3. On a relation of the Chipman approach with the continuous multi-utility
representation problem of preorders
Let be an arbitrarily chosen topological space. In the focus of this section is the problem of determining (characterizing) all preorders ≾ on X the order topology of which is coarser than t that admit a continuous multi-utility representation. It is well known (cf, for instance, Bosi and Herden [3]) that the assumption ≾ to admit a continuous multi-utility representation implies that ≾ must be closed and, therefore, also semi-closed. Hence, the following lemmas provide a first important step towards a complete solution of the just mentioned characterization problem. As in the proof of Theorem 2 throughout this section we may assume without loss of generality that .
In order to proceed let us denote for every point by the set of all points such that . This notation allows us to verify the validity of the following lemma.
Lemma 1. Let be arbitrarily chosen. Then ≾ satisfies the following conditions:
HD: Let ≾ have a continuous multi-utility representation. Then is a Hausdorff space.
OC: Let be a Hausdorff space and let ≾ be closed. Then is open (and closed) for every point that is maximal with respect to .
Proof. HD: Let ≾ have a continuous multi-utility representation and let and be arbitrarily chosen points such that . Then there exists a continuous and increasing function such that . Hence, the desired conclusion follows.
OC: Let be a maximal element of , which means that . Then the assumption to be Hausdorff implies with help of condition SB that . Hence, we may distinguish between the cases to have a maximal element and to have no maximal element. Let us, therefore, assume at first that to have a maximal element m. Then the interval is empty. This means, in particular, that there exists no net of points that converges to y. Hence, the set must be closed and we may conclude that is open and closed. We, thus, proceed by showing that both sets as well as are open and closed. In order to verify these properties of and respectively it suffices to prove that is closed and that is open. Let, therefore, in a first step some point be arbitrarily chosen. Then we have to show that . We, thus, consider some net of points that converges to p. Since ≾ is closed and for all it follows that and it remains to verify that the equivalence can be excluded. Indeed, if then the just proved property of to be open (and closed) implies that there exists some index such that for all that are at least as great as . This contradiction implies that must be closed. For later use, in particular in the proof of Theorem 3, we abbreviate this conclusion by (*). Since is open and is open it follows in a second step that is open, which completes the discussion of the case to have a maximal element. We now still must think of the situation to coincide with . Let, in this situation, be some sub-chain of such that Because of (*) we may assume without loss of generality to be not open (and closed). We, thus, may arbitrarily choose some point in order then consider some net of points that converges to y. Because of the maximality of y with respect to it follows that for every there exist points and such that . Indeed, otherwise the definition of implies that is the meet of two open intervals and, thus, open (and closed), which contradicts our assumption to not being open (and closed). This argument will be abbreviated it by (M). But this consideration allows us to conclude that for every point the set is an open neighborhood of y. Hence, it follows for every point that is open (and closed), which still was to be shown. □
As it already has been announced in the introduction we now characterize those semiclosed preorders ≾ on for which is Hausdorff.
Lemma 2.
Let ≾ be semi-closed. Then in order for to be Hausdorff it is necessary and sufficient that ≾ satisfies the conditions SB and LR.
Proof. As it already has been mentioned in the introduction the validity of the conditions SB and LR are necessary in order to guarantee to be Hausdorff. Hence, we may concentrate on the sufficiency part of the lemma. In order to verify that the assumption ≾ to be semi-closed implies in combination with validity of the conditions SB and LR that is a Hausdorff topology on we notice at first that condition SB is equivalent to condition LU which states that for every point at least one of the sets or is not empty. Let now points and such that be arbitrarily chosen. Then the cases and are possible. Therefore, we have to distinguish between these possible cases.
Case 1: . In this situation we distinguish between two more cases.
Case 1.1: There exist points and such that the interval is empty and . In this case is an open set that contains y and is an open set that contains x. Therefore, the equation settles 1.1.
Case 1.2: The closed interval does not contain any jump. In this situation there exists some point such that . Hence and respectively are disjoint open sets that contain y and x respectively.
