Rigidity of quasisymmetric mappings on self-affine carpets

We show that the class of quasisymmetric maps between horizontal self-affine carpets is rigid. Such maps can only exist when the dimensions of the carpets coincide, and in this case, the quasisymmetric maps are quasi-Lipschitz. We also show that horizontal self-affine carpets are minimal for the conformal Assouad dimension.


Introduction
We consider the following two general questions: (1) By understanding the fine structure of sets, is it possible to say anything about quasisymmetic mappings between the sets? (2) What kind of sets are minimal for the conformal dimension? In the class of horizontal self-affine carpets we can answer both of the questions. Our method to prove the results builds on the analysis of weak tangent sets and mappings. This is done in general metric spaces and therefore, should hold an independent interest and also provide a framework for possible further applications.
If (X, d) and (Y, ) are metric spaces and η : [0, ∞) → [0, ∞) is a homeomorphism, then a homeomorphism f : ≤ η d(x, y) d(x, z) for all x, y, z ∈ X with x = z. Quasisymmetric mappings are a non-trivial generalization of bi-Lipschitz mappings. While bi-Lipschitz maps shrink or expand the diameter of a set by no more than a multiplicative factor, quasisymmetric maps satisfy the weaker geometric property that they preserve the relative sizes of sets: if two sets A and B have diameters t and are no more than distance t apart, then the ratio of their sizes changes by no more than a multiplicative constant.
The Assouad dimension of a set E ⊂ X, denoted by dim A (E), is the infimum of all t satisfying the following: there exists a constant C ≥ 1 such that each set E ∩ B(x, R) can be covered by at most C(r/R) −t balls of radius r centered at E for all 0 < r < R. The conformal Assouad dimension of E is Cdim A (E) = inf{dim A (E ) : E is a quasisymmetric image of E}. It is worth emphasizing that the codomains of the quasisymmetric mappings used in the definition can be any metric spaces. A set E is minimal for the conformal Assouad dimension if dim A (E) = Cdim A (E). We remark that conformal dimension and minimality can similarly be defined also for other set dimensions.
Concerning the question (1), Bonk and Merenkov [3, Theorem 1.1] have recently shown that every quasisymmetric self-map on the standard 1 3 -Sierpiński carpet is an isometry. The 1 3 -Sierpiński carpet is a planar self-similar set satisfying the open set condition obtained by a repetitive process where, at each step, the cube is divided in nine subcubes and the middle one is removed. Concerning the question (2), Mackay [11,Theorem 1.4] has shown that a Gatzouras-Lalley carpet E is minimal when its projection onto the horizontal coordinate is a line segment; otherwise Cdim A (E) = 0. Gatzouras-Lalley carpets are self-affine sets constructed by a repetitive process in which, using the same pattern at each step, the rectangle is first partitioned into vertical tubes and then from each tube a collection of disjoint subrectangles is chosen in such a way that the vertical length of each subrectangle is strictly smaller than the horizontal length which equals the width of the tube.
Theorems A and B below give an answer for both of the questions in the class of horizontal self-affine carpets. A set in this class is constructed by a repetitive process in which, using the same pattern at each step, in a given rectangle, we choose a collection of disjoint subrectangles having width longer than height such that every vertical line going through the rectangle intersects at least two such subrectangles. The precise definition and the proofs of the results are given in §5.
Theorem A. If E and F are horizontal self-affine carpets, then any quasisymmetric mapping f : E → F is quasi-Lipschitz.
uniformly as d(x, y) → 0. The class of quasi-Lipschitz mappings is strictly more general than the class of bi-Lipschitz mappings. For example, f : [0, 1] → R defined by f (0) = 0 and f (x) = x log x for all x ∈ (0, 1] is quasi-Lipschitz, but not Lipschitz. In Lemma 3.5, we show that quasi-Lipschitz mappings preserve the Hausdorff dimension and the upper and lower Minkowski dimensions. Therefore, two horizontal self-affine carpets either have the same dimension or there does not exist a quasisymmetric mapping between them. This is in a huge contrast to the self-similar case. Wang, Wen, and Zhu [14] have shown that all self-similar sets satisfying the strong separation condition are quasisymmetrically equivalent.
Theorem B. Horizontal self-affine carpets are minimal for the conformal Assouad dimension.
In Example 5.11, we show that a self-affine carpet can be minimal even if the projection onto the horizontal coordinate is not a line. Furthermore, in Remark 5.9, we point out that, to obtain Theorem B, instead of the vertical line condition it suffices to assume that the projection of the carpet onto the horizontal coordinate is a line segment. Therefore, under the strong separation condition, Theorem B strictly generalizes the result of Mackay [11,Theorem 1.4].
The proof of Theorem A has three essential steps. At first, in Theorem 3.1, we show that a quasisymmetric mapping is quasi-Lipschitz provided that its quasisymmetric weak tangent maps are bi-Lipschitz. The second step is to find a geometric condition for the weak tangent sets under which quasisymmetric weak tangent maps are bi-Lipschitz. This is done in Lemma 4.2 by modifying the result of Le Donne and Xie [10, Theorem 1.1] for our purposes. We show that quasisymmetric maps between finite unions of fibered spaces are bi-Lipschitz. It is worth emphasizing that both of these results hold in general metric spaces. Finally, in Theorem 5.1, we show that the weak tangent sets of horizontal self-affine carpets are finite unions of fibered spaces. This generalizes the result of Bandt and Käenmäki [1,Theorem 1]. It is also an essential ingredient in the proof of Theorem B. To prove Theorem B, it therefore remains to show that any compact set, whose weak tangents are minimal, is minimal for the conformal Assouad dimension.

