Do we agree? Measuring the cohesiveness of preferences

Abstract

The closeness (or cohesiveness) of preferences in a preference profile has mainly been measured by aggregating the distances between each pair of preferences. We argue in this paper that some important information is lost in this process and we opt for considering the profile as a whole when constructing such a measure. With this idea in mind, we propose axioms a cohesiveness measure should satisfy and show that these properties fully characterize a new family of measures.

Appendix

1.1 Proof of Theorem 1

It is easy to see that the measures belonging to $\Phi$ satisfy REP, NEU, MON, FR, RAN, and CON. Hence, we concentrate on showing that these six properties indeed imply the measure to belong to $\Phi$. The proof will be developed in three main parts:

• 1. It will be shown that, if $K$ only consists of two objects, then the cohesiveness has to be measured by $\sigma _{x,y}(P_N)$, where $K=\{x, y\}$.

• 2. It will be shown that, if $K$ consists of any arbitrary number of objects, the difference in cohesiveness between two preference profiles of the same situation $(N, K), P_N^K$ and $\bar{P}_N^K$, has to take the form of the difference between the formulae of any cohesiveness measure of the family $\Phi$.

• 3. It will be shown that the minimal cohesiveness is always attained when $\sigma _{x,y}= 0$ for all $\{x, y\} \in \bar{K}$. Plugging this result into step 2 yields the theorem.

Part 1: We show first in Lemma 1 that the cohesiveness for a subset of objects $K=\{x,y\}$ is zero if and only if $x \sim _{P_N} y$. Afterwards, in Lemma 2, we establish that the cohesiveness is maximal if and only if all individuals in the society agree on how to order $x$ and $y$. Finally, Proposition 2 establishes that, for all preference profiles $P_N^{\{x,y\}}, M(P^{\{x,y\}}_N)=\sigma _{x,y}(P_N)$.

Lemma 1

Suppose that the cohesiveness measure $M$ satisfies REP, MON, and FR. Then, for all situations $(N,\{x,y\})$ and all profiles $P \in \mathcal{{P}}^{{\mathbb {N}}}, M(P^{\{x,y\}}_N)=0$ if, and only if, $x \sim _{P_N} y$.

Proof

The proof of the lemma is divided into two parts.

$\Rightarrow$ Suppose otherwise, that is, there is some situation $(N,\{x,y\})$ and some profile $P \in \mathcal{{P}}^{{\mathbb {N}}}$ such that $\lnot (x \sim _{P_N} y)$ and $M(P_N^{\{x,y\}})=0$. Suppose first that $x \succ _{P_N} y$. Consider a society $A$ (of size $a=3n$) and a preference profile $\bar{P}_A^{\{x,y\}}$ such that $\bar{P}_A^{\{x,y\}}$ consists of the union of three isomorphic and disjoint copies of $P_N^{\{x,y\}}$. Since $x \succ _{P_N} y$ is equivalent to $n_{x,y}(P_N)>0$ by definition, it must be the case that $n_{x,y}(\bar{P}_A)\ge 3$. Now, let the preference profile $\tilde{P}_A^{\{x,y\}}$ be such that $\bar{P}_A^{\{x,y\}}$ is $(x,y)$-different from $\tilde{P}_A^{\{x,y\}}$ for some individual $i \in A$. We have that $n_{\{x,y\}}(\tilde{P}_A)>0$ and, therefore, $M(\bar{P}_A^{\{x,y\}}) > M(\tilde{P}_A^{\{x,y\}})$ by MON. Since $M(\bar{P}_A^{\{x,y\}})=M(P_N^{\{x,y\}})$ by REP and $M(P_N^{\{x,y\}})=0$ by assumption, it follows that $M(\tilde{P}_A^{\{x,y\}}) <0$. This contradicts the definition of $M$. If $y \succ _{P_N} x$, a similar reasoning applies.

$\Leftarrow$ Take any situation $(N,\{x,y\})$ and any profile $P \in \mathcal{{P}}^{{\mathbb {N}}}$ with the property that $x \sim _{P_N} y$. By FR, there is some society $B$ (of size $b$) and some preference profile $\bar{P}_B^{\{x,y\}}$ for which $M(\bar{P}_B^{\{x,y\}})=0$. By the first part of this proof, $x \sim _{\bar{P}_B} y$. Consider now the society $C$ (of size $c=b \cdot n$) and the preference profile $\tilde{P}_C^{\{x,y\}}$ which consists of the union of $n$ isomorphic and disjoint copies of $\bar{P}_B^{\{x,y\}}$. Then, $M(\tilde{P}_C^{\{x,y\}}) = M(\bar{P}_B^{\{x,y\}}) = 0$ by REP. Since $n_{x,y}(\tilde{P}_C)=n_{x,y}(P_N)=0$ by construction, $\tilde{P}_C^{\{x,y\}}$ consists of the union of $b$ isomorphic and disjoint copies of $P_N^{\{x,y\}}$. Hence, $M(\tilde{P}_C^{\{x,y\}})=M(P_N^{\{x,y\}})$ by REP. Thus, $M(P_N^{\{x,y\}}) = 0$.

This concludes the proof of the lemma. $\square$

Lemma 2

Suppose that the cohesiveness measure $M$ satisfies REP, NEU, FR, and MON. Then, for all situations $(N,\{x,y\})$ and all profiles $P \in \mathcal{{P}}^{{\mathbb {N}}}, M(P_N^{\{x,y\}})=1$ if, and only if, $|n_{x,y}(P_N)|=n$.

Proof

The proof of the lemma is divided into two parts.

