all x, y ∈ G . 2. The p acking indices of σ -comp act sets in Polish groups In this section we show that the packing index o f a σ -compact subset in a Polish groups canno t take an in termediate v alue b etw een ω and c . First we prov e a useful Lemma 1. L et A b e a su bset of a Polish gr oup G . If ind P ( A ) < c , then the closur e of AA − 1 c ont ains a neighb orho o d of the neutr al element e of G . Pr o of. Fix any complete metric ρ generating the top ology o f the P olish g roup G . Assuming that AA − 1 is not a n eighbor ho o d of e , w e shall construct a p erfect subset K ⊂ G suc h that ( xA ) x ∈ K is disjo in t, which will imply that ind P ( A ) = | K | = c . T aking into acco unt that the closed subset AA − 1 is not a neighbo rho o d of e in G , for an y o p en neighbor ho o d U of e we can find a p oint b ∈ U \ AA − 1 and an op en neighbor ho o d V of e suc h that V − 1 bV ⊂ U \ AA − 1 . Using this fact, by induction construct a sequence ( b n ) n ∈ ω of points in G and sequences ( U n ) n ∈ ω and ( V n ) n ∈ ω of op en neigh b orho o ds of e in G suc h that P ACKING INDEX OF SUBSETS IN POLIS H GR OUPS 3 (1) b n ∈ U n = U − 1 n ; (2) V − 1 n +1 b n V n +1 ∩ AA − 1 = ∅ ; (3) b n / ∈ V n +1 V − 1 n +1 ; (4) diam ρ ( bV n +1 ) < 2 − n for any p oint b ∈ B n =  b ε 0 0 · · · b ε n n : ε 0 , . . . , ε n ∈ { 0 , 1 }  ; (5) U 3 n +1 ⊂ V n +1 ⊂ U n . Define a map f : { 0 , 1 } ω → G assigning to each infinite binary sequence ~ ε = ( ε i ) i ∈ ω the infinite pro duct f ( ~ ε ) = ∞ Y i =0 b ε i i = lim n →∞ f n ( ~ ε ) where f n ( ~ ε ) = Q n i =0 b ε i i . Let us sho w that the latter limit exists. It suffices to chec k that ( f n ( ~ ε )) n ∈ ω is a Ca uch y sequence in ( G, ρ ). The condition (5) implies that U 2 i +1 ⊂ U i for all i . This can be used as the inductive step in the pr o of of the inclusion U n · · · U m ⊂ U 2 n for all m ≥ n . Then for every m ≥ n f m ( ~ ε ) ∈ f n ( ~ ε ) · U n +1 · · · U m ⊂ f n ( ~ ε ) · U 2 n +1 ⊂ f n ( ~ ε ) V n +1 and thus ρ ( f m ( ~ ε ) , f n ( ~ ε )) ≤ diam ρ ( f n ( ~ ε ) · V n +1 ) < 2 − n by the condition (4). Therefore, the sequence ( f n ( ~ ε )) n ∈ ω is Cauch y and the limit f ( ~ ε ) = lim n →∞ f n ( ~ ε ) exists. Moreov er, the u pp er bo und ρ ( f m ( ~ ε ) , f n ( ~ ε )) ≤ 2 − n implies tha t the map f : { 0 , 1 } ω → G is cont inuous. On the other hand, the inclusions f m ( ~ ε ) ∈ f n ( ~ ε ) · U 2 n , m ≥ n , imply that f ( ~ ε ) ∈ f n ( ~ ε ) · U 2 n ⊂ f n ( ~ ε ) · U 3 n ⊂ f n ( ~ ε ) · V n +1 . This inclus ion will b e us ed for the pro of o f the injectivity o f f . W e sha ll pr ov e a little bit more: for an y distinct vectors ~ ε and ~ δ in { 0 , 1 } ω we get f ( ~ ε ) A ∩ f ( ~ δ ) A = ∅ . Find the sma llest num b er n ∈ ω such that ε n 6 = δ n . W e lose no gener ality a ssuming that δ n = 0 and ε n = 1. It follows that f ( ~ ε ) ∈ f n ( ~ ε ) U 3 n +1 = f n − 1 ( ~ ε ) b n V n +1 while f ( ~ δ ) = f n ( ~ δ ) V n +1 = f n − 1 ( δ ) · e · V n +1 = f n − 1 ( ~ ε ) V n +1 . Then  f ( ~ δ )  − 1 f ( ~ ε ) ∈ V − 1 n +1 b n V n +1 ⊂ G \ AA − 1 by the condition (2 ) and hence f ( ~ ε ) A ∩ f ( ~ δ ) A = ∅ . Thu s the family ( xA ) x ∈ K is disjoint wher e K = f ( { 0 , 1 } ω ). The injectivit y of f implies that ind P ( A ) ≥ | K | = c .  