As Liu (2011: 102) points out, “preference describes a state of mind in the same way as belief does”. If an agent prefers alternativex toy, he believes, for some reason or another, thatx is better than y. Therefore, it should be the case that an agent prefers x to y iff he knows and believes that he does. And that he does not preferxtoy iff he knows and believes that he does not. In other words, agents should be fully introspective with respect to their preferences. This section shows that this is indeed the case. In order to prove this, the following lemma is useful:
Lemma 4.22. LetM be a model for preference formation,i∈ N and x∈X. Let w, v∈W be such thatw∼i v. Then the following hold:
1. XiK(w) =XiK(v) 2. XiB(w) =XiB(v)
Proof. For the ⊆-inclusion of item 1, let J ∈ XiK(w). That is, there exists a w0 ∈ W such that w ∼i w0 and J = ri(w0, x). As ∼i is an equivalence relation, w ∼i v and w ∼i w0 imply v ∼i w0.
Hence,J ∈XiK(v). The proof of⊇-inclusion is the same. For the⊆-inclusion of item 2, letJ ∈XiB(w). That is, there exists aw0 ∈W such that w∼i w0 and for all w00∈W such thatw0≤i w00, there exists a w000 ∈W such that w00 ≤i w000 and ri(w000, x) =J. Notice that w ∼i v and w ∼i w0 imply v ∼i w0.
Thus, there exists aw0 ∈W such that v∼i w0 and for all w00∈W such that w0 ≤iw00, there exists a
w000 ∈ W such that w00 ≤i w000 and ri(w000, x) =J. Thus, J ∈XiB(v). The proof of the ⊇-inclusion is
similar.
Theorem 4.23. Let M be a model for preference formation, i ∈ N and x, y ∈ X. Let Rwi be defined as knowledge or belief-based weak dominance, maximin, leximin, maximax or leximax and let Q:={w∈W |xRwi y}. Then Q=KiQ=BiQ.
Proof. For starters, suppose that Rwi is defined as knowledge-based weak dominance. We prove the following three inclusions:
1. Q⊆KiQ. Supposew∈Q. That is, for allw0 ∈W such thatw∼iw0 it holds thatri(w0, x)w
0
i
ri(w0, y). In order to show that w∈KiQ, we need to show that for allv∈W such thatw∼i v,
it holds that v ∈ Q. Let v ∈ W be such that w ∼i v. Let v0 ∈ W be such that v ∼i v0.
Notice thatw ∼i v ∼i v0 implies that w∼i v0. Since w∈Q, it holds that ri(v0, x)v
0
i ri(v0, y).
Consequently, v∈Qand w∈KiQ.
2. KiQ⊆BiQ. This follows from the fact that knowledge implies belief.
3. BiQ ⊆Q. According to Observation 2.10, this means that for all w0 ∈ W such that w ∼i w0
there exists aw00∈W such thatw0 ≤iw00and for allw000∈W such that w00≤iw000, it holds that
w000 ∈Q. (∗) In order to show that w∈Q, we need to show that for allv∈W such thatw∼iv
it holds that ri(v, x)vi ri(v, y). Let v∈W be such that w∼i v. By (∗), there exists av0 ∈W
such that v ≤i v0 and for all v00 ∈ W such that v0 ≤i v00, it holds that v00 ∈ Q. In particular,
the reflexivity of≤i gives us that v0 ∈Q. That is, for allu∈W such that v0 ∼i uit holds that
ri(u, x)ui ri(u, y). Using the fact that≤i ⊆ ∼i and that ∼i is an equivalence relation, we can
conclude from v≤iv0 that v0 ∼i v. Consequently, ri(v, x)vi ri(v, y). Thus,w∈Q.
In conclusion, from the fact that Q⊆KiQ⊆BiQ⊆Q, it follows that Q=KiQ=BiQ.
Now, suppose that Rwi is defined as belief-based weak dominance. Again, we prove the following three inclusions:
1. Q⊆KiQ. Supposew∈Q. That is, for allw0 ∈W such thatw∼i w0there exists aw00∈W such
that w0 ≤i w00 and for allw000 ∈W such that w00 ≤i w000, it holds that ri(w000, x)w
000
i ri(w000, y).
In order to show thatw ∈KiQ, we need to show that for all v∈W such that w∼i v, it holds
that v ∈ Q. Let v ∈ W be such that w ∼i v. By definition of belief-based weak dominance, it remains to be shown that for all v0 ∈ W such that v ∼i v0 there exists a v00 ∈W such that
v0 ≤i v00 and for all v000 ∈W such thatv00≤i v000, it holds that ri(v000, x)v
000
i ri(v000, y). This can
easily be proven by using the fact thatw∈Q and that∼i is an equivalence relation. 2. KiQ⊆BiQ. This follows from the fact that knowledge implies belief.