$\Rightarrow$ Suppose otherwise, that is, there is some situation $(N,\{x,y\})$ and some profile $P \in \mathcal{{P}}^{{\mathbb {N}}}$ such that $| n_{x,y}(P_N)| <n$ and $M(P_N^{\{x,y\}})=1$. By NEU, assume without loss of generality that $0 \le n_{x,y}(P_N) < n$. Hence, there is some individual $i \in N$ who prefers $y$ to $x$, while, at the same time, $x\succsim _{P_N}y$. If $x\succ _{P_N}y$, consider the preference profile $\bar{P}_N^{\{x,y\}}$ such that $\bar{P}_N^{\{x,y\}}$ is $(x,y)$-different from $P_N^{\{x,y\}}$ for individual $i$. It follows from MON that $M(\bar{P}_N^{\{x,y\}})>M(P_N^{\{x,y\}})=1$. This contradicts the definition of $M$. On the other hand, if $x\sim _{P_N}y$, it follows from Lemma 1 that $M(P_N^{\{x,y\}})=0$. This contradicts that $M(P_N^{\{x,y\}})=1$.

$\Leftarrow$ Take any situation $(N,\{x,y\})$ and any profile $P \in \mathcal{{P}}^{{\mathbb {N}}}$ with the property that $|n_{x,y}(P_N)|=n$. By NEU, assume without loss of generality that $n_{x,y}(P_N)=n$. By FR, there is some society $B$ (of size $b$) and some preference profile $\bar{P}_B^{\{x,y\}}$ for which $M(\bar{P}_B^{\{x,y\}})=1$. By the first part of this proof, $\bar{P}_B^{\{x,y\}}$ must be unanimous; that is, $|n_{x,y}(\bar{P}_B)|=b$. By NEU, assume without loss of generality that $n_{x,y}(\bar{P}_B)=b$. Consider now the society $C$ (of size $c=b \cdot n$) and the preference profile $\tilde{P}_C^{\{x,y\}}$ which consists of the union of $n$ isomorphic and disjoint copies of $\bar{P}_B^{\{x,y\}}$. Then, $M(\tilde{P}_C^{\{x,y\}})=M(\bar{P}_B^{\{x,y\}})=1$ by REP. Finally, since $\tilde{P}_C^{\{x,y\}}$ consists of the union of $b$ isomorphic and disjoint copies of $P_N^{\{x,y\}}$, we have that $M(\tilde{P}_C^{\{x,y\}})=M(P^{\{x,y\}}_N)$ by REP. Hence, $M(P_N^{\{x,y\}})=1$.

This concludes the proof of the lemma. $\square$

Now, we are ready to provide a full characterization of the measure in case the subset only contains two objects.

Proposition 2

If the cohesiveness measure $M$ satisfies REP, NEU, MON, FR, RAN, and CON, then, for all $x, y \in X$, all $N \subset {\mathbb {N}}$ and all profiles $P \in \mathcal{{P}}^{{\mathbb {N}}}$,

$$\begin{aligned} M\left( P^{\{x,y\}}_N\right) =\sigma _{x,y}(P_N). \end{aligned}$$

Proof

Take any situation $(N,\{x,y\})$ and any profile $P \in \mathcal{P}^{{\mathbb {N}}}$. Since REP implies that $M^{N,\{x,y\}}$ is independent of the identity of the individuals (the classical Anonymity property), the measure $M^{N,\{x,y\}}$ can be written as a function of the number of individuals who prefer $x$ over $y$ [i.e., $n_{x,y}(P_N)$] and the number of individuals who prefer $y$ over $x$ [i.e., $n_{y,x}(P_N)=n-n_{x,y}(P_N)$]. Given that $n$ is fixed for $M^{N, \{x, y\}}$ and that the measure is also symmetric in the objects (by NEU), we have that $M^{N,\{x,y\}}$ only depends on $|n_{x,y}(P_N)|$. The remainder of the proof is now divided into two parts, depending on whether $n$ is even or odd.

• a. Suppose that $n$ is even. If $n=2$, the values of $M^{N,\{x,y\}}$ are entirely determined by the first two lemmas. Suppose that $n=4$. Let the preference profiles $P^{\{x,y\}}_N, \bar{P}^{\{x,y\}}_N$, and $\hat{P}^{\{x,y\}}_N$ be such that $|n_{x,y}(P_N)|=0, |n_{x,y}(\bar{P}_N)|=2$, and $|n_{x,y}(\hat{P}_N)|=4$. Given that $M(P_N^{\{x, y\}})=0$ by Lemma 1, it follows from CON that $M(\hat{P}^{\{x,y\}}_N)=2 \cdot M(\bar{P}^{\{x,y\}}_N)$. We also know from Lemma 2 that $M(\hat{P}^{\{x,y\}}_N) = 1$. Then, we obtain that $M(\bar{P}^{\{x,y\}}_N)=\frac{1}{2}$. This concludes the proof for the case when $n=4$.

  Finally, suppose from now on that $n \ge 6$. Let the preference profiles $P^{\{x,y\}}_N, \bar{P}^{\{x,y\}}_N, \hat{P}^{\{x,y\}}_N$, and $\tilde{P}^{\{x,y\}}_N$ be such that $|n_{x,y}(P_N)|=0, |n_{x,y}(\bar{P}_N)|=2, |n_{x,y}(\hat{P}_N)|=4$, and $6 \le |n_{x,y}(\tilde{P}_N)|=b \le n$. We know from Lemma 1 that $M(P_N^{\{x, y\}})=0$. Then, it follows from CON that $M(\hat{P}^{\{x,y\}}_N)=2 \cdot M(\bar{P}^{\{x,y\}}_N)$. The iterative application of RAN implies in addition that $M(\tilde{P}^{\{x,y\}}_N)= \frac{b}{2} \cdot M(\bar{P}^{\{x,y\}}_N)$. By Lemma 2, we have that if $b=n, M(\tilde{P}^{\{x,y\}}_N)= 1$. Therefore, we have that $M(\bar{P}^{\{x,y\}}_N)=\frac{2}{n}=\sigma _{x,y}(\bar{P}_N)$. Then, we obtain that $M(\hat{P}^{\{x,y\}}_N)= \frac{4}{n}=\sigma _{x,y}(\hat{P}_N)$ and that, for any $b \le n, M(\tilde{P}^{\{x,y\}}_N)=\frac{b}{n}=\sigma _{x,y}(\tilde{P}_N)$.