Now we can pr ov e the main result of this section. Theorem 1. If A is a σ -c omp act s u bset A in a Po lish gr oup G , then ind P ( A ) ∈ N ∪ {ℵ 0 , c } . Mor e over, if the set A is c omp act, then (1) ind P ( A ) = c if G is not lo c al ly c omp act; (2) ind P ( A ) ∈ {ℵ 0 , c } if G is lo c al ly c omp act but not c omp act; (3) ind P ( A ) ∈ N ∪ { c } if G is c omp act. Pr o of. If A is σ -c ompact, then so is the set AA − 1 = { xy − 1 : x, y ∈ A } and then the set ( G \ AA − 1 ) ∪ { e } is a G δ -set in G . In its turn, the subset X = { ( x, y ) ∈ G × G : y − 1 x ∈ ( G \ AA − 1 ) ∪ { e }} 4 T ARAS BANAKH, NADY A L Y ASKO VSKA, DU ˇ SAN REPOV ˇ S is of type G δ in G × G , b eing the preima ge of the G δ -subset ( G \ AA − 1 ) ∪ { e } under the contin uous map g : G × G → G , g : ( x, y ) 7→ y − 1 x . Assuming tha t ind P ( A ) > ℵ 0 , we could find a n uncountable subs et S ⊂ G with disjoint family { xA } x ∈ S , which implies that S × S ⊂ X . Since the Polish space X contains the uncoun table square S × S , we can a pply the Shelah’s result [Sh, 1.14] to conclude tha t X co ntains the square P × P of a p erfect subset P ⊂ G (the latter means that P is clo sed in G and ha s no is olated p oint). It follows from P × P ⊂ X that the family { xA } x ∈ P is disjo in t and th us c = | P | ≤ ind P ( A ) ≤ | G | = c . Now assuming that A ⊂ G is compact w e shall prov e the items (1)–(3) of the theorem. The compactness of A implies the compa ctness of AA − 1 . If AA − 1 is not a neighbo rho o d o f e , then we can apply Lemma 1 to conclude that ind P ( A ) = c . This is so if the g roup G is not lo c ally compac t. So, next w e as sume that AA − 1 is a neighborho o d of e . In this case the group is lo cally compact and we can take a neighborho o d U ⊂ G of e with U U − 1 ⊂ AA − 1 . Then for every B ⊂ G with B − 1 B ∩ AA − 1 = { e } w e get B − 1 B ∩ U U − 1 = { e } , which implies t hat t he f amily ( x U ) x ∈ B is disjoint a nd the set B is at most co untable, being dis crete in the Polish s pace G . This prov es the upp er b ounds ind P ( A ) ≤ ℵ 0 . If the group G is not compac t, then using the Zorn Lemma, we ca n find a maximal set B ⊂ G with B − 1 B ∩ AA − 1 = { e } . W e claim that B AA − 1 = G . Assuming the conv erse, we co uld find a p oint b ∈ G \ B AA − 1 . Then the set bA is disjoint from the set B A and hence we ca n enlarge the s et B to the set ˜ B = B ∪ { b } s uch that ( xA ) x ∈ ˜ B is disjoint. The latter is equiv alent to ˜ B − 1 ˜ B ∩ AA − 1 = { e } and this contradicts the max imality of B . The compactness of AA − 1 and non-co mpactness of G = B AA − 1 implies that B is infinite and th us ind P ( A ) ≥ | B | ≥ ℵ 0 . This completes the proo f of the second item of the theorem. T o prov e the third item, assume that G is compact. In this case G carries a Haar measure µ whic h is a unique pro bability inv ar iant σ - additive Borel measur e on G . If AA − 1 is not a neighbo rho o d of e , then ind P ( A ) = c b y a preceding c ase. So w e assume that AA − 1 is a neighborho o d of e and take another neighbo rho o d U of e with U U − 1 ⊂ AA − 1 . Since finitely man y shifts of U cover the group G , we get µ ( U ) > 0. Now given a ny subset B ⊂ G with B − 1 B ∩ AA − 1 = { e } , w e get B − 1 B ∩ U U − 1 = { e } . The latter equalit y implies that the family ( xU ) x ∈ B is disjoin t and then 1 = µ ( G ) ≥ µ ( B U ) = | B | µ ( U ) implies that | B | ≤ 1 /µ ( U ). Consequently , the packing index ind P ( A ) ≤ 1 /µ ( U ) is finite.  In light of this theorem tw o op en q uestions ar ise naturally . Question 1. Is ther e a c omp act gr oup G and a σ -c omp act subset A with ind P ( A ) = ℵ 0 ? Question 2. Is ther e a Polish gr oup G and a Bor el subset A ⊂ G with ℵ 0 < ind P ( A ) < c ? The latter question do es not reduce to the σ -co mpact case b ecause of the fol- lowing e xample (in whic h T = R / Z s tands for the circle). Prop ositio n 1. The c ountable pr o duct G = T ω c ont ains a G δ -subset A ⊂ G such that ind P ( A ) = c but e ach σ -c omp act subset B ⊂ T ω c ont aining A has ind P ( B ) < ℵ 0 . P ACKING INDEX OF SUBSETS IN POLIS H GR OUPS 5 Pr o of. Let q : Z → T denote the quo tien t ma p, J = q  [0 , 1 2 )  be the half-circle , I = J = q ([0 , 1 2 ]) b e its clo sure, and D = q ( { 0 , 1 2 } ) b e t w o o ppo site p oints on T . It is clear that