aa r X i v : . [ m a t h . GN ] N ov STRONG MEASURE ZERO IN POLISH GROUPS
MICHAEL HRUˇS ´AK AND ONDˇREJ ZINDULKA
Abstract.
The notion of strong measure zero is studied in the context ofPolish groups. In particular, the extent to which the theorem of Galvin, My-cielski and Solovay holds in the context of an arbitrary Polish group is studied.Hausdorff measure and dimension is used to characterize strong measure zero.The products of strong measure zero sets are examined. Sharp measure zero,a notion stronger that strong measure zero, is shown to be related to meager-additive sets in the Cantor set and Polish groups by a theorem very similar tothe theorem of Galvin, Mycielski and Solovay. Introduction
All spaces and topological groups considered are separable and metrizable.
A natural extension of a definition due to Borel (1919) [6] asserts that a metricspace X has strong measure zero ( Smz ) if for any sequence h ε n : n ∈ ω i of positivereal numbers there is a cover { U n : n ∈ ω } of X such that diam U n ε n for all n .In the same paper Borel conjectured that every strong measure zero set of realsis countable. This was shown to be independent of the usual axioms of set theoryby Sierpi´nski (1928) [44] and Laver (1976) [28]. Later it was observed by Carlson[9] that the Borel Conjecture actually implies a formally stronger statement thatall separable
Smz metric spaces are countable.We shall investigate the behaviour of strong measure zero sets in arbitrary Polishgroups. In a sense we shall investigate the world, where the Borel conjecture fails,as most if not all of our results are trivial if the Borel Conjecture holds.The subject of inquiry of this work starts with the theorem of Galvin, Mycielski,and Solovay [13, 14] who, confirming a conjecture of Prikry, proved that a set A ⊆ R is of strong measure zero if and only if A + M = R for every meager set M ⊆ R .Relatively recently Kysiak [26] and Fremlin [12], independently, showed thatan analogous theorem is true for all locally compact metrizable groups (see also[50]). We present a proof of Kysiak and Fremlin’s result based on [19] and considerthe natural question as to how far the result can be extended. The fact that thetheorem does not in general hold for all Polish groups was established in [19] and[50] and extended in [20]. This depends on further set-theoretic axioms, as theresult obviously holds for all Polish groups assuming, e.g., the Borel Conjecture.Cardinal invariants associated with strong measure zero sets on R , ω ω , and 2 ω have been studied rather extensively in recent decades [1, 15, 51, 8]. We review someof these, concentrating on the uniformity invariant of the σ -ideal Smz ( G ) of strong Mathematics Subject Classification.
Key words and phrases. strong measure zero, Polish group, Galvin-Mycielski-Solovay theorem.The first author gratefully acknowledges support from PAPIIT grant IN 100317. The secondauthor was supported from European Regional Development Fund-Project “Center for AdvancedApplied Science” (No. CZ.02.1.01/0.0/0.0/16 019/0000778). measure zero subsets of a Polish group G . A version of the Galvin-Mycielski-Solovaytheorem links this study to the investigation of the so-called transitive coefficient cov ∗ ( M ) in Polish groups [1, 31, 11].It was probably the aforementioned result of Prikry, Galvin, Mycielski and Solo-vay that inspired a few notions of smallness on the real line and Cantor set akin tostrong measure zero. E.g., a set S ⊆ R is strongly meager if S + N = R for eachLebesgue null set N ; it is null-additive if S + N is Lebesgue null for each Lebesguenull set N ; and it is meager-additive if S + M is meager for each meager set M .These notions easily extend to other Polish groups.We will study the latter notion, which is obviously a strengthening of strongmeasure zero. Since the early nineties, meager-additive sets in the Cantor set receivequite some attention. Let us single out the remarkable paper of Shelah [42] thatprovides a proof that each null-additive set in the Cantor set 2 ω is meager-additiveand also the underlying combinatorial characterizations of null-additive and meager-additive sets in 2 ω (cf. 7.7 below), and Pawlikowski’s paper [36] providing finecombinatorics and study of the so called transitive coefficients mentioned abovethat are actually cardinal invariants of strong measure zero, meager-additive andnull-additive sets, and of course Bartoszy´nski’s book [1]. However, all nontrivialresults on meager-additive sets depended heavily on the combinatorial and groupstructure of 2 ω . In 2009 Weiss [47, 48] found a method that made the theorytransferrable to the real line. Only very recently in [54, 52] it was noted thatthere is a description of meager-additive sets that resembles very much the Borel’sdefinition of strong measure zero. Metric spaces having this property were termedto have sharp measure zero . This allowed for the theory of meager-additive sets toextend to other Polish groups. We provide some highlights of the rather new theoryof sharp measure in metric spaces and meager-additive sets, and sharp measure zeroon 2 ω and on Polish groups, including calculation of the uniformity number of sharpmeasure zero and meager-additive sets.Our set-theoretic notation is standard and follows e.g. [25, 21]. In particular, theset of finite ordinals is identified with the set of non-negative integers and denotedinterchangeably by ω and N . In the same vein, the non-negative integers themselvesare identified with the set of smaller non-negative integers, in particular 2 = { , } .All spaces considered are separable and metrizable, often endowed with a com-patible metric denoted d . We denote by B ( x, ε ) the closed ball with radius ε centered at x , the corresponding open ball will be denoted by B ◦ ( x, ε ).The product spaces of the type A ω for some finite or countable set A are con-sidered with the metric of least difference defined by d ( f, g ) = 2 −| f ∧ g | , where f ∧ g = f ↾ n for n = min { k : f ( k ) = g ( k ) } . The clopen balls in the space A ω arerepresented by nodes of the tree A <ω , given s ∈ A <ω , we let h s i = { f ∈ A ω : s ⊆ f } .Given a subtree T of A <ω , we let [ T ] = { f ∈ A ω : ∀ n ∈ ω f ↾ n ∈ T } be the (closed)set of branches of T . A metric space is analytic if it is a continuous image of ω ω ,and it is Borel ( absolutely G δ , resp.) if it is Borel ( G δ , resp.) in its completion.A Polish group is a separable, completely metrizable topological group. A com-patible metric d on a separable metrizable group G is left-invariant if d ( zx, zy ) = d ( x, y ) for any x, y, z ∈ G . TRONG MEASURE ZERO IN POLISH GROUPS 3
A separable group G is a CLI group if it admits a complete left-invariant com-patible metric. Abelian and locally compact Polish groups are
CLI , while, e.g., thegroup S ∞ of all permutations of ω is not.A separable group G is a TSI group if it admits a (both-sided) invariant compat-ible metric. Not every Polish group admits an invariant metric, but if it is compactor abelian, then it does. Also, any invariant metric on a Polish group is complete.2.
Strong measure zero in Polish groups
The notion of strong measure zero is in general neither a topological nor a metricproperty, but a uniform property; in particular, a uniformly continuous image of a
Smz set is
Smz , and if X uniformly embeds into Y , then any set A ⊆ X that isnot Smz in X is not Smz in Y either.As all left-invariant (equiv right-invariant) metrics on a separable metrizablegroup are uniformly equivalent the notion of strong measure zero becomes seemingly“topological”: a subset S of a topological group G is Rothberger bounded if for everysequence h U n : n ∈ ω i of neighbourhoods of 1 G there is a sequence h g n : n ∈ ω i ofelements of the group G such that the family h g n · U n : n ∈ ω i covers S . It follows[12] that a subset of a Polish group G is Rothberger bounded if and only if it isstrong measure zero w.r.t. some (any) left-invariant metric on G .Many of the results stated here could be phrased in the language of uniformitiesand/or in terms of the property of being Rothberger bounded (see [12] for suchtreatment).Whenever G is a Polish group, Smz ( G ) denotes the family of strong measure zerosets with respect to any left-invariant metric (i.e., the (left) Rothberger boundedsets as described above).Of course, the choice of left-invariant over right-invariant is arbitrary, one beingisomorphic to the other via the inverse map of the group in question. In fact,both the left Rothberger bounded and right Rothberger bounded set form a σ -idealwhich is invariant under both left and right translations. Proposition 2.1.
Smz ( G ) is a bi-invariant σ -ideal.Proof. To see that
Smz ( G ) is a σ -ideal let { X n : n ∈ ω } ⊆ Smz ( G ) and a sequence { U n : n ∈ ω } of open subsets of G be given. Let { I n : n ∈ ω } be a partitionof ω into infinite sets. As each X n is of strong measure zero, there is a sequence { g i : i ∈ I n } ⊆ G such that X n ⊆ S i ∈ I n g i · U i . Then S n ∈ ω X n ⊆ S i ∈ ω g i · U i .Now, let X ∈ Smz ( G ) and g ∈ G be given.To see that g · X ∈ Smz ( G ), note that if { U n : n ∈ ω } is a sequence of open subsetsof G and { g i : i ∈ ω } ⊆ G is such that X ⊆ S n ∈ ω g n · U n , then g · X ⊆ S n ∈ ω g · g n · U n .To show that X · g ∈ Smz ( G ), let { U n : n ∈ ω } be a sequence of open subsets of G . Consider the open sets { U n · g − : n ∈ ω } . As X ∈ Smz ( G ) here is a sequence { g n : n ∈ ω } ⊆ G such that X ⊆ S n ∈ ω g n · ( U n · g − ). Then X · g ⊆ S n ∈ ω g n · U n . (cid:3) Now, assuming Borel conjecture, or assuming that the group G has an invariantmetric, the left and right Rothberger bounded sets coincide. This is not true ingeneral, though: Example 2.2.
Assuming CH , there is a left Rothberger bounded subset of thegroup of permutations S ∞ of ω which is not right Rothberger bounded. TRONG MEASURE ZERO IN POLISH GROUPS 4
Proof.
Denote by Ω the set of all finite partial injective functions from some n ∈ ω to ω . Enumerate all sequences of elements of S ∞ as { y α : α < ω } , and all increasingfunctions from ω to ω as { f α : α < ω } .We shall recursively construct { g α : α < ω } ⊆ S ∞ and { z α : α < ω } ⊆ S ω ∞ sothat(1) ∀ β < α < ω ∃ n ∈ ω g α ↾ f β ( n ) = z β ( n ) ↾ f β ( n ), while(2) ∀ β < α < ω ∀ n ∈ ω g − α ↾ n + 1 = y − α ( n ) ↾ n + 1, and(3) ∀ s ∈ Ω and m < m the first two elements of ω \ rng( s ) ∃ a ∈ [ ω ] m +1 (a) ∀ n ∈ a s ⊆ z α ( n ) ↾ f α ( n ),(b) ∀ n ∈ a m ∈ rng( z α ( n ) ↾ f α ( n )),(c) ∀ n ∈ a m rng( z α ( n ) ↾ f α ( n )), and(d) ∀ i = j ∈ a z α ( i ) − ( m ) = z α ( j ) − ( m ).It should be clear, that if this can be accomplished then (1) guaranties that the set X = { g α : α < ω } is of strong measure zero, while (2) makes sure that X − is not.The condition (3) is there for the construction not to prematurely terminate.Assume that g β , z β for β < α have been constructed. First choose z α ∈ S ω ∞ satisfying (3). Then enumerate α = { β i : i ∈ ω } and recursively find { n i : i ∈ ω } so that s i = z β i ( n i ) ↾ f β i ( n i ) satisfy(i) s i ⊆ s i +1 ,(ii) if m i = min( ω \ rng( s i )) then m i ∈ rng( s i +1 ),(iii) ∀ n m i ∃ k n n k n ∈ rng( s i ) s − i ( k n ) = y − α ( n )( k n ).Then let g α = S i ∈ ω s i . Then g α ∈ S ∞ satisfying (1) by (i) and (ii), and (2) by(iii).To construct the sequence h s n : n ∈ ω i start with s − = ∅ . Having found s i ,let m < k be the first two elements of ω \ rng( s i ). By (3), there is n i +1 ∈ ω suchthat s i +1 = z β i +1 ( n i +1 ) ↾ f β i +1 ( n i +1 ) is such that { m, k } ∩ rng( s i +1 ) = { m } , and s − i +1 ( n ) = y α ( n ) − ( n ) for every m n < k . (cid:3) There is a close relation between strong measure zero and Geometric measuretheory which shall be explored later on in the text, in section 5. The first result inthis direction is due to Besicovitch [4, 5] who showed that a set X of reals has strongmeasure zero if and only if every uniformly continuous image of X has Hausdorffdimension 0.Here we shall characterize strong measure sets in Polish groups as exactly thesets of universal invariant submeasure zero , a result due to J. Greb´ık.It is a classical result of Haar [16] that every locally compact Polish group ad-mits an (essentially unique) left-invariant, countably additive, outer regular Borelmeasure. In a similar vein, we shall prove here that every Polish group admits anon-trivial countably subadditive, outer regular, left-invariant diffuse submeasure,a result used in the next section.Recall that a function µ : P ( G ) → R + ∪ {∞} is a submeasure if µ ( ∅ ) = 0, and µ ( A ∪ B ) µ ( A ) + µ ( B ) whenever A, B are subsets of G . A submeasure µ on G is • σ -subadditive if µ ( S n ∈ ω A n ) P n ∈ ω µ ( A n ), for any { A n : n ∈ ω } ⊆ P ( G ), • outer regular if µ ( A ) = inf { µ ( U ) : A ⊆ U, U open in X } , for any A ⊆ G , • left-invariant if µ ( A ) = µ ( g · A ), for any A ⊆ G and g ∈ G , • non-atomic or diffuse if µ ( { x } ) = 0 for every x ∈ G , and • non-trivial if µ ( G ) > TRONG MEASURE ZERO IN POLISH GROUPS 5
Lemma 2.3.
