Universal spaces for asymptotic dimension · 2016-12-16 · Universal spaces for asymptotic...

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Topology and its Applications 140 (2004) 203–225

www.elsevier.com/locate/topo

Universal spaces for asymptotic dimension

A. Dranishnikova,∗,1, M. Zarichnyib

a Department of Mathematics, University of Florida, 358 Little Hall, P.O. Box 118105,Gainesville, FL 32611-8105, USA

b Department of Mechanics and Mathematics, Lviv NationalUniversity, Universytetska 1, 79000 Lviv, Ukrain

Received 23 September 2002; received in revised form 3 March 2003

Dedicated to Jed Keesling on the occasion of his 60th birthday

Abstract

We construct a universal space for the class of proper metric spaces of bounded geometrgiven asymptotic dimension. As a consequence of this result, we establish coincidence of asymptodimension with asymptotic inductive dimension. 2003 Elsevier B.V. All rights reserved.

MSC:54F45

Keywords:Asymptotic dimension; Universal space; Asymptotic inductive dimension

1. Introduction

Asymptotic dimension asdim of a metric space was defined by Gromov for stuasymptotic invariants of discrete groups[14]. Then a successful application of tasymptotic dimension was found by Yu [19]. He proved the Higher Novikov Signaconjecture for finitely presented groupsΓ with a finite asymptotic dimension consideras metric spaces with the word metric. The word metricdS on a groupΓ depends on agenerating setS ⊂ Γ . The distancedS(x, y) in the word metric is the minimal length opresentations of the wordx−1y in the alphabetS. It turns out that the metric spaces(Γ, dS)

* Corresponding author.E-mail addresses:[email protected] (A. Dranishnikov), [email protected], [email protected]

(M. Zarichnyi).1 The first author was partially supported by NSF grant DMS-9971709. The paper was written duri

second author’s visit to the University of Florida. He thanks the Department of Mathematics for hospitality

0166-8641/$ – see front matter 2003 Elsevier B.V. All rights reserved.
doi:10.1016/j.topol.2003.07.009

204 A. Dranishnikov, M. Zarichnyi / Topology and its Applications 140 (2004) 203–225

and (Γ, dS ′) are coarsely equivalent (see Section 2 for the exact definition) for any two

mtsfinite

ps.hsrved

ofmanyhich isas. The

kinds.

-

e

-

tlyin thenall

n be

isvery

hetz–n a

finite generating setsS andS′, and hence asdim(Γ, dS) = asdim(Γ, dS ′). Thus, in the caseof a finitely generated groupΓ , one can speak about its asymptotic dimension asdiΓ

without referring to a generating set. In view of Yu’s theorem, finite-dimensionality resulare of a particular interest. In [14] Gromov proved that all the hyperbolic groups haveasymptotic dimension. In [11] finite dimensionality was established for all Coxeter grouIn [2,3] finite dimensionality theorems were proved for the fundamental group of grapof groups. In particular, it was shown that asymptotic finite dimensionality is preseunder amalgamated product and HNN-extension. In [3] an upper estimate for asdimthe fundamental group of a graph of groups is given which turns out to be exact incases. Strangely, it did not give an exact estimate in the case of the free product wseemingly the simplest case of the graph of groups. Only recently the exact formula hbeen established in [4] with the use of inductive dimension asInd introduced in [10]equality asdim= asInd is proven in the present paper by means of universal spaces.

In classical dimension theory there are many embedding theorems of differentHere we mention three:

(1) Nöbeling–Pontryagin Theorem: Every compactumX with dimX � n can be embedded inR

2n+1.(2) Bowers Theorem [6]: Every compactumX with dimX � n can be embedded in th

product ofn + 1 dendrites.(3) Lefschetz–Menger Theorem [5]: For everyn there exists a universal Menger com

pactumµn, i.e., every compactumX with dimX � n can be embedded inµn.

We are interested in the asymptotic analogsof these theorems. The interest is parmotivated by a topological approach to Yu’s theorem taken in [9]. First, we note thatlarge scale world there are direct analogs to (1)–(3) forn = 0. Here we give a descriptioof a spaceM0 analogous to the Cantor setµ0. We recall that the Cantor set is the set ofnumbers 0� x � 1 which satisfy the following property:

(∗) x can be written without use of1 in the tercimal system(based on{0,1,2}).

ThenM0 can be described as the set of allx ∈ R+ satisfying(∗). Every proper metricspaceX with asdimX = 0 and of bounded geometry (see the definition below) caembedded in the coarse sense inM0 and hence inR.

The direct analogy between classic and asymptotic dimension theories ends forn > 0.It is easy to show that the free groupF2 of two generators has dimension asdimF2 = 1.It cannot be embedded intoRN , sinceF2 has an exponential volume growth. So thereno direct asymptotic analogy to (1). An asymptotic analogy of (2) was found in [9]: Emetric space of bounded geometry can be coarsely embedded in the product ofn+1 locallyfinite trees. Since the above product of trees can be embedded in a(2n + 2)-dimensionalnonpositively curved manifold, it gives some hope for analogy in the case of (1).

The main goal of this paper is an attempt to find a coarse version of the LefscMenger theorem, i.e., to construct universal spaces for asymptotic dimension. We obtairather partial result (Theorem 3.10): For everyn there is a separable metric spaceMn with

A. Dranishnikov, M. Zarichnyi / Topology and its Applications 140 (2004) 203–225 205

asdimMn = n which is universal for proper metric spacesX with bounded geometry and

lingrlishese will

orkst for

scaleucts

t thereicallyse.

e

ity

with asdimX � n. Our universal space is neither proper nor of bounded geometry.The construction of the spaceMn is analogous to those for the Menger and Nöbe

spaces. Besides the asymptotic strategy, themain difference is that we are building oufractals not inR2n+1 as in the classical case but in the product of trees. If one accompthe local (classical) construction in the product of (finite) trees or even dendrites hget the Menger space according to Bestvina’s criterion [5]. We construct our spacesMn

out of the product of locally finite trees. For locally finite trees this construction weven better but we have problems with proving universality. Anyway, we construcevery proper metric space of bounded geometryX with asdimX � n an embeddingX ↪→ ∏n

i=0 Ti into the product ofn + 1 finite trees in such a way thatX lies in the ‘n-skeleton’M({Ti}) of

∏ni=0 Ti . This spaceM({Ti}) has the property asdimM({Ti}) = n.

Moreover, asIndM({Ti}) = n. This allows one to prove the inequality asIndX � asdimX

for proper metric spaces. For everyX the constructed spaceM({Ti}) has the Higsonproperty (see Section 4). There were hopes that the mentioned result on largeembeddings of asymptoticallyn-dimensional spaces of bounded geometry into prodof n + 1 nonpositively curved surfaces could be strengthened so thatone could take thehyperbolic planes as these manifolds. It turns out, however, (see Theorem 3.15) thais no proper metric space of finite asymptotic dimension into which every asymptot1-dimensional space of bounded geometry can be embedded in the large scale sen

2. Preliminaries

The generic metric will be denoted byd . If A is a subset of a metric spaceX andr ∈ R,thenNr(A) is defined as{x ∈ X | d(x,A) < r} if r > 0 andA \ {x ∈ X | d(x,A) < −r}otherwise. The closedr-ball centered atx ∈ X will be denoted asBr(x).

A map f :X → Y of metric spaces is called(λ, s)-Lipschitz if d(f (x), f (y)) �λd(x, y) + s for everyx, y ∈ X. The (λ,0)-Lipschitz maps are also calledλ-Lipschitz;the 1-Lipschitz maps are calledLipschitzor short. An asymptotically Lipschitzmap is amap which is(λ, s)-Lipschitz for someλ > 0, s > 0.

For a coverU of a metric spaceX, we denote byL(U) its Lebesgue number,L(U) =inf{sup{d(x,X \ U) | U ∈ U} | x ∈ X}, and for a familyU of subsets of a metric space wdenote by mesh(U) the least upper bound of the diameters of the elements ofU . A family Aof subsets of a metric space is calleduniformly boundedif there exists a numberC > 0 suchthat diam(A) < C for everyA ∈A.

For r > 0, a metric spaceX is calledr-discreteif d(a, b) � r for everya, b ∈ X, a �= b.A metric spaceX is calleddiscreteif X is r-discrete for somer > 0.