3. BiQ⊆Q. Supposew∈BiQ. That is, for all w0 ∈W such that w∼i w0 there exists aw00∈W
such thatw0 ≤i w00 and for allw000 ∈W such thatw00≤iw000, it holds thatw000 ∈Q. (∗) In order
to show thatw∈Q, we need to show that for all v∈W such thatw∼i vthere exists av0 ∈W
such thatv≤iv0 and for allv00∈W such thatv0≤i v00, it holds thatri(v00, x)v
00
i ri(v00, y). Let
v∈W be such thatw∼i v. From (∗), it follows that there exists a u∈W such thatv≤i u and for all u0 ∈W such that u ≤i u0, it holds that u0 ∈Q. Due to the reflexivity of ≤i, it follows
thatu∈Q. By item 1, it follows thatu∈KiQ. Furthermore, w∼i v≤iu implies thatw∼i u.
Combining this with the fact thatu∈KiQgives us thatw∈Q.
From these inclusions, we can conclude thatQ=KiQ=BiQ.
For the knowledge and belief-based version of maximin, leximin, maximax and leximax, the fact that Q = KiQ = BiQ follows immediately from the definitions and the fact that for all w, w0 ∈ W
such that w∼i w0 it holds (i) thatwi =w
0
i and (ii) thatXiK(w) =XiK(w0) and XiB(w) =XiB(w0),
as was shown in Lemma 4.22.
Corollary 4.24. Let M be a model for preference formation, i ∈ N and x, y ∈ X. Let Rwi be defined as knowledge or belief-based weak dominance, maximin, leximin, maximax or leximax and let Q0 :={w∈W | ¬(xRwi y)}. Then Q0=KiQ0=BiQ0.
Proof. LetRwi be defined as knowledge or belief-based weak dominance, maximin, leximin, maximax or leximax preference. Notice that for all w ∈ W, it is the case that xRwi y or ¬(xRwi y). Hence Q0 = Q, where Q is defined as in Theorem 4.23. Therefore, we need to show thatQ =KiQ=BiQ.
Theorem 4.23 gives us thatQ=KiQ=BiQand, consequently, that Q=KiQ=BiQ. Therefore, it
suffices to show the following: 1. KiQ=KiQ
2. BiQ=BiQ
For item 1, notice that KiQ = {w ∈ W | ∀w0 ∈ W(w ∼i w0 ⇒ w0 ∈/ Q)} and that KiQ = {w ∈
W | ∃w0∈W(w∼iw0∧w0 ∈/ Q}. For the ⊆-inclusion, suppose that w∈KiQ. Due to the reflexivity
of ∼i, we havew∼i w. Sincew ∈KiQ, it follows that w /∈Q. Thus, w∈KiQ. For the⊇-inclusion,
suppose that w∈KiQ. That is, there exists aw∗ ∈W such that w∼i w∗ and w∗ ∈/ Q. Let w0 ∈W
be such that w ∼i w0. Suppose, for contradiction, that w0 ∈ Q. From the fact that w ∼i w0 and w ∼i w∗, it follows that w0 ∼i w∗. From Theorem 4.23, we know that Q = KiQ and, thus, that
w∗ ∈Q. Contradiction. Thus, w0 ∈/ Q. Since for all w0 ∈W such that w∼i w0 it holds thatw0 ∈/ Q,
it is the case thatw∈KiQ.
For item 2, notice the following:
• BiQ={w∈W | ∀w0∈W(w∼i w0 ⇒ ∃w00∈W(w0 ≤i w00∧ ∀w000 ∈W(w00≤i w000 ⇒w000∈/ Q)))}
• BiQ={w∈W | ∃v∈W(w∼iv∧ ∀v0∈W(v≤iv0 ⇒ ∃v00∈W(v0 ≤i v00∧v00 ∈/ Q)))}
For the⊆-inclusion, letw∈BiQ. Since∼i is reflexive, the definition of BiQgives us that there exists
a w00 ∈ W such that w ≤i w00 and for all w000 ∈W such that w00 ≤i w000, it holds that w000 ∈/ Q. (∗) In order to show that w∈ BiQ, let v =w00. Now let v0 ∈ W be such that v≤i v0. It remains to be
shown that there exists av00∈W such thatv0 ≤iv00 and v00∈/Q. Take v00=v0. The reflexivity of ≤i
gives us that v0 ≤iv0. Sincew00=v≤iv0, (∗) gives us thatv0∈/ Q. Thus, w∈BiQ.
For the ⊇-inclusion, let w ∈ BiQ. That is, there exists a v ∈ W such that w ∼i v and for all
v0 ∈ W such that v ≤i v0 there exists a v00 ∈ W such that v0 ≤i v00 and v00 ∈/ Q. Notice that since
v≤iv, it follows that there exists av∗ ∈W such thatv≤i v∗and v∗ ∈/ Q. Suppose, for contradiction, that w /∈ BiQ. That is, there exists a w0 ∈ W such that w ∼i w0 and for all w00 ∈ W such that
w0 ≤i w00 there exists a w000 ∈W such that w00≤i w000 and w000 ∈Q. Since w0 ≤i w0, this means that
there exists aw∗∈W such thatw0 ≤iw∗ and w∗∈Q. From Theorem 4.23, we know that Q=KiQ.
Consequently,w∗ ∈KiQ. Moreover, since ≤i⊆ ∼i and ∼i is an equivalence relation, we can conclude
from w ∼i v ≤i v∗ and w ∼i w0 ≤i w∗ that w∗ ∼i v∗. Since w∗ ∈ KiQ, it follows that v∗ ∈ Q.
Contradiction. Thus, it must be the case thatw∈BiQ.6