In every Polish group G there is a decreasing local basis { U n : n ∈ ω } of open sets at G such that for every m ∈ ω and { a n : n > m } ⊆ P ( G ) such that | a n | = n for every n > m , U m S n>m a n · U n .Proof. Let d be a left invariant compatible metric on G , and let e be a completemetric on G . Recursively choose the open sets U n , n ∈ ω , together with finite sets b n ⊆ U n of size n + 1 so that:In d : The points of b n are 3 diam U n +1 apart, and also 3 diam U n +1 apart fromthe complement of U n , whileIn e : ∀ m < n ∀{ g i : m < i < n } with g i ∈ b i diam Q m m } ⊆ P ( G ) such that | a n | = n for every n > m are given. Recursively choose g n ∈ b n so that the set Q m m form a decreasing sequence of setsof e -diameter converging to 0, hence by completeness of e their intersection is asingleton x ∈ U m which is not in S n>m a n · U n . (cid:3) Theorem 2.4 ([20]) . There is a non-trivial, left-invariant, outer regular, σ -sub-additive diffuse submeasure on every Polish group.Proof. Fix a sequence { U n : n ∈ ω } as in Lemma 2.3 and define for A ⊆ G : µ ( A ) = inf (X i ∈ ω n i : A ⊆ [ i ∈ ω g i · U n i ) . It is immediate from the definition that µ is a diffuse σ -additive, left invariant,outer regular submeasure on G . To see that µ non-trivial it suffices to note that µ ( U m ) = m . To see that µ ( U m ) is not less than m , note that by the key propertyof { U n : n ∈ ω } , if U m ⊆ S i ∈ ω g i · U n i then P i ∈ ω n i > m . (cid:3) The promised characterization is the following:
Theorem 2.5 (J. Greb´ık, see [20]) . A subset A of a Polish group G is of leftstrong measure zero if and only if µ ( A ) = 0 for every left-invariant, outer regular,countably additive diffuse submeasure on G .Proof. Assume first that X ∈ Smz ( G ), let µ be a left-invariant, outer regular,countably additive diffuse submeasure on G , and let ε > µ isnon-atomic and outer regular, there is a sequence { U n : n ∈ ω } of neighborhoodsof 1 G such that P n ∈ ω µ ( U n ) < ε . Now, as X ∈ Smz ( G ), there is a sequence { g n : n ∈ ω } ⊆ G such that X ⊆ S n ∈ ω g n · U n . By left invariance of µ , µ ( X ) P n ∈ ω µ ( U n ) < ε . Hence µ ( X ) = 0.On the other hand, assume that X ⊆ G has µ ( X ) = 0 for every invariant, non-atomic, outer regular submeasure µ on G , and let { V n : n ∈ ω } be a sequence ofopen neighbourhoods of 1 G in G . Let { U n : n ∈ ω } be a decreasing local basis as inLemma 2.3, that is such that for every m ∈ ω and { a n : n > m } such that | a n | = n for every n > m , U m S n>m a n · U n , by passing on to a subsequence, we mayassume that U n ⊆ V n for every n ∈ ω . Let { n i : i ∈ ω } ⊆ ω be such that n i +1 > n i TRONG MEASURE ZERO IN POLISH GROUPS 6 for every i ∈ ω , and let W = { U n i +1 : i ∈ ω } , and w ( U n i +1 ) = n i . Then define asubmeasure µ by putting for A ⊆ G µ ( A ) = inf (X i ∈ ω w ( W i ) : A ⊆ [ i ∈ ω g i · W i ) with each W i ∈ W and g i ∈ G . This is again a left-invariant, σ -subadditive,non-atomic, outer regular submeasure, with µ ( U n i +1 ) = n i . Hence µ ( X ) = 0, inparticular, there is a sequence { W j : j ∈ ω } and a sequence { q j : j ∈ ω } such that X ⊆ S j ∈ ω g i · W j and P j ∈ ω w ( W j ) <
1. This means that every U n i +1 appears fewerthat n i -many times as one of the W j , so there is permutation π ∈ S ∞ such that W π ( n ) ⊆ V n for every n ∈ ω , hence X ⊆ S n ∈ ω g π ( n ) · V n . Hence X ∈ Smz ( G ). (cid:3) In the general context of a metric space, Szpilrajn [45] proved that every
Smz set X has universal measure zero, i.e. has measure zero for every finite diffuse Borelmeasure on X . It should be noted that unlike strong measure zero sets, uncountableuniversal measure zero sets exist in ZFC as shown by Sierpi´nski and Szpilrajn[43].
Proposition 2.6 (Szpilrajn [45]) . Strong measure zero sets in separable metricspaces have universal measure zero.Proof.
Aiming towards contradiction, suppose that X is Smz yet there is a diffusedBorel measure µ on X such that µ ( X ) = 1. Define a function f : (0 , ∞ ) → [0 ,
1] by f ( r ) = sup { µ ( E ) : diam E r } . We claim that lim r → f ( r ) = 0. Otherwise there is ε > E n such that diam E n ց µ ( E n ) > ε . Let E = T n ∈ ω S m > n E n . Thenclearly µ ( E ) > ε >
0. In particular E = ∅ , i.e., there is I ∈ [ ω ] ω such that T n ∈ I E n = ∅ . Suppose without loss of generality that I = ω . Since any two sets E n , E m have a common point, we have diam( S m > n E n ) E n . Thereforediam E n → diam E n = 0, which contradicts µ ( E ) >
0. We proved thatlim r → f ( r ) = 0. Therefore there is, for each n ∈ ω , ε n > P n f ( ε n ) < X is Smz , there are sets U n such that diam U n < ε n that cover X . It followsthat 1 = µ ( X ) X n µ ( U n ) X n f (diam U n ) X n f ( ε n ) < , the desired contradiction. (cid:3) The Galvin-Mycielski-Solovay Theorem in Polish groups
In this section we study the Galvin-Mycielski-Solovay theorem in the context ofan arbitrary Polish group G . We denote by M ( G ), or simply by M if there is nodanger of confusion, the ideal of meager subsets of G . Much of this section existsthanks to the following simple yet crucial observation due to Prikry: Proposition 3.1 (Prikry [38]) . Let G be a separable group, and let S ⊆ G be suchthat S · M = G for all M ∈ M ( G ) . Then S ∈ Smz ( G ) .Proof. Let S be as above, and let { U n : n ∈ ω } be a family of open neighbourhoodsof 1 in G . Let { g n : n ∈ ω } ⊆ G be such that U = S n ∈ ω g n · U n is dense openin G . Then U − is dense open in G , the inverse being a homeomorphism, so M = G \ U − is nowhere dense in G . As S · M = G , there is x ∈ G \ S · M , that is S ⊆ x · U = S n ∈ ω x · g n · U n . Hence, S ∈ Smz ( G ). (cid:3) TRONG MEASURE ZERO IN POLISH GROUPS 7
As mentioned in the introduction, Galvin, Mycielski, and Solovay [13, 14] an-swered Prikry’s question by showing that the reverse inclusion holds for R . Thesame was recently proved for all locally compact groups by Kysiak [26] and Frem-lin [12], independently. We shall present a proof of their theorem (the converse ofPrikry’s result for locally compact groups) here. Our proof follows [19].Call a subset N of a topological group G uniformly nowhere dense if for everyneighborhood U of 1 there is a neighborhood V of 1 such that for every x ∈ G thereis a g ∈ G such that g · V ⊆ x · U \ N . A set M ⊆ G is uniformly meager if it canbe written as a union of countably many uniformly nowhere dense sets. We denotethe family of all uniformly meager subsets of G by UM ( G ) (or simply UM ). Thefollowing generalizes [14, Theorem 4]. Proposition 3.2 ([19]) . Let G be a Polish group which is either locally compact or TSI , and let S ∈ Smz ( G ) . Then S · M = G for all M ∈ UM ( G ) .Proof. Assume first that G admits a invariant metric d . Recall that every invariantmetric on a Polish group is complete. Let N be uniformly nowhere dense subset of G . Note that for every y ∈ G and an open set U ⊆ G , y · U · N = U · y · N , and y · N is uniformly nowhere dense. It follows that for every uniformly nowhere dense N ⊆ G ∀ U open ∃ V open ∀ x, y ∈ G ∃ z ∈ G z · V ⊆ x · U \ V · y · N. (1)Now, fix a Smz set S and a uniformly meager set M written as the union of anincreasing sequence h N n : n ∈ ω i of uniformly nowhere dense sets. Then there is asequence h U n : n ∈ ω i of open subsets of G , such that for every n > ∀ x, y ∈ G ∃ z ∈ G z · U n ⊆ x · U n − \ U n · y · N n . (2)As S is Smz , for every sequence { U n : n ∈ ω } of open sets (of diameter convergingto 0) there is a sequence { g n : n ∈ ω } ⊆ G such that each s ∈ S is contained ininfinitely many of the sets g n · U n . Applying (2) recursively there is a sequence h x n : n ∈ ω i of elements of G such that for every n ∈ ωx n +1 · g n +1 · U n +1 ⊆ x n · g n · U n \ ( g n +1 · U n +1 · N n +1 ) . The sequence h x n : n ∈ ω i is Cauchy, let x be its limit, i.e., { x } = T n ∈ ω x n · g n · U n .Then x S n ∈ ω g n · U n · N n ⊇ S · M , as the sequence h N n : n ∈ ω i is increasingand every element of S is contained in infinitely many of the g n · U n (for every( s, m ) ∈ S × M there is an n ∈ ω such that s ∈ g n · U n and m ∈ N n ).Now if G is locally compact, the proof proceeds along similar lines. Only (1) isreplaced by the following lemma. Lemma 3.3.
Let G be a locally compact Polish group equipped with a completemetric d , and let U ⊆ G be an open set with compact closure C = U and P ⊆ G becompact nowhere dense. Then ∀ ε > ∃ δ > ∀ x ∈ C ∀ y ∈ K ∃ z ∈ C B ( z, δ ) ⊆ B ( x, ε ) \ ( B ( y, δ ) · P ) . Proof.