For r > 0, ther-capacityof a subsetY of a metric space is the maximal cardinalKr(Y ) of an r-discrete subset ofY . A metric spaceX is of bounded geometryif thereexists a numberr > 0 and a functionc : [0,∞) → [0,∞) such that ther-capacity of everyε-ball Bε(x) does not exceedc(ε).

A map f :X → Y between metric spaces is calleduniformly coboundedif for everyr > 0 there existsC > 0 such that for everyy ∈ Y the diameter of the setf −1(Br(y)) doesnot exceedC.


For pointsx, y ∈ X in a metric spaceX an isometric embeddingj : [0, d(x, y)] → X of

cthe

o

2

t

tegoryer

ysisace

the interval[0, d(x, y)] ⊂ R into X is called ageodesic segmentjoining pointsx andy ifj (0) = x andj (d(x, y)) = y.

A spaceX is said to be ageodesic metric spaceif for every two pointsx, y ∈ X thereis a geodesic segment joining them. A geodesic segment that connects pointsx, y in ageodesic metric space will be denoted by[x, y].

A metric space is calleduniformly arcwise connectedif for every ε > 0 there existsδ > 0 such that for every two pointsx, y ∈ X with d(x, y) < ε there exists a path inXconnectingx andy and of diameter� δ.

The (finite) products of metric spaces are always endowed with the sup-metric.Let C a decomposition of a proper metric spaceX. Define thequotient pseudo-metri

� on X by the following rule:�(x, y) is the greatest lower bound of the sums ofform

∑k−1i=1 d(x2i, x2i+1), wherex = x1, y = x2k, and for everyi = 1, . . . , k there exists

an elementCi ∈ C such that{x2i−1, x2i} ⊂ Ci . Obviously, the identity map id :(X,d) →(X,�) is short.

A mapf :X → Y between metric spaces is calledcoarsely uniformif for every C > 0there isK > 0 such that for everyx, x ′ ∈ X with d(x, x ′) < C we haved(f (x), f (x ′)) <

K. A mapf :X → Y is calledmetric properif the preimagef −1(B) is bounded for everybounded setB ⊂ Y . A map iscoarseif it is both coarse uniform and metric proper. Twmapsf,g :X → Y between metric spaces are calledbornotopic(see [17]) if there isC > 0such thatd(f (x), g(x)) < C for everyx, y ∈ X. Two metric spacesX andY arecoarseequivalentif there are coarse mapsf :X → Y , g :Y → X such that the compositionsfg

andgf are bornotopic to the corresponding identity maps.

Lemma 2.1 [8]. Let X be a geodesic metric space andf :X → Y be a coarse uniformmap. Thenf is an asymptotically Lipschitz map.

We will need the following simple result whose proof mimics that of Propositionfrom [10].

Lemma 2.2. For every proper metric spaceX and everyC > 0 there is aC-discrete subseY of X such thatd(x,Y ) � C for everyx ∈ X.

A mapf :X → Y of metric spaces is called alarge scale embedding(see [17]) if thereexist increasing functionsϕ1, ϕ2 : [0,∞) → [0,∞) with limt→∞ ϕ1(t) = limt→∞ ϕ2(t) =∞ such that

ϕ1(d(x, y)

)� d

(f (x), f (y)

)� ϕ2

(d(x, y)

)for everyx, y ∈ X. Since large scale embeddings are embeddings in the coarse ca[18] we will call them alsocoarse embeddings. In the literature they appeared first undthe nameuniform embeddings[14, p. 211], which was used in the classical (local) anallong before for the classical uniform embeddings. It is easy to see that every metric spis coarsely equivalent to its image under a coarse embedding.


2.1. Asymptotic dimension

n

ry

s.areandard

The notion of asymptotic dimension was introduced by Gromov [14].

Definition 2.3. Theasymptotic dimensionof a metric spaceX does not exceedn (writtenasdimX � n) if for everyD > 0 there exists a uniformly bounded coverU of X such thatU = U0 ∪ · · · ∪Un, where allU i areD-disjoint.

(Here a family of subsets of a metric space is said to beD-disjoint if for every its distinctelements,A andA′, we haved(A,A′) = inf{d(a, a′) | a ∈ A, a′ ∈ A′} > D.)

A piecewise-euclideann-dimensional complex of meshD is a complex which is a unioof isometric copies of the standardn-dimensional simplex of meshD in R

n+1.

Lemma 2.4. For a proper metric spaceX the following two conditions are equivalent:

(1) asdimX � n;(2) for anyD > 0 there is a uniformly cobounded short proper map ofX into a piecewise-

euclidean complex of meshD and dimensionn.

Proof. See [14]. �A family of metric spacesXα satisfies the inequality asdimXα � n uniformly(see [2]) if

for arbitrary largeD > 0 there existsR > 0 andR-boundedD-disjoint familiesU0α, . . . ,Un

α

such that⋃n

i=0U iα is a cover ofXα .

Theorem 2.5 [2]. Assume thatX = ⋃α Xα andasdimXα � n uniformly. Suppose that fo

arbitrary large numberR there isYR ⊂ X with asdimYR � n and such that the famil{Xα \ YR} is R-disjoint. ThenasdimX � n.

3. Asymptotic embeddings into product of trees

3.1. Trees

A geodesic metric spaceT is called anR-tree(a real tree) if:

(1) every two points inT are connected by a unique geodesic segment;(2) if [x, y] ∩ [y, z] = {y} then[x, y] ∪ [y, z] = [x, z] for all x, y, z ∈ T .

Recall that in this definition[a, b] stands for a geodesic segment connectinga andb. Wewill also use self-explaining notations[a, b) and(a, b) for (half)open geodesic segment

Every tree (connected acyclic graph) is anR-tree. We assume that the graphsendowed with a geodesic metric whose restriction to every edge is isometric to the stunit segment.

A (half)open segment inT is free if it is an open subset ofT .


pect

f

e

m

s. For

e

l

Fig. 1.

Themeshof T is the greatest lower bound of the diameters of the maximal (with resto the inclusion) free (half)open segments inT .

Definition 3.1. An R-treeT is said to beregular if there exists a sequence ofR-treesT = T0 ⊃ T1 ⊃ T2 ⊃ · · · with the following properties (see Fig. 1):

(1)⋂∞

i=0 Ti = ∅;(2) for every i, the setTi \ Ti+1 is a disjoint union of a uniformly bounded family o

maximal free inTi half-open segments;(3) D0 < D1 < D2 < · · · and limi→∞ Di = ∞, where Di denotes the mesh of th

R-treeTi .

If in the above definitionDi = 2i , we say thatT is binary regular.Suppose thatT (0), . . . , T (n) is a sequence of regularR-trees. For everyj , a sequence

of R-treesT (j) = T(j)0 ⊃ T

(j)1 ⊃ T

(j)2 ⊃ · · · is given such that properties (1)–(3) fro

Definition 3.1 are satisfied withTi = T(j)i . For everyi andj denote byr(j)

i :T (j) → T(j)i

the retraction that maps every componentC of the setT (j) \ T(j)i onto its boundary∂C

(which is a singleton). We will refer to these retractions as the canonical retractionk � i we haver(j)

i = r(j)ki r

(j)k . Let

Mj ={

(x0, . . . , xn) ∈n∏

i=0

T (i) | there existsi such thatri (xi) ∈ ∂(T

(i)j−1 \ T

(i)j

)}andM = M(T (0), . . . , T (n)) = ⋂∞

j=1 Mj .

Denote byMn the class of spaces of the formM(T (0), . . . , T (n)) for some sequencT (0), . . . , T (n) of regularR-trees.

Suppose thatT , S are regularR-trees given with filtrations(Tj )∞j=0, (Sj )

∞j=0 satisfying

the conditions from Definition 3.1. We denote byrj :T → Tj , �j :S → Sj the canonicaretractions.

Definition 3.2. A map f :T → S is said to beregular if the following conditions aresatisfied:


(i) f (Tj ) ⊂ Sj for everyj ;

se

[9].

(ii) f (rj (∂(Tj−1 \ Tj ))) ⊂ �j (∂(Sj−1 \ Sj )) for everyj .

The constructionM(T (0), . . . , T (n)) is functorial in the following sense. Suppothat T i , Si are regularR-trees given with filtrations(T i

j )∞j=0, (Sij )

∞j=0, i ∈ {0,1, . . . , n}

satisfying the conditions from Definition 3.1. We denote byrij :T i → T i

j , �ij :Si → Si

j the

canonical retractions. Suppose thatfi :T i → Si are regular maps.

