Single-crossing choice correspondences

Abstract

We characterize choice correspondences that can be justified by a collection of preferences that satisfy the single-crossing and/or single-peak properties. These are properties vastly used in the applied literature, often in the context of parametric families of objective (utility) functions that satisfy them. Apesteguia et al. (Econometrica 85(2):661–674, 2017) characterize random utility models that satisfy those properties, and we provide similar results in the context of choice correspondences that admit a pseudo-rational or justifiable choice representation. These results might be useful for researchers in situations where it is not possible to work with random utility models or parametric families of objective (utility) functions. In addition, we use them to discuss some aspects of the connection between deterministic and stochastic choice.

Proofs

1.1 Proof of Theorem 1

[Necessity] Suppose $\mathcal {P}:=\{\succcurlyeq _{1},\dots ,\succcurlyeq _{k}\}$ is a collection of linear orders on X that satisfies the single-crossing property with respect to $\succcurlyeq ^{*}$ and is a pseudo-rational representation of c. This implies that c satisfies Path Independence (see Aizerman and Malishevski 1981; Moulin 1985). Now consider distinct $x,y,z\in X$ such that $x\succcurlyeq ^{*}y\succcurlyeq ^{*}z$ and $y\in c(\{x,y,z\})$, and an arbitrary $A\subseteq X$, such that $y\in A$. If $x\in c(A\cup \{x\})$, by the representation of c, we must have $x\succcurlyeq _{j}w$, $\forall w\in A$ (in particular $x\succcurlyeq _{j}y$), for some $\succcurlyeq _{j}\in \mathcal {P}$. Also, because of $y\in c(\{x,y,z\})$ and the single-crossing property, we must have $y\succcurlyeq _{i}x$ and $y\succcurlyeq _{i}z$ for some $i<j$. But then, again because of the single-crossing property, we must have $y\succcurlyeq _{j}z$, implying, by transitivity of $\succcurlyeq _{j}$, that $x\succcurlyeq _{j}z$ and, by the representation of c, that $x\in c(A\cup \{x,z\})$.

Similarly, if $z\in c(A\cup \{z\})$, then we must have $z\succcurlyeq _{i}w$, $\forall w\in A$ (in particular $z\succcurlyeq _{i}y$), for some $\succcurlyeq _{i}\in \mathcal {P}$. Also, because of $y\in c(\{x,y,z\})$ and the single-crossing property, we must have $y\succcurlyeq _{j}z$ and $y\succcurlyeq _{j}x$ for some $j>i$. Then, by the single-crossing property, we must have $y\succcurlyeq _{i}x$, implying, by transitivity of $\succcurlyeq _{i}$, that $z\succcurlyeq _{i}x$ and, by the representation of c, that $z\in c(A\cup \{x,z\})$.

[Sufficiency] Suppose c is a choice correspondence that satisfies Path Independence and Centrality. Define a binary relation $\succcurlyeq$ by $x\succcurlyeq y$ if, and only if, either $x\succcurlyeq ^{*}y$ and $x\in c(\{x,y\})$ or $\{x\}=c(\{x,y\})$. We begin with the following claim:

Claim 1

The relation $\succcurlyeq$ is a linear order such that c satisfies Centrality with respect to $\succcurlyeq$.

Proof

We shall first show that $\succcurlyeq$ is a linear order. Take distinct $x,y\in X$ and suppose $x\succcurlyeq ^{*}y$. If $\{y\}=c(\{x,y\})$ then $y\succcurlyeq x$, if otherwise $x\in c(\{x,y\}),$then $x\succcurlyeq y$. The case when $y\succcurlyeq ^{*}x$ is symmetrical and, given that $\succcurlyeq ^{*}$ is complete, so must be $\succcurlyeq$. It is also easy to check that $\succcurlyeq$ is antisymmetric.

To prove transitivity take $x,y,z\in X$ such that $x\succcurlyeq y$ and $y\succcurlyeq z$, which implies $x\in c(\{x,y\})$ and $y\in c(\{y,z\})$. By Path Independence we must have that neither $\{y\}=c(\{x,y,z\})$ nor $\{z\}=c(\{x,y,z\})$ is true. Suppose we have that $\{y,z\}=c(\{x,y,z\})$, which, again by Path Independence, implies $\{x,y\}=c(\{x,y\})$, $\{y,z\}=c(\{y,z\})$ and, by the definition of $\succcurlyeq$, $x\succcurlyeq ^{*}y\succcurlyeq ^{*}z$. If we take then $A=\{y\}$, Centrality towards $\succcurlyeq ^{*}$ implies $x\in c(\{x,y,z\})$, a contradiction. We must then have that $x\in c(\{x,y,z\})$ and $x\in c(\{x,z\})$. If we have $x\succcurlyeq ^{*}z$ or $\{x\}=c(\{x,z\})$, then $x\succcurlyeq z$ and there is nothing left to prove. Suppose, then, that $\{x,z\}=c(\{x,z\})$ and $z\succcurlyeq ^{*}x$. If we have that either $\{x,y\}=c(\{x,y,z\})$ or $\{x,y,z\}=c(\{x,y,z\})$, then $x\succcurlyeq ^{*}y$ and $y\succcurlyeq ^{*}z$, which is a contradiction, as $\succcurlyeq ^{*}$ is transitive. Meaning that if we have $\{x,z\}=c(\{x,z\})$, we must have $x\succcurlyeq ^{*}z$ and, consequently, $x\succcurlyeq z$.