Fix ε > f : C × K → R by f ( x, y ) = sup { t : ∃ z ∈ C B ( z, t ) ⊆ B ◦ ( x, ε ) \ y · P ) } . Then f is positive on C × K and attains its (positive) minimum.To see that, consider, for each z ∈ C , the functions TRONG MEASURE ZERO IN POLISH GROUPS 8 g z ( x ) = d ( z, X \ B ◦ ( x, ε )) , x ∈ Ch z ( y ) = d ( z, y · P ) , y ∈ K and note that f ( x, y ) = sup z ∈ C min( g z ( x ) , h z ( y )) . (3)Using compactness it is easy to see that, for each z ∈ C , the function h z islower semicontinuous and that while g z does not have to be, it has the followinglower-semicontinuity property: if x n → x and g z ( x n ) →
0, then g z ( x ) = 0.Now suppose that there are ( x n , y n ) ∈ C × K such that f ( x n , y n ) →
0. Since
C, K are compact, passing to subsequences we may assume ( x n , y n ) → ( x, y ) ∈ C × K .Use (3) and the semicontinuity properties of g z and h z to conclude that since f ( x n , y n ) →
0, for any z either g z ( x n ) → g z ( x ) = 0, or else h z ( y n ) → h z ( y ) = 0. Use (3) again to conclude that f ( x, y ) = 0, the desiredcontradiction proving that there is η > f ( x, y ) > η for all x, y . Itfollows that ∀ x ∈ C ∀ y ∈ K ∃ z ∈ C B ( z, η ) ⊆ B ( x, ε ) ∧ B ( z, η ) ∩ y · P = ∅ . The latter of course yields B ( z, η ) ∩ B ( y · P, η ) = ∅ . On the other hand, there is ξ > ∀ y ∈ K B ( y, ξ ) · P ⊆ B ( y · P, η ) . It follows that B ( z, η ) ∩ B ( y, ξ ) · P = ∅ . Thus letting δ = min { η , ξ } yields thelemma. (cid:3) To conclude, write G as the union of an increasing sequence of open sets withcompact closures K n , and write a meager set M as the union of an increasingsequence of compact nowhere dense sets P n . Choose x ∈ G and ε > B ( x , ε ) is compact. Let C = B ( x , ε ). By the above lemma there is a sequence h ε n : n ∈ ω i ∈ (0 , ∞ ) ω such that for every n > ∀ x ∈ C ∀ y ∈ K n ∃ z ∈ C B ( z, ε n ) ⊆ B ( x, ε n − ) \ B ( y, ε n ) ∩ K n ) · P n . We may of course suppose that ε n →
0. Since S is Smz , there is a cover { E n } of S such that diam E n < ε n for all n such that each point of S is covered by infinitelymany E n ’s).For each n there is y such that E n ⊆ B ( y, ε n ). Therefore, using repeatedly (4),there is a sequence h x n : n ∈ ω i in C such that for all n ∈ ωB ( x n +1 , ε n +1 ) ⊆ B ( x n , ε n ) \ ( E n +1 ∩ K n +1 ) · P n +1 . Let x be the unique point of T n ∈ ω B ( x n , ε n ) (there is one, since B ( x , ε ) is compactand is unique as ε n → x / ∈ S n ∈ ω ( E n ∩ K n ) · P n .Thus, to prove that x is not covered by S · M it suffices to show that S × M ⊆ S n ∈ ω ( E n ∩ K n ) × P n . Let ( s, m ) ∈ S × M . There is k such that ( s, m ) ∈ K k × P k .Since there are infinitely many n such that s ∈ E n , there is n > k such that s ∈ E n , hence s ∈ E n ∩ K k ⊆ E n ∩ K n . Also m ∈ P k ⊆ P n . Therefore ( s, m ) ∈ ( E n ∩ K n ) × P n . The desired inclusion is proved. (cid:3) The Galvin-Mycielski-Solovay/Fremlin/Kysiak result follows from the fact thatin a locally compact group every meager set is uniformly meager:
TRONG MEASURE ZERO IN POLISH GROUPS 9
Proposition 3.4 ([19]) . A Polish group G is locally compact if and only if M ( G ) = UM ( G ) .Proof. We shall see first that M ( X ) = UM ( X ) for every locally compact metricspace X .To that end it suffices to see that every nowhere dense subset of a compact spaceis, in fact, uniformly nowhere dense. Let N be a nowhere dense subset of a compactspace X , and let ε >
0. Let F be a finite subset of X such that Z = S x ∈ F B ( x, ε ).For every x ∈ F let y x ∈ B ( x, ε ) and δ x > B ( y x , δ x ) ⊆ B ( x, ε ) \ N .Then δ = min { δ x : x ∈ F } works as B ( x, ε ) ⊆ B ( z, ε ) whenever z ∈ B ( x, ε ).On the other hand, M ( X ) = UM ( X ) for every nowhere locally compact completemetric space X .To see this let X be nowhere locally compact with a complete metric d . Then forevery U with non-empty interior there is an ε U > { V Uk : k ∈ ω } of open balls of radius ε U contained in U .We shall construct a nowhere dense set N which is not uniformly meager. To dothat we recursively construct a family { U s : s ∈ ω <ω } of non-empty regular closed sets so that(1) diam U s −| s | for every s ∈ ω <ω ,(2) S n ∈ ω U s a n ⊆ U s for every s ∈ ω <ω ,(3) U s a n ∩ U s a m = ∅ for every s ∈ ω <ω and any two distinct m, n ∈ ω ,(4) int( U s \ S n ∈ ω U s a n ) = ∅ for every s ∈ ω <ω ,(5) for all s ∈ ω <ω and k ∈ ω and x ∈ V U s k there is a n ∈ ω such that U s a n ⊆ B ( x, − k ).To do this is straightforward.Having constructed such a family, let N = T j ∈ ω S | s | = j U s . This is the requiredset:It is nowhere dense as a non-empty open set U is either disjoint from N , orcontains U s for some s ∈ ω <ω . Then, however, ∅ 6 = int( U s \ S n ∈ ω U s a n ) ⊆ U \ N by the property (4) above.Now we will prove that N is not uniformly meager in X . The set N is naturallyhomeomorphic to ω ω (see properties (1)–(3) above), hence satisfies the Baire Cat-egory Theorem. Aiming toward a contradiction assume that N ⊆ S l ∈ ω N l , whereeach N l is a closed uniformly nowhere dense subset of X . By the Baire CategoryTheorem applied to N there is an s ∈ ω <ω and an l ∈ ω such that U s ∩ N ⊆ N l ,hence U s ∩ N is uniformly nowhere dense. So, there is a δ > ε U s . Consider V U s k , for 2 − k < δ .Then, on the one hand there is an x ∈ V U s k such that B ( x, − k ) ⊆ V U s k \ N ,and on the other hand, there is (see property (5) above) an n ∈ ω such that ∅ 6 = N ∩ U s a n ⊆ B ( x, − k ), which is a contradiction.The result follows as every Polish group is either locally compact or nowherelocally compact. (cid:3) And finally:
Theorem 3.5 (Fremlin [12], Kysiak [26]) . Let G be a locally compact Polish group.A set A ⊆ G is of strong measure zero if and only if A · M = G for every meagerset M ⊆ G Recall that a set U is regular closed if U is the closure of the interior of U . TRONG MEASURE ZERO IN POLISH GROUPS 10
Proof.
The theorem follows directly from Propositions 3.1, 3.2 and 3.4. (cid:3)
Next we shall discuss the possibility of extending the Galvin-Mycielski-Solovaytheorem to a larger class of Polish groups. First, one needs to realize that assumingthe Borel conjecture, the theorem holds trivially for every Polish group G , as strongmeasure zero sets in all Polish groups are exactly the countable subsets ([9]), hence S · M is meager for every strong measure zero set S and meager set M , hence S · M = G .On the other hand, it was shown in [19], that the Galvin-Mycielski-Solovaytheorem fails for the Baer-Specker group Z ω , assuming cov ( M ) = c . We conjecturethat assuming a strong failure of the Borel conjecture the locally compact Polishgroups are exactly the ones for which the theorem holds. Conjecture 3.6 (CH) . The Galvin-Mycielski-Solovay theorem holds in a Polishgroup G if and only if G is locally compact. The Continuum Hypothesis is optimal in the sense that under CH the Galvin-Mycielski-Solovay Theorem fails for as many Polish groups as possible follows fromthe logical complexity of the problem. The statement G satisfies the Galvin-Mycielski-Solovay Theorem is a Π -statement with G as a parameter, and hence isdecided by the Ω-logic under the Continuum hypothesis. Moreover, if the statementis true assuming CH it is true in ZFC (see e.g. [27]).We shall verify (following [20]) that the conjecture is true for Abelian Polishgroups, in fact, it is true for all groups with a complete (both-sided)-invariantmetric, and also for closed subgroups of the permutation group S ∞ . Whether it istrue in general remains open.There is a, perhaps an even more interesting, stronger ZFC conjecture on thestructure of Polish groups. The following concept was introduced in [19] and theterm coined in [20]: A nonempty subset C of a Polish group G is said to be anti-GMS if it is nowhere dense and for every sequence { U n : n ∈ ω } of open neighborhoodsof 1 there is a sequence { g n : n ∈ ω } of elements of G such that for every g ∈ G ,the set g · S n ∈ ω g n · U n is dense in C .The reason for introducing anti-GMS sets is the following: Proposition 3.7 ([19]) . Assuming cov ( M ) = c , if C ⊆ G is anti-GMS, then thereis a strong measure zero set S such that S · C = G .Proof. Enumerate G = { g α : α < c } and enumerate all sequences of open sets in G as {h U αn : n ∈ ω i : α < c } . Let C ⊆ G be anti-GMS, and for every α < c let h g αn : n ∈ ω i be such that for all g ∈ G , ( g · S n ∈ ω g αn · U αn ) ∩ M is comeager in M .Let U α = S n ∈ ω g αn · U αn .As cov ( M ) = c , the intersection of fewer than c relatively dense open subsets of M is not empty. In particular, for every α < c , there is an m α ∈ M ∩ ( g − α · \ β α U β ) . There is then an x α ∈ T β α U β such that m α = g − α · x α . That is g α = x α · m − α .Let X = { x α : α < c } . Then G = X · M − . Let us see that X ∈ Smz ( G ):Given a sequence { U n : n ∈ ω } of neighbourhoods of 1 G , consider the subsequence { U n : n ∈ ω } . It is listed as h U αn : n ∈ ω i for some α < c . By the construction,every x γ ∈ U α for γ > α . On the other hand, X \ U α ⊆ { x β : β < α } , hence has size TRONG MEASURE ZERO IN POLISH GROUPS 11 less than cov ( M ) = c . Committing the sin of forward reference, by Theorem 4.2(i), X \ U α is a Smz set, hence can be covered by the sets { U n +1 : n ∈ ω } . Hence, X has strong measure zero. (cid:3) The anti-GMS sets are our only tool for disproving the Galvin-Mycielski-SolovayTheorem in non-locally compact groups. Hence the
Strong Conjecture is:
Conjecture 3.8.
Exactly one of the following holds for any Polish group G : Either G is locally compact, or it contains an anti-GMS set. It is immediate from Proposition 3.7 that the Strong conjecture, indeed, solvesthe GMS conjecture stated above. It is, in fact the Strong conjecture we haveverified for the aforementioned classes of groups:
Theorem 3.9 ([20]) . Let G be a non-locally compact TSI
Polish group. Then G contains an anti-GMS set.Proof. Let µ be a non-trivial left-invariant countably subadditive, diffuse outerregular submeasure given by Theorem 2.4. Claim 3.10.