Lemma 3.3. Under these conditions

(f0 × · · · × fn)(M

(T (0), . . . , T (n)

)) ⊂ M(S(0), . . . , S(n)

).

Proof. Obvious. �Proposition 3.4. For everyM ∈Mn we haveasdimM � n.

Proof. We apply Lemma 2.4. Note that for everyj the map

rj =n∏

i=0

r(i)j :

n∏i=0

T (i) →n∏

i=0

T (i)

mapsM onto ann-dimensional piecewise-euclidean polyhedron of meshDj . Sincerj isuniformly cobounded, we are done.�Theorem 3.5. For every proper metric spaceX of asdimX � n there exist locally finitebinary regular treesT 0, . . . , T n such thatX is large scale embeddable intoM(T 0, . . . ,

T n).

The proof is a modification of the corresponding result from [9].

Proposition 3.6. LetX be a proper metric space with base pointx0 andasdimX � n. Thenthere exists a sequence(Uk)

∞k=1 of open covers ofX such that everyUk splits into the union

Uk = U0k ∪ · · · ∪ Un

k of dk-disjoint families and the following conditions are satisfied:

(1) L(Uk) > dk and for everyi and everyU ∈ U ik we haveN−dk (U) �= ∅;

(2) dk+1 = 22k+2mk , wheremk is the mesh ofUk;(3) for everym ∈ N and for everyi there isk ∈ N and an elementU ∈ U i

k such thatNm(x0) ⊂ U ;

(4) for everyk, l, k < l and everyU ∈ U ik , V ∈ Ui

l , if U �⊂ V , thend(U,V ) � dk/2.

Proof. We start with repeating the construction from the proof of Proposition 1 fromWe proceed by induction. LetU0 be an open cover ofX which is d0-discrete withd0 > 2.We enumerate the partitionU0 = U 0

0 ∪ · · · ∪ U n0 in such a way thatd(x0,X \ U) > d0 for

someU ∈ U 00 .


Assume that familiesUk are constructed for allk � l. Define dl+1 = 2l+2ml anda

e ofs

consider a uniformly bounded cover�Ul+1 with L(�Ul+1) > 2dl+1 and such that there issplitting �Ul+1 = �U 0

l+1∪· · ·∪ �U nl+1, where�U i

l+1 aredl+1-disjoint andN−2dl+1(U) �= ∅ for all

U ∈ �Ul+1. The families�U 0l+1, . . . ,

�U nl+1 are enumerated in such a way thatd(x0,X \ U) >

2dl+1 for someU ∈ �U il+1, wherei = l + 1 modn + 1. For everyU ∈ �U i

l+1 let

U = U \ {N4(V ) | V ∈ U i

k, k � l, V �⊂ U}

andU il+1 = {U | U ∈ �U i

l+1}.It is proved in [9] that the sequence(Uk) satisfies the following properties:

(1) L(Uk) > dk and for everyi and for everyU ∈ U ik we haveN−dk (U) �= ∅;

(2) dk+1 = 2kmk , wheremk is the mesh ofUk ;(3) for everym ∈ N and for everyi there isk ∈ N and an elementU ∈ U i

k such thatNm(x0) ⊂ U ;

(4) for everyU ∈ U ik , V ∈ U i

l , U �⊂ V we haved(U,V ) � 4.

We are going to modify the sequence(Uk)∞k=1. Passing, if necessary, to a subsequenc

the sequence(Uk)∞k=1 we may assume that the sequence(Uk)

∞k=1 itself satisfies condition

(1), (3), (4) and the following condition:

(2∗) dk+1 � 24k+6mk .

Define, by induction, numbersdk and familiesU ik . Let d0 = d0 andU0 = U0 and suppose

dj andUj have been defined for everyj < k. For everyU ∈ U ik apply induction byp to

define the setsU(p). Let U(0) = U and suppose setsU(p) have been defined for allp < q ,for someq � k. Let

U(q) = U(q − 1) ∪⋃{

V ∈ U iq | d(

U(q − 1),V)< dq/2

}.

Then, by the definition,U = U(k − 1) andU ik = {U | U ∈ U i

k}.We are going to verify conditions (1)–(4) (withUk , dk, andmk replaced byUk , dk, and

mk , respectively). Conditions (1) and (3) are obvious.Denote bymk the mesh ofU i

k and let

dk = dk − 2k+2mk−1. (3.1)

Note that for everyU, V ∈ U ik , U �= V we have

d(U , V

) = d(U(k − 1), V (k − 1)

)� d

(U(k − 2), V (k − 2)

) − dk−1 − 2mk−1

� d(U(k − 2), V (k − 2)

) − 3mk−1

� d(U(k − 3), V (k − 3)

) − 3mk−1 − 3mk−2...

� d(U,V ) − 3k−1∑i=0

mi � dk − 3kmk−1 � dk,

and therefore the familyU ik is dk-disjoint.


Condition (2). LetU ∈ U i , thenU = U(k − 1) and

r

)

ily

ums

k

diamU � diamU(k − 2) + 2mk−1 + dk−1 � diamU(k − 2) + 3mk−1...

� diamU(0) + 3k−1∑j=0

mj � mk + 3kmk−1 � mk + 2k+2mk−1

(here we used an obvious equalitymj � dj , for everyj ), whence

mk � mk + 2k+2mk−1. (3.2)

Then

mk � mk

k−1∑j=0

2j (k+2) � mk22k+3

and using condition (2∗) we obtain

dk = dk − 2k+2mk−1 � dk − 22k+3mk−1

� 24k+6mk−1 − 22k+3mk−1 � 24k+5mk−1

� 22k+2mk−1. (3.3)

Let us verify condition (4). Suppose the contrary. Then there existU ∈ U ik , V ∈ U i

l ,U �⊂ V , andk � l such thatd(U , V ) < dk/2. We also suppose thatl is the minimal numbewith that property. Sinced(U, V ) < dk/2, there existsx ∈ V such thatd(x, U) < dk/2. Letj � 0 be the minimal integer with the propertyx ∈ V (j). It follows from the definition ofU i

l that j � l − 1. Sincex ∈ V (j) \ V (j − 1), it follows from the definition ofV (j) thatthere existsW ∈ U i

j such thatx ∈ W . Sincej < l, this contradicts to the choice ofl. �For every elementU ∈ Ui denote byψ(U) the minimal (with respect to inclusion

V ∈ Ui such thatU is a proper subset ofψ(U).

Proposition 3.7. LetX be a proper metric space and(Uk) a sequence of open covers ofX

satisfying properties(1)–(4)from Proposition3.6. LetU ∈ Uk and� be the quotient metricon U with respect to the partition ofU into singletons and the elements of the famV = ψ−1(U). For everyV ∈ V ∩ U i

j , W ∈ V ∩U il , wherej � l, we have�(V,W) � 22j .

Proof. The assertion is obvious forj = 0. LetV ∈ V ∩U ij , W ∈ V ∩U i

l , where 1� j � l.

Then�(V,W) is the greatest lower bound of the sums of the form∑p−1

i=1 d(x2i, x2i+1),wherex ∈ V , x2p ∈ W , and for everyi = 1, . . . , p there exists an elementCi ∈ ψ−1(U)

such that{x2i−1, x2i} ⊂ Ci . Without loss of generality, we may assume that in the saboveCi ∈ ψ−1(U) \ Vj for all j < p. Then (p − 2)mj−1 + (p − 1)dj−1 � dj andtherefore, by condition (2) of Proposition 3.6,

p − 1 � dj − dj−1

mj−1 + dj−1� dj

4mj−1� 22j

and∑p−1

i=1 d(x2i, x2i+1) � (p − 1)d0 � 22j . Hence�(V,W) � 22j . �


Proposition 3.8. Let X, U , and � be as in Proposition3.7. There existsx ∈ U with

i-

.

9]w

ence

,

�(x,X \ U) � 22k.

