The structure of incomplete preferences

Leandro Gorno

doi:10.1007/s00199-017-1057-9

Economic Theory

pp 1–27 | Cite as

The structure of incomplete preferences

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Leandro GornoEmail author

Research Article

First Online: 09 June 2017

Received: 14 June 2016

Accepted: 23 May 2017

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Abstract

I study incomplete preferences as a means to represent indecisiveness. A decomposition into maximal domains of comparability is characterized and used to link optimization of incomplete preferences with maximization of local utility functions. Larger maximal domains are shown to correspond to more decisive preferences. The decomposition can be uniquely recovered from choice data under standard assumptions. Applications to different models within decision theory are discussed.

Keywords

Incomplete preferences Indecisiveness Maximal elements Best elements Revealed preference

JEL Classification

D11 D81

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Notes

Acknowledgements

I thank the editor, a very helpful anonymous referee, Carlos da Costa, Bruno Furtado and Gil Riella for their comments. I am also grateful to Ana Luiza Perdigão and Rodrigo Naumann for excellent research assistance.

Appendix: Proofs

Proof of Lemma 1

To prove the first claim, it suffices to apply Hausdorff maximal principle to the quotient space $X / {\sim }$.

The second claim is also straightforward. If A is a maximal $\succsim $-domain, then A must be a $\succsim $-domain, so Property (a) holds trivially. Moreover, suppose Property (b) was violated. Then, there would be $y \in X {\setminus } A$ such that $y {{\mathrm{\triangleleft \triangleright }}}x$ for all $x \in A$. Hence, $A \cup \{y\} \supset A$ would still be a $\succsim $-domain, obtaining the desired contradiction. For the other direction, suppose that $A \in \mathscr {P}_X$ satisfies Properties (a) and (b). Then, A is a $\succsim $-domain by Property (a). To see that A is indeed a maximal $\succsim $-domain, suppose that there exists another $\succsim $-domain $B \supset A$. Then, any $y \in B{\setminus }A \ne \emptyset $ would contradict Property (b). $\square $

Proof of Theorem 1

For each $x \in X$, define $S_x := S_x^+ \cup S_x^-$. Since Open image in new window

, it follows that $\bigcap _{x \in A} S_x = \xi (A,\succsim )$ for every $A \in \mathscr {P}_X$. I claim that every maximal $\succsim $-domain $D \in \mathscr {D}_X(\succsim )$ satisfies

$$\begin{aligned} D = \bigcap _{x \in D} S_x. \end{aligned}$$

(5)

To see $D \subseteq \bigcap _{x \in D} S_x$, pick any $y \in D$ and suppose that there exists $x \in D$ such $y \notin S_x$. This would imply $y \bowtie x$ and $\succsim $ would not be complete on D, a contradiction. To see that $\bigcap _{x \in D} S_x \subseteq D$, pick y such that $y \in S_x$ for all $x \in D$ and assume that $y \notin D$. Since $x {{\mathrm{\triangleleft \triangleright }}}y$ for every $x \in D$, $\succsim $ would be complete on the set $D \cup \{y\}$, contradicting the maximality of D.

I have shown that maximal $\succsim $-domains satisfy Eq. 5. I will now show that if $D \in \mathscr {P}_X$ satisfies Eq. 5, then D is a maximal $\succsim $-domain. To see this, assume that D satisfies Eq. 5 and pick any two alternatives $x,y \in D$. Since $x \in D \subseteq S_y$, it follows that $x {{\mathrm{\triangleleft \triangleright }}}y$, showing that D is a $\succsim $-domain. To prove that D is in fact a maximal $\succsim $-domain, consider some $y \in X {\setminus } D$ and suppose, seeking a contradiction, that there is no $x \in D$ such that $x \bowtie y$. It follows that $y \in S_x$ for all $x \in D$, which means that $y \in \bigcap _{x\in D} S_x$. Equation 5 thus implies $y \in D$, providing the desired contradiction.

The argument above shows that Eq. 5 fully characterizes maximal $\succsim $-domains, completing the proof of the first claim.

It is time to prove the properties of $\mathscr {D}_X(\succsim )$. Property (i) is easy to establish. The inclusion $\succsim \supseteq \bigcup _{D \in \mathscr {D}_X(\succsim )} \succsim _D$ follows from the fact that $\mathscr {D}_X(\succsim )$ covers X. To see $\succsim \subseteq {\bigcup _{D \in \mathscr {D}_X(\succsim )} \succsim _D}$, pick $x,y \in X$ such that $x \succsim y$. Since $\{x,y\}$ is a $\succsim $-domain, Lemma 1 implies that there must be a maximal $\succsim $-domain $D \in \mathscr {D}_X(\succsim )$ satisfying $D \supseteq \{x,y\}$. Hence, $x \succsim _D y$, proving the desired inclusion.

Since every $D \in \mathscr {D}_X(\succsim )$ satisfies Eq. 5, Property (ii) is a direct consequence of the fact that, whenever $\succsim $ is continuous at D, $S_x$ is closed for every $x \in D$.

