A Metric Space with Transfinite Asymptotic Dimension 2ω+1*

Yan WU Jingming ZHU

Abstract The authors construct a metric space whose transfinite asymptotic dimension and complementary-finite asymptotic dimension are both 2ω+1， where ω is the smallest infinite ordinal number.Therefore， an example of a metric space with asymptotic property C is obtained.

Keywords Transfinite asymptotic dimension， Complementary-finite asymptotic dimension， Asymptotic property C

1 Introduction

M.Gromov introduced the notion of asymptotic dimension to study finitely generated groups in [1].In 1998， Guoliang Yu discovered a successful application of asymptotic dimension.He proved that a group with finite asymptotic dimension satisfies the higher Novikov signature conjecture (see [2]).In 2000， N.Higson and J.Roe proved that metric space with bounded geometry and finite asymptotic dimension has property A(see[3]).There is a large class of groups with finite asymptotic dimension， such as finite generated commutative groups， finite rank free groups， Gromov hyperbolic groups and so on.In [4]， A.Dranishnikov introduced asymptotic property C which is a natural extension of asymptotic dimension.To classify the metric spaces with infinite asymptotic dimension， T.Radul defined the transfinite asymptotic dimension(trasdim)and found that asymptotic property C can be characterized by transfinite asymptotic dimension.i.e.， a metric space X has asymptotic property C if and only if trasdim(X)＜∞(see [5]).There are examples of metric spaces with trasdim=∞， and with trasdim=ω as well，where ω is the smallest infinite ordinal number (see [5]).In [6]， we constructed a metric space X with trasdim(X)=ω+1 which is the first example we found out with transfinite asymptotic dimension greater than ω.By the technique developed in [6]， we constructed metric space Xω+kwith trasdim(Xω+k) = ω+k in [7]， which generalized the results in [6].In this paper，we construct a metric space X2ω+1with coasdim(X2ω+1)=trasdim(X2ω+1)=2ω+1.

This paper is organized as follows: In Section 2，we recall some definitions and properties of transfinite asymptotic dimension and complementary-finite asymptotic dimension.In Section 3， we construct a concrete metric space X2ω+1， whose transfinite asymptotic dimension and complementary-finite asymptotic dimension are both 2ω+1， where ω is the smallest infinite ordinal number.

2 Preliminaries

Let (X，d) be a metric space and U，V ⊆X，

Let R ＞0 and U be a family of subsets of X.U is said to be R-bounded if

In this case， U is said to be uniformly bounded.Let r ＞0， a family U is said to be r-disjoint if

In this paper， we denote ∪{U |U ∈U} by ∪U， denote {U |U ∈U1or U ∈U2} by U1∪U2and denote {Nδ(U)|U ∈U} by Nδ(U) for some δ ＞0.Letting A be a subset of X， we denote{x ∈X |d(x，A)＜∈} by N∈(A) for some ∈＞0.

Definition 2.1(see [8])The asymptotic dimension of a metric spaceXdoes not exceedn(denoted byasdim(X)≤n)which means if there existsn ∈N， such that for everyr ＞0， there exists a sequence of uniformly bounded families{Ui}ni=0of subsets ofXsuch thatcoversXand eachUiisr-disjoint fori=0，1，··· ，n.In this case， we say thatXhas finite asymptotic dimension.

In[5]，T.Radul generalized asymptotic dimension of a metric space X to transfinite asymptotic dimension denoted by trasdim(X).

Definition 2.2(see [5])LetFinNdenote the collection of all finite， nonempty subsets ofN，and letM ⊆FinN.Forσ ∈{Ø}∪FinN， let

LetMabe the abbreviation forM{a}fora ∈N.Define the ordinal numberOrdMinductively as follows:

Lemma 2.1(see[9])LetM ⊆FinNandk ∈N，OrdM ≤ω+kif and only ifOrdMτ＜ωfor everyτ ∈FinNwith|τ|=k+1.

Definition 2.3(see [5])Given a metric spaceX， define the following collection:

The transfinite asymptotic dimension ofXis defined astrasdim(X)=OrdA(X).

Definition 2.4Letbe a sequence of subspaces of a metric space(Z，dZ).Let

For everyx，y ∈X， there exist uniquel，k ∈N， xl∈Zlandyk∈Zk，such thatx =(0，··· ，0，xl，0，···)andy = (0，··· ，0，yk，0，···).Assume thatl ≤k.Letc = 0ifl = kandc=l+(l+1)+···+(k-1)ifl ＜k.Define a metric onXby

The metric space(X，d)is said to be an asymptotic union ofwhich is denoted byasAnd we denoteasas a subspace ofasfor everyn ∈N.

For every k，n ∈N， let

Lemma 2.2(see [7]) coasdim(Xω+k)=ω+kfor everyk ∈N.

where Yω+kis a subspace of the metric space asfor each k ∈N.

Lemma 2.3(see [7])For everyk ∈N，trasdim(Yω+k)=ω+kandtrasdim(Y2ω)=2ω.

