Proving Completeness of the Hausdorff Induced Metric Space

PROVING COMPLETENESS OF THE HAUSDORFF INDUCED METRIC SPACE Abstract Given a metric space (X, d), we may define a new metric space with Hausdorff metric h on the set K of the collection of all nonempty compact subsets of X. We show that if (X, d) is complete, then the Hausdorff metric space (K, h) is also complete. Introduction The Hausdorff distance, named after Felix Hausdorff, measures the distance between subsets of a metric space. Informally, the Hausdorff distance gives the largest length out of the set of all distances between each point of a set to the closest point of a second set. Given any metric space, we find that the Hausdorff distance defines a metric on the space of all nonempty compact subsets of the metric space. We find that there are many interesting properties of this metric space, which will be our focus in this paper. The first property is that the Hausdorff induced metric space is complete if our original metric space is complete. Similarly, the second property we explore is that if our original metric space is compact, then our Hausdorff induced metric space is also compact. 1 2 In the next section, we provide some definitions and theorems necessary for understanding this paper. We then define the Hausdorff distance in the following section, and examine its properties through some examples and short proofs. We find that the Hausdorff distance satisfies the conditions for a metric on a space of nonempty compact subsets of a metric space. Finally, in our last section, we prove that if our original metric space is complete then the Hausdorff induced metric space is also complete. We further show that (K, h) is compact when (X, d) is compact. Preliminaries The concepts in this paper should be familiar to anyone who has taken a course in Real Analysis. The notation and terminology in this paper will come from Gordon’s Real Analysis: A First Course [1]. Therefore, we expect the reader to be familiar with the following concepts regarding metric spaces and real numbers. Definition 2.1 Let S be a nonempty set of real numbers that is bounded below. The number α is the infimum of S if α is a lower bound of S and any number greater than α is not a lower bound of S. We will write α = inf S. The definition of the supremum of S is analogous and will be denoted by sup S. Completeness Axiom Each nonempty set of real numbers that is bounded below has an infimum. Similarly, any nonempty set of real numbers that is bounded above has a supremum. The reader may be more familiar with the following definitions when applied to the metric space (R, d), where d(x, y) = |x − y|. However, with the exclusion of some examples, for the majority of this paper we will be working in a general metric space. Thus our definitions will be given with respect to any metric space (X, d). PROVING COMPLETENESS OF THE HAUSDORFF INDUCED METRIC SPACE 3 Definition 2.2 A metric space (X, d) consists of a set X and a function d : X × X → R that satisfies the following four properties. (1) d(x, y) ≥ 0 for all x, y ∈ X. (2) d(x, y) = 0 if and only if x = y. (3) d(x, y) = d(y, x) for all x, y ∈ X. (4) d(x, y) ≤ d(x, z) + d(z, y) for all x, y, z ∈ X. The function d, which gives the distance between two points in X, is called a metric. For example, a metric on the set of real numbers is d(x, y) = |x − y|. It is easily verified that d satisfies the four properties listed above. For the next set of definitions, let (X, d) be a metric space. Definition 2.3 Let v ∈ X and let r > 0. The open ball centered at v with radius r is defined by Bd (v, r) = {x ∈ X : d(x, v) < r}. Definition 2.4 A set E ⊆ X is bounded in (X, d) if there exist x ∈ X and M > 0 such that E ⊆ Bd (x, M ). Definition 2.5 A set K ⊆ X is totally bounded if for each  > 0 there is a finite n [
