On metric spaces with given transfinite asymptotic dimensions
aa r X i v : . [ m a t h . GN ] A ug On metric spaces with given transfinite asymptoticdimensions
Yan Wu ∗ Jingming Zhu ∗∗ Taras Radul ∗∗∗ ∗
Abstract.
For every countable ordinal number ξ , we construct a metric space X ξ whose trans-finite asymptotic dimension and complementary-finite asymptotic dimension are both ξ . We alsoprove that the metric space X ξ has finite decomposition complexity. Keywords
Asymptotic dimension, Transfinite asymptotic dimension, Complementary-finiteasymptotic dimension, Finite decomposition complexity; In coarse geometry, asymptotic dimension of a metric space is an important concept which was defined byGromov for studying asymptotic invariants of discrete groups [1]. This dimension can be considered as anasymptotic analogue of the Lebesgue covering dimension. As a large scale analogue of W.E. Havers property C in dimension theory, A. Dranishnikov introduced the notion of asymptotic property C in [2]. It is wellknown that every metric space with finite asymptotic dimension has asymptotic property C . But the inverseis not true, which means that there exists some metric space X with infinite asymptotic dimension andasymptotic property C . Therefore how to classify the metric spaces with infinite asymptotic dimension intosmaller categories becomes an interesting problem.In [3], T. Radul defined the transfinite asymptotic dimension (trasdim) which can be viewed as a trans-finite extension of the asymptotic dimension and proved that for a metric space X , trasdim( X ) < ∞ if andonly if X has asymptotic property C. He also gave examples of metric spaces with trasdim= ∞ and withtrasdim= ω , where ω is the smallest infinite ordinal number (see [3]). But whether there is a metric space X with ω < trasdim( X ) < ∞ (stated as“omega conjecture”in [4] by M. Satkiewicz) was unknown until recently.A metric space X with trasdim X = ω + 1 was constructed in [5]. Some examples of metric spaces withhigher trasdim were constructed in [6]. Let us remark that trasdim takes only countable values [3]. In thispaper, for every countable ordinal number ξ , we construct a metric space X ξ with trasdim( X ξ ) = ξ andcoasdim( X ξ ) = ξ , which generalized the results in [5] and [6].As another generalization of asymptotic dimension, E.Guentner, R.Tessera and G.Yu introduced thenotion of finite decomposition complexity to study topological rigidity of manifolds in [7]. And they provedthat every metric space with finite asymptotic dimension has finite decomposition complexity in [8]. Therelation between asymptotic property C and finite decomposition complexity was studied by A.Dranishnikovand M.Zarichnyi in [9]. But till now, there is no example of metric space known which make a differencebetween asymptotic property C and finite decomposition complexity. The metric space X ξ constructed ∗∗ College of Mathematics Physics and Information Engineering, Jiaxing University, Jiaxing , 314001, P.R.China.E-mail: [email protected] ∗ ∗ College of Mathematics Physics and Information Engineering, Jiaxing University, Jiaxing , 314001, P.R.China.E-mail: [email protected] ∗∗∗
Institute of Mathematics, Casimirus the Great University of Bydgoszcz, Poland;Department of Mechanics and Mathematics, Ivan Franko National University of Lviv, Universytetska st., 1.79000 Lviv, Ukraine.E-mail: [email protected] This research was supported by the National Natural Science Foundation of China under Grant(No.11871342,11801219,11301224,11326104) X ξ has both asymptotic property C and finitedecomposition complexity.The paper is organized as follows: In Section 2, we recall some definitions and properties of transfiniteasymptotic dimension and complementary-finite asymptotic dimension. In Section 3, we introduce a concretemetric space X ξ with trasdim( X ξ ) = ξ and coasdim( X ξ ) = ξ for every countable ordinal number ξ . Finally,we prove the metric space X ξ has finite decomposition complexity. Our terminology concerning the asymptotic dimension follows from [10] and for undefined terminologywe refer to [3] and [6].Let (
X, d ) be a metric space and
U, V ⊆ X , letdiam U = sup { d ( x, y ) | x, y ∈ U } and d ( U, V ) = inf { d ( x, y ) | x ∈ U, y ∈ V } . Let
R > U be a family of subsets of X . U is said to be R -bounded ifdiam U . = sup { diam U | U ∈ U} ≤ R. In this case, U is said to be uniformly bounded . Let r >
0, a family U is said to be r -disjoint if d ( U, V ) ≥ r for every U, V ∈ U with U = V. In this paper, we denote S { U | U ∈ U} by S U , denote { U | U ∈ U or U ∈ U } by U ∪ U . Let A bea subset of X and ǫ >
0. We denote { x ∈ X | d ( x, A ) < ǫ } by N ǫ ( A ) and denote { x ∈ X | d ( x, A ) ≤ ǫ } by N ǫ ( A ). We denote { N ǫ ( U ) | U ∈ U} by N ǫ ( U ) and denote { N ǫ ( U ) | U ∈ U} by N ǫ ( U ). Definition 2.1. ([1]) A metric space X is said to have finite asymptotic dimension if there exists n ∈ N ,such that for every r >
0, there exists a sequence of uniformly bounded families {U i } ni =0 of subsets of X suchthat the family S ni =0 U i covers X and each U i is r -disjoint for i = 0 , , · · · , n . In this case, we say that the asymptotic dimension of X less than or equal to n , which is denoted by asdim( X ) ≤ n .We say that asdim( X ) = n if asdim( X ) ≤ n and asdim( X ) ≤ n − X to transfinite asymptotic dimensionwhich is denoted by trasdim( X ) (see [3]). We will need the set-theoretical function Ord from [11] whichclassifies families of finite non-empty subsets of a set T . We will consider only the particular case when T = N . Definition 2.2. ([11]) Let
F in N denote the collection of all finite, nonempty subsets of N and let M ⊆ F in N .For σ ∈ { ∅ } ∪ F in N , let M σ = { τ ∈ F in N | τ ∪ σ ∈ M and τ ∩ σ = ∅ } . Let M a abbreviate M { a } for a ∈ N . Define the ordinal number Ord M inductively as follows:Ord M = 0 ⇔ M = ∅ , Ord M ≤ α ⇔ ∀ a ∈ N , Ord M a < α, Ord M = α ⇔ Ord M ≤ α and Ord M < α is not true , Ord M = ∞ ⇔ Ord M ≤ α is not true for every ordinal number α. We call a family M ⊆ F in N inclusive if and only if for each σ, τ ∈ F in N such that τ ⊆ σ and σ ∈ M , wehave τ ∈ M . We consider only inclusive families in the following.The following lemmas are particular cases of the corresponding lemmas from [11]. Lemma 2.1. ([11]) Let M ⊆ F in N and n ∈ N . ThenOrd M ≤ n if and only if | σ | ≤ n for every σ ∈ M . Lemma 2.2. ([11]) If
M, N ⊆ F in N and M ⊆ N , then Ord M ≤ Ord N. emma 2.3. ([11]) If M, N ⊆ F in N , then Ord( M ∪ N ) ≤ max { Ord M, Ord N } . Lemma 2.4. ([11]) Let L and L ′ be sets. Let F inL and
F inL ′ denote collections of all finite, nonemptysubsets of L and L ′ , respectively. Let M ⊆ F inL , M ′ ⊆ F inL ′ and φ : L → L ′ be a function such that forevery σ ∈ M , we have φ ( σ ) ∈ M ′ and | φ ( σ ) | = | σ | . Then Ord M ≤ Ord M ′ .Let M ⊆ F in N and K is a infinite subset of N . Then there is a standard bijection ϕ from N to K which keeps the order. We define M [ K ] = n { ϕ ( k ) , ϕ ( k ) , ..., ϕ ( k m ) } (cid:12)(cid:12)(cid:12) { k , k , ..., k m } ∈ M o . Note that M = M [ N ]. By Lemma 2.4, we obtain the following result. Corollary 2.1.
Let M ⊆ F in N and K is a infinite subset of N , then Ord M =Ord M [ K ]. Definition 2.3. ([3]) Given a metric space X , define the following collection: A ( X, d ) = { σ ∈ F in N | there are no uniformly bounded families U i for i ∈ σ such that each U i is i -disjoint and [ i ∈ σ U i covers X } and A ( X, d ) = { σ ∈ F in N | there are no uniformly bounded families V i for i ∈ σ such that each V i is 2 i -disjoint and [ i ∈ σ V i covers X } . The transfinite asymptotic dimension of X is defined as trasdim X =Ord A ( X, d ) .
