A CONTRIBUTION TO THE STUDY OF THE PROBABILISTIC HAUSDORFF METRIC

The purpose of the present paper is to study several important properties of the probabilistic Hausdorff metric. Indeed, we prove that the collection of nonempty compact subsets of a probabilistic metric space is complete with respect to the probabilistic Hausdorff metric if and only if the probabilistic metric space is complete . As a consequence, the relationship between completion of the probabilistic Hausdorff metric for a given PM space and probabilistic Hausdorff metric of its completion is examined. Finally, Deterministic counterpart of these results is deduced. Our work extend some results of [2] and [12]. AMS Subject Classification: 54E70, 54A40, 54B20


Introduction and Preliminaries
The concept of probabilistic metric space was first introduced and studied in 1942 by K.Menger [7].It is a probabilistic generalization of metric space in which the distance d(x, y) between two point x, y was replaced by a real function F xy whose value F xy (t), for any positive real t, is interpreted as the probability that the distance between x and y is less than or equal to t.The study of these spaces was performed extensively by B.Schweizer and A.Sklar [9].
We briefly recall some definitions and known results about probabilistic metric spaces, for more detail we refer the reader to [9], Recall that a nonnegative real function f defined on IR + = IR + ∪ {+∞} is known as a distance distribution function (briefly, a d.d.f) if it is non-decreasing, left continuous on ]0, +∞[, with f (0) = 0 and f (+∞) = 1.The set of all d.d.f will be denoted by ∆ + .
A simple example of distribution function is Heavyside function defined by: In the sequel we will define some functions, on IR + and consider them automatically extended to IR + .According to [9] a commutative, associative and non-decreasing mapping τ : A triangle function τ is called sup-continuous, if for any family Definition 1.1.Let f and g be in ∆ + , let h be in (0, 1], and let (f ,g;h) denote the condition The modified Lévy distance is the function The pair (∆ + , d L ) is a compact metric space.
(2) For any t > 0 By a probabilistic metric space we mean a triplet (M, F, τ ) such that M is nonempty set,τ is a triangle function and F is a mapping from M × M into ∆ + satisfying the following conditions for all p, q, r in M : For a given PM space (M, F, τ ), B. Schweizer and A.Sklar ( [9]) point out that If τ is continuous, then F generates a first countable Hausdorff topology T F on M which has as a base the family of sets of the form {N x (t) : x ∈ M, t > 0}, where N x (t) = {y ∈ M : F xy (t) > 1 − t}.On the other hand {U t : t > 0} is a base for a uniformity U (F ) on M compatible with T F , where U t = {(x, y) ∈ M × M : F xy (t) > 1 − t} for all t > 0 .U (F ) is called the strong uniformity induced by F .In the sequel, when we speak about a PM space (M, F, τ ), we always assume that τ is continuous.
In virtue of T F , a sequence (x n ) n∈IN in M, is said to be convergent to x (we write x n → x or lim The PM space (M, F, τ ) is said to be complete, if every Cauchy sequence in M converges to some point in M .
Following [10] (see also, [9]) a completion of a PM space (M, F, τ ) is a pair ((N, G, σ), Φ), where (N, G, σ) is a complete PM space and Φ is an isometry from M into N (i.e.F pq = G Φ(p)Φ(q) for all p, q ∈ M ) such that Φ(M ) is dense in N .It is well known that every metric space has a completion which is unique up to isometry see for instance [4].Sherwood in [10] established the following result (see also [9]) [10]) Every PM space (M, F, τ ) with a continuous triangle function has a completion that is unique up to isometry.
Indeed, The construction of the completion of a PM space (M, F, τ ) with continuous triangle function follows the line as in the classical metric case.
In 1968 R.J. Egbert [3] extended the classical Hausdorff distance to the probabilistic setting by introducing the notion of probabilistic Hausdorff metric in the case of Menger space.Thereafter, Tardiff [11] enlarged this notion to the context of general probabilistic metric spaces (see also [9]).Let (M, F, τ ) be a PM space and denote by P 0 (M ) the family of all nonempty subsets of M .Given p ∈ M , B ∈ P 0 (M )."The probabilistic distance" form p to B is defined as Given A, B ∈ P 0 (M ) and defined and The probabilistic Hausdorff distance between A and B is the d.d.f: In [9] it was proved that If the triangle function τ is sup-continuous.Then the triplet (C 0 (M ), H F , τ ) is a PM space, where C 0 (M ) denote the collection of nonempty closed subset of M .Moreover H F satisfies the following properties, where cl T F denotes the closure of with respect to the strong topology.