subset {xi : 1 ≤ i ≤ n} of K such that K ⊆ Bd xi ,  . i=1 For the following definitions, let {xn } be a sequence in a metric space (X, d). Definition 2.6 The sequence {xn } converges to x ∈ X if for each  > 0 there exists a positive integer N such that d(xn , x) <  for all n ≥ N . We say {xn } converges if there exists a point x ∈ X such that {xn } converges to x. Definition 2.7 The sequence {xn } is a Cauchy sequence if for each  > 0 there exists a positive integer N such that d(xn , xm ) <  for all m, n ≥ N . It is easy to verify that every convergent sequence is a Cauchy sequence. Definition 2.8 A metric space (X, d) is complete if every Cauchy sequence in (X, d) converges to a point in X. An example of a metric space that is not complete is the space (Q, d) of rational numbers with the standard metric given by d(x, y) = |x − y|. However, the space 4 R of real numbers and the space C of complex numbers under the same metric d(x, y) = |x − y| are complete. Definition 2.9 A set K ⊆ X is sequentially compact in (X, d) if each sequence in K has a subsequence that converges to a point in K. Note that by Theorem 8.59 in [1], a subset of a metric spaces is compact if and only if it is sequentially compact; therefore, we will use the concepts of sequentially compact and compact interchangeably throughout this paper. Definition 2.10 The point x is a limit point of a set E if for each r > 0, the set E ∩ Bd (x, r) contains a point of E other than x. As an alternative to the definition, Theorem 8.49 in [1] states that x is limit point of the set E if and only if there exists a sequence of points in E\{x} that converges to x. This theorem provides us with the opportunity to choose a sequence converging to x, which will be useful in proving that a set is closed. Definition 2.11 A set E is closed in (X, d) if E contains all of its limit points. Definition 2.12 The closure of E, denoted E, is the set E ∪ E 0 , where E 0 is the set of all limit points of E. The following two results and lemma are placed in this section to be referred to in later proofs. In addition they will serve as an introduction to proofs that use the definition of convergent sequences and the triangle inequality. Result 1: Let {xn } and {yn } be sequences in a metric space (X, d). If {xn } converges to x and {yn } converges to y, then {d(xn , yn )} converges to d(x, y). Proof. Let  > 0. Since {xn } converges to x, by definition there exists a positive integer N1 such that d(xn , x) <  2 for all n ≥ N1 . Similarly, since {yn } converges to y, there exists a positive integer N2 such that d(yn , y) <  2 for all n ≥ N2 . Choose N = max{N1 , N2 }. Then for all n ≥ N , we find that d(xn , yn ) ≤ d(xn , x) + d(x, y) + d(y, yn ) <   + d(x, y) + = d(x, y) + , 2 2 PROVING COMPLETENESS OF THE HAUSDORFF INDUCED METRIC SPACE 5 and d(x, y) ≤ d(x, xn ) + d(xn , yn ) + d(yn , y) <   + d(xn , yn ) + = d(xn , yn ) + . 2 2 Together these inequalities imply |d(xn , yn ) − d(x, y)| <  for all n ≥ N . Therefore, {d(xn , yn )} converges to d(x, y).  Result 2: If {zk } is a sequence in a metric space (X, d) with the property that d(zk , zk+1 ) < 1/2k for all k, then {zk } is a Cauchy sequence. Proof. Let  > 0 and choose a positive integer N such that 1 < . Then for all 2N −1 n > m ≥ N we find that d(zm , zn ) ≤ d(zm , zm+1 ) + d(zm+1 , zm+2 ) + · · · + d(zn−1 , zn ) 1 1 1 + m+1 + · · · + n−1 2m 2 2 ∞ X 1 1 1 = m−1 ≤ N −1 < . < k 2 2 2 < k=m It follows that {zk } is a Cauchy sequence.  Lemma 1: Let (X, d) be a metric space and let A be a closed subset of X. If {an } converges to x and an ∈ A for all n, then x ∈ A. Proof. Suppose {an } is a sequence that converges to x and an ∈ A for all n. There are two cases to consider. If there exists a positive integer n such that an = x, then it is clear x ∈ A. If there does not exist a positive integer n such that an = x, then x is a limit point of A by Theorem 8.49 in [1]. Since A is closed, x ∈ A.  Construction of the Hausdorff Metric We now define the Hausdorff metric on the set of all nonempty, compact subsets of a metric space. Let (X, d) be a complete metric space and let K be the collection of all nonempty compact subsets of X. Note that K is closed under finite unions and nonempty intersections. For x ∈ X and A, B ∈ K, define r(x, B) = inf{d(x, b) : b ∈ B} and ρ(A, B) = sup{r(a, B) : a ∈ A}. 