Continuous Utility Functions Through Scales

Abstract

total preorders, continuous utility representations, separating scales, decreasing scales, C 60, D 11, 54 F 05, 06 A 06,

Unified Representability of Total Preorders and Interval Orders through a Single Function: The Lattice Approach

“…In the case of [0, 1] this is nothing but the separability in the usual sense. We shall see what it means in the second case in Example 3 (2). This is precisely the key fact to be put in relation with the results in [25], where this kind of maps were considered and the notion of scale was extended to this more general case.…”

“…With different nuances, the concept of a scale 1 (or similar notions, as R-separable systems) has already been used to get continuous or semicontinuous representations of total preorders (see e.g. [2,8,9,11,14,29,30]) and interval orders (see e.g. [6,12]).…”

“…Usually, one scale of a certain special type is necessary to characterize the continuous representability of total preorders (see e.g. [2]), whereas two intertwined scales satisfying a list of restrictions are needed to characterize the continuous representability of interval orders, as in [6]. (By the way, even for the case of total preorders, other constructions of continuous representations based on particular variations of the notion of a scale need more than one scale, indeed a countably infinite amount of scales, to be achieved, as discussed in [2]).…”

Characterization of the existence of a weak qualitative probability

“…Alcantud and Rodríguez-Palmero [1] and Alcantud et al [2]). Further, when we consider a preorder on a group, it is possible to characterize the existence of an additive utility function by imposing suitable conditions on a lower scale (see e.g.…”

“…By condition (2) in the definition of a countable lower scale (see Definition 2.6), it is clear that p(∅) = 0 and p(Ω) = 1. Further, we have that 0 ≤ p(A) ≤ 1 directly from the definition of p. In order to prove that p is additive, assume that there exist A, B ∈ A Ω such that A ∩ B = ∅ and p (A ∪ B) = p(A) + p(B).…”