. In other words, it is possible to move from (ψ0, φ0) to (ψ1, φ1) along a path which, at every instant, is regarded as an ‘infinitesimal improvement’. We refer to γ as an improvement path.
The relation (
❀) may violate (IP1) unless further conditions are imposed on the system of order relations X ={
(ψ,φ)}(ψ,φ)∈Ψ×Φ. The systemX issmoothif there exists
an open cover{Oj}j∈J of Ψ (for some indexing setJ), and for eachj ∈ J, a smooth function uj :Oj ×Φ−→Rsuch that:
(Sm1) For each j ∈ J, each ψ1, ψ2 ∈ Oj, we have uj({ψ1} ×Φ) =uj({ψ2} ×Φ).24 (Sm2) For each j ∈ J, each (ψ, φ) ∈ Oj ×Φ, and each ~v ∈
T
(ψ,φ)(Ψ ×Φ), if~v
(ψ,φ)
~0(ψ,φ), then ∇ui(ψ, φ)[~v]≥0.25
(Sm3) For any j, k ∈ J, if Oj ∩ Ok 6= ∅, then uj and uk are ‘ordinally equivalent’
on their domain overlap: for all ψ, ψ′ ∈ Oj ∩ Ok and all φ, φ′ ∈ Φ, we have ui(ψ, φ)≤ui(ψ′, φ′) if and only if uj(ψ, φ)≤uj(ψ′, φ′).
Theorem 8.1 IfΨis simply connected andX is smooth, then(
❀)is an interpersonal preorder.
Appendix: Proofs
Proof of Theorem 3.1. Let f : R−→(0,1) be a strictly increasing bijection (for example: f(x) = (tanh(x)+1)/2), and letv :=f◦u. Thenvalso satisfies statements (1) and (2).
For all x ∈ X, letYx :={y∈ X ; yx}, and define wx : X −→R by wx(y) := 1 for all y∈ Yx, whilewx(y) := 0 for all other y∈ X. Then define ux :=v+wx. Claim 1: ux is a utility function for().
24These two sets are intervals inR, because Φ is connected andujis a continuous function. Thus, it is equivalent to simply require uj({ψ1} ×Φ) anduj({ψ2} ×Φ) to have the same maximum (or supremum) and minimum (or infimum).
25∇ui(ψ, φ) is the gradient of ui at (ψ, φ) (a linear functional on
T
(ψ,φ)(Ψ×Φ)). The quantity∇u(ψ, φ)[~v] is the ‘instantaneous rate of change’ ofu when moving from (ψ, φ) in the direction~v.
Proof. We must check statements (1) and (2). Let y, z ∈ X. Suppose yz. Then wx(y)≥wx(z) (because ifz ∈ Yx, then also y∈ Yx, by transitivity).
Ifyz, then v(y)≥v(z) because v satisfies (1). Thus, ux(y)≥ux(z).
Ify≻z, thenv(y)> v(z) becausevsatisfies (2). Thus,ux(y)> ux(z). ✸Claim 1 Claim 2: Letx, y, z ∈ X. If ux(y)≥ux(z), then wx(y)≥wx(z).
Proof. (by contrapositive) Suppose wx(y)< wx(z). Then we must have wx(y) = 0 and wx(z) = 1. But v(y) < 1 and v(z) > 0, because v ranges over (0,1) by construction. Thus, ux(y) = wx(y) +v(y) = 0 +v(y) < 1 = wx(z) < ux(z), so
ux(y)< ux(z). ✸ Claim 2
Claim 3: For anyy, z ∈ X,
yz
⇐⇒
ux(y)≥ux(z) for all x∈ X . Proof. “=⇒” follows from Claim 1. To see “⇐=”, suppose ux(y) ≥ ux(z) for all
x ∈ X. Then Claim 2 implies that wx(y) ≥ wx(z) for all x ∈ X. In particular, wz(y) ≥ wz(z) = 1, so wz(y) = 1, which means y ∈ Yz, which means y z.26
✸ Claim 3
Thus, the setU :={ux ; x∈ X }provides a multiutility representation for (). ✷
Proof of Theorem 6.2. Clearly, (
& ) is reflexive. We must show that (
& ) is transitive and satisfies properties (IP1) and (IP2).