Lemma 2.5.
Ord A ( X, d ) =Ord A ( X, d ). Proof.
Since A ( X, d ) ⊆ A ( X, d ), Ord A ( X, d ) ≤ Ord A ( X, d ) by Lemma 2.2. The inverse inequality followsfrom Lemma 2.4 by considering the exponential function φ : N → N defined by φ ( n ) = 2 n . Definition 2.4. ([12]) Every ordinal number γ can be represented as γ = λ ( γ ) + n ( γ ), where λ ( γ ) is thelimit ordinal or 0 and n ( γ ) ∈ N ∪ { } . Let X be a metric space, we define complementary-finite asymptoticdimension of X (coasdim( X )) inductively as follows: • coasdim( X ) = − ⇔ X = ∅ , • coasdim( X ) ≤ λ ( γ ) + n ( γ ) ⇔ for every r >
0, there exist r -disjoint uniformly bounded families U , ..., U n ( γ ) of subsets of X such that coasdim( X \ S S n ( γ ) i =0 U i ) < λ ( γ ), • coasdim( X ) = γ ⇔ coasdim( X ) ≤ γ and coasdim( X ) ≤ β is not true for any β < γ , • coasdim( X ) = ∞ ⇔ coasdim( X ) ≤ γ is not true for any ordinal number γ . X is said to have complementary-finite asymptotic dimension if coasdim( X ) ≤ γ for some ordinal number γ . Remark 2.1.
It is easy to see that for every n ∈ N ∪
0, coasdim( X ) ≤ n if and only if asdim( X ) ≤ n . Lemma 2.6. ([12]) Let X be a metric space with X , X ⊆ X . Thencoasdim( X ∪ X ) ≤ max { coasdim( X ) , coasdim( X ) } .Let X and Y be metric families. A map of families from X to Y is a collection of functions F = { f } , eachmapping some X ∈ X to some Y ∈ Y and such that every X ∈ X is the domain of at least one f ∈ F . Weuse the notation F : X → Y and, when confusion could occur, write f : X f → Y f to refer to an individualfunction in F . Definition 2.5.
A map of metric space families F : X → Y is uniformly expansive if there exists anon-decreasing function θ : [0 , ∞ ) → [0 , ∞ ) such that for every f ∈ F and every x, y ∈ X f d ( f ( x ) , f ( y )) ≤ θ ( d ( x, y ));3t is effectively proper if there exists a proper non-decreasing function δ : [0 , ∞ ) → [0 , ∞ ) such that for every f ∈ F and every x, y ∈ X f d ( f ( x ) , f ( y )) ≥ δ ( d ( x, y ));A map of metric space families F : X → Y is a coarse embedding if it is both uniformly expansive andeffectively proper.
Definition 2.6. ([7],[8]) A metric family X is r -decomposable over a metric family Y if every X ∈ X admitsa decomposition X = X ∪ X , X i = G r -disjoint X ij , where each X ij ∈ Y . It is denoted by X r → Y . Definition 2.7. ([7],[8])(1) Let D be the collection of uniformly bounded families: D = {X : X is uniformly bounded } .(2) Let α be an ordinal number greater than 0, let D α be the collection of metric families decomposableover [ β<α D β : D α = {X : ∀ r > , ∃ β < α, ∃ Y ∈ D β , such that X r → Y} . Remark 2.2. • We view a metric space X as a singleton family { X } . • D β ⊆ D α for every β < α . • It is known that X has finite asymptotic dimension if and only if X belongs to D n for some n ∈ N . Moreover, asdim( X ) ≤ n implies X ∈ D n +1 ([7],[8]). Lemma 2.7. ([8]) Let α and β be ordinal numbers and α < β . Let X and Y be metric spaces such that X ∈ D α and Y ∈ D β . Then X ∪ Y ∈ D β +1 . Definition 2.8. ([7],[8]) A metric family X has finite decomposition complexity if there exists a countableordinal number α such that X ∈ D α . Lemma 2.8. (Coarse invariance, [8]) Let X and Y be two metric families such that there is a coarseembedding φ : X → Y . If
Y ∈ D α for some countable ordinal number α , then X ∈ D α . For each τ = { k , k , ..., k s } ∈ F in N , we choose an indexation such that k < k < · · · < k s . For every n ∈ N ∪ { } , let K ( τ, n ) = { k n +1 , k n +2 ..., k s } if n < s and K ( τ, n ) = ∅ if n ≥ s ,and let i ( τ, n ) = max { k , k ..., k n } = k n if n ≤ s and i ( τ, n ) = k s if n > s .For each limit ordinal α , we fix an increasing sequence { ζ i ( α ) + i } of ordinals such thateach ζ i ( α ) is a limit ordinal or 0 and α = sup i ( ζ i ( α ) + i ).When α = sup i ( β + i ) for some limit ordinal β , we put ζ i ( α ) = β for each i ∈ N .For each countable ordinal number ξ , we write ξ = γ ( ξ ) + n ( ξ ), where γ ( ξ ) is a limit ordinal or 0 and n ( ξ ) ∈ N ∪ { } . We build a family S ξ ⊆ F in N by induction. Definition 3.9.
Let n ∈ N and let ξ be a countable ordinal number , we define4 S n = { σ ∈ F in N | | σ | ≤ n } . • S ξ = S γ ( ξ )+ n ( ξ ) = n σ ∈ F in N (cid:12)(cid:12)(cid:12) K ( σ, n ( ξ )) ∈ S ζ l ( γ ( ξ ))+ l ∪ ∅ for some l ∈ { , , · · · , i ( σ, n ( ξ )) } o . Remark 3.3. • It follows from the definition of the family S ξ that σ ∈ S ξ for each σ ∈ F in N with | σ | ≤ n ( ξ ) + 1. • Note that if σ = { k , ..., k s } ∈ S ξ , then { l, k , ..., k s } ∈ S ξ +1 for every l < k . Lemma 3.1.
Let ξ = γ ( ξ ) + n ( ξ ) be an infinite ordinal number and τ = { k , ..., k s } ∈ S ξ . Then we have τ ∈ S ζ l ( γ ( ξ ))+ l + n ( ξ )+1 for some l ∈ { , , · · · , i ( τ, n ( ξ )) } . Proof.
Since τ = { k , ..., k s } ∈ S ξ , we have K ( τ, n ( ξ )) ∈ S ζ l ( γ ( ξ ))+ l ∪ ∅ for some l ∈ { , , · · · , i ( τ, n ( ξ )) } . • If K ( τ, n ( ξ )) ∈ S ζ l ( γ ( ξ ))+ l , then τ = { k , ..., k n ( ξ ) , k n ( ξ )+1 , ..., k s } ∈ S ζ l ( γ ( ξ ))+ l + n ( ξ )+1 by Remark 3.1. • If K ( τ, n ( ξ )) = ∅ , then K ( τ, l + n ( ξ ) + 1) ⊆ K ( τ, n ( ξ )) implies K ( τ, l + n ( ξ ) + 1) = ∅ . So τ ∈ S ζ l ( γ ( ξ ))+ l + n ( ξ )+1 . Lemma 3.2.
The family S ξ is inclusive for each countable ordinal ξ . Proof. • It follows from the definition of S ξ that the result is true when ξ is finite. • Assume that the result holds for every infinite ordinal number ξ < α . Now for ξ = α , let τ = { k j , k j , ..., k j t } ⊆ σ = { k , ..., k s } ∈ S ξ . Note that t ≤ s and K ( τ, n ( ξ )) ⊆ K ( σ, n ( ξ )). If t ≤ n ( ξ ),then we have τ ∈ S ξ by definition. Consider the case t > n ( ξ ). Then we have s > n ( ξ ) and K ( σ, n ( ξ )) ∈ S ζ l ( γ ( ξ ))+ l for some 1 ≤ l ≤ i ( σ, n ( ξ )) = k n ( ξ ) ≤ k j n ( ξ ) = i ( τ, n ( ξ )) . By inductive assumption, we have K ( τ, n ( ξ )) ∈ S ζ l ( γ ( ξ ))+ l . So τ ∈ S γ ( ξ )+ n ( ξ ) . Lemma 3.3. S γ + n ⊆ S γ + m , where γ is a limit ordinal and m ∈ N , n ∈ N ∪ { } such that n < m . Proof.