Convergence of metric spaces and mappings
The purpose of this section is to introduce weak tangents for metric spaces and quasisymmetric mappings. To that end, let us start fixing some notation. We say that a metric space (X, d) is doubling with a constant N ∈ N if any closed ball B X (x, r) with center x ∈ X and r > 0 can be covered by N balls of radius r/2. If the underlying metric space is clear from the context, we write B(x, r) instead of B X (x, r). Recall that if a metric space (X, d) is doubling, then, according to the Assouad embedding theorem, for any 0 < α < 1, (X, d α ) can be mapped onto a subset of some Euclidean space by a bi-Lipschitz mapping. We say that a metric space (X, d) is uniformly perfect with a constant D ≥ 1 if for all x ∈ X and r > 0 we have B(x, r) \ B(x, r/D) = ∅ whenever X \ B(x, r) = ∅.
We continue by recalling some of the definitions from the book of David and Semmes [4, §8]. Let (F i ) be a sequence of non-empty closed subsets of R d . We say that F i converges to a nonempty for all R > 0. In this case, if (X, d) is a metric space, and ϕ i : F i → X and ϕ : F → X, then we say that ϕ i converges to ϕ if for each sequence ( Note that the limit function ϕ is unique. Recall that a pointed metric space is a triple (X, d, p), where (X, d) is a complete doubling metric space and p ∈ X is a fixed basepoint. Suppose that (X, d, p) and (X i , d i , p i ) are pointed metric spaces for all i such that all the metric spaces are doubling with the same constant. By the Assouad embedding theorem, choose 0 < α ≤ 1 and bi-Lipschitz embeddings h i : [4,Lemma 8.12], the limit (X, d, p) is unique up to an isometry that respects the basepoint. Observe also that, by [4,Lemma 8.13], a sequence of pointed metric spaces has a converging subsequence.
A mapping package is a triplet ((X, d, p), (Y, , q), f ) consisting of two pointed metric spaces and a mapping f : X → Y for which f (p) = q. Suppose that P = ((X, d, p), (Y, , q), f ) and where h, g, h i , g i are the Assouad embeddings of X, Y, X i , Y i , respectively, as above. By [4,Lemma 8.21], the limit P is unique up to isometries that respect the basepoints. Observe also that, by [4,Lemma 8.22], if P i = ((X i , d i , p i ), (Y i , i , q i ), f i ) are mapping packages for all i such that the mappings f i are equicontinuous and uniformly bounded on bounded sets and all the metric spaces are doubling with the same constant, then P i has a converging subsequence. Recall that {f i } is equicontinuous on bounded sets if for every R > 0 and ε > 0 there is δ > 0 such that is a mapping package for all i ∈ N such that all the metric spaces are doubling with the same constant. Assume further that there exists a constant C ≥ 1 and a sequence (w i ) such that w i ∈ X i , for all i ∈ N. Then, after passing to a subsequence, the mapping packages P i converges to a mapping package ((X, d, p), (Y, , q), f ), where f : X → Y is η-quasisymmetric. Moreover, there exists w ∈ X such that 1/C ≤ d(p, w) ≤ C, and 1/C ≤ (q, f (w)) ≤ C.
Proof. As remarked above, the convergence of mapping packages follow once we verify that the family {f i } is equicontinuous and uniformly bounded on bounded sets. The argument here can be compared to [13,Theorems 2.21 and 3.4] and [5,Corollary 10.30]. Let us start with the equicontinuity. Fix R, ε > 0. Let δ > 0 be so that δ < (4C) −1 and η (4Cδ) η ((C + R)C) C < ε.
In the case d i (x i , p i ) > (4C) −1 we get the same estimate just by switching the roles of p i and w i above. We have thus verified the equicontinuity on bounded sets. The uniform boundedness on bounded sets follows immediately since for fixed R > 0 and Therefore, there exists a subsequence along which the mapping packages converge. In what follows, we keep denoting the subsequence by the original sequence. It remains to show that f : X → Y is η-quasisymmetric. Since the mapping packages P i converge there are Assouad embeddings h, g, h i , g i of X, Y , X i , Y i , respectively, such that Furthermore, since both pointed metric spaces of P i converge we also have Fix three different points x, y, z ∈ X. Pick three sequences h i (x i ), h i (y i ), h i (z i ) converging to h(x), h(y), h(z), respectively. By (2.1), we see that Therefore, since This implies that f is a continuous injection. If it were a bijection, then it is elementary to see that f −1 satisfies (2.3) with 1/η −1 (t −1 ) in place of η(t). Hence f −1 is continuous and f is η-quasisymmetric. So the only thing left to prove is that f is surjective.
To that end, fix y ∈ Y . Pick a sequence g i (y i ) converging to g(y). By (2.2), there is i 0 such that Similarly as in the beginning of the proof, we see that {f −1 i } is uniformly bounded on bounded sets. Thus there exists C ∈ R such that d for all i ≥ i 0 . Since the sequence (h i (x i )) is contained in a compact subset of some Euclidean space it has a converging subsequence. Denoting the limit point of this subsequence by z, we have, by (2.1), that f (h −1 (z)) = y. This is what we wanted to show.
