aa r X i v : . [ m a t h . GN ] F e b A METRIC SPACE WITH TRANSFINITEASYMPTOTIC DIMENSION ω T.Radul
Abstract.
We build an example of a metric space with transfinite asymptotic di-mension 2 ω .
0. Introduction.
The asymptotic dimension asdim of a metric space wasdefined by Gromov for studying asymptotic invariants of discrete groups [1]. Thisdimension can be considered as an asymptotic analogue of the Lebesgue coveringdimension dim.A transfinite extensions trasdim for the asymptotic dimensions asdim was intro-duced in [2]. There was constructed a proper metric space X such that trasdim X = ω . Metric spaces X ω + k with trasdim X ω + k = ω + k were constructed in [3] for eachnatural k .
1. Preliminaries.
A family A of subsets of a metric space is called uniformly bounded if there existsa number C > A ≤ C for each A ∈ A ; A is called r - disjoint forsome r > d ( A , A ) ≥ r for each A , A ∈ A such that A = A .The asymptotic dimension of a metric space X does not exceed n ∈ N ∪ { } (written asdim X ≤ n ) iff for every D > U of X such that U = U ∪ · · · ∪ U n , where all U i are D -disjoint [1].Since the definition of asdim is not inductive, we cannot immediately extend thisdimension to transfinite numbers. We need some set-theoretical construction usedby Borst to extend the covering dimension and metric dimension [4,5].Let L be an arbitrary set. By F inL we shall denote the collection of all finite,non-empty subsets of L . Let M be a subset of F inL . For σ ∈ {∅} ∪ F inL we put M σ = { τ ∈ F inL | σ ∪ τ ∈ M and σ ∩ τ = ∅} . Mathematics Subject Classification . 54F45, 54D35.
Key words and phrases.
Asymptotic dimension,transfinite extension. Typeset by
AMS -TEX T.RADUL
Let M a abbreviate M { a } for a ∈ L .Define the ordinal number Ord M inductively as followsOrd M = 0 iff M = ∅ ,Ord M ≤ α iff for every a ∈ L , Ord M a < α ,Ord M = α iff Ord M ≤ α and Ord M < α is not true, andOrd M = ∞ iff Ord M > α for every ordinal number α .Given a metric space ( X, d ), let us define the following collection: A ( X, d ) = { σ ∈ F in N | there is no uniformly bounded families V i for i ∈ σ such that ∪ i ∈ σ V i covers X and V i is i − disjoint } . Let (
X, d ) be a metric space. Then put trasdim X = Ord A ( X, d ) [2].We will need some properties of trasdim proved in [2].
Proposition A,[2].
Let X be a metric space and Y ⊂ X . Then trasdim Y ≤ trasdim X . We denote by N R ( A ) = { x ∈ X | d ( x, A ) ≤ R } for a metric space X , A ⊂ X and R > Proposition B,[2].
Let Y be a metric space and X ⊂ Y . Then trasdim N n ( X ) =trasdim X for each n ∈ N . We also will use another transfinite extension coasdim of asdim defined in [3].We represent every ordinal number γ = λ ( γ )+ n ( γ ) where λ ( γ ) is the limit ordinalnumber or 0 and n ( γ ) ∈ N ∪ { } . For a metric space X we define complementary-finite asymptotic dimension coasdim( X ) inductively as follows [3]:coasdim( X ) = − X = ∅ ,coasdim( X ) ≤ γ iff for every r > r -disjoint uniformly boundedfamilies V , . . . , V n ( γ ) of subsets of X such that coasdim( X S ( S n ( γ ) i =0 )) < λ ( γ ),coasdim( X ) = γ iff coasdim( X ) ≤ γ and coasdim( X ) ≤ β is not true for each β < γ , andcoasdim( X ) = ∞ iff coasdim( X ) ≤ α is not true for every ordinal number α .It was proved in [3] that inequality coasdim( X ) ≤ ω + k implies trasdim( X ) ≤ ω + k for each metric space. By routine checking we can generalize this fact. METRIC SPACE WITH TRANSFINITE ASYMPTOTIC DIMENSION 2 ω Proposition 1.
We have trasdim( X ) ≤ coasdim( X ) for each metric space X . Using similar arguments as in the proof of Proposition B, we can prove thefollowing statement.
Proposition 2.
Let Y be a metric space and X ⊂ Y . Then coasdim N n ( X ) =coasdim X for each n ∈ N .
3. The main result.
We are going to construct a metric space X ω such that trasdim( X ω ) = coasdim( X ω ) =2 ω . We develop the ideas from [2] and [3]. Firstly let us recall the construction ofthe spaces X ω + k for k ∈ N from [3].We consider R n with the sup-metric defined as follows d ∞ (( k , . . . , k n ) , ( l , . . . , l n )) =max {| k − l | , . . . , | k n − l n |} . For k ≤ n we consider the natural isometrical embed-ding i kn : R k → R n defined by the formula i kn ( x , . . . , x k ) = ( x , . . . , x k , , . . . , i, k ∈ N we define X ( i ) k = { x ∈ R i | |{ l ∈ { , . . . , i } | x l / ∈ i Z }| ≤ k } . Put X ω + k = F ∞ i =1 X ( i ) k . Define a metric d k on X ω + k as follows. Consider any x , y ∈ X ω + k . Let x ∈ X ( m ) k and y ∈ X ( n ) k with m ≤ n . Put c k = 0 if n = m and c k = k ( m +( m +1)+ · · · +( n − m < n . Define d k ( x, y ) = max { d ∞ ( i mn ( x ) , y ) , c k } .It is proved in [3] that trasdim( X ω + k ) = coasdim( X ω + k ) = ω + k for each k ∈ N .Let us remark that we slightly modified the metric d k compared with [3], but thismodification does not affect on the statements from [3].It is easy to see that for each pair n , m ∈ N such that m ≤ n the naturalincluding j mn : X ω + m → X ω + n is an isometrical embedding.Put Y ( m ) k = X ( m ) k ∩ (2 k Z ) m and Y ω + k = F ∞ i =1 Y ( i ) k . It is easy to see that N k ( Y ω + k ) = X ω + k , hence trasdim( Y ω + k ) = ω + k for each k ∈ N . Put X ω = F ∞ i =1 Y ω + i . Define a metric d on X ω as follows. Consider any x , y ∈ X ω . Let x ∈ X ( m ) k and y ∈ X ( n ) k with m ≤ n . Put c = 0 if n = m and c = m +( m +1)+ · · · +( n − m < n . Define d ( x, y ) = max { d n ( j mn ( x ) , y ) , c } . Theorem 1. trasdim( X ω ) = 2 ω .Proof. The inequality trasdim( X ω ) ≥ ω follows from the fact that trasdim( Y ω + k ) = ω + k for each k ∈ N and Proposition A. T.RADUL
Let us show that coasdim( X ω ) ≤ ω . Let r >
0. Choose k ∈ N such that k ≥ r . Consider the family V = {{ x }| x ∈ ∪ ∞ i = k Y ω + i } . The family V is evidently r -disjoint. We have X ω \ ∪V = ∪ k − i =1 Y ω + i . But ∪ k − i =1 Y ω + i ⊂ N c ( Y ω + k − ), where c = 1 + · · · + k −
1. Hence coasdim( X ω \ ∪V ) ≤ ω + k − < ω .Now, the theorem follows from Proposition 1.Certainly, the presented result is only one step ahead. The general problem stillremains: for each countable ordinal ξ find a metric space X ξ with trasdim( X ξ ) = ξ . References
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