Cardinal invariants of Haar null and Haar meager sets

A subset $X$ of a Polish group $G$ is \emph{Haar null} if there exists a Borel probability measure $\mu$ and a Borel set $B$ containing $X$ such that $\mu(gBh)=0$ for every $g,h \in G$. A set $X$ is \emph{Haar meager} if there exists a compact metric space $K$, a continuous function $f : K \to G$ and a Borel set $B$ containing $X$ such that $f^{-1}(gBh)$ is meager in $K$ for every $g,h \in G$. We calculate (in $ZFC$) the four cardinal invariants ($\rm add$, $\rm cov$, $\rm non$, $\rm cof$) of these two $\sigma$-ideals for the simplest non-locally compact Polish group, namely in the case $G = \mathbb{Z}^\omega$. In fact, most results work for separable Banach spaces as well, and many results work for Polish groups admitting a two-sided invariant metric. This answers a question of the first named author and Z. Vidny\'anszky.


Introduction
Small sets play a fundamental role in many branches of mathematics. Perhaps the most important example is the family of nullsets of a natural invariant measure. Such a natural measure is the Lebesgue measure on R d , or more generally the Haar measure on a locally compact group. However, on larger groups such as C[0, 1] or S ∞ there is no such measure (a Polish group carries a Haar measure iff it is locally compact, see e.g. [8]). Therefore J. P. R. Christensen [5] introduced the following notion. Definition 1.1. A subset X of a Polish group G is Haar null if there exists a Borel probability measure µ and a Borel set B containing X such that µ(gBh) = 0 for every g, h ∈ G.
The family of Haar null sets is denoted by HN (G) or simply HN .
Christensen proved that these sets form a proper σ-ideal which coincides with the family of sets of Haar measure zero in the locally compact case. This notion turned out to be very useful in various branches of mathematics, see e.g. the survey paper [8].

Definition 1.2.
A subset X of a Polish group G is Haar meager if there exists a compact metric space K, a continuous function f : K → G and a Borel set B containing X such that f −1 (gBh) is meager in K for every g, h ∈ G.
The family of Haar meager sets is denoted by HM(G) or simply HM.
Analogously to the Haar null case, Darji proved that these sets form a proper σ-ideal which coincides with the family of meager sets in the locally compact case. For more information see e.g. the survey paper [8].
When investigating a notion of smallness, a fundamental concept is that of cardinal invariants. The four most notable ones are the following. These invariants are called the additivity, covering number, uniformity, and cofinality of I, respectively.
For more information on cardinal invariants see e.g. the monograph [3].
The goal of this paper is to determine these cardinal invariants of HN and HM. The case of HN was asked in [9,Question 5.7]. Before we proceed, let us describe the most important results on this topic so far. First, note that if G is locally compact then the Haar null sets agree with the sets of Haar measure zero and Haar meager sets agree with the meager sets [8]. Moreover, it is also well-known that the four invariants of the measure zero sets do not depend on the underlying measure space as long as it is a Polish space equipped with a continuous σ-finite Borel measure, e.g. a locally compact non-discrete Polish group equipped with the (left) Haar measure. Similarly, the four invariants of the meager sets do not depend on the underlying space as long as it is a Polish space without isolated points, e.g. a non-discrete Polish group [3]. Therefore, if G is locally compact and non-discrete then the four invariants of HN (G) agree with the respective invariants of N (the family of Lebesgue nullsets of R), and the four invariants of HM(G) agree with the respective invariants of M (the family of meager subsets of R).
Recall that a set is universally measurable if it is measurable with respect to the completion of every Borel probability measure.

