For unknowns uzi ∈ R, and dz ∈ Rd, one for each z ∈ K, consider the following set of equalities and inequalities:
uzi − uyi > 0, for all z, y ∈ K with z i y,
uzi − uyi = 0, for all z, y ∈ K with z ∼i y,
uzi − uyi − (dz)T(z − y) > 0, for all z ∈ K, all y ∈ K, y 6= z.
Since preferences over K can be represented by strictly concave ui, there exists a solution
to this system (by setting uz
i = ui(z), and dz equal to a supergradient of ui at z). If we set
dxˆ = 0, uxˆ
i = ui(ˆx) + η for small enough η > 0, and maintain the remaining values of the
original solution, we obtain a solution to the modified system
uzi − uyi > 0, for all z, y ∈ K \ {ˆx} with z i y,
uxiˆ− uz
i > 0, for all z ∈ K \ {ˆx},
uzi − uyi = 0, for all z, y ∈ K \ {ˆx} with z ∼i y,
uzi − uyi − (dz)T(z − y) > 0, for all z ∈ K, all y ∈ K, y 6= z.
This solution to the latter system produces a strictly concave utility function ˜ui : Rd → R
(as in Richter and Matzkin, 1991, or Richter and Wong (2004)) defined as ˜ ui(x) = min z∈Ku z i + (d z)T(x − z) − ε(x − z)T(x − z) ,
Proof of Theorem
4
We have( bSq0)⇒[{(yj, zj, vji)}j∈M ∪ ˆM satisfies(S
0
q)]⇔( bS0), the latter equivalence by The-
orem 2. Furthermore, by Lemma 1, we also have ( bS0)⇒( bS
c). Since ( bSc)⇒( bSq)⇒( bSq0) and
( bSc)⇒( bSc0)⇒( bSq0), we conclude ( bS0)⇔( bSc)⇔( bSq)⇔( bSc0)⇔( bSq0)⇔( bS0). Since ( bS0)⇒( bS), it re-
mains to show ( bS)⇒( bS0), which follows from cases 1 and 2 in the proof of Theorem 5.
Cases 1 and 2 in the proof of Theorem 5 applied on subsets M0 ⊆ M ∪ ˆM with |M0| = 2 also prove that ( bS
1)⇒[{(yj, zj, vji)}j∈M ∪ ˆM satisfies (S1)]. Thus, if d = 1, (S1)⇔(S)
by Theorem 2, hence( bS1)⇒( bS), since we have shown (S)⇔( bS) when d = 1. Thus, we have
shown that if d = 1, ( bS1)⇔( bS).
Proof of Theorem
5
We have (cWq)⇒[{(yj, zj, vji)}j∈M ∪ ˆM satisfies (Wq)]⇔( cW0), the latter by Theorem 3.
Furthermore, by Lemma 1, we also have ( cW0)⇒(cW
c). Thus, ( cW0)⇔(cWc)⇔(cWq). Since
( cW0)⇒(cW ), it remains to show(cW )⇒( cW0), i.e., our goal is to show that {(y
j, zj, vji)}j∈M ∪ ˆM
satisfies condition (W ) when (cW ) holds. Consider any M0 ⊆ M ∪ ˆM and distinguish three possibilities:
Case 1, M0 ⊆ ˆM : Then Yi
M0 = {ˆxi} and ˆxi ∈ N/ Mi 0 so that there exists x ∈ E (XM0)
such that x /∈ Yi M0.
Case 2, M0∩M 6= ∅, and there exists x ∈ E(XM0∩M∪{ˆxi}) such that x /∈ Yi
M0∩M∪{ˆxi}:
Since x 6= ˆxi, we must have x ∈ XM0∩M\ C(Yi
M0∩M∪ {ˆxi}). We also have XM0 ⊇ XM0∩M and
Yi
M0 ⊆ YMi 0∩M∪ {ˆxi}, since Y{j}i = {ˆxi} for all j ∈ M0\ M . Hence we have x ∈ XM0\ C(Yi
M0).
We conclude that there exists x0 ∈ E(XM0) such that x0 ∈ Y/ i
M0.
Case 3, M0∩ M 6= ∅, and there exists non-empty M00 ⊆ M0 ∩ M such that Yi
M00 =
NMi 00 ⊆ E(XM0∩M ∪ {ˆxi}), and Yi
M00∩ Y(Mi 0∩M )\M00 = ∅: We distinguish two subcases. First,
if Ni
M0\M \ C(XM0∩M ∪ {ˆxi}) 6= ∅ then, since Yi
M0\M = {ˆxi} given that M0\ M 6= ∅, there
exists x ∈ E (XM0) ∩ Ni
M0\M such that x /∈ YMi0. Second, if NMi 0\M ⊂ C(XM0∩M ∪ {ˆxi}),
then E (XM0 ∪ {ˆxi}) = E(XM0∩M ∪ {ˆxi}). Note that we must have ˆxi ∈ N/ i
M00, otherwise
violating condition (cW ). Thus, we conclude that for M00 ⊆ M0 we have Yi
M00 = NMi 00 ⊆
E(XM0∩M∪{ˆxi})\{ˆxi} ⊆ E(XM0). Furthermore, since ˆxi ∈ Y/ i
M00, YMi0\M00 ⊆ Y(Mi 0∩M )\M00∪{ˆxi}
and YMi 00∩ Y(Mi 0∩M )\M00 = ∅, we conclude that YMi00∩ YMi 0\M00 = ∅.