Finally, to prove Property (iii), first pick $x \in \mathscr {M}(X, \succsim )$. Since $\mathscr {D}_X(\succsim )$ covers X, there is $D \in \mathscr {D}_X(\succsim )$ such that $x \in D$. Since D is a $\succsim $-domain, $x {{\mathrm{\triangleleft \triangleright }}}y$ holds for all $y \in D$. Considering the definition of $\mathscr {M}(X, \succsim )$, this implies $x \succsim y$ for all $y \in D$, so $x \in \mathscr {B}(D, \succsim )$. It follows that $\mathscr {M}(X,\succsim ) \subseteq \bigcup _{D \in \mathscr {D}_X(\succsim )} \mathscr {B}(D, \succsim )$. To establish the opposite inclusion, pick $D \in \mathscr {D}_X(\succsim ), x \in {\mathscr {B}(D,\succsim )}$ and suppose, seeking a contradiction, that there exists some $y \in X$ such that $y \succ x$. By definition of ${\mathscr {B}(D,\succsim )}$, $y \not \in D$ must hold, so $y \in X {\setminus } D$. Since D is a maximal $\succsim $-domain, there exists $z \in D$ such that $z \bowtie y$. Note that $x \in {\mathscr {B}(D,\succsim )}$ and $z \in D$ imply $x \succsim z$. Transitivity thus implies $y \succsim z$, providing the desired contradiction. $\square $

Proof of Theorem 2

For simplicity, define $\mathscr {S}_A := \mathscr {M}\left( A \cap \mathscr {D}_X(\succsim ),\supseteq \right) $ for each $A {\in } \mathscr {P}_X$. I claim that $\mathscr {S}_A \subseteq {\mathscr {D}_A(\succsim _A)}$. Pick $\tilde{D} \in \mathscr {S}_A$. This means that there exists $D \in {\mathscr {D}_X(\succsim )}$ such that $\tilde{D} = A \cap D$ and that every $D' \in \mathscr {D}_X(\succsim )$ such that $D' \supseteq \tilde{D}$ must satisfy $A \cap D' = \tilde{D}$. I will show that $\tilde{D} = \bigcap _{x\in \tilde{D}}S^A_x$, where $S^A_x := \left( S^+_x \cup S^-_x\right) \cap A$, which is equivalent to $\tilde{D} \in \mathscr {D}_A(\succsim _A)$ by applying Theorem 1 on the set A. To see $\tilde{D} \subseteq \bigcap _{x\in \tilde{D}}S^A_x$, note that, since $D \in \mathscr {D}_X(\succsim )$, Eq. 1 implies that $D = \bigcap _{x \in D}S_x^X$. Therefore,

$$\begin{aligned} \tilde{D} = A \cap D = A \cap \left( \bigcap _{x\in D}S^X_x\right) \subseteq A \cap \left( \bigcap _{x\in {A \cap D}}S^X_x\right) = \bigcap _{x\in \tilde{D}}S^A_x. \end{aligned}$$

For the reverse inclusion, pick $y \in \bigcap _{x\in \tilde{D}}S^A_x$. I need to show that $y \in \tilde{D}$. Clearly $y \in A$ and $y {{\mathrm{\triangleleft \triangleright }}}x$ for all $x \in \tilde{D}$. Then, by Lemma 1, there exists $D' \in \mathscr {D}_X(\succsim )$ such that $D' \supseteq \tilde{D} \cup \{y\}$. Note that $y \in A \cap D'$. Moreover, since $D' \supseteq \tilde{D}$, the $\supseteq $-maximality in the definition of $\mathscr {S}_A$ implies $A \cap D' = \tilde{D}$. It follows that $y \in \tilde{D}$, as desired.

Now I shall prove that $\mathscr {D}_A(\succsim _A) \subseteq \mathscr {S}_A$. So, pick $\tilde{D} \in \mathscr {D}_A(\succsim _A)$. Since $x {{\mathrm{\triangleleft \triangleright }}}y$ for all $x,y \in \tilde{D}$, there must be $D \in \mathscr {D}_X(\succsim )$ such that $D \supseteq \tilde{D}$. Obviously, $A \cap D \supseteq A \cap \tilde{D} = \tilde{D}$. Moreover, since Theorem 1 implies $D = \bigcap _{x\in D}S^X_x$,

$$\begin{aligned} A \cap D = A \cap \left( \bigcap _{x\in D}S^X_x\right) = \bigcap _{x\in D}S^A_x \subseteq \bigcap _{x\in \tilde{D}}S^A_x = \tilde{D}. \end{aligned}$$

Thus, $A \cap D = \tilde{D}$. This proves that $\mathscr {D}_A(\succsim _A) \subseteq A \cap \mathscr {D}_X(\succsim )$. To show $\tilde{D} \in \mathscr {S}_A$, pick any $D' \in \mathscr {D}_X(\succsim )$. Since $\succsim $ is complete on $D'$, $\succsim _A$ is complete on $A \cap D'$. This means that $A \cap D'$ is a $\succsim _A$-domain. Note that $\tilde{D}$ is a maximal $\succsim _A$-domain, so $A \cap D' \supseteq \tilde{D}$ implies $\tilde{D} = A \cap D'$. It follows that $\tilde{D} \in \mathscr {S}_A$. $\square $

For the proof of Theorem 3, the following lemma is required:

Lemma 2

If $\succsim $ is a branching preference, then

$$\begin{aligned} \mathscr {B}\left( A\cap D,\succsim \right) = \mathscr {B}\left( A\cap D',\succsim \right) \end{aligned}$$

for all $A \in \mathscr {P}_X$ and $D,D' \in \mathscr {D}_X(\succsim )$ such that $A \cap D \cap D' \ne \emptyset $.