There is a nowhere dense set C ⊆ G such that for every open set O intersecting C there is an open set U and { g m : m ∈ ω } ⊆ G such that for every m ∈ ω g m · U ⊆ O and lim m ∈ ω µ ( g m · U \ C ) = 0 .Proof. To construct the set C let { B n : n ∈ ω } be a basis for the topology of G . Recursively construct an increasing sequence of open sets { W k : k ∈ ω } and asequence { A k : k ∈ ω } of countable sets of pairs of the form h U, ε i , where U is anopen subset of G and ε > B k ∩ W k +1 = ∅ ,(2) ∀ g ∈ G |{h U, ε i ∈ A k : g ∈ U }| k ,(3) ∀h U, ε i ∈ A k µ ( W k +1 ∩ U \ W k ) < ε k , and if k is the least such that h U, ε i ∈ A k then W k ∩ U = ∅ , and(4) (a) either B k ⊆ W k +1 ,(b) or, there is an open neighborhood V of 1 G and there are distinct { g i : i ∈ ω } ⊆ G and { ε i : i ∈ ω } such that for all i ∈ ω g i · V ⊆ B n , h g i · V, ε i i ∈ A k and lim i ∈ ω ε i = 0.To carry out the construction, put first W = A = ∅ . Having constructed W k and A k , see first whether B k ⊆ W k . If so, let A k +1 = A k and let W k +1 = W k ∪ B k .If B k W k , let U , U be disjoint open subsets of B k \ W k . By (2) there is anopen set U ⊆ U contained or disjoint from every U appearing in A k , by non-atomicity of µ we may require µ ( U ) to be so small that W k +1 = W k ∪ U satisfies(3) for all h U, ε i ∈ A k such that U ⊆ U . Finally, as U is not compact, there areinfinitely many balls of the same diameter (i.e. translates of the same open set)with disjoint closures contained in U , add them to A k +1 paired with some realnumbers converging to 0. It is clear that (1)-(4) are satisfied.Now, let C = G \ S k ∈ ω W k . Then C is a closed nowhere dense set by (1). By(3), µ ( U \ C ) < ε for every h U, ε i ∈ S k ∈ ω A k , and if B k ∩ C = ∅ then by (4) B k contains infinitely many sets with the required properties. (cid:3) The set C from the claim is anti-GMS . In order to verify this let a sequence { U m : m ∈ ω } of open neighborhoods of 1 G be given, without loss of generality ofdiameter shrinking to 0. Let again { B n : n ∈ ω } be a basis for the topology of G , TRONG MEASURE ZERO IN POLISH GROUPS 12 and let { A n : n ∈ ω } be a partition of ω into infinite sets. For every B n such that B n ∩ C = ∅ let W n be an open neighborhood of 1 G such that there are distinct { g i : i ∈ ω } ⊆ G such that g i · W n ⊆ B n and lim i ∈ ω µ ( g i · W n \ C ) = 0.Now, let V n be an open neighborhood of 1 G such that V n · U j ⊆ W n for all butfinitely many j ∈ A n , and let g j , h j ∈ G be such that h j · W n ⊆ B n , µ ( h j · W n \ C ) <µ ( U j ), and G = S j ∈ A n h j · V n · g − j (here is where we use the invariance of the metric).The sequence { g i : i ∈ ω } ⊆ G witnesses (for the sequence { U i : i ∈ ω } that C is anti-GMS. Indeed, of g ∈ G and B n ∩ C = ∅ then there is a j ∈ A n such that g ∈ h j · V n · g − j , i.e. g · g j ∈ h j · V n , hence g · g j · U j ⊆ h j · V n · U j ⊆ h j · W n ⊆ B n . As µ ( h j · W n \ C ) < µ ( U j ), we get C ∩ B n ∩ g · g j · U j = ∅ , as required. (cid:3) Corollary 3.11.
The strong conjecture is true for Abelian groups.
We do not know, whether the strong conjecture is true for all Polish groups, butwe can confirm it for another important class of groups – the automorphism groupsof countable structures:
Theorem 3.12 ([20]) . Let G be a non-locally compact closed subgroup of S ∞ . Then G contains an anti-GMS set.Proof. As G is not locally compact, there is an infinite A ⊆ ω such that for all n ∈ A there are infinitely many m ∈ ω for which there are g ∈ G such that g ↾ n = 1 G ↾ n and g ( n ) = m . Given n, m ∈ A , n < m define R n,m = { ( g, h ) ∈ G × G : g ( n ) rng( h ↾ m ) and h ( n ) rng( g ↾ m ) } . Note that the relation is left-invariant. Let B = { u ∈ ω <ω : U is one-to-one and dom ( u ) ∈ A and u ⊆ g for some g ∈ G } . Claim 3.13.
For every n < m ∈ A and u ∈ B such that | u | = n there are g, h ∈ G both extending u such that ( g, h ) ∈ R n,m .Proof. By the left-invariance of R n,m , we may assume that u is the identity on n = dom ( u ). Let { g k : k ∈ ω } ⊆ G be such that g k ↾ n = u and { g k ( n )) : k ∈ ω } are pairwise distinct, by further shrinking we may assume that for every l ∈ [ n, m )either { g k ( n )) : k ∈ ω } are pairwise distinct or all equal. In particular, for every k ∈ ω the set { k ∈ ω : g k ( n ) ∈ rng( g k ↾ [ n, m ) } is finite, hence ( g , g k ) ∈ R n,m for almost all k ∈ ω . (cid:3) Given a subset H of B let T ( H ) = { u ∈ B : ∀ k ∈ ω u ↾ k H } . Construct D ⊆ B such that(1) ∀ u ∈ B ∃ v ∈ D u ⊆ v ,(2) ∀ v, v ′ ∈ D rng( v ) ⊆ rng( v ′ ) or rng( v ′ ) ⊆ rng( v ), and(3) ∀ u ∈ T ( D ) ∀ m ∈ ω ∃ v ∈ T ( D ) m ∈ rng( v ).It is easy to construct such a set using a simple bookkeeping argument. Havingdone so, let C = [ T ( D )] . To see that C is anti-GMS, consider an infinite set Z ⊆ A . By the Claim thereis a sequence { g n : n ∈ Z } ⊆ G such that ∀ u ∈ B ∃ n o < n ∈ Z u ⊆ g n ∩ g n and ( g n o , g n ) ∈ R | u | ,n . To finish the proof it suffices to show that g · S n ∈ Z h g n ↾ n i∩ C is dense in C for every g ∈ G . To that end fix g ∈ G and v ∈ B such that h v i ∩ C = ∅ , and let k = dom ( v ) TRONG MEASURE ZERO IN POLISH GROUPS 13 and u = g − · v . Choose n < n ∈ Z such that u ⊆ g n ∩ g n and ( g n o , g n ) ∈ R k,n .Then either h g · g n ↾ n i ∩ C = ∅ or h g · G n ↾ n i ∩ C = ∅ : If not, then there are s , s ∈ D such that s ⊆ g · g n ↾ n and s ⊆ g · g n ↾ n . Neither s ⊆ v nor s ⊆ v as C ∩ h v i 6 = ∅ , so g · g n ( k ) ∈ rng( s ) and g · g n ( k ) ∈ rng( s ). This, however,contradicts the assumption that rng( s ) ⊆ rng( s ) or rng( s ) ⊆ rng( s ). (cid:3) Cardinal invariants of
Smz in Polish groups
Given an ideal I of subsets of a set X the following are the standard cardinalinvariants associated with I : non ( I ) = min {| Y | : Y ⊆ X ∧ Y / ∈ I} , add ( I ) = min {|A| : A ⊆ I ∧ S A / ∈ I} , cov ( I ) = min {|A| : A ⊆ I ∧ S A = X } , cof ( I ) = min {|A| : A ⊆ I ∧ ( ∀ I ∈ I )( ∃ A ∈ A )( I ⊆ A ) } . We denote by M , N the ideals of meager and Lebesgue null subsets of 2 ω , respec-tively. For f, g ∈ ω ω , we say that f ∗ g if f ( n ) g ( n ) for all but finitely many n ∈ ω (the order of eventual dominance ). A family F ⊆ ω ω is bounded if there isan h ∈ ω ω such that f ∗ h for all f ∈ F ; and F is dominating if for any g ∈ ω ω there is f ∈ F such that g ∗ f . The cardinal invariants related to eventual domi-nance are b (the minimal cardinality of an unbounded family) and d (the minimalcardinality of a dominating family).We shall briefly review the results (not necessarily in the chronological order)concerning other cardinal invariants of Smz ( G ), after which we give a more detailedaccount of the non ( Smz ) in Polish groups. We shall denote by
Smz the ideal ofstrong measure zero subsets of R . Concerning additivity of Smz , Carlson [9] in effectshowed that add ( N ) add ( Smz ), that add ( Smz ) non ( N ) is a triviality, whileGoldstern, Judah and Shelah [15] showed that consistently cof ( M ) < add ( Smz ),and, of course, Laver [28] that consistently add ( Smz ) < b and Baumgartner [2] thatconsistently add ( Smz ) < non ( N ).For cofinality of Smz , there are lower bounds cov ( N ) and cov ( M ) (see below),and it is folklore fact that assuming CH cof ( Smz ) > c , while the Borel conjectureproduces models where cof ( Smz ) = c . Yorioka [51] and more recently Cardona[7] produced models of ZFC where cof ( Smz ) < c . According to our knowledge,it has not been subject to study if add ( Smz ( G )), and/or cof ( Smz ( G )) may varydepending on the Polish group in question.It is, however, known that the uniformity numbers may differ depending on thegroup. There is also a surprising asymmetry between cov ( Smz ) and non ( Smz ).The trivial lower bound for cov ( Smz ) is cov ( N ) (every Smz -set has Lebesgue mea-sure zero) and it also seems to be the best one. Pawlikowski [37] showed that cov ( Smz ) < add ( M ) is consistent. On the other hand, Cardona, Mej´ıa and Riera-Marid [8] recently showed that cov ( Smz ) = ω = c in the iterated Sacks model,hence, there seems to be very little in terms of upper bounds on cov ( Smz ), inparticular, consistently cof ( N ) < cov ( Smz ).Before moving on we shall mention some of the open problems concerning theseinvariants:
Question 4.1. (1) ([8]) Is it consistent that all four of the cardinal invariantscorresponding to
Smz have simultaneously different values?
TRONG MEASURE ZERO IN POLISH GROUPS 14 (2) ([8]) Is it consistent that add ( Smz ) < min { cov ( Smz ) , non ( Smz ) } ?(3) Do any of add ( Smz ( G )), cov ( Smz ( G )), cof Smz ( G )) depend on which Polishgroup one considers?Finally, we arrive at the uniformity of Smz ( G ) which we shall discuss in consid-erably more detail. Two more invariants are required here (see [31, 29, 1, 19]): eq = min {| F | : F ⊆ ω ω bounded, ∀ g ∈ ω ω ∃ f ∈ F ∀ n ∈ ω f ( n ) = g ( n ) } . eq ∗ = min {| F | : F ⊆ ω ω bounded, ∀ g, h ∈ ω ω ∃ f ∈ F ∃ ∞ n ∀ k ∈ [ h ( n ) , h ( n + 1)) f ( n ) = g ( n ) } . It is a theorem of Bartoszy´nski [1, 2.4.1] that omitting “bounded” from the defini-tion of eq yields cov ( M ), i.e. cov ( M ) = min {| F | : F ⊆ ω ω ∀ g ∈ ω ω ∃ f ∈ F ∀ n ∈ ω f ( n ) = g ( n ) } . The following diagram (see [1, 18] for proofs) summarizes the provable inequalitiesbetween the cardinal invariants mentioned (the arrows point from the smaller tothe larger cardinal). b / / d add ( M ) / / O O % % ❑❑❑❑❑❑❑❑❑❑ cov ( M ) / / O O eq / / non ( N ) eq ∗ ; ; ①①①①①①①①① In addition, add ( M ) = min { b , eq } = min { b , eq ∗ } , while cov ( M ) < min { d , eq } is consistent with ZFC by a theorem of Goldstern, Judah and Shelah [15]. Forany separable metric space X there are upper and lower bounds for non ( Smz ( X ))given by Rothberger [40] and Szpilrajn [46], respectively. The uniformity invariant non ( Smz ( X )) for X = 2 ω and X = ω ω was calculated by Bartoszy´nski [1], andFremlin and Miller [30], respectively. Theorem 4.2.