In sum, in all three cases, either there exists x ∈ E (XM0) such that x /∈ Yi
M0, or there
exists non-empty M00 ⊆ M0 such that Yi
M00 = NMi 00 ⊆ E(XM0) and YMi 00 ∩ YMi 0\M00 = ∅, i.e.,
condition (W ) is satisfied for the augmented voting record {(yj, zj, vji)}j∈M ∪ ˆM as we wished
to show.
We also conclude from cases 1 to 3 above applied to subsets M0 ⊆ M ∪ ˆM with |M0| = 2 that (cW
1)⇒[{(yj, zj, vji)}j∈M ∪ ˆM satisfies (W1)]. Thus, if d = 1, (W1)⇔(W ) by
Theorem 3, hence(cW1)⇒(cW ), since we have shown (W )⇔(cW ).
Proof of Theorem
6
Assume d = 1. Without loss of generality, let ˆx1 ≤ ˆx2 ≤ ... ≤ ˆxn and (by swapping
alternatives yj, zj if necessary) let vj1 = yes for all j ∈ M . The proof is by construction.
If ˆx1 < ˆx2, position alternatives y1, ..., ym in the interval [ˆx1, ˆx2), so that ˆx1 ≤ y1 < ... <
ym < ˆx2, otherwise set y1 < ... < ym < ˆx1. Position alternatives z1, ..., zm in (ˆxn, +∞) so
that ˆxn < zm < ... < z1. Now, for every pair of voting items h, j ∈ M , and for every voter
i, Ni
{j}∩ (Y{j,h}i ∪ {ˆxi}) = ∅. Furthermore, if h > j, we have yj < yh < zh < zj. Thus,
for every voter i and for every h, j ∈ M , x ∈ N{j}i (where x = zj in the case of voter 1)
is such that x ∈ E (X{j}∪ {ˆxi}) and x /∈ Y{j}i ∪ {ˆxi}, and, if h > j, x ∈ E(X{j,h}∪ {ˆxi})
and x /∈ Yi
{j,h}∪ {ˆxi}. We conclude that condition ( bS1) is satisfied and, by Theorem 4, for
every voter i the voting record {(yj, zj, vji)}j∈M is rationalizable by a strictly concave utility
function ui with ideal point ˆxi.
To see that the Theorem also obtains in d0 > 1 dimensions, note that the constructed voting records, {(yj, zj, vji)}j∈M, in d = 1 dimension are strictly rationalizable satisfying(N ).
Then, by part (ii) of Theorem 7, for every ideal points ˆxi ∈ Rd
0
, i ∈ N , d0 > 1, there exist voting alternatives y0j, zj0 ∈ Rd0 such that for every voter i, voting record {(y0j, zj0, vji)}j∈M is
Proof of Theorem
7
Let ˆX = {ˆx1, ..., ˆxn} and construct a finite set XM0 ⊂ Rd
0
such that |XM0 | = |XM| and
XM0 = E (XM0 ∪ ˆX). Since d0 > 1, such an XM0 trivially exists. Consider any onto function f : XM → XM0 , and for every M0 ⊆ M , denote the image of XM0 under f by XM0 0 = f (XM0).
Fix arbitrary voter i, and consider the voting record {f (yj), f (zj), vji}j∈M. We shall show:
Claim: If E (XM0 0 ∪ {ˆxi}) ⊆ YMi0 ∪ {ˆxi} for some non-empty M0 ⊆ M , then there exists
non-empty M00 ⊆ M0 such that Ni
M00 = YMi 00 ⊆ E(XM0 0 ∪ {ˆxi}) and YMi 00 ∩ YMi 0\M00 = ∅.
Since XM0 = E (XM0 ∪ ˆX), we have XM0 0 ⊆ E(XM0 0 ∪ {ˆxi}) and XM0 0 ∩ {ˆxi} = ∅ for all
voters i. Thus, we have Ni
M0 ⊆ YMi 0. As in the proof that (Wq)⇒(W ) for Theorem 3,
inductively define a nested sequence of nonempty subsets of M0 by setting M00 = M0, and Mt+10 = {j ∈ Mt0 : Yi
{j} ⊆ NMi t0}. We analogously conclude that there exists integer k such
that Mk−10 = Mk0 = M00 6= ∅. Furthermore, Ni
M00 = YMi 00 and YMi00 ∩ YMi 0\M00 = ∅, the latter
because otherwise there exists j ∈ M0\M00 such that Yi
{j}⊆ YMi 00 = NMi 00, which is impossible
by the definition of the sequence {Mt0}.
From the Claim we conclude that the voting record {(f (yj), f (zj), vji)}j∈M satisfies (cW ). Since the Claim obtains for every voter i, the conclusion in part (i) follows by Theorem
5. Under the additional assumption of part (ii), the voting record {(f (yj), f (zj), vji)}j∈M
satisfies(A), since the original record {(yj, zj, vij)}j∈M does and f is one to one. Thus, since
the voting record {(f (yj), f (zj), vij)}j∈M satisfies(cW )and (A), it also satisfies( bS), and part
(ii) now follows from Theorem 4.
Proof of Theorem
8
We have already shown parts (i) and (ii), so it remains to show part (iii). By Corol- lary 1, a voting record {(yj, zj, vji)}mj=0, with y0 = x violates (W ) if it violates (S) and
{(yj, zj, vji)}mj=0 satisfies (A). But, since {(yj, zj, vji)}j∈M satisfies (A), {(yj, zj, vij)}mj=0 must
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