Non-separability of the Lipschitz distance

Let $X$ be a compact metric space and $\mathcal M_X$ be the set of isometry classes of compact metric spaces $Y$ such that the Lipschitz distance $d_L(X,Y)$ is finite. We show that $(\mathcal M_X, d_L)$ is not separable when $X$ is a closed interval, or an infinite union of shrinking closed intervals.


Introduction
For compact metric spaces (X, d X ) and (Y, d Y ), the Lipschitz distance d L (X, Y ) is defined to be the infimum of ǫ ≥ 0 such that an ǫ-isometry f : X → Y exists. Here a bi-Lipschitz homeomorphism f : X → Y is called an ǫ-isometry if where dil(f ) denotes the smallest Lipschitz constant of f , called the dilation of f : Let M be the set of isometry classes of compact metric spaces. It is well-known that (M, d L ) is a complete metric space. See, e.g., [4,Appendix A] for the proof of the completeness and see, e.g., [1,2] for details of the Lipschitz distance.
Then the following question arises: (Q) : Is the metric space (M, d L ) separable?
The answer is no, which can be seen easily by the following facts: (a) if d L (X, Y ) < ∞, the Hausdorff dimensions of X and Y must coincide; (b) for any non-negative real number d, there is a compact metric space X whose Hausdorff dimension is equal to d.
See, e.g., [1,Proposition 1.7.19] for (a) and [3] for (b). The fact (b) indicates that (M, d L ) is too big to be separable. Then we change the question (Q) to the following more reasonable one (Q'): For a compact metric space X, let M X be the set of isometry classes of compact metric spaces Y such that d L (X, Y ) < ∞. Any elements of M X have a common Hausdorff dimension by (a). Then the following question arises: The main results of this paper give the negative answer for this question for several X. To be more precise, we give two examples for X such that (M X , d L ) is not separable: (i) Infinite unions of shrinking closed intervals with zero We would like to stress that (M X , d L ) becomes non-separable even when X are the above elementary cases. We note that the non-separability of the first example follows from the non-separability of the second example. The first example, however, is easier to show the non-separability than the second example.
The present paper is organized as follows: In the first section, we show that the set of isometry classes of the infinite unions of shrinking closed intervals with zero is not separable. In the second section, we show that the set of isometry classes of the closed interval is not separable.

The first example
Let Z >0 = {n ∈ Z : n > 0} denote the set of positive integers. For n, m ∈ Z >0 , let I(n, m) be an interval in R defined as follows: For each u = (u n ) n∈Z >0 ∈ {1, 2} Z >0 , we define the following subset in R: We equip X u with the usual Euclidean metric in R: Then it is easy to check that (X u , d) is a compact metric space.
such that all components are equal to one. Let X 1 be the set defined in (1) for the element 1. Let M X 1 denote the set of isometry classes of compact metric spaces X whose Lipschitz distances from X 1 are finite, that is, d L (X, X 1 ) < ∞. Then we have the following result: Proof. It is enough to find a certain discrete subset X ⊂ M X 1 with the continuous cardinality. We introduce a subset X ⊂ M, which is the set of isometry classes of all X u for u ∈ Z >0 : It is clear that the cardinality of X is continuum. We show that X ⊂ M X 1 and X is discrete (i.e., every point in X is isolated).
We first show that x ∈ I(n, u n ) .
Then f u,v is a bi-Lipschitz continuous function from X u to X v and for x, y ∈ X u , Therefore the Lipschitz distance between X u and X v is bounded by Thus we have that X ⊂ M X 1 . Second we show that X is discrete: Since f is homeomorphic, any intervals must be mapped to intervals by f . That is, there exists a bijection P : To show u n = v n for all n ∈ Z >0 , we have two steps: (ii) P (n) = n.
We start to show (i) by contradiction. Assume there exists n 0 ∈ Z >0 such that n 0 + u n 0 = P (n 0 ) + v P (n 0 ) . Since f | I(n 0 ,un 0 ) is homeomorphic, the endpoints of I(n 0 , u n 0 ) must be mapped to the endpoints of I(P (n 0 ), v P (n 0 ) ) by f | I(n 0 ,un 0 ) . Therefore Thus the dilation of f is at least bigger than By the assumption of n 0 + u n 0 = P (n 0 ) This contradicts the inequality (2). Hence we have n + u n = P (n) + v P (n) for all n ∈ Z >0 . We start to show (ii) by contradiction. Assume there exists n 0 ∈ Z >0 such that P (n 0 ) = n 0 . Let us define Then P (n * ) > n * by definition. Since we know that n + u n = P (n) + v P (n) by the first step (i), and that u n and v P (n) are in {1, 2}, thus the possibility of values of P (n) is that P (n) = n − 1, n or n + 1. This implies that P (n * ) = n * + 1, P (n * + 1) = n * and P (n * + 2) = n * + 2, or n * + 3. Since the endpoints of intervals must be mapped to the endpoints of intervals by f , the possibility of values of f (1/2 n * +1 ) and f (1/2 n * +2 ) is , 1 2 n * +3 , or 1 2 n * +3 + 1 2 n * +3+v (n * +3) .
By the above two steps, we have that u n = v n for all n ∈ Z >0 , and we have completed the proof of Lemma 2.
We resume the proof of Theorem 1. Proof of Theorem 1. By using Lemma 2, we know that (X, d L ) is discrete. Since the cardinality of X is continuum and X ⊂ M X 1 , we have that (M X 1 , d L ) is not separable. We have completed the proof.