Definition 1.4.
A subset X of a Polish group G is generalized Haar null if there exists a (completed) Borel probability measure µ and a universally measurable set B containing X such that µ(gBh) = 0 for every g, h ∈ G.
The family of generalized Haar null sets is denoted by HN gen (G) or simply HN gen .
The following results were proved by T. Banakh [1]. For the definition of the so called bounding number b and dominating number d see e.g. [3].
This last statement is really peculiar, since all the usual cardinal invariant are at most the continuum. Now we turn to the results concerning HN . The following theorem will answer [9,Question 5.7], and will show a surprising contrast to the above results of Banakh.
For the sake of completeness we list the values of all four invariants, but note that additivity was already calculated by the first named author and Z. Vidnyánszky [9]. Theorem 1.6.
Remark 1.7. In fact, additivity and cofinality works for all non-locally compact Polish groups admitting a two-sided invariant metric.
Now we turn to the case of HM. Again, we also list additivity here, which was already calculated by M. Doležal an V. Vlasák [7]. Let M denote the family of meager subsets of R.
add(HM(Z ω )) = ω 1 , cov(HM(Z ω )) = cov(M), non(HM(Z ω )) = non(M), cof(HM(Z ω )) = c. Remark 1.9. In fact, additivity and cofinality works for all non-locally compact Polish groups admitting a two-sided invariant metric, and covering number and uniformity works for non-locally compact Polish groups G admitting a continuous surjective homomorphism onto a non-discrete locally compact Polish group. This latter holds e.g. for Z ω and for separable infinite dimensional Banach spaces (indeed, Z ω admits a continuous homomorphism onto (Z/2Z) ω , and Banach spaces admit continuous homomorphisms onto their finite dimensional subspaces).

Proofs
2.1. Covering number and uniformity. The results of this section build heavily on [1], in fact, most results of the section are already present in Banakh's paper in some form. However, one of the key points of the present paper is the sharp contrast between the cardinal invariants of HN and HN gen , so we need to be careful and repeat many familiar argument using this more restrictive definition of Haar nullness. The next lemma is also known.
is lower semi-continuous (and closed-valued and nonempty-valued). Therefore by the zero-dimensional Michael Selection Theorem [11,Theorem 2] there is a continuous selection, which in this case means a continuous function s : 2 ω → G such that ϕ • s = f . We claim that then s is a witness function for ϕ −1 (A). Indeed, an easy calculation shows that gϕ −1 (A)g ′ = ϕ −1 (ϕ(g)Aϕ(g ′ )), and hence which is null (resp. meager) in 2 ω since f was a witness function for A.
Also recall that in locally compact groups Haar null sets agree with the sets of Haar measure zero and Haar meager sets agree with the meager sets [8]. Moreover, it is also well-known that cov(N ) and non(N ) do not depend on the underlying space as long as it is a Polish space equipped with a continuous σ-finite Borel measure, e.g. a locally compact non-discrete Polish group equipped with the left Haar measure. Similarly, cov(M) and non(M) do not depend on the underlying space as long as it is a Polish space without isolated points, e.g. a non-discrete Polish group [3]. Proof. Let ϕ : G → H be a continuous surjective homomorphism onto a nondiscrete locally compact Polish group H. By the above remarks, H can be covered by cov(N ) many sets of Haar measure zero, which are Haar null in this case, since H is locally compact. Similarly, H can be covered by cov(M) many Haar meager sets. But then the preimages under ϕ of these sets clearly cover G, and these preimages are Haar null (resp. Haar meager) by the previous lemma, finishing the proof of the first two inequalities.
Recall that a set A ⊂ G is called o-bounded if for each sequence {U n } n∈ω of neighborhoods of the identity there is a sequence {F n } n∈ω of finite subsets of G such that A ⊂ ∪ n∈ω F n U n . Proof. This is essentially [1,Lemma 4], just note that when this paper proves oB(G) ⊂ HN gen (G) the constructed set is in fact Borel.