6 Note that r is nonnegative and exists by the Completeness Axiom since d(a, b) ≥ 0 by the definition of a metric space. Since r exists and is nonnegative, then both ρ(A, B) and ρ(B, A) exist and are nonnegative. In addition, we define the Hausdorff distance between sets A and B in K as h(A, B) = max{ρ(A, B), ρ(B, A)}. Before proving that h defines a metric on the set K, let us consider a few examples to get a grasp on how these distances work. Consider the following example of closed interval sets in (R, d), where d(x, y) = |x − y|. Example Let A = [0, 20] and let B = [22, 31]. We find that r(x, B) is going to be the infimum of the set of distances from each a ∈ A to the closest point in B. As an example of one of these distances, consider a = 12. Then r(12, B) = inf{d(12, b) : b ∈ B} = d(12, 22) = 10. We can note that for each a ∈ A, the closest point in B that gives the smallest distance will always be b = 22. Therefore, we find that ρ(A, B) = sup{d(a, 22) : a ∈ A}. The point a = 0 in A maximizes this distance. Therefore ρ(A, b) = d(0, 22) = |22 − 0| = 22. Similarly, we find that ρ(B, A) = sup{d(b, 20) : b ∈ B}, since the point a = 20 will give the smallest distance to any point in B. The point b = 31 in B maximizes this distance, so we have ρ(B, A) = d(20, 31) = |20 − 31| = 11. We will note that ρ(B, A) and ρ(A, B) are not always equal. It follows that h(A, B) = max{ρ(A, B), ρ(B, A)} = 22. Let us consider another example in (R, d) of discrete sets where the metric d(x, y) = |x − y|. Example Let A = {5n : 0 ≤ n ≤ 19} and B = {p : p < 100, p prime}. Since we are working with discrete, finite subsets of the real number line, we find that r(p, A) is equal to the minimum distance from any prime number p ∈ B to a multiple of 5 in the set A. That is, for any p ∈ B, r(p, A) = min{|p − 5n| : 0 ≤ n ≤ 19}. PROVING COMPLETENESS OF THE HAUSDORFF INDUCED METRIC SPACE 7 Note that the minimum distance from any prime number to the nearest multiple of 5 is either 2 or 1. For example, if p = 17, then we find that r(17, A) = min{|17 − 5n| : 0 ≤ n ≤ 19} = 2, when n = 3. If p = 71, then r(71, A) = min{|71 − 5n| : 0 ≤ n ≤ 19} = 1, where n = 14. Therefore we find that, for any point a ∈ A, ρ(B, A) = max{r(b, A) : b ∈ B} = 2. In other words, ρ(B, A) is equal to the maximum of the minimum distances from a prime p in the set B to the multiples of 5 in the set A. Similarly, we find that ρ(A, B) is equal to the maximum of the minimum distances from each multiple of 5 in the set A to the set of primes in the set B. Thus r(a, B) is equal to the minimum distance from any multiple of 5 in the set A to a prime number p ∈ B. There is no efficient way to do this, except by looking at the distances between each point in A to each point in B. . By looking at the distance between all of the multiples of 5 to a prime number we find that the largest of these minimum distances occurs at the point 50 ∈ A and 47, 53 ∈ B. Therefore h(A, B) = max{ρ(A, B), ρ(B, A)} = max{3, 2} = 3. Now consider the following example of r, ρ, and h in the complete metric space p (R2 , d), where d((x1 , y1 ), (x2 , y2 )) = (x1 − x2 )2 + (y1 − y2 )2 . Example Let A and B be subsets of R2 defined by A = {(x, y) : x2 + y 2 ≤ 1} and B = {(x, y) : 0 ≤ x ≤ 3, 0 ≤ y ≤ 1} [see Figure 1]. By definition, r(x, A) = inf{d(x, a) : a ∈ A}. So r(x, A) is the set of all distances from each x ∈ B to the “closest” point a ∈ A to x. Note that x will be an ordered pair, as are a and b. If b ∈ A ∩ B, then it is clear that r(b, A) = 0. If b ∈ B\A, then r(b, A) is found using the line from b to the origin [see Figure 2]. We find that the point that yields the largest distance is the upper right vertex of the rectangle at the point (3, 1).   1 3 on the Therefore, ρ(B, A) is equal to the distance from the point √ , √ 10 10 circle to the point (3, 1). 8 Figure 1. Graph of A and B in (R2 , d). Figure 2. Area shaded contains all points that give the infimum distance r(a, √ B) = 0. Additionally, we find ρ(A, B) = 1 and ρ(B, A) = 10 − 1. Thus,  ρ(B, A) = d 3 1 √ ,√ 10 10   , (3, 1) = √ 10 − 1. Now we must find ρ(A, B). We find that we can use any of the points in the bottom lower left quadrant on the unit circle. Let us choose the point (−1, 0). Then
ρ(A, B) = d (−1, 0) , (0, 0) = 1. √ √ Therefore, the Hausdorff distance is h(A, B) = max{1, 10 − 1} = 10 − 1. The next example is also in the metric space (R2 , d) with the same metric d from the previous example. However, this time we will consider two subsets of the plane that do not intersect. PROVING COMPLETENESS OF THE HAUSDORFF INDUCED METRIC SPACE 9 Example Let A and B be sets defined by A = {(x, y) : 0 ≤ x ≤ 1, 0 ≤ y ≤ 1} and let B = {(x, y) : 3 ≤ x ≤ 5, 0 ≤ y ≤ 4} [see Figure 3]. Figure 3. The sets A and B in (R2 , d).

If (a1 , a2 ) ∈ A, then r (a1 , a2 ), B) = d (a1 , a2 ), (3, a2 ) = 3 − a1 . Since 0 ≤ a1 ≤ 1, we find that ρ(A, B) = 3.