• b. Suppose that $n$ is odd. Consider any society $A$ (of size $a=2n$) and the preference profile $\bar{P}_A^{\{x,y\}}$ that consists of the union of two isomorphic and disjoint copies of $P_N^{\{x,y\}}$. By REP, $M(\bar{P}_A^{\{x,y\}})=M(P_N^{\{x,y\}})$. Since $|n_{x,y}(\bar{P}_A)|= 2 \cdot |n_{x,y}(P_N)|$ by construction and $M(\bar{P}_A^{\{x,y\}})=\frac{|n_{x,y}(\bar{P}_A)|}{a}$ by Case (a), it follows that $M(P_N^{\{x,y\}})=\frac{|n_{x,y}(\bar{P}_A)|}{a}=\frac{|n_{x,y}(P_N)|}{n} = \sigma _{x,y}(P_N)$.

This concludes the proof of the proposition. $\square$

Part 2: We focus now on the general case where $K$ is allowed to be any subset of $X$ with at least two objects. Our particular objective will be to calculate the difference in the cohesiveness between any two preference profiles of the same situation $(N,K)$. In the main step of the proof, we are going to show that this difference can be expressed as a weighted average over the values of $\sigma$ for all possible subsets of two objects, where the weight $\omega _k$ assigned to the subset $\{x,y\}$ in the preference profile $P_N^K$ depends on $r_{x,y}(P^K_N)=p_x(P^K_N)+p_y(P^K_N)+\frac{1}{2}(i_x(P^K_N)+i_y(P^K_N))$. Finally, we will see that $\omega _k(r_{x,y}(P_N^K)) = 1 + 2\,\alpha _k \,(r_{x,y}(P_N^K) - 1)$ for some $\alpha _k \ge 0$. To be more concrete, the objective is to prove the following proposition.

Proposition 3

If the cohesiveness measure $M$ satisfies REP, NEU, MON, FR, RAN, and CON, then there exists a sequence $\alpha \equiv \{\alpha _k\}_{k\in {\mathbb {N}}, k\ge 2}$, with $\alpha _k \ge 0$ for all $k \in {\mathbb {N}}$, such that for all situations $(N,K)$ and all profiles $\bar{P}, P \in \mathcal{{P}}^{{\mathbb {N}}}$, the difference between $M(\bar{P}_N^K)$ and $M(P_N^K)$ is equal to

$$\begin{aligned}&\frac{\sum \limits _{\{x,y\} \in \bar{K}} \sigma _{x,y}(\bar{P}_N) \cdot \left[ 1\,+\, 2\,\alpha _k\,\left( r_{x,y}\left( \bar{P}^K_N\right) -1\right) \right] }{\sum \limits _{\{x,y\} \in \bar{K}} \left[ 1+ 2\,\alpha _k\,\left( r_{x,y}\left( \bar{P}^K_N\right) -1\right) \right] }\\&\qquad \qquad -\, \frac{\sum \limits _{\{x,y\} \in \bar{K}} \sigma _{x,y}(P_N) \cdot \left[ 1+ 2\,\alpha _k\,\left( r_{x,y}\left( P^K_N\right) -1\right) \right] }{\sum \limits _{\{x,y\} \in \bar{K}} \left[ 1+ 2\,\alpha _k\,\left( r_{x,y}\left( P^K_N\right) -1\right) \right] }. \end{aligned}$$

The proof of this proposition is rather lengthy as we will have to establish several claims as we proceed. In order to maintain the overall readability, the argument will therefore be developed in three different subparts.

Part 2.1: We decompose the difference ${M(\bar{P}_N^K)-M(P_N^K)}$ into different parts

Observe that it is possible to arrive at $\bar{P}^K_N$ starting from $P^K_N$ by means of a (not necessarily unique) succession of $m$ changes of contiguous pairs of objects. During the proof, we are going to decompose the difference $M(\bar{P}^K_N)-M(P^K_N)$ into $\frac{k (k-1)}{2}$ quantities, $\left\{ A_{x,y}\right\} _{\{x,y\} \in \bar{K}}$, where we accumulate the differences that accrue during this succession of changes. Suppose that, in the first change, the preference profile $P^K_N$ is changed to $\tilde{P}^K_N$, where $\tilde{P}^K_N$ is $(x,y)$-different from $P^K_N$ for some individual $i \in N$. There are four possible cases:

• a. If $x \succ _{P_N} y$ (and, then, $x \succ _{\tilde{P}_N} y$), then $M(\tilde{P}^K_N)>M(P^K_N)$ by MON. By RAN, the change in the measure only depends on the numbers $p_x(P^K_N), p_y(P^K_N), i_x(P^K_N)$, and $i_y(P^K_N)$. Consequently, we have that $M(\tilde{P}^K_N)-M(P^K_N)= \delta ^K_N(p_x(P^K_N), p_y(P^K_N), i_x(P^K_N), i_y(P^K_N))$. We also define the function $\rho ^K_N$ as $\rho ^K_N(\cdot ) = \frac{n}{2} \,\delta ^K_N(\cdot )$. We classify the difference between $M(\tilde{P}^K_N)$ and $M(P^K_N)$ as follows: $A_{x,y}^1=\frac{2}{n} \, \rho ^K_N(p_x(P^K_N), p_y(P^K_N), i_x(P^K_N), i_y(P^K_N))$ and $A^1_{w,z}=0$ for all $\{w,z\} \in \bar{K} \setminus \{\{x,y\}\}$.