D − 1 D ∩ J J − 1 = { e } while I · I − 1 = T . It follows that A = J ω is a G δ -subset o f T ω with ind P ( A ) = | D ω | = c b ecaus e ( D ω ) − 1 D ω ∩ AA − 1 = { e } . Now given an y σ -compact subset B ⊃ A in T ω , we should chec k that ind P ( B ) < ℵ 0 . Replacing B b y B ∩ I ω , if necessa ry , w e ca n assume that B ⊂ I ω . Since B ⊂ I ω contains the dense G δ -subset J ω of I ω , the standard application o f the Baire Theo rem yields an non-empt y open subset U ⊂ I ω with U ⊂ B . W e lose no generality assuming that U is of bas ic form U = V × I ω \ n for some n ∈ ω and so me op en set V ⊂ I n . O bserve that U − 1 U = V V − 1 × I ω \ n ( I ω \ n ) − 1 = V V − 1 × T ω \ n is an o pen neigh b orho o d of e in T ω . Co nsequently , B B − 1 ⊃ U U − 1 is also an op en neighborho o d o f e in T ω . P ro ceeding a s in the pro of of the first item of Theor em 1, we can s ee that ind P ( B ) ≤ 1 /µ ( V × T ω − n ) < ℵ 0 .  3. The p acking indices of anal ytic sets in Polish groups In this section we shall give a partial answ er to Question 2 related to the packing indices of Bo rel subsets in Polish groups. It is well-known that each Borel subset of a P olish space is a nalytic. W e r ecall that a metr izable space X is analytic if X is a co nt inuous image of a Polish space. A spa ce X is c o analytic if for s ome Polish space Y con taining X the c omplement Y \ X is analytic. The classes of analytic and coanalytic s paces are deno ted by Σ 1 1 and Π 1 1 , resp ectively . It is well-known that the intersection ∆ 1 0 = Σ 1 1 ∩ Π 1 1 coincides with the class of a ll absolute Borel (metrizable separ able) spaces. By Σ 0 2 and Π 0 2 we denote the clas ses of σ -compact and Polish spa ces, res pec tively . W e shall sa y that a subset X ⊂ 2 ω × 2 ω contains a square of size κ if there is a subset A ⊂ 2 ω with A × A ⊂ X a nd | A × A | = κ . Given a class C of spaces denote b y sq ( C ) t he smallest cardina l κ s uch that each subspace X ∈ C of 2 ω × 2 ω that contains a square of size κ co nt ains a squa re of size c . The Shela h’s result [Sh] (applied in the pr o of of Theor em 1) guarantees that sq (Π 0 2 ) = ℵ 1 . F o r o ther descr iptive classes C the v alue sq ( C ) is not so definite and dep ends o n Set-T heoretic Axioms. In particular, the Contin uum Hyp othesis implies that sq (Σ 0 2 ) = sq (Σ 1 1 ) = sq (Π 1 1 ) = c . On one hand, the strict inequality sq (Π 1 1 ) < c is co nsistent with ZFC+MA, see [Sh, 1.9, 1 .10]. How ever, there is a subs tant ial difference b et ween the classes Π 0 2 and Σ 0 2 of Polish and σ -co mpact spaces . B y [Sh] each Polish space X ⊂ 2 ω × 2 ω containing an uncountable squar e contains a Perfect s quare. On the other hand, there is a ZFC-example of a σ -c ompact subspac e X ⊂ 2 ω × 2 ω that co n tains a s quare of size ℵ 1 but not the p erfect one, s ee [K]. Prop ositio n 2. L et A b e an analytic subset of a Polish gr oup G . If ind P ( A ) ≥ sq (Π 1 1 ) , then ind P ( A ) = c . Pr o of. Using the fact tha t each Polish space is a contin uous one-to-o ne ima ge of a zero-dimensio nal P olish space, we can show that sq (Π 1 1 ) coincides with the smallest 6 T ARAS BANAKH, NADY A L Y ASKO VSKA, DU ˇ SAN REPOV ˇ S cardinal κ such that for an y P olish space X a coanalytic subse t C ⊂ X × X con tains a squar e o f size c provided C co nt ains a sq uare o f size ≥ κ . Given an analytic subse t A of a Polish group G we can s ee that b oth the sets AA − 1 and AA − 1 \ { e } are a nalytic and thus the set C = { ( x, y ) ∈ G × G : y − 1 x / ∈ AA − 1 \ { e }} is co analytic. Assuming that ind P ( A ) ≥ sq (Π 1 1 ), we could find a subset S ⊂ G of size | S | ≥ sq (Π 1 1 ) such that the family { xA } x ∈ S is disjoint. The latter is equiv alent to S − 1 · S ⊂ G \ ( AA − 1 \ { e } ) and to S × S ⊂ C . By the eq uiv alent definition of sq (Π 1 1 ) (with 2 ω replaced by any Polish space), the coanalytic s ubset C ⊂ G × G c ontains a square K × K of size c (b ecause it cont ains the square S × S of ca rdinality | S × S | ≥ sq (Π 1 1 )). It follows fro m K × K ⊂ C that the family { xA } x ∈ K is dis joint and thus ind P ( A ) ≥ | K | = c .  