Proof

By symmetry, it suffices to establish the inclusion ${\mathscr {B}(A{\cap } D,\succsim )} {\subseteq } \mathscr {B}(A\cap D',\succsim )$. Note that, by assumption, there must be $y {\in } A \cap D \cap D'$. Let $x {\in } \mathscr {B}\left( A \cap D, \succsim \right) $. Obviously, $x \succsim y$.

I claim that $x \in D'$. Suppose, seeking a contradiction, that $x \not \in D'$. Then, by Lemma 1, there exists $w \in D'$ such that $w \bowtie x$. Since $D'$ is $\succsim $-domain and $y,w \in D'$, it must be the case that $w {{\mathrm{\triangleleft \triangleright }}}y$. But, given that $w \bowtie x\succsim y$, $w \succsim y$ contradicts that $\succsim $ is branching, while $y \succ w$ leads to $x \succ w$ and, so, to $w {{\mathrm{\triangleleft \triangleright }}}x$. Since all possibilities lead to contradictions, it can be concluded that $x \in D'$.

Pick $z \in A \cap D'$ and suppose, seeking a contradiction, that $x \not \succsim z$. It follows that either $z \bowtie x$ or $z \succ x$. Assume $z \bowtie x$. Since $x \succsim y$ and $\succsim $ is branching, $z \not \succsim y$. This leads to a contradiction because it follows that either $y \succ z$ (which implies $x \succ z$) or $y \bowtie z$ (which implies that $D'$ is not a $\succsim $-domain). Assuming $z \succ x$ instead, then $z \not \in D \in \mathscr {D}_X(\succsim )$. By Lemma 1, there exists $w \in D$ such that $w \bowtie z$. It must be the case that $w \succ x$ (since otherwise $z \succ x \succsim w$ leads to a contradiction). But, if so, $w \succsim x$, $z \succsim x$ and $w \bowtie z$ contradict that $\succsim $ is branching. I conclude that $x \succsim z$ for all $z \in A \cap D'$. This means that $x \in \mathscr {B}\left( A \cap D', \succsim \right) $. $\square $

Proof of Theorem 3

Let $\succsim $ be a branching preference. Suppose, seeking a contradiction, that Eq. 2 does not hold. Then, there must be $A \in \mathscr {P}_X$ and $D \in {\mathscr {D}_X(\succsim )}$ such that ${\mathscr {B}\left( A \cap D,\succsim \right) } {\setminus } {\mathscr {M}\left( A,\succsim \right) } \ne \emptyset $. Pick $x \in \mathscr {B}\left( A \cap D,\succsim \right) {\setminus } \mathscr {M}\left( A,\succsim \right) $. Since $x \not \in \mathscr {M}\left( A,\succsim \right) $, there must be some $y \in A$ satisfying $y \succ x$. By the first part of Lemma 1, there exists $D' \in \mathscr {D}_X(\succsim )$ such that $D' \supseteq \{x,y\}$. Since $y \in A \cap D'$ and $y \succ x$, $x \not \in \mathscr {B}\left( A \cap D', \succsim \right) $. Moreover, since $x \in A \cap D \cap D'$ and $\succsim $ is branching, Lemma 2 implies that $\mathscr {B}\left( A\cap D,\succsim \right) = \mathscr {B}\left( A\cap D',\succsim \right) $. It follows that $x \not \in \mathscr {B}\left( A \cap D, \succsim \right) $, a contradiction.

Now suppose that $\succsim $ is not branching. Then, there must be $x,y,z \in X$ such that $x \succsim z$, $y \succsim z$ and $x \bowtie y$. By Lemma 1, there exists $D \in {\mathscr {D}_X(\succsim )}$ such that $\left\{ x,z\right\} \subseteq D$. Consider the subproblem $A = \left\{ y,z\right\} $. Clearly, ${\mathscr {B}\left( A \cap D, \succsim \right) } = {\mathscr {B}\left( \{z\}, \succsim \right) } = \{z\}$. However, $z \not \in {\mathscr {M}\left( A, \succsim \right) }= \left\{ y\right\} $. I conclude that Eq. 2 fails. This completes the proof. $\square $

Proof of Proposition 1

By Property (iii) in Theorem 1,

$$\begin{aligned} \mathscr {M}(X,\succsim ) =\bigcup _{D \in \mathscr {D}_X(\succsim )} \mathscr {B}(D, \succsim ) \supseteq \bigcup _{D \in \mathscr {D}} \mathscr {B}(D, \succsim ). \end{aligned}$$

Suppose, seeking a contradiction, that there is $x \in \mathscr {M}(X,\succsim )$ such that $x \not \in \mathscr {B}(D,\succsim )$ for every $D \in \mathscr {D}$. Because $x \in \mathscr {M}(X,\succsim )$, $x \not \in \mathscr {B}(D,\succsim )$ implies $x \not \in D$. It follows that $x \not \in \bigcup _{D \in \mathscr {D}} D$. As a consequence, $(x,x) \not \in \bigcup _{D \in \mathscr {D}} \succsim _D$. Since $\succsim $ is reflexive, ${\bigcup _{D \in \mathscr {D}} \succsim _D} \not =\,\, \succsim $. This means that $\mathscr {D}$ is not a decomposition, a contradiction. $\square $