2.2.
Cofinality. The main goal of this section is to prove the following. Theorem 2.7. If G is a non-locally compact Polish group admitting a two-sided invariant metric then cof(HN (G)) = cof(HM(G)) = c.
We will need the following definitions.
Definition 2.8. A set A ⊂ G is compact catcher if for every compact set K ⊂ G there are g, h ∈ G such that gKh ⊂ A. Let us say that A is left compact catcher if for every compact set K ⊂ G there exists g ∈ G such that gK ⊂ A.
It is easy to see that if A is compact catcher then it is neither Haar null nor Haar meager, see e.g. [8]. Clearly, the same holds for left compact catcher sets, since left compact catcher sets are compact catcher.
Recall that for a Polish space G the Effros standard Borel space of G is denoted by F (G). This space consists of the non-empty closed subsets of G, and the Borel structure on it is the σ-algebra generated by the sets of the form {F ∈ F(G) : The main technical tool will be the following. Definition 2.9. We say that the Polish group G is nice if there exists a Borel map ϕ : if P ⊂ 2 ω is non-empty perfect then ∪ x∈P ϕ(x) is left compact catcher.
Theorem 2.7 will immediately follow from the next two results.

Theorem 2.10. Every non-locally compact Polish group admitting a two-sided invariant metric is nice.
Theorem 2.11. If G is a nice Polish group then cof(HN (G)) = cof(HM(G)) = c.
Since the main technical difficulty lies in the proof of Theorem 2.10, we prove Theorem 2.11 first.
which is a coprojection, hence it suffices to check that {(x, y) ∈ 2 ω × G : y / ∈ ϕ(x)} and {(x, y) ∈ 2 ω × G : y ∈ B} are coanalytic. The latter one is clearly Borel, so we just need to check that the complement of the former one, In fact, these sets are already Borel. Indeed, the former one is the graph of a Borel function multiplied by G, hence Borel. As for the latter one, is Borel by the definition of the Effros space.

Proof. (Theorem 2.11)
First we show that it suffices to prove that for every Haar null (resp. Haar meager) Borel set B ⊂ G we have Indeed, first note that if the Continuum Hypothesis holds than we are clearly done, since less than c many, in other words countably many nullsets cannot be cofinal, since they cannot even cover G, so even a suitable singleton will show that the family is not cofinal. Otherwise, assume that κ < c and {B α : α < κ} is a cofinal family of Haar null (resp. Haar meager) sets, then by the definition of Haar null (resp. Haar meager) sets without loss of generality we can assume that these sets are Borel. Then every B α can contain ϕ(x) for at most ω 1 many x ∈ 2 ω , hence there are at most κ · ω 1 < c many x ∈ 2 ω such that ϕ(x) is contained in some of the B α 's, contradicting that the family was cofinal. Now we prove (2.1). Let B be a Haar null (resp. Haar meager) Borel set. Assume to the contrary that |{x ∈ 2 ω : ϕ(x) ⊂ B}| > ω 1 . This set is coanalytic by Lemma 2.12, and it is well-known that a coanalytic set of cardinality greater than ω 1 contains a non-empty perfect set P . (Indeed, this easily follows from the facts that every coanalytic set is the union of ω 1 many Borel sets, and every uncountable Borel set contains a non-empty perfect set.) But then is non-Haar null (resp. non-Haar meager) by the definition of niceness, and B is Haar null (resp. Haar meager), a contradiction.
It remains to prove Theorem 2.10.

Proof. (Theorem 2.10)
First, it is easy to see that a closed Haar null set is Haar meager. (Indeed, just restrict the witness measure to a compact set such that each relatively open non-empty subset of this set is of positive measure.) Therefore instead of checking For the proof we will need a lot of preparation. For a finite sequence s ∈ S n ⊂ S <ω , |s| will denote n, the length of s. The natural numbers are the finite ordinals and we can consider them as von Neumann ordinals, i.e. for every n ∈ ω n = {0, 1, . . . , n − 1}.
The following notion will have great importance. Definition 2.13. A function ψ with dom(ψ) = l≤i j<l n j for some i ∈ ω, n 0 , n 1 , . . . , n i−1 ∈ ω \ {0} is an element of T (and we call ψ a "labeled tree"), if ψ : dom(ψ) → 2 <ω is an injective mapping, such that i.e. ψ maps end-extensions to end-extensions, and incomparable elements to incomparable elements. (Under j<0 n j we mean the set containing exactly the empty sequence ∅, thus ∅ ∈ dom(ψ). Moreover, for the function π defined only on ∅, mapping it to ∅, we have π ∈ T .) Note that T is countable. We partition T as follows.
that is the unique element of T 0 .
Now we define a partial order on T .
Definition 2.17. T is a poset with the following partial order: Combining this with Remark 2.16 we have the following. In particular, if ψ = π ∈ T i , then ψ and π are incomparable w.r.t. the partial order on T . Definition 2.20. We define an embedding of (T , ≤) into (ω <ω , ⊂) as follows. To each ψ ∈ T we assign by induction on |n ψ | a sequence m ψ ∈ ω <ω such that First, fix a compatible two-sided invariant metric d, i.e. for which (Such a metric is also automatically complete.) The following two lemmas state well-known facts using the invariance of d, we leave the proof to the reader.
The next technical step is similar to the one in the proof of the main theorem in [12] used for constructing 2 ω -many pairwise disjoint compact-catcher sets.
such that

