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Assouad dimension, Nagata dimension, and uniformly close metric tangents

We study the Assouad dimension and the Nagata dimension of metric spaces. As a general result, we prove that the Nagata dimension of a metric space is always bounded from above by the Assouad dimension. Most of the paper is devoted to the study of when these metric dimensions of a metric space are locally given by the dimensions of its metric tangents. Having uniformly close tangents is not sufficient. What is needed in addition is either that the tangents have dimension with uniform constants independent from the point and the tangent, or that the tangents are unique. We will apply our results to equiregular subRiemannian manifolds and show that locally their Nagata dimension equals the topological dimension.


Introduction
Assouad dimension, Nagata dimension, and metric tangents are relevant in the program of doing analysis in the metric space setting. The Assouad dimension is a quantification of the doubling property. It plays a role, for example, in the study of spaces that are quasisymmetrically embeddable in Euclidean spaces, of fractal sets, and of boundaries of groups, [Ass79, Ass83,Luu98,KL04,Mac11]. The Nagata dimension, introduced in [Nag58,Ass82], is a local-and-global metric version of the topological dimension. The bounded-scale version is called linearly controlled dimension and the large-scale version is called asymptotic Assouad-Nagata dimension. Such dimensions are relevant for embeddings, in particular in Geometric Group Theory, [BDS07]. Nagata dimension links with quasisymmetric embedding into metric trees and with Lipschitz extension properties, [LS05,WY10].
In [LS05] it has been shown that doubling metric spaces have finite Nagata dimension. In this paper, we prove the sharp bound: Theorem 1.1. For all metric spaces X, the Nagata dimension of X is less than or equal to the Assouad dimension of X.
Metric tangents provide a way of studying the infinitesimal properties of metric spaces. Limits of metric spaces were introduced by Gromov in the setting of Geometric Group Theory to study asymptotic cones of groups of polynomial growth [Gro81]. However, many of the results hold in the context of metric spaces with finite Assouad dimension and have applications in different areas of mathematics, for instance in the study of limits of Riemannian manifolds with curvature bounds and Reifenberg-flat metric spaces, see [CC97], [DT99], and subsequent work. Additional results regarding tangents of general metric spaces can be found in [HH00,LD11,Her11].
In this paper we study how and when one can deduce the Nagata dimension or the Assouad dimension of a space, knowing the respective dimensions of its tangents. A concrete application of our results is given by the Lipschitz extension problem for subRiemannian manifolds. Lang and Schlichenmaier gave a connection of the Lipschitz extension problem with the Nagata dimension and the Lipschitz connection property. Lipschitz connectivity and Lipschitz homotopy groups have recently been studied in [DHLT11,WY10,HS13]. By the results in this paper, knowing now the Nagata dimension of equiregular subRiemannian manifolds, one can deduce for example that a partially defined Lipschitz map f : A → Y from a subset A of an equiregular subRiemannian manifold M can be Lipschitz extended on compact sets of M if Y is Lipschitz m-connected for all m strictly smaller than the topological dimension of M . For such applications see [LS05, Theorem 1.5 and Theorem 1.6]. Another property one deduces for a space (X, d) with Nagata dimension at most some number n ∈ N is that for sufficiently small p ∈ (0, 1) there exists a bi-Lipschitz embedding of (X, d p ) into the product of n + 1 metric trees, see [LS05,Theorem 1.3].
Let us now present in detail our results on tangent spaces. Let X be a metric space. For each x ∈ X, let Tan(X, x) be the collection of all the metric spaces tangent to X at x, in the pointed Gromov-Hausdorff sense. We say that X has uniformly close tangents if, for each x ∈ X, the convergence of the dilated spaces of (X, x) toward Tan(X, x) is uniform. In other words and more generally, we say that on a subset K ⊂ X the convergence to tangents is uniform if, for all > 0, there exists λ > 0 such that, for all k ∈ K and all λ > λ , there exists a tangent Y of X at k with Dist GH ((λX, k), Y ) < .
Here Dist GH is a specific distance that we fix in Section 2.3 to induce the Gromov-Hausdorff topology on pointed metric spaces and λX is the metric space (X, λd X ). The condition of uniform convergence to tangents is motivated by the fact that this is what happens on equiregular subRiemannian manifolds, see Theorem 4.1.
Assuming uniform convergence towards unique tangents, our first result for the Nagata dimension, which we denote by dim N , is the following: Theorem 1.2. Let X be a metric space that at every point admits a single tangent space. Let Y ⊆ X be a relatively compact set with dim N Y < ∞. Assume that the convergence toward the tangents is uniform on the closure of Y (as just defined). Then we have Here int Y and cl Y denote the interior and the closure of the set Y , respectively. From Example 3.3 we see that the assumption on uniqueness of tangents is necessary in Theorem 1.2. Relative compactness of Y is needed already to handle the large scales, and the necessity of the interior of Y is seen by taking Y = {0} and X = R. We shall prove Theorem 1.2 at the end of Section 3.1 as a consequence of Theorem 3.4, with assumption (ii). An application of Theorem 1.2 is the following result, which was actually our initial goal (the notion of equiregular subRiemannian manifold is recalled in Section 4). Corollary 1.3 relies also on a result by Urs Lang and the first-named author which states that the Nagata dimension of a Carnot group equals its topological dimension. For completeness, we include a short proof of this fact in Section 4.
We prove the following analog of Theorem 1.2 for the Assouad dimension, which we denote by dim A .
Theorem 1.4. Let X be a metric space that at every point admits a single tangent space. Let Y ⊆ X be a relatively compact set. Assume that the convergence toward the tangents is uniform on Y (as defined above). Then we have Theorem 1.4 is a consequence of Theorem 3.9. These results will be proved in Section 3.2. Theorem 1.4 gives an alternative proof of a fact essentially proven in [NSW85]. Namely, the Assouad dimension of an equiregular subRiemannian manifold equals the Assouad dimension of its tangents.
One direction of research where the above results can be used is the study of generalizations of Reifenberg vanishing-flat metric spaces where the model space R n is replaced by any fixed doubling metric space. Namely, we say that a metric space X is vanishing-flat modeled on a metric space Y if Y is the tangent space at any point x of X and the convergence toward the tangents is uniform. The case when Y is a Euclidean space is called Reifenberg vanishing-flat metric spaces, and it has been manly considered in [CC97,DT99]. It is a natural problem to study what properties of X can then be deduced from the ones of Y .
We shall also provide results when the tangents are not assumed to be single spaces. However, without such an assumption, it is necessary to require uniformity in values of the constants appearing in the definition of the dimension of the tangents. Such uniformity of constants is true for instance for many self-similar spaces and these spaces usually have more than one tangent at every point. Let us briefly recall that the linearly controlled dimension is the bounded-scale version of the Nagata dimension.
