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Chapter 5. Hausdorff Distance

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Chapter 5. Hausdorff Distance Chapter 5 Hausdorff distance. Schedule. Hausdorff distance, equivalence of three definitions, triangle inequality for Hausdorff distance, Hausdorff distance is a metric on the set of all closed bounded nonempty subsets, coincidence of Vietoris topology and metric topology generated by Hausdorff distance on the set of all compact subsets, limits theory for nonempty subsets, definition of lim sup and some its properties (equivalent definitions), definition of lim inf and some its properties (equivalent definitions), convergence w.r.t. Hausdorff distance (Hausdorff convergence) and calculating lim inf, Hausdorff convergence of singletons, definition of lim, Hausdorff convergence implies existence of lim, the cases of decreasing and increasing sequences that are Hausdorff converging, equivalence of Hausdorff convergence and existence of lim in compact spaces, the cases of decreasing and increasing sequences in compact spaces, convergence in complete metric spaces, simultaneous completeness (total boundness, compactness) of the original space and the hyperspace of all closed bounded nonempty subsets, inheritance of the property to be geodesic for compact space and the hyperspace of all its closed nonempty subsets. Let X be an arbitrary metric space. For arbitrary nonempty sets A; B ⊂ X we put 1 j j 2 j j 2 (5.1) dH (A; B) = max sup aB : a A ; sup Ab : b B ; (5.2) d2 (A; B) = inf r 2 [0; 1]: A ⊂ B (B)& B (A) ⊃ B ; H r r 3 2 1 ⊂ ⊃ (5.3) dH (A; B) = inf r [0; ]: A Ur(B)& Ur(A) B : ⊂ 1 2 3 Proposition 5.1. For nonempty subsets A; B X of a metric space X we have dH (A; B) = dH (AB) = dH (A; B). i 1 j j 2 1 Proof. Put ri = dH (A; B). First, let r1 = . Without loss of generality, we assume that sup aB : a A = , but then neither A ⊂ Ur(B), nor A ⊂ Br(B) holds for any finite r > 0, therefore r2 = r3 = 1. Now suppose that r1 < 1, then for any r > r1 all inclusions in the definitions of r2 and r3 take place, therefore r2 and r3 are also finite. So, we have shown that either all three ri are infinite at the same time, or all of them are finite. It remains to analyze the case of finite ri. First, we show that r1 = r2 and r1 = r3. Let us note that r2 ≤ r3 because A ⊂ Ur(B) implies A ⊂ Br(B) (and the same for A and B swaped). By definition of r1, we have jaBj ≤ r1 for all a 2 A, and jAbj ≤ r1 for all b 2 B, therefore for all r > r1 it holds A ⊂ Ur(B) and B ⊂ Ur(A), hence r3 ≤ r. Since r > r1 is arbitrary, we have r2 ≤ r3 ≤ r1. On the other hand, for any r > r2 we have A ⊂ Br(B) and B ⊂ Br(A), so for any a 2 A and b 2 B it holds jaBj ≤ r and jAbj ≤ r, therefore, r1 ≤ r and since r > r2 is arbitrary, we get r1 ≤ r2 ≤ r3. i The value dH (A; B) from Proposition 5.1 is denoted by dH (A; B). It is easy to see that dH is non-negative, symmetric, and dH (A; A) = 0 for any nonempty A ⊂ X, thus, dH is a generalized distance on the family P0(X) of all nonempty subsets of a metric space X. The function dH is called the Hausdorff distance. Proposition 5.2. For an arbitrary metric space X, the function dH is a generalized pseudometric on P0(X). Proof. It remains to prove the triangle inequality. Choose arbitrary A; B; C 2 P0(X) and set c = dH (A; B), a = dH (B; C), b = dH (A; C). We have to show that b ≤ c + a. If either c = 1, or a = 1, then the equality holds. Suppose now that the both c and a are finite. Choose arbitrary finiter > c and s > a, then A ⊂ Ur(B) and B ⊂ Us(C) implies, by virtue of Item (4) of Problem 2.2, that A ⊂ Ur Us(C) ⊂ Ur+s(C). Similarly, Ur+s(A) ⊃ C. Thus, b ≤ r + s. Since r > c and s > a are arbitrary, we obtain what is required. Denote by H(X) ⊂ P0(X) the set of all nonempty closed bounded subsets of a metric space X. Theorem 5.3. For an arbitrary metric space X, the generalized pseudometric dH is a metric on H(X). 