Lobachevskii Journal of Mathematics Vol. 8, 2001, 185 – 189 \begin{matrix}c \\ ◯ \end{matrix}E.N.Sosov

E.N.Sosov

ON HAUSDORFF INTRINSIC METRIC

(submitted by B.N.Shapukov)

Abstract. In this paper we prove that in the set of all nonempty bounded closed subsets of a metric space (X, ρ) the Hausdorff metric is the Hausdorff intrinsic metric if and only if the metric ρ is an intrinsic metric. In a space with an intrinsic metric we obtain the upper bound for the Hausdorff distance between generalized balls.

1. Preliminary

We shall use the following notation. R_{+} is the set of all nonnegative real numbers. B(X) (B[X]) is the set of all nonempty bounded (closed) subsets of a metric space (X, ρ).

ρ(x, W) = inf_{y∈W} ρ(x, y).

B[x, r] (B(x, r)) is a closed (open) ball with the center x ∈ X and radius r > 0. B[M, r] = {x ∈ X : ρ(x, M) ≤ r} is a generalized closed ball for M ⊂ X ([1], p. 219). β(M, W) = sup_{x∈M} ρ(x, W) is the deviation of a set M ∈ B(X) from a set W ⊂ X

([1], c. 223). α : B(X) × B(X) → R_{+}, α(M, W) = max{β(M, W); β(W, M)} is the Hausdorff pseudometric in B(X) ([1], p. 223). (Σ(X), α) is a set of all finite subsets in (X, ρ) with the induced Hausdorff metric α. (Σ_{2}(X), α) is a set of all nonempty subsets in (X, ρ) including at most two points with the induced Hausdorff metric α. B_{B}[M, r] is a closed ball in a space (B(X), α) with center M ∈ B(X) and with radius r > 0. P_{M} : X × R_{+} → B(X), P_{M}(x, ε) = M ∩ B[x, ρ(x, M) + ε] is a metric ε - projection operator (if ε = 0 and M ∩ B[x, ρ(x, M)] ≠ ∅ for each point x ∈ X, then it is a metric projection operator) in a nonempty set M ⊂ X [2].

Supported by the Russian Fund of Basic Research, Grant N 00 - 01 - 00308.

Let us recall necessary definitions. A set M ⊂ X is called an existential set, if M ∩ B[x, ρ(x, M)] is nonempty for each point x ∈ X [3]. A metric space (X, ρ) is called a space with intrinsic metric if for each points x, y ∈ X and for each ε > 0 there exists a finite sequence z_{0} = x, z_{1}, …, z_{k} = y in X such that ρ(z_{i}, z_{i+1}) < ε (0 ≤ i ≤ k − 1) and ρ(z_{0}, z_{1}) + … + ρ(z_{k−1}, z_{k}) < ρ(x, y) + ε [4]. In the paper ([5], p. 121) the following condition (A) for a metric space (X, ρ) is used. A. For all x, y ∈ X, ε > 0 the set ω(x, y, ε) = {z ∈ X : 2 max[ρ(x, z), ρ(z, y)] < ρ(x, y) + ε} is a nonempty set. Now we will formulate our results.

Lemma 1. A metric space X is a space with intrinsic metric if and only if the metric space X satisfies (A).

Theorem 1. A metric space (B[X], α) is a space with intrinsic metric if and only if the metric space (X, ρ) is a space with an intrinsic metric.

Let us consider a condition (A_{0}) for a metric space (X, ρ). A_{0}. For all x, y ∈ X the set ω(x, y) = {z ∈ X : 2 max[ρ(x, z), ρ(z, y)] ≤ ρ(x, y)} is nonempty. Recall that a complete metric space with (A_{0}) is called a geodesic space [6].

Theorem 2. A metric space (X, ρ) satisfies (A_{0}) if and only if for every bounded existential sets M, W ⊂ X there exists a set Ω ∈ B(X) such that α(M, Ω) = α(Ω, W) = α(M, W)/2.

Corollary 1. Let a metric space (X, ρ) satisfy (A_{0}).

a) If for all x, y ∈ X the set ω(x, y) is finite, then the metric space (Σ(X), α) satisfies (A_{0}). b)If for all x, y ∈ X the set ω(x, y) is a point, then the metric space (Σ_{2}(X), α) satisfies (A_{0}).

Lemma 2. If a metric space (X, ρ) is a space with intrinsic metric, then α(B[M, r], B[W, R]) ≤ α(M, W) + |R − r| for all M, W ∈ B(X), R > 0,

r > 0.