Transitive. Suppose (ψ1, φ1)
&(ψ2, φ2) and (ψ2, φ2)
&(ψ3, φ3). We must show that
(ψ1, φ1)
&(ψ3, φ3).
We have (ψ2, φ2)
ψ2,ψ3(ψ3, φ3), and (ψ2, φ2)
ψ3,ψ2(ψ3, φ3), while (ψ1, φ1)
ψ1,ψ2(ψ2, φ2), so consistency requires that (ψ1, φ1)
ψ1,ψ3(ψ3, φ3).
Likewise, (ψ2, φ2)
ψ1,ψ2(ψ1, φ1), and (ψ2, φ2)
ψ2,ψ1(ψ1, φ1), while (ψ3, φ3)
ψ3,ψ2(ψ2, φ2), so consistency requires that (ψ3, φ3)
ψ3,ψ1(ψ1, φ1).
Thus, (ψ1, φ1)
ψ1,ψ3(ψ3, φ3) and (ψ1, φ1)
ψ3,ψ1(ψ3, φ3), so (ψ1, φ1)
&(ψ3, φ3), as de-sired.
(IP1) Fix ψ ∈ Ψ and φ, φ′ ∈ Φ, with φ1
ψ φ′1. By hypothesis, (
ψ,ψ) is an interpersonal preorder on{ψ} ×Φ, so it agrees with (
ψ ). Thus, settingψ1 =ψ2 =ψ in definition
(4), we get
(ψ, φ)
& (ψ, φ′)
⇐⇒
(ψ, φ)
ψ,ψ(ψ, φ′)
(5)
26The argument of the last sentence is from Proposition 1 of Evren and Ok (2009).
By similar logic, we have con-trapositive of statement (5) yields
(ψ, φ) 6
ψ1,ψ2) is an interpersonal preorder, so it satisfies (IP2), so there is some φ′2 ∈ Φ such that (ψ1, φ1)
Through an identical construction, we can obtain some φ2 ∈ Φ such that (ψ1, φ1)
& (ψ2, φ2). This works for all ψ1, ψ2 ∈ Ψ and φ1 ∈ Φ; thus, (
& )
satis-fies (IP2). ✷
To prove Theorem 7.3 we need some technical preliminaries. A preference chain is a sequence (ψ0, φ0)
Proof. It suffices to prove this when ψ
eǫ ψ′ (the general case follows by induction).
There are two cases: either the elementary homotopyremovesa link from the chain, or it addsa link.
Case 2. (Link addition) Suppose ψ:= (ψ0, ψ1, . . . , ψn−1, ψn+1,· · ·, ψN) carries the
Proof. By hypothesis, the preference chain (6) contains at least one strict preference;
say (ψm, φm)≺
ψm(ψm+1, φm+1) for some m ∈ [1. . . N−1]. It suffices to show that we can ‘shift’ this strict preference backwards or forwards in the chain.
Backwards shift. Semicontinuity of (
ψm) yields some φ′m ∈ Φ such that
Forwards shift. Semicontinuity of (
ψm) yields some φ′m+1 ∈ Φ such that
(IP1) Case 1. First suppose that each local relation ( and by Lemma A.1, this homotopy transforms the preference chainξ into a prefer-ence chain (ψ1, φ′1) at least one of these preferences is strict. Thus, transitivity of (
ψ1) forces (ψ1, φ′1)≺
ψ }ψ∈Ψ be arbitrary. By hypothesis, we can extend each local interpersonal preorder (
ψ ) to some semicontinuous interpersonal pre-order (c
ψ ) is an interpersonal preorder on Nψ ×Φ. Next, there exists φ2 ∈ Φ with (ψ1, φ1)
ψ1(ψ2, φ2), because (
ψ1) is an interpersonal preorder on Nψ
1 ×Φ. Proceeding inductively, we obtain a preference chain (ψ, φ)
RO) is an interpersonal preorder. It remains only to show that (
RO) is complete if each (
ψ ) is continuous.