A similar result holds for the a lmost packing index Ind P ( A ) = sup {| S | : S ⊂ G and { xA } x ∈ S is almost disjoin t } o f A. W e r ecall that { xA } x ∈ S is almost disjoi nt if | xA ∩ y A | < | A | for any distinct p oints x, y ∈ S . In the proo f of the following theorem we shall use a kno wn fact o f the Descr iptiv e Set Theory saying that for a B orel subset A ⊂ X × Y in the pro duct of tw o Polish spaces the set { y ∈ Y : | A ∩ ( X × { y } ) | ≤ ℵ 0 } is co analytic in Y , see [Ke, 18 .9]. Prop ositio n 3. L et A b e a Bor el subset of a Polish gr oup G . If Ind P ( A ) ≥ sq (Π 1 1 ) , then Ind P ( A ) = c . Pr o of. It follows from Ind P ( A ) ≥ sq (Π 1 1 ) > ℵ 0 that the space G is uncoun table. If A is c ountable, then trivially , I nd P ( A ) = ind P ( A ) = c . So w e assume that A is uncoun table. Firs t we sho w that the subset C = { x ∈ G : | A ∩ xA | ≤ ℵ 0 } is coana lytic. Consider the homeomor phism h : G × G → G × G , h : ( x, y ) 7→ ( x, y − 1 x ), a nd the Borel subset B = h ( A × A ) ⊂ G × G . Since C = { z ∈ G : | B ∩ ( G × { z } ) | ≤ ℵ 0 } , w e may apply the mentioned result [Ke, 18 .9] to conclude that the set C is coanalytic. Then the set D = { ( x, y ) ∈ G × G : y − 1 x ∈ C } is coanalytic as the preimag e o f the coanalytic subset under a contin uous map betw een Polish spa ces. Assuming that Ind P ( A ) ≥ sq (Π 1 1 ), we could find a subset S ⊂ X such that | S | ≥ sq (Π 1 1 ) such that the family { xA } x ∈ S is almost disjoin t. Then for an y distinct x, y ∈ S the in tersection xA ∩ y A , b eing a B orel subse t of cardinality | xA ∩ y A | < | A | ≤ c , is at most coun table. Consequently , y − 1 x ∈ C and thus S × S ⊂ D . By the equiv alent definition of sq (Π 1 1 ), the coanalytic se t D contains a square K × K of size c . It follows from K − 1 K ⊂ C that the family { xA } x ∈ K is almost disjoint. Consequently , c = | K | ≤ Ind P ( A ) ≤ | G | = c .  Question 3. L et A b e a Bor el subset of a Polish gr oup G . I s t her e an at most c oun table subset C ⊂ A such that ind P ( A \ C ) = Ind P ( A ) ? The other pro blem concer ns the cardina ls sq ( C ) for v arious descr iptive classes C . If such a class C co nt ains the square of a co un table metriza ble space, then ℵ 1 ≤ sq ( C ) ≤ c and thus sq ( C ) falls into the categor y of the so -called sma ll uncountable cardinals, see [V]. How ever unlike to other typical small unco un table cardinals, sq ( C ) do es no collapse to c under the Martin Axiom, see [Sh]. P ACKING INDEX OF SUBSETS IN POLIS H GR OUPS 7 Problem 1. Explor e p ossible values and ine qualities b etwe en classic al smal l un- c oun table c ar dinals and the c ar dinals sq ( C ) for various descriptive classes C . 4. Rela tion of the p acking index to other no tions of sm allness T aking into account that a subset A with large packing index ind P ( A ) is geo- metrically s mall, it is na tural to consider the relation of the packing index to some other known concepts of smallness, in par ticular, the smallness in the sense of Baire category and the measure. W e recall that a subset A of a top olog ical space X is me ager if A can be written as the countable union of nowhere dense subsets. W e shall need the following class ical fact. Prop ositio n 4 (Bana ch-Kuratowski-P ettis) . F or any analytic non-me ager su bset A of a Po lish gr oup G the set AA − 1 c ont ains a neighb orho o d of the n eutr al element of G . A simila r r esult ho lds for analytic