Case 2: . In this situation condition LU implies that the lemma will be shown if the cases or successfully have been handled. Since both cases can be settled by completely analogous arguments it suffices to concentrate on the case to be not empty. The inequality implies with help of condition LR that there exists some such that in case that or that there exist points and such that in case that . Since ≾ is semi-closed it, thus, follows that and respectively or and respectively are disjoint open sets that contain the point x and the point y respectively, which still was to be shown. □
It is well known that a closed preorder ≾ on X is semi-closed. On the other side, however, a semi-closed preorder, in general, is not closed. Indeed, in Bosi and Herden [2, Theorem 2.2] very restrictive necessary and sufficient conditions for a semi-closed preorder to be closed have been presented. Because of this theorem it is somewhat surprising that the following proposition holds, which surely is worth to be stated separately.
Proposition 1. Let be Hausdorff and let ≾ be semi-closed. Then ≾ is closed.
Proof. In order to verify the proposition we must show that for any two points and such that there exist (open) neighborhoods U of x and V of y such that for every point and every point the relation holds. Indeed, having proved the existence of U and V it follows that and we are done. An analysis of the proof of Lemma 2 allows us to concentrate on the case that also the relation holds and that neither nor is empty. Let us, therefore, assume in contrast that every open neighborhood of x and every open neighborhood of y that is disjoint from contains points and respectively such that . In order to proceed we set and . Since is Hausdorff we may conclude that and . Then we distinguish between the cases as well as to be open and closed, to be open and closed and to only be closed, to only be closed and to be open and closed and as well as to only be closed. The case as well as to be open and closed is trivial. Indeed, in this case we may set and . The remaining three cases can be done by analogous arguments. Hence, we may concentrate without loss of generality on the case as well as to only be closed. Let now in every interval and every interval points and such that be arbitrarily chosen. Then we may assume without loss of generality that for all intervals such that or that for all intervals such that and that for all intervals such that or that for all intervals such that . The symmetry of the cases to be considered now allows us to concentrate on the case for all intervals such that and for all intervals such that and on the case for all intervals such that and for all intervals such that . Since ≾ is assumed to be semi-closed it follows in the first case that which means that and, thus, contradicts our assumption x to be not smaller or equivalent to y. The assumptions of the second case imply that But ≾ is semi-closed. Hence, we may conclude that the smallest closed increasing set that contains is . It, thus, follows that which again implies that and, therefore, contradicts the relation . This conclusion, finally, proves the validity of the proposition. □
In combination with Lemma 1, Lemma 2 and Proposition 1, the following theorem now presents a complete solution of the characterization problem that is in focus of this section and, in this way, (in opinion of the authors) also allows an interesting comparison of the Chipman approach on the one side and the real-valued approach on the other side to mathematical utility theory (cf. Theorem 4).
Theorem 3. Let be a preordered topological space the topology t of which is finer than the order topology . Then the following assertions are equivalent:
(i)≾ admits a continuous multi-utility representation.
(ii) is Hausdorff and ≾ is closed.
(iii) is Hausdorff and ≾ is semi-closed.
Proof. (i) ⇒ (ii): It already has been mentioned above that a preorder ≾ that admits a continuous multi-utility representation must be closed. Therefore, Lemma 1 guarantees the validity of the implication (i) ⇒ (ii)".
(ii) ⇒ (iii): Since a closed preorder ≾ is semi-closed nothing has to be proved.
(iii) ⇒ (ii): Proposition 1.
(ii) ⇒ (i): Let assertion (ii) be valid and let points and such that be arbitrarily chosen. Then we must prove that there exists some continuous and increasing function such that . In order to verify the existence of we distinguish between the cases y to be contained in and y to be contained in .
