Characterising SUBSET

Recall that SUBSETwas defined as follows:

Definition 3.8. (SUBSET) SUBSET(P, i) =      N\{i} ifPi=∅ {i} ifPi∈ L(A) {j∈N\{i} |Pi ⊆Pj} otherwise

Theorem 3.9. SUBSET is the unique proxy mechanism satisfying Proxy Availability,Independence of Irrelevant Proxies,Zero Regret andPreference Monotonicity.

Proof. Clearly,SUBSETsatisfiesPA,IIPandZR. To see thatSUBSETsatisfies

PM, suppose thatj ∈SUBSET(P, i) andj =6 i, for some P ∈ P(A)n, i, j ∈

N. So Pi ⊆ Pj. Suppose that there is k ∈ N such that Agree(Pi, Pj) ⊆

Agree(Pi, Pk) and Disagree(Pi, Pk) ⊆ Disagree(Pi, Pj). Then we must havePi ⊆Pk, sinceAgree(Pi, Pj) =Pi, sincePi⊆Pj. Sok∈SUBSET(P, i), as required.

For the other direction (i.e. to show uniqueness), I prove the contrapos- itive. Suppose g 6= SUBSET is a proxy mechanism, and suppose g satisfies

PA,IIP andPM. I will show thatg does not satisfyZR. It will help to prove the following intermediate claim.

Lemma 3.10. LetP ∈ P(A)n and i, j ∈N, such that Pi ∈ L/ (A). Then if

Pi ⊆Pj, and g satisfiesPA,IIP and PM, we havej ∈g(P, i).

Proof. Since g satisfies PA, there must be some P0 ∈ P(A)n such that

P_i0 =Pi and g(P0, i)=6 ∅. Supposek∈ g(P0, i), for some k∈N. Then we can construct a new profileP00 where

P_i00=P_i0 =Pi

P_j00=Pj

P_k00=P_k0

By IIP, we must have k∈g(P00, i), since P_i00 =P_i0 and P_k00 =P_k0. But then byPM, we must have j∈(P00, i), since

P_i00=Pi ⊆Pj =Pj00

implying thatj must agree at least as much withi askin profile P00. But then by another application ofIIP, we must have j∈g(P, i), sinceP_i00=Pi and P_j00 =Pj.

We are now ready to prove the uniqueness of SUBSET. Since g 6= SUBSET, there must be some P ∈ P(A)n and i, j ∈ N with Pi ∈ L/ (A) such that either

Pi *Pj and j∈g(P, i) or

Pi ⊆Pj and j /∈g(P, i)

But note that Lemma 3.10 rules out this latter case. So we only need to consider the case where

SincePi *Pj, there must be some ab∈Pi such that ab /∈Pj. But now consider a profileP0 where, for some k∈N:

P_i0=Pi

P_j0 =Pj

P_j0∪ {ba} ⊆P_k0, andP_k0 ∈ L(A)

Note that this profile is well defined; sinceab /∈Pj =Pj0, we must have that P_j0∪ {b a} is still anti-symmetric, and thus can be extended to a linear orderP_k0.

Since P_i0 =Pi and Pj0 =Pj, we must havej ∈g(P0, i), by IIP. But then we must have k ∈ g(P0, j) by Lemma 3.10, since P_j0 ⊆ P_k0. So then if i

picksj as her proxy and jpickskas her proxy, thenkwill be i’s guru. But

P_i0 *Pk0, sinceab∈Pi =Pi0 and ab /∈Pk0 (because ba∈Pk0). So g is notZR.

Showing each condition is necessary

I have characterised SUBSET as the conjunction of four conditions: Surjec- tivity, IIP,ZR and PM. We can also show that each of these conditions is individually necessary for characterisingSUBSET, by showing that the other three conditions are not jointly sufficient.

Proposition 3.11. PAis necessary for characterising SUBSET.

Proof. Recall that TRIV was defined as follows.

TRIV(P, i) =      N\{i} ifPi=∅ {i} ifPi∈ L(A) ∅ otherwise

TRIV is a proxy mechanism which does not satisfy PA. But note thatTRIV does satisfyIIP,ZR andPM.

Proposition 3.12. ZR is necessary for characterisingSUBSET.

Proof. Recall that UNIV was defined as follows:

UNIV(P, i) =      N\{i} ifPi=∅ {i} ifPi∈ L(A) N\{i} otherwise

UNIV is a proxy mechanism which does not satisfyZR. But note thatUNIV does satisfyPA,IIP andPM.

Proof. Considerg defined as follows: g(P, i) =      N\{i} ifPi =∅ {i} ifPi ∈ L(A) {j∈N|Pj ∈ L(A) andPi⊂Pj} otherwise

g is a proxy mechanism which does not satisfy PM (just consider some

Pj ∈ L/ (A) such that Pi ⊆ Pj). But note that g does satisfy PA, IIP and

ZR.

Proposition 3.14. IIP is necessary for characterising SUBSET.

Proof. Considerg defined as follows:

g(P, i) =                N\{i} ifPi=∅ {i} ifPi∈ L(A) {j∈N |Pj ∈ L(A)} ifPi∈ L/ (A) andPi ⊆Pj,∀j ∈N\{i} such that Pj ∈ L(A) ∅ otherwise

g is a proxy mechanism which does not satisfy IIP. But note that g does satisfyPA,PM and ZR.