X− ∪
β∈AαIβ
∪ Iα
i.e., Qx = Qα. We also have Qxclosed in X . By the following Proposition2.2.1, the graph G defining the preference order p is closed in X× X. The proof would then be completed.
(AII) Proof of Theorem1(Uniform approximation theorem)
Step 1 Let Wc ≡ X − W. Evidently, Wcis also a compact set. Let L1≡ Wc∩ Qx(po), L2≡ Wc∩ Rx(po).
Then L1∪ L2= Wc. Clearly, L1and L2are disjoint and complement to each other.
But they are open in Wcand hence closed in Wc. As X is compact, L1and L2are also compact. Also, L1 ≺ x ≺ L2in po. By the argument used in Step 1 of the proof of Proposition2.2.1,∃ x1∈ L1and x2∈ L2with L1 x1and x2 L2 in po. Let V be a neighborhood of x such that V is compact and V ⊂ Rx1 ∩ Qx2. Thus we have
L1 x1≺ V ≺ x2 L2 in po.
Now po ∈ (L1, V ) ∩ (V , L2). By the definition of pn → poin, there exists N such that pn ∈ (L1, V ) ∩ (V , L2), ∀ n > N. This means that ∀ n > N and
∀ v ∈ V,
L1≺ v ≺ L2 in pn.
In particular, givenv ∈ V, any point of X indifferent to v in pnwith n> N and v ∈ V is disjoint from Wc≡ L1∪ L2. It says that
Iv(pn) ⊂ W, ∀ n > N and ∀ v ∈ V.
Step 2 When constructing the proof of the second statement, we keep in mind the fact mentioned in Example24where K is not connected. Given a connected set K with x ∈ K ⊂ I nt Ix(po). Since X is a locally connected, locally compact, T3-space,
∃ V open in X with its compact closure V connected such that
x∈ K ⊂ V ⊂ V ⊂ I nt Ix(po).
Evidently,∀ v ∈ V,
v ∈ V ⊂ I nt Iv(po).
We have po∈ (V , X). By Definition 2.5.2, ∃ N1such that pn∈ (V , X), ∀ n > N1; i.e., V ⊂ o(pn), ∀ n > N1. For any fixed n> N1and each givenv ∈ V we claimed that
V ⊂ I nt Iv(pn).
Let Ko ≡ V ∩ Iv(pn). The indifference set Iv(pn) is closed in X. Clearly, Ko= φ sincev ∈ Ko. Also, Kois closed in V , We then claim that Kois also open in V. It suffices to show that Ko⊂ I nt Iv(pn). For any y ∈ Ko, we have y∈ V and y ∼ v in pndue to the definition of Ko. Since V ⊂ o(pn), it holds that y ∈ o(pn). By the definition of the interior singular seto(pn), ∃ U open in X with
y∈ U ⊂ Iy(pn), i.e., y ∈ I nt Iy(pn).
However, Iy(pn) = Iv(pn). Therefore, y ∈ I nt Iv(pn) and hence Ko⊂ I nt Iv(pn).
In other words, Ko= V ∩ I nt Iv(pn). It follows that Kois open in V . By the connec-tedness of V , V contains no proper subset which is both open and closed. We have V = Ko, and consequently,
K ⊂ V ⊂ I nt Iv(pn), which completes the proof.
(AIIa)Proof of Corollary3.1.1
The necessity has been proved in Sect.2.5and in Theorem1. We show the suffi-ciency. Given pn, po∈ P(X), satisfying (A1) and (A2), let po∈ (K, L)∩(J, W), we claim∃No pn∈ (K, L) ∩ (J, W), ∀n > No. The proof is straightforward.
Step 1 For x ∈ K , y ∈ L, we have x ≺ y in po. Note that K and L are compact.
By transitivity, Ix ≺ Iyin po. Then (A1) implies Ix ≺ Iyin pn,∀n > some no. Hence x≺ y in pn,∀n > no. By Proposition2.5.1and applying the argument of finite open covering on K and L, we have pn∈ (K, L), ∀n > no.