If (b1 , b2 ) ∈ B, then r (b1 , b2 ), A) = d (b1 , b2 ), (1, a2 ) , where 0 ≤ a2 ≤ 1, which varies with our choice of (b1 , b2 ). We find that the point that maximizes r is
b = (5, 4) such that ρ(B, A) = d (5, 4), (1, 1) = 5. Therefore, the Hausdorff distance is given by h(A, B) = ρ(B, A) = 5. Now that we have gained a knowledge on how r, ρ, and h work in a few special cases, we prove some basic properties of r and ρ. Theorem 1. Let x ∈ X and let A, B, C ∈ K. (1) r(x, A) = 0 if and only if x ∈ A. (2) ρ(A, B) = 0 if and only if A ⊆ B. (3) There exists ax ∈ A such that r(x, A) = d(x, ax ). (4) There exists a∗ ∈ A and b∗ ∈ B such that ρ(A, B) = d(a∗ , b∗ ). (5) If A ⊆ B, then r(x, B) ≤ r(x, A). (6) If B ⊆ C, then ρ(A, C) ≤ ρ(A, B). (7) ρ(A ∪ B, C) = max{ρ(A, C), ρ(B, C)}. (8) ρ(A, B) ≤ ρ(A, C) + ρ(C, B). 10 Proof. We first prove Property (1). Suppose x ∈ A. Then the infimum distance is d(x, a) = 0, where a = x. Now suppose that r(x, A) = 0. Then for each positive integer n, there exists an ∈ A such that d(x, an ) < 1 n. By definition {an } converges to x. Since A is compact, it is closed. By Lemma 1 it follows that x ∈ A. We now prove Property (2). Suppose A ⊆ B. Let a ∈ A. Since a ∈ B, by Property (1) then r(a, B) = 0. Therefore ρ(A, B) =sup{0} = 0. To prove the converse, suppose ρ(A, B) = 0. Let a ∈ A. Then 0 ≤ r(a, B) ≤ ρ(A, B) = 0, and thus by Property (1), we find a ∈ B. It follows that A ⊆ B. To prove Property (3), by definition of an infimum we can let {an } be a sequence in A such that d(x, an ) < r(x, A) + n1 . We know A is sequentially compact, so there exists a subsequence {ank } of {an } that converges to an element ax ∈ A. Then we find that r(x, A) ≤ d(x, ax ) ≤ d(x, ank ) + d(ank , ax ) ≤ r(x, A) + Since lim k→∞ 1 + d(ank , ax ). nk
1 + d(ank , ax ) = 0, it follows that d(x, ax ) = r(x, A). nk To prove Property (4), by definition of a supremum we can let {an } be a sequence in A such that ρ(A, B) is the limit of r(an , B). By Property (3) there exists a sequence {bn } in B such that r(an , B) = d(an , bn ). Since A is sequentially compact, there exists a subsequence {ank } of {an } that converges to an element a∗ ∈ A. Since B is sequentially compact, there exists a subsequence {bnkj } of {bnk } that converges to b∗ . Since {bnkj } converges to b∗ and {ankj } converges to a∗ , by Result 1, we know that d(ankj , bnkj ) converges to d(a∗ , b∗ ). Therefore, it follows that ρ(A, B) = lim r(ankj , B) = lim r(ankj , bnkj ) = d(a∗ , b∗ ). j→∞ j→∞ Thus it follows that d(a∗ , b∗ ) = ρ(A, B). Now we will prove Property (5). Suppose A ⊆ B and x ∈ X. Let a ∈ A. Since A is a subset of B, we find that a ∈ B. It follows that d(x, a) ≥ inf{d(x, b) : b ∈ B} = r(x, B). PROVING COMPLETENESS OF THE HAUSDORFF INDUCED METRIC SPACE 11 Since this is true for all a ∈ A, we find that r(x, A) = inf{d(x, a) : a ∈ A} ≥ r(x, B). For Property (6), suppose B ⊆ C. By Property (5) then r(a, C) ≤ r(a, B) for all a ∈ A. It follows that sup{r(a, C) : a ∈ A} ≤ sup{r(a, B) : a ∈ A} and thus ρ(A, C) ≤ ρ(A, B). For Property (7), by the definitions of r and ρ we see that ρ(A ∪ B, C) = sup{r(x, C) : x ∈ A ∪ B} = max{sup{r(x, C) : x ∈ A}, sup{r(x, C) : x ∈ B}} = max{ρ(A, C), ρ(B, C)}. We now turn to Property (8). Property (3) guarantees that for each a ∈ A there exists ca ∈ C such that r(a, C) = d(a, ca ). We then have r(a, B) = inf{d(a, b) : b ∈ B} ≤ inf{d(a, ca ) + d(ca , b) : b ∈ B} = d(a, ca ) + inf{d(ca , b) : b ∈ B} = r(a, C) + r(ca , B) ≤ ρ(A, C) + ρ(C, B). Since a ∈ A was arbitrary, taking the supremum, we find that ρ(A, B) ≤ ρ(A, C) + ρ(C, B). This completes the proof.  Note that by Property (4) of Theorem 1, there exist points a1 ∈ A and b1 ∈ B such that ρ(A, B) = d(a1 , b1 ) and alternatively there exist points a2 ∈ A and b2 ∈ B such that ρ(B, A) = d(a2 , b2 ). Since h(A, B) is just the maximum of ρ(A, B) and ρ(B, A), it follows that there exist points a∗ ∈ A and b∗ ∈ B such that h(A, B) = d(a∗ , b∗ ). 