Proof.
Let ε 0 = 1, and choose a countable dense subset {q will satisfy conditions (iv), (vi). Also, the set {q 2 ) can be chosen so that condition (iii) holds.
Definition 2.25. For each i ∈ ω and j > 0 let By our previous lemma we can deduce the following.
and fix an enumeration of ω × ω: Let j ∈ ω j > 0 be fixed, and assume that the t (i) n l ,m l 's are determined for l < j.
nj −1 }. Now assume on the contrary that there is no t ∈ B(ε i ) such that This means that for each t ∈ B(ε i ) there is an element q ∈ Q, and l < j such that And because for each t ∈ B(ε i ) there exist q ∈ Q, and l < j such that (2.9) holds, we obtain that Using the finiteness of Q = Q Lemma 2.26 yields the following.
Proof. Since the two products have the first i − 1 coordinates, i.e.
by the invariance of the metric d it is enough to prove that Now, since s i−1 ∈ n i−1 , and s ′ i−1 ∈ n ′ i−1 by our assumptions, using Definition 2.25 q . This yields using Lemma 2.26 and Now we can turn to the construction of ϕ from Proposition 2.10. Fix c ∈ 2 ω .
Note that condition (2.3) from Definition 2.13 implies that (for a fixed ψ ∈ T ) at most one t ∈ j<|n ψ | n ψ j can exist for which ψ(t) ∈ 2 <ω is an initial segment of c.
The following is an easy observation, the proof is left to the reader.
For i = 0 and the unique element We are ready to define ϕ(c).
The following lemmas will ensure that ϕ(c) is closed.
Proof. W.l.o.g. we can assume that k = i − 1. Then by Definition 2.27 Hence by the invariance of d it is enough to show that
It is worth mentioning the following consequence of Lemma 2.36. Corollary 2.39. For a fixed c ∈ 2 ω , i ≤ j ∈ ω implies that ψ∈Ti(c) F ψ c ⊇ ψ∈Tj(c) F ψ c . Lemma 2.40. Let c ∈ 2 ω be given. Then Proof. It is enough to show that for a fixed i ∈ ω the set is closed. Recall the fact that for a system of closed sets {F i : i ∈ I} with a constant d 0 > 0 such that d 0 ≤ d(F i , F j ) whenever i = j ∈ I, the union i∈I F i is closed. Therefore (by the invariance of d) it suffices to show that there is a constant r i > 0 such that for ψ = π ∈ T i (c) . But for fixed ψ = π ∈ T i (c) applying Corollary 2.38 we have , i.e. we proved (2.25) with r i = 7δ i−1 .