Let X be a separable metric space that is not Smz . (i) ([40]) cov ( M ) non ( Smz ( X )) , (ii) ([46]) if X is not of universal measure zero , then non ( Smz ( X )) non ( N ) , (iii) ([1]) non ( Smz ( ω ω )) = cov ( M ) and (iv) ([30]) non ( Smz (2 ω )) = eq .Proof. (i) by now is standard: Given a separable metric space X and a sequence { ε n : n ∈ ω } , pick a countable dense set { d n : n ∈ ω } ⊆ X . For every one to onefunction f ∈ ω ω let U f = S n ∈ ω B ( x n , ε f n ). Now, assume that | X | < cov ( M ). Tofinish the proof it suffices to note that for every x ∈ X the set N x = { f ∈ ω ω : f isone-to-one and x U f } is nowhere dense in the (closed) subspace of ω ω consistingof one-to-one functions.To see (ii) recall that every Smz set is of universal measure zero by Proposition2.6, so it suffices to show that non ( N ) is the minimal size of a space which is Recall that a metric space X is of universal measure zero if there is no probability Borelmeasure on X vanishing on singletons. TRONG MEASURE ZERO IN POLISH GROUPS 15 not of universal measure zero. To see this note that any diffuse Borel probabilitymeasure µ on a separable metric space X extends to a finite Borel diffuse probabilitymeasure µ on its completion ˆ X by putting µ ( A ) = µ ( A ∩ X ). Now, by a theoremof Parthasarathy [35] there is a measure preserving Borel isomorphism between ˆ X with µ and [0 ,
1] equipped with the Lebesgue measure, hence | X | > non ( N ).For (iii) it suffices to see that non ( Smz ( ω ω )) cov ( M ). Let F ⊆ ω ω be suchthat | F | = cov ( M ) and ∀ g ∈ ω ω ∃ f ∈ F ∀ n ∈ ω f ( n ) = g ( n ). We shall showthat F Smz ( ω ω ). Let h s n : n ∈ ω i be a sequence of elements of 2 <ω such that | s n | = n + 1. Define g ∈ ω ω by putting g ( n ) = s n ( n ). Then there is an f ∈ F suchthat f ( n ) = g ( n ) for all n ∈ ω . This, however, means, that s n f for any n ∈ ω .That is no sequence of open sets of diameter n +1 covers F .The proof of (iv) is similar. First we shall show that non ( Smz (2 ω )) eq . Tothat end let X ⊆ ω be of size less that eq , and let a sequence h ε n : n ∈ ω i ofpositive real numbers be given. Let h ∈ ω ω be such that h ( n ) ε n for every n ∈ ω . For each n ∈ ω enumerate 2 n – the set of binary sequences of length n - as { s nm : m < ω } . For every x ∈ X let f x ∈ ω ω be defined by f x ( n ) = m if and only if x ↾ h ( n ) = s h ( n ) m . Then f x ( n ) h ( n ) for every x ∈ X and n ∈ ω . As | X | < eq , there is a g ∈ ω ω suchthat f x ∩ g = ∅ for every x ∈ X , and without loss of generality, g ( n ) h ( n ) for every n ∈ ω (values above are irrelevant, and can be changed). Then hh s h ( n ) g ( n ) i : n ∈ ω i covers X .On the other hand, assume that F ⊆ ω ω is a bounded family of size less than non ( Smz (2 ω )). Let h ∈ ω ω be such that f ( n ) h ( n ) for every f ∈ F and n ∈ ω .Let { s nm : m < ω } enumerate 2 n as above. For every f ∈ ω ω let x f = s h (0) f (0) a s h (1) f (1) a s h (2) f (2) a . . . and consider the set X = { x f : f ∈ F } . As X is Smz , there is a sequence h t n : n ∈ ω i such that(1) t n ∈ P i n h ( n ) , and(2) ∀ f ∈ F ∃ n ∈ ω t n ⊆ x f .Let g ∈ ω ω be such that g ( n ) h ( n ) for all n ∈ ω , and g ( n ) = m whenever t n +1 = t n a s h ( n ) m . Note that g ( n ) = f ( n ) whenever t n ⊆ x f . (cid:3) In particular, the theorem evaluates non ( Smz ) for two groups: the compactboolean group 2 ω and the Baer-Specker group Z ω . In order to extend these to awider class of groups we shall need two easy observations:As mentioned before, strong measure zero is a uniform property, in particular,a uniformly continuous image of a Smz set is
Smz and, on the other hand, if X uniformly embeds into Y , then any set A ⊆ X that is not Smz in X is not Smz in Y either. It follows that: Lemma 4.3. (i)
If a space Y is a uniformly continuous image of a space X then non ( Smz ( X )) non ( Smz ( Y )) . (ii) If X uniformly embeds into Y , then non ( Smz ( X )) > non ( Smz ( Y )) . Lemma 4.4. A CLI
Polish group is either locally compact, or contains a uniformcopy of ω ω . TRONG MEASURE ZERO IN POLISH GROUPS 16
Proof.
Let G be a group equipped with a complete, left-invariant metric d . Assum-ing G is not locally compact no open set is totally bounded, hence for every ε > δ > B (1 , ε ) contains an infinite set of points that arepairwise at least δ apart. Using this fact construct, for each n ∈ ω , ε n > { x in : i ∈ ω } ⊆ B (1 , ε n ) such that if i = j then d ( x in , x jn ) > ε n +1 . For s ∈ ω n let y s = x s (0)0 · x s (1)1 · x s ( n − n − · · · · · x s ( n − n − . The construction ensures that forany f ∈ ω ω the sequence h y f ↾ n : n ∈ ω i is Cauchy. Let z f be its limit. It is easy tocheck that since d is left-invariant, the mapping f z f is a uniform embedding of ω ω into G . (cid:3) Theorem 4.5.
Let G be a CLI
Polish group. (i) If G is locally compact, then non ( Smz ( G )) = eq , (ii) if G is not locally compact, then non ( Smz ( G )) = cov ( M ) .Proof. As every Polish group contains a (uniform) copy of 2 ω , cov ( M ) non ( Smz ( G )) eq for every Polish group G . Similarly, non ( Smz ( G )) cov ( M ) whenever G containsa uniform copy of ω ω by Theorem 4.2. In particular, non ( Smz ( G )) = cov ( M ) if G is a non-locally compact CLI group.So all that remains to be seen is that eq non ( Smz ( G )) for any locally compactPolish group G . Write G as an increasing union of compact subsets K n , n ∈ ω . Aseach K n is a uniformly continuous image of 2 ω (see e.g. [22, Theorem 4.18]), everysubset of K n which is not of strong measure zero has size at least eq by Lemma 4.3(i)and Theorem 4.2(iii). On the other hand, as Smz ( G ) is a σ -ideal (Proposition 2.1),every subset of G which is not Smz has a non-
Smz intersection with one of the k n ’s, hence eq non ( Smz ( G )). (cid:3) It is, of course, a natural question whether the result of the Theorem (or of thepreceding lemma) remains true also for Polish groups which are not
CLI .The final remark of this section deals with transitive covering for category (con-sidered by Bartoszy´nski [1, 2.7] for 2 ω , and Miller and Stepr¯ans [31] for generalPolish group and further studied in [19]): cov ∗ ( M ( G )) = min {| A | : A ⊆ G and A · M = G for some meager set M ⊆ G } By Theorem 3.5, for a locally compact group G strong measure zero sets co-incide with the sets whose meager translates do not cover G , hence, in partic-ular, non ( Smz ( G )) = cov ∗ ( M ( G )) for every locally compact group G . It fol-lows from Prikry’s Proposition 3.1 that cov ∗ ( M ( G )) non ( Smz ( G )), for everyPolish group G , hence cov ( M ) cov ∗ ( M ( G )) eq for any Polish group. As non ( Smz ( G )) = cov ( M ) for all CLI groups which are not locally compact, we canconclude that:
Corollary 4.6. non ( Smz ( G )) = cov ∗ ( M ( G )) for any CLI group.
The conjecture is that the two numbers coincide for any Polish group.
Question 4.7. Is cov ∗ ( M ( G )) = cov ( M ) for all Polish groups which are not locallycompact? TRONG MEASURE ZERO IN POLISH GROUPS 17 Strong measure zero and Hausdorff measures
As mentioned above, there is a profound connection between strong measurezero and Hausdorff measures. The following characterizations of strong measurezero in terms of Hausdorff measures and dimensions were proved in [54]. They arebased on a classical Besicovitch result [4, 5].
Hausdorff measure.
Before getting any further we need to review Hausdorff mea-sure and dimension. We set up the necessary definitions and recall relevant facts.A non-decreasing, right-continuous function h : [0 , ∞ ) → [0 , ∞ ) such that h (0) =0 and h ( r ) > r > gauge . Gauges are often ordered as follows, cf. [39]: g ≺ h def ≡ lim r → h ( r ) g ( r ) = 0 . Notice that for any sequence h h n : n ∈ ω i of gauges there is a gauge h such that h ≺ h n for all n .Given δ >
0, call a cover A of a set E ⊆ X a δ -fine cover if ∀ A ∈ A diam A δ .If h is a gauge, the h -dimensional Hausdorff measure H h ( E ) of a set E ⊆ X isdefined thus: For each δ > H hδ ( E ) = inf (X n ∈ ω h (diam E n ) : { E n } is a countable δ -fine cover of E ) and let H h ( E ) = sup δ> H hδ ( E ) . In the common case of h ( r ) = r s for some s >
0, we write H s for H h , andlikewise for H hδ .Properties of Hausdorff measures are well-known. The following, including thetwo lemmas, can be found e.g. in [39]. The restriction of H h to Borel sets is a G δ -regular Borel measure. Recall that a sequence of sets h E n : n ∈ ω i is termed a λ -cover of E ⊆ X if every point of E is contained in infinitely many E n ’s. Theorem 5.1 ([4, 5]) . A metric space X is Smz if and only if H h ( X ) = 0 for eachgauge h .Proof. Suppose first that X is Smz . Let h be a gauge. For each δ > h ε n i such that 0 < ε n < δ and h ( ε n ) < δ − n . There is a cover { U n } of X such that diam U n < ε n for all n . Obviously P h (diam U n ) < δ and it follows that H hδ ( X ) δ . Let δ → H h ( X ) = 0.Now suppose that H h ( X ) = 0 for every gauge. Let h ε n i be a sequence of positivenumbers. Choose a gauge h such that h ( ε n ) > n . Since H h ( X ) = 0, there is acountable cover { U n } such that P h (diam U n ) <
1. As h is right-continuous, thereare δ n > diam U n such that P h ( δ n ) <
1. Since δ n >