subsets that ar e not Haar null. W e recall that a subset A of a topolog ical g roup G is called Haar nul l if there is a Borel p roba bilit y measure µ o n G such that µ ( xAy ) = 0 for a ll x, y ∈ G . This notion was introduced by J.Chris tensen [C] and thoroug hly studied in [THJ]. In particular , a subset A of a lo cally compact gr oup G is Haa r n ull if and only if A ha s z ero Haa r measur e. Y et, Haar null se ts e xists in non-lo cally compact gr oups (admitting no inv ariant measure). Prop ositio n 5 (Chistensen) . If an analytic subset A of a Polish gr oup G is not Haar nul l, then AA − 1 c ont ains a neighb orho o d of the neutr al element of G . W e shall use those pr op ositions to prov e Theorem 2. L et A b e an analytic subset of a Polish gr oup G . If ind P ( A ) > ℵ 0 , then A is m e ager and Haar nul l. Pr o of. O therwise, w e can apply Prop os itions 4 or 5 to conclude t hat AA − 1 contains a neighbo rho o d U of the neutral element e of G . Find another neig h b orho o d V ⊂ G of e with V V − 1 ⊂ U ⊂ AA − 1 . Since ind P ( A ) > ℵ 0 there is an uncountable subset S ⊂ X such that the family { xA } x ∈ S is disjoint, which is equiv alent to S − 1 S ∩ AA − 1 = { e } . It follows from the choice of V that S − 1 S ∩ V V − 1 ⊂ S − 1 S ∩ AA − 1 = { e } and thu s the family { xV } x ∈ S is dis joint . Since V is an op en neighbor ho o d of e , the s et S , b eing discre te in G , is at most co un table. This c ontradiction c ompletes the proo f.  5. The p acking index and unions It is k nown that the countable union of meager (r esp. Haar null) subse ts of a Polish group is mea ger (resp. Haar null) . In contrast, the union of tw o s ubsets A, B ⊂ G with lar ge packing index need no t ha ve lar ge packing index . A simplest example is g iven by the s ets A = R × { 0 } and B = { 0 } × R on the plane R 2 . They ha ve packing indices ind P ( A ) = ind P ( B ) = c but ind P ( A ∪ B ) = 1 . In fact, this situation is t ypical. Accor ding to [L2], each infinite group G co nt ains two sets A, B ⊂ G such that ind P ( A ) = ind P ( B ) = | G | but Ind P ( A ∪ B ) = 1. The following theorem is a top o logical version of this result. 8 T ARAS BANAKH, NADY A L Y ASKO VSKA, DU ˇ SAN REPOV ˇ S Theorem 3. Each non-discr ete Polish Ab elian gr oup G c ontains two close d subsets A, B ⊂ G such that ind P ( A ) = ind P ( B ) = c and Ind P ( A ∪ B ) = 1 . Pr o of. Fix an inv a riant metric ρ generating the topo logy of G . This metric is complete bec ause G is P olish. Since G is Abelian, w e use the additive notation for the gr oup op eration o n G . The neutr al element of G will b e denoted b y 0 . W e define a subset D of G to be ε -sep ar ate d if ρ ( x, y ) ≥ ε for a n y distinct p oints x, y ∈ D . By t he Zorn le mma, eac h ε -separa ted subse t can b e enla rged to a maxima l ε -separa ted subset of G . Put ε − 1 = ε 0 = 1 and cho ose a maximal 2 ε 0 -separa ted subset H 0 ⊂ G con taining zero. Pr o ceeding by induction w e shall define a sequence ( h n ) n ∈ N ⊂ G o f p oints, a sequence ( ε n ) n ∈ ω of p os itive real num b ers and a s equence ( H n ) n ∈ ω of s ubsets o f G such that for every n > 0 (i) B (0 , ε n − 1 ) \ B (0 , 33 ε n ) is not empty and ε n < 2 − 6 ε n − 1 ; (ii) k h n k = 5 ε n , (iii) H n ⊃ { 0 , h n } is a maximal 2 ε n -separa ted subset in B (0 , 8 ε n − 1 ). It follows fro m (i) t hat the se ries P n ∈ ω ε n is conv ergent and thus for any sequence x n ∈ H n , n ∈ ω , the se ries P n ∈ ω x n is convergen t (because k x n k < 8 ε n − 1 for a ll n ∈ N according to (iii)). So it is leg al to consider the sets of sums Σ 0 =  X n ∈ ω x 2 n : ( x 2 n ) n ∈ ω ∈ Y n ∈ ω H 2 n  , Σ 1 =  − X n ∈ ω x 2 n +1 : ( x 2 n +1 ) n ∈ ω ∈ Y n ∈ ω H 2 n +1  . Let A a nd B b e the closur es of the sets Σ 0 and Σ 1 in G . It rema ins to