Finally, since the Assouad embeddings h i are bi-Lipschitz with the same constant, the points h i (w i ) are contained in an annulus centered at the origin. Therefore there exists a converging subsequence. By definitions, this finishes the proof.
We will next apply the notions of convergence to define weak tangents for metric spaces and quasisymmetric mappings. Let (X, d) be a complete doubling metric space. We say that (X,d, p) is a weak tangent of (X, d) if there are a sequence (p i ) of points in X and a sequence (r i ) of positive reals converging to zero such that (X, d/r i , p i ) converges to (X,d, p). Note that from each given (p i ) and (r i ) one can extract a subsequence along which there exists a weak tangent. If (p i ) is a constant sequence, then weak tangents are called tangents.
Furthermore, let (Y, ) be another complete doubling metric space and f : X → Y . If (X,d, p) and (Ŷ ,ˆ , q) are weak tangents of (X, d) and (Y, ), respectively, then we say thatf :X →Ŷ is a weak tangent mapping of f if the mapping packages Observe that if f is η-quasisymmetric, then by choosing a sequence (w i ) of points in X such that d(p i , w i ) converges to zero and setting r i = d(p i , w i ) and t i = (f (p i ), f (w i )) for all i, Lemma 2.1 guarantees that there exists a subsequence along which the weak tangent mappingf of f is η-quasisymmetric and a point w = p inX such that (f (p),f (w)) =d(p, w). (2.4)

Rigidity of quasisymmetric mappings
In this section, we prove a rigidity result for quasisymmetric mappings between two metric spaces. We show that if the weak tangent mappings are bi-Lipschitz, then the original quasisymmetric mapping is quasi-Lipschitz. This result allows us to transfer the tangent level rigidity information back to the original quasisymmetric mapping. This is a useful observation since it is often the case that tangents are more regular than the original object. In §5, we will exhibit this phenomenon in a concrete setting.
Theorem 3.1. Suppose that (X, d) and (Y, ) are uniformly perfect compact doubling metric spaces. If f : X → Y is η-quasisymmetric such that any η-quasisymmetric weak tangent mappingf of f satisfying (2.4) for some p and w is L-bi-Lipschitz with L depending only on η, then f is quasi-Lipschitz.
Let us sketch the main idea of the proof. Assuming that f is not quasi-Lipschitz, we find a sequence of pairs of points in which f obeys a true Hölder behavior. In Lemma 3.3, we show that for each pair we find a triplet of points, with comparable distances, in which f still obeys a true Hölder behavior. These triplets then allow us to define weak tangets so that the limiting maps are quasisymmetric and the convergence of distances ensures that the true Hölder behavior is visible at the limit. This contradicts the assumption that the weak tangent mappings are bi-Lipschitz.
We will first recall a general lemma about the Hölder behavior of quasisymmetric mappings.
Lemma 3.2. Suppose that (X, d) and (Y, ) are uniformly perfect bounded metric spaces. If f is η-quasisymmetric, then there exist exponents Λ ≥ λ > 0 and constants C ≥ c > 0 depending only on η, the diameters of the spaces, and uniform perfectness constants such that Proof. The lemma is stated in [5,Corollary 11.5]. By [5,Theorem 11.3], there are constants Q ≥ 1 and 0 < β ≤ 1 depending only on η and uniform perfectness constants such that f is η-quasisymmetric whereη Note that the same estimates work even if we have to choose z = y. The other direction follows by using similar estimates for the inverse of f .
The above result is needed in the proof of the following lemma which is a key observation to prove Theorem 3.1.
Proof. Observe that, by Lemma 3.2, there exist exponents Λ ≥ λ > 0 and constants C ≥ c > 0 such that Fix ε 0 > 0. Let us first show that for every 0 < ε < ε 0 and m ∈ N there exist triplets (a, b, c) satisfying (1) and (2). To that end, fix 0 < ε < ε 0 and m ∈ N. Let m 0 ≥ m be such that Set z 0 i = y i and, relying on the uniform perfectness, choose points z 1 i , . . . , z n i i inductively so that Figure 1 for an illustration. Note that, by (3.3), such a number n i exists. Therefore, . . , n i } and the uniform perfectness condition is applicable. It now follows immediately that for each j ∈ {1, . . . , n i } the triplet (x i , z j−1 i , z j i ) satisfies the conditions (1) and (2).
. Thus we are basically saying that the is roughly a constant and thus the constant C is independent of i. On the other hand, our original assumption was that (f ( Let us next assume to the contrary that for every 0 < ε < ε 0 there exists m ∈ N such that for every triplet (a, b, c) satisfying (1) and (2) which is a contradiction; see Figure 2 for an illustration. Thus, for some j ∈ {1, . . . , n i }, the triplet (x i , z j−1 i , z j i ) satisfies the conditions (1), (2), and (3).
Remark 3.4. To prove Theorem 3.1, it seems to be essential to use weak tangents instead of tangents. Even though the sequence (x i ) converges to some point x (along some subsequence), it might happen that when looking at triplets (x, x i , y i ), the relative distance d(x i , y i )/d(x, y i ) tends to zero and we would not see anything relevant about f at the tangent at x. For this reason we need the triplets found in Lemma 3.3.
We are now ready to prove Theorem 3.1.