12 The following theorem shows that the Hausdorff distance defines a metric on K. Theorem 2. The set K with the Hausdorff distance h define a metric space (K, h). Proof. To prove that (K, h) is a metric space, we need to verify the following four properties. (1) h(A, B) ≥ 0 for all A, B ∈ K. (2) h(A, B) = 0 if and only if A = B. (3) h(A, B) = h(B, A) for all A, B ∈ K. (4) h(A, B) ≤ h(A, C) + h(C, B) for all A, B, C ∈ K. To prove the first property, since ρ(A, B) and ρ(B, A) are nonnegative, it follows that h(A, B) ≥ 0 for all A, B ∈ K. For the second property, suppose A = B. Therefore A ⊆ B and B ⊆ A. By Property (2) of Theorem 1 we find that ρ(A, B) = 0 and ρ(B, A) = 0, and thus h(A, B) = 0. Now suppose h(A, B) = 0. This implies ρ(A, B) = ρ(B, A) = 0. By Property (2) of Theorem 1, we see that A ⊆ B and B ⊆ A and it follows that A = B. The third property can be proved from the symmetry of the definition since h(A, B) = max{ρ(A, B), ρ(B, A)} = max{ρ(B, A), ρ(A, B)} = h(B, A). The final property follows from the definition of ρ and h and from Property (8) of Theorem 1. We find that ρ(A, B) ≤ ρ(A, C) + ρ(C, B) ≤ max{ρ(A, C), ρ(C, A)} + max{ρ(C, B), ρ(B, C)} =h(A, C) + h(C, B). Similarly, ρ(B, A) ≤ ρ(B, C) + ρ(C, A) ≤ max{ρ(B, C), ρ(C, B)} + max{ρ(C, A), ρ(A, C)} =h(C, B) + h(A, C). Therefore, h(A, B) = max{ρ(A, B), ρ(B, A)} ≤ h(A, C) + h(C, B).  PROVING COMPLETENESS OF THE HAUSDORFF INDUCED METRIC SPACE 13 Therefore we know that h defines a metric on K. In the next section, we will look at examples of what this metric space might look like, and then proceed to prove that if the metric space (X, d) is complete, then the metric space (K, h) is also complete. Examples of the Hausdorff Metric Space Given a complete metric space (X, d), we have now constructed a new metric space (K, h) from the nonempty, compact subsets of X using the Hausdorff metric. Now it remains for us to prove that (K, h) is also complete. To be a complete metric space, every Cauchy sequence in the space must converge to a point also in the space. Therefore, when we think about the metric space (K, h), we are choosing a sequence of nonempty, compact sets and showing that this sequence converges to another nonempty, compact set. To better visualize the abstract mathematics we are doing, consider the following two examples that demonstrate metric spaces of nonempty, compact sets with the Hausdorff metric. Example Let (R, d0 ) be the complete metric space, where d0 is the discrete metric, d0 (x, y) =     0, when x = y.    1, when x 6= y. Since K is the set of all nonempty, compact subsets of (R, d0 ), we find that K is the set of all nonempty finite subsets of R. The infinite sets are not in K because they are not totally bounded and are thus not compact. Furthermore, we may notice that r(x, B) = inf{d0 (x, b) : b ∈ B} = d0 (x, b) =     0, when x ∈ B.    1, when x ∈ / B. 14 Therefore, ρ(A, B) = sup{r(a, B) : a ∈ A} =     0, when a ∈ B.    1, when a ∈ / B. . So it follows that h(A, B) =     0, when A = B.    1, when A 6= B. Therefore we have a metric space with the set K of the discrete subsets of R with the Hausdorff metric as the discrete metric. It is easy to verify that our newly created space is not totally bounded. However, we know all discrete metric spaces are complete, so (K, h) is complete. Therefore, the space (K, h) of finite sets with the discrete metric is an example of our Hausdorff induced metric space (K, h). To illustrate our notion of completeness, now briefly consider a sequence of nonempty compact sets that converges to the unit circle in R2 . This is an example a converging Cauchy sequence in the Hausdorff induced metric space that converges to a set also in the space. Example Let (R2 , d) be the complete metric space where for x = (x1 , x2 ) and for p y = (y1 , y2 ), then d((x1 , y1 ), (x2 , y2 )) = (x1 − x2 )2 + (y1 − y2 )2 . Let K be the set of all nonempy, compact subsets of R2 , or in other words let K be the set of all nonempty closed and bounded sets of R2 . As we will later prove, we know that since (R2 , d) is complete, then the metric space (K, h) is complete. To see an example of a Cauchy sequence in this space that converges to something in the set, let us consider a sequence of sets converging to the unit circle. For each positive integer k, let Ak = {(r, θ) : r = 1 + k1 cos(kθ), 0 ≤ θ ≤ 2π}, and let A be the unit circle in R2 . It is easy to see that each Ak is in K. Examining the Hausdorff distance between sets, we see that h(Ak , A) = k1 . PROVING COMPLETENESS OF THE HAUSDORFF INDUCED METRIC SPACE 15 We can note that as k increases, the sets converge to the unit circle [See figures 6-8]. Therefore, {Ak } is an example of a Cauchy sequence that converges to A ∈ K. Figure 4. The set A5 defined by r = 1 + Figure 5. The set A20 defined by r = 1 + 1 5 1 20 cos(5θ). cos(20θ). Proving that the Metric (K, h) is Complete As previously stated, to be a complete metric space, every Cauchy sequence in (K, h) must converge to a point in K. Therefore, in order to prove that the metric space (K, h) is complete, we will choose an arbitrary Cauchy sequence {An } in K and show that