prov e t hat the sets A, B have the desired pro pe rties: ind P ( A ) = ind P ( B ) = c and Ind P ( A ∪ B ) = 1. This will be done in three s teps. 1. Fir st we prov e that ind P ( A ) = c . By Lemma 1, this equalit y will follo w a s so on as we chec k that A − A is not a neighbor ho o d of the ne utral element 0 in G . It suffices for every k ∈ ω to find a p oint a p oint g ∈ B (0 , ε 2 k ) \ AA − 1 . By condition (i), there is a po int g ∈ G with 33 ε 2 k +1 ≤ k g k < ε 2 k . W e c laim that g / ∈ A − A = Σ 0 − Σ 0 . Mo re precis ely , dist( g , A − A ) = dist( g , Σ 0 − Σ 0 ) ≥ min { ε 2 k +1 , ε 2 k / 2 } . T ake a ny t wo distinct po int s x, y ∈ Σ 0 and find infinite sequence s ( x 2 n ) n ∈ ω , ( y 2 n ) n ∈ ω ∈ Q n ∈ ω H 2 n with x = P n ∈ ω x 2 n and y = P n ∈ ω y 2 n . Let m = min { n ∈ ω : x 2 n 6 = y 2 n } . If m ≥ k + 1, then k x − y k = k X n ≥ m x 2 n − y 2 n k ≤ X n ≥ m k x 2 n k + k y 2 n k ≤ ≤ 2 X n ≥ m 8 ε 2 n − 1 ≤ 32 ε 2 m − 1 ≤ 32 ε 2 k +1 < k g k − ε 2 k +1 and hence ρ ( x − y , g ) ≥ ε 2 k +1 . P ACKING INDEX OF SUBSETS IN POLIS H GR OUPS 9 If m ≤ k , then k x − y k = k ( x 2 m − y 2 m ) + X n>m ( x 2 n − y 2 n ) k ≥ ≥ k x 2 m − y 2 m k − X n>m ( k x 2 n k + k y 2 n k ) ≥ ≥ 2 ε 2 m − 32 ε 2 m +1 ≥ 3 2 ε 2 m > k g k + 1 2 ε 2 m and aga in ρ ( x − y , g ) ≥ 1 2 ε 2 m ≥ 1 2 ε 2 k . 2. In the same ma nner we can pr ov e that ind P ( B ) = c . 3. It remains to prove that Ind P ( A ∪ B ) = 1 . First we reca ll some sta ndard notation. Denote by 2 <ω = S n ∈ ω 2 n the set o f finite binar y sequences. F or a n y sequence s = ( s 0 , . . . , s n − 1 ) ∈ 2 <ω and i ∈ 2 = { 0 , 1 } by | s | = n we denote the length of s and b y s ˆ i = ( s 0 , . . . , s n − 1 , i ) the conca tenation of s a nd i . F or a finite or infinite binary sequence s = ( s i ) i ε n ; (3) k g − P t ≤ s x t ) k < 7 ε | s | . W e star t c ho osing a po in t x ∅ ∈ H 0 with ρ ( x ∅ , g ) < 2 ε − 1 = 2 ε 0 . Suc h a po int x ∅ exists b ecause H 0 is a ma ximal (0 , 2 ε 0 )-separated set in G . Next we pr o ceed by induction. Supp ose that for some n the p o in ts x s , s ∈ 2 > 5 ε n − 2 ε n − 2 ε n = ε n . The condition (3) fo llows fro m the estima tes k g − X t ≤ s ˆ0 x t k = k g − x s ˆ0 − X t ≤ s x t k = k g s − x s ˆ0 k < 2 ε n = 2 ε | s ˆ0 | and k g − X t ≤ s ˆ1 x t k = k g − x s ˆ1 − X t ≤ s x t k = k g s + h n − x s ˆ1 − h n k ≤ ≤ k g s + h n − x s ˆ1 k + k h n k < 2 ε n + 5 ε n = 7 ε | s ˆ1 | . After co mpleting the inductive constructio n, we ca n use the condition (3) to se e that for ev ery infinite binary sequence s ∈ 2 ω we get g = X n ∈ ω x s | n = X n ∈ ω x s | 2 n + X n ∈ ω x s | 2 n +1 . W e claim that the set D 0 =  X n ∈ ω x s | 2 n : s ∈ 2 ω  10 T ARAS BANAKH, NADY A L Y ASKO VSKA, DU ˇ SAN REPOV ˇ S lies in the intersection A ∩ ( g + B ). It is clear that D 0 ⊂ Σ 0 ⊂ A . T o see that D 0 ⊂ g + B , take an y p oint x ∈ D 0 and find a n infinite binary sequence s ∈ 2 ω with x = P n ∈ ω x s | 2 n . Then x = X n ∈ ω x s | 2 n + X n ∈ ω x s | 2 n +1 − X n ∈ ω x s | 2 n +1 ∈ g + Σ 1 ⊂ g + B . It remains to prove that | D 0 | ≥ c . Note that the s et D 0 , b eing a contin uous image of the Cantor cub e 2 ω , is co mpact. Now the equa lit y | D 0 | = c will follow a s so on as we c heck that D 0 has no isola ted po in ts. Giv en any sequence s ∈ 2 ω and δ > 0 we should find a sequence t ∈ 2 ω such that 0 < k X n ∈ ω x s | 2 n − X n ∈ ω x t | 2 n k < δ. Find even n umber 2 m ∈ ω such that P n ≥ m 20 ε 2 n − 1 < δ and ta ke any sequence t ∈ 2 ω such tha t t | 2 m − 1 = s | 2 m − 1 but t | 2 m 6 = s | 2 m . Then k X n ∈ ω x s | 2 n − X n ∈ ω x t | 2 n k = k X n ≥ m x s | 2 n − X n ≥ m x t | 2 m k ≤ ≤ X n ≥ m k x s | 2 n k + k x t | 2 n k ≤ X n ≥ m 32 ε 2 n − 1 < δ. On the other hand the low er b ound k x s | 2 m − x t | 2 m k > ε 2 m supplied by (2) implies k X n ∈ ω x s | 2 n − X n ∈ ω x t | 2 n k = k X n ≥ m x s | 2 n − X n ≥ m x t | 2 n k ≥ ≥ k x s | 2 m − x t | 2 m k − k X n>m ( x s | 2 n − x t | 2 n ) k > > ε 2 m − X n>m 16 ε 2 n − 1 > ε 2 m − 32 ε 2 m +1 > 0 (the latter tw o ineq ualities follow from (i)). Now w e s ee that | D 0 | = c and thus | A ∩ ( g + B ) | ≥ | D 0 | = c , whic h implies that Ind P ( A ∪ B ) = 1.  