Proof of Theorem 3.1. Suppose to the contrary that there is an η-quasisymmetric mapping f : X → Y which is not quasi-Lipschitz. Then there is κ > 0 such that for every δ > 0 there exist x, y ∈ X with d(x, y) < δ such that We may assume that (f (x), f (y)) < d(x, y) 1+κ since otherwise we can consider the inverse of f . Observe that the inverse of a weak tangent mapping of f is a weak tangent mapping of f −1 . Thus there exist sequences (x i ) and (y i ) of points in X such that Let L ≥ 1 be the common bi-Lipschitz constant of η-quasisymmetric weak tangent mappings given by the assumption and let D ≥ 1 be the uniform perfectness constant of X. By Lemma 3.3, there exist 0 < ε < (2DL) −2/κ and a sequence of triplets By Lemma 2.1, the mapping packages ,f ) along some subsequence (which we keep denoting as the original sequence). By the convergence of the spaces, there are Assouad embeddings Hence, by the assumption,f is a bi-Lipschitz mapping with the constant L. Since the embeddings g and g m are bi-Lipschitz with the same constants there exists m 0 ∈ N such that for all m ≥ m 0 . Furthermore, by the convergence of the distances and (2), we have εD −1d (a, c) ≤ d(a, b). Sinced(a, c) = 1 andf is L-bi-Lipschitz we get, by (3), This contradiction finishes the proof.
To finish this section, we show that quasi-Lipschitz mappings preserve dimension. We denote the Hausdorff, upper Minkowski, and lower Minkowski dimensions by dim H , dim M , and dim M , respectively.
Proof. Note that since a quasi-Lipschitz mapping is locally invertible, X is separable, and the Hausdorff dimension is countably stable, we may, without loss of generality assume that f is invertible. Thus it suffices to show that dim H f (X) ≤ dim H X. Let t > dim H X and choose ε > 0 such that (1 − ε)t > dim H X. It suffices to show that H t (f (X)) < ∞. By the quasi-Lipschitz assumption, there is δ 0 > 0 such that Since this holds for all 0 < δ < δ 0 we have shown that H t (f (X)) < ∞.
To show the second claim, observe that since X is compact and the Minkowski dimension is finitely stable, we may again assume that f is invertible. Let t > dim M X and choose ε > 0 such that (1 − ε)t > dim M X. By the quasi-Lipschitz assumption, there is δ 0 > 0 such that and similarly for dim M . Since f was assumed to be invertible the other inequalities follow by the same estimate for f −1 .

Fibered spaces
In this section, we present a condition for weak tangent sets that guarantees that the assumptions of Thorem 3.1 are satisfied. The condition is a modification of a result of Le Donne and Xie [10, Theorem 1.1] for our purposes. We show that if the weak tangent sets are unions of fibered spaces, then the weak tangent mappings that map fibers onto fibers are bi-Lipschitz with the same constant. To this end, we introduce the following definition. If (X, d) and (Y, ) are metric spaces, f : X → Y is η-quasisymmetric, and there are points p, w ∈ X and a constant C ≥ 1 such that then we say that f is (C, η)-quasisymmetric. Note that a given quasisymmetric mapping clearly satisfies (4.1) for some C. The idea here is to show that the restriction of a (1, η)-quasisymmetric weak tangent mapping to each fibered space is a (C, η)-quasisymmetric mapping with the same constant C. This will then guarantee the existence of a uniform bi-Lipschitz constant in the assumptions of Theorem 3.1.
Let us first recall the definition of the fibered space from [10]. Let (X, d) be a metric space and F, G ⊂ X closed sets. We say that d(γ(t i−1 ), γ(t i )) : n ∈ N and 0 = t 0 < t 1 < · · · < t n−1 < t n = 1 .
A metric space (X, d) is called fibered if there are an index set I and closed sets F i ⊂ X, i ∈ I, called fibers, with X = i∈I F i so that the following properties are satisfied: (F1) Fibers are unbounded geodesic metric spaces:  .1) give for all x, y ∈ X, which is what we wanted to show.
In §5, we will need the lemma in the form where the spaces are unions of fibered spaces. The task is thus to show that each restriction of the (C, η)-quasisymmetric mapping to the fibered space is (C , η)-quasisymmetric for some C depending only on η and C. The claim of the lemma then follows by applying Lemma 4.1 on each fibered space and gluing the obtained bi-Lipschitz mappings together.
Lemma 4.2. Suppose that (X, d) and (Y, ) are metric spaces such that X = i∈I F i and Y = j∈J G i , where F i and G j are fibered spaces (in the restriction metrics). If f : X → Y is (C, η)quasisymmetric such that it maps fibers of each F i homeomorphically onto fibers of some G j , then f is L-bi-Lipschitz, where L depends only on η and C.
Proof. By the assumption, there exist points p, w ∈ X so that C −1 d(p, w) ≤ (f (p), f (w)) ≤ Cd(p, w). Let us first assume that p, w ∈ F i for some i ∈ I. Since f | F i is (C, η)-quasisymmetric and fibers of F i are mapped homeomorphically onto fibers of G j , we get, by Lemma 4.1, that f | F i is L 1 -bi-Lipschitz, where L 1 -depends only on η and C. To show that also f | F i is bi-Lipschitz, we seek points u, v ∈ F i so that f | F i satisfies condition (4.1) for u, v and some C depending on only η and C. Fix x ∈ F i and u ∈ F i . Since fibers of F i and F i are unbounded geodesic spaces, we may choose y ∈ F i and v ∈ F i so that d(y, and, similarly, Thus f | F i is (C , η)-quasisymmetric where C = η(1) 2 L 1 depends only on η and C. Lemma 4.1 implies now that f | F i is L 2 -bi-Lipschitz for some L 2 ≥ 1 depending only on η and C.