Proof of Theorem 4

Let $\mathbb {D}_X(\succsim )$ be the class of all decompositions for $\succsim $. Since $\mathscr {D}_X(\succsim )$ is a decomposition by Theorem 1, the class $\mathbb {D}_X(\succsim )$ is nonempty. Moreover, $\mathbb {D}_X(\succsim )$ is partially ordered by $\subseteq $. By Hausdorff maximal principle, there exists a maximal $\subseteq $-totally ordered subset of $\mathbb {D}_X(\succsim )$. Let $\{\mathscr {D}_\alpha \}$ be this subset ($\alpha $ is an index) and define $\mathscr {D}^* := \bigcap _\alpha \mathscr {D}_\alpha $. I claim that $\mathscr {D}^*$ is the desired minimal decomposition.

To prove this, it suffices to show $\mathscr {D}^* \in \mathbb {D}_X(\succsim )$. Property (i) in Theorem 1 implies

$$\begin{aligned} \bigcup _{D \in \mathscr {D}^*} \succsim _D \, \subseteq \bigcup _{D \in \mathscr {D}_X(\succsim )} \succsim _D\, =\,\, \succsim . \end{aligned}$$

To establish the opposite inclusion, let $x,y \in X$ be any two alternatives such that $x \succsim y$. To show $(x,y) \in \bigcup _{D \in \mathscr {D}^*} \succsim _D$, it suffices to construct $D^* \in \mathscr {D}^*$ such that $\{x,y\} \subseteq D^*$. By definition, either $x \succ y$ or $x \sim y$.

Suppose first that $x \succ y$. Then, since $\succsim $ is locally finite, each decomposition $\mathscr {D}_\alpha $ must have a positive and finite number of maximal $\succsim $-domains containing $\{x,y\}$. But this implies that there must exist $D^* \in \mathscr {D}^*$ containing $\{x,y\}$.

Suppose now that $x\sim y$. One the one hand, if there is $z \in X$ such that either $x \succ z$ or $z \succ x$, then the previous reasoning applies to $\{x,z\}$. As a result, there exists $D^* \in \mathscr {D}^*$ containing $\{x,z\}$. Since $x \sim y$ and $D^*$ is a maximal $\succsim $-domain, $y \in D^*$. It follows that $\{x,y\} \subseteq D^*$. On the other hand, if there is no $z \in X$ such that either $x \succ z$ or $z \succ x$, then Open image in new window is a maximal $\succsim $-domain. Moreover, $D^* \cap D = \emptyset $ for all $D \in \mathscr {D}_X(\succsim ) {\setminus } \{D^*\}$. It follows from this property that $D^* \in \mathscr {D}^*$ (otherwise $\mathscr {D}_\alpha \not \in \mathbb {D}_X(\succsim )$ for some $\alpha $). Clearly, $\{x,y\} \subseteq D^*$, completing the proof. $\square $

Proof of Theorem 5

The result follows immediately from Theorem 3.1 in Fishburn (1970). $\square $

Proof of Proposition 2

On the one hand, if $\mathscr {D}_X(\succsim )=\{X\}$, X is a $\succsim $-domain. Thus, $\succsim $ is complete. On the other hand, if $\succsim $ is complete, X is a $\succsim $-domain. Since, $X \supseteq A$ holds for every $A \in \mathscr {P}_X$, X is the only maximal $\succsim $-domain and $\mathscr {D}_X(\succsim ) = \{X\}$. $\square $

Proof of Theorem 6

The “if” direction in the first claim is easy. Suppose, seeking a contradiction, that there are $x,y \in X$ such that $x {{\mathrm{\triangleleft \triangleright }}}_2 y$ holds but $x {{\mathrm{\triangleleft \triangleright }}}_1 y$ does not. Since $x {{\mathrm{\triangleleft \triangleright }}}_2 y$, there must be $D_2 \in \mathscr {D}_X(\succsim _2)$ such that $\{x,y\} \subseteq D_2$. The properties assumed on $\psi $ imply $\{x,y\} \subseteq D_2 \subseteq \psi (D_2) \in \mathscr {D}_X(\succsim _1)$. It follows that $x {{\mathrm{\triangleleft \triangleright }}}_1 y$, yielding the desired contradiction.

For the “only if” part, assume that $\succsim _1$ is more decisive than $\succsim _2$ and pick any $D_2 \in \mathscr {D}_X(\succsim _2)$. Since $D_2$ is a $\succsim _2$-domain and ${{\mathrm{\triangleleft \triangleright }}}_1 \supseteq {{\mathrm{\triangleleft \triangleright }}}_2$, $D_2$ is also a $\succsim _1$-domain. By Lemma 1, there exists a maximal $\succsim _1$-domain $D_1 \in \mathscr {D}_X(\succsim _1)$ such that $D_1 \supseteq D_2$. Thus, by the axiom of choice, there is a function $\psi : \mathscr {D}_X(\succsim _2) \rightarrow \mathscr {D}_X(\succsim _1)$ with the desired property.