• b. If $x \sim _{P_N} y$ (and, then, $x \succ _{\tilde{P}_N} y$), the properties of RAN and CON imply together that the change in the measure, $M(\tilde{P}^K_N)-M(P^K_N)$, is equal to $\delta ^K_N(p_x(\tilde{P}^K_N), p_y(\tilde{P}^K_N), i_x(\tilde{P}^K_N),i_y(\tilde{P}^K_N)) + t^K_N \, \sum \limits _{z \in K \setminus \{x, y\}} (M(P^{\{x, z\}}_N) - M(P^{\{y, z\}}_N))$. It can be seen that the sum consists of two parts: The function $\delta ^K_N(\cdot )$ refers to the change in the measure in case $x$ would win against $y$ in the pairwise comparison ceteris paribus, while the second part adapts $\delta ^K_N(\cdot )$ to the fact that the two objects tie at $P^K_N$. Also, by Proposition 2, for all $z \in K \setminus \{x, y\}, M(P^{\{x, z\}}_N) = \sigma _{x, z}(P_N)$ and $M(P^{\{y, z\}}_N) = \sigma _{y, z}(P_N)$. If we define the function $\rho ^K_N(\cdot )$ in the same way as before, the difference between $M(\tilde{P}^K_N)$ and $M(P^K_N)$ can be classified as follows: $A_{x,y}^1=\frac{2}{n} \, \rho ^K_N(p_x(\tilde{P}^K_N), p_y(\tilde{P}^K_N), i_x(\tilde{P}^K_N),i_y(\tilde{P}^K_N)), A_{x,z}^1 = t^K_N \, \sigma _{x, z} (P_N)$, and $A_{y,z}^1 = -t^K_N \, \sigma _{y, z} (P_N)$ for all $z \in K \setminus \{x, y\}$, and $A_{v, w}^1 = 0$ for all $\{v, w\} \in \bar{T}$, where $T = K \setminus \{x, y\}$.

• c. If $y \succ _{\tilde{P}_N} x$ (and, then, $y \succ _{P_N} x$), $M(P^K_N)>M(\tilde{P}^K_N)$ by MON. It follows then from RAN that $M(\tilde{P}^K_N)-M(P^K_N)=\, - \, \delta ^K_N(p_y(\tilde{P}^K_N),p_x(\tilde{P}^K_N),i_y(\tilde{P}^K_N), i_x(\tilde{P}^K_N))$. Since $p_z(\tilde{P}^K_N) = p_z(P^K_N)$ and $i_z(\tilde{P}^K_N) = i_z(P^K_N)$ for all $z \in \{x, y\}$, we can classify the difference between $M(\tilde{P}^K_N)$ and $M(P^K_N)$ in the following way: $A_{x,y}^1=-\frac{2}{n} \, \rho ^K_N(p_y(P^K_N), p_x(P^K_N), i_y(P^K_N), i_x(P^K_N))$ and $A^1_{w,z}=0$ for all $\{w,z\} \in \bar{K} \setminus \{\{x,y\}\}$.

• d. If $x \sim _{\tilde{P}_N} y$ (and, then, $y \succ _{P_N} x$), the properties of RAN and CON imply together that $M(\tilde{P}^K_N) - M(P^K_N) = \, - \, \delta ^K_N(p_y(P^K_N), p_x(P^K_N), i_y(P^K_N), i_x(P^K_N)) - t^K_N \, \sum \limits _{z \in K \setminus \{x, y\}} (M(\tilde{P}^{\{y, z\}}_N) - M(\tilde{P}^{\{x, z\}}_N))$. Also, by Proposition 2, $M(\tilde{P}^{\{w, z\}}_N) = M(P^{\{w, z\}}_N) = \sigma _{w,z}(P_N)$ for all $w \in \{x,y\}$ and $z \in K \setminus \{x,y\}$. Consequently, the difference in the cohesiveness between the two preference profiles can be decomposed as follows: $A_{x,y}^1 = - \frac{2}{n} \, \rho ^K_N(p_y(P^K_N), p_x(P^K_N), i_y(P^K_N), i_x(P^K_N)), A_{x,z}^1 = -t^K_N \, \sigma _{x, z} (P_N)$, and $A_{y,z}^1 = t^K_N \, \sigma _{y, z} (P_N)$ for all $z \in K \setminus \{x, y\}$, and $A_{v, w}^1 = 0$ for all $\{v, w\} \in \bar{T}$, where $T = K \setminus \{x, y\}$.

Next, if we repeat the same steps for the remaining $m-1$ changes in order to arrive at the profile $\bar{P}^K_N$, we see that the difference $M(\bar{P}^K_N) - M(P^K_N)$ can be decomposed as follows: $\sum \limits _{\{x, y\} \in \bar{K}} \, \sum \limits _{i=1}^m A^i_{x,y}$. This concludes Part 2.1.

Part 2.2: Four symmetry properties of the function $\underline{\rho }$ are established

Since we have calculated the difference $M(\bar{P}^K_N) - M(P^K_N)$ in a very direct way, the formula obtained does not provide many insights on how to calculate the cohesiveness. In four claims, we are going to establish symmetry properties of the function $\rho ^K_N$ that will help us derive the exact weights and thereby to enrich the expression obtained at the end of Part 2.1.

Claim 1

For all $K \subset X$, all $N, N' \subset {\mathbb {N}}$ and any possible $(a, b, c, d), \rho ^K_N(a,b,c, d) = \rho ^K_{N'}(a,b,c,d)$.

Proof

Consider any preference profiles $P^K_N$ and $\bar{P}^K_N$ and any two objects $x, y \in K$ satisfying that $x \succ _{P_N} y, p_x(P^K_N) = a, p_y(P^K_N) = b, i_x(P_N^K) = c, i_y(P^K_N) = d$ and such that $\bar{P}^K_N$ is $(x, y)$-different from $P^K_N$ for some individual $i \in N$. Then, by Case (a) in the proof of Part 2.1, $M(\bar{P}^K_N) - M(P^K_N) = \frac{2}{n} \cdot \rho ^K_N(a,b,c,d)$. Consider now any other two preference profiles $Q^K_A$ and $\bar{Q}^K_A$, corresponding to a society of size $\# A = n \cdot n'$, that consist of the union of $n'$ isomorphic and disjoint copies of $P^K_N$ and $\bar{P}^K_N$, respectively. It is easy to see that $M(\bar{Q}^K_A) - M(Q^K_A) = n' \cdot \frac{2}{n \cdot n'} \cdot \rho ^K_A(a,b,c,d) = \frac{2}{n} \cdot \rho ^K_A(a,b,c,d)$. Given that both $M(\bar{Q}^K_A) = M(\bar{P}^K_N)$ and $M(Q^K_A) = M(P^K_N)$ by REP, we obtain that $\rho ^K_N(a,b,c,d) = \rho ^K_A(a,b,c,d)$.