Let us then assume that p ∈ F i and w ∈ F i for some Thus f | F i is (C , η)-quasisymmetric where C = η(1)C depends only on η and C. Now, continuing as in the first part of the proof, we conclude that there exists L ≥ 1 depending only on η and C such that f | F i is L-bi-Lipschitz for all i ∈ I. It remains to glue the mappings f | F i together. Fix x ∈ F i and y ∈ F i . Choose z ∈ F i so that d(x, y) = d(x, z). Now we have Therefore f is bi-Lipschitz with a constant max{L, η(1)L} which only depends on η and C.

Horizontal self-affine carpets
In this section, we apply Theorem 3.1 in a concrete setting. We consider horizontal self-affine carpets on the plane. We will first show, generalizing the result of Bandt and Käenmäki [1,Theorem 1], that all weak tangents of such sets are unions of fibered spaces. We remark that [1, Theorem 1] is recently generalized in another direction by Käenmäki, Koivusalo, and Rossi; see [7, Theorem 3.1]. It is essential for us that we get the result for all weak tangents, not just for almost all tangents as in the above mentioned results. Recall that a set E ⊂ R 2 is porous with a constant 0 < α < 1 if for every x ∈ E and r > 0 there exists y ∈ R 2 such that B(y, αr) ⊂ B(x, r) \ E.
Theorem 5.1. If E is a horizontal self-affine carpet, then weak tangents of E are of the form where C left and C right are uniformly perfect porous sets, and at least one of them is nonempty.
Relying on this result, Theorem 3.1 together with Lemma 4.2 unveils that the class of quasisymmetric mappings between two horizontal self-affine carpets is rigid and the proof of Theorem A follows. In particular, by Lemma 3.5, such a quasisymmetric mapping can exist only when E and F have the same dimension. Let us now start adding details for these claims. We will first define horizontal self-affine carpets. Let Φ = {ϕ i } N i=1 be a collection of contractive self-maps on R 2 . The collection Φ is called an iterated function system (IFS). By Hutchinson [6], there exists a unique non-empty compact set E, called the invariant set of the iterated function system, satisfying Since we are interested in dimensional properties of E we may assume that diam(E) = 1. If the images ϕ i (E) are pairwise disjoint, then the IFS is said to satisfy the strong separation condition (SSC). In the study of iterated function systems, it is often convenient to use the following notation. Let N ≥ 2 be an integer and let Σ n = {1, . . . , N } n be the collection of all sequences of length n formed from the set {1, . . . , N }. If i = (i 1 , . . . , i n ) ∈ Σ n , then we write |i| = n and i − = (i 1 , . . . , i n−1 ). The set of all finite sequences n∈N Σ n is denoted by Σ * and the set of all infinite sequences {1, . . . , N } N is denoted by Σ. If i = (i 1 , i 2 , . . .), then we write i| n = (i 1 , . . . , i n ) ∈ Σ n for all n ∈ N. The concatenation of two sequences i and j is denoted by ij.
The correspondence between the invariant set E and the set Σ is given by the surjective mapping π : Σ → E, defined by the relation where K is any non-empty compact set satisfying N i=1 ϕ i (K) ⊂ K. If i = (i 1 , . . . , i n ) ∈ Σ n for some n, then we write ϕ i = ϕ i 1 • · · · • ϕ in and E i = ϕ i (E). The sets E i , and the corresponding sets [i] = {j ∈ Σ : j| |i| = i}, are the cylinder sets of level i.
We assume that all the mappings of the IFS are invertible and affine such that the linear parts are diagonal. Denoting In this case, the invariant set E is called a self-affine carpet. It is horizontal if the corresponding IFS satisfies the SSC and the following two conditions hold: (H1) For each i ∈ {1, . . . , N } we have α 1 (i) > α 2 (i). (H2) Every vertical line that intersects X, the convex hull of E, intersects ϕ i (X) for at least two different i ∈ {1, . . . , N }.
Observe that any Gatzouras-Lalley carpet satisfying the SSC and (H2) is a horizontal self-affine carpet. By the iterative structure and compactness, (H2) implies that each vertical line that intersects X, also intersects E. Thus the projection of E onto the horizontal coordinate is a line segment. For a more detailed argument, see, for example, [7,Remark 3.3]. Furthermore, if then the condition (H1) implies that 0 < β < 1. Finally, the SSC guarantees that To finish the preliminaries on horizontal self-affine carpets, let us introduce some more notation.
Lemma 5.2. For every i ∈ Σ and 0 < t < 1 it holds that Proof. (1) If α n < t, then ϕ i|n (E) ⊂ B(π(i), t). Thus B(π(i), t) contains two cylinders of level n + 1 and therefore, n(i, t) ≤ n. To show the other inequality, observe first that the distances between cylinders of level n + 1 are at least α n δ. If α n δ > t, then B(π(i), t) intersects E i|ni for some i ∈ {1, . . . , N } but it can not intersect another cylinder of level n + 1. Thus n ≤ n(i, t).