0, rearranging the sequencewe may suppose that δ n decrease. Therefore nh ( δ n ) P i 1. It followsthat h ( δ n ) < n < h ( ε n ) and consequently δ n < ε n and in particular diam U n < ε n ,as required. (cid:3) We will need a cartesian product inequality. Given two metric spaces X and Y with respective metrics d X and d Y , provide the cartesian product X × Y with themaximum metric(5) d (cid:0) ( x , y ) , ( x , y ) (cid:1) = max( d X ( x , x ) , d Y ( y , y )) . TRONG MEASURE ZERO IN POLISH GROUPS 18 A gauge h satisfies the doubling condition or h is doubling if lim r → h (2 r ) h ( r ) < ∞ . Lemma 5.2 ([23, 17]) . Let X, Y be metric spaces, g a gauge and h a doublinggauge. Then H h ( X ) H g ( Y ) H hg ( X × Y ) . The following lemma on uniformly continuous mappings is well-known, see, e.g.,[39, Theorem 29]. Lemma 5.3. Let f : ( X, d X ) → ( Y, d Y ) be a uniformly continuous mapping and g its modulus, i.e., d Y ( f ( x ) , f ( y )) g ( d X ( x, y )) for all x, y ∈ X . Then H h ( f ( X )) H h ◦ g ( X ) for any gauge h . Recall that the Hausdorff dimension of X is defined bydim H X = sup { s > H s ( X ) = ∞} = inf { s > H s ( X ) = 0 } . Properties of Hausdorff dimension are well-known. In particular, dim H X = 0 if X is countable; and if f : X → Y is Lipschitz, then dim H f ( X ) dim H X . Corollary 5.4 ([54]) . Let X be a metric space. The following are equivalent. (i) X is Smz , (ii) dim H f ( X ) = 0 for each uniformly continuous mapping f on X , (iii) dim H ( X, ρ ) = 0 for each uniformly equivalent metric ρ on X .Proof. (i) ⇒ (ii) Let s > f : X → Y be uniformly continuous andlet g be the modulus of f . Define h ( x ) = ( g ( x )) s . Lemma 5.3(i) yields H s ( f ( X )) H h ( X ). By the above theorem H h ( X ) = 0 and thus H s ( f ( X )) = 0. Since thisholds for all s > 0, it follows that dim H f ( X ) = 0.(ii) ⇒ (iii) is trivial.(iii) ⇒ (i) Denote by d the metric of X . Let h be a gauge. Choose a strictlyincreasing, convex (and in particular subadditive) gauge g such that g ≺ h . Theproperties of g ensure that ρ ( x, y ) = g ( d ( x, y )) is a uniformly equivalent metric on X . The identity map id X : ( X, ρ ) → ( X, d ) is of course uniformly continuous and itsmodulus is g − , the inverse of g . Hence by Lemma 5.3(i) H h ( X, d ) H h ◦ g − ( X, ρ ).Since g ≺ h , we have H h ◦ g − ( X, ρ ) H ( X, ρ ) and H ( X, ρ ) = 0 by (ii). Thus H h ( X, d ) = 0. (cid:3) Our next goal is to characterize Smz by behavior of cartesian products. Recallthat for p ∈ <ω , h p i = { x ∈ ω : p ⊆ x } denotes the cone determined by p and for T ⊆ <ω we let h T i = S p ∈ T h p i .The coordinatewise addition modulo 2 makes 2 ω a compact topological group.Routine proofs show that in the metric of the least difference (defined in the intro-duction) the 1-dimensional Hausdorff measure H coincides on Borel sets with itsHaar measure, i.e., the usual product measure on 2 ω . In particular H (2 ω ) = 1.We consider the important σ -ideal E on 2 ω generated by closed null sets, i.e.,the ideal of all subsets of 2 ω that are contained in an F σ set of Haar measure zero. Lemma 5.5. (i) For each I ∈ [ ω ] ω , the set C I = { x ∈ ω : x ↾ I ≡ } is in E . (ii) For each h ≺ there is I ∈ [ ω ] ω such that H h ( C I ) > .Proof. (i) Let I ∈ [ ω ] ω . For each n ∈ ω , the family {h p i : p ∈ C I ↾ n } is obviouslya 2 − n -fine cover of C I of cardinality 2 | n \ I | . Therefore H − n ( C I ) | n \ I | − n =2 −| n ∩ I | . Hence H ( C I ) lim n →∞ −| n ∩ I | = 0. TRONG MEASURE ZERO IN POLISH GROUPS 19 (ii) h ≺ h (2 − n )2 − n → ∞ . Therefore there is I ∈ [ ω ] ω sparse enoughto satisfy 2 | n ∩ I | h (2 − n )2 − n , i.e., 2 −| n \ I | h (2 − n ) for all n ∈ ω . Consider theproduct measure λ on C I given as follows: If p ∈ n and h p i ∩ C I = ∅ , put λ ( h p i ∩ C I ) = 2 −| n \ I | . Straightforward calculation shows that h (diam E ) > λ ( E )for each E ⊆ C I . Hence P n h (diam E n ) > P n λ ( E n ) > λ ( C I ) = 1 for each cover { E n } of C I and H h ( C I ) > (cid:3) Theorem 5.6. The following are equivalent. (i) X is Smz , (ii) H h ( X × Y ) = 0 for every gauge h and every compact metric space Y such that H h ( Y ) = 0 , (iii) H ( X × E ) = 0 for every E ∈ E .Proof. (i) ⇒ (ii): Suppose X is Smz and let Y be compact. Let δ > 0. Since H h ( Y ) = 0, for each j ∈ ω there is a finite cover U j of Y such that P U ∈U j h (diam U ) < − j − δ . We may also assume that diam U < δ for all U ∈ U j .Let ε j = min { diam U : U ∈ U j } . Since X is Smz , there is a cover { V j } of X such that diam V j ε j . Define W = { V j × U : j ∈ ω, U ∈ U j } . W is obviously a cover of X × Y . The choice of ε j yields diam( V j × U ) = diam U for all j and U ∈ U j . Therefore X W ∈W h (diam W ) = X j ∈ ω X U ∈U j h (diam U ) < X j ∈ ω − j − δ = δ. It follows that H hδ ( X × Y ) < δ , and H h ( X × Y ) = 0 obtains by letting δ → ⇒ (iii) ⇒ (iv) is trivial.(iv) ⇒ (i): Suppose X is not Smz . We will show that H ( X × E ) > E ∈ E . By assumption and Theorem 5.4 there is a gauge h such that H h ( X ) > h be concave and h ( r ) > √ r . In particular, by concavity of h thefunction g ( r ) = r/h ( r ) is increasing. Moreover, h ( r ) > √ r yields lim r → g ( r ) = 0,i.e., g is a gauge, and g ≺ 1. Further, g (2 r ) = 2 r/h (2 r ) r/h ( r ) = 2 g ( r ), i.e., g is doubling.Use Lemma 5.5(ii) to find I ∈ [ ω ] ω such that H g ( C I ) > E = C I . ByLemma 5.5(i), E ∈ E . Since g is doubling, Lemma 5.2 applies: H ( X × C I ) = H h · g ( X × C I ) > H h ( X ) · H g ( C I ) > . (cid:3) Corollary 5.7. If X is Smz , then dim H X × Y = dim H Y for every compact metricspace Y . Note that this consistently fails when we drop the assumption that Y is compact:By a classical example (cf. [12, 534P]), if cov ( M ) = c , then there is a Smz set X ⊆ R such that X + X = R . Since X + X is a Lipschitz image of X × X , we havedim H X × X > dim H X + X = 1, while X is Smz and dim H X = 0.6. Sharp measure zero Recall that a set S in a Polish group is called meager-additive if S · M is meagerfor every meager set M . This is obviously a strengthening of the “algebraic” charac-terization of Smz in the Galvin-Mycielski-Solovay Theorem. Meager-additive sets,in particular in 2 ω , have received a lot of attention, e.g., in [42, 1, 33, 47]. TRONG MEASURE ZERO IN POLISH GROUPS 20 Very recently it was shown that meager-additive sets are characterized by acombinatorial condition very similar to the definition of Smz and also in terms ofHausdorff measures. In this section we will have a look at these descriptions. Definition 6.1. A set S ⊆ X in a complete metric space X has sharp measurezero if for every gauge h there is a σ -compact set K ⊇ S such that H h ( K ) = 0.We first work towards an intrinsic definition equivalent to the one above. Thefollowing variation of Hausdorff measure seems to be the right notion for that. Let h be a gauge. For each δ > H hδ ( E ) = inf ( N X n =0 h (diam E n ) : { E n : n N } is a finite δ -fine cover of E ) and let H h ( E ) = sup δ> H hδ ( E ) . Note the striking similarity with the Hausdorff measure. The only difference isthat only finite covers are taken into account. It is easy to check that H h is finitelysubadditive. However, it is not a measure, since it may (due to the finite covers)lack σ -additivity. To turn it into a measure we need to apply the operation knownas Munroe’s Method I construction (cf. [32] or [39]): H h ( E ) = inf nX n ∈ ω H h ( E n ) : E ⊆ [ n ∈ ω E n o . Thus the defined set function H h is indeed an outer measure whose restriction toBorel sets is a Borel measure. We will called it h -dimensional upper Hausdorffmeasure .Upper Hausdorff measures behave much like Hausdorff measures. We list someimportant properties of upper Hausdorff measures. We refer to [54] for details.Denote N σ ( H h ) the smallest σ -additive ideal that contains all sets E with H h ( E ) = 0. Note that while E ∈ N σ ( H h ) ⇒H h ( E ) = 0, the reverse implica-tion in general fails. Write E n ր E to denote that h E n : n ∈ ω i is an increasingsequence of sets with union E . The following lists some basic features of H h and H h . Lemma 6.2 ([54]) . Let h be a gauge and E a set in a metric space. (i) If H h ( E ) < ∞ , then E is totally bounded. (ii) H h ( E ) = H h ( E ) . (iii) H h ( E ) = H h ( E ) if E is compact. (iv) If E ∈ N σ ( H h ) , then H h ( E ) = 0 . (v) If X is complete, E ⊆ X and E ∈ N σ ( H h ) , then there is a σ -compact set K ⊇ E such that H h ( K ) = 0 . (vi) If X is complete and E ⊆ X , then H h ( E ) = inf {H h ( K ) : K ⊇ E is σ -compact } . (vii) In particular H h ( E ) = H h ( E ) if E is σ -compact. (viii) If g ≺ h and H g ( E ) < ∞ , then E ∈ N σ ( H h ) ; in particular, H h ( E ) = 0 . Lemma 6.3 ([54]) . E ∈ N σ ( H h ) if and only if E has a γ -groupable cover h U n : n ∈ ω i such that P n ∈ ω h (diam U n ) < ∞ . TRONG MEASURE ZERO IN POLISH GROUPS 21 Proof. Suppose that E ∈ N σ ( H h ). Let E n ր E be such that H h ( E n ) = 0. Foreach n let G n be a finite cover of E n such that P G ∈G n h (diam G ) < − n . Since thefamily G = S n G n is obviously a γ -groupable cover, we are done.In the opposite direction, suppose that h U n : n ∈ ω i is a γ -groupable cover h U n : n ∈ ω i such that P n ∈ ω h (diam U n ) < ∞ with witnessing families G j . Let E k = T j > k S G j . Then E = S k ∈ ω E k . For each k , the set E k is covered by each G j , j > k , and P G ∈G j h (diam G ) is as small as needed for j large enough. Hence H h ( E k ) = 0 and consequently E ∈ N σ ( H h ). (cid:3) It is straightforward from Lemma 6.2 that the following intrinsic definition ofsharp measure zero is consistent with the one above. Definition 6.4. A metric space X has sharp measure zero if H h ( X ) = 0 for everygauge h . Sharp measure zero is abbreviated as Smz ♯ .It is no surprise that Theorem 5.4 has a counterpart for Smz ♯ , with basically thesame proof. Upper Hausdorff dimension is defined as expected:dim H X = sup { s > H s ( X ) = ∞} = inf { s > H s ( X ) = 0 } , see [54, 53] for more on the Hausdorff dimension. Theorem 6.5 ([54]) . Let X be a metric space. The following are equivalent. (i) X is Smz ♯ , (ii) dim H f ( X ) = 0 for each uniformly continuous mapping f on X , (iii) dim H ( X, ρ ) = 0 for each uniformly equivalent metric ρ on X . It is straightforward from the definition and Theorem 6.5 that Smz ♯ is a σ -additiveproperty and that it is preserved by uniformly continuous mappings.Sharp measure zero can be described in terms of covers. The description isstrikingly similar to the Borel’s definition of strong measure zero.A countable cover { U j } of X is a called a γ -cover if each x ∈ X belongs to allbut finitely many U j .The following notion was studied, e.g., in [24]. A sequence h W n i of sets in X is called a γ -groupable cover if there is a partition ω = I ∪ I ∪ I ∪ . . . intoconsecutive finite intervals (i.e. I j +1 is on the right of I j for all j ) such that thesequence h S n ∈ I j W n : j ∈ ω i is a γ -cover. The partition h I j i will be occasionallycalled a witnessing partition and the finite families { U n : n ∈ I j } will be occasionallycalled witnessing