Take now any other two preference profiles $\hat{P}^K_{N'}$ and $\tilde{P}^K_{N'}$ such that $x \succ _{\hat{P}_{N'}} y, p_x(\hat{P}^K_{N'}) = a, p_y(\hat{P}^K_{N'}) = b, i_x(\hat{P}^K_{N'}) = c, i_y(\hat{P}^K_{N'}) = d$ and such that $\tilde{P}^K_{N'}$ is $(x, y)$-different from $\hat{P}^K_{N'}$ for some individual $j \in N'$. Then, by Case (a) in the proof of Part 2.1, $M(\tilde{P}^K_{N'}) - M(\hat{P}^K_{N'}) = \frac{2}{n'} \cdot \rho ^K_{N'}(a,b,c,d)$. Consider now other two preference profiles $\hat{Q}^K_A$ and $\tilde{Q}^K_A$ that consist of the union of $n$ isomorphic and disjoint copies of $\hat{P}^K_{N'}$ and $\tilde{P}^K_{N'}$, respectively. It is easy to see that $M(\tilde{Q}^K_A) - M(\hat{Q}^K_A) = n \cdot \frac{2}{n \cdot n'} \cdot \rho ^K_A(a,b,c,d) = \frac{2}{n'} \cdot \rho ^K_A(a,b,c,d)$. Given that both $M(\hat{Q}^K_A) = M(\hat{P}^K_{N'})$ and $M(\tilde{Q}^K_A) = M(\tilde{P}^K_{N'})$ by REP, we obtain that $\rho ^K_{N'}(a,b,c,d) = \rho ^K_A(a,b,c,d)$. Since we have already shown before that $\rho ^K_N(a,b,c,d) = \rho ^K_A(a,b,c,d)$, we can finally conclude that $\rho ^K_N(a,b,c, d) = \rho ^K_{N'}(a,b,c,d)$. $\square$

As a consequence of Claim 1, we can drop from now on the subscript $N$ from the $\rho$ function. In the next two claims, it will be shown that $\rho$ has to treat the two objects within each subset of two objects neutrally; that is, the change in the measure does not depend on which of the two objects wins against or ties with a third object, and it only matters against how many of the other objects the two objects win against and tie with in total.

Claim 2

For all $K \subset X$ and all possible vectors $(a,b,c,d)$ and $(a, b, c+1, d-1), \rho ^K(a,b,c,d)=\rho ^K(a,b,c+1,d-1)$.

Proof

Fix any set $K \subset X$. Consider any preference profile $P^K_N$, corresponding to the electorate $N$ with $n \ge 3$, that satisfies the following conditions for some $x, y, z \in K$: (a) $x \succ _{P_N} y$ and $(y,x)$ is a contiguous pair of objects for individual $3$; (b) $n_{z,x}(P_N)-n_{x,z}(P_N)=2$ and $(z,x)$ is a contiguous pair of objects for individual $1$; (c) $y \sim _{P_N} z$ and $(y,z)$ is a contiguous pair of objects for individual $2$; and (d) $p_x(P^K_N)=a, p_y(P^K_N)=b, i_x(P_N^K)=c$, and $i_y(P^K_N)=d$.⁷

We also introduce the preference profile $\bar{P}^K_N$ which is equal to $P^K_N$ apart from that the three individuals mentioned before change their rankings on one pairwise comparison each: individual $3$ switches the contiguous pair $(y,x)$, individual $1$ switches the contiguous pair $(z,x)$, and individual $2$ switches the contiguous pair $(y,z)$. Consequently, we have at this preference profile that $(p_x(\bar{P}^K_N),p_y(\bar{P}^K_N),$ $i_x(\bar{P}^K_N),i_y(\bar{P}^K_N))=(a,b,c+1,d-1)$.

The proof now consists in exploiting the differences in the cohesiveness that accrue along two different paths from $P^K_N$ to $\bar{P}^K_N$. This is done in such a way that all changes along the paths apart from $\rho ^K(a,b,c,d)$ and $\rho ^K(a,b,c+1,d-1)$ cancel out. To provide the necessary intuition, individual $3$ changes her/his preferences on the contiguous pair $(y,x)$ first in the first path and last in the second path. The other two individuals change preferences in the same order: individual $1$ changes preferences before individual $2$ in both paths. In the following, we sometimes use the notation $P_{K,N}$ alternatively for $P^K_N$.

The first path

• 1. The preference profile $P^{1,1}_{K,N}$ is $(x,y)$-different from $P_{K, N}^{1,0} \equiv P_N^K$ for individual $3$. By step (a) of Part 2.1, $M(P^{1,1}_{K,N})-M(P_N^K)=\frac{2}{n} \, \rho ^K(a,b,c,d)$.

• 2. The preference profile $P^{1,2}_{K,N}$ is $(x,z)$-different from $P^{1,1}_{K,N}$ for individual $1$. By step (d) of Part 2.1, $M(P^{1,2}_{K,N})-M(P^{1,1}_{K,N})=-\frac{2}{n} \, \rho ^K(p_z(P^{1,1}_{K,N}),a, i_z(P^{1,1}_{K,N}),c)- t^K_N \sum \limits _{w \in K \setminus \{x, z\}} (\sigma _{z, w}(P^{1,1}_N) - \sigma _{x, w}(P^{1,1}_N))$.