As for the second claim, it is sufficient to establish that the existence of functions $\psi _{1} : \mathscr {D}_X(\succsim _1) \rightarrow \mathscr {D}_X(\succsim _2)$ and $\psi _{2} : \mathscr {D}_X(\succsim _2) \rightarrow \mathscr {D}_X(\succsim _1)$ such that $\psi _1(D) \supseteq D$ for all $D\in \mathscr {D}_X(\succsim _1)$ and $\psi _2(D) \supseteq D$ for all $D\in {\mathscr {D}_X(\succsim _2)}$ implies ${\mathscr {D}_X(\succsim _1)} = {\mathscr {D}_X(\succsim _2)}$. To prove this implication, suppose that such $\psi _1, \psi _2$ exist. Now pick an arbitrary $D_1 \in {\mathscr {D}_X(\succsim _1)}$ and define $D_2 := \psi _1(D_1)$ and $D'_1 := \psi _2(D_2)$. The assumptions on $\psi _1$ and $\psi _2$ imply $\psi _2(\psi _1(D_1)) \supseteq \psi _1(D_1) \supseteq D_1$, which means that $D'_1 \supseteq D_2 \supseteq D_1$. Thus, $D'_1 \supseteq D_1$. Since $\mathscr {D}_X(\succsim _1)$ is the set of all $\succsim _1$-maximal domains, it must be that $D'_1 = D_1$. It follows that $D_1 \supseteq D_2 \supseteq D_1$ and so $D_1 = D_2 \in \mathscr {D}_X(\succsim _2)$, showing that ${\mathscr {D}_X(\succsim _1)} \subseteq {\mathscr {D}_X(\succsim _2)}$. A symmetric argument yields ${\mathscr {D}_X(\succsim _2)} \subseteq {\mathscr {D}_X(\succsim _1)}$, completing the proof. $\square $

Proof of Proposition 3

By Lemma 2 in Ribeiro and Riella (2017), there exists a unique regular preference $\succsim '$ such that $\succsim \subseteq \succsim '$ and $\succ = \succ '$. Because $\succsim $ is not regular and $\succsim '$ is, it must be that $\succsim \subset \succsim '$. It follows that ${{\mathrm{\triangleleft \triangleright }}}= {\succsim \cup \prec } \subset {\succsim ' \cup \prec '} = {{\mathrm{\triangleleft \triangleright }}}'$. Thus, by Theorem 6, ${\mathscr {D}_X(\succsim )} \ne {\mathscr {D}_X(\succsim ')}$. Finally, $\succ = \succ '$ also implies that Open image in new window for all $A \in \mathscr {A}$. $\square $

Proof of Theorem 7

Let $\succsim $ be a preference such that $c(A)={\mathscr {M}(A,\succsim )}$ for all $A \in \mathscr {A}$. Suppose first that $\succsim $ is regular. Considering the construction of $\hat{\mathscr {D}}(c)$, it suffices to show that $D = X {\setminus } \left( \bigcup _{x \in D} R_{c,x}\right) $ holds if and only if $D \in \mathscr {D}_X(\succsim )$. The characterization of $\mathscr {D}_X(\succsim )$ in Theorem 1 implies that it is enough to verify that $S_x = X {\setminus } R_{c,x}$ for all $x \in X$.

To show that $S_x \subseteq X {\setminus } R_{c,x}$ suppose, seeking a contradiction, that there exists $y \in S_x \cap R_{c,x}$. Since $y \in R_{c,x}$, one would have $y \in Q_{c,x}$. This means that $c(\left\{ x,y\right\} )=\left\{ x,y\right\} $, which implies that either $x \bowtie y$ or $x \sim y$. Moreover, because $y \in S_x$, either $x \succsim y$ or $y \succsim x$. These two observations combined imply that $x \sim y$. But $y \in R_{c,x}$ also implies that either $Q_{c,x} {\setminus } Q_{c,y} \ne \emptyset $ or $Q_{c,y} {\setminus } Q_{c,x} \ne \emptyset $. Suppose first that $Q_{c,x} {\setminus } Q_{c,y} \ne \emptyset $. Then, there exists $z \in X$ such that $c(\left\{ x,z\right\} ) = \left\{ x,z\right\} $ and $c(\left\{ y,z\right\} ) \ne \left\{ y,z\right\} $. This means that either $x \bowtie z$ or $x \sim z$, while at the same time $y \left( \succ \cup \prec \right) z$. This contradicts $x \sim y$.

A similar argument yields a contradiction in the case $Q_{c,y} {\setminus } Q_{c,x} \ne \emptyset $. It follows that $S_x \cap R_{c,x} = \emptyset $.

I will now show that $X {\setminus } R_{c,x} \subseteq S_x$. Suppose, seeking a contradiction, that there exists $y \in X {\setminus } (R_{c,x} \cup S_x)$. On the one hand, $y \notin S_x$ implies $x \bowtie y$, and so ${c(\left\{ x,y\right\} )} =\left\{ x,y\right\} $. It follows that $y \in Q_{c,x}$. On the other hand, since $y \notin R_{c,x}$ but $y \in Q_{c,x}$, it must be that $Q_{c,y} {\setminus } Q_{c,x} = Q_{c,x} {\setminus } Q_{c,y} = \emptyset $. This implies that, for every $z \in X$, ${c(\left\{ y,z\right\} )} =\left\{ y,z\right\} $ holds if and only if ${c(\left\{ x,z\right\} )} =\left\{ x,z\right\} $. Hence, there is no $z \in X$ such that either $x\left( \succ \cup \prec \right) z$ or $y \left( \succ \cup \prec \right) z$. Since $\succsim $ is regular, there is a contradiction.