6. Haar and universall y null sets with unit p acking index In this s ection in eac h non-discrete Polish gr oup G w e shall co nstruct geometr i- cally lar ge subsets which are sma ll in the sense of measure. Theorem 4. Each non-discr ete Polish gr oup G c ontains a close d nowher e dense Haar nul l su bset C with Ind P ( C ) = 1 and thus C C − 1 = G . Pr o of. By Theor em 3, the group G contains tw o clos ed subsets A, B ⊂ G with ind P ( A ) = ind P ( B ) = c and Ind P ( A ∪ B ) = 1. By Theorem 2, the sets A, B are Haar null. Then the unio n C = A ∪ B is Haar null, and, b eing clo sed in G , is nowhere dense.  Under the set-theoretic assumption cov( K ) = c in ea ch non-discrete Polish gr oup G we can construct an universally null subset A ⊂ G with Ind P ( A ) = 1. W e r ecall that a subset A of a to po logical spa ce X is c alled universal ly nu l l if µ ( A ) = 0 for every co nt inuous probability Bo rel measure µ on X . A mea sure µ o n X is c ont inu ous if µ ( { x } ) = 0 for all x ∈ X . It is clea r that ea ch universally null s ubset of a no n-discrete Polish group in Haar null. P ACKING INDEX OF SUBSETS IN POLIS H GR OUPS 11 By cov( K ) we denote the smallest cardina lit y of a cover of the real line by meager subsets. It is known [JW, 1 9.4] that the equalit y cov( K ) = c is equiv alent to MA countable , the Martin Axiom for countable p os ets. In par ticular, cov( K ) = c holds under MA, the Martin Axiom. Theorem 5. Under cov( K ) = c e ach non-discr ete Polish gr oup G c ontains a uni- versal ly nu l l subset A with Ind P ( A ) = 1 . Pr o of. Fix a countable dense subset { x n } n ∈ ω ⊂ G and a dec reasing sequence ( U n ) n ≥ 0 of op en neighborho o ds o f the unit e of G with e = T n ≥ 0 U n . T o ea ch function f : ω → N w e can assig n a dens e G δ -subset D f = T n ∈ ω S k ≥ n x k U f ( k ) of G . Let N ω = { f α : α < c } b e any enumeration of the set N ω . The gro up G , b eing Polish and non-disc rete, has s ize | G | = c . Let G × G = { ( g α , g ′ α ) : α < c } be any enumeration of the pro duct G × G . This enumeration induces an en umeratio n G = { g α : α < c } of G such that |{ α : g α = g }| = c fo r each element g ∈ G . Put G α = T β <α D f β for α < c . Observe that the equality cov( K ) = c implies that | G α ∩ g α G α | = c . O therwise G can be presen ted as the union of < c many meager subsets: G = ( G α ∩ g α G α ) ∪ [ β <α ( G \ D f β ) ∪ [ β <α ( G \ g α D f β ) . The equality | G α ∩ g α G α | = c , α < c , allows us to construct inductively a transfinite sequenc e of points { a α : α < c } ⊂ G such that a α ∈ G α ∩ g α G α \ { a β , g β a β : β < α } for all α < c . The choice of the enu meratio n G = { g α : α < c } implies that the set A = { a α , g α a α : α < c } ha s Ind P ( A ) = 1. It remains to chec k that A is universally n ull. Given a σ -additive Borel probabil- it y co nt inuous measure µ on G , for every n ∈ ω find a n umber f ( n ) ∈ N suc h that µ ( a n U f ( n ) ) < 2 − n . It follows that the dense G δ -subset D f has mea sure µ ( D f ) = 0. Find an ordinal α suc h that f α = f and observe that a β , g β a β ∈ G f α ⊂ D f for all β > α . The inequality cov( K ) ≤ non( L ) following fro m the Cichon’s diagram (see [V]) guar antees that µ ( { a β , g δ a β : β ≤ α } ) = 0 and hence µ ( A ) ≤ µ ( { a β , g β a β : β ≤ α } ) + µ ( D f ) .  