Suppose now c satisfies Centrality with respect to $\succcurlyeq ^{*}$ and $x,y,z\in X$ are distinct alternatives such that $x\succcurlyeq y\succcurlyeq z$ and $y\in c(\{x,y,z\})$. Then, by Path Independence, $y\in c(\{x,y\})$ and $y\in c(\{y,z\})$ which implies that $x\succcurlyeq ^{*}y$ . If $\{y,z\}=c(\{y,z\})$, then we also have $y\succcurlyeq ^{*}z$ and we obtain the desired conclusion from the fact that c satisfies Centrality with respect to $\succcurlyeq ^{*}$. If $\{y\}=c(\{y,z\})$, then Path Independence implies that for any $A\subseteq X$ with $y\in A$, $z\notin c(A\cup \{z\})$ and $c(A)=c(A\cup \{z\})$. We must then have that $x\in c(A\cup \{z\})$ if $x\in c(A)$. We conclude that c satisfies Centrality with respect to $\succcurlyeq$. $\square$

From now on, whenever we refer to Centrality, we mean Centrality with respect to $\succcurlyeq$.

Define a binary relation $\succcurlyeq _{1}\subseteq X\times X$ by $x\succcurlyeq _{1}y$ if, and only if, $x\succcurlyeq y$ and $\{x\}=c(\{x,y\})$ or $y\succcurlyeq x$ and $x\in c(\{x,y\})$. We note that $\succcurlyeq _{1}$ is complete and anti-symmetric. We need the following claim:

Claim 2

The relation $\succcurlyeq _{1}$ is a linear order.

Proof

We only need to show that $\succcurlyeq _{1}$ is transitive. Suppose, thus, that $x\succcurlyeq _{1}y$ and $y\succcurlyeq _{1}z$, for some $x,y,z\in X$. If $x=y$ or $y=z$, there is nothing to prove, so suppose that $x\ne y$ and $y\ne z$. If $y\notin c(\{x,y\})$ and $z\notin c(\{y,z\})$, then Path Independence implies that $c(\{x,y,z\})=c(\{x,z\})=\{x\}$ and, consequently, $x\succcurlyeq _{1}z$. If $y\in c(\{x,y\})$ and $z\in c(\{y,z\})$, then we must have $y\succcurlyeq x$, $z\succcurlyeq y$, $x\in c(\{x,y\})$ and $y\in c(\{y,z\})$. By the transitivity of $\succcurlyeq$, we learn that $z\succcurlyeq x$. If $x\notin c(\{x,y,z\})$, Path Independence implies that $c(\{x,y,z\})=\{y,z\}$. But then Centrality plus $x\in c(\{x,y\})$ implies that $x\in c(\{x,y,z\})$, which is a contradiction. We learn that $x\in c(\{x,y,z\})$. Now Path Independence implies that $x\in c(\{x,z\})$ and, since $z\succcurlyeq x$, we learn that $x\succcurlyeq _{1}z$. Now suppose that $y\notin c(\{x,y\})$, but $z\in c(\{y,z\})$. By Path Independence, $y\notin c(\{x,y\})$ implies that $y\notin c(\{x,y,z\})$. By the definition of $\succcurlyeq _{1}$, $z\in c(\{y,z\})$ implies that $z\succcurlyeq y$ and $y\in c(\{y,z\})$. Now, by Path Independence, we must have $x\in c(\{x,y,z\})$, which, again by Path Independence, gives us that $x\in c(\{x,z\})$. We cannot have $x\succcurlyeq z$ and $z\in c(\{x,y,z\})$, otherwise Centrality would imply that $y\in c(\{x,y,z\})$ which is not true. By Path Independence, $\{x\}=c(\{x,z\})$ if $x\succcurlyeq z$, which implies that $x\succcurlyeq _{1}z$. Since $x\in c(\{x,z\})$, we also get that $x\succcurlyeq _{1}z$ if $z\succcurlyeq x$. We are left with the case $y\in c(\{x,y\})$ and $z\notin c(\{y,z\})$. In this case, we must have $y\succcurlyeq x$ and $x\in c(\{x,y\})$. Also, Path Independence implies that $z\notin c(\{x,y,z\})$. Now Path Independence implies that $c(\{x,y,z\})=\{x,y\}$ and another application of Path Independence implies that $x\in c(\{x,z\})$. If $z\succcurlyeq x$, we immediately get $x\succcurlyeq _{1}z$. If $x\succcurlyeq z$, then Centrality would imply that $z\in c(\{x,y,z\})$ if $z\in c(\{x,z\})$. Since $z\notin c(\{x,y,z\})$, we learn that $z\notin c(\{x,z\})$ and, consequently, $x\succcurlyeq _{1}z$. We conclude that $\succcurlyeq _{1}$ is a linear order. $\square$

We can now prove the following claim:

Claim 3

For every A, if $x\in \max (A,\succcurlyeq _{1})$, then $x\in c(A)$.