• 3. The preference profile $\bar{P}_N^K \equiv P^{1,3}_{K,N}$ is $(z,y)$-different from $P^{1,2}_{K,N}$ for individual $2$. It can be concluded from step (b) of Part 2.1, $M(P^{1,3}_{K,N})-M(P^{1,2}_{K,N})=\frac{2}{n} \, \rho ^K(p_z(P^{1,3}_{K,N}), b, i_z(P^{1,3}_{K,N}), d) + t^K_N \sum \limits _{w \in K \setminus \{z,y\}} (\sigma _{z, w}(P^{1,2}_N) - \sigma _{y, w}(P^{1,2}_N))$. This concludes the description of the first path.

The second path

• 1. The preference profile $P^{2,1}_{K,N}$ is $(x,z)$-different from $P^K_N\equiv P^{2,0}_{K,N}$ for individual $1$. It can then be concluded from step (d) of Part 2.1 that $M(P^{2,1}_{K,N})-M(P_N^K)=-\frac{2}{n} \, \rho ^K(p_z(P^K_N),a, i_z(P^K_N),c)-t^K_N \sum \limits _{w \in K \setminus \{x, z\}} (\sigma _{z, w}(P_N) - \sigma _{x, w}(P_N))$.

• 2. The preference profile $P^{2,2}_{K,N}$ is $(z,y)$-different from $P^{2,1}_{K,N}$ for individual $2$. By step (b) of Part 2.1, $M(P^{2,2}_{K,N})-M(P^{2,1}_{K,N})=\frac{2}{n} \, \rho ^K(p_z(P^{2,2}_{K,N}),b, i_z(P^{2,2}_{K,N}),d)+ t^K_N \sum \limits _{w \in K \setminus \{z,y\}} (\sigma _{z, w}(P^{2,1}_N) - \sigma _{y, w}(P^{2,1}_N))$.

• 3. The preference profile $P^{2,3}_{N,K} \equiv \bar{P}^K_N$ is $(x,y)$-different from $P^{2,2}_{K,N}$ for individual $3$. By step (a) of Part 2.1, $M(\bar{P}_{K,N})-M(P^{2,2r}_{K,N})= \frac{2}{n} \, \rho ^K(a,b,c+1,d-1)$. This concludes the description of the second path.

Next, we compare the differences that accrue along the two paths. We have that $\rho ^K(p_z(P^{1,1}_{K,N}),a,i_z(P^{1,1}_{K,N}),c)$ is equal to $\rho ^K(p_z(P^K_N),a, i_z(P^K_N),c)$ because $z$ has not been involved in any change in preferences until that point. Similarly, $\rho ^K(p_z(P^{1,3}_{K,N}), b, i_z(P^{1,3}_{K,N}), d)$ is equal to $\rho ^K(p_z(P^{2,2}_{K,N}),b, i_z(P^{2,2}_{K,N}),d)$ because $z$ has been switched along both paths in exactly the same way: It has only been made indifferent with $x$. Taking into account these relationships, one can then see that

$$\begin{aligned} M\left( P_{K, N}^{1, 2}\right) - M\left( P_{K, N}^{1, 1}\right) = M\left( P_{K, N}^{2, 1}\right) - M\left( P_{K, N}^{2, 0}\right) + t^K_N \cdot \frac{2}{n}\\ M\left( P_{K, N}^{1, 3}\right) - M\left( P_{K, N}^{1, 2}\right) = M\left( P_{K, N}^{2, 2}\right) - M\left( P_{K, N}^{2, 1}\right) - t^K_N \cdot \frac{2}{n}. \end{aligned}$$

Consequently, all changes apart from $M(P^{1,1}_{K,N})-M(P_{K,N})$ and $M(\bar{P}_{K,N})-M(P^{2,2}_{K,N})$ cancel out along the two paths. Finally, since the starting and ending profiles in both paths are the same, it has to be the case that $M(P^{1,1}_{K,N})-M(P_{K,N}) = M(\bar{P}_{K,N})-M(P^{2,2}_{K,N})$. This equation is equivalent to $\rho ^K(a,b,c,d)=\rho ^K(a,b,c+1,d-1)$, which concludes the proof of the claim. $\square$

Claim 3

For all $K \subset X$ and all possible vectors $(a, b, c, d)$ and $(a-1, b+1, c, d), \rho ^K(a, b, c, d) = \rho ^K(a-1, b+1, c, d)$

Proof

Fix any set $K \subset X$. Consider any preference profile $P^K_N$, corresponding to the electorate $N$ with $n \ge 5$, that satisfies the following conditions for some $x, y, z \in K$: (a) $p_x(P^K_N)=a, p_y(P^K_N)=b, i_x(P^K_N)=c$, and $i_y(P^K_N)=d$; (b) $x \succ _{P_N} y$ and $(y,x)$ is a contiguous pair of objects for individual $1$; (c) $n_{z,x}(P_N)-n_{x,z}(P_N)=n_{y,z}(P_N)-n_{z,y}(P_N)=2$; (d) the ordered pair $(z,x)$ is contiguous for individuals $2$ and $4$; and (e) the ordered pair $(y,z)$ is contiguous for individuals $3$ and $5$.⁸

Consider the preference profile $\bar{P}^K_N$ that differs from $P^K_N$ only because the individuals 1 to 5 change the order of the above-mentioned contiguous pairs of objects in their preferences. We consider again two different paths from $P^K_N$ to $\bar{P}^K_N$. Individuals invert preferences along the first path following the order of the natural numbers. The second path differs from the first only because individual 1 now inverts her/his preference on the subset $\{x,y\}$ last instead of first. In the next step, we collect the differences in the cohesiveness that accrue along the two different paths.