Theorem 1.5. Let X be a metric space of finite linearly controlled dimension. Assume that X has uniformly close tangents (as defined above). Then the linearly controlled dimension of X equals the infimum of all integers n for which, for all x ∈ X, the linearly controlled dimension of any Y ∈ Tan(X, x) is at most n with constants independent from Y and x.
See Theorem 3.4, with assumption (i), for a more explanatory statement of the upper bound. The lower bound follows from Corollary 2.19.
We also have the analogue of Theorem 1.5 for the Assouad dimension.
Theorem 1.6. Let X be a metric space. Assume that X has uniformly close tangents (as defined above). Then the Assouad dimension of X equals the infimum of all α ≥ 0 for which, for all x ∈ X, the Assouad dimension of any Y ∈ Tan(X, x) is at most α with constants independent from Y and x.
A more explanatory statement of the upper bound is given in Theorem 3.7, which will be an immediate consequence of Proposition 3.8. The lower bound is given by Corollary 2.17.
The paper is organized as follows. Section 2 is devoted to preliminaries. We recall the definitions and basic properties of Assouad dimension, Nagata dimension, and locally controlled dimension. In Section 2.3, we give the definition of Gromov-Hausdorff distance for pointed metric spaces and we define the set of tangents. We provide some remarks about the lower semicontinuity of Assouad dimension and Nagata dimension. In particular, Corollary 2.17 (resp. Corollary 2.19) gives the lower bound for Theorem 1.4 and Theorem 1.6 (resp. Theorem 1.2 and Theorem 1.5). In Example 2.20, we show that in general, even for compact subsets of R, the dimension of all the tangents could be strictly smaller than the dimension of the set. However, in Proposition 2.21 we prove that, if a doubling space has Nagata dimension equal to one, it has some weak tangent with Nagata dimension one.
In Section 3 we study metric spaces with uniformly close tangents. In Example 3.3 and Example 3.10 we show that having uniformly close tangents does not imply an upper bound for the dimension of the space in terms of the dimensions the tangents. In Theorem 3.4 we provide such a bound with the additional assumption that either the tangents have linearly controlled dimension less than n with respect to a uniform constant c, or the tangents are unique. Such theorem gives the missing upper bound for Theorem 1.5 and Theorem 1.2. In Section 3.2 we consider Assouad dimension. We prove Theorem 3.7 (resp. Theorem 3.9) giving the upper bound needed for concluding the proof of Theorem 1.6 (resp. Theorem 1.4).
In Section 4 we apply Theorem 1.2 to prove Corollary 1.3, after recalling some results on subRiemannian geometry and Carnot groups. In Section 5 we prove Theorem 1.1 and Theorem 5.1, which is a bounded-scale version of Theorem 1.1.

Preliminaries
2.1. Assouad dimension and Nagata dimension. In this paper we will consider the notion of Assouad dimension. It is also known with other names such as: metric covering dimension, uniform metric dimension, or doubling dimension. We recall here the definition from [Hei01, page 81]. The Assouad dimension of a metric space X is denoted by dim A X and is defined as the infimum of all numbers β > 0 with the property that there exists some C > 1 such that, for every > 0, every set of diameter D can be covered by using no more than C −β sets of diameter at most D. In this case, we say that the Assouad dimension is less than or equal to β with constant C.
We will need a quantified and local version of the definition that makes explicit the constants involved.
Definition 2.1 (Assouad dimension up to a scale). LetR > 0 and C > 1. We say that a metric space X has Assouad dimension at most β up to scaleR with constant C if, for all 0 < r < R <R, any ball of radius R in X can be covered with C (R/r) β or less balls of radius r in X. In this case we write dim A (X, C,R) ≤ β.
Metric spaces with finite Assouad dimension are precisely the doubling metric spaces. We recall that a metric space is doubling with constant L, for some L > 0, if, for every s > 0, every subset of the metric space with diameter at most 2s can be covered by L or fewer sets of diameter at most s.
Other notion of metric dimension that we will consider is the Nagata dimension. Before giving the definition, let us recall some basic terminology, following [LS05]. Two subsets A, B of a metric space are s-separated, for some constant s ≥ 0, if dist(A, B) := inf{d(a, b) ; a ∈ A, b ∈ B} ≥ s. A family of subsets is called s-separated if each distinct pair of element in it is s-separated. Let B be a cover of a metric space X. Then, for s > 0, the s-multiplicity of B is the infimum of all n such that every subset of X with diameter at most s meets at most n members of the family B. Furthermore, B is called D-bounded, for some constant D ≥ 0, if diam B := sup{d(x, x ) ; x, x ∈ B} ≤ D, for all B ∈ B.
Definition 2.2 (Nagata dimension). Let X be a metric space. The Nagata dimension, or Assouad-Nagata dimension, of X is denoted by dim N X and is defined as the infimum of all integers n with the following property: there exists a constant c > 0 such that, for all s > 0, X admits a cs-bounded cover with s-multiplicity at most n + 1.
As shown in [LS05, Proposition 2.5], the Nagata dimension can be defined equivalently as the infimum of all integers n with the following property: there exists a constant c > 0 such that for all s > 0, the metric space admits an s-bounded cover of the form B = n k=0 B k where each B k is cs-separated. (2.1) The constant c in (2.1) can be different from the constant c in Definition 2.2.
We should notice that the notion of Nagata dimension is global. Moreover, it is both for small and large scales. We will need the bounded-scale version. For a better formulation of our statements, we give a definition that points out both the scale and the constant.
Definition 2.3 (Nagata dimension at a scale). We say that a metric space X has Nagata dimension bounded from above by n ∈ N := {0, 1, 2, . . .} with constant c > 0 at scale s > 0 and write dim N (X, c, s) ≤ n, if the metric space admits an s-bounded cover of the form B = n k=0 B k where each B k is cs-separated.
Definition 2.4 (Linearly controlled dimension). We say that a metric space X has linearly controlled dimension bounded from above by n, written as dim LC X ≤ n, provided that there exists c > 0 and s 0 > 0 such that dim N (X, c, s) ≤ n, for all 0 < s < s 0 .
The notation -dim is also used for the linearly controlled dimension, see [BS07, Section 9.1.4] Notice that the space X has Nagata dimension bounded from above by n exactly when we can take s 0 = ∞, in other words if there exists c > 0 such that dim N (X, c, s) ≤ n for all s > 0.
Since any cover for a space X restricts to a cover to any of its subsets Y , we have the easy (2.2)

Basic facts about the dimensions.