(2) Observe first that we have αα 2 (i| − t ) ≤ α 2 (i| t ). Since any vertical line through E i|t intersects at least two of its sub-cylinders we have t < α 2 (i| − t ) yielding the first inequality. Since B(π(i), t) intersects at least two sub-cylinders of E i|t we have α 2 (i| t )δ < t. Thus also the second inequality holds.
be the smallest closed rectangle containing E with sides parallel to the coordinate axis. Observe that, since we assumed diam(E) = 1, the height of Q is at most 1. We denote Q i = ϕ i (Q) and call Q i a construction rectangle of level |i|. By horizontal endings of a rectangle Q i we mean the vertical line segments . Note that even though the SSC means that sets ϕ i (E) and ϕ j (E) are disjoint for i and j with [i] ∩ [j] = ∅, it does not imply that Q i and Q j are disjoint. This is not a problem since the crucial thing is that Q i approximates ϕ i (E) well. Lemma 5.3. For every K ∈ N there exists t K > 0 such that for any 0 < t < t K and i ∈ Σ the ball B(π(i), t) intersects at most one vertical line containing a horizontal ending of some rectangle Q i|tj with |j| = K.
Proof. Fix K ∈ N. Observe first that the construction rectangles of level K have 2N K horizontal endings. Denote the smallest positive horizontal distance between them by δ K > 0 and choose 0 < t K < 1 so that δ K αβ −n * (t) ≥ 3 for all 0 < t < t K . Fix i ∈ Σ and 0 < t < t K . Since α 2 (i| t )/α 1 (i| t ) ≤ β n(i,t) Lemma 5.2 gives Observe that the level n(i, t) + K horizontal endings in Q i|t have horizontal separation either zero or at least δ K α 1 (i| t ). Therefore we conclude that B(x, t) can intersect at most one vertical line containing such a ending.
Since V y (E) does not depend on y 2 we denote it also by V y 1 (E). Lemma 5.4. If E is a horizontal self-affine carpet, then every vertical slice of E is porous with a constant min{δ, 1}/4 and uniformly perfect with a constant δ −1 α −k−1 , where k is the smallest integer with α k < δ.
Proof. Let us first show that the vertical slices are porous. Let z ∈ E and fix t > 0. Let y = (y 1 , y 2 ) = π(i) be so that y 1 = z 1 and y 2 ∈ V z (E). If α 2 (i| t ) < t/2, then by the definition of n(i, t) we have V z (E) ∩ (y 2 + t/2, y 2 + t) = ∅ and (y 2 + t/2, y 2 + t) ⊂ (y 2 − t, y 2 + t). Furthermore, if α 2 (i| t ) ≥ t/2, then by the SSC, the distances of level n(i, t) + 1 cylinder sets are at least δα 2 (i| t ) ≥ δt/2. Thus there exists x 2 so that and (x 2 − δt/4, x 2 + δt/4) ⊂ (y 2 − t, y 2 + t). Thus V z (E) is porous with the constant min{δ, 1}/4. Let us then show the uniform perfectness. Let z ∈ E and fix y = (y 1 , y 2 ) = π(i) so that y 1 = z 1 and y 2 ∈ V z (E). Let t > 0 be so that Let k be the smallest integer with α k < δ and let j ∈ Σ k be such that y ∈ E i|tj . By the condition (H2), a vertical line through y intersects two sub-cylinders of E i|tj . Thus there exists a point x = (y 1 , x 2 ) ∈ E i|tj so that it is not contained in the same n(y, t) + k + 1 level cylinder with y. Now, by Lemma 5.2(2) and the choice of k, we have In other words, x ∈ V z (E) ∩ (y 2 − t, y 2 + t) \ (y 2 − δα k+1 t, y 2 − δα k+1 t) and V z (E) is uniformly perfect with the constant δ −1 α −k−1 .
Let E ⊂ R 2 be closed, p ∈ E, and t > 0. Note that the pointed metric space (E, | · |/t, p) is homothetic to ((E − p)/t, | · |, 0) via the homothety x → (x − p)/t. Here (E − p)/t = {(x − p)/t ∈ R 2 : x ∈ E}. Therefore, whenever we consider weak tangents of subsets of R 2 , we can always choose the associated bi-Lipschitz embeddings to be this homothety. Recall that the weak tangents are unique up to an isometry. We say that T ⊂ R is a weak vertical slice tangent of E if there exists a sequence (y i ) of points in E and a sequence (t i ) of positive reals converging to zero such that ((V y i (E) − proj 2 y i )/t i , | · |, 0) converges to (T, | · |, 0). Here proj 2 is the orthogonal projection onto the vertical axis. We make the corresponding choice for the embeddings also in this case.
If D, F, G ⊂ R 2 , then we write (4.2). Furthermore, if I is a collection of sets, then we will slightly abuse notation and write I to denote also the union I∈I I.
Proof. Fix K ∈ N, let t K > 0 be as in Lemma 5.3, and choose 0 < t < t K . Let i ∈ Σ be such that x = π(i). If |j| = K, then, by Lemma 5.2(2), the height of Q i|tj is By the condition (H2), we have d H (E i|tj , Q i|tj ) ≤ δ −1 tα K . Furthermore, by Lemma 5.3, there is a point w ∈ R such that all the horizontal endings of rectangles Q i|tj intersecting B(x, t) are contained in the line {w} × R. Define This immediately means that where E(ε) is the ε-neighborhood of E. Now fix u and v so that w − t < u < w < v < w + t and consider the vertical slices V u (E) and V v (E). Recall that the vertical lines containing the horizontal endings of the rectangles Q i|tj are at least 3t apart. Therefore, in the above inequalities E can be replaced by the set This proves the claim in the case rectangles Q i|tj have horizontal endings intersecting B(x, t). If there are no such endings, then we may choose u = v = w = proj 1 (x), where proj 1 is the orthogonal projection onto the horizontal axis.