The preceding argument establishes that the equality $\hat{\mathscr {D}}(c) = \mathscr {D}_X(\succsim )$ is satisfied for all regular $\succsim $.

Now suppose that $\succsim $ is not regular. By Lemma 2 in Ribeiro and Riella (2017), there is a unique regular preference $\succsim ^R$ such that $\succsim \subseteq \succsim ^R$ and $\succ = \succ ^R$. It follows that $\mathscr {M}(A,\succsim ^R)=\mathscr {M}(A,\succsim )=c(A)$ for all $A \in \mathscr {A}$. Thus, the argument used above implies $\mathscr {D}_X(\succsim ^R) = \hat{\mathscr {D}}(c)$. Moreover, $\succsim \subseteq \succsim ^R$ implies ${{\mathrm{\triangleleft \triangleright }}}\subseteq {{\mathrm{\triangleleft \triangleright }}}^R$. Therefore, by Theorem 6, $\mathscr {D}_X(\succsim ^R)$ is coarser than $\mathscr {D}_X(\succsim )$. It follows that $\hat{\mathscr {D}}(c)$ is coarser than $\mathscr {D}_X(\succsim )$.

It only remains to show that $\hat{\mathscr {D}}(c) \ne \mathscr {D}_X(\succsim )$. To see this, suppose, seeking a contradiction, that $\hat{\mathscr {D}}(c)=\mathscr {D}_X(\succsim )$. Since $\mathscr {D}_X(\succsim ^R) = \hat{\mathscr {D}}(c)$, the second part of Theorem 6 implies that ${{\mathrm{\triangleleft \triangleright }}}= {{\mathrm{\triangleleft \triangleright }}}^R$. Moreover, $\prec \,\, = \,\, \prec ^R$ obviously holds. Finally, because $\succsim $ is not regular and $\succsim ^R$ is, it must be that $\succsim \,\, \subset \,\, \succsim ^R$. Since $\{\succsim , \prec \}$ and $\{\succsim ^R, \prec ^R\}$ are partitions of ${{\mathrm{\triangleleft \triangleright }}}$ and ${{\mathrm{\triangleleft \triangleright }}}^R$, respectively, the previous comparisons imply that

$$\begin{aligned} {{\mathrm{\triangleleft \triangleright }}}= \left( \succsim \cup \prec \right) \subset \left( \succsim ^R \cup \prec ^R\right) = {{{\mathrm{\triangleleft \triangleright }}}}^R. \end{aligned}$$

Hence, ${{\mathrm{\triangleleft \triangleright }}}\ne {{\mathrm{\triangleleft \triangleright }}}^R$, obtaining the desired contradiction. $\square $

Proof of Proposition 4

Define $x {:=} \lim _{n \rightarrow \infty } x_n$. For each $n {\in } \mathbb {N}$, pick $D_n {\in } \mathscr {D}_{A_n}(\succsim _{A_n})$ such that $x_n \in D_n$. Note that $D_n$ is closed in $A_n$ by Property (ii) in Theorem 1. Since $A_n$ is compact, $D_n \in \mathscr {K}_X$. Because $\mathscr {K}_X$ is compact in the Hausdorff metric topology, the sequence $(D_n)_{n \in \mathbb {N}}$ has a subsequence $(D_{n_h})_{h \in \mathbb {N}}$ converging to some $D \in \mathscr {K}_X$.

Condition 4 implies that $D \in \mathscr {D}_A(\succsim _A)$. Since $x_{n_h} \in \mathscr {B}(D_{n_h},\succsim )$ for all $h \in \mathbb {N}$, $\lim _{h \rightarrow \infty } D_{n_h} = D$, $\lim _{h \rightarrow \infty } x_{n_h} = x$ and $\succsim $ is strongly continuous, it follows that $x \in {\mathscr {B}(D,\succsim )} = {\mathscr {B}(D,\succsim _A)}$. Thus, $x \in {\mathscr {M}(A,\succsim _A)} = {\mathscr {M}(A,\succsim )}$ by Property (iii) of Theorem 1. $\square $