Proof. By condition (1) of Proposition 3.6, there existsx ∈ U with d(x, ∂U) � dk.Arguing like in the proof of Proposition 3.7 we conclude that�(x,X \ U) � 22k. �

Fix a sequence(Uk) of open covers ofX satisfying properties (1)–(4) from Propostion 3.6. GivenU ∈ U i

k define a Lipschitz mapfU : �U → IU = [0,22k ] by the followingprocedure. Let� denote the pseudometric generated by the partition ofU into singletonsand the elements of the familyψ−1(U) and letVj = ({V ∈ U i

l | j � l < k}) ∩ ψ−1(U).

We are going to define by induction with respect toj mapsθj , θj : ∂U ∪ ⋃Vj → IU .

Define the mapθ1 : ∂U ∪ ⋃Vk−1 → IU by the formulaθ1(x) = 2−k�(x,X \ U). Let

θ1(x) = 2k−1[2−k+1θ1(x)]. The mapθ1 is a 2−k+1-Lipschitz map intoIU defined on theclosed subset∂U ∪ ⋃

Vk−1 of �U .Suppose the mapsθj , θj are defined for allj , j0 < j < k and the mapsθj are 2−j -

Lipschitz. By the theorem on extension of Lipschitz maps (see [8]) there exists a 2−j0+1-Lipschitz extension ofθj0−1 onto the set∂U ∪ ⋃

Vj0. Let θi0(x) = 2i0[2−i0θi0(x)].As a final result, we obtain a 1-Lipschitz map

θ0 : ∂U ∪⋃

V0 = ∂U ∪⋃

ψ−1(U) → IU .

Denote byfU its 1-Lipschitz extension onto�U .We will follow [9] and construct a locally finite treeT i by the following procedure

Assign to everyU ∈ U ik, k = 0,1, . . . , a line segmentIU = [0,2k]. The treeT i will be the

quotient space of the disjoint union⊔{IU | U ∈ ⋃∞

k=0U ik} with respect to the following

equivalence relation∼: supposeU ∈ U ik , V ∈ U i

l , wherek < l andU ∈ ψ−1(V ). Then0 ∈ IU is identified withfV (∂U) ∈ IV . We argue like in the proof of Theorem 3 from [to show that the quotient space

⊔{IU | U ∈ ⋃∞k=0U i

k}/∼ is a tree. We are going to shothatT I is locally finite. Assume the contrary and lety be a vertex of infinite order inT i .Then there exists an infinite familyV ⊂ U i such thaty = q(0U) for everyU ∈ V . Since thesetψ−1(U) is finite for everyU ∈ U , we conclude that there exists an increasing sequ(k(j))∞j=1 of positive integers and setsUk(j) ∈ U i

k(j) such thatUk(j) ∈ ψ−1Uk(j+1) andfUk(j+1)

(Uk(j)) = 0. By conditions (3) and (4), there existsj > 1 such thatx0 ∈ Uk(j)

and Uk(1) ⊂ Uk(j). Then, by condition (3),Uk(j+1) ⊃ Nmk(j)+22k(j)+2dk(x0) and, since

diam(Uk(j)) � mk(j), by condition (3),d(Uk(j),X \ Uk(j+1)) � 22k(j)+2dk and, thereforeby Proposition 3.8,�(Uk(j),X \ Uk(j+1)) � 2k(j). HencefUk(j+1)

(Uk(j)) � 1 �= 0 and weobtain a contradiction.

Let q :⊔{IU | U ∈ ⋃∞

k=0U ik} → T i be the quotient map. Define the mapf = f i :X →

T i as follows. Letx ∈ X. By condition (3) of Proposition 3.6, there exists minimalk suchthat x ∈ U for someU ∈ U i

k. Putf i(x) = qfU(x). It can be easily seen thatf i is well-defined and is Lipschitz.

The diagonal mapf = (f i)ni=0 :X → ∏ni=0 T i is Lipschitz. Show thatf is a large-

scale embedding. Assuming the contrary we can find sequences(xl) and(yl) in X suchthat d(xl, yl) → ∞ while d(f (xl), f (yl)) < C for someC > 0. Let k be a positive


integer such that 2k > C. There exists an integerl such thatd(x, y) > mk. There exists

totic

nit

ey

ly3, wee

i ∈ {0, . . . , n} such thatxl ∈ U andd(x,X \ U) > dk for someU ∈ U ik . ThenfU (x) =

2k, by Proposition 3.8. Sinced(x, y) > mk , we see thaty /∈ U and fU (y) = 0. Thus,d(f (x), f (y)) � 2k > C, which gives a contradiction.

Note that everyT i is a regular tree. Indeed, for everyj denote byT ij the subspace

q(⊔{IU | U ∈ ⋃∞

k=j U ik}). Let us verify conditions (1)–(3) of Definition 3.1 (withT and

Tj replaced byT i andT ij , respectively). Condition (1) is obvious. The setT i

j+1 \ T ij is

a disjoint union of isometric copies of the half-open segment(0,2j+1] and is thereforeuniformly bounded. To verify condition (3) note that the mesh of the treeT i

j is 2j , by theconstruction.

The treesT 0, . . . , T n together with filtrationsT i0 ⊃ T i

1 ⊃ T i2 ⊃ · · · determine the

subspaceM(T 0, . . . , T n) of∏n

i=1 T i . We are going to show thatf (X) ⊂ M(T 0, . . . , T n).Supposex ∈ X. For everyk = 0,1, . . . there isi(k) ∈ {0,1, . . . , n} such thatx ∈ U for someU ∈ U i(k)

k . Thenri(k)k+1(f (x)) ∈ ∂T

i(k)k+1 \ T

i(k)k , i.e.,f (x) ∈ Mk+1. Therefore,f (x) ∈ Mk+1

for everyk, i.e.,f (x) ∈ M(T 0, . . . , T n).

3.2. Universal space

A class of metric spacesC is said to be universal for a class of metric spacesD if foreveryD ∈D there is a large scale embedding ofC into someC ∈ C.

Theorem 3.5 can be reformulated as follows.

Theorem 3.9. The classMn is universal for the class of proper metric spaces of asympdimension� n.

A metric spaceX is said to beuniversalfor a class of metric spacesD if for everyD ∈D there is a large scale embedding ofD into X.

Construct anR-treeT by the following procedure. Define inductively a sequence(Ti)

of R-trees and retractionsri :Ti → T0. Let T0 = R+. Denote byT1 an R-tree obtainedby attaching to every integer point inR+ a countable set of isometric copies of the usegment[0,1]. Denote byr1 a retraction ofT1 onto T0 that sends every componentC

which is constant on every component of the complementT1 \ T0. Suppose thatR-treesTi

and retractionsri are defined for alli < j . Denote byS the subtreer−1j−1([0,2j ]) with base

point 2j . TheR-treeTj is obtained fromTj−1 by attaching to every point of the formk2j ,k ∈ N, of R+ ⊂ Tj−1 a countable family of isometric copies ofS by the base point. Denotby rj a retraction ofTj ontoT0 that sends every componentC which is constant on evercomponent of the complementTj \ T0.

Let T = ⋃∞i=1 Ti . It is easy to see thatT is a regularR-tree that contains every local

finite binary regular tree so that the inclusion preserves the filtration. By Lemma 3.conclude that for every sequenceT 0, T 1, . . . , T n of locally finite binary regular trees whave

M(T 0, T 1, . . . , T n

) ⊂ Mn = M(S0, S1, . . . , Sn

),

whereSi is isomorphic toT for everyi = 0,1, . . . , n.


We therefore obtain the following:

hh

m 0’s

ily

Theorem 3.10. There exists a separable metric spaceMn with asdimMn = n universal forthe class of proper metric spacesX with asdimX � n.

For the spaces of asymptotic dimension zero the result can be improved.

Theorem 3.11. There exists a proper metric spaceM0 of bounded geometry witasdimM0 = 0 universal for the class of proper metric spacesX of bounded geometry witasdimX � 0.

Proof. Let M0 denote the set of integers whose tercimal expansion consists only froand 1’s. Obviously, asdimM0 = 0 andM0 is of bounded geometry. For everyk ∈ N the setM0 can be represented as the union of 3k-discrete family of the intervals of length 3k inthe set of natural numbers.

Let X be a proper metric space of bounded geometry with asdimX � 0. Let x0 ∈ X

be a base point. There exists a sequence(Uk) of uniformly bounded covers ofX with thefollowing properties:

(1) Uk refinesUk+1;(2) Uk is k-discrete;(3)

⋃{U ∈ ⋃k Uk | x0 ∈ U} = X.