Let (ψ, φ),(ψ′, φ′)∈Ψ×Φ; we must show these two points are comparable. Since N chain-connects Ψ, there is an N-chain ψ =ψ0, ψ1, ψ2, . . . , ψN =ψ′ connecting ψ
toψ′. Now, for all n∈[0...N) the interpersonal preorder ( Oj contains an open ball aroundψ, and if this open ball is small enough, it is simply connected (because Ψ is a manifold). Thus, letNψ ⊂ Oj be some simply connected open neighbourhood ofψ, and let uψ be the restriction of uj toNψ×Φ. This yields a simply connected neighbourhood system N={Nψ}ψ∈Ψ, as in Example 7.2(a).
For everyψ ∈Ψ, define a ‘local’ interpersonal preorder (
ψ ) onNψ×Φ as follows:
❀) is obtained by taking the join of all the local interpersonal preorders { ψ }ψ∈Ψ, exactly as in the definition of (
RO) in §7. Thus, it suffices to show that the system{
ψ }ψ∈Ψ is consistent, and then invoke Theorem 7.3(a).
Letψ ∈Ψ. Define (c
ψ }ψ∈Ψ satisfies property(RO) from§7.
Proof. Let ψ1, ψ2 ∈Ψ. Suppose Nψ1∩ Nψ2 6=∅, and the relations (c
ψ1) and (c
ψ2) are defined by (7). Supposeuψ1 is the restriction ofuj toNψ1 anduψ2 is the restriction ofuk toNψ2, for somej, k ∈ J. Thus, Oj∩ Ok 6=∅ (since it containsNψ1∩ Nψ2), and then property (Sm3) ensures that (c
ψ1) and (c
ψ2) agree onNψ1∩Nψ2. ✸Claim 2 Claim 3: For anyψ ∈Ψ, the preorder (c
ψ ) extends (
ψ ).
Proof. Let (ν0, φ0),(ν1, φ1) ∈ Nψ ×Φ, with (ν0, φ0)
ψ(ν1, φ1); we must show that (ν0, φ0)b
ψ(ν1, φ1). But if (ν0, φ0)
ψ(ν1, φ1), then there is some improvement path γ : [0,1]−→Nψ×Φ with γ(0) = (ν0, φ0) and γ(1) = (ν1, φ1). Thus,
uψ(ν1, φ1) = uψ◦γ(1)
(∗) uψ◦γ(0)+
Z 1 0
(uψ◦γ)′(t)dt ≥
(†)
uψ◦γ(0) = uψ(ν0, φ0), so (ν0, φ0)b
ψ(ν1, φ1), as desired.
Here, (∗) is the Fundamental Theorem of Calculus. Inequality (†) is because (uψ ◦γ)′(t) (c) ∇uψ(γ(t))[γ′(t)] ≥
(⋄)
0 for all t∈ [0,1]. Here, (c) is by the Chain Rule, and (⋄) is by (Sm2) and the fact that γ′(t)
γ(t)
~0γ(t) for allt ∈[0,1] (because
γ is an improvement path). ✸ Claim 3
Thus, the system {
ψ }ψ∈Ψ is consistent, so Theorem 7.3(a) implies that (
❀) is an
interpersonal preorder. ✷
Remark. In the proof of Theorem 8.1, the inequality uψ(ν0, φ0)≤ uψ(ν1, φ1) is necessary, but not sufficient to conclude that (ν0, φ0)
❀(ν1, φ1). Thus, assuming the existence of a function uψ :Nψ×Φ−→R is not tantamount to assuming some ‘local’
form of ‘ordinal, fully comparable’ utility functions —it is a much weaker assumption.
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