Remark 1. Theorem 5 ca nnot b e proved in ZF C b eca use in Lav er’s mo del of Z F C each universally n ull subset A of a Polish group G has size | A | < c , which implies that ind P ( A ) = Ind P ( A ) = c . 7. Constructing small subsets with a given (sha rp) p acking index In this sec tion we shall s how that Theorem 2 ca nnot b e r eversed: no where dense Haar n ull sets ca n hav e a rbitrary pa cking index. In fact, we sha ll constr uct s uc h sets A with an arbitr ary (sharp) pa cking index ind ♯ P ( A ) = sup {| S | + : S ⊂ G a nd { xA } x ∈ S is disjoint } , Ind ♯ P ( A ) = sup {| S | + : S ⊂ G a nd { xA } x ∈ S is almost disjoin t } . 12 T ARAS BANAKH, NADY A L Y ASKO VSKA, DU ˇ SAN REPOV ˇ S The formulas ind P ( A ) = sup { κ : κ < ind ♯ P ( A ) } and Ind P ( A ) = sup { κ : κ < Ind ♯ P ( A ) } show that the shar p packing indices ca rry more infor mation a bo ut a s et A compar- ing to the usual packing indices. All p oss ible v alues of the sharp packing indices o f subsets of a g iven Abe lian group are determined by the following r esult proved in [L2]. Prop ositio n 6. L et G b e an infin it e Ab elian gr oup and L ⊂ G b e a su bset with Ind P ( L ) = 1 . F or a c ar dinal κ ∈ [2 , | G | + ] the fo l lowing c onditions ar e e quivalent: (1) ther e is a subset A ⊂ G with ind ♯ P ( A ) = κ ; (2) ther e is a subset A ⊂ L with ind ♯ P ( A ) = Ind ♯ P ( A ) = κ ; (3) if   G/ [ G ] 2   ≤ 2 , then κ 6 = 4 and if G = [ G ] 3 , then κ 6 = 3 . Here [ G ] p = { x ∈ G : x p = e } for p ∈ { 2 , 3 } . Combining Pr op osition 6 w ith Theorems 4 and 5, w e obtain the main result of this section. Theorem 6. L et G b e a non-discr ete Polish Ab elian gr oup and κ b e a c ar dinal such that (i) 2 ≤ κ ≤ | G | + , (ii) k 6 = 3 if G = [ G ] 3 , and (iii) k 6 = 4 if   G/ [ G ] 2   ≤ 2 . (1) The gr oup G c ontains a nowher e dense Haar nul l subset A such that ind ♯ P ( A ) = Ind ♯ P ( A ) = κ ; (2) If cov( K ) = c , then G c ont ains a universal ly nul l subset A with ind ♯ P ( A ) = Ind ♯ P ( A ) = κ . T aking in to accoun t that ind P ( A ) = sup { κ : κ < ind ♯ P ( A ) } , we can a pply Theo- rem 6 to deduce the follo wing co rollary . Corollary 1. L et G b e a n on-discr ete Polish Ab elian gr oup and κ b e a c ar dinal such that (i) 1 ≤ κ ≤ | G | , (ii) k 6 = 2 if G = [ G ] 3 , and (iii ) k 6 = 3 if   G/ [ G ] 2   ≤ 2 . (1) The gr oup G c ontains a nowher e dense Haar nul l subset A such that ind P ( A ) = Ind P ( A ) = κ ; (2) If cov( K ) = c , then G c ont ains a universal ly nul l subset A with ind P ( A ) = Ind P ( A ) = κ . 8. Acknow ledgements The authors express their sincere thanks to Heik e Milden b erger , Andrzej Ros lanowski, and Lub omyr Zdomskyy for their help in understanding Shela h’s pap er [Sh]. References [BL1] T. Banakh, N. Ly asko vsk a, We akly P-smal l not P- smal l subsets in Ab elian gr oups , Alge- bra and Discrete M athematics, N o.3 (2006), 29-34. [BL2] T. Banakh, N. Lyask o vsk a, Wea kly P-smal l not P-smal l subsets in gr oups , In tern. J. of Algebra and Computations, (to app ear). [C] J. Christensen, On sets of Haar mea sur e zer o in ab elian Polish gr oups , Israel J. 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T opsøe, J.Hoffman-Jørgensen, Analytic sp ac es and t heir applic ations , in: C.Rogers et al. Analytic sp ac es (Academic Pr ess, London, 1980), pp. 317–402. [V] J. V aughan, Smal l unc ountable c ar dinals and top olo gy , in: Op en Problems in T op ology (J. v an Mill, G.Reed eds.), Elsevier, Am sterdam, 1990. Dep a r tmen t of Ma thema tics, Iv an Franko Na tional Un iversity of L viv, Ukraine E-mail addr e ss : tbanakh@yah oo.com E-mail addr e ss : lyaskovska@ yahoo.com Institute of Ma thema tics, Physics and Mechanics, and F a cul ty of Educa tion, Uni- versity of Ljubljan a, P.O.B.2964 , Ljubljan a, S lovenia E-mail addr e ss : dusan.repov s@guest.arnes.si