For every σ = { k , ..., k s } ∈ S γ + n . • If s ≤ m , then we have σ ∈ S γ + m by definition. • If s > m , then s > n . It follows that K ( σ, n ) ∈ S ζ l ( γ )+ l for some 1 ≤ l ≤ i ( σ, n ) = k n < k m = i ( σ, m ) . Since n < m , we have K ( σ, m ) ⊆ K ( σ, n ) . Since S ζ l ( γ )+ l is inclusive by Lemma 3.2, we have K ( σ, m ) ∈ S ζ l ( γ )+ l for some l ∈ { , , · · · , i ( σ, m ) } . Hence σ ∈ S γ + m . .2 Transfinite asymptotic dimension of X ξ Lemma 3.4.
For each countable ordinals ξ , Ord S ξ ≤ ξ . Proof.
By Lemma 2.1, the result is true when ξ is finite. Assume that the result holds for every ξ < α . Now ξ = α . For every a ∈ N and σ ∈ S aξ , { a } ⊔ σ ∈ S ξ . It follows that K ( { a } ⊔ σ, n ( ξ )) ∈ S ζ l ( γ ( ξ ))+ l ∪ ∅ for some l ∈ { , , · · · , i ( { a } ⊔ σ, n ( ξ )) } . • When n ( ξ ) ∈ N . – If i ( { a } ⊔ σ, n ( ξ )) > a , then K ( σ, n ( ξ ) −
1) = K ( { a } ⊔ σ, n ( ξ )) ∈ S ζ l ( γ ( ξ ))+ l ∪ ∅ . Note that1 ≤ l ≤ i ( { a } ⊔ σ, n ( ξ )) = i ( σ, n ( ξ ) −
1) which implies that σ ∈ S γ ( ξ )+ n ( ξ ) − . – If i ( { a } ⊔ σ, n ( ξ )) ≤ a , then l ≤ a and K ( σ, n ( ξ )) ⊆ K ( { a } ⊔ σ, n ( ξ )) . Since S ζ l ( γ ( ξ ))+ l is inclusive, K ( σ, n ( ξ )) ∈ S ζ l ( γ ( ξ ))+ l ∪ ∅ . So by Remark 3.1, σ ∈ S ζ l ( γ ( ξ ))+ l + n ( ξ )+1 for some l ∈ { , , · · · , a } .Therefore, S aξ ⊆ S γ ( ξ )+ n ( ξ ) − [ ( a [ l =1 S ζ l ( γ ( ξ ))+ n ( ξ )+ l +1 ) . Then by Lemma 2.3 and inductive assumption,Ord( S aξ ) ≤ γ ( ξ ) + n ( ξ ) − < ξ. • When n ( ξ ) = 0. – If i ( { a } ⊔ σ,
0) = a , then σ = K ( { a } ⊔ σ, ∈ S ζ l ( γ ( ξ ))+ l ∪ ∅ . Since σ = ∅ , σ ∈ S ζ l ( γ ( ξ ))+ l for some l ∈ { , , · · · , a } . – If i ( { a } ⊔ σ, < a , then K ( σ, ⊆ K ( { a } ⊔ σ, ∈ S ζ l ( γ ( ξ ))+ l ∪ ∅ for some l ∈ { , , · · · , i ( { a } ⊔ σ, } ,which implies σ ∈ S ζ l ( γ ( ξ ))+ l +1 for some l < a .So by Lemma 3.3, S aξ ⊆ a [ l =1 S ζ l ( γ ( ξ ))+ l +1 . Then by Lemma 2.3 and inductive assumption,Ord( S aξ ) ≤ ζ a ( γ ( ξ )) + a + 1 < ξ. So in both cases, we have Ord S ξ ≤ ξ .Let σ = { k , ..., k s } ∈ F in N and τ = { p , ..., p s } ∈ F in N . We denote σ ≤ τ if for each i ∈ { , , · · · , s } , k i ≤ p i . Lemma 3.5.
Let ξ be an countable ordinal number and σ ∈ S ξ . Then for each τ ∈ Fin N such that σ ≤ τ ,we have τ ∈ S ξ . Proof.
It is easy to see the result is true when ξ is finite. Assume that the result holds for every countableordinal number ξ < α . For ξ = α . Let σ = { k , ..., k s } ∈ S ξ and τ = { p , ..., p s } such that k i ≤ p i for i ∈ { , , · · · , s } . • If s ≤ n ( ξ ), then τ ∈ S ξ by Remark 3.1. 6 If s > n ( ξ ), then K ( σ, n ( ξ )) ∈ S ζ l ( ξ )+ l for some 1 ≤ l ≤ i ( σ, n ( ξ )) = k n ( ξ ) ≤ p n ( ξ ) = i ( τ, n ( ξ )). Notethat K ( σ, n ( ξ )) ≤ K ( τ, n ( ξ )). By inductive assumption, K ( τ, n ( ξ )) ∈ S ζ l ( ξ )+ l for some l ∈ { , , · · · , i ( τ, n ( ξ )) } .So τ ∈ S ξ . Proposition 3.1.
Ord S ξ = ξ for each countable ordinal number ξ . Proof.
By Lemma 2.1, it is easy to obtain that the result is true when ξ is finite. Assume that the result holdsfor every countable ordinal number ξ < α . For ξ = α . By Lemma 3.4, it suffices to show that Ord S ξ ≥ ξ . • n ( ξ ) = 0. It means that ξ is a limit ordinal. For every k ∈ N , let S ζ k ( ξ )+ k ( k ) = { τ ∈ S ζ k ( ξ )+ k | t > k for each t ∈ τ } .Then for each τ ∈ S ζ k ( ξ )+ k ( k ), { k } ⊔ τ ∈ S ξ by definition. Since S ξ is inclusive by Lemma 3.2, τ ∈ S ξ .So S ζ k ( ξ )+ k ( k ) ⊆ S ξ .Note that Ord S ζ k ( ξ )+ k = Ord S ζ k ( ξ )+ k ( k ) . Indeed, by Lemma 2.2, S ζ k ( ξ )+ k ( k ) ⊆ S ζ k ( ξ )+ k implies Ord S ζ k ( ξ )+ k ( k ) ≤ Ord S ζ k ( ξ )+ k . There is afunction φ : N → N defined by φ ( i ) = i + k . Then by Lemma 3.5, φ ( σ ) ∈ S ζ k ( ξ )+ k ( k ) whenever σ ∈ S ζ k ( ξ )+ k and | φ ( σ ) | = | σ | . By Lemma 2.4, Ord S ζ k ( ξ )+ k ≤ Ord S ζ k ( ξ )+ k ( k ) . So Ord S ζ k ( ξ )+ k = Ord S ζ k ( ξ )+ k ( k ) . By Lemma 2.2 and inductive assumption,Ord S ξ ≥ Ord S ζ k ( ξ )+ k ( k ) = Ord S ζ k ( ξ )+ k = ζ k ( ξ ) + k for each k ∈ N . Hence Ord S ξ ≥ ξ . • n ( ξ ) ∈ N . For every k ∈ N , let S ξ − ( k ) = { τ ∈ S ξ − | t > k for each t ∈ τ } .Then for each τ ∈ S ξ − ( k ), by Remark 3.1, { k } ⊔ τ ∈ S ξ which implies τ ∈ S kξ . Then S ξ − ( k ) ⊆ S kξ .By the similar argument and inductive assumption,Ord S kξ ≥ Ord S ξ − ( k ) = Ord S ξ − = ξ − . Hence Ord S ξ ≥ ξ . Definition 3.10. ([13]) Let X be a metric space and let A, B be a pair of disjoint subsets of X . We saythat a subset L ⊆ X is a partition of X between A and B if there exist open sets U, W ⊆ X satisfying thefollowing conditions A ⊆ U, B ⊆ W and X = U ⊔ L ⊔ W. Definition 3.11. ([6]) Let X be a metric space and let A, B be a pair of disjoint subsets of X . For every ǫ >
0, we say that a subset L ⊆ X is an ǫ -partition of X between A and B if there exist open sets U, W ⊆ X satisfying the following conditions A ⊆ U, B ⊆ W, X = U ⊔ L ⊔ W, d ( L, A ) > ǫ and d ( L, B ) > ǫ Clearly, an ǫ -partition L of X between A and B is a partition of X between A and B .7 emma 3.6. ([13], Lemma 1.8.14) Let F + i , F − i , where i = 1 , , . . . , n , be the pairs of opposite faces of I n . = [0 , n . If I n = L ⊃ L ⊃ . . . ⊃ L n is a decreasing sequence of closed sets such that L i +1 is a partitionof L i between L i ∩ F + i +1 and L i ∩ F − i +1 for i ∈ { , , , . . . , n − } , then L n = ∅ . Lemma 3.7.