Main Results
We begin this section by studying completeness property of the probabilistic Hausdorff metric.The first result is a direct consequences of [ [6], Th 12].
Theorem 2.1.A PM space (M, F, τ ) with sup-continuous triangle function is complete if and only if the PM space (C 0 (M ), H F , τ ) is complete.
To state our next result we recall that a subset Y of an uniform space (X, U ) is said to be totally bounded if for every U ∈ U there exists a finite subset A subset Y of a uniform space (X, U ) is compact if and only if Y is complete and totally bounded ([ [5], Ch. 6, Th. 32]).Let K 0 (M ) denote the collection of nonempty compact subset of PM space (M, F, τ ).
Theorem 2.2.A PM space (M, F, τ ) with sup-continuous triangle function is complete if and only if PM space (K 0 (M ), H F , τ ).
Proof.At first suppose that (K 0 (M ), H F , τ ) is a complete PM space.Let (x n ) n∈IN be a Cauchy sequence in M and consider the sequence (A n ) n∈IN where A n = {x n } for all n ∈ IN.Then, from Lemma 1.3 (2) (A n ) n∈IN is Cauchy sequence in (K 0 (M ), H F , τ ), so there exists A ∈ K 0 (M ) such that A n → A with respect to H F . it is easy to show that (x n ) n∈IN converge to every x ∈ A. hence (M, F, τ ) is complete.
Conversely, by Theorem 2.1 The convergence of (A n ) n∈IN and (2) of Lemma 1.1 implies that there exists Now, let p ∈ A, by (5) of Lemma 1.3 and (2.2) of Lemma 1.1, there is q ∈ A N such that since A N is compact, so A N is totally bounded, Then there exists a finite subset Z ⊂ M such that A N ⊂ U µt (Z), which implies the existence of z ∈ Z such that Hence form Lemma 1.1 (1), ( 2) and the inequalities (2.1), (2.3) and (2.4) we have This shows that A ⊆ U t (Z).Thus, A is totally bounded and since A is closed in the complete PM space (M, F, τ ), it follows that A is complete.Hence, A ∈ K 0 (M ).
In the sequel we will refer to ( M , F , τ ) as the completion of the PM space (M, F, τ ).Our next object is to find the relationship between ( C 0 (M ), H F , τ ) and (C 0 ( M ), H F , τ ) (resp, ( K 0 (M ), H F , τ ) and (K 0 ( M ), H F , τ ) ). Proof.Let (M, F, τ ) be a PM space and consider the mapping Φ from C 0 (M ) into C 0 ( M ) defined by Φ(A) = cl T F (A) for all A ∈ C 0 (M ).We will claim that Φ is an isometry from (C 0 (M ), H F , τ ) to (C 0 ( M ), H F , τ ).Let A, B ∈ C 0 (M ), from Lemma 1.3 (4) we have for all t > 0 In the next step we show that φ(C 0 (M )) is dense in C 0 ( M ).Equivalently we need to show that (2.5) Putting t 1 = min{s, µ s }, since M is dense in M , then for each a ∈ A there exists (5).To this end take a ∈ A then there exists x a such that Next, let c ∈ C, then there exists x a ∈ B such that Then we have and Combining (2.5) with (2.7) and (2.8), we obtain Hence,from (2.6) and (2.9) we conclude that ).Thus, by Theorem 1.2 we conclude that (C 0 ( M ), H F , τ ) and Now we will show that Theorem 2.3 remain valid if we replace C 0 (M ) by K 0 (M ).
Lemma 2.1.Let (M, F, τ ) be a PM space with sup-continuous triangle function.If K is a dense subset of M , then K 0 (K) is dense in the PM space (K 0 (M ), H F , τ ).
Proof.Let A ∈ K 0 (M ), t > 0 and 0 < s < t.Since τ is uniformly continuous, then there is µ s > 0 such that for all G, R ∈ ∆ Putting t 1 = min{s, µ s }.Since A is compact, then A is totally bounded, so there exists a finite subset Y = {y 1 , . . ., y n } such that A ⊂ 1≤i≤n U t (y i ) = 1≤i≤n N y i (t).By hypothesis K is dense in M , then for each i ∈ {1, . . ., n} there exists z i ∈ K such that F z i y i (t 1 ) > 1 − t 1 .Putting Z = {z 1 , . . ., z n }.Applying the same arguments as in Theorem 2.3 we show that Theorem 2.4.Let (M, F, τ ) be a PM space with sup-continuous triangle function.Then, (K 0 ( M ), H F , τ ) and ( K 0 (M ), H F , τ ) are isometric.
Proof.Using the fact the restriction to K 0 (M ) of the map Φ defined in Theorem 2.3 is the identity, Lemma 2.1 and Theorem 2.2 we deduce that (K 0 ( M ), H F , τ ) is a completion of (K 0 (M ), H F , τ ).The conclusion follows immediately from Theorem 1.2.