Here is a first lemma showing that, in the local version of the Assouad dimension, we have no problem if the radius a priori depends on the exponent.
Proof. Fix η > α. We can take R := R η . Indeed, pick any β > α. Fix 0 < r < R < R η and x ∈ X. If R < R β , we are done. So we assume R ≥ R β . Cover B(x, R η ) with C η (R η /R β ) η or less balls of radius R β . Cover each of these balls with at most C β (R β /r) β balls of radius r. Hence, the ball B(x, R), which is inside B(x, R η ), needs no more than C η (R η /R β ) η C β (R β /r) β balls of radius r to cover it. The proof is concluded by putting C β : Regarding the Nagata dimension, we shall study what happens when we consider spaces that are the union of spaces of which we know their Nagata dimension. We start with an easy statement, whose proof is straightforward.
Lemma 2.6. Let X = i∈I X i with X i that are r-separated and dim N (X i , c, s) ≤ n, for all i ∈ I. Then dim N X, min c, r s , s ≤ n.
Regarding finite unions, we show the following Lemma 2.7, which is essentially the quantified version of [LS05, Theorem 2.7].
Proof. Let s ∈ [s 0 , s 1 ] and take an s-bounded cover n j=0 Using K j write new collections of sets covering X ∪Y as follows. For all i ∈ I j and j = 0, . . . , n set under the assumptions that we have made for s, namely, The claim then follows.
Iterating the above lemma we get the analogous statement for finite unions.
By the above results (or just by [LS05, Theorem 2.7]) finite unions of sets with Nagata dimension n have Nagata dimension n. The same is not valid for a countable union. Indeed, the space Z has dimension one but it is the countable union of points, which have dimension zero. However, we can conclude that the Nagata dimension of a space is n, if we know that each ball B(x, r), at a fixed point x ∈ X, has Nagata dimension equal to n for the same constant c. In the case of separable metric spaces, we have the following more general fact. Suppose that there exists an increasing sequence of subsets X 1 ⊂ X 2 ⊂ · · · ⊂ X with ∞ i=1 X i = X separable and such that all X i have Nagata dimension equal to n with the same constant c > 0 in (2.1). Then one can show that dim N X = n. Without the assumption of separability, we can show the following fact, which is the one that we will use to give an upper bound for the Nagata dimension of limits of metric spaces.
Proof. By taking a subsequence if necessary, we may assume that r m+1 − r m > cs 1 . Let Notice that Similarly, dim N (Z, c, s) ≤ n, for all 0 < s ≤ s 1 .
Therefore the claim follows from Lemma 2.7.
Next lemma was observed together with Urs Lang. We will use the lemma and the example afterward to show that the Nagata dimension of Carnot groups is equal to the topological one.
Lemma 2.10 (See also [LS05, Proposition 2.8]). Let X be a metric space. Suppose there exist positive constants r, c and integers L, n such that every ball B(x, r), x ∈ X, is doubling with constant L and has Nagata dimension at most n with constant c. Then X has linearly controlled dimension at most n.
Proof. Put s 0 := r/3 and choose a set Z ⊆ X that is maximal, with respect to inclusion, subject to the condition that distinct points in Z are at distance more than s 0 from each other. Then the family of balls B(z, s 0 ), z ∈ Z, covers X. By the doubling condition, there is an integer N such every ball B(z, 3s 0 ) can be covered by N or fewer sets of diameter at most s 0 , so Z ∩ B(z, 3s 0 ) has cardinality at most N . It follows that there is a coloring of Z by N colors, k : Z → {1, . . . , N }, such that k(z) = k(z ) whenever 0 < d(z, z ) ≤ 3s 0 (see [Ass83, Lemma 2.4]). Let C k be the union of all balls B(z, s 0 ) with k(z) = k, for k = 1, . . . , N . Clearly every C k has linearly controlled dimension at most n. Now use finite gluing (Corollary 2.8) to show that X has linearly controlled dimension at most n.
Example 2.11. Assume G is a Lie group equipped with a left-invariant Riemannian distance. If n is the topological dimension of G, then G has linearly controlled dimension n. Indeed, fixing r > 0 small enough, all balls B(x, r), for x ∈ G, are isometric to each other and biLipschitz to an open set of the n-dimensional Euclidean space. In particular, such balls are doubling and have linearly controlled dimension at most n, with uniform constants. Hence, Lemma 2.10 gives one of the bounds. Recall that it is a general fact that the topological dimension does not exceed the linearly controlled dimension, see [LS05, Theorem 2.2].
2.3. Gromov-Hausdorff distance of pointed metric spaces. For defining limits of unbounded metric spaces, we need to consider pointed metric spaces. A pointed metric space is a pair (X, x) of a metric space X = (X, d) and a point x ∈ X. Recall that, when we are considering a metric space (X, d), we tend to refer to it simply as X whenever it is not necessary to specify the distance. Moreover, we denote by d the distance of any metric space that we are considering. Likewise, we will denote simply by X a pointed metric space when it is not important to specify the base point. In case we need to denote the base point (and the metric space in discussion is clear) we will use the symbol .
For a set A in a metric space X we denote the δ-neighborhood of A as Given two metric spaces X = (X, d X ) and Y = (Y, d Y ), we say that d is an extension of the distances on Y X if d is a semidistance on the set Y X (i.e., d might vanish on distinct points) and it coincides with d X when restricted to X and coincides with d Y when restricted to Y . We shall use the Gromov-Hausdorff convergence for pointed metric spaces. We will actually want to have a precise distance giving such a topology. Inspired by the definitions of Gromov and Gabber, see [Gro81, Section 6], we define the modified Gromov-Hausdorff measurement: for all pointed metric spaces X = (X, X ) and Y = (Y, Y ), we set ∃ an extension of the distances on Y X : Probably, the function Dist GH is not a distance, since it is not clear whether it satisfies the triangle inequality, as Gromov already pointed out. However, using the following Lemma 2.12 we can easily modify it to be a distance.
Lemma 2.12. Let X, Y, Z be pointed metric spaces.
Proof. The claim follows by considering distances of the form We define the Gromov-Hausdorff distance of two pointed metric spaces X, Y as Other simple properties that hold for Dist GH and thus for Dist GH are the following.
Lemma 2.13. Let X be a metric space.
(i) For all λ 1 > λ 2 > 0 and all x ∈ X, Proof. To easily obtain (i) one can isometrically embed X into the Banach space L ∞ (X). The dilations of the ambient Banach space give a straightforward calculation of the Hausdorff distance of λ 1 X and λ 2 X.
To see (ii) use the original distance as the extension. The claim (iii) follows from (i) and (ii).