If we denote by $P_{K,N}^{i,j}$ the profile obtained after the $j-$th change in the $i-$th path, for $i \in \{1, 2\}$ and $j \in \{1, \ldots , 5\}$,⁹ we have that

$$\begin{aligned}&M\left( P_{K, N}^{1, 2}\right) - M\left( P_{K, N}^{1, 1}\right) = M\left( P_{K, N}^{2, 1}\right) - M\left( P_{K, N}^{2, 0}\right) + t^K_N \cdot \frac{2}{n}\\&M\left( P_{K, N}^{1, 3}\right) - M\left( P_{K, N}^{1, 2}\right) = M\left( P_{K, N}^{2, 2}\right) - M\left( P_{K, N}^{2, 1}\right) - t^K_N \cdot \frac{2}{n}\\&M\left( P_{K, N}^{1, 4}\right) - M\left( P_{K, N}^{1, 3}\right) = M\left( P_{K, N}^{2, 3}\right) - M\left( P_{K, N}^{2, 2}\right) + t^K_N \cdot \frac{2}{n}\\&M\left( P_{K, N}^{1, 5}\right) - M\left( P_{K, N}^{1, 4}\right) = M\left( P_{K, N}^{2, 4}\right) - M\left( P_{K, N}^{2, 3}\right) - t^K_N \cdot \frac{2}{n}. \end{aligned}$$

Then, since the starting and ending profiles in both paths are the same, we have that $M(P_{K, N}^{1, 1}) - M(P_{K, N}^{1, 0}) = M(P_{K, N}^{2, 5}) - M(P_{K, N}^{2, 4})$. Given that $M(P_{K, N}^{1, 1}) - M(P_{K, N}^{1, 0}) = \frac{2}{n} \, \rho ^K(a,b,c,d)$ and that $M(P_{K, N}^{2, 5}) - M(P_{K, N}^{2, 4}) = \frac{2}{n} \, \rho ^K(a-1, b+1, c, d)$, we have that $\rho ^K(a,b,c,d) = \rho ^K(a-1,b+1,c,d)$, which concludes the proof of the claim. $\square$

Claims 2 and 3 imply that we can now introduce the function $\beta ^K$, which satisfies the condition that $\beta ^K(a+b, c+d)$ equals the former $\rho ^K(a, b, c, d)$. In the final claim of this part, we show that a win at the social level counts double an indifference.

Claim 4

For all $K \subset X$ and all possible vectors $(a, b)$ and $(a+1, b-2), \beta ^K(a,b)=\beta ^K(a+1,b-2)$.

Proof

Fix any set $K \subset X$. Consider any preference profile $P^K_N$, corresponding to the electorate $N$ with $n \ge 3$, that satisfies the following conditions for some $x, y, z \in K$: (a) $x \succ _{P_N} y$ and $(y,x)$ is a contiguous pair of objects for individual $1$; (b) $x \sim _{P_N} z$ and $(z,x)$ is a contiguous pair of objects for individual $2$; (c) $y \sim _{P_N} z$ and $(y,z)$ is a contiguous pair of objects for individual $3$; and (d) $p_x(P^K_N)=c, p_y(P^K_N)=d, i_x(P^K_N)=e$, and $i_y(P^K_N)=f$ such that $c+d = a$ and $e+f = b$.¹⁰ Consider also the preference profile $\bar{P}^K_N$ which is equal to $P^K_N$ apart from that individuals $1, 2$, and $3$ switch the preferences on the contiguous pairs of objects mentioned before. The proof consists again in exploiting the differences in the cohesiveness that accrue along two different paths from $P^K_N$ to $\bar{P}^K_N$.

The first path

• 1. The preference profile $P^{1,1}_{K,N}$ is $(x,y)$-different from $P^K_N$ for individual 1. By step (a) of Part 2.1 and Claims 2 and 3, $M(P^{1,1}_{K,N})-M(P^K_N)=\frac{2}{n} \, \beta ^K(a,b)$.

• 2. The preference profile $P^{1,2}_{K,N}$ is $(x,z)$-different from $P^{1,1}_{K,N}$ for individual 2. By step (b) of Part 2.1 and Claims 2 and 3, $M(P^{1,2}_{K,N})-M(P^{1,1}_{K,N})=\frac{2}{n} \, \beta ^K(c+p_{z}(P^{1,1}_{K,N})+1,e+i_z(P^{1,1}_{K,N})-2)+ t^K_N \, \sum \limits _{w \in K \setminus \{x, z\}} (\sigma _{x,w}(P^{1,1}_N) - \sigma _{z, w}(P^{1,1}_N))$.

• 3. The preference profile $\bar{P}^K_N$ is $(z,y)$-different from $P^{1,2}_{K,N}$ for individual 3. By step (b) of Part 2.1 and Claims 2 and 3, $M(\bar{P}^K_N)-M(P^{1,2}_{K,N})=\frac{2}{n} \, \beta ^K(p_z(\bar{P}^K_N)+d+1, i_z(\bar{P}^K_N)+f-2)+ t^K_N \, \sum \limits _{w \in K \setminus \{z,y\}} (\sigma _{z,w}(P^{1,2}_N) - \sigma _{y, w}(P^{1,2}_N))$.

The second path

• 1. The preference profile $P^{2,1}_{K,N}$ is $(x,z)$-different from $P^K_N$ for individual 2. By step (b) of Part 2.1 and Claims 2 and 3, $M(P^{2,1}_{K,N})-M(P^K_N)=\frac{2}{n} \, \beta ^K(c+p_z(P^K_N)+1,e+i_z(P^K_N)-2)+t^K_N \, \sum \limits _{w \in K \setminus \{x, z\}} (\sigma _{x,w}(P_N) - \sigma _{z, w}(P_N))$.

• 2. The preference profile $P^{2,2}_{K,N}$ is $(z,y)$-different from $P^{2,1}_{K,N}$ for individual 3. From step (b) of Part 2.1 and Claims 2 and 3, we get $M(P^{2,2}_{K,N})-M(P^{2,1}_{K,N})=\frac{2}{n} \, \beta ^K(p_z(P^{2,2}_{K,N})+d+1, i_z(P^{2,2}_{K,N})+f-2)+ t^K_N \, \sum \limits _{w \in K \setminus \{z,y\}} (\sigma _{z,w}(P^{2,1}_N) - \sigma _{y, w}(P^{2,1}_N))$.