families . Theorem 6.6 ([54]) . A metric space X is Smz ♯ if and only if it has the followingproperty: for every sequence h ε n : n ∈ ω i of positive real numbers there is a γ -groupable cover { U n : n ∈ ω } of X such that diam U n ε n for all n .Proof. The pattern of the proof is the same as that of the proof of the Besicovitch’stheorem 5.1, but there are details that make it much more involved.The forward implication is easy. Let h be a gauge. Pick ε n > P n h ( ε n ) < ∞ . By assumption, there is a γ -groupable cover h G n i such thatdiam G n ε n . Therefore P n h (diam G n ) P n h ( ε n ) < ∞ . Now apply Lemma 6.3to conclude that X ∈ N σ ( H h ) and in particular H h ( X ) = 0.The reverse implication: Let h ε n i ∈ (0 , ∞ ) ω . Choose a gauge g such that g ( ε n ) > n for all n > h ≺ g . Since H h ( X ) = 0, Lemma 6.2(viii) yields TRONG MEASURE ZERO IN POLISH GROUPS 22 X ∈ N σ ( H g ), which in turn yields, with the aid of Lemma 6.3, a γ -groupable cover h G n : n ∈ ω i such that P n g (diam G n ) < ∞ . Let { I j : j ∈ ω } be the witnessingpartition and G j = { G n : n ∈ I j } the witnessing families.We want to permute the cover so that diameters decrease. Some of the diametersmay be 0. Also, the permutation may break down the witnessing families. We thushave to exercise some care.For each n choose δ n > diam G n so that P n g ( δ n ) < ∞ . Next choose an increas-ing sequence h j k : n ∈ ω i satisfying for all k ∈ ω (a) P { g ( δ n ) : n ∈ I j k } < − k − ,(b) max { δ n : n ∈ I j k +1 } < min { δ n : n ∈ I j k } .Let I = S k ∈ ω I j k . Rearrange G n ’s within each group G j k so that h δ n : n ∈ I j k i form a non-increasing sequence. Together with (b) this ensures that the sequence h δ n : n ∈ I i is non-increasing.For each n ∈ ω let b n ∈ I be the unique index such that n = | I ∩ b n | and define H n = G b n . It follows, with the aid of (a) and the definition of g , that for all n ∈ ωg (diam H n ) = g (diam G b n ) g ( δ b n ) n X { g ( δ m ) : m ∈ I, m b n } n X { g ( δ m ) : m ∈ I } n < g ( ε n )and thus diam H n ε n for all n . Moreover, the families G j k , k ∈ ω , witness that h H n : n ∈ ω i is a γ -groupable cover. (cid:3) Theorem 6.7 ([54]) . Let X be a metric space. The following are equivalent. (i) X is Smz ♯ , (ii) H h ( X × Y ) = 0 for each gauge h and Y such that H h ( Y ) = 0 , (iii) H ( X × E ) = 0 for each E ∈ E .Proof. (i) ⇒ (ii): Suppose X is Smz ♯ . Let h be a gauge and H h ( Y ) = 0. ByLemma 6.3 there is a γ -groupable cover U of Y such that P U ∈U h (diam U ) < ∞ .For each U ∈ U there is δ U > diam U such that P U ∈U h ( δ U ) < ∞ . Denote by U j the witnessing families and let ε j = min { δ U : U ∈ U j } . Since X is Smz ♯ ,Theorem 6.6 yields a γ -groupable cover h V j : n ∈ ω i of X such that diam V j ε j .Denote by V k the witnessing families. Define a family of sets in X × Y W = { V j × U : j ∈ ω, U ∈ U j } . It is routine to check that W is a γ -groupable cover of F × Y . Since diam( V j × U ) δ U for all j and U ∈ U j by the choice of ε j , we have X W ∈W h (diam W ) X U ∈U h ( δ U ) < ∞ . Thus it follows from Lemma 6.3 that X × Y ∈ N σ ( H h ) and in particular H h ( X × Y ) = 0.(ii) ⇒ (iii) is trivial. The proof of (iii) ⇒ (i) is very much like that of Theorem 5.6.Suppose X is not Smz ♯ . We need to find E ∈ E such that H ( X × E ) > 0. Byassumption there is a gauge h such that H h ( X ) > 0. We may suppose h is concave,and find a doubling gauge g ≺ g ( r ) h ( r ) = r . Then use Lemma 5.5(ii)to find I ∈ [ ω ] ω such that H g ( C I ) > 0. We now need a product inequality on upperHausdorff measures analogous to Lemma 5.2 proved in [54, 3.5,7.4]. TRONG MEASURE ZERO IN POLISH GROUPS 23 Lemma ([54]) . Let X, Y be metric spaces and g a gauge and h a doubling gauge.Then H h ( X ) H g ( Y ) H hg ( X × Y ) . Using this lemma, we get H ( X × C I ) = H h · g ( X × C I ) > H h ( X ) · H g ( C I ) > . (cid:3) As we already mentioned, under cov ( M ) = c there is an example ([12, 534P]) ofa Smz set X ⊆ R such that X × X is not Smz . Scheepers [41] examines thoroughlyconditions imposed on a Smz set X that would ensure that a product of X withanother Smz set is Smz . A recent roofing result claims that if one of the factors is Smz ♯ , then the product is Smz . Theorem 6.8 ([54]) . (i) If X and Y are Smz ♯ , then X × Y is Smz ♯ . (ii) If X is Smz and Y is Smz ♯ , then X × Y is Smz .Proof. Suppose Y is Smz ♯ . By Lemma 6.2(viii), Y ∈ N σ ( H h ) for all gauges h .(i) If X is Smz ♯ , then Theorem 6.7(ii) yields H h ( X × Y ) = 0 for all gauges h .(ii) Let h be a gauge. Since Y is Smz ♯ , it is σ -totally bounded and therefore thereis a σ -compact set K ⊇ Y in the completion of Y such that H h ( K ) = 0. Since X is Smz , Theorem 5.6(ii) yields H h ( X × Y ) = 0, which is by Theorem 5.4(ii)enough. (cid:3) This theorem, together with the above example, provides an easy argument thatshows that consistently not every Smz set is Smz ♯ : The Smz set X such that X × X is not Smz cannot be Smz ♯ .We illustrate the power of the theorem by the following Corollary 6.9. Let X ⊆ R . The following are equivalent. (i) X is Smz ♯ , (ii) all orthogonal projections of X on lines are Smz ♯ , (iii) at least two orthogonal projections of X on lines are Smz ♯ .Proof. Since orthogonal projections are uniformly continuous, (i) ⇒ (ii) from preser-vation of Smz ♯ by uniformly continuous mappings. (ii) ⇒ (iii) is trivial. (iii) ⇒ (i):Let L , L be two nonparallel lines and π , π the corresponding orthogonal projec-tions. Mutatis mutandis we may suppose that L is the x -axis and L is the y -axis.Thus X ⊆ π X × π X . Theorem 6.8(ii) thus concludes the proof. (cid:3) Theorem 6.8(ii) also raises the question whether a space whose product with any Smz set of reals is Smz has to be Smz ♯ . As shown in [54], the answer is consistentlyno: The forcing extension constructed by Corazza in [10] we have the following. Asimilar observation was noted without proof in [33] and also in [49]. Proposition 6.10. In the Corazza model there is a set X ⊆ ω that is not Smz ♯ and yet X × Y is Smz for each Smz set Y ⊆ ω . Meager additive sets and sharp measure zero We now look at the meager-additive sets in Polish groups and establish theirsurprising and profound connection with Smz ♯ sets. The theory nicely parallelsthe Galvin-Mycielski-Solovay Theorem. Most of the material of this section comesfrom [52] and [54]. TRONG MEASURE ZERO IN POLISH GROUPS 24 Whenever G is a Polish group, Smz ♯ ( G ) denotes the family of sharp measurezero sets with respect to any left-invariant metric. The notion od Smz ♯ is of coursea uniform invariant – it is neither a topological, nor a metric property. Therefore itdoes not matter which left-invariant metric we choose. The same proof shows that Smz ♯ sets, just like Smz sets, form a bi-invariant σ -ideal. Proposition 7.1. Smz ♯ ( G ) is a bi-invariant σ –ideal. Recall that if G is a Polish group, we denote by M ( G ), or simply by M if thereis no danger of confusion, the ideal of meager subsets of G . Definition 7.2. A set S ⊆ G is called meager-additive (or M -additive ) if SM is meager for every meager set M ⊆ G . The family of all meager-additive sets isdenoted by M ∗ ( G ).It is straightforward from the definition that Proposition 7.3. M ∗ ( G ) is a bi-invariant σ –ideal. The hard implication of Galvin-Mycielski-Solovay Theorem claims that if S is Smz set in a locally compact group, then S · M = G . The analogous statement for Smz ♯ and meager-additive sets is about as hard as that. Theorem 7.4 ([52]) . Let G be a locally compact Polish group. Then every Smz ♯ set S ⊆ G is M -additive, i.e., Smz ♯ ( G ) ⊆ M ∗ ( G ) .Proof. The proof utilizes Lemma 3.3. Suppose S ⊆ G is a Smz ♯ set and let M ⊆ G be meager. Let K n be compact sets in G with K n ր G and let P n be compactnowhere dense sets with P n ր M . Let { U k } be a countable base of G . For each k choose x k ∈ G and ε k > B ( x k , ε k ) ⊆ U k is compact, and let C k = B ( x k , ε k ). Use Lemma 3.3 to recursively construct a sequence h ε n : n ∈ ω i ofpositive numbers such that ∀ n ∀ i n ∀ x ∈ C i ∀ y ∈ K n ∃ z ∈ C i B ( z, ε n ) ⊆ B ( x, ε n − ) \ (( B ( y, ε n ) ∩ K n ) · P n ) . (6)Since S is Smz ♯ , there is an γ -groupable cover { E n } of S such that diam E n < ε n for all n . Hence for each n there is y such that E n ⊆ B ( y, ε n ). Therefore we mayuse (6) to construct for each k a sequence h x kn : n ∈ ω i such that(7) B ( x kn +1 , ε n +1 ) ⊆ B ( x kn , ε n ) \ (( E n +1 ∩ K n +1 ) · P n +1 ) . It is easy to check that since { E n } is a γ -groupable cover of S and K n ր G , thefamily { E n ∩ K n } is also a γ -groupable cover of S . Thus we might have supposedthat E n ⊆ K n , and also that all E n ’s are closed. Therefore (7) simplifies to(8) B ( x kn +1 , ε n +1 ) ⊆ B ( x kn , ε n ) \ ( E n +1 · P n +1 ) . In particular, B ( x kn , ε n ) is a decreasing sequence of compact balls for all k and thusthere is a point x k ∈ U k such that(9) x k / ∈ [ n ∈ ω ( E n · P n ) . Now construct a set b S as follows: Let G j be the groups of E n ’s witnessing to the γ -groupability of { E n } . Put G n = T n ∈G j E n and let F n = T i TSI groups. Compact or abelian Polish groups are TSI ; any invariant metric on a TSI group is complete. Theorem 7.5 ([52]) . Let G be a TSI Polish group. If S ⊆ G is an M -additive set,then S is Smz ♯ (in any metric on G ). Theorem 7.6. Let G be a locally compact TSI Polish group. Then Smz ♯ ( G ) = M ∗ ( G ) . It takes several pages to prove Theorem 7.5, in contrast to the five lines of theproof of Proposition 3.1. We present a proof of the particular case of G = 2 ω thattakes advantage of the regular combinatorial structure of the Cantor set and a deepcharacterization by Shelah of M -additive sets ([42] or [1, Theorem 2.7.17]) and isthus much shorter. Lemma 7.7 ([42]) . X ⊆ ω is M -additive if and only if ∀ f ∈ ω ↑ ω ∃ g ∈ ω ω ∃ y ∈ ω ∀ x ∈ X ∀ ∞ n ∃ kg ( n ) f ( k ) < f ( k + 1) g ( n + 1) & x ↾ [ f ( k ) , f ( k + 1)) = y ↾ [ f ( k ) , f ( k + 1)) . Theorem 7.8. Smz ♯ (2 ω ) = M ∗ (2 ω ) .Proof. Let S ⊆ ω be M -additive. Let h be a gauge. We will show that H h ( S ) = 0.Define recursively f ∈ ω ↑ ω