Requisitos funcionales

Denition 5.3. Let (X, d) be a metric space, and let U be a collection of subsets of X. The collection U is uniformly bounded, if supU ∈Ud(U )is nite.

This denition can be extended to coarse spaces as follows: Let X be a coarse space, and let U be a collection of subsets of X. The collection U is uniformly bounded if the set SU ∈UU

2 _{is controlled. The aforementioned set will be called the bounding set of U.}

Asymptotic dimension has many equivalent denitions. As the initial denition, the most clear counterpart to Denition 5.2 is used.

Denition 5.4. Let X be a metric space or a coarse space. The asymptotic dimension of X, denoted by asdim X, is at most n, if the following condition is true: For every uniformly bounded cover U of X, there is a uniformly bounded cover V with an order of at most n + 1, and the initial cover U is a renement of V.

The asymptotic dimension asdim X is the smallest nonnegative integer for which asdim X ≤ n holds. If no nonnegative integer fullls said inequality, asdim X is in- nite.

The following Proposition yields a convenient reformulation for the asymptotic dimen- sion of metric spaces. Recall that a cover U of a metric space X has a Lebesgue number of λ, if for every element x of X, the open ball BX(x, λ) is contained within some set of

U.

Proposition 5.5. Let X be a metric space, and let n be a nonnegative integer. The following conditions are equivalent:

1. The asymptotic dimension of X is at most n.

2. For every positive real number λ, there exists a uniformly bounded cover V of X with Lebesgue number λ and an order of at most n + 1.

Proof. Assume that asdim X is at most n, and x a positive real number λ. Let U be the cover of X consisting of all open balls of radius λ. Hence, U is a renement of a uniformly bounded cover V with an order of at most n + 1. Since every open ball of radius λ is contained in a renement of V, the cover V has a Lebesgue number of λ. Hence, the rst condition implies the second one.

Next, assume that the second condition holds, and x a uniformly bounded cover U of X. One may assume that U does not contain the empty set. Denote by R the value of the nite bound supU ∈Ud(U ). Using the second condition, select a cover V with the value

λ = R + 1. Note that V is uniformly bounded and has an order of at most n+1. For every set U in the cover U, select an element xu from U. Now, since d(U) is less than R + 1, U

is contained within the open ball BX(xu, R + 1), which in turn is contained within some

set of V. Due to this, U is a renement of V. Therefore, the second condition implies the rst one.

Next, the main alternate denition of asymptotic dimension is formulated. The de- nition is based on uniformly bounded families where all sets are suciently far from each other. This is formalized by the following denition.

Denition 5.6. Let (X, d) be a metric space, and x a positive real number r. Let U be a family of subsets of X. The family U is r-disjoint if, for every pair U1, U2 of two

dierent elements of U, the inequality d(U1, U2) > r holds.

Alternatively, let (X, E) be a coarse space, and x a controlled set E. Let U be a family of subsets of X. The family U is E-disjoint if, for every pair U1, U2 of two dierent

elements of U, the intersection E ∩ (U1× U2) is empty.

For a metric space X, denote by Er the controlled set {(x, y) ∈ X2 | d(x, y) ≤ r}.

In this case, r-disjointness is equivalent to Er-disjointness. If r and r0 are two positive

reals fullling r0 _{≤ r, r-disjointness implies r}0_{-disjointness. Similarly, if E and E}0 _{are two}

controlled sets fullling E0 _{⊂ E, E-disjointness implies E}0_{-disjointness.}

Proposition 5.7. Let X be a coarse space, and let n be a nonnegative integer. The following conditions are equivalent:

1. The asymptotic dimension of X is at most n.

2. For every controlled set E, there exist n + 1 uniformly bounded families U0, . . . , Un,

where every family Ui is E-disjoint, and the union S n

i=0Ui is a cover of X.

Furthermore, if X is a metric space, the following condition is also equivalent with the above:

3. For every positive real number r, there exist n + 1 uniformly bounded families U0, . . . , Un, where every family Ui is r-disjoint, and the union Sn_i=0Ui is a cover

of X.

Proof. Let X be a metric space, and again denote the controlled set {(x, y) ∈ X2_{|d(x, y) ≤}

r} by Er. Since r-disjointness and Er-disjointness are equivalent, condition 2 implies

condition 3. Furthermore, every controlled set E is contained in the set Er for some r,

and in this case, an r-disjoint family is also E-disjoint. Therefore, conditions 2 and 3 are equivalent for metric spaces.

Next, condition 1 is proven from condition 2. Let X be a coarse space, and assume the second condition. Let U be a uniformly bounded cover of X, and let E be the bounding set of U. Use condition 2 to select uniformly bounded (E ◦ E)-disjoint families V0

0, . . . V 0 n.

Denote by Fi the bounding sets of the covers Vi0, and let F be the union S n

i=0Fi of

bounding sets. The union V0 ₌Sn

i=0V 0

i is a uniformly bounded cover with bounding set

F. Dene the sets Vi and the cover V as follows:

VV0 =U ∈ U U ∩ V0 6= ∅ for each V0 _{∈ V}0 , Vi =∪VV0  V0 ∈ V_i0

for each i ∈ {0, . . . , n}, and V =

n

[

i=0

Vi.