Corollary 2 follows from the theorem 1 and the lemma 2.

Corollary 2. If a metric space (X, ρ) is a space with intrinsic metric, then α_{B}(B_{B}[M, r], B_{B}[W, R]) ≤ α(M, W) + |R − r| for all M, W ∈ B(X), R > 0,

r > 0, where α_{B} : B[B(X)] × B[B(X)] → R_{+} is the Hausdorff metric.

Lemma 3. If F is a subset in a metric space (B[X], α) with the induced metric, then

α(H, Q) ≤ α_{F} (H, Q) for all H, Q ⊂ F, where α_{F} : B(F) × B(F) → R_{+} is the Hausdorff pseudometric.

2. Proofs

Proof of Lemma 1. Necessity. Let for each points x, y ∈ X and for each ε > 0 there exist a finite sequence z_{0} = x, z_{1}, …, z_{k} = y in X such that ρ(z_{i}, z_{i+1}) < ε/2 (0 ≤ i ≤ k −1) and ρ(z_{0}, z_{1})+…+ρ(z_{k−1}, z_{k}) < ρ(x, y)+ε/2. If x = y, then the proof is trivial. Assume x ≠ y. We may suppose that points z_{0} = x, z_{1}, …, z_{k} = y are different pairwise. Then there exists a point z_{i} (i ∈ {0, …, k − 1}) such that ρ(z_{0}, z_{i}) ≤ ρ(x, y)/2, ρ(z_{0}, z_{i+1}) ≥ ρ(x, y)/2. From the condition of Lemma 1 and the triangle inequality we get ρ(z_{0}, z_{i+1}) + ρ(z_{i+1}, z_{k}) ≤ ρ(z_{0}, z_{1}) + … + ρ(z_{k−1}, z_{k}) < ρ(x, y) + ε/2 (i ∈ {0, …, k −1}). Then ρ(z_{i+1}, z_{k}) < ρ(x, y)+ε/2−ρ(z_{0}, z_{i+1}) ≤ ρ(x, y)/2+ε/2, ρ(z_{0}, z_{i+1}) ≤ ρ(z_{0}, z_{i}) + ρ(z_{i}, z_{i+1}) < ρ(x, y)/2 + ε/2. Hence, z_{i+1} ∈ ω(x, y, ε). Sufficiency. If x, y ∈ X and ε > 0, then there exists a natural number n such that ρ(x, y) < ε(2^{n} − 1). Set δ = ε/(2^{n} − 1), z_{00} = x, z_{01} = y. Using the condition (A), on the first step we find points : z_{10} = z_{00}, z_{11} ∈ ω(z_{00}, z_{01}, δ), z_{12} = z_{01}, where the first digit of the index is the number of a step and the second digit is the number of a point at the step. If we continue using condition (A), then at the n-th step we get 2^{n} + 1 points : z_{n0} = z_{n−1,0}, …, z_{n,2j} = z_{n−1,j}, z_{n,2j+1} ∈ ω(z_{n−1,j}, z_{n−1,j+1}, δ), …, z_{n,2m} = z_{n−1,m}, where m = 2^{n−1}, 0 ≤ j < m. Then ρ(z_{n,2j}, z_{n,2j±1}) < ρ(z_{n−1,j}, z_{n−1,j±1})/2 + δ/2 < … < ρ(x, y)2^{−n} + δ(1 − 2^{−n}) where 0 ≤ j < m in the case of plus sign in the index and 1 ≤ j ≤ m in the case of the minus sign in the index. Therefore, ρ(z_{n0}, z_{n1})+…+ρ(z_{n,2m−1}, z_{n,2m}) < 2^{n}[ρ(x, y)2^{−n} + δ(1 − 2^{−n})] = ρ(x, y) + ε. Thus Lemma 1 is proved.