Proof

This is true by definition when $|A|=2$. We proceed by induction. Suppose the claim is true whenever $|A|\le n$ and fix A such that $|A|=n+1$ and $x\in \max (A,\succcurlyeq _{1})$. If there is $y\in A{\setminus }\{x\}$ such that $x\succcurlyeq y$, then the definition of $\succcurlyeq _{1}$ implies that $y\notin c(\{x,y\})$ and, by Path Independence, $c(A)\subseteq A{\setminus }\{y\}$. Now Path Independence and the induction hypothesis imply that $x\in c(A{\setminus }\{y\})\subseteq c(A)$. The only remaining case is when $y\succcurlyeq x$ for every $y\in A{\setminus }\{x\}$. Let $z\in A$ be such that $z\succcurlyeq w$ for every $w\in A$. If $c(A)=\{z\}$, then Path Independence implies that $c(\{x,z\})=\{z\}$, which contradicts the definition of $\succcurlyeq _{1}$, since $x\succcurlyeq _{1}z$. We conclude that $c(A)\ne \{z\}$. If there exists $y\in A{\setminus }\{x,z\}$ such that $y\in c(A)$, then Centrality and the induction hypothesis imply that $x\in c(A)$. If there does not exist such a y, then we must also have $x\in c(A)$, because $c(A)\ne \{z\}$. $\square$

Now fix $n\in \mathbb {N}$ and make the following induction hypothesis:

Induction hypothesis. The collection $\{\succcurlyeq _{1},\dots ,\succcurlyeq _{n}\}$ is an ordered set of linear orders on X that satisfies single-crossingness with respect to $\succcurlyeq$ and such that

$$\begin{aligned} \bigcup _{i=1}^{n}\max (A,\succcurlyeq _{i})\subseteq c(A) \end{aligned}$$

for every choice problem A.

Enumerate the elements of X according to $\succcurlyeq _{n}$. That is, $x_{1}\succcurlyeq _{n}\dots \succcurlyeq _{n}x_{k}$. Let $i\in \mathbb {N}$ be the first natural number such that $x_{i+1}\succ x_{i}$ and $x_{i+1}\in c(\{x_{i},\dots ,x_{k}\})$. Define $\succcurlyeq _{n+1}:=(\succcurlyeq _{n}{\setminus }\{(x_{i},x_{i+1})\})\cup \{(x_{i+1},x_{i})\}$. It is easy to see that $\succcurlyeq _{n+1}$ is a linear order such that $\max (A,\succcurlyeq _{n+1})\subseteq c(A)$ for every $A\in 2^{X}{\setminus }\{\emptyset \}$ and $(\succcurlyeq _{1},\dots ,\succcurlyeq _{n+1})$ satisfies single-crossingness. This observation gives us an inductive procedure to build an ordered set $\{\succcurlyeq _{1},\dots ,\succcurlyeq _{n}\}$ of linear orders that satisfies single-crossingness and such that $\max (A,\succcurlyeq _{i})\subseteq c(A)$ for every $A\in 2^{X}{\setminus }\{\emptyset \}$ and every $i\in \{1,\dots ,n\}$.

Let $\{\succcurlyeq _{1},\dots ,\succcurlyeq _{n}\}$ be the collection of linear orders constructed by the inductive procedure above. We need the following claim:

Claim 4

If $\{\succcurlyeq _{1},\dots ,\succcurlyeq _{n}\}$ is the collection of linear orders constructed by the inductive procedure above, then $\succcurlyeq _{n}=\succcurlyeq$.

Proof

Suppose the claim is not true. Let x be the $\succcurlyeq _{n}$-minimal element for which there exists $y\in X$ with $x\succ _{n}y$, but $y\succ x$. In fact, by the minimality of x, there exists such y with x being the $\succcurlyeq _{n}$-successor⁹ of y. If there exists $z\in L(y,\succ _{n})$ with $y\succ z\succ x$ and $z\in c(\{x,y,z\})$, then $y\in c(L(x,\succcurlyeq _{n}))$ whenever $y\in c(L(x,\succcurlyeq _{n}){\setminus }\{x\})$, by Centrality. But $L(x,\succcurlyeq _{n}){\setminus }\{x\}=L(y,\succcurlyeq _{n})$ and we know that $y\in c(L(y,\succcurlyeq _{n}))$. Otherwise, let $A:=\{z\in L(y,\succ _{n}):z\notin c(L(x,\succcurlyeq _{n}))\}$. By Path Independence, $c(L(x,\succcurlyeq _{n}))=c(L(x,\succcurlyeq _{n}){\setminus } A)$. Define also $B:=\{z\in L(y,\succ _{n}):y\succ x\succ z\}$. Successive applications of Centrality give us that $y\in c(L(x,\succcurlyeq _{n}){\setminus } A)$ if $y\in c(L(x,\succcurlyeq _{n}){\setminus }(A\cup B))$. But $L(x,\succcurlyeq _{n}){\setminus }(A\cup B)=\{x,y\}$ and we know that $y\in c(\{x,y\})$, by the definition of $\succcurlyeq$. We have already seen this implies that $y\in c(L(x,\succcurlyeq _{n}){\setminus } A)=c(L(x,\succcurlyeq _{n}))$. But then we should have a relation $\succcurlyeq _{n+1}$ with $y\succ _{n+1}x$, which is a contradiction. $\square$

We now need the following claim:

Claim 5

For any distinct $x,y,z\in X$ with $x\succcurlyeq y\succcurlyeq z$ and $y\in c(\{x,y,z\})$, and any $i\in \{1,\dots ,n\}$, it cannot be true that $z\succcurlyeq _{i}x\succcurlyeq _{i}y$.