Case 1:. In this case we must distinguish between the situations y being a maximal element of and y to not being a maximal element of. In the first situation we may apply property OC in order to setting

for all . In the second situation, i.e. y is not maximal element of there exists some point such that . Of course, u may be a maximal element of . In this case, however, we may apply the argument that has been applied in the first situation. Hence, we may assume without loss of generality that there exist points and such that . We proceed by assuming at first that v and, thus, also u is contained in . We abbreviate this assumption by (**). In addition, we assume both (open) intervals and to be empty. These assumptions imply that there does not exist any open interval of that contains u and is completely contained in . Hence, there cannot exist any net of points that converges to u, which implies that the set of all points the indifference classes of which are different from must be closed. This means that we now may apply the conclusion that in the proof of Lemma 1 has been abbreviated by (*) in order to conclude that is open and closed. Since both points u and v are contained in these considerations, finally, allow us to define the desired continuous and increasing function by setting

for all . In addition, the above considerations imply that for the moment we may assume without loss of generality that there exist no points , and such that and both intervals and are empty. With help of this assumption we now verify that the preordered set is not scattered. Indeed, the assumption implies by complete induction, each induction step of which may be settled by some straightforward indirect argument that contains some order-dense subset, i.e. a subset that does not contain any jumps, or that the set that is ordered by setting or for all intervals and is order-dense. Let be the real unit interval. Then our considerations allow us to conclude that in any case there exists an order-embedding . We, thus, proceed by showing that for all rationals and such that the inclusion holds. Since ≾ is closed and, therefore, also semi-closed it follows that . The validity of the strong inequality , thus, implies the desired inclusions . These considerations imply that the assumptions of Peleg’s Theorem (cf. Peleg [10]) are satisfied or, equivalently, that the family is a (decreasing) separable system in the sense of Herden [8]. Peleg’s theorem or Theorem 4.1 in Herden [8], therefore, implies the existence of some continuous and increasing function such that and . Let us abbreviate these arguments by (***). In order to finish the first case we still must consider the situation v to be not contained in , i.e. the situation v to be contained in . Of course, it is possible that , which means that also . In this situation, however, is an open and closed subset of X. Hence, we may define the desired continuous and increasing function by setting

for all . These considerations now allow us to assume that but and . Since it already has been shown that we may assume without loss of generality that at least one of the (open) intervals or is not empty we first briefly discuss the case the case to be not empty. In this case, however, we may apply the arguments that have been summarized by (***) in order to guarantee the existence of some continuous and increasing function such that and . Hence, we now may assume that the (open) interval is not empty. In this situation reiteration of the just used argument in combination with an analysis of the arguments that have been summarized by (***) imply that there exists some net of points that converges to u. We now proceed by applying an indirect argument. This means precisely that we assume for all indexes each point to be contained in . Since ≾ is a closed preorder this assumption allows us to conclude, however, that , which contradicts our assumption u to be an element of . This contradiction guarantees the existence of some point such that . Since now the same situation is given as has been described in (**). This reduction to assumption (**), finally, settles the first case.
Case 2: In this case we distinguish between the sub-case to coincide with and the sub-case to be properly contained in . But since the sub-case to coincide with already has been discussed in the second part of the first case we only have to consider that strong inclusion . Here also the arguments having been used in discussing the first case apply. Indeed, let in this sub-case some point be arbitrarily chosen. Then or . If then the situation is given and we may apply the arguments having been used above when discussing the situation . If then the inclusion implies and we may apply the arguments that already have been applied when considering the last part of the first case in order to also handle this situation. This last conclusion completes the proof of the theorem. □
In the second section of this paper the universal character of the Chipman approach to mathematical utility has been demonstrated (cf. Theorem 1 and Theorem 2). Concentrating on continuous multi-utility representation in Theorem 3, however, it could be shown that the Chipman approach, at least formally, is not as far away from the real-valued approach as it seems at first sight. Therefore, we now still discuss the relations between the Chipman approach, the continuous multi-utility approach and the Richter-Peleg approach to mathematical utility theory in more detail (cf. sub-section 1.2 of the introduction). The relation between the Chipman approach and the continuous multi-utility approach can be described by combining Theorems 1 and Theorem 2 in order to state the following theorem.
Theorem 4. Let, ≾ be a semi-closed that is defined on some topological space , and let us assume, in addition that is Hausdorff and that the order topology is coarser than t. Then the following assertions hold:
(i) There exists some cardinal number κ and a preorder ≲ on that is coarser than for which there exists a continuous order-embedding ;
(ii) ≾ admits a continuous multi-utility representation.
Proof. Let now
be the cardinality of the set
of all pairs
such that
. Then we consider the direct sum
of ordered topological spaces
as well as the direct sum
of preordered topological spaces
. As it already has been shown in sub-section 1.2 of the introduction it follows that there exists a continuous order-preserving function
. This consideration already clarifies the relation of the continuous multi-utility approach for representing a preorder with the Richter-Peleg approach for representing a preorder (cf. Evren and Ok [7]). Let, finally,
and
. Then Theorem 2 implies the existence of a continuous order-embedding
and of a continuous order-embedding
such that the following diagram commutes

□
This last theorem completely clarifies the relations between the Chipman-approach, the continuous multi-utility approach and the Richter-Peleg approach to mathematical utility theory (cf. sub-section 1.2 of the introduction).