Let L . = [0 , B ] n for some B > F + i , F − i , where i = 1 , , · · · , n , be the pairs of oppositefaces of L and let 0 < ǫ < B . For k = 1 , , · · · , n , let U k be an ǫ -disjoint and B -bounded family ofsubsets of [0 , B ] n . Then there exists an ǫ -partition L k +1 of L k between F + k +1 ∩ L k and F − k +1 ∩ L k such that L k +1 ⊆ L k ∩ ( S U k +1 ) c for k = 0 , , , · · · , n −
1. Moreover, L n = ∅ . Proof.
For k = 1 , , · · · , n , let A k . = { U ∈ U k | d ( U, F + k ) ≤ ǫ } and B k . = { U ∈ U k | d ( U, F + k ) > ǫ } . Notethat A k ∪ B k = U k . Let A k = [ { N ǫ ( U ) | U ∈ A k } and B k = [ { N ǫ ( U ) | U ∈ B k } . Then d ( A k , F − k ) > B − B − ǫ − ǫ > ǫ and d ( B k , F + k ) > ǫ − ǫ > ǫ . It follows that( A k ∪ N ǫ ( F + k )) ∩ ( B k ∪ N ǫ ( F − k )) = ∅ . Let L k +1 . = L k \ (( A k +1 ∪ N ǫ ( F + k +1 )) ∪ ( B k +1 ∪ N ǫ ( F − k +1 ))) for k = 0 , , , · · · , n − F + k +1 ∩ L k = ∅ and F − k +1 ∩ L k = ∅ for k = 0 , , · · · , n − . Indeed, • Clearly, F +1 ∩ L = ∅ and F − ∩ L = ∅ . • For k = 1 , , · · · , n − L k = L k − \ (( A k ∪ N ǫ ( F + k )) ∪ ( B k ∪ N ǫ ( F − k )))= L k − \ (( A k − ∪ N ǫ ( F + k − )) ∪ ( B k − ∪ N ǫ ( F − k − )) ∪ ( A k ∪ N ǫ ( F + k )) ∪ ( B k ∪ N ǫ ( F − k )))= · · · = L \ (( A ∪ N ǫ ( F +1 )) ∪ ( B ∪ N ǫ ( F − )) ∪ · · · ∪ ( A k ∪ N ǫ ( F + k )) ∪ ( B k ∪ N ǫ ( F − k ))) . Note that F + k +1 \ ( N ǫ ( F +1 )) ∪ N ǫ ( F − ) ∪ · · · ∪ N ǫ ( F + k ) ∪ N ǫ ( F − k )) is a nonempty connected set withdiameter greater than B and ( A ∪ B ∪ · · · A k ∪ B k ) is a union set of a ǫ -disjoint and B -boundedfamily. Then F + k +1 \ (( A ∪ N ǫ ( F +1 )) ∪ ( B ∪ N ǫ ( F − )) ∪ · · · ∪ ( A k ∪ N ǫ ( F + k )) ∪ ( B k ∪ N ǫ ( F − k ))) = ∅ . It follows that F + k +1 ∩ L k = ∅ . Similarly, F − k +1 ∩ L k = ∅ .Therefore, L k +1 is an ǫ -partition of L k between F + k +1 ∩ L k and F − k +1 ∩ L k such that L k +1 ⊂ L k ∩ ( S U k +1 ) c .By Lemma 3.6, L n = ∅ .Let L = { n + 2 | n ∈ N } and S ξ [ L ] = n { k + 2 , k + 2 , · · · , k s + 2 } (cid:12)(cid:12)(cid:12) { k , k , · · · , k s } ∈ S ξ o . Then byCorollary 2.1, Ord S ξ [ L ]=Ord S ξ = ξ for each countable ordinal number ξ . Definition 3.12.
For τ = { k + 2 , k + 2 , ..., k m + 2 } ∈ S ξ [ L ], we define X τ = n ( x i ) mi =0 ∈ (2 k Z ) m +1 (cid:12)(cid:12)(cid:12) |{ j | x j / ∈ k p Z }| ≤ p for every p ∈ { , , · · · , m } o .We consider X τ with sup-metric. Proposition 3.2. τ ∈ A ( X τ , d ) for every τ = { k + 2 , k + 2 , ..., k m + 2 } ∈ S ξ [ L ].8 roof. Suppose that τ = { k + 2 , k + 2 , ..., k m + 2 } / ∈ A ( X τ , d ), then there are B -bounded families U , U , ..., U m such that U j is 2 k j +2 -disjoint for every j ∈ { , , · · · , m } and S mi =0 U i covers [0 , B ] m +1 ∩ X τ for some B > k m +1 .Assume that p = B km ∈ N . Take a bijection ψ : { , , · · · , p m +11 } → { , , , · · · , p − } m +1 . Let Q ( t ) = m +1 Y j =1 [2 k m ψ ( t ) j , k m ( ψ ( t ) j + 1)] , where ψ ( t ) j is the j th coordinate of ψ ( t ) . Let Q = { Q ( t ) | t ∈ { , , · · · , p m +11 }} , then [0 , B ] m +1 = S Q ∈Q Q . Note that[0 , B ] m +1 ∩ X τ ⊆ [ Q ∈Q ∂ m Q , where ∂ m Q is the m -dimensional skeleton of Q .Let L = [0 , B ] m +1 . Since N km ( U m ) is 2 k m +1 -disjoint and (2 k m +1 + B )-bounded, by Lemma 3.7, thereexists a 2 k m +1 -partition L of [0 , B ] m +1 between F +1 and F − such that L ⊆ ( [ N km ( U m )) c ∩ [0 , B ] m +1 and d ( L , F + / − ) > k m +1 ,where F + / − is a pair of opposite facets of [0 , B ] m +1 . Since L is a partition of [0 , B ] m +1 between F +1 and F − , [0 , B ] m +1 = L ⊔ A ⊔ B such that A , B are open in [0 , B ] m +1 and A , B contain two oppositefacets F − , F +1 respectively.Let M = { Q ∈ Q | Q ∩ L = ∅} and M = S M . Since L is a 2 k m +1 -partition of [0 , B ] m +1 between F +1 and F − , then M is a partition of [0 , B ] m +1 between F +1 and F − , i.e., [0 , B ] m +1 = M ⊔ A ′ ⊔ B ′ such that A ′ , B ′ are open in [0 , B ] m +1 and A ′ , B ′ contain two opposite facets F − , F +1 respectively. Let L ′ = ∂ m M = S { ∂ m Q | Q ∈ M } , then [0 , B ] m +1 \ ( L ′ ⊔ A ′ ⊔ B ′ ) is the union of some disjoint open m + 1-dimensional cubes with length of edge = 2 k m . So L ′ is a partition of [0 , B ] m +1 between F +1 and F − and L ′ ⊆ ( S U m ) c ∩ [0 , B ] m +1 .Since N km − ( U m − ) is 2 k m − +1 -disjoint and (2 k m − +1 + B )-bounded, there exists a 2 k m − +1 -partition L of L ′ between F +2 ∩ L ′ and F − ∩ L ′ such that L ⊆ ( [ N km − ( U m − )) c ∩ L ′ and d ( L , F + / − ) > k m − +1 .Since L is a partition of L ′ between F +2 ∩ L ′ and F − ∩ L ′ , L ′ = L ⊔ A ⊔ B such that A , B are openin L ′ and A , B contain two opposite facets F − ∩ L ′ , F +2 ∩ L ′ respectively.Assume that p = B km − ∈ N . Take a bijection ψ : { , , · · · , p m +12 } → { , , , · · · , p − } m +1 . Let Q ′ ( t, l ) = l Y j =1 (2 k m − ψ ( t ) j , k m − ( ψ ( t ) j + 1)) × k m − ψ ( t ) l × m +1 Y j = l +1 (2 k m − ψ ( t ) j , k m − ( ψ ( t ) j + 1)) , where ψ ( t ) j is the j -th coordinate of ψ ( t ). Let Q = { Q ′ ( t, l ) ∩ L ′ | t ∈ { , , ..., p m +12 } , l ∈ { , ..., m + 1 }} be a family of m -dimensional cubes with length of edges=2 k m − , then L ′ = S Q ∈Q Q . Let M = { Q ′ ∈ Q | Q ′ ∩ L = ∅} and M = [ M .Since L is a 2 k m − +1 -partition of L ′ between F +2 ∩ L ′ and F − ∩ L ′ , then M is a partition of L ′ between F +2 ∩ L ′ and F − ∩ L ′ , i.e., L ′ = M ⊔ A ′ ⊔ B ′ such that A ′ , B ′ are open in L ′ and A ′ , B ′ containtwo opposite facets F − ∩ L ′ , F +2 ∩ L ′ respectively. Let L ′ = ∂ m − M = S { ∂ m − Q ′ | Q ′ ∈ M } . Then L ′ \ ( L ′ ⊔ A ′ ⊔ B ′ ) is the union of some disjoint open m -dimensional cubes with length of edge = 2 k m − . So L ′ is a partition of L ′ between F +2 ∩ L ′ and F − ∩ L ′ and L ′ ⊆ ( S ( U m − ∪ U m )) c ∩ [0 , B ] m +1 .After m + 1 steps, we obtain L ′ m +1 to be a partition of L ′ m between F + m +1 ∩ L ′ m and F − m +1 ∩ L ′ m and L ′ m +1 ⊆ ( S ( S mi =0 U i )) c ∩ [0 , B ] m +1 . Note that L ′ m +1 is 0-dimensional cubes with length of edge = 2 k . Bythe construction, L ′ m +1 ⊆ X τ . Since S mi =0 U i covers [0 , B ] m +1 ∩ X τ , L ′ m +1 = ∅ which is a contradictionwith Lemma 3.7. 9 efinition 3.13. Let M Z = { ( x i ) | x i ∈ Z and there exists k ∈ N such that x j = 0 for each j ≥ k } with the sup-metric ρ . For every τ = { k + 2 , k + 2 , ..., k m + 2 } ∈ S ξ [ L ], we define an isometric embedding i τ : X τ → L Z by for every x = ( x , x , · · · , x m ) ∈ X τ , i τ ( x ) i = x i ∀ i ∈ { , , · · · , m } and i τ ( x ) i = 0 ∀ i / ∈ { , , · · · , m } , where i τ ( x ) i is the i -th coordinate of i τ ( x ).For any countable ordinal number ξ , we define X ξ as the disjoint union of X τ with τ ∈ S ξ [ L ], i.e., X ξ = G τ ∈ S ξ [ L ] X τ with the metric d ξ which is defined as d ξ ( x, y ) = ( ρ ( i τ ( x ) , i τ ( y )) if x, y ∈ X τ , max { s ( τ ) , s ( τ ) , ρ ( i τ ( x ) , i τ ( y )) } if x ∈ X τ , y ∈ X τ and τ = τ , where s ( σ ) = 2 max σ for any σ ∈ S ξ [ L ]. Theorem 3.1. trasdim( X ξ ) ≥ ξ . Proof.
By Proposition 3.2, for every τ ∈ S ξ [ L ], τ ∈ A ( X τ , d ) ⊆ A ( X ξ , d ). i.e., S ξ [ L ] ⊆ A ( X ξ , d ) . Sotrasdim X ξ = Ord A ( X ξ , d ) = Ord A ( X ξ , d ) ≥ Ord S ξ [ L ] = ξ. For every n, i, k ∈ N , let Y i,n,k = n ( x , . . . , x i ) ∈ Z i (cid:12)(cid:12)(cid:12) |{ j | x j / ∈ n Z }| ≤ k o , which is considered as a subspace of the metric space ( L Z , ρ ). Lemma 3.8.
For every r ∈ N with r ≥
4, there exist n = r ∈ N and r -disjoint, 2 n -bounded families U , U such that U ∪ U covers Y i,n, . Proof.
For every r ∈ N and r ≥
4, choose n = r ∈ N . Let U = n ( i Y t =1 ( n t n − r, n t n + r )) ∩ Y i,n, (cid:12)(cid:12)(cid:12) n t ∈ Z o , U = n ( j − Y t =1 ( n t n − r, n t n + r ) × [ n j n + r, ( n j +1)2 n − r ] × i Y t = j +1 ( n t n − r, n t n + r )) ∩ Y i,n, (cid:12)(cid:12)(cid:12) n t ∈ Z , j ∈ { , , · · · , i } o It is easy to see that U and U are r -disjoint and 2 n -bounded families. Now for every x = ( x , . . . , x i ) ∈ Y i,n, \ ( S U ), there exists unique j ∈ { , , · · · , i } such that x j ∈ [ n j n + r, ( n j + 1)2 n − r ]. It follows that x ∈ U . Therefore, U ∪ U covers Y i,n, . Lemma 3.9.
For every i, r ∈ N , r ≥ k ∈ N , there exist n = 3 k − r ∈ N and r -disjoint, 2 n + k − -boundedfamilies U , U , . . . , U k such that U ∪ U ∪ . . . ∪ U k covers Y i,n,k .10 roof. We will prove it by induction on k . By Lemma 3.8, the result is true for k = 1. Assume that the resultis true for k = m , then for every r ∈ N and r ≥
4, there exist n = 3 m r ∈ N and 3 r -disjoint, 2 n + m − -boundedfamilies V , V , . . . , V m such that V ∪ V ∪ . . . ∪ V m covers Y i,n,m . Now for k = m + 1, let U = { N r ( V ) | V ∈ V } , · · · , U m = { N r ( V ) | V ∈ V m } . Then U , U , · · · , U m are r -disjoint and 2 n + k -bounded families such that U ∪ U ∪ . . . ∪ U m covers N r ( Y i,n,m ).Let U m +1 = n { x t } j − t =1 × [ n j n + r, ( n j + 1)2 n − r ] × ( x t ) j − t = j +1 × [ n j n + r, ( n j + 1)2 n − r ] × { x t } j − t = j +1 ×· · · × { x t } j m +1 − t = j m +1 × [ n j m +1 n + r, ( n j m +1 + 1)2 n − r ] × { x t } it = j m +1 +1 (cid:12)(cid:12)(cid:12) x t ∈ n Z , n j k ∈ Z o . It is easy to see that U m +1 is r -disjoint and 2 n -bounded.Note that Y i,n,m +1 \ [ U m +1 ⊆ N r ( Y i,n,m ) . Indeed, for any x = ( x t ) it =1 ∈ Y i,n,m +1 \ S U m +1 , ( x t ) it =1 ∈ Y i,n,m +1 implies that there exists at most m + 1coordinates x t such that x t / ∈ n Z and x / ∈ S U m +1 implies that, among all the x t with x t / ∈ n Z , thereexists at least one x t such that d ( x t , n Z ) < r . It follows that x ∈ N r ( Y i,n,m ).Then Y i,n,m +1 ⊆ ( [ U m +1 ) ∪ N r ( Y i,n,m ) . Therefore, U ∪ U ∪ . . . ∪ U m +1 covers Y i,n,m +1 . So the result is true for k = m + 1. Lemma 3.10. asdim( X n ) ≤ n for every n ∈ N . Proof.