Proof

Suppose the claim is not true and let $i^{*}$ be the minimal $i\in \{1,\dots ,n\}$ such that there exist $x,y,z\in X$ with $x\succ y\succ z$, $y\in c(\{x,y,z\})$ and $z\succ _{i^{*}}x\succ _{i^{*}}y$. By the minimality of $i^{*}$ and the construction of the collection $\{\succcurlyeq _{1},\dots ,\succcurlyeq _{n}\}$, we must have that

• 1. $z\succ _{i^{*}-1}y\succ _{i^{*}-1}x$;

• 2. there exist no $w,\hat{w}\in X$ with $w\succcurlyeq _{i^{*}-1}y$, $w\succ \hat{w}$, $\hat{w}$ being the successor of w with respect to $\succcurlyeq _{i^{*}-1}$ and $w\in c(L(\hat{w},\succcurlyeq _{i^{*}-1}))$;

• 3. $x\in c(L(y,\succcurlyeq _{i^{*}-1}))$.

We claim that there must exist $w\in X$ such that $w\succ z$, $z\succ _{i^{*}-1}w\succcurlyeq _{i^{*}-1}y$ and $w\in c(L(z,\succcurlyeq _{i^{*}-1}))$. To see that, suppose that there does not exist w such that $w\succ z$, $z\succ _{i^{*}-1}w\succ _{i^{*}-1}y$ and $w\in c(L(z,\succcurlyeq _{i^{*}-1}))$. Let $A:=\{w\in X:z\succ _{i^{*}-1}w\succ _{i^{*}-1}y\text { and }w\notin c(L(z,\succcurlyeq _{i^{*}-1}))\}$. By Path Independence, $c(L(z,\succcurlyeq _{i^{*}-1}))=c(L(z,\succcurlyeq _{i^{*}-1}){\setminus } A)$. Now let $B:=\{w\in X:z\succ w\text { and }z\succ _{i^{*}-1}w\succ _{i^{*}-1}y\}$. Since $z\in c(L(z,\succcurlyeq _{i^{*}-1}))$, $z\in c(\{y,w,z\})$ for every $w\in B$. Now successive applications of Centrality give us that $y\in c(L(z,\succcurlyeq _{i^{*}-1}){\setminus } A)$ if $y\in c(L(z,\succcurlyeq _{i^{*}-1}){\setminus }(A\cup B))=c(L(y,\succcurlyeq _{i^{*}-1})\cup \{z\})$. Since $x\in c(L(y,\succcurlyeq _{i^{*}-1}))$ and $y\in c(\{x,y,z\})$, Centrality implies that $x\in c(L(y,\succcurlyeq _{i^{*}-1})\cup \{z\})$. Now let $C:=\{\hat{w}\in X{\setminus }\{y\}:x\succ _{i^{*}-1}\hat{w}\text { and }\hat{w}\notin c(L(y,\succcurlyeq _{i^{*}-1})\cup \{z\})\}$ and $D:=\{\hat{w}\in X{\setminus }\{y\}:x\succ _{i^{*}-1}\hat{w}\text { and }\hat{w}\in c(L(y,\succcurlyeq _{i^{*}-1})\cup \{z\})\}$. By Path Independence, $c(L(y,\succcurlyeq _{i^{*}-1})\cup \{z\})=c((L(y,\succcurlyeq _{i^{*}-1})\cup \{z\}){\setminus } C)$. Now, fix $\hat{w}\in D$. By Path Independence, this implies that $\hat{w}\in c(\{x,\hat{w},z\})$, so, the minimality of $i^{*}$ implies we cannot have that $x\succ \hat{w}\succ z$. That is, either $\hat{w}\succ x\succ y$ or $y\succ z\succ \hat{w}$. Now successive applications of Centrality give us that $y\in c((L(y,\succcurlyeq _{i^{*}-1})\cup \{z\}){\setminus } C)$ if $y\in c((L(y,\succcurlyeq _{i^{*}-1})\cup \{z\}){\setminus }(C\cup D))$. But notice that $(L(y,\succcurlyeq _{i^{*}-1})\cup \{z\}){\setminus }(C\cup D)=\{x,y,z\}$, so we know that $y\in c((L(y,\succcurlyeq _{i^{*}-1})\cup \{z\}){\setminus }(C\cup D))$. As we have argued above, this implies that $y\in c((L(y,\succcurlyeq _{i^{*}-1})\cup \{z\}){\setminus } C)=c(L(y,\succcurlyeq _{i^{*}-1})\cup \{z\})$. In turn, this implies that $y\in c(L(z,\succcurlyeq _{i^{*}-1}){\setminus } A)=c(L(z,\succcurlyeq _{i^{*}-1}))$ and, consequently, it is always true that there exists $w\in X$ such that $w\succ z$, $z\succ _{i^{*}-1}w\succcurlyeq _{i^{*}-1}y$ and $w\in c(L(z,\succcurlyeq _{i^{*}-1}))$, as we may have $w=y$. Fix, then, a $w\in X$ with $w\succ z$, $z\succ _{i^{*}-1}w\succcurlyeq _{i^{*}-1}y$ and $w\in c(L(z,\succcurlyeq _{i^{*}-1}))$. Let $\hat{w}$ be the successor of w with respect to $\succcurlyeq _{i^{*}-1}$. We cannot have $\hat{w}\succ w$, otherwise we would have a contradiction to the minimality of $i^{*}$. But then we arrive at a contradiction to 2 above. This proves the claim. $\square$