To define tangent metric spaces, we dilate a metric space and consider the accumulation points of such sequences of dilated spaces. Recall that whenever the distance and the base point are clear we just write X to denote the pointed metric space (X, d, ), or we write d = d X to emphasize that d is the distance on X. Moreover, we denote by λX the pointed metric space obtained by dilating the distance by λ > 0. Namely, we set λX := (X, λd X , ).
Definition 2.14 (Tangent metric spaces). Let (X, d X ) be a metric space and x a point of it. Then a pointed metric space (Y, d Y , y) is said to be a tangent of (X, d X ) at x if (X, λd X , x) accumulate to (Y, d Y , y) in the pointed Gromov-Hausdorff topology, as λ → ∞. Namely, there exists a diverging sequence λ j such that We denote by Tan(X, x) the collection of all tangents of (X, d X ) at x.
It is easy to come up with examples of metric spaces with more than one tangent at a given point. Here is an easy criterion (which follows from Lemma 2.13(i)) to conclude that the tangent at a point is unique.
Lemma 2.15. Let X and Y be two pointed metric spaces.
(ii) More generally, let a n be a diverging sequence such that a n+1 /a n → 1, as n → ∞. If a n X → Y , as n → ∞, then λX → Y , as λ → ∞, λ ∈ R.
2.4. Some remarks on the dimensions of limits. In this section we prove that the dimensions of the tangents bound from below the dimension of a space. Regarding the Assouad dimension, this fact is a quantified version of the well-known result that limits of doubling metric spaces, with uniform constants, are doubling.
Lemma 2.16. Suppose that a sequence of pointed metric spaces X j converges to a pointed metric space Proof. Take 0 < r < R <R and x ∈ X ∞ . For all j ∈ N, let d j be a distance on X j X ∞ extending the distances on X j and X ∞ such that d( Indeed, take any y ∈ B X∞ (x, R) and y j ∈ X j with d j (y, y j ) < j . Now and so y j ∈ B X j (x j,i , r − 2 j ) for some i. Because we have y ∈ B X∞ (x j,i , r).
Since j → 0 as j → ∞, with large enough j we obtain a cover of B X∞ (x, R) with no more than C (R/r) β balls of radius r.
The following consequence can be also found in [MT10, Proposition 6.1.5].
Corollary 2.17. The Assouad dimension of any tangent space of a metric space X does not exceed the Assouad dimension of X.
Regarding the Nagata dimension, the similar bound is slightly less trivial and is based on Lemma 2.9.
Proposition 2.18. Suppose that a sequence of pointed metric spaces X j coverges to a pointed metric space X ∞ . Let 0 < s 1 < ∞ and 0 < c < 1. Assume that dim N (X j , c, s) ≤ n, for all 0 < s ≤ s 1 and all j = 1, 2, . . . .
Proof. Our aim is to show that for all 0 < c < c we have dim N (B X∞ ( X∞ , k), c , s) ≤ n, for all 0 < s < s 1 and all k = 1, 2, . . . .

(2.3)
Once we have obtained this, the claim will follow from Lemma 2.9. Fix an integer k > 0 and 0 < c < c. For all j ∈ N, let d j be a distance on X j X ∞ extending the distances on X j and X ∞ such that d( X j , X∞ ) ≤ j , B X j ( X j , 1/ j ) ⊆ B X∞ X j (X ∞ , j ), and B X∞ ( X∞ , 1/ j ) ⊆ B X∞ X j (X j , j ) for some sequence j going to 0.
Take 0 < s < s 1 and j ∈ N so that j < 1 k and cs − 2(1 + c) j ≥ c s.
(2.4) Let n i=0 B i be an (s − 2 j )-bounded cover of X j where each B i is c(s − 2 j )-separated. Define the new collections of sets B i as Now, for each x ∈ B X∞ ( X∞ , k), there exists x j ∈ X with d j (x, x j ) < j . Because x j ∈ B for some i = 0, . . . , n and B ∈ B i , we have x ∈ B for some i = 0, . . . , n and B ∈ B i . Therefore n i=0 B i is an s-bounded cover of B X∞ ( X∞ , k) where each B i is c s-separated, by (2.4). Corollary 2.19. The Nagata dimension of any tangent space of a metric space X does not exceed the Nagata dimension of X.
As shown by the next example, in general one cannot hope to deduce an upper bound for the dimension of the space by simply looking at the dimensions of the tangents.
Example 2.20. Define a set X ⊂ R as Let the distance be induced by the Euclidean distance on R. Then X is compact, dim A X = dim N X = 1 and Tan(X, where dim is either the Nagata or the Assouad dimension. Notice that the space in Example 2.20 has multiple tangents at 0, the convergence to the tangents is not uniform, and that in the definition of a tangent space we keep the base point fixed. Hence, if we want to obtain the Nagata dimension of the space as the maximum of the dimensions of the tangents, we need to either change the notion of tangents (e.g., consider weak tangents) or impose more restrictions on the convergence to the tangents. Regarding this last option, we will study the case of uniformly close tangent and, in the subsequent Section 4, we will apply our results to subRiemannian manifolds.
Observe also that in Example 2.20, if we take a larger class of tangents where we allow the change of the the base point, we obtain equality in (2.5). Simply let r i = i2 −i 3 and x i = 2 −i 2 . This observation is true in compact doubling metric spaces when dim N X = 1. This is the content of the following proposition, where we consider weak tangents, i.e., limits of the forms ( 1 r i X, x i ). Proposition 2.21. Let X be a doubling metric space with dim LC X ≥ 1. Then there exists a sequence of points x i ∈ X and a sequence r i 0 such that the sequence ( 1 r i X, x i ) converges to a space with linearly controlled dimension at least 1.
Proof. For each x ∈ X and δ > 0, define the iterated δ-neighborhoods of x by setting N 0 Since dim LC X ≥ 1, we claim that there exist a point x i ∈ X and a radius r i < 1 i such that diam(N ∞ X (x i , r i /i)) > r i . Indeed, suppose that this is not true. Then for every r < 1 i the collection {N ∞ X (x, r/i) : x ∈ X} would be an r-bounded r 2i -separated cover of X. This would mean that dim LC X = 0. By Gromov Theorem, see [BBI01, Theorem 8.1.10], because of the doubling assumption the sequence (r −1 i X, x i ) has a subsequence converging to a pointed metric space (Z, z). Now, for any 0 < < 1 2 , there is i > 1 such that Dist GH (r −1 i X, x i ), (Z, z) < .
Consequently, there exists a distance d extending the distances on r −1 i X and Z such that We do not know if Proposition 2.21 is true with higher lower bound on the dimension. We were unable to find a way to generalize our argument to the higher dimensional case.