Proof of Proposition 5

Suppose, seeking a contradiction, that Condition 4 in Proposition 4 is violated. This means that there exists a convergent sequence $(D_n)_{n \in \mathbb {N}}$ such that $D_n \in \mathscr {D}_{A_n}(\succsim _{A_n})$ for all $n \in \mathbb {N}$, and nevertheless, $D := \lim _{n \rightarrow \infty } D_n \not \in \mathscr {D}_A(\succsim _A)$. Since $\succsim $ is strongly continuous, D is a $\succsim $-domain and, thus, also a $\succsim _A$-domain. By Lemma 1, there exists $D' \in \mathscr {D}_A(\succsim _A)$ such that $D' \supseteq D$. Since $D \not \in \mathscr {D}_A(\succsim _A)$, it must be the case that $D' \supset D$. Pick $x \in D'{\setminus } D$. Since $x \in A \supseteq D'$ and $\lim _{n \rightarrow \infty } A_n = A$, there must be a sequence $(x_n)_{n \in \mathbb {N}}$ such that $x_n \in A_n$ for all $n \in \mathbb {N}$ and $\lim _{n \rightarrow \infty } x_n = x$. Since $\lim _{n \rightarrow \infty } x_n \not \in D$ and $\lim _{n \rightarrow \infty } D_n = D$, there must be a subsequence $(x_{n_k})_{k \in \mathbb {N}}$ such that $x_{n_k} \not \in D_{n_k}$ for all $k \in \mathbb {N}$. But $D_{n_k} \in \mathscr {D}_{A_{n_k}}(\succsim _{A_{n_k}})$, so Lemma 1 implies that there exists $y_k \in D_{n_k}$ such that $y_k \bowtie x_{n_k}$ for every $k \in \mathbb {N}$. Due to the compactness of X, there is no loss of generality in assuming that $y := \lim _{k \rightarrow \infty } y_k$ exists. Since $y_k \in D_{n_k}$ for all $k \in \mathbb {N}$ and $\lim _{k \rightarrow \infty } D_{n_k} = D$, it follows that $y \in D$. On the one hand, $y \in D \subset D'$ implies that $(x,y) \not \in \bowtie $. On the other hand, $\lim _{k \rightarrow \infty } (x_{n_k}, y_k) = (x,y)$ and $(x_{n_k},y_k) \in \,\, \bowtie $ for all $k \in \mathbb {N}$. Since $\bowtie $ is closed, it follows that $(x,y) \in \,\, \bowtie $, a contradiction. $\square $

Proof of Proposition 6

To prove the first property, I need to show that $\succsim _{(u,\mathscr {D}_X(\succsim ))} \, =\,\, \succsim $. If $x \succsim _{(u,\mathscr {D}_X(\succsim ))} y$, then there is $D \in \mathscr {D}_X(\succsim )$ such that $x,y \in D$, so $x \succsim y$ or $y \succ x$. Since $y \succ x$ would imply $u(y) > u(x)$ and $u(x) \ge u(y)$ holds, it must be that $x \succsim y$. Now suppose that $x \succsim y$. Then, by Property (i) in Theorem 1, there is $D \in \mathscr {D}_X(\succsim )$ such that $x,y \in D$. Moreover, $u(x) \ge u(y)$ since u is a Richter–Peleg representation of $\succsim $. Hence, $x \succsim _{(u,\mathscr {D}_X(\succsim ))} y$.

The second property follows directly from Theorem 2 and the fact that ${\mathscr {B}(D,\succsim )} = {{\mathrm{arg~max}}}_{x \in D}u(x)$ whenever $\succsim $ is complete on D. $\square $

Proof of Proposition 7

Without loss of generality, assume that $\succsim $ is antisymmetric (otherwise, the following argument can be applied to its indifference sets). Suppose first that $\succsim $ is near-complete. In order theoretic terms, this means that the width of $\succsim $ (i.e., the supremum over the cardinality of all $\succsim $-antichains) is some finite number n. Dilworth’s theorem asserts that a poset has width $n \in \mathbb {N}$ if and only if it can be partitioned in n chains (see section 2.5 in Harzheim (2005) for an exposition of this result). It follows that there are $\succsim $-chains $A_1,...,A_n$ such that $\{A_1,...,A_n\}$ is a partition for X. For each $k=1,...,n$, $A_k$ is a $\succsim $-domain, so Lemma 1 implies the existence of $D_k \in {\mathscr {D}_X(\succsim )}$ such that $D_k \supseteq A_k$. Since $\bigcup _{k=1}^n D_k \supseteq \bigcup _{k=1}^n A_k = X$, $\{D_1,...,D_n\}$ is a finite subcover of ${\mathscr {D}_X(\succsim )}$, establishing one direction.

For the other direction, suppose $\{D_1,...,D_n\} \subseteq \mathscr {D}_X(\succsim )$ covers X. Define $A_1 := D_1$ and, recursively, $A_k := D_k {\setminus } \bigcup _{h=1}^{k-1} D_h$ for each $k=2,...,n$. Clearly, $\{A_1,...,A_n\}$ is a collection of $\succsim $-chains that partitions X. Thus, Dilworth’s theorem implies that the width of $\succsim $ cannot exceed $n<+\infty $. It follows that $\succsim $ is near-complete. $\square $

Proof of Proposition 8

Suppose that $\succsim $ is branching and not locally finite. Then, there must be $x,y\in X$ and an injective sequence $(D_n)_{n \in \mathbb {N}}$ on $\mathscr {D}_X(\succsim )$ such that $x \succ y$ and $\{x,y\} \subseteq D_n$ for all $n \in \mathbb {N}$. Note that, since $\succsim $ is branching, $v \bowtie w$ implies $y \succ v$ and $y \succ w$ for all $v,w \in \bigcup _{n \in \mathbb {N}} D_n$.