We can now prove the following result:

Claim 6

For any $A\in 2^{X}{\setminus }\{\emptyset \}$, if $y\in c(A),$ then there exists $\succcurlyeq _{i}\in \{\succcurlyeq _{1},\dots ,\succcurlyeq _{n}\}$ with $y\in \max (A,\succcurlyeq _{i})$.

Proof

Let $B:=\{x\in A:x\succ _{n}y\}=\{x\in A:x\succ y\}$ and $C:=\{z\in A:y\succ _{n}z\}=\{z\in A:y\succ z\}$. By Path Independence, $y\in c(\{x,y,z\})$ and $y\in c(\{x,y\})$ for every $x\in B$ and $z\in C$, and by Claim 5 for no $i\in \{1,\dots ,n\}$ it will be true that $z\succcurlyeq _{i}x\succcurlyeq _{i}y$ for some $x\in B$ and $z\in C$. By the definition of $\succcurlyeq _{1}$, we have $y\succcurlyeq _{1}x$ for every $x\in B$. If $B=\emptyset$, we could take $\succcurlyeq _{i}=\succcurlyeq _{n}$ and we would have nothing left to prove, so suppose $B\ne \emptyset$. Let $i^{*}$ be the first $i\in \{1,\dots ,n\}$ such that $x\succ _{i^{*}}y$ for some $x\in B$. By Claim 5, the minimality of $i^{*}$ and the construction of the collection $\{\succcurlyeq _{1},\dots ,\succcurlyeq _{n}\}$, we must have $y\succ _{i^{*}}w$ for every $w\in A{\setminus }\{x,y\}$. Since the only difference between $\succcurlyeq _{i^{*}-1}$ and $\succcurlyeq _{i^{*}}$ is that $y\succ _{i^{*}-1}x,$ we see that $\{y\}=\max (A,\succcurlyeq _{i^{*}-1})$. $\square$

This shows that c has the desired representation, except that, for now, we only know that $\{\succcurlyeq _{1},\dots ,\succcurlyeq _{n}\}$ satisfies single-crossingness with respect to $\succcurlyeq$. We conclude the proof with the following claim:

Claim 7

The collection $\{\succcurlyeq _{1},\dots ,\succcurlyeq _{n}\}$ satisfies single-crossingness with respect to $\succcurlyeq ^{*}$.

Proof

Suppose $x,y\in X$ are such that $x\succ ^{*}y$ and $x\succ _{i}y$ for some $i\in \{1,\dots ,n-1\}$. By the construction of the collection $\{\succcurlyeq _{1},\dots ,\succcurlyeq _{n}\}$, we must have $x\succ y$.¹⁰ Since $\{\succcurlyeq _{1},\dots ,\succcurlyeq _{n}\}$ satisfies single-crossingness with respect to $\succcurlyeq$, we have $x\succ _{j}y$ for every $j\ge i$. This shows that $\{\succcurlyeq _{1},\dots ,\succcurlyeq _{n}\}$ satisfies single-crossingness with respect to $\succcurlyeq ^{*}$. $\square$

This concludes the proof of the theorem.

1.2 Proof of Theorem 2

It is routine to show that the axioms are necessary for the representation, so we will show only that they are sufficient. For that, suppose c satisfies Strong Centrality, Path Independence, Neighbors and Betweenness Centrality. Take any $w_{1}\in c(X)$ and let $N:=|X|$. For each $n\in \{1,2,\dots ,N-1\}$, choose $w_{n+1}\in c(X{\setminus }\{w_{1},\dots ,w_{n}\})\cap N(X{\setminus }\{w_{1},\dots ,w_{n}\},w_{n})$. Notice that Neighbors guarantees there will always be a $w_{n+1}$ to be chosen this way. Let $\succcurlyeq$ be the linear order defined by $w_{i}\succcurlyeq w_{j}$ if, and only if, $i\le j$. By construction, $\succcurlyeq$ is single-peaked with respect to $\succcurlyeq ^{*}$. Moreover, Path Independence implies that, for every choice problem A, $\max (A,\succcurlyeq )\in c(A)$. Let $\mathcal {P}$ be the collection of all linear orders that may be defined by the procedure above.