Question 2.22. Let X be a metric space. Does there exist a sequence of points x i ∈ X and a sequence r i 0 such that the weak tangent of X along the sequences x i , r i has exactly the same linearly controlled dimension as X?

Dimension of uniformly close tangents
As we saw in Example 2.20, both the Nagata dimension and the Assouad dimension of a space can in general be strictly larger than the supremum of the dimensions of its tangents. We wonder when such a supremum equals the dimension of the space. A key assumption that we will make, which will still not be enough, is that the convergence to the tangents is uniform, as we now explain. Recall that Dist GH is the distance defined in Section 2.3.
Let X be a metric space. We say that the metric space X has uniformly close tangents in K ⊂ X if we have lim λ→∞ Dist GH ((λX, x), Tan(X, x)) = 0 uniformly in x ∈ K. If X has uniformly close tangents in X, we simply say that X has uniformly close tangents. Before investigating what other assumptions we need to make, we list some basic properties of uniformly close tangents.
Lemma 3.1. Assume that a metric space X has uniformly close tangents. If, for all x ∈ X, there exists only one element T x ∈ Tan(X, x), then the function x → T x is continuous.
We omit the easy proof of the above fact, which is proved exactly as one proves that the uniform limit of continuous maps is continuous, via Lemma 2.13(ii). In general, without the uniqueness assumption on the tangents, one cannot even conclude that the "graph" {(x, T ) : x ∈ X, T ∈ Tan(X, x)} is closed. Look for example at Example 3.3.
From [LD11] we know that if a complete metric space admits a doubling measure µ and if the space has unique tangents, then at µ-almost every point the unique tangent is an isometrically homogeneous space admitting dilations. We have the following version of this result.
Proposition 3.2. Let X be a complete doubling metric space with uniformly close tangents such that, for all x ∈ X, there exists only one element T x ∈ Tan(X, x). Then, for all x ∈ X, the tangent T x is an isometrically homogeneous space admitting dilations.
Proof. Since X is a complete doubling metric space there exists a doubling measure µ on X (see for instance [VK84], [LS98], or [KRS12]). By [LD11, Theorem 1.4] we already know that at µ-almost every x ∈ X the tangent T x is an isometrically homogeneous space. By Lemma 3.1 we know that all the tangents T x are Gromov-Hausdorff limits of isometrically homogeneous spaces. Since being isometrically homogeneous is stable under Gromov-Hausdorff convergence, all the tangents are isometrically homogeneous. The fact that the tangents admit dilations follows directly from the fact that Tan(X, x) are singletons.
3.1. Nagata dimension and uniformly close tangents. It turns out that having uniformly close tangents is not enough to tie the Nagata dimension of the space to the Nagata dimension of its tangents. This is shown by the next example. The set S n consists of n equally distributed points on the unit circle in R 2 .
Using the sets S n define for all n ∈ N the set The construction of X is illustrated in Figure 3.3. As tangents for every x ∈ cl (E n + 2 −n 2 ) we have and for the origin Tan(X, (0, 0)) = {{0, t} : t ∈ [0, ∞)} , We claim that the space X (with the distance induced by the Euclidean distance on R 2 ) has uniformly close tangents. To see this, take λ > 0 and x ∈ X. Let i ∈ N be such that 2 i 2 −i < λ ≤ 2 (i+1) 2 −i−1 . Then diam(λ2 −j 2 A j ) ≤ 2 −i for all j ≥ i + 1, so down to scale 2 −i we can consider the sets 2 −j 2 A j , j ≥ i + 1, to be just points in the λ-dilated distance. On the other hand, since any two distinct points in A i−1 have distance at least 1 i−1 between them and λ 1 i−1 2 −(i−1) 2 ≥ 1 i−1 2 i−1 , there is at most one point of 2 −(i−1) 2 A i−1 inside a ball of radius 1 1−i 2 i−2 in the λ dilated distance. In particular, Dist GH ((λX, x), Tan(X, x)) ≤ (1 − i)2 2−i for all x ∈ X. Hence X has uniformly close tangents. The Nagata dimension of any of the tangents of X is zero. However, the Nagata dimension and the linearly controlled dimension of the space are one.
In Example 3.3 the tangents of the space have Nagata dimension zero with smaller and smaller constant c as we move the base point towards (0, 0). On the other hand, the set of tangents is more than a singleton at every point in the space. In the following Theorem 3.4 we show that if we rule out one of the above mentioned properties, namely if we require either uniformity of the constant c or the uniqueness of tangents, the linearly controlled dimension of the space is bounded above by the supremum of the dimensions of its tangents.
Theorem 3.4. Let X be a metric space. Let K ⊂ X be a subset on which the convergence to tangents is uniform. Assume that dim LC K < ∞ and that one of the following two situation holds: (i) dim LC (B Y ( , 1)) ≤ n, for all x ∈ K and Y ∈ Tan(X, x), with uniform constants c and s, or (ii) K is compact, for every x ∈ K, the tangent is unique, i.e., the set Tan(X, x) is a singleton denoted by T x X, and dim LC (B TxX ( , 1)) ≤ n. Then dim LC K ≤ n.
For proving the above theorem, we need the following two lemmas.
Lemma 3.5. Let (X, x) and (Y, y) be two pointed metric spaces. Suppose that there exist some , r, c, s, n such that Dist GH ((X, x), (Y, y)) < and dim N (B Y (y, r), c, s) ≤ n.
Proof. Since (X, x) and (Y, y) have distance < , we can see them as a subset of a metric space Z such that d(x, y) ≤ , and B X (x, 1/ ) ⊆ B Z (Y, ).
We claim that Let U be an s-bounded cover of B Y (y, r) such that U = U 0 . . . U n with each U j cs-separated. Set V j := {B Z (U, ) ∩ X : U ∈ U j }. Clearly, V j are (s + 2 )-bounded and (cs − 2 )-separated. Since (3.1), the family V 0 ∪ . . . ∪ V n gives a cover of B X (x, r ).
and that there exists r > 0 such that, for all x ∈ K, for all r ∈ (0, r ), and for all s ∈ (0, s 0 ), Proof. Take 0 < r < min{r , s}. From (3.2), we have an r 2 -bounded cover For all j = 0, . . . , n, setting U j := U ∈U j U , from Lemma 2.6 we get , r(as + b ) ≤ n, for all 0 < s < s 0 .