Relative Results
In this section we will show a counterpart of Theorem 2.3 and Theorem 2.4 in the case of metric spaces.In the sequel ( M , d) denote the completion of given metric space (M, d).It is well known (see, [1]) that for a given metric space (M, d) the Hausdorff metric on the collection af all closed subsets C 0 (M ) is defined as the infinite valued metric In [8] Michael proved that a metric space (M, d) is complete if and only if (C 0 (M ), H d ) is complete.Let (M, d) be a metric space and consider the mapping for all p, q ∈ M, t > 0.
It is easy to check that (M, F d , π) is a PM space where π(f, ) = f.g for all f, g ∈ ∆ + and that (M, F d , π) is complete if and only if (M, d) is complete.In addition we have (ii) Let p, q ∈ M and (p n ) n∈IN , (q n ) n∈IN two Cauchy sequences in (M, d) such that (p n ) n∈IN ∈ p and (q n ) n∈IN ∈ q Then for each t > 0 we have F d p q (t) = lim n F p n q n (t) =

Lemma 3 . 1 .
Let (M, d) be a metric space then(i) H F d and F H d coincide in C 0 (M ).(ii) F d and F d coincide in M .Proof.(i) Let A, B ⊂ C 0 (M ) and s > 0, we can easily show that F d pB (,B) .ThusH − F d (A, B)(t) = sup s<t s s + sup a∈A d(a, B) = t t + H − d (A, B) (t) = F H − d AB (t) similarly we show that H + F d (A, B) = F H + d AB .Hence, H F d = F H d .

Theorem 3 . 1 .
t).Hence, F d and F d coincide in M .Let (M, d) be a metric space.Then(i) (C 0 ( M ), H d ) and ( C 0 (M ), H d ) are isometric.(ii) (K 0 ( M ), H d ) and ( K 0 (M ), H d ) are isometric.Proof.From Lemma 3.1 we have H F d = H F d = F H d on C 0 ( M ) andH F d = F H d = F H d on C 0 (M ).from Theorem 2.3 there is and isometry Φ from(C 0 ( M ), F H d ) into ( C 0 (M ), F H d ).hence Φ is an isometry from (C 0 ( M ), H d ) into ( C 0 (M ),H d ) .(ii) follows similarly by replacing Theorem 2.3 by Theorem 2.4.