The second example
In this section, we show the non-separability of M [0,1] : Proof. It is enough to find a certain discrete subset Y ⊂ M [0,1] with the continuous cardinality.
For each u = (u n ) n∈Z >0 ∈ {0, 1} Z >0 , let Y u be a subset in R 2 as an infinite union of flat parts and pulse parts with the origin: See the figure below: 1 0 We equip Y u with the usual Euclidean distance in R 2 : It is easy to check that (Y u , d) is a compact metric space. Let Y be the set of isometry classes of Y u for all u ∈ {0, 1} Z >0 : 1] . For u ∈ {0, 1} Z >0 , let f u be the projection from Y u to [0, 1] such that x = (x 1 , x 2 ) → x 1 . Then it is easy to see that f u is bi-Lipschitz continuous and, for x, y ∈ Y u , Therefore the Lipschitz distance between [0, 1] and Y u is bounded by Thus we have Y ⊂ M [0,1] . Now we show that Y is discrete: We show that u n = v n for all n ∈ Z >0 . By the assumption, there exists a bi-Lipschitz function f from Y u to Y v such that Let us define a subset in Z >0 as follows: Without loss of generality, we may assume that P u is not empty. That is, Y u has at least one pulse. The pulse part J(n, u n ) of Y u for n ∈ P u is called n-pulse of Y u . We note that, by the definition of the pulse parts, the peak of the n-pulse is attained at 5/2 n+2 in x-axis. It is enough for the desired result to show that P u = P v . We show that there is a bijection F : P u → P v such that F (n) = n. To show this, we have the following three steps: (i) The first step: for n ∈ P u , (ii) The second step: for n ∈ P u , (iii) The third step: for n ∈ P u , In fact, if we show the above three statements, each maximizers 5/2 n+2 of n-pulses of Y u are mapped to the maximizers 5/2 n+2 of n-pulses of Y v by u . This correspondence of n-pulses defines the map F : P u → P v such that F (n) = n.
The proof of the all three steps (i)-(iii) are governed by the same scheme: (A) Assume that the statements do not hold (proof by contradiction); (B) Estimate lower bounds of the dilations of f and f −1 ; (C) The lower bounds obtained in (B) contradict the inequality (5).
We start to show the first step (i). Since f is homeomorphic, the maximizer 5/2 n+2 of the pulse cannot be mapped to the endpoints of [0, 1] by and prove (i) by contradiction. By the continuity of f , there exists 0 < δ < 1/2 n+3 such that, for any x ∈ [5/2 n+2 − δ, 5/2 n+2 + δ], we have Therefore we have Here we use the fact that the three points f •f −1 u (5/2 n+2 −δ), f •f −1 u (5/2 n+2 ) and f • f −1 u (5/2 n+2 + δ) are on the same line. By using the inequality (4), the dilation of f is estimated as follows: In the above last line, we just calculated the distance following the Euclidean distance (3) in the n-pulse J(n, 1). This implies that dil(f ) ≥ 2 This contradicts the inequality (5). Therefore we have, for any n ∈ P u , We start to show the second step (ii) by contradiction. Assume that, for some n ∈ P u , Then there exists n 1 ∈ Z >0 such that On the other hand, there exist x 0 ∈ [ 1 2 n 1 − δ, 1 2 n 1 ) and y 0 ∈ ( 1 2 n 1 , 1 2 n 1 + δ] such that Thus the triangle determined by the three vertices is an isosceles right triangle, and we can calculate By (6), (7) and (8), we have a bound for the dilation of f : Here we used the equality (8) in the second line, the equality (6) and the definition of the dilation in the third line, and the inequality (7) in the last line. The inequality (9) implies that dil(f ) ≥ 2 This contradicts the inequality (5). Therefore we have, for any n ∈ P u , By the same argument as above, we have, for any n ∈ P u , Now we start to show the third step (iii). By the above two steps (i) and (ii), we have that, for any n ∈ P u , there exists p f (n) ∈ Z >0 such that By the same argument as the first step (i), we can check that p f (n) ∈ P v , that is, v p f (n) = 1. Also for the inverse function f −1 , we have that, for any n ∈ P v , there exists p f −1 (n) ∈ P u such that Since f is a bijection, the map p f is a bijection from P u to P v and p −1 f = p f −1 . Now it suffices to show that p f (n) = n for all n ∈ P u . We assume that there exists l ∈ P u such that p f (l) = l. Without loss of generality, we may assume p f (l) > l. We first show that To show this, it suffices to show that where the above inequality means that the point f Then, by the similar proof to that of Theorem 3, we obtain (i) For every ǫ > 0, the set Y ǫ = {Y ǫ u : u ∈ {0, 1} Z >0 }/isometry is discrete with cardinality of the continuum.