By decreasing s and to be small enough, we may assume that 2(as + b ) < 1. Set s := b(2+c) ac . Then one can easily check that, if s > s , Indeed, one just needs to verify that, by substituting the value of in terms of s , the above inequality is equivalent to s > s . Therefore, setting c 1 = min c 2 , c we have that, for all s between s and s 0 and all j = 0, . . . , n, we have dim N U j , c 1 , r(as + b ) ≤ n. In order to have dim LC X ≤ n it now suffices to select the constant so that there exists some s satisfying (3.4). This is the case if we take Proof of Theorem 3.4. We shall prove the claims by checking that the assumptions of Lemma 3.6 hold. First of all, the assumption dim LC X < ∞ implies that there exist n, s, c such that (3.2) is satisfied in both cases of the theorem. Since by assumption the tangents are uniformly close in K, for any > 0 there exists λ such that, for all λ ≥ λ and all x ∈ K, Dist GH ((λX, x), Tan(X, x)) < . (3.5) We now consider separately the two situations assumed in the hypothesis of the theorem. In the first case we will show that (3.3) holds. So let ∈ (0, 1 8 ) be fixed. In this case we are assuming that we already have c and s 0 such that, for all x ∈ K and for all Y ∈ Tan(X, x), for all 0 < s < s 0 .
With respect to the distance of X, the equation reads as for all λ ≥ λ and s ∈ (0, s 0 ), which is (3.3) with r := 1 λ , (put r = 1 λ ). By Lemma 3.6, the first case then follows.
Regarding the second case, we have that for some c x depending on x ∈ K and for all s > 0, since unique tangents admit dilations. Take now x ∈ K and let x ∈ (0, 1 8 ) be the in Lemma 3.6 given by the constants s 0 = ∞, a = 4, b = 12, and c = c x . We abbreviate λ x := λ x . From (3.5) and (3.7), Lemma 3.5 implies that, since for all λ ≥ λ x and all s > 0.
Finally, multiplying the distance by r, we get (3.3) for x replaced by any y ∈ B X (x, 1 4λx ) and with r := 1 4λ x , s 0 = ∞, a = 4, b = 12, and c = c x . By Lemma 3.6 we then have By compactness of K there exists a finite collection of balls {B X (x, 1 4λx )} x covering K. Using Corollary 2.8 we then conclude that which finishes the proof of the second case.
Proof of Theorem 1.2. Since Y is relatively compact, the set K := cl Y is compact. Note that dim N K = dim N Y . By Theorem 3.4 with assumption (ii), Take now a point x in the interior of Y . Then by Corollary 2.
3.2. Assouad dimension and uniformly close tangents. Let us now state the analog of the first part of Theorem 3.4 for Assouad dimension.
Theorem 3.7. Let X be a metric space with uniformly close tangents. Let C > 1 and α ≥ 0. Assume that, for all x ∈ X and for all (Y, y) ∈ Tan(X, x), the Assouad dimension of Y is less than or equal to α with constant C. Then there exists R > 0 such that for all x ∈ X.
In fact, Theorem 3.7 is an immediate consequence of the following proposition where we only need to have a cover of the balls in the tangents centered at the base point.
Proposition 3.8. Let X be a metric space with uniformly close tangents. Assume that there are constants C > 1 and α ≥ 0 such that, for all x ∈ X, for all (Y, y) ∈ Tan(X, x), and for all δ ∈ (0, 1), we need at most C(2/δ) α balls of radius δ/2 to cover the ball B(y, 2). Then there exists a constant R > 0 such that, for all β > α, there exists C > 1 for which for all x ∈ X.
The analog of the second part of Theorem 3.4 is the following.
Theorem 3.9. Let X be a metric space with uniformly close unique tangents in a compact subset K ⊂ X. Then Again, as for the Nagata dimension, we have to make one of the two assumptions on the tangents, given in the two previous theorems, in order to be able to get a bound on the Assouad dimension of the space from the Assouad dimensions of its tangents. The following example, similar to Example 3.3, shows that the assumptions are necessary.
Example 3.10. In Example 3.3 we defined the space X as a subset of the Euclidean plane. The construction had three stages: first we defined S n , then using it E n and finally X. Here we replace S n by a set {0, 1} n . We want the points in this set to be equidistant and so we consider {0, 1} n ⊂ R n with the maximum norm.
We define E n ⊂ R n as 1} n if i odd, and a i ∈ {0, 1} if i even and using it we define X ⊂ R N as where the embedding of different dimensional R n to R N are understood by identifying x ∈ R n with (x, 0, 0, . . . ) ∈ R N . We equip X with the supremum distance of R N . We have for all n ∈ N and x ∈ E n , and As in Example 3.3, we have uniformly close tangents essentially because for all x ∈ X. This set of tangents takes care of the scales where other sets of the form {0, 1} n can not be yet seen. Take any α > 0. For any x ∈ X and any Y ∈ Tan(X, x) the tangent Y has only finite number of points and hence dim A Y ≤ α. However, for any n ∈ N the set {0, 1} n needs 2 n balls of radius r < 1 to cover it. Therefore for any C > 0 there are arbitrarily small balls B(x, r) in X that can not be covered by less than C2 α balls of radius r 2 . Thus dim A X ≥ α. This was true for any α > 0 and so, dim A X = ∞ and sup{dim A Y : x ∈ X, Y ∈ Tan(X, x)} = 0.
Hence, the two assumptions in Theorem 3.7 and in Theorem 3.9 are necessary.
Before proving Proposition 3.8 we provide a lemma.
Lemma 3.11. Assume that, for some δ ∈ (0, 1), we have Dist GH ((X, x), (Y, y)) < δ/4 and that, for some L ∈ N, we have that the ball B Y (y, 2) can be covered with L balls of radius δ/2. Then the ball B X (x, 1) can be covered with L balls of radius δ.
By assumption, there exists λ 0 > 1 such that, for all λ > λ 0 and for all x ∈ X, there exists (Y, y) ∈ Tan(X, x) such Dist GH ((λX, x), (Y, y)) < δ/4. Let R and r such that 0 < r < R < 1/λ 0 . Fix N such that We will apply Lemma 3.11 N times. First, since 1/R > λ 0 , by Lemma 3.11, we have that, in the contracted metric space 1 R X, we need at most L := C(4/δ) α balls of radius δ to cover B 1 R X (x, 1). Such balls are of the form 1). Each of these balls needs at most L balls of radius δ to cover it in the space 1 δR X. Iterating, we need at most L N balls of radius δ, with respect to the distance of 1 δ N −1 R X, to cover B 1 R X (x, 1) = B X (x, R). Such sets are balls of radius δ N R for X. In other words, we covered B X (x, R) with balls of radius r, since r > δ N R. The number of these balls is bounded by Applying Lemma 2.5 with C β := 1/δ β , we conclude.