ForU ∈ Uk , denote byψ(U) the uniqueV ∈ Uk+1 that containsU .Since X is of bounded geometry, for everyk ∈ N there existsCk > 0 such that

|ψ−1(U)| � Ck for everyU ∈ Uk+1.Using the mentioned decomposition ofM0 into the union of intervals, we can eas

construct a sequence of covers(Vk) satisfying the properties:

(4) Vk refinesVk+1;(5) Vk is k-discrete;(6) |ψ−1(V )| > Ck for everyV ∈ Vk+1 (as above, for everyW ∈ Vk by ψ(V ) we denote

the unique element ofVk+1 that containsV ).

For everyU ∈ U1 andV ∈ V1 denote bygUV arbitrary constant map fromU to V .Suppose that for everyi � k and everyU ∈ Ui andV ∈ Vi a mapgUV :U → V is defined.Given U ∈ Uk+1, V ∈ Vk+1, consider arbitrary injective mapα :ψ−1(U) → ψ−1(V ).DefinegUV :U → V as follows:gUV |W = gWα(W) for everyW ∈ ψ−1(U).

Now we are ready to define a mapf :X → M0. For everyk ∈ N denote byUk the uniqueelement ofUk that containsx0. By induction, define a sequence of mapsfk :Uk → M0 withthe following properties:

(1′) fk+1|Uk = fk , for everyk;(2′) f (Uk) is contained in an element ofUk+1.


Let f1 = gU1V :U1 → V ⊂ M0, for someV ∈ V1. Suppose thatfi are defined for every

etric

s of

i � k. Let V ′ ∈ Vk be such thatfk(Uk) ⊂ V ′. Denote byα :ψ−1(ψ(Uk)) → ψ−1(ψ(V ′))an embedding such thatα(Uk) = V ′. Definefk+1 by the conditionsfk+1|Uk = fk andfk+1|W = gWα(W) for everyW ∈ ψ−1(ψ(Uk)) \ {Uk}.

By condition (1′), the sequence of maps(fk) uniquely determines a mapf :X → M0.Using properties (1)–(6) it is easy to see thatf is a coarse embedding.�3.3. Nonexistence results

There is no counterpart of Theorem 3.11 in higher dimensions.

Theorem 3.12. There is no proper metric space universal for the class of proper mspacesY with asdimY � 1.

Proof. Suppose the contrary and letX be a proper metric space universal for the clasproper metric spacesY with asdimY � 1. Letx0 ∈ X be a base point. Denote byα(r) the1-capacity of the ballNr(x0), i.e., the numberK1(Nr(x0)). Note that for everyx ∈ X wehave

K1(Nr(x)

)� K1

(Nr+d(x,x0)(x0)

)� α

(r + d(x, x0)

).

Let T be a locally finite tree with a base pointt0 and the index functionϕ(r) whichwe specify later. Suppose thatf :T → X is a coarse embedding. SinceT is a geodesicspace, the mapf is proper and(λ, s)-Lipschitz for someλ > 0, s � 0. There existsa > 0 such thatd(x, y) � a implies d(f (x), f (y)) � 1. There existλ′, s′ > 0 such thatf (Nr(t0)) ⊂ Nλ′r+s ′(x0). Therefore

Ka

(Nr(t0)

)� K1

(Nλ′r+s ′(x0)

) = α(λ′r + s′).

Now suppose thatϕ :N → N satisfies the following property: for everyn ∈ N thereexistsm ∈ N with α(nr + n) < ϕ(r) for all r � m. DefineT as follows: LetT0 = R+.For everym ∈ N denote byTm a tree obtained by attaching to every integer pointn ∈ R+,n � m, ϕ(n) copies of the segment[0,m] by its endpoints. LetT = ⋃{Tn | n ∈ N ∪ {0}}.

Now

Ka

(Nr(t0)

)� α

(λ′r + s′) � α(nr + n),

wheren ∈ N, n � max{λ′, s′}. Let m ∈ N, m � a. Then for everyr � 2m + 1 we obtain

Ka

(Nr(t0)

)� Km

(Nr(t0)

)� ϕ

([r]),a contradiction with the choice ofϕ. �

We show that there is no universal proper metric space of asymptotic dimensionn � 1in the class of proper metric spaces of bounded geometry with asdim� n.

We fix naturaln � 1.For naturalk, denote byX(1, k) the set{

(x0, . . . , xn) ∈ [0, k]n+1 | xi /∈ Z for at most onei = 0, . . . , n}.

For naturalm, let X(m,k) = {mx | x ∈ X(1, k)}.


Denote byV the set of coversV of X(1, k) such thatV = ⋃ni=0V i , whereVi are 3-

ry

s a

rh

t

h

overs

discrete,i = 0,1, . . . , n. For V ∈ V, denote byµV (k) the maximal 1-capacity ofV ∈ Vand letµk = min{µV (k) | V ∈ V}.

Givenλ > 0, we say that a subsetA of a metric spaceX is aλ-componentof X if A isa maximal (with respect to inclusion) subset ofB with respect to the property that evetwo points ofA can be connected by aλ′-chain, for someλ′ < λ.

Lemma 3.13. µk → ∞ ask → ∞.

Proof. Assume the opposite, i.e., there exists a constantS such thatµk � S, for all k.GivenV ∈ V with V∞

i=0 = ⋃V i , whereV i are 3-discrete,i = 0, . . . , n, for everyV ∈ V ,

denote byV the family of < 3-connected components ofV . Let Vi = ⋃{V | V ∈ Vi},i = 0, . . . , n, andV = ⋃n

i=0 Vi . From our assumption it easily follows that there existconstantC > 0 such that for everyW ∈ V we have diam(W) � C.

Now, for everyW ∈ V defineW as the 1-neighborhood ofW in [0, k]n+1. The families{W | W ∈ Vi}, i = 0,1, . . . , n, are 1-discrete, of mesh� C, and together form a coveof [0, k]n+1. Then the families{(1/k)W | W ∈ Vi}, i = 0,1, . . . , n, are discrete, of mes� C/k, and together form a cover of[0,1]n+1. Because of the arbitrariness ofk, we obtaina contradiction with Ostrand’s characterization of the covering dimension [16].�

Passing to the images of spaces under the homothety with coefficientm and centered athe origin, we derive from Lemma 3.13 the following statement.

Proposition 3.14. For everym,s ∈ N there existsr(m, s) ∈ N such that for anyr � r(m, s)

the following holds: given a coverV of X(m, r) that splits into union ofn + 1 3m-disjointfamilies, there isV ∈ V with Km(V ) � s.

Theorem 3.15. There is no proper metric spaceY of bounded geometry,asdimY = n,with the following property: for every proper metric spaceX of bounded geometry witasdimX = n, there exists a large scale embedding ofX into Y .

Proof. Suppose the opposite and letY be such a space. There exists a sequence of c(Uk) of Y with the following properties:

(1) Uk is uniformly bounded;(2) Uk = ⋃n

i=0U ik , whereU i

k aredk-discrete withdk � k2.

SinceY is of bounded geometry,ck = max{K1(U) | U ∈ Uk} < ∞.For everym ∈ N, chooseRm so that for anyr � r(m,Rm) the following holds: given a

coverV of X(m, r) that splits into union of two 3m-disjoint families, there isV ∈ V withKm(V ) �

∑mi=1 ci + 1.

Define a spaceX as follows: attach toR+ at every pointm ∈ N ⊂ R+ a copy of spaceX(m,Rm), with the geodesic metric on the adjunction space. Note that asdimX = 1.


Suppose thatf :X → Y is a large scale embedding, i.e., there exist monotone,

e

t

rsely

, it is

l

increasing to infinity functionsϕ1, ϕ2 :R+ → R+ such that

ϕ1(d(x, y)

)� d

(f (x), f (y)

)� ϕ2

(d(x, y)

)for everyx, y ∈ X.

SinceX is a geodesic space, by Lemma 2.1 one may chooseϕ2 linear,ϕ2(t) = at + b,a > 0. There existsm0 ∈ N such thatd(f (x), f (y)) � 1 asd(x, y) � m0.

For anym � m0, find minimalk(m) ∈ N such that the preimagef −1(A) of anydk(m)-discrete family is 3m-discrete. We have forx ∈ A, y ∈ B, A,B ∈A, A �= B:

dk(m) � d(f (x), f (y)

)� ϕ2

(d(x, y)

) = ad(x, y) + b

whenced(x, y) � (dk(m) − b)/a and if (dk(m) − b)/a � 3m, we are done. Therefore, thminimalk(m) is found from the inequalitydk(m) � 3ma + b.