it converges to some A ∈ K. Define A to be the set of all points x ∈ X such that there is a sequence {xn } that converges to x and satisfies xn ∈ An 16 Figure 6. The set A50 defined by r = 1 + 1 50 cos(50θ). for all n. We will eventually show that the set A is an appropriate candidate. However, we must begin with some important theorems regarding A. Given a set A ∈ K and a positive number , we define the set A +  by {x ∈ X : r(x, A) ≤ }. We need to show that this set is closed for all possible choices of A and . To do this, we will begin by choosing an arbitrary limit point of the set A + , and then showing that it is contained in the set. Proposition 1: A +  is closed for all possible choices of A ∈ K and  > 0. Proof. Let A ∈ K and  > 0. Additionally, let x be a limit point of A + . Then there exists a sequence {xn } of points in (A + )\{x} that converges to x. Since xn ∈ A +  for all n, by definition r(xn , A) ≤  for all n. Property (3) of Theorem 1 guarantees that for each n there exists an ∈ A such that r(xn , A) = d(xn , an ). Thus d(xn , an ) ≤  for all n. The set A is sequentially compact, so it follows from the definition that each sequence {an } has a subsequence {ank } that converges to a point a ∈ A. Since {xn } converges to x, we know that any subsequence of {xn } converges to x. Therefore, the subsequence {xnk } converges to x. By Result 1, then d(xnk , ank ) converges to d(x, a). Note that {xnk } and {ank } are subsequences of {xn } and {an }, respectively, so d(xnk , ank ) ≤  for all k. Therefore, we find that d(x, a) ≤ . By the definition of r(x, A), then r(x, A) ≤ , so x ∈ A +  by our PROVING COMPLETENESS OF THE HAUSDORFF INDUCED METRIC SPACE 17 definition of A + . Since x was an arbitrary limit point, then A +  is a closed set since it contains all of its limit points.  As an example where the set A +  is not compact, consider the set R with the discrete metric. Let A be any nonempty finite set and choose  > 1. Then the set A +  = R is closed, but not totally bounded and is therefore is not compact. To show (K, h) is complete, we will need to show that A ∈ K, and that {An } converges to A. By our definition of convergence, we must show that there exists a positive integer N such that h(An , A) <  for all n ≥ N . However, the following theorem gives us an alternative way of proving convergence. Theorem 3. Suppose that A, B ∈ K and that  > 0. Then h(A, B) ≤  if and only if A ⊆ B +  and B ⊆ A + . Proof. By symmetry, it is sufficient to prove ρ(B, A) ≤  if and only if B ⊆ A + . Suppose B ⊆ A + . By definition of the set A + , for every b ∈ B then r(b, A) ≤ . It follows that ρ(B, A) ≤ . Now suppose ρ(B, A) ≤ . Then for every b ∈ B then r(b, A) ≤ . It follows by definition of the set A + , that B ⊆ A + .  Extension Lemma: Let {An } be a Cauchy sequence in K and let {nk } be an increasing sequence of positive integers. If {xnk } is a Cauchy sequence in X for which xnk ∈ Ank for all k, then there exists a Cauchy sequence {yn } in X such that yn ∈ An for all n and ynk = xnk for all k. Proof. Suppose {xnk } is a Cauchy sequence in X for which xnk ∈ Ank for all k. Define n0 = 0. For each n that satisfies nk−1 < n ≤ nk , use Property 3 to choose yn ∈ An such that r(xnk , An ) = d(xnk , yn ). Then we find, using the definitions of ρ and r that d(xnk , yn ) = r(xnk , An ) ≤ ρ(Ank , An ) ≤ h(Ank , An ). Note that since xnk ∈ Ank , then d(xnk , ynk ) = r(xnk , Ank ) = 0. It follows that ynk = xnk for all k. 18 Let  > 0. Since {xnk } is a Cauchy sequence in X, there exists a positive integer K such that d(xnk , xnj ) <  3 for all k, j ≥ K. Since {An } is a Cauchy sequence in K, by definition there exists a positive integer N ≥ nK such that h(An , Am ) <  3 for all n, m ≥ N . Suppose that n, m ≥ N . Then there exists integers j, k ≥ K such that nk−1 < n ≤ nk , and nj−1 < m ≤ nj . Then we find that d(yn , ym ) ≤ d(yn , xnk ) + d(xnk , xnj ) + d(xnj , ym ) = r(xnk , An ) + d(xnk , xnj ) + r(xnj , Am ) ≤ ρ(Ank , An ) + d(xnk , xnj ) + ρ(Anj , Am ) ≤ h(Ank , An ) + d(xnk , xnj ) + h(Anj , Am ) <    + + = . 3 3 3 Therefore, by definition and from our earlier set up, {yn } is a Cauchy sequence in X such that yn ∈ An for all n and ynk = xnk for all k. This completes the proof.  