Proof of Theorem 3.9. Denote by (T x X, ) the unique element in Tan(X, x), for x ∈ K. Take β > α > sup{dim A T x X : x ∈ K}. Fix some x ∈ K and let 1 < C x < ∞ be such that for any 0 < r < R < ∞ and y ∈ T x X we need at most C x (R/r) α balls of radius r to cover the ball B TxX (y, R). Let δ x ∈ (0, 1) be such that ( 1 16δx ) β = C x (4/δ x ) α =: L. By assumption, there exists λ x > 1 such that, for all λ > λ x and for all y ∈ K we have Dist GH ((λX, y), (T y X, )) < δ x /4.
Our aim is to show that for any z ∈ B(x, 1 4λx ) ∩ K and 0 < r < 1 4λx we need at most L balls of radius 8δ x r to cover the ball B(z, r). Fix such z and r. By the definition of C x the ball B TxX ( , 2) needs at most L balls of radius δx 2 to cover it. Since 2λ x > λ x we have Dist GH ((2λ x X, x), (T x X, )) < δ x /4, and so by Lemma 3.11 we need at most L balls of radius δx 2λx to cover the ball B(x, 1 2λx ). Since B(z, 1 4λx ) ⊂ B(x, 1 2λx ), the same collection of balls covers B(z, 1 4λx ). Using the fact that the tangent of X at z is unique and the triangle inequality, we get Dist GH ((r −1 X, z), (8λ x X, z)) ≤ Dist GH ((r −1 X, z), (T z X, )) + Dist GH ((T z X, ), (8λ x X, z)) < δ x /2.
Since the ball B 8λxX (z, 2) = B X (z, 1 4λx ) needed at most L balls of the type B 8λxX (x , 4δ x ) = B X (x , δx 2λx ) to cover it, by Lemma 3.11 we need no more than L balls of radius 8δ x r to cover the ball B(z, r).
Now we continue like in the proof of Proposition 3.8. Let R and r such that 0 < r < R < First, since R < 1 4λx , by the above considerations we need at most L balls of radius 8δ x R to cover the set B(x, R) ∩ K, and hence at most L balls of radius 16δ x R centered at B(x, R) ∩ K. Each of these balls, since they are centered at B(x, R) ∩ K, need at most L balls of radius (16δ x ) 2 R centered at B(x, R) ∩ K to cover them. Continuing inductively N times, we obtain a cover of B(x, R) ∩ K with at most L N balls of radius (16δ x ) N R. Again we estimate the number By compactness of K we then have dim A K ≤ β. Letting α and β tend to sup{dim A T x X : x ∈ K, Tan(X, x) = {T x X}} finishes the proof.
Proof of Theorem 1.4. Since Y is relatively compact, K := cl Y is compact. Note that dim A K = dim A Y . By Theorem 3.9, Take now a point x in the interior of Y . Then by Corollary 2.

Nagata dimension of Carnot groups and equiregular subRiemannian manifolds
Let M be a smooth manifold. Let ∆ be a smooth subbundle of the tangent bundle of M . Denote by Γ(∆) ⊂ Vec(M ) the C ∞ (M )-module of the smooth sections of ∆. One says that ∆ satisfies the bracket-generating condition if (4.1) A subRiemannian manifold is a triple (M, ∆, g), where M is a connected smooth manifold, ∆ is a bracket-generating subbundle of the tangent bundle of M and g is a Riemannian metric tensor restricted to ∆.
A subRiemannian manifold has a natural structure of metric space, where the distance is the so-called Carnot-Carathéodory distance As a consequence of Chow-Rashevsky Theorem such a distance is always finite and induces on M the original topology.
The subRiemannian manifold is called equiregular if the dimensions of the spaces ∆ i q , i ∈ N, as defined in (4.1), do not depend on the point q.
Very particular subRiemannian manifolds are the Carnot groups. Let us briefly recall that a Carnot group is a stratified Lie group endowed with a left-invariant Carnot-Carathéodory metric where the subbundle is the first stratum of the Lie algebra. See [LD13] for an introduction to Carnot groups from a point of view of metric geometry.
Theorem 4.1 (Mitchell-Bellaïche [Bel96]). Let M be an equiregular subRiemannian manifold equipped with its Carnot-Carathéodory distance d cc . Then, at every point the tangent space is a Carnot group of same topological dimension. Moreover, the convergence to the tangents is uniform on compact sets.
We will be able to deduce the local Nagata dimension of an equiregular subRiemannian manifold, since we know the Nagata dimension of Carnot groups, by a result originally proved by Urs Lang and the first-named author. Many thanks go to Lang for giving the permission to include here a short proof.
Remark 4.3. In the case that the subRiemannian manifold is not equiregular, one can still use the fact that the set of regular points for the subbundle is an open dense subset of the manifold. Here, a point p is called regular if it has a neighborhood on which the subbundle is equiregular. Hence, we have that there exists a neighborhood of p whose Nagata dimension equals the topological dimension of the manifold.
Remark 4.4. Probably the most famous Carnot-Carathéodory space that is not an equiregular subRiemannian manifold is the Grushin plane. In this example the dimension of ∆ q varies with q. Also in this case the Nagata dimension is equal to the topological one. Indeed, recently Meyerson gave simple examples of quasi-symmetric maps between the Grushin plane and the Euclidean plane. In the usual coordinates, an example of such maps is F (x, y) = (x|x|, y), see [Mey11]. On the other hand, Lang and Schlichenmaier in [LS05] proved that the Nagata dimension is a quasi-symmetric invariant. Therefore, the Nagata dimension of the Grushin plane is two, like the topological dimension.
On compact sets, a Carnot-Carathéodory distance d C and a Riemannian distance d R on G satisfy the inequalities 2) for some m ∈ N and k ∈ (0, 1), see [NSW85]. Because snowflaking a distance preserves the Nagata dimension, it is natural to wonder if partial snowflaking as in (4.2) preserves it as well, giving an alternative method of proof for Theorem 4.2 and Corollary 1.3. We give an example showing that this is not the case.
Example 4.5. We introduce a topological space X with two proper distances d 1 and d 2 with the property that The space X is of the form {p n,j : n ∈ N, j = 0, . . . , n} endowed with the discrete topology. The distances are d 1 (p n,j , p n ,j ) = if n = n , 1 2 n , if n = n , j = j 0, if n = n , j = j and d 2 (p n,j , p n ,j ) = One can easily check that these are distance functions, that satisfy d 2 1 ≤ d 2 ≤ d 1 , and that dim N (X, d 1 ) = 0 but dim N (X, d 2 ) = 0. The only calculation that is not completely straightforward is that dim N (X, d 2 ) = 0. For doing this, one assumes that the dimension is 0 with respect to some constant c. Then we fix n ∈ N and let s = 1 n2 n . Take a cs-bounded cover with s-multiplicity at most 1. This last property implies that p n,0 and p n,1 , which have distance s, need to be contained in the same element U of the cover. Likewise, p n,j ∈ U , for all j = 0, . . . , n. Thus 1 2 n ≤ diam U ≤ cs = c 1 n2 n . So c > n, for any n ∈ N, a contradiction.