We see that there existsm1 ∈ N such thatdm � (3m − b)/a for everym � m1.Now considerm > max{m0,m1}. The coverVm splits into the union ofdm-disjoint

familiesV im, i = 0,1, . . . , n.

The cover f −1(Vm) of X(m,Rm) splits into the union of 3m-disjoint familiesf −1(V i

m), i = 0, . . . , n. By Proposition 3.14, there exists an elementV ∈ Vm such thatKm(f −1(V )) �

∑mi=1 ci + 1 > cm. As m > m0, we see thatK1(V ) > cm, a contradic-

tion. �

4. Higson property

A metric spaceX with asdimX � n is said to satisfy theHigson propertyif there existsC > 0 such that for everyD > 0 there exists a coverU of X with mesh(U) < CD andsuch thatU = U0 ∪ · · · ∪ Un, whereU0, . . . ,Un areD-disjoint. Equivalently,X has theHigson property if there exists a sequence(Uk) of uniformly bounded covers ofX andC > 0 such that mesh(Uk) → ∞ and every ball inX of radius� C mesh(Uk) intersects amostn + 1 elements of(Uk). The latter is a large scale analog of theNagata dimensionN-dim defined by Assouad [1] as follows: for a metric spaceX we have N-dimX � n ifthere is a constantC such that, for eachr > 0, there is an open coverU(r) of X by sets ofdiameter� Cr such that each open ball of radiusr meets at mostn + 1 members ofU(r).

Proposition 4.1. Every proper metric space of finite asymptotic dimension is coaequivalent to a proper metric space that satisfies the Higson property.

Proof. By Theorem 3.5, there is a large scale embedding ofX into a space of the formM(T 0, . . . , T n), for a sequence of regular locally finite treesT 0, . . . , T n together withfiltrationsT i

0 ⊃ T i1 ⊃ T i

2 ⊃ · · · . Since the Higson property is preserved by subspacessufficient to show that the spaceM(T 0, . . . , T n) satisfies it.

Note that the setKj = rj (M(T 0, . . . , T n)) is a cubic piecewise-euclideann-dimensionacomplex with mesh(Kj ) = Dj = 2j . Passing to the second barycentric subdivision ofKj


we are able to produceDk/3-discrete familiesAl with mesh(Al ) � 2Dk/3 � Dk , l ∈

ce4.1,

t

(see

{0, . . . , n}. LetU lk = {r−1

k (A) | A ∈ Al}, then

mesh(U l

k

)� 2(Dk + Dk−1 + · · · + D0) � 4Dk.

Since the maprk is short, the familyU lk is alsoDk/3-discrete. �

A metric spaceX satisfies then-dimensional Nagata property(briefly denoted by(Pn))if for every r > 0, everyx ∈ X, and everyy1, . . . , yn+2 such thatd(yi,Nr(x)) < 2r forevery i = 1, . . . , n + 2, there existi, j ∈ {1, . . . , n + 2}, i �= j , such thatd(yi, yj ) < 2r

(see [15]).

Theorem 4.2. For a proper metric spaceX the following are equivalent:

(1) asdimX � n;(2) X is large scale equivalent to a metric space with property(Pn).

Proof. (1) ⇒ (2) By Proposition 4.1,X is coarsely equivalent to a proper metric spawith the Higson property. Moreover, as we can see from the proof of Propositionwithout loss of generality, we may assume that there exists a sequence(Uk) of uniformlybounded open covers ofX andC > 0 such that mesh(Uk) = 2k andUk splits into the unionof n + 1 families that areC2k-discrete. Equivalently, every ball of radiusC2k−1 intersectsat mostn + 1 element of the coverUk . By Lemma 2.2, there exists a 1-discrete subseX′of X which is coarsely equivalent toX. For everyr ∈ (0,∞), put

U ′(r) ={{{x} | x ∈ X′} whenever 0< r < 1,{

U ∩ X′ | U ∈ U[r]}

wheneverr � 1,

andC′ = C/4. Then every ball inX′ of radius� C′r intersects at mostn + 1 elementof the coverU ′(r), r ∈ (0,∞). The latter means that N-dimX′ � n. It is proved in [1,Proposition 2.2] that there exists a metricδ on X′ such thatδ satisfies(Pn) and δp isLipschitz equivalent to the original metric,d , onX′. This means that there existA,B > 0such that

Aδ(x, y)p � d(x, y) � Bδ(x, y)p, x, y ∈ X′,

i.e., the spaces(X′, d) and(X′, δ) are coarsely equivalent.(2) ⇒ (1) The proof is completely parallel to that in the case of dimension dim

[15]). Let X be a proper metric space whose metric,d , satisfies property(Pn). GivenD > 0, find a maximal (with respect to inclusion) setY ⊂ X which is 2D-discrete. ThenU = {N2D(y) | y ∈ Y } is a uniformly bounded cover ofX.

Show that for everyx ∈ X the setND(x) intersects at mostn+1 elements ofU . Indeed,suppose the contrary; then there arey1, . . . , yn+2 ∈ Y such thatND(x) ∩ N2D(yi) �= ∅ foreveryi = 1, . . . , n + 2. Property(Pn) implies that there existi, j ∈ {1, . . . , n + 2}, i �= j ,such thatd(yi, yj ) < 2D. This is a contradiction with the choice ofY . Because of thearbitrariness ofD, asdimX � n. �


5. Coincidence of asymptotic dimensions

ry

ded

pace

e

d

y

ed

torhe

thelass

r

5.1. Higson compactification and Higson corona

Let ϕ :X → R be a function defined on a metric spaceX. For everyx ∈ X and everyr > 0 letVr(x) = sup{|ϕ(y) − ϕ(x)| | y ∈ Nr(x)}. A functionϕ is calledslowly oscillatingwhenever for everyr > 0 we haveVr(x) → 0 asx → ∞ (the latter means that for eveε > 0 there exists a compact subspaceK ⊂ X such that|Vr(x)| < ε for all x ∈ X \ K). Let�X be the compactification ofX that corresponds to the family of all continuous bounslowly oscillating functions. TheHigson coronaof X is the remainderνX = �X \ X of thiscompactification.

It is known that the Higson corona is a functor from the category of proper metric sand coarse maps into the category of compact Hausdorff spaces. In particular, ifX ⊂ Y ,thenνX ⊂ νY .

For any subsetA of X we denote byA′ its trace onνX, i.e., the intersection of thclosure ofA in �X with νX. Obviously, the setA′ coincides with the Higson coronaνA.

5.2. Asymptotic inductive dimensions

The notion of asymptotic inductive dimension asInd was introduced in [10].Recall (see, e.g., [12]) that a closed subsetC of a topological spaceX is a separator

between disjoint subsetsA,B ⊂ X if X \ C = U ∪ V , whereU,V are open subsets inX,U ∩ V = ∅, A ⊂ U , V ⊂ B.

Let X be a proper metric space. A subsetW ⊂ X is called anasymptotic neighborhooof a subsetA ⊂ X if lim r→∞ d(X \Nr(x0),X \W) = ∞. Two setsA,B in a metric spaceareasymptotically disjointif lim r→∞ d(A \Nr(x0),B \Nr(x0)) = ∞. In other words, twosets are asymptotically disjoint if the tracesA′, B ′ onνX are disjoint.

A subsetC of a metric spaceX is an asymptotic separatorbetween asymptoticalldisjoint subsetsA1,A2 ⊂ X if the traceC′ is a separator inνX betweenA′

1 andA′2.

By the definition, asIndX = −1 if and only if X is bounded. Suppose we have definthe class of all proper metric spacesY with asIndY � n− 1. Then asIndX � n if and onlyif for every asymptotically disjoint subsetsA1,A2 ⊂ X there exists an asymptotic separaC betweenA1 andA2 with asIndC � n − 1. The dimension function asInd is called tasymptotic inductive dimension.

We can similarly define the small inductive asymptotic dimension asind. Bydefinition, asindX = −1 if and only if X is bounded. Suppose we have defined the cof all proper metric spacesY with asindY � n − 1. Then asindX � n if and only if foreveryA ⊂ X and everyx ∈ νX \A′ there exists a subsetC of X such thatC′ is a separatobetweenA′ and{x} in νX and asindC � n − 1.