Definition 2.5(see [9])Every ordinal numberγcan be represented asγ = λ(γ)+n(γ)，whereλ(γ)is the limit ordinal or0andn(γ) ∈N.LettingXbe a metric space， we define the complementary-finite asymptotic dimension ofX (coasdim(X))inductively as follows:

· coasdim(X)=-1 ⇔X =Ø.

· coasdim(X) ≤λ(γ)+n(γ) ⇔for everyr ＞0，there existr-disjoint uniformly bounded familiesU0，··· ，Un(γ)of subsets ofXsuch thatcoasdim

· coasdim(X)=γ ⇔coasdim(X)≤γandcoasdim(X)βfor everyβ ＜γ.

· coasdim(X)=∞⇔coasdim(X)γfor every ordinalγ.Xis said to have complementary-finite asymptotic dimension ifcoasdim(X) ≤γfor some ordinal numberγ.

Lemma 2.4(see [9])LetXbe a metric space withX1，X2⊆X.Then

Lemma 2.5(see[10])LettingXbe a metric space， ifXhas complementary-finite asymptotic dimension， thentrasdim(X)≤coasdim(X).

3 Main Result

Let

where (p1，··· ，pn)，(q1，··· ，qn) ∈Nnand p1≤··· ≤pn.Then= X((k，n)，(k，n-k))when n ≥k.Since for every k ∈N，

We have

is not true.

Lemma 3.1For everyr ∈N， there arer-disjoint uniformly bounded familiesU0andU1，such thatU0∪U1coversas

ProofFor every r ∈N， k，n ∈N and k ≥2r， n ≥k.Let

For every x ∈X((0，k)，(1，n))，without loss of generality，we assume that x=(x1，··· ，xn)∈Z×2kZ×···×2kZ.Then xi∈2kZ for i=2，3，··· ，n and x1is in one of the following cases.

· x1∈[2ki-r，2ki+r] for some i ∈Z， it is easy to see that x ∈∪.

· x1∈V for some V ∈， it is easy to see that x ∈∪.

· x1∈V for some V ∈， it is easy to see that x ∈∪.covers X((0，k)，(1，n)).Let

Since for every n，m ≥k and n/=m，

U0，k，U1，kare r-disjoint and 2r-bounded families such that U0，k∪U1，kcovers as

U0，U1are r-disjoint and 2r-bounded families such that U0∪U1covers

Proposition 3.1Let

Thencoasdim(X)≤2ω+1.

ProofSince for k ≤n， X((0，k，n)，(1，k，n-k))⊆X((0，k)，(1，n)) and by Lemma 3.1， for any r ∈N， there are r-disjoint uniformly bounded families U0and U1such that

Note that

Then by Lemma 2.2 and Lemma 2.4，

Proposition 3.2trasdim

ProofIt can be obtained easily by Lemma 2.5 and Proposition 3.1.

Definition 3.1(see[11])LetXbe a metric space and letA，Bbe a pair of disjoint subsets ofX.We say that a subsetL ⊂Xis a partition ofXbetweenAandB， if there exist open setsU，W ⊂Xsatisfying the following conditions

Definition 3.2(see [7])LetXbe a metric space and letA，Bbe a pair of disjoint subsets ofX.For any∈＞0， we say that a subsetL ⊂Xis an∈-partition ofXbetweenAandB， if there exist open setsU，W ⊂Xsatisfying the following conditions

Clearly， an ∈-partition L of X between A and B is a partition of X between A and B.

Lemma 3.2(see [7])LetL0.= [0，B]nfor someB ＞0，be the pairs of oppositefaces ofL0， wherei = 1，2，··· ，nand let0 ＜∈＜Fork = 1，2，··· ，n， letUkbe an∈-disjoint andB-bounded family of subsets of[0，B]n.Then there exists an∈-partitionLk+1ofLkbetweensuch thatandLk+1⊂Lkfork=0，1，2，··· ，n-1.

ProofFor everyClearly， Ak∪Bk=Uk.Let

Let

for k =0，1，2，··· ，n-1.Therefore，

And Lk+1is an ∈-partition of Lkbetweensuch that Lk+1⊂Lk∩(∪Uk+1)c.

Lemma 3.3(see [11， Lemma 1.8.19])Letbe the pairs of opposite faces ofIn.= [0，1]n， wherei ∈{1，··· ，n}.IfIn= L0⊃L1⊃··· ⊃Lnis a decreasing sequence of closed sets such thatLiis a partition ofLi-1betweenLi-1∩F+iandLi-1∩F-ifori ∈{1，2，··· ，n}， thenLn/=Ø.

Proposition 3.3Let

Thentrasdim(X)≤2ωis not true.