Assouad dimension bounds Nagata dimension
Denoting by dim X the topological dimension of a metric space X and by dim H X its Hausdorff dimension, recall that one has the chain of inequalities From the bound proven in this section, we will conclude that we also have the inequalities where the first inequality is obtained in [LS05]. In [LS05], Lang and Schlichenmaier also proved the inequality dim N X ≤ 3 dim A X − 1. We optimally improve it. We should also point out that in general there is no relation between dim N and dim H . As examples, on one hand dim N Q = 1 > 0 = dim H Q, and on the other hand dim N (R, | · | 1 2 ) = 1 < 2 = dim H (R, | · | 1 2 ). After the proof, in Theorem 5.1 we show how the argument can be modified to give a boundedscale version of the same result.
Proof of Theorem 1.1. Let X be a metric space. We shall prove that dim N X ≤ dim A X. Without loss of generality we may assume dim A X < ∞. Take α so that dim A X < α < dim A X + 1. From the definition of the Assouad dimension we know that there is some constant C > 0 such that, for any x ∈ X and 0 < r < R < ∞, we need at most C R r α balls of radius r to cover the set B(x, R) ∩ X. (5.1) We shall fix r R to be determined later in terms of C and α only. The idea of the proof is the following. First decompose X =: X 0 as with the properties that: diam(Y 1 n ) < 2R, dist(Y 1 n , Y 1 m ) > r, for all n = m, and all balls B(x, R) ∩ X 1 , with x in X 1 , need at most C 1 R r α−1 balls of radius r to cover them. Here C 1 is a constant depending only on α and C. Hence, roughly speaking, the subset X 1 has codimension one on scales R and r. Then we will iterate the decomposition X k−1 = X k Y k 1 Y k 2 · · · with similar properties for Y k n . As soon as k > α and both R and r are properly chosen, the subset X k is empty. The collection {Y k n : k = 1, . . . , α + 1, n = 1, . . . , N k } gives a cover showing that the Nagata dimension is less than α.
Take R > 0 and let {x i } N 1 i=1 ⊂ X be a maximal R 4 -separated net of points. (It might be that N 1 = ∞. However, recall that metric spaces with finite Assouad dimension are separable and hence separated nets are countable.) Notice that in particular the balls B(x i , R 2 ) cover the set X. Take 0 < r < R 4 . For each x n ∈ {x i } there exists by (5.1) a collection B = {B(y i , r)} of at most C R r α balls covering the larger ball B(x n , R). Let k ∈ N be such that kr < R 2 ≤ (k + 1)r. Consider the annular regions A n,i := B(x n , R − ir) \ B(x n , R − (i + 1)r).
The collection {A n,i } k−1 i=0 is disjointed and Observe that any ball B ∈ B intersects at most 3 of the annular regions A n,i .
Therefore there exists some 0 ≤ k n ≤ k − 1 such that the annulus A n,kn meets at most 3 k C R r α balls of the collection B. Notice that we can bound Then, in particular, for all n, the number of balls needed to cover A n,kn is at most 12C R r α−1 .
(5.2) Set X 1 := n A n,kn . Now, take a point x ∈ X 1 . We claim that the ball B(x, R) intersects at most C16 α annular regions A i,k i . What we have now obtained is a decomposition of X into disjointed collection {Y 1 n } N 1 n=1 ∪{X 1 } with Y 1 n having diameter less than 2R and the property that dist(Y 1 i , Y 1 j ) ≥ r for any i = j. Repeating the above argument we can decompose X 1 into disjointed collection {Y 2 n } N 2 n=1 ∪ {X 2 } with Y 2 n again having diameter less than 2R and the property that dist(Y 2 i , Y 2 j ) ≥ r for any i = j, and X 2 such that for any point x ∈ X 2 we need at most C 2 R r α−2 balls of radius r to cover the set X 2 ∩ B(x, R), where C 2 depends only on C and α.
We continue this for m = α +1 steps so that we have a decomposition of X into disjointed collection with all Y k n having diameter less than 2R and dist(Y k i , Y k j ) ≥ r for any i = j. But now at the last iteration, step m, when we have arrived to (5.2) we notice that there exists an annular region that intersects at most 12C m−1 R r α−m balls of the cover. Provided that we have chosen the ratio r R to be small enough from the beginning, this means that the annular region is empty, since α − m < 0. The conclusion is that X m = ∅.

Hence, we have
Now, given any x ∈ X, for any k = 1, . . . , m, the ball B(x, r/2) intersects at most one set from each collection {Y k n } N k n=1 . Thus, dim N X ≤ α ≤ dim A X.
The previous proof generalizes to the case of local dimensions of metric spaces that admit well-ordered separated nets. For example, this is the case in separable metric spaces. Nevertheless, if we assume the Axiom of Choice, any net can be well-ordered.
Theorem 5.1. Let X be a metric space. Let α ≥ 0, C > 0 and R > 0. Assume that, for all x ∈ X, we have dim A (B(x, R), C, R) ≤ α. Then dim LC X ≤ α.
Proof. The strategy of the proof is very similar to the one of Theorem 1.1. Hence we only explain the differences. Instead of having a countable maximal R 4 -separated net of points {x i } i , we now have just a maximal R 4 -separated net {x j } j∈J where J is a general set of indices.
By the Well-Ordering Theorem the set J admits a total order such that every nonempty subset of J has a least element for this ordering. (This result is also known as Zermelo's theorem. It follows easily from Zorn's lemma and it is actually equivalent to it.) For all j ∈ J, we select an annulus A j,k j as in the proof of Theorem 1.1. Then we set X := j∈J A j,k j and The only nontrivial property to check is that {Y 1 j } j∈J together with X 1 is a cover. Pick x ∈ X. Set E x := {j ∈ J : x ∈ B(x j , R − k j r)}.
The set E x is nonempty, since R − k j r > R 2 . By the well-ordering, we have a least element j x ∈ E x . Thus x is in B(x jx , R − k jx r) but not in any of the B(x i , R − k i r) for i < j x . If x is not in X 1 (and hence not in A jx,k jx ), we have that x is in Y 1 jx . The rest of the proof is exactly the same.