The following Lemma makes use of the Extension Lemma to guarantee that A is closed and nonempty. We will need this fact in proving that A is in K, since we must show that A is a nonempty, compact subset of K. This Lemma gives us that A is closed and nonempty. Since closed and totally bounded sets are compact, it remains to show that A is totally bounded. Lemma 2: Let {An } be a sequence in K and let A be the set of all points x ∈ X such that there is a sequence {xn } that converges to x and satisfies xn ∈ An for all n. If {An } is a Cauchy sequence, then the set A is closed and nonempty. Proof. We begin by proving that A is nonempty. Since {An } is a Cauchy sequence, there exists an integer n1 such that h(Am , An ) < 1 21 = 1 2 for all m, n ≥ n1 . Sim- ilarly, there exists an integer n2 > n1 such that h(Am , An ) < 1 22 = 1 4 for all m, n ≥ n2 . Continuing this process we have an increasing sequence {nk } such that h(Am , An ) < 1 2k for all m, n ≥ nk . Let xn1 be a fixed point in An1 . By Property 2 PROVING COMPLETENESS OF THE HAUSDORFF INDUCED METRIC SPACE 19 of Theorem 2, we can choose xn2 ∈ An2 such that d(xn1 , xn2 ) = r(xn1 , An2 ). Then by definition of r, ρ, and h we find that d(xn1 , xn2 ) = r(xn1 , An2 ) ≤ ρ(An1 , An2 ) ≤ h(An1 , An2 ) < 21 . Similarly we can choose xn3 ∈ An3 such that d(xn2 , xn3 ) = r(xn2 , An3 ) ≤ ρ(An2 , An3 ) ≤ h(An2 , An3 ) < 41 . Continuing this process we can construct a sequence {xnk } where each xnk ∈ Ank and for all k, d(xnk , xnk+1 ) = r(xnk , Ank+1 ) ≤ ρ(Ank , Ank+1 ) ≤ h(Ank , Ank+1 ) < 1 . 2k By Result 2 {xnk } is a Cauchy sequence. Therefore, since {xnk } is a Cauchy sequence and xnk ∈ Ank for all k, by the Extension Lemma there exists a Cauchy sequence {yn } in X such that yn ∈ An for all n and ynk = xnk for all k. Since X is complete, the Cauchy sequence {yn } converges to a point y ∈ X. Since yn ∈ An for all n, then by definition of the set, y ∈ A. Therefore A is nonempty. Now we will prove that A is closed. Suppose a is a limit point of A. Then there exists a sequence ak ∈ A\{a} that converges to a. Since each ak ∈ A, there exists a sequence {yn } such that {yn } converges to ak and yn ∈ An for each n. Consequently, there exists an integer n1 such that xn1 ∈ An1 and d(xn1 , a1 ) < 1. Similarly, there exists an integer n2 > n1 and a point xn2 ∈ An2 such that d(xn2 , a2 ) < 1 2. Continuing this process we can choose an increasing sequence {nk } of integers such that d(xnk , ak ) < 1 k for all k. Then it follows that d(xnk , a) ≤ d(xnk , ak ) + d(ak , a). Note that as we take k to infinity, the distance between {xnk } and a converges to zero, so it follows that {xnk } converges to a. Every convergent sequence is Cauchy, so it follows that {xnk } is a Cauchy sequence for which xnk ∈ Ank for all k. The Extension Lemma guarantees that there exists a Cauchy sequence {yn } in X such that yn ∈ An for all n and ynk = xnk . Therefore a ∈ A, so A is closed.  20 With the previous lemma, to prove A ∈ K, it only remains to show that A is totally bounded. The following lemma will allow us to do so. Lemma 3: Let {Dn } be a sequence of totally bounded sets in X and let A be any subset of X. If for each  > 0, there exists a positive integer N such that A ⊆ DN + , then A is totally bounded. Proof. Let  > 0. Choose a positive integer N so that A ⊆ DN + 4 . Since DN is totally bounded, by definition we can choose a finite set {xi : 1 ≤ i ≤ q} where q [  xi ∈ DN such that DN ⊆ Bd (xi , ). By reordering the xi ’s, we may assume 4 i=1 that Bd (xi , 2 ) ∩ A 6= ∅ for 1 ≤ i ≤ p and Bd (xi , 2 ) ∩ A = ∅ for p < i. Then for p [  each 1 ≤ i ≤ p, let yi ∈ Bd (xi , 2 ) ∩ A. We claim that A ⊆ Bd (yi , ). Let a ∈ A. i=1 Then a ∈ DN + 4 , so r(a, DN ) ≤ 4 . By Theorem 1 Property (3), then there exists x ∈ DN such that d(a, x) = r(a, DN ). Then we find that d(a, xi ) ≤ d(a, x) + d(x, xi ) ≤    + = . 4 4 2 So x ∈ Bd (xi , 2 ) for some 1 ≤ i ≤ p. Thus we have yi ∈ Bd (xi , 2 ) ∩ A such that d(xi , yi ) < 2 . It follows that d(a, yi ) ≤ d(a, xi ) + d(xi , yi ) <   + = . 2 2 Therefore since for each a ∈ A we found yi for 1 ≤ i ≤ p such that a ∈ Bd (yi , ), p [ then it follows that A ⊆ Bd (yi , ). Thus by definition, A is totally bounded. i=1 This completes the proof.  