Proposition 5.1. For every metric spaceX we haveasindX � asIndX.

Proof. Induction by asIndX. By the definition, the properties asindX = −1 and asIndX =−1 are equivalent. Suppose asIndX � n and A ⊂ X, x ∈ νX \ A′. There exists acontinuous functionϕ : �X → [−1,1] such thatϕ|�A = −1, ϕ(x) = 1. The setsA andB =


(ϕ|X)−1([0,1]) = ϕ−1([0,1]) ∩ X are asymptotically disjoint subsets ofX ad therefore

s,

et

n

e

there exists an asymptotic separatorC betweenA and B with asIndC � n − 1. Thenobviously, C′ is a separator betweenA′ and x in νX and, by inductive hypothesiasindC � n − 1. Therefore asindX � n. �

The following is a counterpart of Theorem 1 in [10].

Proposition 5.2. LetX be a proper metric space. ThenasindX � indνX.

Proof. Induction by asindX. Obviously, asindX = −1 if and only if indνX = −1.Suppose that asindX � n and show that indνX � n. Let A be a closed subset ofνX

andx ∈ νX \ A. There exists a continuous functionϕ : �X → [−1,1] such thatϕ|A = −1,ϕ(x) = 1. LetB = ϕ−1([−1,0]). Thenx /∈ B ′ and, since asindX � n, there exists a subsC ⊂ X with asindC � n − 1 and such thatC′ is a separator betweenB ′ andx in νX. Bythe inductive hypothesis,

indC′ = indνC � asindC � n − 1.

SinceC is a separator betweenB ′ andx in νX, it is also a separator betweenA andx.Therefore, for every closed subsetA of νX and x ∈ νX \ A we are able to find a

separatorK with indK � n − 1. This means that indνX � n. �It is proved in [10] (see Proposition 1 therein) that asIndX = asIndY for coarsely equiv-

alent spacesX andY .

Proposition 5.3. Let X and Y be coarsely equivalent metric spaces. ThenasindX =asindY .

Proof. Analogous to that of Proposition 1 in [10].�Lemma 5.4. Suppose thatA andB are asymptotically disjoint subsets ofX ⊂ Y andC isan asymptotic separator inY betweenA andB with asdimC � m. Then there exists aasymptotic separatorC betweenA andB in X with asdimC � m.

Proof. Define subsetsDk of X by induction. Let D1 = N1(C) ∩ X and Dk+1 =(Nk+1(C) ∩ X) \ Nk(Dk). By Lemma 2.2, there exists a(k + 1)-netDk+1 in Dk+1 whichis k-discrete. LetC = ⋃∞

k=1 Dk .We first show that asdimC � m. Remark that for everyk > 0 the complement to th

subset⋃{Dj | j � k} is the union of ak-disjoint family {Dj | j > k}. To apply Theo-

rem 2.5, we have to verify that asdimDk � m uniformly. If k > 0, then for someR > 0there exists anR-bounded coverU of the set

⋃{Dj | j � k} that splits into the unionU0 ∪ · · · ∪ Um of k-discrete families. LetV0 = U0 ∪ {{x} | x ∈ ⋃{Dj | j > k}}, then thefamily V0 ∪ U1 ∪ · · · ∪ Um is a min{R,k}-uniformly bounded familythat splits into theunion ofm + 1 k-disjoint families.

Show that for every integerk > 0 there existsR = R(k) > 0 such thatNk(C) ∩ X ⊂NR(k)(C). Indeed, ifx ∈ Nk(C) ∩ X, then eitherx ∈ Dk ⊂ N2k(Dk) or


x ∈ Nk(C) \ Dk ⊂ Nk−1(Dk−1) ⊂ Nk−1(Nk−1

(Dk−1

))

t

at for

t for

te

f the

lex

r

⊂ N2k−2(Dk−1

) ⊂ N2k−2(C

)and we can chooseR(k) = 2k.

Now suppose thatC is an asymptotic separator betweenA andB in Y . Show thatC isan asymptotic separator betweenA andB in X. It is sufficient to show thatC′ ⊃ C′ ∩ νX.Suppose thata ∈ C′ ∩ νX andU is a closed neighborhood ofa in �Y . Then there exissequences(xi) in X ∩ U and(ci) in C ∩ U andk ∈ N such thatd(xi, ci) � k for everyi.

Case(1). For an infinite number ofi (passing to a subsequence we then assume thall i) xi ∈ Dk . Then there existsxi ∈ Dk with d(xi, xi) � k.

Case(2). For infinite number ofi (passing to a subsequence we then assume thaall i) xi ∈ Nk−1(Dk−1). Then there existsyi ∈ Dk−1 such thatd(xi, yi) � k − 1 and thereexistsxi ∈ Dk−1 such thatd(yi, xi) � k − 1.

In both cases,xi ∈ C for every i andd(xi, xi) � max{k,2k − 1} = 2k and thereforexi ∈ U ∩ X for all but finitely many i. This means that the setsC and U are notasymptotically disjoint and thereforeC′ ∩ U �= ∅. Because of arbitrariness ofU weconclude thata ∈ C′. �Theorem 5.5. For every spaceX ∈Mn we haveasIndX � n.

We need some auxiliary results. LetT 0, . . . , T n be a sequence of regular locally finitrees.

Proposition 5.6. For every disjoint asymptotically disjoint subsetsA, B in M(T 0, . . . , T n)

there exists a separatorC between them such thatasdimC � n − 1 and the setsA, B, andC are asymptotically disjoint.

Proof. Let

U = {x ∈ M

(T 0, . . . , T n

) | d(x,A) � (1/2)d(x,B)},

V = {x ∈ M

(T 0, . . . , T n

) | d(x,B) � (1/2)d(x,A)},

thenU , V are closed, disjoint, asymptotically disjoint asymptotic neighborhoods osetsA,B, respectively. For everyj let Rj be a number such that

d(rj

(U \ NRj (x0)

), rj

(V \ NRj (x0)

))� 2j+2

(such a number exists becauseU,V are asymptotically disjoint and the retractionrj is2j+1-close to the identity map). We will assume thatRj+1 � 2Rj . Put Uj = rj (U ∩(NRj+1(x0) \ NRj (x0))).

Denote byUj the union of all n-dimensional cubic simplices in the subcomprj (M(T 0, . . . , T n)) that intersectUj . Let Sj = r−1(∂(N1/2(Uj ))) andS = ⋃∞

i=0 Sj .We are going to show thatS contains a separator betweenU andV and asdim(S) �

n − 1.Put S′ = ⋃{S2k | k ∈ N}, S′′ = ⋃{S2k−1 | k ∈ N}. By the finite sum theorem fo

asymptotic dimension (see, e.g., [2]), it is sufficient to prove that asdimS′ � n − 1,asdimS′′ � n − 1 and, because of complete analogy, we only prove that asdimS′ � n − 1.


Prove that the family{S2k} has asymptotic dimension� n − 1 uniformly. LetD > 1.

n

d,

There exists a naturalj such that every(n−1)-dimensional cubic complex of mesh 2j canbe covered by a 2j -bounded family of its subsets that splits into the union ofn familieswhich are 2D-disjoint.

Claim. r−1jj ′ (∂(N1/2(Uj ′))) ⊂ N3/4(r

−1jj ′ (Uj ′))(n−1).

Proof. To prove the Claim, suppose thatrjj ′(x) ∈ ∂(N1/2(Uj ′)). Then there is a cubicn-dimensional simplexK in Uj ′ such thatrjj ′(x) ∈ ∂(N1/2(K)). The simplexK is of theform K = ∏n

i=0 Ki , whereKi is a 1-dimensional simplex inT ij ′ and there isl ∈ {0, . . . , n}

such thatKl is a singleton,Kl = {al}. Then∂N1/2(K) = (⋃n

k=0 Rk) ∩ Mj ′ , where

Rk =( ∏

p∈{0,...,n}\{k}N1/2(Kp)

)∂N1/2(Kk).

Thenr−1jj ′ (∂N1/2(Kk)) = ∂N1/2(Kk) and we obtain

r−1jj ′ (Rk) ∩ Mj ′ =

( ∏p∈{0,...,n}\{k}

(rp

jj ′)−1(

N1/2(Kp)))

∂N1/2(Kk) ∩ Mj ′ .