Step 2 As po ∈ (J, W), by definition of , J is compact and non-empty, W is open in X and J ⊂ o(po) ⊂ (po) ⊂ W. Let (po) = ∞∪Ixi(po) and {xi1, . . . , xik, . . .} be a subsequence of {xi} such that Jik ≡ Ixik(po) ∩ J = φ. Since Jik ⊂ o(po), Jik ⊂ IntIxik(po) and Jikis open in J . Hence{Jik} is an open covering of J . By J compact, there are only a finite number l of Jik. Denote Ck ≡ Jik, k= 1, . . . , l.
For each k, consider
C ⊂ IntI (p ) ⊂ I (p ) ⊂ W.
By(A2) (the precise definition is given in Theorem1),∃Nksuch that the last formula is valid if pois replaced by pn,∀n > Nk. Let N≡ max{N1, . . . , Nl}. As J = ∪l
k=1Ck, we obtain pn∈ (J, W), ∀n > N.
Step 3 Choose No= max{no, N}. Then the claim is proved.
(AIIIa) Proof of LemmaA
(i) The proof ofo(p) ⊂ I nt (p) is trivial. We claim the reverse. Let y ∈ I nt (p)in X.Then there exists a neighborhood B of y in X, such that B ⊂
(p). We may let B be an open ball, so that B is connected, as X = In. Since the topology of X has a base of countably many open sets,o(p)has at most countably many components, Thus(p)can be expressed by a countable union of Ixi(p) with each xian interior singular point, i.e.,(p) = ∪∞i=1Ixi(p), But each Ixi(p) is closed in X. It is observed that any two distinct Ixi(p) and Ixj(p) are disjoint (Suppose otherwise, ∃z ∈ Ixi(p) ∩ Ixj(p), then ∀v ∈ Ixi(p) andw ∈ Ix j(p), v ∼ z ∼ w, which implies Ixi(p) = Ixj(p), contradiction).
Denote Ii ≡ Ixi(p) ∩ B, then each Ii is a closed subset of B such that they are disjoint from each other. Now B ⊂ (p). It means that B is a disjoint union of at most countably many closed sets Iis, which is possible only when there is exactly one non-empty Ii, i.e., B = Ii (Even in the weird example16, looking at y ≡ 13, B ≡ 1
6,12
, there are countably many indifference sets Ixi(p) in (p) which accumulate around y, but B is still a countable union of such corresponding Iis.) Remark that in the last statement we have used the fact that B is not a union of countable disjoint and non-empty closed sets, which is equivalent to the theorem that Cantor’s set in the unit interval has uncountable points in its complement. Therefore y∈ B = Ii ⊂ Ixi(p) and y is an interior point of Ixi(p). i.e., y ∈ o(p).
(ii) That∂((p)) ⊃ ∪∞i=1∂(Ixi(p)) follows directly from (i).
(iii) When(p) is assumed closed in X, we have ∂((p)) ⊂ (p) = ∪∞i=1Ixi(p).
Given y∈ ∂((p)), we have y ∈ Ixi(p) for some i. However, y /∈ Int(Ixi(p)), otherwise it violates y∈ ∂((p). Then it is necessary that y ∈ ∂(Ixi(p)). Thus
∂((p) ⊂ ∪∞i=1∂(Ixi(p)).
By (ii), we conclude that∂((p) = ∪∞i=1∂(Ixi(p)).
(AIIIb) Proof of Theorem2(The lifting theorem )
Step 1 Given p∈ P, we first claim that the lifting function f (x) is continuous in x. Letε > 0, choose a µ-measurable compact set L ⊂ Qx ≡ Qx(p)such that
µ
Qx− L
< ε 4 and choose an open set W ⊃ Qx ≡ Qx(p) such that
µ (W − Qx) < ε 4
Let the notation “+” applied between two sets denotes the “disjoint union” of them, where the two sets have been assumed disjoint. Correspondingly, the summation nota-tion Proposition2.2.2, we may choose V , open in X, with x ∈ V ⊂ the compact closure
¯V ⊂ W − L, such that L ≺ ¯V ≺ X − W. Then ∀y ∈ ¯V , we have which proves (3). Hence f(x) is continuous in x.
Step 2 ConsiderF(K, U) where K is a compact set of X and U is open in the real line R1. In order to show the continuity ofξ, it suffices to show that ξ−1(F(K, U)) is open in P, . Let p∈ ξ−1(F(K, U)), i.e., the function f ≡ ξ(p) maps K into U .
