We want to show that Cj2(X2) = Cj2(K2)- It is sufficient to show that
Y Γ ll'2e Cp2(X2); that is,
X2 U (У ∩ B2c) ∈ Kf .
By condition (??), and since B2 = B1 U Bc,
X2U(YnBf) = (X ∩ B2) U (У ∩ B≈)
= (X ∩ B2) U (У ∩ B1) U (У ∩ Bc).
Claim 1 (X ∩ B1) U (X ∩ B2) U (У ∩ Bc) ∈ Kf.
Proof Since Y ∈ ‰ (y∩B)u(y∩Bc) ∈ 7⅞. Therefore, y∩Bc ∈ Cj(B)
for some B ∈ AC(B). Moreover, since B is a section and X ∩ B ∈ AC(B),
Remark 2 implies that (X ∩ B) U (Y ∩ Bc) ∈ Kf. Hence, (X ∩ B1) U (X ∩
B2) U (Y ∩ Bc) ∈ Kf , which is the statement of the claim.
Therefore, by Claim 1 and the hypothesis that B1 is a section,
(X ∩ B2) U (У ∩ Bc) ∈ Cj1 (Bfl)
for all B'l ∈ AC(Bi). Because (Y ∩ B1) ∈ AC(Bi) we have, by Remark 2,
(X ∩ B2) U (У ∩ B1) U (У ∩ Bc) ∈ Kf . Hence, (У ∩ B≈) ∈ Cp2(X2). ■
Definition 9 A partition {B1, ...,Bq} of Rf is a Cylindric decomposition
of Rf if for all p = 1, ...,q, Bp is a section of Rf. A cylindric decomposition
is called minimal if there is no finer cylindric decomposition of Rf .
Remark 3 Let {B1,...,B9} be a partition of Rf. Then, {B1,...,B9} is a
cylindric decomposition of Rf if and only if
Kf = AC (Bi) + ... + AC (Bq).
We want to show (Proposition 3 below) that, given any social choice
function F, its corresponding set Rf has a unique minimal cylindric decom-
position. In the proof of Proposition 3 we will use the following Lemma.
13
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