If x is an element of X, it is contained in some Ux ∈ U and some Vx0 ∈ V0. Hence, the

intersection V0

x∩ Ux is nonempty, and due to this, x is an element of ∪VV0

x. Therefore, V

is a cover. Since V0 _{is a cover, every element U of U intersects some V}0 U of V

0_{. Due to this,}

U is a renement of V.

Let V = ∪VV0 be an element of V, and let (x, y) be an element of V2. There exist sets

Ux and Uy of U which intersect V0 and contain x and y respectively. Let vx0 and v 0 y be

elements of the respective nonempty intersections Ux∩ V0 and Uy∩ V0. Since the sets Ux

and Uy are elements of U which has the bounding set E, the pairs (x, vx0) and (v 0

y, y) are

elements of E. Similarly, since V0 _{is an element of the cover V}0 _{with the bounding set F ,}

the pair (v0

x, vy0)is an element of F . Using this, one can conclude that

[

V ∈V

V2 ⊂ E ◦ F ◦ E,

and therefore the cover V is uniformly bounded.

It remains to verify that V has an order of at most n + 1. The claim follows from showing that the families Vi are disjoint. Let x be an element of X, and let V1 = ∪VV10

and V2 = ∪VV0

is contained in both V1 and V2. Hence, there exist sets U1 and U2 which contain X and

intersect V0

1 and V 0

2 respectively. Choose points u1 and u2 from these intersections. Now

(u1, x) and (x, u2) are elements of E. Therefore, (u1, u2) is contained in both E ◦ E and

V₁0× V0

2. This is a contradiction, since the family Vi0 was selected to be (E ◦ E)-disjoint.

The proof that condition 2 implies condition 1 is now nished.

Finally, condition 2 is proven from condition 1. This proof follows the one presented in the article [6]. For a given controlled set E of E, point x of X, and subset A of X, dene the E-ball at x and the E-interior of A as follows:

BE(x, E) = {x0 ∈ X | (x, x0) ∈ E}

intE(A) = {a ∈ A | BE(a, E) ⊂ A} .

Let E be a controlled set, and denote by F the set E ∪ E−1_{∪ ∆}

X. In this case, F is a

controlled symmetric set containing the diagonal. Dene the sets Fi for positive natural

numbers i as follows:

Fi = F ◦ F ◦ . . . ◦ F

| {z }

icopies

.

The sets Fi are symmetric and contain ∆X. Next, let V0 be the family of Fn+1-balls

{BE(x, Fn+1) | x ∈ X}. Since Fn+1 contains the diagonal ∆X, V0 is a cover. Further-

more, if x is an element of X, using the fact that Fn+1 is symmetric, one concludes that

(BE(x, Fn+1))2 is contained within Fn+1◦ Fn+1. Hence, V0 is a uniformly bounded cover.

By the denition of asymptotic dimension, one may select a uniformly bounded cover V which V0 _{renes and which has an order of at most n + 1.}

Using the previously dened cover V, one can dene for values i = 0, . . . , n the collec- tions

Vi = {V0∩ . . . ∩ Vi| Vj ∈ V, Vj 6= Vk for j 6= k} ,

Wi =intFn+1−i(V ) | V ∈ Vi , and

Ui = {W \ ∪Wi+1| W ∈ Wi} , where Wn+1 = {∅}.

Note that every set U ∈ Ui is a subset of some V ∈ V. Hence, the collections Ui are

uniformly bounded. Let x be an element of X. There is a set V ∈ V for which BE(x, Fn+1)

is contained in V . This, on the other hand, means that x is an element of intFn+1(V ),

which is a set of W0. Hence, x is an element of Ui, where i is the largest integer for which

x is an element of a set of Wi. In conclusion, U = Sn_i=0Ui is a uniformly bounded cover

of X.

It remains to check that the collections Ui are E-disjoint. For this, it is enough to show

that they are F -disjoint. Assume to the contrary that U1 and U2 are distinct elements of

V1,0∩ . . . ∩ V1,i and V2,0∩ . . . ∩ V2,i containing BE(x, Fn+1−i)and BE(y, Fn+1−i)respectively.

Note that there are at least i + 2 dierent Vj,k sets in total, since the intersections are

distinct.

Because (x, y) is in F , BE(x, F ) contains y. In the case where i equals n, this means

that the sets V1,j all contain y. This is a contradiction, since y would be contained in an

intersection of n + 2 dierent sets of V. On the other hand, if i is less than n, the ball BE(x, Fn+1−i) contains the ball BE(y, Fn−i). Hence, the ball BE(y, Fn−i) is contained in

all sets Vj,k. This implies that y is an element of ∪Wi+1, which is also a contradiction.

Hence, the families Ui are F -disjoint, which concludes the proof.