55 5.1. Vietoris topology and Hausdorff metric 56 Proof. Choose arbitrary A; B 2 H(A). Since they are bounded, for some r > 0 we have A ⊂ Ur(B) and Ur(A) ⊃ B, hence dH (A; B) < 1. Thus, dH is finite. If A =6 B, then without loss of generality we can assume that there exists a 2 A n B, but since the set X n B is open, there exists r > 0 such that Ur(a) \ B = ;, in particular, jaBj ≥ r and, therefore, dH (A; B) ≥ r. Thus, dH is non-degenerate and, therefore, positively defined. Recall that by K(X) we denoted the collection of all nonempty compact subsets of a topological space X. Since each compact subset of a metric space is closed and bounded, we have K(X) ⊂ H(X) and, thus, we get Corollary 5.4. For an arbitrary metric space X, the function dH is a metric on K(X). In what follows, when speaking about the distance in H(X), we will always have in mind the Hausdorff metric, and for topology consider the corresponding metric one. Note that different authors use different notations for this hyperspace. We introduced the notation H(X) by virtue of the fact that this is the largest natural set of subsets of a metric space on which the Hausdorff distance is defined. We present a few Hausdorff distance properties in the next exercise. Problem 5.1. Prove the following statements for an arbitrary metric space X. (1) Let f : X !P0(X) be given by the formula f : x 7! fxg, then f is an isometric embedding. (2) For any A; B 2 P0(X) we have dH (A; B) = dH (A; B¯) = dH (A;¯ B) = dH (A;¯ B¯). (3) For any A; B 2 P0(X) we have dH (A; B) = 0 if and only if A¯ = B¯. (4) If Y ⊂ X is an "-net in A ⊂ X, then dH (A; Y ) ≤ ". Proposition 5.5. Let X be an arbitrary metric space, and A; B 2 P0(X), r = dH (A; B). Then A ⊂ Br(B), B ⊂ Br(A), and for less r one of these inclusions fails. Thus, for A; B 2 P0(X) we can change inf to max in equality (5:2). Proof. The fact that smaller r do not fit follows directly from the definition of the Hausdorff distance. Let us prove the first part. Suppose the contrary, and let, say, B 6⊂ Br(A). This means that there exists b 2 B for which R := jbAj > r, thus for r < s < R we have B 6⊂ Bs(A), which contradicts the definition of dH (A; B). Problem 5.2. Prove that for A; B 2 K(X) there exist a 2 A and b 2 B such that dH (A; B) = jabj. Is it possible to change K(X) with H(X)? Problem 5.3. Let X be an arbitrary metric space and A; B; A0;B0 2 H(X) such that A0 ⊂ A and B0 ⊂ B. Prove 0 0 that dH (A [ B ;B [ A ) ≤ dH (A; B). Problem 5.4. Let X be an arbitrary metric space and A; B; C 2 H(X) such that C ⊂ B. Prove that dH (A; A[C) ≤ dH (A; B). 5.1 Vietoris topology and Hausdorff metric In Construction 1.5 we defined the Vietoris topology on the set of all nonempty subsets of a topological space X. Recall that a base of this topology is the family of sets h i f ⊂ ⊂ [n \ 6 ; g U1;:::;Un = Y X : Y i=1Ui; and Y Ui = for all i = 1; : : : ; n over all possible finite families U1;:::;Un of open subsets of X. Theorem 5.6. Let (X; d) be an arbitrary metric space, then the metric topology on K(X) defined by the Hausdorff metric dH coincides with the Vietoris topology. In particular, for a boundedly compact X we have H(X) = K(X), thus the above result holds if we change K(X) with H(X). 5.1. Vietoris topology and Hausdorff metric 57 Proof. We use Problem 1.5 to prove that each open set in metric topology generated by dH is also open in Vietoris topology. To do that, it suffices to take arbitrary A 2 K(X), r > 0, and to construct an open neighborhood of A in dH C C Vietoris topology that belongs to Ur (A). Consider the family = Ur=3(a) a2A, then is an open cover of the compact set A, thus we can extract from C a finite subcover fU1;:::;Ung. Notice that A 2 hU1;:::;Uni because ⊂ [n \ 6 ; h i ⊂ dH A i=1Ui and A Ui = for each i = 1; : : : ; n. Further, we claim that U1;:::;Un Ur (A). Indeed, take an 0 0 0 0 arbitrary A 2 hU1;:::;Uni, then A ⊂ Ur=3(A). Further, for every a 2 A there exist Ui 3 a and a 2 Ui \ A , hence 0 0 0 jaa j < 2r=3, so A ⊂ U2r=3(A ), hence dH (A ;A) ≤ 2r=3 < r. h i 2 h i n [n Now, consider an arbitrary U1;:::;Un and any A U1;:::;Un .
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