Now, fix some choice problem A and some $x\in c(A)$. We may show that there exists $\succcurlyeq \in \mathcal {P}$ with $x=\max (A,\succcurlyeq )$. First note that, if $x\in c(X)$, then it is evident that we will have $x=\max (A,\succcurlyeq )$ for some $\succcurlyeq$ constructed in the way described above. If, otherwise, $x\notin c(X)$, then notice that successive applications of Betweenness Centrality allow us to find a set B such that $x\in c(B)$ and $A\cup U(x,\succcurlyeq ^{*})\subseteq B$ or $A\cup L(x,\succcurlyeq ^{*})\subseteq B$. Let us assume that $A\cup U(x,\succcurlyeq ^{*})\subseteq B$. The case $A\cup L(x,\succcurlyeq ^{*})$ is similar and, therefore, omitted. If $B\cap L(x,\succ ^{*})\ne \emptyset$, let $z:=\max (B\cap L(x,\succ ^{*}),\succ ^{*})$. Strong Centrality implies that $x\in c(U(x,\succcurlyeq ^{*})\cup L(z,\succcurlyeq ^{*}))$, with the convention that $L(z,\succcurlyeq ^{*})=\emptyset$ if $B\cap L(x,\succ ^{*})=\emptyset$. By Path Independence, as $x\notin c(X)$ there must exist $w\in X{\setminus }(U(x,\succ ^{*})\cup L(z,\succcurlyeq ^{*}))$ with $w\in c(X)$. Let $w_{1}$ be the maximal such w with respect to $\succcurlyeq ^{*}$. Follow the procedure described above and build a relation $\succcurlyeq \in \mathcal {P}$ by starting from $w_{1}$ and whenever possible choosing $w_{n+1}$ so that $w_{n+1}\succ ^{*}w_{n}$. By construction, we have $x\succcurlyeq y$ for every $y\in U(x,\succcurlyeq ^{*})$, so that $x=\max (U(x,\succcurlyeq ^{*})\cup L(z,\succcurlyeq ^{*}),\succcurlyeq )$ if $L(z,\succcurlyeq ^{*})=\emptyset$. Otherwise, we have that $z\succcurlyeq y$ for every $y\in L(z,\succcurlyeq ^{*})$. It is enough, thus, to show that $x\succcurlyeq z$, when $L(z,\succcurlyeq ^{*})\ne \emptyset$. To see that, suppose that $\{w_{1},\dots ,w_{n+1}\}$ in the construction of $\succcurlyeq$ is such that $w_{n+1}=z$. Let $y\in N(X{\setminus }\{w_{1},\dots ,w_{n}\},w_{n})$ be such that $y\succ ^{*}w_{n}$. By construction, this implies that $y\succ ^{*}w_{i}$ for $i=1,\ldots ,n$ and $y\notin c(X{\setminus }\{w_{1},\dots ,w_{n}\})$. If $x\succcurlyeq ^{*}y$, since $x\in c(U(x,\succcurlyeq ^{*})\cup L(z,\succcurlyeq ^{*}))$, Path Independence implies that there exists $\hat{w}$ such that $x\succcurlyeq ^{*}\hat{w}\succ ^{*}y$ and $\hat{w}\in c(X{\setminus }\{w_{1},\dots ,w_{n}\})$. But then Strong Centrality implies that $\hat{w}\in c(X)$, which contradicts the $\succcurlyeq ^{*}$-maximality of $w_{1}$. We conclude that $y\succ ^{*}x$, which implies that $x\in \{w_{1},\dots ,w_{n}\}$ and, consequently, $x\succ z$. We conclude that $x=\max (U(x,\succcurlyeq ^{*})\cup L(z,\succcurlyeq ^{*}),\succcurlyeq )$. Since $A\subseteq B\subseteq U(x,\succcurlyeq ^{*})\cup L(z,\succcurlyeq ^{*})$, this implies that $x=\max (A,\succcurlyeq )$, as we wished.