Given x ∈ K with f (x) ∈ U . Chooseε > 0, such that the interval ( f (x) − ε, f(x) + ε) ∈ U . We first claim that there exist a compact neighborhood V of x in X and a neighborhood S of p in P such that
ξ(q) maps V into U ,∀q ∈ S (4)
Let y ∈ L1, we have y ≺ x in p. By Proposition2.2.2,∃Uy Vx, compact neighbo-rhoods of y, x, respectively such that
Uy ≺ Vxin p.
Hence p∈ Uy, Vx
. Cover L1by the union of all such Uy, with y∈ L1, Since L1
is compact,∃ finite subcover
Uy1, . . . , Uyk
of L1. Consider the corresponding
V1, . . . , Vk
with x ∈ Vi and Uyi ≺ Vi, ∀i.
Denote
S1≡ ∩hi=1
Uyi, Vi
, V1≡ ∩hi=1Vi
Then p∈ S1and the set S1is open in P. Thus∀q ∈ S1andv ∈ V1, we have
L1⊂ Qv(q) (6)
(In this sense we may say that Qx(p) is “lower semi-continuous in p”, as L1 is arbitrarily close to Qx(p) from the interior. See also (IV) of Introduction.)
(ii) Similarly, choose L2compact in Rx(p), such that µ
Rx(p) − L2
< ε
4 (7)
Let W ≡ X − L2. We now claim that there exist a compact neighborhood V2of x in X and a neighborhood S2of p in P such that V2⊂ W and
Qv(q) ⊂ W, ∀q ∈ S2andv ∈ V2
By the argument given in (i), we can have
L2⊂ Rv(q), ∀q ∈ S2andv ∈ V2
which implies that
Qv(q) ⊂ Qv(q) = X − Rv(q) ⊂ X − L2= W, (8)
∀q ∈ S2andv ∈ V2.
Step 4 Since each singular indifference set Ixi(p) is µ-measurable, (p) = ∪∞i=1 Ixi(p) is µ-measurable. There exist two unions E and F of finitely many closed n-rectangles such that
E⊂ (p) ⊂ I nt F ⊂ F and µ (F − E) < ε
8 (9)
(Note that this is not true in Example17a, where(p) is not µ-measurable). But now I nt E ⊂ I nt (p) = o(p) where the last equality is established by LemmaA.
Step 5 Choose a compact set L3⊂ I nt E such that µ (E − L3) = µ (I nt E − L3) < ε
8 (10)
noting thatµ (E − Int E) = 0. We see that S3≡ (L3, I nt F) is an open neighborhood of p, where∀q ∈ S3.
L3⊂ o(q) ⊂ (q) ⊂ I nt F ≡ Fo (11) By (9) and (10), we have
µ (Fo− L3) < ε 8 +ε
8 = ε
4 (12)
Step 6 By (6), (8) and (11), we have
L1− Fo⊂ Qv(q) − (q) ⊂ W − L3
and hence the difference
µQv(q) − (q)
− µ
Qx(p) − (p)
≤ µ ((W − L3) − (L1− Fo)) ,
∀q ∈ S ≡ S1∩ S2∩ S3and∀v ∈ V ≡ V1∩ V2, since by (6) and (11)
L1− Fo⊂ Qx(p) − (p).
But
µ ((W − L3) − (L1− Fo)) ≤ µ ((W − L3) − (L1− L3)) + µ (Fo− L3)
≤ µ (W − L3− L1) +ε
4 (by(12))
≤ µ (W − (L3∩ Ix(p)) − L1) +ε 4
≤ µ ((X − L2) − (L3∩ Ix(p)) − L1) +ε 4
≤ µ
Rx(p) − L2
∪(Ix(p)−L3)∪
Qx(p)−L1
+ε 4
≤ε +ε
+ε +ε
= ε (by(7), (12), (5))
It shows that
|ξ(q)(v) − ξ(p)(x)| < ε.
In other words,∀q ∈ S,
ξ(q) maps V into U . Thus the claim (4) of step 2 is proved.
Step 7 By Step 6, there exists for each x ∈ K a neighborhood Vx of x in X and a neighborhood Sx of p in P such that
ξ(q) maps Vx into U , ∀q ∈ Sx . By K compact, there exists a finite cover
Vx1, Vx2, . . . , Vxh
of K . Choose
So≡ Sx1∩ Sx2 ∩ · · · ∩ Sxh. We have∀q ∈ So,
ξ(q) maps K into U ,
i.e., ξ(q) ∈ F(K, U ). In other words, the open neighborhood So of p lies in ξ−1(F(K, U)), for any given p ∈ ξ−1(F(K, U)). It shows that ξ−1(F(K, U)) is open in P. The proof of the continuity ofξ is completed.