Proof of Theorem 1. Necessity follows from Lemma 1 and the definition of Hausdorff metric. Sufficiency. Set Ω(M, W, ε) = ⋃{ω(x, v, ε) ∪ ω(y, u, ε) : x ∈ M, y ∈ W, v ∈ P_{W} (x, ε), u ∈ P_{M}(y, ε)}, where ε > 0.

a) Let us prove that 2β(M, Ω(M, W, ε)) ≤ β(M, W) + 2ε. For any δ > 0 there exists a point x ∈ M such that β(M, Ω(M, W, ε)) < ρ(x, Ω(M, W, ε))+δ. Hence, for any δ > 0 there exists a point x ∈ M such that for each v ∈ P_{W} (x, ε), β(M, Ω(M, W, ε)) < ρ(x, v)/2 + ε/2 + δ ≤ ρ(x, W)/2 + ε + δ. Then for each δ > 0, β(M, Ω(M, W, ε)) < β(M, W)/2+ε+δ. Since δ > 0 is arbitrary positive number, we get β(M, Ω(M, W, ε)) ≤ β(M, W)/2 + ε.

b) Let us prove that 2β(Ω(M, W, ε), M) ≤ α(M, W) + 3ε. Assume the contrary. Then there exist sets M, W ∈ B[X] such that 2β(Ω(M, W, ε), M) > α(M, W) + 3ε. Hence, there are points z ∈ Ω(M, W, ε), u ∈ P_{M}(z, ε/2) such that 2ρ(z, u) > α(M, W) + 3ε. Now only two cases are possible. Case 1. There exist points v ∈ W, g ∈ P_{M}(v, ε) such that z ∈ ω(v, g, ε). Then ρ(v, M) ≥ ρ(v, g) − ε > 2ρ(z, g) − 2ε ≥ 2ρ(z, M) − 2ε ≥ 2ρ(z, u) − 3ε > α(M, W). We get a contradiction. Case 2. There exist points t ∈ M, q ∈ P_{W} (t, ε) such that z ∈ ω(t, q, ε). Then ρ(t, W) ≥ ρ(t, q)−ε > 2ρ(z, t)−2ε ≥ 2ρ(z, M)−2ε ≥ 2ρ(z, u)−3ε > α(M, W). We get a contradiction. Thus, inequality 2β(Ω(M, W, ε), M) ≤ α(M, W) + 3ε is proved. By (a), (b) we get 2α(Ω(M, W, ε), M) < α(M, W) + 4ε. Similarly we get 2α(Ω(M, W, ε), W) < α(M, W) + 4ε. Theorem 1 follows from Lemma 1 and the obtained inequalities.

Proof of Theorem 2. Sufficiency of the theorem follows from the definition of Hausdorff metric. Necessity. Set Ω = ⋃{ω(x, v) ∪ ω(y, u) : x ∈ M, y ∈ W, v ∈ P_{W} (x, 0), u ∈ P_{M}(y, 0)}.

a) Let us prove that 2β(M, Ω) ≤ β(M, W). For any δ > 0 there exists a point x ∈ M such that β(M, Ω) < ρ(x, Ω) + δ. Hence, for any δ > 0 there exists a point x ∈ M such that for each v ∈ P_{W} (x, 0) β(M, Ω) < ρ(x, v)/2 + δ = ρ(x, W)/2 + δ. Then, for each δ > 0, β(M, Ω) < β(M, W)/2 + δ. It follows from the arbitrariness of δ > 0 that β(M, Ω) ≤ β(M, W)/2.

b) Now let us prove that 2β(Ω, M) ≤ α(M, W). Assume the contrary. Then there exist bounded existential sets M, W ⊂ X such that 2β(Ω, M) > α(M, W). Hence, there exist points z ∈ Ω, u ∈ P_{M}(z, 0) such that 2ρ(z, u) > α(M, W). Now only two cases are possible. Case 1. There exist points v ∈ W, g ∈ P_{M}(v, 0) such that z ∈ ω(v, g). Then ρ(v, M) = ρ(v, g) ≥ 2ρ(z, u) > α(M, W). We get a contradiction. Case 2. There exist points t ∈ M, q ∈ P_{W} (t, 0) such that z ∈ ω(t, q). Then ρ(t, W) = ρ(t, q) ≥ 2ρ(z, u) > α(M, W). We get a contradiction. Thus inequality 2β(Ω, M) ≤ α(M, W) is proved. By (a), (b) we get 2α(Ω, M) ≤ α(M, W). Similarly we get 2α(Ω, W) ≤ α(M, W). Theorem 1 follows from the triangle inequality and the obtained inequalities.