1.3 Proof of Theorem 3

Suppose c satisfies Strong Centrality, Path Independence, Neighbors and Betweenness Centrality. We have already proved that in this setting c has a single-peaked representation. Let $\mathcal{P}$ be the collection of all linear orders that can be used in a single-peaked representation of c. Now let $\mathcal{P}':=\{\succcurlyeq _{1},\dots ,\succcurlyeq _{m}\}$ be a $\supseteq \text {-maximal}$ subset of $\mathcal {P}$ under the restriction that $\{\succcurlyeq _{1},\dots ,\succcurlyeq _{m}\}$ can be ordered so that the collection satisfies single-crossingness with respect to $\succcurlyeq ^{*}$. In the proof of Theorem 1, we have seen that the relation $\succcurlyeq _{0}$ defined by $x\succcurlyeq _{0}y$ if, and only if, $y\succcurlyeq ^{*}x$ and $x\in c(\{x,y\})$ or $\{x\}=c(\{x,y\})$ is a linear order such that $x\in \max (A,\succcurlyeq _{0})$ implies $x\in c(A)$ for every choice problem A. Moreover, since $\{\succcurlyeq _{1},\dots ,\succcurlyeq _{m}\}\subseteq \mathcal {P}$, for every $x,y\in X$ such that $x\succcurlyeq ^{*}y$ and $x\succcurlyeq _{0}y$, we must have that $x\succcurlyeq _{i}y$ for $i=1,\ldots ,m$. This shows that the collection $\{\succcurlyeq _{0},\succcurlyeq _{1},\dots ,\succcurlyeq _{m}\}$ satisfies single-crossingness with respect to $\succcurlyeq ^{*}$. Suppose that $y\succcurlyeq ^{*}x\succcurlyeq ^{*}\max (X,\succcurlyeq _{0})$. Since $\max (X,\succcurlyeq _{0})\in c(X)$, successive applications of Neighbors plus Path Independence imply that $x\in c(U(x,\succcurlyeq ^{*}))$. By Path Independence, this implies that $x\in c(\{x,y\})$ and, consequently, $x\succcurlyeq _{0}y$. Finally, suppose that $\max (X,\succcurlyeq _{0})\succcurlyeq ^{*}x\succcurlyeq ^{*}y$. If $x=\max (X,\succcurlyeq _{0})$ or $x=y$, we have nothing to prove, so suppose that $\max (X,\succcurlyeq _{0})\succ ^{*}x\succ ^{*}y$. This now implies that $y\notin c(\{y,\max (X,\succcurlyeq _{0})\})$. If we had $y\in c(\{x,y\})$, Strong Centrality would imply that $y\in c(\{x,y,\max (X,\succcurlyeq _{0})\})$, so that, by Path Independence, we would have $y\in c(\{y,\max (X,\succcurlyeq _{0})\})$, a contradiction. We conclude that $y\notin c(\{x,y\})$ and, consequently, $x\succ _{0}y$. This shows that $\succcurlyeq _{0}$ satisfies single-peakedness with respect to $\succcurlyeq ^{*}$. By the maximality of the collection $\{\succcurlyeq _{1},\dots ,\succcurlyeq _{m}\}$, we must have that $\succcurlyeq _{1}=\succcurlyeq _{0}$.

In the proof of Theorem 1, we have seen that the relation $\succcurlyeq _{m+1}$ defined by $x\succcurlyeq _{m+1}y$ if, and only if, $x\succcurlyeq ^{*}y$ and $x\in c(\{x,y\})$ or $\{x\}=c(\{x,y\})$ is a linear order such that $x\in \max (A,\succcurlyeq _{m+1})$ implies $x\in c(A)$ for every choice problem A. Since, for every $x,y\in X$ with $x\succ ^{*}y$, but $y\succ _{m+1}x$, we must have $x\notin c(\{x,y\})$, for every such pair of alternatives we must have that $y\succ _{i}x$ for $i=1,\ldots ,m$. This shows that the collection $\{\succcurlyeq _{1},\dots ,\succcurlyeq _{m},\succcurlyeq _{m+1}\}$ satisfies single-crossingness with respect to $\succcurlyeq ^{*}$. An argument symmetric to the one used in the previous paragraph shows that $\succcurlyeq _{m+1}$ satisfies single-peakedness with respect to $\succcurlyeq ^{*}$. By the maximality of the collection $\{\succcurlyeq _{1},\dots ,\succcurlyeq _{m}\}$ , we must have that $\succcurlyeq _{m}=\succcurlyeq _{m+1}$.

Now suppose we have a choice problem A and some $x\in c(A)$ such that $\{x\}\ne \max (A,\succcurlyeq _{i})$ for every $i\in \{1,\ldots ,m\}$. Let in fact A be a maximal set with respect to this property. Let $i^{*}$ be the first i such that there exists $y^{*}\in A$ with $y^{*}\succ ^{*}x$ and $y^{*}\succ _{i^{*}}x$. By Path Independence, $x\in c(\{x,y\})$ for every $y\in A$, so that the definitions of $\succcurlyeq _{1}$ and $\succcurlyeq _{m}$ imply that $i^{*}$ is well-defined and $i^{*}\in \{2,\ldots ,m\}$. Given the maximality of A and since c satisfies Betweenness Centrality, we must have $U(x,\succcurlyeq ^{*})\subseteq A$ or $L(x,\succcurlyeq ^{*})\subseteq A$. Assume that $U(x,\succcurlyeq ^{*})\subseteq A$. Since $\succcurlyeq _{i^{*}}$ is single-peaked with respect to $\succcurlyeq ^{*}$, we must have that $\max (X,\succcurlyeq _{i^{*}})\succ ^{*}x$. This now implies that $x\succcurlyeq _{i^{*}}y$ for every $y\in L(x,\succcurlyeq ^{*})$. Similarly, since $\succcurlyeq _{i^{*}-1}$ is also single-peaked with respect to $\succcurlyeq ^{*}$ and $x\notin \max (A,\succcurlyeq _{i^{*}-1})$, we must have that $x\succ ^{*}\max (X,\succcurlyeq _{i^{*}-1})$. This now implies that $x\succcurlyeq _{i^{*}-1}y$ for every $y\in U(x,\succcurlyeq ^{*})$. Pick $z\in N(A,x)$ with $x\succ ^{*}z$.¹¹ Strong Centrality and the maximality of A imply that $A=U(x,\succcurlyeq ^{*})\cup L(z,\succcurlyeq ^{*})$. We need the following claim:

Claim 1

For every $y\in X{\setminus } A$, we have $y\succ _{i^{*}-1}\max (A,\succcurlyeq _{i^{*}-1})$.