(AIVa) Let X = [−1, 1] × [−1, 1] ⊂ R2, and an ∈ (1/n, 0) ∈ X. Define preference vector fields pn = (0, 0) at anand(0, 1) elsewhere. Define p = (0, 0) at an, for any n, and (0,1) elsewhere. Then singular set Snof pnis anand that of p is
∪Sn= {a1, a2, . . . , an, . . .}. Note that | pn(x)− p(x) |= 0, ∀x ∈ X −∪Sn. According to the formula (∗∗) in Remark Aof Sect.1.1, which was defined by Chichilnisky (Chichilnisky 1982, pp349), it is clear that { p1, p2, p3, . . . , pn, . . .} converges to p.
But { p2, p4, p6, . . . , p2k, . . .} does not converge to p . In fact, X− ∪∞k=1S2k= {a1,a3,a5. . .}
and| p2k(a1) − p(a1) |=| (0, 1) − (0, 0) |= 1, so the sup norm is 1. Furthermore, a similar argument shows that even sequence { p2, p3, p4, p5, . . . , pn, . . .} does not converge to p. This is weird. It says that the formula (∗∗) does not define a topology, nor the notion of continuity.
(AIVb) We will show that dc∗(p, q) given by a symmetric form, supx∈X−Sp∪Sq | p(x) − q(x) |, is also not a distance:
(1) It is not positive definite, since we may construct an example with dc∗(p, q) = 0 does not imply p= q. In fact, let A B X = I and p = (0, 0) on A, yet
= (0, 1) elsewhere. Let q = (0, 0) on B, yet = (0, 1) elsewhere. It yields that singular sets Sp= A B = Sq. Clearly, p = (0, 1) = q on X − Sp∪ Sq. So dc∗(p, q) = 0, but p = q.
(2) The triangular inequality is not satisfied: Let X = {(x, y); 0 ≤ x ≤ 1, 0 ≤ y≤ 1}. Let e1= (1, 0) and e2 = (0, 1) be the canonical orthonormal frame of R2.Consider three preference vector fields p, q, and r as follows. Let
p= sin πy e1+ cos πy e2
q = sin πy e1− cos πy e2
for 0 ≤ y ≤ 1/2, i.e., as y changes from 0 to 1/2, p turns clockwise from e2 to e1 for 90 degrees, and yet q turns counter clockwise. Furthermore, let r = (0, 0) for 0 ≤ y ≤ 1/2, and p = q = r = e1 for 1/2 ≤ y ≤ 1. Then dc∗(p, q) = 2 > dc∗(p, r) + dc∗(r, q) = 0, where dc∗(p, r) = dc∗(r, q) = 0, since Sr = {0 ≤ y ≤ 1/2}, Sp= Sq= empty set φ, and X − Sp∪ Sr = X − Sr∪ Sq= {1/2 ≤ y ≤ 1} on which p = q = r.
(AIVc) If the preference space of form γ (i.e., it is the quotient topological space F/ ∼ of function space F on X ), or of form δ, is considered, Chichilnisky’s proof of her theorem is invalid: Looking at Sect.1.1, the range space S1∪ {0} of induced map F ": S1× S1→ S1∪ {0} is now a connected set. It is even not a T1-space, i.e., not every point is a close point. In fact, given an equivalent class [ f ] with f ∈ F,we have f ∼ t f for any t > 0, and therefore [ f ]∼ [t f ]. But t f converges to zero function 0, as t→ 0. By the quotient topology ∗con F/ ∼ , [t f ] converges to the null preference o = [0]. Therefore, ∃ no open set in the preference space, which contains o and is disjoint from [ f ]=[t f ]. Hence [ f ] is not a closed point, and S1∪ {0} in this topology
∗c becomes a connected set. This makes the proof of impossibility theorem invalid.
The argument about the preference space of formδ is the same. It is noteworthy to remark that the topology∗cof formγ is equivalent to odefined in Definition 2.5.1.
The equivalence is not difficult to prove (seeHuang 1996).
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