I shall prove that $\succsim $ is not near-complete, by showing that, for each $k \in \mathbb {N}$, there exists a set of k mutually incomparable alternatives in $\bigcup _{n = 1}^k D_n$. The proof of this claim is by induction on k. For $k=1$, the claim holds trivially. Suppose that the claim holds for $k \ge 1$. That is, suppose there is a set $\{z_1,...,z_k\} \subseteq \bigcup _{n = 1}^k D_n$ such that $z_n \bowtie z_m$ for $n,m \in \{1,...,k\}$, $n \ne m$. There is no loss of generality in assuming that $z_n \in D_n$ for every $n \in \{1,...,k\}$. For each $n \in \{1,...,k\}$, pick $\tilde{z}_{k+1,n} \in D_{k+1} {\setminus } D_n$. By Lemma 1, there exists $\hat{z}_n \in D_n$ such that $\hat{z}_n \bowtie \tilde{z}_{k+1,n}$. Let $z'_n$ be a $\succsim $-worst element of $\{z_n, \hat{z}_n\}$. Since $\succsim $ is branching, the alternatives in the set $\{z'_1,...,z'_k\}$ constructed in this way are mutually incomparable. Let $z'_{k+1}$ be a $\succsim $-worst element of $\{\tilde{z}_{k+1,1},...,\tilde{z}_{k+1,k}\}$. Since $\succsim $ is branching, $z'_{k+1} \bowtie z'_n$ for all $n \in \{1,...,k\}$. It follows that $\{z'_1,...,z'_{k+1}\} \subseteq \bigcup _{n=1}^{k+1}D_n$ is a set of $k+1$ mutually incomparable alternatives. This completes the inductive argument and shows that $\succsim $ is not near-complete. $\square $

Proof of Proposition 9

By Proposition 3 in Dubra et al. (2004), $\succsim $ admits a continuous Aumann utility: a continuous function $v:C\rightarrow \mathbb {R}$ such that $x \succ y$ implies $\int _C v dx > \int _C v dy$ and $x \sim y$ implies $\int _C v dx = \int _C v dy$. Obviously, the mapping $V:X \rightarrow \mathbb {R}$ given by $V(x) := \int _C v dx$ is a Richter–Peleg representation of $\succsim $. By Proposition 6, $\left( V,\mathscr {D}_X(\succsim )\right) $ is an extended Richter–Peleg representation of $\succsim $. Thus, the claim follows directly from Definition 2. $\square $

Proof of Proposition 10

Pick a connected set $C \subseteq X$ such that $\succsim $ is not trivial on C. It is easy to check that $\succsim _C$ is a nontrivial continuous preference with open sections on C. Hence, it follows from the Theorem by Schmeidler (1971) that C is a $\succsim $-domain. By Lemma 1, there is a maximal $\succsim $-domain $D \in \mathscr {D}_X(\succsim )$ such that $D \supseteq C$. $\square $

Proof of Proposition 11

Since $\pi $ represents $\succsim $ and $\pi (z) = \mu (z)$ for every $z \in \mathscr {E}$, $\succsim $ is complete on $\mathscr {E}$. I will now show that $\mathscr {E}$ is a maximal $\succsim $-domain. Considering the definition of representation by a capacity, whenever $x \in \mathscr {E}$ and $y \in X$, the statement $x \bowtie y$ is equivalent to $\mu (x)> \pi (y) \wedge \mu (x^C) > \pi (y^C)$ or $\mu (x) \,{<}\, \pi (y) \wedge \mu (x^C) \,{<}\, \pi (y^C)$ (note that $x^C \in \mathscr {E}$ if and only if $x \in \mathscr {E}$ because unambiguous events form an algebra). Since $\mu (x^C) = 1- \mu (x)$, this is equivalent to ${\pi (y)}< {\mu (x)} < {1-\pi (y^C)}$ or ${1-\pi (y^C)}< {\mu (x)} < {\pi (y)}$. This disjunction can be written at once as

$$\begin{aligned} \min \{\pi (y), 1-\pi (y^C)\}< \mu (x) < \max \{\pi (y), 1-\pi (y^C)\}. \end{aligned}$$

So, pick $y \in X {\setminus } \mathscr {E}$ (if no such y exists, maximality holds trivially). Proposition 1 in Gul and Pesendorfer (2014) implies that

$$\begin{aligned} \mathscr {E} = \left\{ z \in X| \pi (z)+\pi (z^C) = 1\right\} . \end{aligned}$$

Thus, $y \not \in \mathscr {E}$ implies $\pi (y)+\pi (y^C) \ne 1$. Hence, defining $a := \min \{\pi (y), 1-\pi (y^C)\}$ and $b:= \max \{\pi (y), 1-\pi (y^C)\}$, it follows that $0 \le a < b \le 1$. It is well known that nonatomic probability measures are convex-ranged. As a result, it is always possible to find $x \in \mathscr {E}$ such that

$$\begin{aligned} \mu (x) = \frac{a + b}{2}. \end{aligned}$$

Because $a< \mu (x) < b$, it must be that $x \bowtie y$. It follows that $\mathscr {E}$ is a maximal $\succsim $-domain and, so, $\mathscr {E} \in \mathscr {D}_X(\succsim )$, as claimed. $\square $

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Leandro Gorno
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1.FGV/EPGE - Escola Brasileira de Economia e FinançasRio de JaneiroBrazil

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