We are now ready to prove Theorem 5.1.  for all R > 0. Fix R > 0, let K ∈ N, and choose i 0 ∈ N such that 0 < t i < t K /2R for all i ≥ i 0 , where t K > 0 is as in Lemma 5.5. Now for each i ≥ i 0 , by Lemma 5.5, there exist The use of double radius here ensures that we do not need to worry about the convergence on the boundary of B(0, R). As i increases we can let K → ∞, and so the above Hausdorff distance converges to zero. Thus by (5.1), where C left and C right are weak vertical slice tangents of E. By Lemma 5.4, vertical slices of E are porous and uniformly perfect. Since these properties are preserved in the limit we have finished the proof.
The following example illustrates that C left and C right can be disjoint.
Example 5.6. Let g : R 2 → R 2 , g(x 1 , x 2 ) = (0.5x 1 , 0.2x 2 ), and then set f 1 = g, f 2 = g + (0.5, 0.25), f 3 = g + (0, 0.55), and f 4 = g + (0.5, 0.8). The invariant set E of the iterated function system Figure 3. Notice that the vertical center line of the unit cube contains left and right endings of the construction rectangles. To obtain a weak tangent so that C left and C right are disjoint and nonempty, we just choose the defining sequences (x i ) and (t i ) so that for each i the vertical center line of an appropriate construction rectangle is in the middle of the ball B(x i , t i ). This ensures that the unit ball of the weak tangent also has a separating vertical line in the middle.
and likewise T left , T right , and T in the codomain side. For topological reasons, it is clear that

Now clearly
and the fibers F y are unbounded geodesic metric spaces. Thus the condition (F1) is satisfied. Since dist(F y , F z ) = |y − z| > 0 for all y, z ∈ T left ∪ T right with y = z, also the condition (F2) is satisfied. Fibers F y and F z are non-parallel only if y ∈ T left and z ∈ T right or vice versa. This implies that d H (F y , F z ) = ∞ and the condition (F3) is satisfied. To check (F4), it suffices to show that there are no isolated points in T left (nor in T right ). Recall that C left and C right are closed sets with no isolated points. Hence for each y ∈ T left there exists a sequence (y i ) in C left which converges to y. Suppose to the contrary that y is an isolated point of T left . By the definition of T left this means that y i ∈ C right for all large enough i. Since C right is closed we conclude that y ∈ C right and hence y / ∈ T left which is a contradiction. Let us then show that R × S is a fibered space, where Let S be the corresponding set in the codomain side. Observe that S ∪ T left ∪ T right is dense in T ∪ T left ∪ T right = C left ∪ C right . As above, the conditions (F1)-(F3) follow immediately. To check (F4), it suffices to show that there are no isolated points in S. Suppose to the contrary that y ∈ S is an isolated point of S. Recalling that C left ∪ C right has no isolated points, there exists a sequence (y i ) in T ∪ T left ∪ T right which converges to y. Since y is an isolated point and, by the definition of S, there does not exist a sequence in T left ∪ T right converging to y, we have y i ∈ T \ S for all large enough i. We may assume that y i ∈ T \ S for all i. For each y i , by the definition of S, there exists a sequence (z i j ) j in T left ∪ T right converging to y i . Therefore, by choosing suitable points from these sequences, we may construct a sequence (z i j i ) i in T left ∪ T right which converges to y. Thus y / ∈ S which is a contradiction.
To finish the proof, we use the denseness and apply Lemma 4.2 in the union of these fibered spaces. Sincef (R × T ) = R × T it follows from the continuity off thatf (R × S) = R × S . Thereforef maps the fibers of R × S homeomorphically onto the fibers of R × S and the fibers of (−∞, w]×T left ∪[w, ∞)×T right homeomorphically onto the fibers of (−∞, w ]×T left ∪[w , ∞)×T right . where L depends only on η. Sincef is L-bi-Lipschitz on a dense set it is L-bi-Lipschitz on the whole set.
We will then turn to the proof of Theorem B. Let Q = [0, 1] 2 . Recall that a set M ⊂ Q is a Furstenberg miniset of E ⊂ Q if M ⊂ (λE + z) ∩ Q = {λx + z : x ∈ E} ∩ Q for some λ ≥ 1 and z ∈ R 2 . The number λ is called the scaling coefficient of the miniset M . A set M ⊂ Q is a Furstenberg microset of a compact set E ⊂ Q if there exists a sequence (M n ) of Furstenberg minisets of E such that d H (M n , M ) → 0 as n → ∞. The sequence (λ n ) n∈N , where each λ n is a scaling coefficient of the miniset M n , is called the scaling sequence of the microset M . Note that a Furstenberg miniset is clearly a Furstenberg microset. Furthermore, if the scaling sequence of a microset M is unbounded, then M is a subset of a weak tangent of E. Fix n ∈ N and 0 < s < t. We will show that there exist k(n, s) ∈ N and a dyadic cube D n,s of level l n ≥ n so that ν k(n,s) (D ) ν k(n,s) (D n,s ) for all l -level dyadic cubes D ⊂ D n,s for all l n < l ≤ l n +n. Here diam(A) = sup{|x−y| : x, y ∈ A} is the diameter of the set A. The claim means that when we look at most n levels further, the measure appears to be quite evenly spread in Q n,s . To show (5.2), we argue by contradiction. So, if the claim does not hold, then for every k, n ∈ N and an n-level dyadic cube D n of positive measure there exists a dyadic cube D l 1 n ⊂ D n of level n < l 1 n ≤ 2n so that We choose D n such that ν k (D n ) ≥ 2 −2n . This can be done since ν k is a probability measure and there are 2 2n dyadic cubes of level n. Since the claim fails also for D l 1 n we find a dyadic cube D l 2 n of level l 1 n < l 2 n ≤ l 1 n + n so that Let s < s < t and observe that there exists k 0 ∈ N such that N k (E) > 2 ks for all k ≥ k 0 . Since D l m n contains at most 2 2n dyadic cubes of level k we have for all k ≥ k 0 . Putting (5.3) and (5.4) together gives 2 −ks 2 ns 2 −2n < 2 2n 2 −ks which is clearly a contradiction if k is chosen to be large enough. Therefore (5.2) holds.