ProofSuppose that trasdimThen for every a ∈N， OrdA(X)a≤ω+m for some m=m(a)∈N.By Lemma 2.1， for every τ ∈FinN satisfying a /∈τ and |τ| = m+1， OrdA(X){a}⊔τ≤n for some n = n(a，τ) ＞1.Then for any σ ∈FinN with |σ|=n+1 and ({a}⊔τ)∩σ =Ø， {a}⊔τ ⊔σ /∈A(X).Let

and

Then there are a-disjoint B-bounded family U，(a+2m+3)-disjoint B-bounded families V1，··· ，Vm+1and (a+2m+n+4+m)-disjoint B-bounded families W1，··· ，Wn+1， such thatcovers X for some B ＞2m+n+4+a+m.It follows thatcovers

and let

where ψ(t)jis the jth coordinate of ψ(t).

Let Q={Q(t)|t ∈{1，2，··· ，pm+n+3}}， then

Let L0=[0，6B]m+n+3.By Lemma 3.2，since N2m+n+2(W1)is(a+m+2m+n+3)-disjoint and(2m+n+3+B)-bounded， there exists a (a+m+2m+n+3)-partition L1of [0，6B]m+n+3betweensuch that

Let M1={Q ∈Q|Q∩L1/=Ø}and M1=∪M1.Since L1is a(a+m+2m+n+3)-partition of [0，6B]m+n+3between， M1is a partition of [0，6B]m+n+3betweeni.e.， [0，6B]m+n+3= M1⊔A1⊔B1such that A1， B1are open in [0，6B]m+n+3and A1， B1contain two opposite facetsrespectively.Let

where ∂m+n+2Q is the set of(m+n+2)-skeleton of Q.Then[0，6B]m+n+3(L′1⊔A1⊔B1)is the union of some disjoint open (m+n+3)-dimensional cubes with length of edge being 2m+n+2.So L′1is a partition of [0，6B]m+n+3between F+1and F-1， and

Similarly，by Lemma 3.2，there exists a(a+m+2m+n+3)-partition L2of L′1between L′1∩F+2and L′1∩F-2such that

Let M2={Q ∈M1|Q∩L2/=Ø}and M2=∪M2.Since L2is a(a+m+2m+n+3)-partition of L′1between L′1∩F+2and L′1∩F-2，M2∩L′1is a partition of L′1between L′1∩F+2and L′1∩F-2，i.e.，L′1=(M2∩L′1)⊔A2⊔B2such that A2，B2are open in L′1and A2，B2contain two opposite facets L′1∩F-2， L′1∩F+2respectively.Let L′2=L′1∩(∂m+n+1M2)=L′1∩∪{∂m+n+1Q|Q ∈M2}，then L′1(L′2⊔A2⊔B2) is the union of some disjoint open (m+n+2)-dimensional cubes with length of edge = 2m+n+2.So L′2is also a partition of L′1between L′1∩F+2and L′1∩F-2and L′2⊂∪(W1∪W2)c∩[0，6B]m+n+3.

After n+1 steps above， we obtain a partitionof L′nbetween L′n∩and L′n∩such that

and L′n+1is m + 2-skeleton.Note that2m+n+2Z}|≤m+2}.

By Lemma 3.2 and since N2m+1(V1) is (a+2m+2)-disjoint and (2m+2+B)-bounded， there exists a (a+2m+2)-partition Ln+2of L′n+1betweensuch that

Similarly to L0， L′n+1can be represented as the union of (m+2)-dimensional cubes with length of edges being 2m+1.Let Q′be a family of(m+2)-dimensional cubes above with length of edges being 2m+1.Let

Since Ln+2is a (a+2m+2)-partition of L′n+1betweenL′n+1is a partition of L′n+1between L′n+1∩and L′n+1∩F-n+2， i.e.， L′n+1= (Mn+2∩L′n+1)⊔An+2⊔Bn+2such that An+2， Bn+2are open in L′n+1and An+2， Bn+2contain two opposite facets L′n+1∩L′n+1∩respectively.Let L′n+2=L′n+1∩(∂m+1Mn+2).Then L′n+1(L′n+2⊔An+2⊔Bn+2) is the union of some disjoint open (m+2)-dimensional cubes with length of edge being 2m+1.So L′n+2is a partition of L′n+1between L′n+1∩and L′n+1∩， and L′n+2⊂(V1)c∩L′n+1.

After m+1 steps above， we have L′m+n+2to be a partition of L′m+n+1between L′m+n+1∩

Since

we have

By Lemma 3.2 and U is a-disjoint and B-bounded， there exists a partition Ln+m+3of L′n+m+2such thatThen

which is a contradiction to Lemma 3.3.

Proposition 3.4Let

Thencoasdim(X)≤2ωis not true.

ProofBy Lemma 2.5 and Proposition 3.3， coasdim(X)≤2ω is not true.

Proposition 3.5Let

Thencoasdim(X)= trasdim(X)=2ω+1.

ProofBy Proposition 3.1 and Proposition 3.4， coasdim(X) = 2ω + 1.Moreover， by Proposition 3.2 and Proposition 3.3， trasdim(X)=2ω+1.