subject to2 f ( k ) · h (cid:0) − f ( k +1) (cid:1) − k , k ∈ ω. By Lemma 7.7 there is g ∈ ω ω and y ∈ ω such that(10) ∀ x ∈ X ∀ ∞ n ∃ kg ( n ) f ( k ) < g ( n + 1) & x ↾ [ f ( k ) , f ( k + 1)) = y ↾ [ f ( k ) , f ( k + 1)) . Define B k = (cid:8) h p a y ↾ [ f ( k ) , f ( k + 1)) i : p ∈ f ( k ) (cid:9) , k ∈ ω, G n = [(cid:8) B k : g ( n ) f ( k ) < g ( n + 1) (cid:9) , n ∈ ω, B = [ k ∈ ω B k = [ n ∈ ω G n . With this notation (10) reads(11) ∀ x ∈ X ∀ ∞ n ∃ G ∈ G n x ∈ G. TRONG MEASURE ZERO IN POLISH GROUPS 26 Since each of the families G n is finite, it follows that G n ’s witness that B is a γ -groupable cover of X . Using Lemma 6.3 (and Lemma 6.2(iv)) it remains to showthat the Hausdorff sum P B ∈B h (diam B ) is finite. Note that the cones forming thefamilies B ∈ B k are actually balls of radius 2 − f ( k +1) , i.e., diam B = 2 − f ( k +1) forall k and all B ∈ B k . Each of the cones is determined by one p ∈ f ( k ) , therefore |B k | = 2 f ( k ) . Overall we have X B ∈B h (diam B ) = X k ∈ ω X B ∈B k h (diam B ) = X k ∈ ω f ( k ) · h (2 − f ( k +1) ) X k ∈ ω − k < ∞ . (cid:3) The most interesting question raised by Theorems 7.4 and 7.5 is of course if theanalogue of Conjecture 3.6 holds for Smz ♯ . Question 7.9 (CH) . Is it true that if the inclusion Smz ♯ ( G ) ⊆ M ∗ ( G ) holds for aPolish group G , then G is locally compact?Another open problem is the rˆole of TSI in Theorem 7.6. Question 7.10. Can the TSI assumption in Theorem 7.6 be dropped, or weakenedto CLI ? Continuous images and cartesian products of meager-additive sets. Sincemeager-additive sets coincide with Smz ♯ sets in TSI locally compact groups, theyare preserved by continuous mappings and by cartesian products. Theorem 7.11. Let G be a TSI Polish group and G a locally compact Polishgroup. Let f : G → G a continuous mapping. If X ⊆ G is M -additive, then sois f ( X ) . Theorem 7.12. Let G , G be TSI locally compact Polish groups. Let X ⊆ G , X ⊆ G . (i) If X is Smz and X is M -additive, then X × X is Smz . (ii) If X and X are M -additive, then so is X × X . Corollary 7.13. Let G , G be TSI locally compact Polish groups. Let X ⊆ G × G .The following are equivalent. (i) X is M -additive, (ii) proj X and proj X are M -additive, (iii) proj X × proj X is M -additive. We conclude this section with a few remarks on meager-additive sets in theCantor set 2 ω . There are a few variations of M -additivity: A set S in 2 ω is • M -additive if ∀ M ∈ M S + M ∈ M , • sharply M -additive if ∀ M ∈ M ∃ F ⊇ S F σ F + M ∈ M , • flatly M -additive if ∀ M ∈ M ∃ F ⊇ S F σ F + M = 2 ω , • E -additive if ∀ E ∈ E S + E ∈ E , • sharply E -additive if ∀ E ∈ E ∃ F ⊇ S F σ F + E ∈ E .The question whether E -additive sets are related to M -additive sets are relatedwas posed by Nowik and Weiss [34]. Their question was answered in [54] by thefollowing theorem. Let us note that while one would expect that, e.g., the proof of(ii) ⇒ (iv) is a matter of routine, it is actually surprisingly difficult. Theorem 7.14 ([54]) . Let S ⊆ ω . The following properties of S are equivalent. TRONG MEASURE ZERO IN POLISH GROUPS 27 (i) S is Smz ♯ , (ii) S is M -additive, (iii) S is sharply M -additive, (iv) S is flatly M -additive, (v) S is E -additive, (vi) S is sharply E -additive. The definitions of these notions extend in a straightforward way to Polish groups(or in case E is considered, locally copact Polish groups). However, the proofs ofthe above theorem depends very much on the fine combinatorial structure of 2 ω andare thus not easily transferable to a context of a Polish group. With quite someeffort the equivalence (i) ⇔ (ii) ⇔ (iii) ⇔ (iv) in TSI locally compact Polish groupswas proved in [52]. (The equivalence (i) ⇔ (ii) is presented in this section as Theo-rem 7.6.) But the equivalences including E are still not understood. Question 7.15 ([52]) . Let G be a locally compact TSI Polish group.(i) Is every M -additive set in G E -additive?(ii) Is every E -additive set in G M -additive?(iii) Is every E -additive set in G sharply E -additive?8. Uniformity number of meager-additive sets Since the notion of sharp measure zero is rather new, not much is known aboutthe cardinal invariants of the σ -ideal of Smz ♯ sets. We will investigate only theuniformity number of Smz ♯ for metric spaces and Polish groups. We refer to section4 for the notation.Bartoszy´nski [1] and Pawlikowski [36] investigated and calculated a related car-dinal – the uniformity number of the ideal of meager-additive sets in 2 ω . Thiscardinal invariant is termed transitive additivity of M and denoted by add ∗ ( M ).By [1, 2.7.14], add ∗ ( M ) = eq ∗ .The two results on Polish groups, Theorem 8.3 and Corollary 8.4, come from[54]. The other results of this section are new. Theorem 8.1. Let X be a separable metric space that is not Smz ♯ . (i) non ( Smz ♯ ( ω ω )) = add ( M ) , (ii) non ( Smz ♯ (2 ω )) = eq ∗ , (iii) add ( M ) non ( Smz ♯ ( X )) , (iv) if X is σ -totally bounded, then eq ∗ non ( Smz ♯ ( X )) , (v) if X is not of universal measure zero, then non ( Smz ♯ ( X )) non ( N ) .Proof. (i) Recall that the metric on ω ω is the least difference metric given by d ( f, g ) = 2 − n ( f,g ) , where n ( f, g ) = min { n : f ( n ) = g ( n ) } .Let S ⊆ ω ω be an unbounded set such that | X | = b . Since X is not bounded,it is not σ -totally bounded and in particular it is not Smz ♯ . It follows that non Smz ♯ ( ω ω ) b .Theorem 4.2(iii) yields a set S ⊆ ω ω that is not Smz and | S | = cov M . This setis clearly not Smz ♯ and thus non Smz ♯ ( ω ω ) cov M .It follows that non Smz ♯ ( ω ω ) min( cov M , b ). By a theorem of Miller [29]min( cov M , b ) = add ( M ) (see also [1, 2.2.9]). Thus we have non Smz ♯ ( ω ω ) add ( M ). TRONG MEASURE ZERO IN POLISH GROUPS 28 In the other direction, suppose that S ⊆ ω ω is not Smz ♯ . By Theorem 7.5, S isnot meager-additive set in the group Z ω . Therefore there is a meager set M ⊆ Z ω such that S + M = S s ∈ S ( M + s ) is not meager and hence | S | > add ( M ). Thus non Smz ♯ ( ω ω ) > add ( M ).(ii) By Theorem 7.4, a set S ⊆ ω is Smz ♯ if and only if it is meager-additive.Thus non Smz ♯ (2 ω ) = add ∗ ( M ) and the latter equals by the aforementioned resultof Bartoszy´nski to eq ∗ .(iii) Let { z m : m ∈ ω } ⊆ X be a dense set in X . To each x ∈ X assign a function b x ∈ ω ω defined by(12) b x ( n ) = min { m : d ( x, z m ) < − n } . We claim that the inverse map b x x is Lipschitz. Indeed, if d ( b x, b y ) = 2 − n , then b x ( n − 1) = b y ( n − 1) and in particular ∃ m d ( x, z m ) < − n +1 and d ( y, z m ) < − n +1 .Therefore d ( x, y ) < · − n +1 , whence d ( x, y ) < d ( b x, b y ). In particular b x x isuniformly continuous. So if S ⊆ X is not Smz ♯ , then b S = { b x : x ∈ S } is not Smz ♯ as well. If follows that non ( Smz ♯ ( X )) > non ( Smz ♯ ( ω ω )) and (iii) follows from (i).(iv) Consider the completion X ∗ of X . Since X is σ -totally bounded, there isa σ -compact set K ⊆ X ∗ that contains X . Let K n ր K be a sequence of compactsets. Suppose that S ⊆ X is a not Smz ♯ set such that | S | = non ( Smz ♯ ( X )). Thereis n such that S ∩ K n is not Smz ♯ , therefore | S ∩ K n | = non ( Smz ♯ ( X )). Since K n iscompact, it is dyadic : there is a uniformly continuous mapping f : 2 ω → K n onto K n . For each x ∈ S ∩ K n pick b x ∈ f − ( x ) and set b S = { b x : x ∈ S ∩ K n } . Then f ( b S ) = S ∩ K n , hence b S is not Smz ♯ , and clearly | b S | = | S ∩ K n | = non ( Smz ♯ ( X )).It follows that non ( Smz ♯ (2 ω )) non ( Smz ♯ ( X )) whence add ∗ ( M ) non ( Smz ♯ ( X ))by (ii).(v) is a trivial consequence of Theorem 4.2(ii). (cid:3) As expected, for analytic metric spaces we can do better. Recall that a metricspace has the small ball property if it admits a base { B n } such that diam B n → Theorem 8.2. Let X be an uncountable analytic metric space. (i) add ( M ) non ( Smz ♯ ( X )) eq ∗ , (ii) if X is σ -totally bounded, then non ( Smz ♯ ( X )) = eq ∗ , (iii) if X does not have the small ball property, then non ( Smz ♯ ( X )) = add ( M ) .Proof. (i) The left-hand inequality is Theorem 8.1(iii). Since X contains a (uniform)copy of 2 ω , Theorem 8.1(ii) yields eq ∗ = non ( Smz ♯ (2 ω )) > non Smz ♯ ( G ).(ii) follows from (i) and Theorem 8.1(iv).(iii) Consider the mapping x b x defined by (12). Let B ⊆ ω ω be an unboundedset such that | B | = b . As shown in [19, 4.4], the set b X = { b x : x ∈ X } is dominatingin ω ω . Therefore for each f ∈ B there is x f ∈ X such that b x f > ∗ f . Set S = { x g : f ∈ B } . Then b S is not bounded, because it dominates B . It follows that S is not σ -totally bounded in X , and in particular it is not Smz ♯ . Since clearly | b S | = | B | = b , we conclude that non ( Smz ♯ ( X )) b . By (i) also non ( Smz ♯ ( X )) add ( M ), so non ( Smz ♯ ( X )) min( eq ∗ , b ) = add ( M ), and the reverse inequalityfollows from (i). (cid:3) TRONG MEASURE ZERO IN POLISH GROUPS 29 We will now calculate uniformity numbers of Smz ♯ ( G ) for CLI Polish groups andof M ∗ ( G ) for TSI Polish groups. Theorem 8.3 ([54]) . Let G be a CLI Polish group. (i) If G is locally compact, then non ( Smz ♯ ( G )) = eq ∗ . (ii) If G is not locally compact, then non ( Smz ♯ ( G )) = add ( M ) .Proof. (i) is a straightforward from Theorem 8.2(ii).(ii) By Lemma 4.3, G contains a uniform copy of ω ω . Therefore non ( Smz ♯ ( G )) non ( Smz ♯ ( ω ω )). Now apply Theorem 8.1(i) to get non ( Smz ♯ ( G )) non ( M ). Thereverse inequality is straightforward from Theorem 8.2(i). (cid:3) The following is a trivial consequence of this theorem and Theorem 7.4. Corollary 8.4 ([54]) . Let G be a TSI Polish group. 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MR 295768654. , Strong measure zero and meager-additive sets through the prism of fractal measures ,Comment. Math. Univ. Carol. (2019), no. 1, 131–155. Centro de Ciencias Matem´aticas, Universidad Nacional Aut´onoma de M´exico, ´CampusMorelia, 58089, Morelia, Michoac´an, M´exico. E-mail address : [email protected] URL : ∼ michael Department of Mathematics, Faculty of Civil Engineering, Czech Technical Univer-sity, Th´akurova 7, 160 00 Prague 6, Czech Republic E-mail address : [email protected] URL ::