By Michel Coste (auth.), J. -D. Boissonnat, J. -P. Laumond (eds.)

**Read Online or Download Geometry and Robotics: Workshop, Toulouse, France, May 26–28, 1988 Proceedings PDF**

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**Additional resources for Geometry and Robotics: Workshop, Toulouse, France, May 26–28, 1988 Proceedings**

**Example text**

To motivate the next definition, let us consider two simple examples from Figure 1. We recall that all sets listed in Figure 1 are ordered with respect to certain order relations. Consider the set IV2 R of interval vectors of dimension 2. The elements of this set are intervals of two dimensional real vectors. Such an element describes a rectangle in the x, y plane with sides parallel to the axes. Such interval vectors are special elements of the powerset PV2 R of two dimensional real vectors. PV2 R is defined as the set of all subsets of two dimensional real vectors.

These properties are made precise by the following two theorems. 27. Let {M, ≤} be a complete lattice and let {S, ≤} and {D, ≤} both be lower (resp. upper) screens of {M, ≤} with the property S ⊂ D ⊆ M . Further, let : M → S, 1 : M → D, 2 : D → S (resp. : M → S, 1 : M → D, : D → S) be the associated monotone downwardly (resp. upwardly) directed 2 roundings. Then a= 2( a= resp. 1 a) a∈M 2( 1 a) . a∈M Proof. S ⊂ D ⇒ L(a)∩S ⊆ L(a)∩D ⇒ a = i(L(a)∩S) ≤ If we apply the mapping 2 to this inequality, we obtain 2( a) = (R1) a ≤ (R2) 2( 1 a).

B) We still have to show that the mapping : M → S with the properties (R1), (R2) and (R3) also fulfills (R). By (R3), a ∈ L(a) ∩ S. Let b be any element of L(a) ∩ S. , a is the greatest element of L(a) ∩ S. 24 {S, ≤} is a lower screen, it is a complete sup-subnet of {M, ≤}. Then there exists the element sup(L(a) ∩ S) ∈ S ⊆ M . 3 Screens and Roundings 31 element of a set is always its supremum we also have (R) a∈M a := sup(L(a) ∩ S) a := inf(U (a) ∩ S) . resp. 24. 25. Since the infimum and the supremum of a subset of a complete lattice are unique, there exists only one monotone downwardly (resp.