For every r ∈ N , let Y ,r = G τ ∈ S n [ L ] s ( τ ) ≤ r X τ , Y ,r = G τ ∈ S n [ L ] s ( τ ) >r X τ . Note that X τ ⊆ R n for every τ ∈ S n [ L ]. Since asdim( R n ) ≤ n , there are r -disjoint, B ( r )-bounded families U ( τ ) , U ( τ ) , · · · , U n ( τ ) such that S ni =0 U i ( τ ) covers X τ . Let U i = (cid:16) [ τ ∈ S n [ L ] s ( τ ) >r U i ( τ ) (cid:17) for i ∈ { , , · · · , n } . Then U , U , · · · , U n are r -disjoint, B ( r )-bounded families such that S ni =0 U i covers Y ,r .Note that { τ | τ ∈ S n [ L ] , s ( τ ) ≤ r } is a finite set. By Lemma 2.6,coasdim( Y ,r ) ≤ max { coasdim( X τ ) | τ ∈ S n [ L ] , s ( τ ) ≤ r } ≤ coasdim( R n ) ≤ n .There are r -disjoint, uniformly bounded families V , V , · · · , V n such that S ni =0 V i covers Y ,r .Let W i = U i [ V i for i ∈ { , , · · · , n } . Since d ( Y ,r , Y ,r ) > r , W , W , · · · , W n are r -disjoint, uniformly bounded families such that S ni =0 W i covers X n . i.e., asdim( X n ) ≤ n . Theorem 3.2. coasdim( X ξ ) ≤ ξ for each countable ordinal number ξ .11 roof. By Lemma 3.10, the result is true for finite ξ . Assume that the result is true for every ξ < α . Now ξ = α . For every r ∈ N , let p = 3 n ( ξ ) r and let T p = { τ ∈ S ξ [ L ] | i ( τ, n ( ξ )) ≥ p + 2 } .For each τ ∈ T p , X τ ⊆ Y i,p,n ( ξ ) for some i ∈ N and the metric on each X τ is the restriction of ρ on L Z .Then by Lemma 3.9, there are 2 p + n ( ξ ) -bounded, r -disjoint families U τ , U τ , · · · , U τn ( ξ ) whose union covers Y i,p,n ( ξ ) ∩ X τ = X τ . For i = 0 , , · · · , n , let U i = { U | U ∈ U τi , τ ∈ T p } . Since the distance between X τ and X σ is greater or equal than 2 p for every τ , σ ∈ T p and τ = σ , the families U , . . . , U n ( ξ ) are 2 p + n ( ξ ) -bounded, r -disjoint and S n ( ξ ) i =0 U i covers the set Y = ` { X τ | τ ∈ T p } .Note that X ξ \ Y = S { X τ | τ ∈ S ξ [ L ] , i ( τ, n ( ξ )) < p + 2 } . By Lemma 3.1, [ { X τ | τ ∈ S ξ [ L ] , i ( τ, n ( ξ )) < p + 2 } ⊆ p +1 [ l =1 ( X ζ l ( γ ( ξ ))+ l + n ( ξ )+1 ∩ X ξ ) ⊆ X ξ . Moreover, the natural embedding i : ( X ζ l ( γ ( ξ ))+ l + n ( ξ )+1 ∩ X ξ , d ξ ) → ( X ζ l ( γ ( ξ ))+ l + n ( ξ )+1 , d ζ l ( γ ( ξ ))+ l + n ( ξ )+1 )is isometric by the definitions of the metrics d ξ and d ζ l ( γ ( ξ ))+ l + n ( ξ )+1 . Then by Lemma 2.6 and inductiveassumption, coasdim (cid:16) p +1 [ l =1 ( X ζ l ( γ ( ξ ))+ l + n ( ξ )+1 ∩ X ξ ) (cid:17) ≤ max l ∈{ , ,...,p +1 } { coasdim( X ζ l ( γ ( ξ ))+ l + n ( ξ )+1 ∩ X ξ ) } ≤ max l ∈{ , ,...,p +1 } n coasdim (cid:16) X ζ l ( γ ( ξ ))+ l + n ( ξ )+1 (cid:17)o ≤ coasdim( X ζ p +1 ( γ ( ξ ))+( p +1)+ n ( ξ )+1 ) ≤ ζ p +1 ( γ ( ξ )) + p + n ( ξ ) + 2 . Then coasdim( [ { X τ | τ ∈ S ξ [ L ] , i ( τ, n ( ξ )) < p + 2 } ) ≤ ζ p +1 ( γ ( ξ )) + p + n ( ξ ) + 2 < γ ( ξ ) . It follows that coasdim( X ξ \ S S n ( ξ ) i =0 U i ) ≤ coasdim( X ξ \ Y ) < γ ( ξ ) . So coasdim X ξ ≤ γ ( ξ ) + n ( ξ ) = ξ . Lemma 3.11. ([6]) Let X be a metric space, if X has complementary-finite asymptotic dimension, thentrasdim( X ) ≤ coasdim( X ). Theorem 3.3. coasdim( X ξ ) = ξ and trasdim( X ξ ) = ξ for every countable ordinal ξ . Proof.
By Theorem 3.2 and Lemma 3.11, trasdim( X ξ ) ≤ ξ . Then trasdim( X ξ ) = ξ by Theorem 3.1. ByLemma 3.11, coasdim( X ξ ) ≥ trasdim( X ξ ) = ξ . Then coasdim( X ξ ) = ξ by Theorem 3.2. X ξ Now we will prove that the metric space X ξ has finite decomposition complexity. More specifically, forany countable infinite ordinal ξ with ξ = γ ( ξ ) + n ( ξ ) where γ ( ξ ) is a limit ordinal and n ( ξ ) ∈ N ∪ { } , X γ ( ξ )+ n ( ξ ) ∈ D γ ( ξ )+ n ( ξ ) . Definition 3.14.
Let ξ = γ ( ξ ) + n ( ξ ) be a countable infinite ordinal number with n ( ξ ) ≥ τ = { k + 2 , k + 2 , · · · , k m + 2 } ∈ S ξ [ L ] such that m ≥ n ( ξ ). For i ∈ { , , · · · , n ( ξ ) } , let X τ,i = n ( x i ) mi =0 ∈ (2 k Z ) m +1 (cid:12)(cid:12)(cid:12) |{ j | x j / ∈ k n ( ξ ) Z }| ≤ i and |{ j | x j / ∈ k p Z }| ≤ p, where p ∈ { , , · · · , m } \ { n ( ξ ) } o . and let X iξ = G τ ∈ S ξ [ L ] | τ |≥ n ( ξ )+1 X τ,i ⊆ ( X ξ , d ξ ) , X ξ , d ξ ). Remark 3.4.
Note that X τ, ⊆ X τ, · · · ⊆ X τ,n ( ξ ) = X τ and X ξ = X n ( ξ ) ξ [ (cid:16) G τ ∈ S ξ [ L ] | τ |≤ n ( ξ ) X τ (cid:17) . Lemma 3.12. X n = G τ ∈ S n [ L ] X τ ∈ D n +1 . Proof.
By Lemma 3.10, asdim ( X n ) ≤ n , which implies that X n ∈ D n +1 . Lemma 3.13. X ω +1 = G τ ∈ S ω +1 [ L ] X τ ∈ D ω . Moreover, X ω ∈ D ω . Proof.
For every r ∈ N , let Y ,r = G τ ∈ S ω +1 [ L ] i ( τ, ≤ r +2 X τ , Y ,r = G τ ∈ S ω +1 [ L ] i ( τ, >r +2 X τ . For τ = { k + 2 , k + 2 , · · · , k m + 2 } ∈ S ω +1 [ L ] and i ( τ, ≤ r + 2, by Lemma 3.1, τ ∈ r +2 [ l =1 S l +2 [ L ] which implies Y ,r ⊆ r +2 [ l =1 X l +2 as a subspace.By Lemma 2.7 and Lemma 3.12, Y ,r ∈ D r +8 .Note that for every τ ∈ S ω +1 [ L ] and i ( τ, > r + 2, X τ ⊆ Y j,r, for some j ∈ N . By Lemma 3.8, thereexist r -disjoint and 2 r -bounded families U ( τ ) , U ( τ ) such that U ( τ ) ∪ U ( τ ) covers X τ . Let U = (cid:16) [ τ ∈ S ω +1 [ L ] i ( τ, >r +2 U ( τ ) (cid:17) ∪ { Y ,r } , U = [ τ ∈ S ω +1 [ L ] i ( τ, >r +2 U ( τ ) . Then U , U are r -disjoint families with U ∈ D r +4 , U ∈ D and U ∪ U covers X ω +1 . Let Y = U S U , then X ω +1 r → Y and Y ∈ D r +4 . Therefore, X ω +1 ∈ D ω .By Lemma 3.3, X ω = G τ ∈ S ω [ L ] X τ ⊆ G τ ∈ S ω +1 [ L ] X τ = X ω +1 ∈ D ω . Lemma 3.14.