For everyx = (x0, . . . , xn) ∈ r−1jj ′ (Rk) ∩ Mj ′ there existsp ∈ {0, . . . , n} \ {k} such

that xp ∈ ∂(Tp

j−1 \ Tpj ). Then d(x, (x1, . . . , xk−1, yk, xk+1, . . . , xn)) = 1/2 for some

yk ∈ ∂Kk (the boundary is considered inT k ; note that thenyk ∈ rkj ′(∂T k

j ′−1 \ T kj ′)) and

(x1, . . . , xl−1, al, xl+1, . . . , xn) ∈ N3/4(r−1jj ′ (Uj ′))(n−1). The claim is proved. �

Denote byV a cover of the cubic simplexMj that splits into the unionV = V0∪· · ·∪Vn

of uniformly 2j -bounded covers which are 2D-discrete. Now for everyk and everyi ∈{0, . . . , n} consider the familyV i

k = {N3/4(V ) ∩ r−1j,2k(U2k) | V ∈ V i

k} of the setr−1j,2k(U2k).

The coverVk = V0k ∪ · · · ∪ Vn

k is uniformly (2j + 3)-bounded and splits into the unioof D-discrete familiesV i

k , i ∈ {0, . . . , n}. Then the familiesW ik = {r−1

j (V ) ∈ V ∈ V ik} are

uniformly (2j+1 + 3)-bounded andD-discrete and, becauseS2k = r−1j r−1

j,2k(U2k), we see

that⋃n

i=0W ik coversS2k .

Applying Theorem 2.5 we conclude that asdimS � n − 1. Finally, note thatS containsa separator betweenA andB, namely, the set∂N1/2(

⋃∞j=1 Uj ). �

Proposition 5.7. Let C be a separator between asymptotically disjoint subsetsA,B in aproper uniformly arcwise connected spaceX and the setsC andA ∪B are asymptoticallydisjoint. ThenC is an asymptotic separator betweenA andB in X.

Proof. Denote byU the set of points that can be connected toA by an arc in theset X \ C. We show that the sets(�U ) ∩ X and B are asymptotically disjoint. Indeeassuming the opposite we obtain sequences(ai) and(bi) in U andB respectively, suchthatd(ai, x0) → ∞, d(bi, x0) → ∞ (herex0 is an arbitrary base point inX) and that there


existsR > 0 for which d(ai, bi) < R, i ∈ N. SinceX is uniformly arcwise connected,of

hat

herets

l

plies

lity

g of

e spacessume

.6,,

there existsS > 0 such that, for everyi, the pointsai andbi can be connected by an arcdiameter� S. There exists naturali0 such that for everyi � i0 we haved(bi,C) � R + S.Then obviously there exists an arc inX \ C that connectsai andbi , i � i0. Therefore,bi ∈ U and we obtain a contradiction.

Thus,B ′ ⊂ (X \ U)′. In order to prove thatC is an asymptotic separator betweenA

andB it remains to prove that�U ∩ �B ∩ νX ⊂ C′. Indeed, suppose the opposite, i.e., tthere exists a pointy ∈ �U ∩ �B ∩ νX \ C′. Then there exist disjoint neighborhoodsV ofy andW of �C in �X and obviouslyU ∩ V ∩ �B �= ∅. Therefore, there exist sequences(ai)

in U ∩ V and(bi) in B such that limi→∞ d(ai, x0) = ∞, limi→∞ d(bi, x0) = ∞, and thesequence(d(ai, bi)) is bounded. Arguing similarly as above we conclude that then texists a sequence(ci) in C for which the sequence(d(ai, ci)) is bounded. This contradicto our choice ofV . �Lemma 5.8. The spaceM(T 0, . . . , T n) is uniformly arcwise connected.

Proof. Note that for everyl and everyz ∈ rl (M(T 0, . . . , T n))\rl+1(M(T 0, . . . , T n)) thereexists a path of diameter� 2l+1 in M(T 0, . . . , T n) connectingz with rl(z).

Let x = (x0, . . . , xn), y = (y0, . . . , yn) ∈ M(T 0, . . . , T n). Let j be the minimal naturanumber such thatd(x, y) < 2j . Thenrj (x) = rj (y). Let J be a path connectingx andy

that consists of paths of diameter� 2l+1 connectingrl−1(x) with rl(x) and of paths ofdiameter� 2l+1 connectingrl−1(y) with rl(y), 1 � l � j (recall thatr0 is the identitymap). Then

diam(J ) �j∑

l=1

d(rl−1(x), rl(x)

) +j∑

l=1

d(rl−1(y), rl(y)

)� 2

j∑l=1

2l+1 � 2j+3 � 16d(x, y). �

Now we are going to prove Theorem 5.5. It is proved in [10] that asIndX � IndνX.This result together with the classical theorem on the comparison of Ind and dim imthe inequality asIndX � dimνX. Suppose asdimX < ∞. Then asdimX = dimνX (see[7]), and we obtain asIndX � asdimX for everyX with asdimX < ∞.

To prove the opposite inequality we apply induction on asdimX. It is known (see [10])that the properties asdimX = 0 and asIndX = 0 are equivalent. Assume that the inequaasIndX � asdimX is proved for everyX with asdimX � n − 1.

Suppose that asdimX � n. Then by Theorem 3.5 there is a large scale embeddinX into the spaceM(T 0, . . . , T n), for some regular locally finite treesT 0, . . . , T n. Sincethe image of any space under a large scale embedding is coarsely equivalent to thand the dimension asInd is a coarse invariant (see [10, Proposition 1]), we may athatX is a subspace ofM(T 0, . . . , T n). Let A,B be asymptotically disjoint subsets ofX.Without loss of generality we assumeA,B to be closed and disjoint. By Proposition 5there exists a separatorC betweenA andB with asdimC � n − 1. Since, by Lemma 5.8


the spaceM(T 0, . . . , T n) is uniformly arcwise connected, it follows from Proposition 5.7

ict

classs factsult onge

er the

stead

uwerin the.

the

thatC is also an asymptotic separator betweenA andB in M(T 0, . . . , T n).By Lemma 5.4, there exists an asymptotic separatorC betweenA andB in X with

asdimC � n − 1. Applying the induction assumption we see that asIndC � n − 1.Since for every pair of asymptotically disjoint subsets inX there exists an asymptot

separator between them whose dimension asInd does not exceedn − 1, we conclude thaasIndX � n.

The equality asIndX = asdimX is therefore proven for allX with asdimX < ∞.One can similarly prove the following result.

Theorem 5.9. For all proper metric spacesX with asdimX < ∞ we haveasindX =asdimX.

We finally obtain that the dimensions asdim, asInd, and asind coincide in theof proper metric spaces that are finite dimensional with respect to asdim. Thican be regarded as a counterpart in the asymptotic topology of the classical recoincidence of the covering dimension, the small inductive dimension, and the larinductive dimension (see [12]).

6. Open problems

The coincidence of the dimension functions asdim and asInd is proved undassumption of finiteness of asdim. This leads to the following natural question.

Question 6.1. Is there a proper metric spaceX with asdimX = ∞ and asIndX < ∞?

A similar question can be formulated for ind.A closed subsetC of a topological spaceX is acut between disjoint subsetsA,B ⊂ X

if every continuum (compact connected space)T ⊂ X that intersects bothA andB alsointersectsC.

The notion of the large inductive dimension can be based on the notion of cut inof separator. A subsetC of a metric spaceX is called anasymptotic cutbetweenasymptotically disjoint setsA andB if for everyr > 0 there existsλ > 0 such that everyr-chain joiningA andB intersects theλ-neighborhoodNλ(C) of C. Here anr-chain joiningA andB is a sequencex0, . . . , xn of points inX with x0 ∈ A, Xn ∈ B andd(xi, xi+1) < r.

It turns out (see [13]) that this dimension in the classical case (it was defined by Browho called it Dimensiongrad) coincides with the classical large inductive dimensionclass of separable metrizable spaces). In the asymptotic case there is an open question

Question 6.2 (E. Shchepin). Does large asymptotic inductive dimension defined onbase of an asymptotic cut coincide with asInd?


References

7–71.

int.380)

th.

aces,

th.

dings of1.

nn,

54 (2

).

5)

th.

.,

2)

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