Let us now use (5.2) to prove the proposition. For every n ∈ N we choose 0 < s n < t such that s n → t as n → ∞. We consider the microsets K n = (M k(n,sn) ) Dn,s n and probability measures µ n = (ν k(n,sn) ) Dn,s n . Let K and µ be such that K n → K in the Hausdorff distance and µ n → µ weakly along some subsequence (which we keep denoting as the original sequence). Observe that spt µ ⊂ K and that K, as a limit of microsets, is a microset of E. Furthermore, if U (x, r) is an open ball centered at x ∈ K with radius r > 0, then Notice also that there is a constant p, depending only on the dimension of the ambient space, such that any ball of radius r can be covered by p many dyadic cubes D i with r/2 < diam(D i ) ≤ r. By (5.2), we have µ n (D i ) = ν k(n,sn) (D i ) −1 S Dn,s n ν k(n,sn) (D i ) ≤ r sn whenever 2 −n < r < 1. Thus we have for all x ∈ K and 0 < r < 1. By the definition of the Hausdorff measure, this implies that dim H (K) ≥ t. The proof is finished.
The following proposition is a rather immediate corollary of Proposition 5.7.
Proposition 5.8. If E ⊂ Q is compact, then there exists a weak tangent W of E such that Proof. By Proposition 5.7, there exists a Furstenberg microset M of E having unbounded scaling sequence such that dim H M ≥ dim A E. Hence, M is a subset of W ∩ Q, where W is a weak tangent of E. By [11, Proposition 2.1], we have dim A W ∩ Q ≤ dim A E. Therefore, which is what we wanted to show.
We remark that in Proposition 5.8 it is essential to use weak tangents -there does not necessarily exist such tangent sets; see [9,Example 2.20]. Together with Remark 3.4, this observation further emphasizes the use of weak tangents in our analysis. We are now ready to prove Theorem B.  To conclude the article, we discuss about the role of the condition (H2) in Theorems A and B.
Remark 5.9. In Theorem B, the condition (H2) can be replaced by the following condition: (H2') The projection of E onto the horizontal coordinate is a line segment. It is clearly equivalent to (H2') to assume that every vertical line that intersects X, the convex hull of E, intersects ϕ i (X) for some i ∈ {1, . . . , N }. Theorem B remains true since with (H2') we can modify the proofs of Lemmas 5.3 and 5.5 to show that the weak tangents have the form (−∞, w] × C left ∪ [w, ∞) × C right , where C left and C right are closed porous sets, which is enough for the conformal Hausdorff minimality of the weak tangents. Observe that the proof of Theorem B actually shows that any compact set E, whose weak tangents are either minimal for the conformal Hausdorff dimension or minimal for the conformal Assouad dimension, is minimal for the conformal Assouad dimension. Note that this is valid in any dimension.
Remark 5.10. In this remark, we consider the following generalization of the condition (H2): (H2") Every vertical line that intersects X, the convex hull of E, either does not intersect any of the sets ϕ i (X) or intersects ϕ i (X) for at least two distinct i ∈ {1, . . . , N } 2 . Note that according to (H2"), the projection of E onto the horizontal coordinate is a finite union of line segments. These line segments are the projections of the first level images of X. All theorems of this section are still true if we replace (H2) by (H2"). The key observation is that even with this weaker condition, the form of the weak tangents remains the same. In Lemmas 5.2-5.5, the only difference is that we have different constants. To see this, let G denote the collection of the vertical lines that intersect X but not E. Note that the images ϕ i (G) divide the rectangle Q i into finitely many rectangles. These rectangles will be then used in place of the rectangles Q i in Lemmas 5.3 and 5.5. Thus there are more, but still finitely many vertical endings to deal with. Since this has an effect only on constants, the results remain the same.
Example 5.11. Mackay [11,Theorem 1.4] showed that for a Gazouras-Lalley carpet E, if the projection of E onto the horizontal coordinate is not a single line segment (in which case it is a porous set and hence, (H2") is not satisfied), then Cdim A E = 0. For more general self-affine carpets, this is not true anymore. If a self-affine carpet E satisfies the SSC, (H1), and (H2"), then, by Remark 5.10 and Theorem B, E is minimal for the conformal Assouad dimension. Let us next define a self-affine carpet which satisfies these assumptions but does not satisfy (H2).