Lemma 2.41. Let c ∈ 2 ω . Then F = ϕ(c) is Haar null.
Proof. First we define a measure µ according to which each two-sided translate of ϕ(c) will be null. For a fixed i ∈ ω and s ∈ 2 i define We will prove that C is a Cantor set, i.e. it is homeomorphic to 2 ω by showing that (which together with (2.27) will imply that (Recall that since diam(C s ) ≤ 2δ |s|−1 by the invariance of d, and the δ i 's converge to 0 by (iv) from Lemma 2.24, therefore the mapping 2 ω ∋ c → ∩ n∈ω C c | n is indeed a homeomorphism, see [10, (6.2)].) First, if i > 0 then since q Moreover, the fact that δ i < δ i−1 /2 (by (iv) from Lemma 2.24) clearly implies that (using Lemma 2.23) After taking closures we obtain (2.29). Now as we already have (2.29), for (2.28) it suffices to show that for a fixed 0 < i ∈ ω if s = s ′ ∈ 2 i , and s | i−1 = s ′ | i−1 (i.e. the i − 1-th is the first coordinate on which s and s ′ differ) then But in this case, since

Now, since
which proves (2.31), thus finishes the proof of (2.28). Observe that (by the definition of the C s 's (2.26)) (2.32) implies the following We define µ to be the standard coin-tossing measure supported by C, i.e. if s ∈ 2 i then µ(C s ) = 1 2 |s| (and µ(G \ C) = 0). The following claim together with its corollary will ensure that each translate of the closed set F = ϕ(c) ⊂ G is null w.r.t. µ, that is, µ witnesses that F = ϕ(c) is Haar null.
Before proving the claim first observe that since µ(C s ) = 1 2 |s| (s ∈ 2 <ω ) Claim 2.42 has the following corollary. ). We will prove for this fixed c ∈ 2 ω , h, h ′ ∈ G that for each i ∈ ω by induction on i. Before proving (2.34), we briefly describe the idea: In the i-th step we will use the induction hypothesis, that is c h ′ and we will prove and use the fact that the hF ψ c h ′ 's (ψ ∈ T i (c)) are small enough (compared to d(C t ′ 0 , C t ′ 1 )) so that each hF ψ c h ′ can intersect only one of the C t ′ j 's. Moreover, we will verify that the hF ψ c h ′ 's are far enough from each other so that whenever η ∈ T i (c) is such that . Now turning to the proof of (2.34), for i = 0, let t = ∅ be the only element of 2 0 (hence C 0 = C ∅ ), and π be the only element of T 0 (c) (Remark 2.31), thus (2.34) obviously holds.
Suppose that i > 0 and t ′ ∈ 2 i−1 , π ′ ∈ T i−1 (c) are such that Using that the C i 's are decreasing (2.30) (2.27) and (2.28)) the following: We can summarize (2.37) and (2.38) in (in this case let π ∈ T i (c) be arbitrary, (2.34) obviously holds). This means that from now on we can assume that (this with (2.39) will complete the proof of (2.34)). Corollary 2.38 gives that for ψ ∈ On the other hand, since π ∈ T i (c), diam(hF π c h ′ ) ≤ 2δ i−1 by Definition 2.33 and the invariance of d, and d(C t ′ 0 , C t ′ 1 ) ≥ 3δ i−1 by (2.33), therefore hF ψ c h ′ can only intersect one of C t ′ 0 , C t ′ 1 . Thus hF π c h ′ ∩ C t = ∅ together with (2.41) verifies (2.40).
Next we prove that perfectly many ϕ(c) form a right compact-catcher set. Lemma 2.44. Let P ⊂ 2 ω be non-empty, perfect, and K ⊂ G be compact. Then there exists h ∈ G such that Proof. We will need the following technical statements. Let T ⊂ 2 <ω denote the downward closed tree corresponding to P , i.e. (2.43) (see [10, (2.4)]). We will need the following Lemma.