For every infinite countable ordinal number ξ = γ ( ξ ) + n ( ξ ) such that n ( ξ ) ≥ X ξ = G τ ∈ S ξ [ L ] | τ |≥ n ( ξ )+1 X τ, ∈ D γ ( ξ ) . Proof.
We prove it by induction on ξ . • Let ξ = ω + 1. By Lemma 3.13 and X ω +1 ⊆ X ω +1 , X ω +1 ∈ D ω . i.e., the result is true for ξ = ω + 1.13 Assume that the result is true for every infinite countable ordinal number α = γ ( α ) + n ( α ) such that α < ξ and n ( α ) ≥
1. For every r ∈ N , let Y ,r = G τ ∈ S ξ [ L ] | τ |≥ n ( ξ )+1 i ( τ,n ( ξ )) ≤ r +2 X τ, , Y ,r = G τ ∈ S ξ [ L ] | τ |≥ n ( ξ )+1 i ( τ,n ( ξ )) >r +2 X τ, , then X ξ = Y ,r ∪ Y ,r .For τ ∈ S ξ [ L ] with i ( τ, n ( ξ )) ≤ r + 2, by Lemma 3.1, τ ∈ S r +2 l =1 S ζ l ( γ ( ξ ))+ l + n ( ξ )+1 [ L ] . Then Y ,r ⊆ r +2 [ l =1 X ζ l ( γ ( ξ ))+ l + n ( ξ )+1 . By inductive assumption, X ζ l ( γ ( ξ ))+ l + n ( ξ )+1 ∈ D ζ l ( γ ( ξ )) ⊆ D ζ l ( γ ( ξ ))+ l for each l ∈ { , , · · · , r + 2 } .Then by Lemma 2.7, Y ,r ∈ D ζ r +2 ( γ ( ξ ))+ r +3 . Note that for every τ ∈ S ξ [ L ] and i ( τ, n ( ξ )) > r + 2, X τ, ⊆ Y j,r, for some j ∈ N . By Lemma 3.8,there exist r -disjoint, 2 r -bounded families U ( τ ) , U ( τ ) such that U ( τ ) ∪ U ( τ ) covers X τ, . Let U = (cid:16) [ τ ∈ S ξ [ L ] | τ |≥ n ( ξ )+1 i ( τ,n ( ξ )) ≤ r +2 U ( τ ) (cid:17) [ { Y ,r } , U = [ τ ∈ S ξ [ L ] | τ |≥ n ( ξ )+1 i ( τ,n ( ξ )) >r +2 U ( τ ) , then U ∈ D ζ r ( γ ( ξ ))+ r +1 and U ∈ D . Let Y = U S U , then X ξ r → Y and Y ∈ D ζ r ( γ ( ξ ))+ r +1 .Therefore, X ξ ∈ D γ ( ξ ) . Lemma 3.15.
Let ξ = γ ( ξ ) + n ( ξ ) be a countable infinite ordinal number with n ( ξ ) ≥ τ = { k + 2 , k + 2 , · · · , k m + 2 } ∈ S ξ [ L ] such that m ≥ n ( ξ ). Then for each i ∈ { , , · · · , n ( ξ ) } , X τ,i ∩ N r ( X τ,i − ) = n ( x i ) mi =0 ∈ (2 k Z ) m +1 (cid:12)(cid:12)(cid:12) |{ j | x j / ∈ N r (2 k n ( ξ ) Z ) }| ≤ i − o ∩ X τ,i . Proof. • First we will prove that X τ,i ∩ N r ( X τ,i − ) ⊆ n ( x i ) mi =0 ∈ (2 k Z ) m +1 (cid:12)(cid:12)(cid:12) |{ j | x j / ∈ N r (2 k n ( ξ ) Z ) }| ≤ i − o ∩ X τ,i . For every x = ( x i ) mi =0 ∈ X τ,i ∩ N r ( X τ,i − ), there exists y = ( y i ) mi =0 ∈ X τ,i − ⊆ X τ,i such that d ( x, y ) < r , which implies that d ( x i , y i ) < r for i ∈ { , , · · · , m } . Note that |{ j | y j / ∈ k n ( ξ ) Z }| ≤ i − y j ∈ k n ( ξ ) Z implies x j ∈ N r (2 k n ( ξ ) Z ) . Then |{ j | x j / ∈ N r (2 k n ( ξ ) Z ) }| ≤ |{ j | y j / ∈ k n ( ξ ) Z }| ≤ i − . So x ∈ n ( x i ) mi =0 ∈ (2 k Z ) m +1 (cid:12)(cid:12)(cid:12) |{ j | x j / ∈ N r (2 k n ( ξ ) Z ) }| ≤ i − o ∩ X τ,i . • Now we will prove that n ( x i ) mi =0 ∈ (2 k Z ) m +1 (cid:12)(cid:12)(cid:12) |{ j | x j / ∈ N r (2 k n ( ξ ) Z ) }| ≤ i − o ∩ X τ,i ⊆ X τ,i ∩ N r ( X τ,i − ) . Assume that x = ( x i ) mi =0 such that x ∈ X τ,i and |{ j | x j / ∈ N r (2 k n ( ξ ) Z ) }| ≤ i − . Let I i = { j | x j ∈ N r (2 k i Z ) } for i ∈ { , . . . , m } and I m +1 = ∅ .14ince k < k < · · · < k m , I m ⊆ I m − ⊆ · · · ⊆ I ⊆ I . x ∈ X τ,i and |{ j | x j / ∈ N r (2 k n ( ξ ) Z ) }| ≤ i − | I p | ≥ m + 1 − p, ∀ p ∈ { , , · · · , m } \ { n ( ξ ) } and | I n ( ξ ) | ≥ m + 1 − ( i − . In particular, I m = ∅ and I = { , , · · · , m } . Note that I = m G i =0 ( I i \ I i +1 (cid:17) and if j ∈ (cid:16) I i \ I i +1 (cid:17) , then there exists y j ∈ k i Z such that d ( x j , y j ) < r .Let y = ( y i ) mi =0 . Note that ∀ p ∈ { , , · · · , m } \ { n ( ξ ) } , |{ j | y j ∈ k p Z }| = | I p | ≥ m + 1 − p and |{ j | y j ∈ k n ( ξ ) Z }| = | I n ( ξ ) | ≥ m + 1 − ( i − . Hence ∀ p ∈ { , , · · · , m } \ { n ( ξ ) } , |{ j | y j / ∈ k p Z }| ≤ p and |{ j | y j / ∈ k n ( ξ ) Z }| ≤ i − , which implies that y ∈ X τ,i − . So x ∈ N r ( X τ,i − ). Corollary 3.1.
Let ξ = γ ( ξ ) + n ( ξ ) be a countable infinite ordinal number with n ( ξ ) ≥ τ = { k + 2 , k + 2 , · · · , k m + 2 } ∈ S ξ [ L ] such that m ≥ n ( ξ ). Then for each i ∈ { , , · · · , n ( ξ ) } , X τ,i ⊆ N r ( X τ,i − ) ∪ n ( x i ) mi =0 ∈ (2 k Z ) m +1 (cid:12)(cid:12)(cid:12) |{ j | x j / ∈ N r (2 k n ( ξ ) Z ) }| = i o . Proof.