Finally, we have the foundation to prove our final result. Given a complete metric space (X, d), we constructed the metric space (K, h) from the nonempty compact subsets of X using the Hausdorff metric. After examining important theorems and results, we can now prove our main goal. PROVING COMPLETENESS OF THE HAUSDORFF INDUCED METRIC SPACE 21 Theorem 4. If (X, d) is complete, then (K, h) is complete. Proof. Let {An } be a Cauchy sequence in K, and define A to be the set of all points x ∈ X such that there is a sequence {xn } that converges to x and satisfies xn ∈ An for all n. We must prove that A ∈ K and {An } converges to A. By Lemma 2, the set A is closed and nonempty. Let  > 0. Since {An } is Cauchy then there exists a positive integer N such that h(An , Am ) <  for all m, n ≥ N . By Theorem 3 then Am ⊆ An +  for all m > n ≥ N . Let a ∈ A. Then we want to show a ∈ An +. Fix n ≥ N . By definition of the set A, there exists a sequence {xi } such that xi ∈ Ai for all i and {xi } converges to a. By Proposition 1 we know that An +  is closed. Since xi ∈ An +  for each i, then it follows that a ∈ An + . This shows that A ⊆ An + . By Lemma 3, the set A is totally bounded. Additionally, we know A is complete, since it is a closed subset of a complete metric space. Since A is nonempty, complete and totally bounded, then A is compact and thus A ∈ K. Let  > 0. To show that {An } converges to A ∈ K, we need to show that there exists a positive integer N such that h(An , A) <  for all n ≥ N . To do this, Theorem 3 tells us that we need to show two conditions, that A ⊆ An +  and An ⊆ A + . From the first part of our proof, we know there exists N such that A ⊆ An +  for all n ≥ N . To prove An ⊆ A + , let  > 0. Since {An } is a Cauchy sequence, we can choose a positive integer N such that h(Am , An ) <  2 for all m, n ≥ N . Since {An } is a Cauchy sequence in K, there exists a strictly increasing sequence {ni } of positive integers such that n1 > N and such that h(Am , An ) < 2−i−1 for all m, n > ni . We can use Property (3) of Theorem 1 to get the following: since An ⊆ An1 + 2 , there exists xn1 ∈ An1 such that d(y, xn1 ) ≤ 2 . since An1 ⊆ An2 + 4 , there exists xn2 ∈ An2 such that d(xn1 , xn2 ) ≤ 4 . since An2 ⊆ An3 + 8 , there exists xn3 ∈ An3 such that d(xn2 , xn3 ) ≤ 8 . By continuing this process we are able to obtain a sequence {xni } such that for all positive integers i then xni ∈ Ani and d(xni , xni+1 ) ≤ 2−i−1 . By Result 2 we find 22 {xni } is a Cauchy sequence, so by the Extension Lemma the limit of the sequence a is in A. Additionally we find that d(y, xni ) ≤ d(y, xn1 ) + d(xn1 , xn2 ) + d(xn2 , xn3 ) + · · · + d(xni−1 , xni ) ≤     + + + ··· + i <  2 4 8 2 Since d(y, xni ) ≤  for all i, it follows that d(y, a) ≤  and therefore y ∈ A + . Thus we know that there exists N such that An ⊆ A+, so it follows that h(An , A) <  for all n ≥ N and thus {An } converges to A ∈ K. Therefore, if (X, d) is complete, then (K, h) is complete.  We will now prove that if X is compact, then K is compact. Note that a metric space is compact if and only if it is complete and totally bounded. Theorem 5. If (X, d) is compact, then (K, h) is compact. Proof. By previous result we know that K is complete. Since we know that a set is compact if and only if it is complete and totally bounded, we must prove that K is totally bounded. Let  > 0. Since X is totally bounded, there exists a finite n [
set {xi : 1 ≤ i ≤ n} such that X ⊆ Bd xi , 3 and xi ∈ X for each i. Let i=1 {Ck : 1 ≤ k ≤ 2p − 1} be the collection of all possible nonempty unions of the closures of these balls. Since X is compact, the closure of each ball is a compact set. Therefore, each Ck is a finite union of compact sets and thus compact, so p 2[ −1 Ck ∈ K. We want to show that K ⊆ Bh (Ck , ). k=1 To do this, let Z ∈ K. Then we want to show that Z ∈ Bh (Ck , ) for some k. Choose SZ = {i : Z ∩ Bd (xi , ) 6= ∅}. Then choose an index j so that [
Cj = Bd xi , 3 . Since Z ⊆ Cj , then by Property (2) of Theorem 1 then we i∈SZ know ρ(Z, Cj ) = 0. Now let c be an element in Cj . Then there exists some i ∈ SZ
and z ∈ Z such that c, z ∈ Bd xi , 3 . This implies that r(c, Z) ≤ 23 . Since our choice of c was arbitrary, then we find that ρ(Cj , Z) ≤ 2 3 . Therefore, we find PROVING COMPLETENESS OF THE HAUSDORFF INDUCED METRIC SPACE 23 that h(Z, Cj ) = ρ(Cj , Z) < , and thus Z ⊆ Bh (Cj , ), so K is totally bounded. Therefore, we have proved that if (X, d) is compact, then (K, h) is compact.  Conclusions The Hausdorff distance is a measure that assigns a nonnegative real number as the distance between sets. We investigated this distance through several examples. Then, given a metric space (X, d), we found that the Hausdorff distance defines a metric on the space K of all nonempty compact subsets of X. We explored some of the nice qualities of this metric. Most importantly, that if (X, d) is complete, then (K, h) is complete. In addition, we proved that if (X, d) is compact, then (K, h) is totallycompact, which is a truly remarkable result. References [1] Russ Gordon, Real Analysis: A First Course. Walla Walla, Washington, 1st Edition, 2011.