Proof. For i = 0 let ψ 0 : {∅} → {∅} be the unique element of T 0 . It is straightforward to check that (i) − (vi) hold. Suppose that i > 0 and ψ j (j < i) and m j , n j (j < i − 1) are already constructed satisfying (i) − (vi). By (vi) The definition of the q (where the last equality is due to Lemma 2.23). Now because translations of a fixed open set by a dense set cover we have that (for a fixed s ∈ j<i−1 n j ) Therefore by (2.45) and the compactness of K, there exist l ∈ ω such that (2.46) Kt (This can be done since ran(ψ i−1 ) ⊂ T , and T is a perfect tree, i.e. each of its element has incomparable extensions, see [10, (6.14)].) Define It remains to check (vi). As we have already defined n i−1 to be such that l = n i−1 satisfies (2.46), we can formulate it as (2.49) Using that for s ∈ j<i n j (by Definition 2.27) the multiplication of (2.49) by t ni−1,mi−1 from the right gives But ε l + δ l < δ l−1 /2 by (iv) from Lemma 2.24, from which (by induction for j ≥ i, always replacing δ j by the smaller ε j+1 + δ j+1 ) From this (2.51) follows.
The following two lemmas will ensure that (2.42) holds.
Lemma 2.48. With (n i ) i∈ω , (m i ) i∈ω given by Lemma 2.45 and h as in (2.52), Proof. (Lemma 2.48) Fix i ∈ ω, i > 0. By (vi) we know that (2.53) Kt Now Claim 2.46 states that if k ≥ i, then Observe that Using (2.55), (2.53) and finally (2.54) and (since by Lemmas 2.23, 2.22 we are done. Proof. (Lemma 2.49) Recall that T ⊂ 2 <ω is defined so that the perfect set P ⊂ 2 ω is the body of T ((2.43)), and the sequences (n i ) i∈ω , (m i ) i∈ω ∈ ω ω , (ψ) i∈ω fulfill the criteria (i) − (vi). Fix an element x ∈ i>0 s∈ j<i nj cl(B(δ i−1 ))g n | i ,m | i ,s , and define B ⊂ i>0 j<i n j such that First observe that using Lemma 2.35 B is downward closed. B is an infinite tree by the choice of x, and it has finitely many nodes on each level, thus by Kőnig's Lemma there is an infinite branch through it. Let r ′ ∈ i<ω n i be the union of that branch, i.e. the corresponding infinite sequence, for which Define (2.57) c ′ = ∪{ψ i (r ′ | i ) : i ∈ ω}.
We need that for each Borel set B ⊂ F(G) (w.r.t. the Effros Borel structure) its preimage, ϕ −1 (B) ∈ B(2 ω ). For this it suffices to show that for each fixed open set U ⊂ G the preimage of {F ∈ F(G) : F ∩ U = ∅} under ϕ is a Borel subset of 2 ω , since those sets form a generator system of the Effros Borel structure on F (G). This is provided by the following Lemma. Now by Corollary 2.38 for a fixed i the F ψ c 's (ψ ∈ T i (c)) are disjoint, hence for each i > 0 there is a unique ψ ∈ T i (c) for which x ∈ F ψ c . As diam(F ψ c ) ≤ 2δ i−1 for ψ ∈ T i (c) and δ i tends to 0 as i tends to ∞ ((iv) from Lemma 2.24), we can fix an index i ′ (and the unique π ∈ T i ′ (c)) so that (2.58) x ∈ F π c ⊂ U. Observe that if l witnesses that π ∈ T i ′ (c) according to Definition 2.29, i.e. c | l = π(b c,π ) (Definition 2.30), then also for each y ∈ 2 ω with y | l = c | l we have π ∈ T ′ i (y), and b y,π = b c,π . Define i 0 = l. Fixing such a y it follows from this that (2.59) F π c = F π y (Definition 2.33). Now we show that for each y ∈ 2 ω the tree T (y) ⊂ T is pruned, i.e. for each i ∈ ω, ψ ∈ T i (y) there exists ψ ′ ∈ T i+1 (y) with ψ ≤ ψ ′ (w.r.t. Definition 2.17). First we show that this completes the proof of Claim 2.51. Indeed, an infinite branch π (0) = π ≤ π (1) ≤ π (2) ≤ . . . ≤ π (k) ≤ . . . (π (j) ∈ T i ′ +j (y)) in T (y) would yield an infinite decreasing chain of closed sets F π (0) y = F π y ⊇ F π (1) y ⊇ . . . ⊇ F π (k) y ⊇ . . .