Proof of Corollary 1. a) It follows from Theorem 2 because each finite subset in X is a bounded existential set.

b) Let A = {a_{1}, a_{2}}, B = {b_{1}, b_{2}} ∈ Σ_{2}(X) and ρ(a_{i}, b_{j}) = max(ρ(a_{1}, b_{1}), ρ(a_{1}, b_{2}), ρ(a_{2}, b_{1}), ρ(a_{2}, b_{2})), i, j ∈ {1, 2}. If i = j, we set Ω = {ω(a_{1}, b_{2}), ω(a_{2}, b_{1})}. If i ≠ j, we set Ω = {ω(a_{1}, b_{1}), ω(a_{2}, b_{2})}. The equalities α(A, Ω) = α(Ω, B) = α(A, B)/2 can be proved by direct calcuation. Thus Corollary 1 is proved.

Proof of Lemma 2.

a) Let us prove that ρ(x, B[W, R]) = max(ρ(x, W) − R, 0) for each x ∈ X. It follows from the triangle inequality for β [2] that ρ(x, W)−R ≤ ρ(x, B[W, R])+ β(B[W, R], W) − R ≤ ρ(x, B[W, R]) for each x ∈ X. Hence, ρ(x, B[W, R]) ≥ max(ρ(x, W) − R, 0) for each x ∈ X. We will use a proof by contradiction. Assume the equality fails. Then there exists a point x ∈ X\B[W, R] such that ρ(x, W) − R < ρ(x, B[W, R]). Therefore, there exists δ ∈ (0, R) such that ρ(x, W) + δ − R < ρ(x, B[W, R]). Then there is a point y ∈ W such that ρ(x, y) + δ − R < ρ(x, B[W, R]). It follows from the definition of the intrinsic metric that for each ε ∈ (0, δ/2) there exists a finite sequence z_{0} = x, z_{1}, …, z_{k} = y in X such that ρ(z_{i}, z_{i+1}) < ε (0 ≤ i ≤ k − 1) ρ(z_{0}, z_{1}) + … + ρ(z_{k−1}, z_{k}) < ρ(x, y) + ε. Then there exists a point z_{i} ∈ {z_{1}, …, z_{k−1}} such that z_{i} ∈ B[W, R], ρ(z_{i}, y) ≥ R − ε. Consequently, ρ(x, z_{i}) < ρ(x, y) − ρ(z_{i}, y) + ε ≤ ρ(x, y) − R + 2ε < ρ(x, B[W, R]) + 2ε − δ < ρ(x, B[W, R]). We get a contradiction.

b) Using (a), we get

α(B[M, r], B[W, R]) = max{ sup_{z∈B[M,r]} ρ(z, B[W, R]); sup_{w∈B[W,R]} ρ(w, B[M, r])} =

max{ sup_{z∈B[M,r]} max{ρ(z, W) − R, 0}; sup_{w∈B[W,R]} max{ρ(w, M) − r, 0}} ≤

max{max{β(B[M, r], W) − R, 0}; max{β(B[W, R], M) − r, 0}} ≤ max{max{β(B[M, r], M)+β(M, W)−R, 0}; max{β(B[W, R], W)+β(W, M)− r, 0}} ≤

ON HAUSDORFF INTRINSIC METRIC 189

max{max{β(M, W)+r−R, 0}; max{β(W, M)+R−r, 0}} ≤ α(M, W)+|R−r|. Thus Lemma 2 is proved.

Proof of Lemma 3. Let \underline{∈} be the sign of belonging of element for the metric space (F, α) and x ∈ H. Then there exists an element Z\underline{∈}H such that x ∈ Z. Moreover

ρ(x, Q) ≤ inf_{Y \underline{∈}Q} ρ(x, Y) ≤ inf_{Y \underline{∈}Q} α(Z, Y) ≤ sup_{Z\underline{∈}H} inf_{Y \underline{∈}Q} α(Z, Y) = β_{F} (H, Q).

Hence, β(H, Q) = sup_{x∈H} ρ(x, Q) ≤ β_{F} (H, Q).

Similarly we get β(Q, H) ≤ β_{F} (Q, H). Thus Lemma 3 is proved.

References

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[3] Vlasov L.P., Approximative properties of sets in normed linear spaces, Uspekhi Mat. Nauk, V. 28, No. 6 (174), 3–66 (1973). (Russian)

[4] Burago Yu.D., Gromov M.L., Perel’man G.D., A.D. Alexandrov’s spaces with curvatures bounded from below, Uspekhi Mat. Nauk, V. 47, No. 2 (284), 3-51, (1992). (Russian)

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Research Institute of Mathematics and Mechanics, Kazan State University, Universitetskaya Str. 17, Kazan, Tatarstan, 420008 Russia E-mail address: Evgenii.Sosovksu.ru

Received April 3, 2001