For every x = ( x i ) mi =0 ∈ X τ,i \ N r ( X τ,i − ), x ∈ X τ,i implies x ∈ (2 k Z ) m +1 and |{ j | x j / ∈ N r (2 k n ( ξ ) Z ) }| ≤ |{ j | x j / ∈ k n ( ξ ) Z }| ≤ i. Suppose that { j | x j / ∈ N r (2 k n ( ξ ) Z ) }| ≤ i −
1, then by Lemma 3.15, x ∈ N r ( X τ,i − ), which is a contradiction.Then |{ j | x j / ∈ N r (2 k n ( ξ ) Z ) }| = i . Therefore, X τ,i \ N r ( X τ,i − ) ⊆ n ( x i ) mi =0 ∈ (2 k Z ) m +1 (cid:12)(cid:12)(cid:12) |{ j | x j / ∈ N r (2 k n ( ξ ) Z ) }| = i o and hence X τ,i ⊆ N r ( X τ,i − ) ∪ n ( x i ) mi =0 ∈ (2 k Z ) m +1 (cid:12)(cid:12)(cid:12) |{ j | x j / ∈ N r (2 k n ( ξ ) Z ) }| = i o . Proposition 3.3.
For every countable infinite ordinal number ξ = γ ( ξ ) + n ( ξ ) such that n ( ξ ) ≥ i ∈ { , , · · · , n ( ξ ) } , X iξ = G τ ∈ S ξ [ L ] | τ |≥ n ( ξ )+1 X τ,i ∈ D γ ( ξ )+ i − . In particular, X n ( ξ ) ξ ∈ D ξ − . Proof.
By Lemma 3.13, X ω +1 ∈ D ω . By Lemma 3.14, X ξ ∈ D γ ( ξ ) . Assume that X ˜ ξ ∈ D ˜ ξ − for everyinfinite countable ordinal number ˜ ξ such that ˜ ξ < ξ , n ( ˜ ξ ) ≥ X i − ξ ∈ D γ ( ξ )+ i − . Now we will show that X iξ ∈ D γ ( ξ )+ i − .For every r ∈ N , let Y ,r = G τ ∈ S ξ [ L ] | τ |≥ n ( ξ )+1 i ( τ,n ( ξ )) ≤ r +2 X τ,i , Y ,r = G τ ∈ S ξ [ L ] | τ |≥ n ( ξ )+1 i ( τ,n ( ξ )) >r +2 X τ,i , then X iξ = Y ,r ∪ Y ,r . 15or τ ∈ S ξ [ L ] with i ( τ, n ( ξ )) ≤ r + 2, by Lemma 3.1, τ ∈ S r +2 l =1 S ζ l ( γ ( ξ ))+ l + n ( ξ )+1 [ L ] . Then Y ,r ⊆ r +2 [ l =1 X iζ l ( γ ( ξ ))+ l + n ( ξ )+1 . Note that for each l ∈ { , , · · · , r + 2 } , X iζ l ( γ ( ξ ))+ l + n ( ξ )+1 ⊆ X ζ l ( γ ( ξ ))+ l + n ( ξ )+1 ∈ D ζ l ( γ ( ξ ))+ l + n ( ξ ) by inductive assumption. Then by Lemma 2.7, Y ,r ∈ D ζ r +2 ( γ ( ξ ))+ r + n ( ξ )+3 . For τ = { k + 2 , k + 2 , · · · , k m + 2 } ∈ S ξ [ L ], let U ( τ ) = n { x t } j − t =0 × [ n j k n ( ξ ) + r, ( n j + 1)2 k n ( ξ ) − r ] × ( x t ) j − t = j +1 × [ n j k n ( ξ ) + r, ( n j + 1)2 k n ( ξ ) − r ] × { x t } j − t = j +1 × · · · × { x t } j m +1 − t = j m +1 × [ n j i k n ( ξ ) + r, ( n j i + 1)2 k n ( ξ ) − r ] × { x t } mt = j i +1 (cid:12)(cid:12)(cid:12) x t ∈ k n ( ξ ) Z , n j k ∈ Z , k ∈ { , , · · · , i } o . It is easy to see that U ( τ ) is r -disjoint. By Corollary 3.1, X τ,i ⊆ N r ( X τ,i − ) ∪ n ( x i ) mi =0 ∈ (2 k Z ) m +1 (cid:12)(cid:12)(cid:12) |{ j | x j / ∈ N r (2 k n ( ξ ) Z ) }| = i o ⊆ N r ( X τ,i − ) ∪ ( [ U ( τ )) . Note that [ τ ∈ S ξ [ L ] | τ |≥ n ( ξ )+1 i ( τ,n ( ξ )) >r +2 N r ( X τ,i − ) ⊆ N r (cid:16) [ τ ∈ S ξ [ L ] | τ |≥ n ( ξ )+1 i ( τ,n ( ξ )) >r +2 X τ,i − (cid:17) ⊆ N r (cid:16) X i − ξ (cid:17) ∈ D γ ( ξ )+ i − and there is a natural coarse embedding F : [ τ ∈ S ξ [ L ] | τ |≥ n ( ξ )+1 i ( τ,n ( ξ )) >r +2 U ( τ ) → { Z i } such that for every f ∈ F and every x, y ∈ X f , d ( f ( x ) , f ( y )) = d ( x, y ) . By Lemma 2.8 and { Z i } ∈ D i , [ τ ∈ S ξ [ L ] | τ |≥ n ( ξ )+1 i ( τ,n ( ξ )) >r +2 U ( τ ) ∈ D i . Let U = { [ τ ∈ S ξ [ L ] | τ |≥ n ( ξ )+1 i ( τ,n ( ξ )) >r +2 N r ( X τ,i − ) } , U = (cid:16) [ τ ∈ S ξ [ L ] | τ |≥ n ( ξ )+1 i ( τ,n ( ξ )) >r +2 U ( τ ) (cid:17) [ { Y ,r } . Let Y = U S U , then X iξ r → Y and Y ∈ D γ ( ξ )+ i − . Therefore, X iξ ∈ D γ ( ξ )+ i − . Corollary 3.2. X ξ ∈ D ξ for every infinite countable ordinal number ξ . Proof. • If n ( ξ ) ≥
1, let Y = G τ ∈ S ξ [ L ] | τ |≤ n ( ξ ) X τ , Y = G τ ∈ S ξ [ L ] | τ | >n ( ξ ) X τ . Y ⊆ F τ ∈ S n ( ξ ) [ L ] X τ = X n ( ξ ) ∈ D n ( ξ )+1 by Lemma 3.12. Note that Y = G τ ∈ S ξ [ L ] | τ |≤ n ( ξ ) X τ = G τ ∈ S ξ [ L ] | τ |≤ n ( ξ ) X τ,n ( ξ ) = X n ( ξ ) ξ ∈ D ξ − by Proposition 3.3. Therefore, X ξ = Y ∪ Y ∈ D ξ . • If n ( ξ ) = 0, thenFor every r ∈ N , let Y ,r = G τ ∈ S ξ [ L ] || τ |≥ n ( ξ )+1 i ( τ, ≤ r +2 X τ,i , Y ,r = G τ ∈ S ξ [ L ] || τ |≥ n ( ξ )+1 i ( τ, >r +2 X τ,i , then X ξ = Y ,r ∪ Y ,r .For τ ∈ S ξ [ L ] with i ( τ, ≤ r + 2, by Lemma 3.1, τ ∈ S r +2 l =1 S ζ l ( γ ( ξ ))+ l +1 [ L ] . Then Y ,r ⊆ r +2 [ l =1 X ζ l ( γ ( ξ ))+ l +1 . Note that X ζ l ( γ ( ξ ))+ l +1 ∈ D ζ l ( γ ( ξ ))+ l +1 for each l ∈ { , , · · · , r + 2 } . Then by Lemma 2.7, Y ,r ∈ D ζ r +2 ( γ ( ξ ))+ r +4 . Let U = {{ x } | x ∈ Y ,r } , then U is a r -disjoint, uniformly bounded family. Let Y = U ∪ { Y ,r } , then X ξ r → Y and Y ∈ D ζ r +2 ( γ ( ξ ))+ r +4 . Therefore, X ξ ∈ D γ ( ξ ) = D ξ . The authors wish to thank the reviewers for careful reading and valuable comments. This work was supportedby NSFC grant of P.R. China (No. 11871342, 11801219, 11301224, 11326104). And the authors (Yan Wuand Jingming Zhu) want to thank V.M. Manuilov and Benyin Fu for helpful discussions. The third author(Taras Radul) is grateful to Taras Banakh for valuable and stimulating discussions.
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