Proof

Suppose that there exists $y\in X{\setminus } A$ with $\max (A,\succcurlyeq _{i^{*}-1})\succ _{i^{*}-1}y$. Note that any such y must satisfy that $x\succ _{i^{*}}y$, since $x\succ ^{*}y$ and $x\succ _{i^{*}-1}y$. If there exists such a y with $y\in c(L(\max (A,\succcurlyeq _{i^{*}-1}),\succcurlyeq _{i^{*}-1}))$, this contradicts the maximality of A, because single-crossingness implies that $\max (A,\succcurlyeq _{i^{*}-1})\succ _{j}y$ for every $j\le i^{*}-1$, since $y\succ ^{*}\max (A,\succcurlyeq _{i^{*}-1})$, and $x\succ _{j}y$ for every $j\ge i^{*}$. If $y\notin c(L(\max (A,\succcurlyeq _{i^{*}-1}),\succcurlyeq _{i^{*}-1}))$ for every such y, then Path Independence implies that $x\in c(L(\max (A,\succcurlyeq _{i^{*}-1}),\succcurlyeq _{i^{*}-1}))=c(A)$ and we again have a contradiction to the maximality of A. $\square$

The claim above and the single-peakedness of $\succcurlyeq _{i^{*}-1}$ imply that $x\succ ^{*}\max (X,\succcurlyeq _{i^{*}-1})\succcurlyeq ^{*}y$ for every $y\in A$ with $x\succ ^{*}y$.¹² Now let $z^{*}$ be the worst element in X, according to $\succcurlyeq ^{*}$, with $z^{*}\succ _{i^{*}-1}x$. Note that $z^{*}\in A$ and $x\succ ^{*}\max (X,\succcurlyeq _{i^{*}-1})\succcurlyeq ^{*}z^{*}$. Moreover, $L(z^{*},\succcurlyeq _{i^{*}-1})\subseteq A$, which implies that $x\in c(L(z^{*},\succcurlyeq _{i^{*}-1})\cup \{x\})$. Let $\hat{\succcurlyeq }:=(\succcurlyeq _{i^{*}-1}{\setminus }\{(z^{*},x)\})\cup \{(x,z^{*})\}$. It is easy to check that $\hat{\succcurlyeq }\in \mathcal {P}$. Moreover, the ordered collection $\{\succcurlyeq _{1},\ldots ,\succcurlyeq _{i^{*}-1},\hat{\succcurlyeq },\succcurlyeq _{i^{*}},\ldots ,\succcurlyeq _{m}\}$ satisfies single-crossingness with respect to $\succcurlyeq ^{*}$. Since $\max (X,\succcurlyeq _{i^{*}})\succ ^{*}x$, we also know that $\hat{\succcurlyeq }\ne \succcurlyeq _{i^{*}}$. This contradicts the maximality of $\{\succcurlyeq _{1},\ldots ,\succcurlyeq _{m}\}$, so we conclude that $\{\succcurlyeq _{1},\ldots ,\succcurlyeq _{m}\}$ is a representation of c that satisfies single-crossingness and single-peakedness with respect to c. The case when $L(x,\succcurlyeq ^{*})\subseteq A$ is similar.

1.4 Proof of Theorem 4

It is routine to show that the axioms are necessary for the representation, so we will show only that they are sufficient. Suppose c satisfies Path Independence and Extremality. We already know Path Independence is sufficient to guarantee that c has a pseudo-rational representation. We will show such a representation must satisfy single-dippedness towards $\succcurlyeq ^{*}$. Let $\mathcal{R}=\{\succcurlyeq _{1},\dots ,\succcurlyeq _{n}\}$ be a pseudo-rational representation of c and take $x,y\in X$, such that $x\succ ^{*}y$. Take an arbitrary $\succcurlyeq _{i}\in \mathcal{R}$ and let $w:=\min (X,\succcurlyeq _{i})$. If we have $w\succcurlyeq ^{*}x\succcurlyeq ^{*}y$, Extremality implies $x\notin c(\{w,x,y\})$ and, therefore, we cannot have $x\succcurlyeq _{i}y$. If we have $x\succcurlyeq ^{*}y\succcurlyeq ^{*}w$, then Extremality gives us that $y\notin c(\{x,y,w\})$ and, therefore, we cannot have $y\succcurlyeq _{i}x$, and $\succcurlyeq _{i}$ must satisfy single-dippedness towards $\succcurlyeq ^{*}$. As we chose $\succcurlyeq _{i}$ arbitrarily, we must have that $\mathcal{R}$ satisfies single-dippedness towards $\succcurlyeq ^{*}$.