• 3. The preference profile $\bar{P}^K_N$ is $(x,y)$-different from $P^{2,2}_{K,N}$ for individual 1. By step (a) of Part 2.1 and Claims 2 and 3, $M(\bar{P}^K_N)-M(P^{2,2}_{K,N})=\frac{2}{n} \, \beta ^K(a+1,b-2)$.

As in the previous claims, it is easy to see that

$$\begin{aligned}&M\left( P^{1,2}_{K,N}\right) -M\left( P^{1,1}_{K,N}\right) = M\left( P^{2,1}_{K,N}\right) -M\left( P^K_N\right) + t^K_N \cdot \frac{2}{n}\\&M\left( \bar{P}^K_N\right) -M\left( P^{1,2}_{K,N}\right) = M\left( P^{2,2}_{K,N}\right) -M\left( P^{2,1}_{K,N}\right) - t^K_N \cdot \frac{2}{n}. \end{aligned}$$

Then, $\beta ^K(a+1,b-2)=\beta ^K(a,b)$, which concludes the proof. $\square$

Claim 4 establishes that the weight of a subset $\{x, y\} \in \bar{K}$ depends on $r_{x,y}(P^K_N)$; that is, we can introduce the function $\omega ^K$ which is such that, for all $(a,b), \omega ^K(a+\frac{b}{2})=\beta ^K(a,b)$. This concludes Part 2.2.

Part 2.3: We derive the exact expression of ${M(\bar{P}^K_N)-M(P^K_N)}$

Remember that the objective of Part 2.1 has been to classify the differences accumulated in the process of arriving at $\bar{P}_N^K$ from $P_N^K$ in subsets of two objects. In effect, we have seen that $M(\bar{P}^K_N) - M(P^K_N) = \sum \limits _{\{x, y\} \in \bar{K}} \, \sum \limits _{i=1}^m A^i_{x,y}$. Using the results from Part 2.2, we then get that the accumulated difference in the subset $\{x,y\} \in \bar{K}$ along all changes of contiguous pairs equals $\sum \limits _{i=1}^m A^i_{x,y} = \sigma _{x, y}(P_N) \, (\omega ^K(r_{x,y}(\bar{P}^K_N)) - \omega ^K(r_{x,y}(P^K_N))) + (\sigma _{x, y}(\bar{P}_N) - \sigma _{x, y}(P_N)) \, \omega ^K(r_{x,y}(\bar{P}^K_N))$. Summing these differences over all possible subsets of two objects, we obtain that

$$\begin{aligned} M\left( \bar{P}_N^K\right) -M\left( P_N^K\right)= & {} \frac{\sum \limits _{\{x,y\} \in \bar{K}} \omega ^K\left( r_{x,y}\left( \bar{P}_N^K\right) \right) \cdot \sigma _{x,y}(\bar{P}_N)}{\sum \limits _{\{x,y\} \in \bar{K}} \omega ^K\left( r_{x,y}\left( \bar{P}_N^K\right) \right) }\\&- \frac{\sum \limits _{\{x,y\} \in \bar{K}} \omega ^K\left( r_{x,y}\left( P_N^K\right) \right) \cdot \sigma _{x,y}(P_N)}{\sum \limits _{\{x,y\} \in \bar{K}} \omega ^K\left( r_{x,y}\left( P_N^K\right) \right) }. \end{aligned}$$

Then, by NEU we can easily deduce that, for all $S, T \subset X$ such that $\# S = \# T$, the functions $\omega ^S$ and $\omega ^T$ are equal. Then, we will write $\omega _k$ instead of $\omega ^K$. To fully complete the proof of the proposition, it remains to be shown that, for each $k \in {\mathbb {N}}$, there exists $\alpha _k \ge 0$ such that $\omega _k(v) = 1 + 2\,\alpha _k\,(v-1)$ for all possible $v$.

First, since the weights have to be positive by MON, it is possible to normalize and fix $\omega _k(1)=1$ for all $k \in {\mathbb {N}}$. We now prove the following claim that completes the proof of the proposition.

Claim 5

For all $k \in {\mathbb {N}}$, there exists $\alpha _k \ge 0$ so that for all $v \in \{\frac{3}{2},2,...,2k-3\}, \omega _k(v)=\omega _k(v-\frac{1}{2})+\alpha _k$

Proof

We will only show that $\omega _k(\frac{3}{2})=\omega _k(1)+ \alpha _k$, with $\alpha _k \ge 0$, because all other equations can be derived using similar arguments. Consider a preference profile $P^K_N$ which satisfies the following conditions for some $x, y, z \in K$: (a) $x \succ _{P_N} y$ and $(y,x)$ is contiguous pair of objects for individual $1$; (b) $n_{x,z}(P_N)+2=n_{z,x}(P_N)$ and $(z,x)$ is a contiguous pair of objects for individual $2$; and (c) for all $w \in K \setminus \{x, y\}, w \succ _{P_N} x$ and $w \succ _{P_N} y$.¹¹ Consider also the preference profile $\bar{P}^K_N$ which is equal to $P^K_N$ apart from that individuals 1 and 2 have changed their preferences over the contiguous pairs $(y, x)$ and $(z, x)$, respectively. The proof consists in exploiting the differences in the cohesiveness that accrue along two different paths from $P^K_N$ to $\bar{P}^K_N$.

The first path

• 1. The preference profile $\hat{P}^K_N$ is $(x,y)$-different from $P^K_N$ for individual 1. Hence, by step (a) in Part 2.1 and Claim 4, $M(\hat{P}^K_N)-M(P^K_N)=\frac{2}{n} \, \omega _k(1)$.

• 2. The preference profile $\bar{P}^K_N$ is $(x,z)$-different from $\hat{P}^K_N$ for individual 2. Hence, it follows from step (d) in Part 2.1 and Claim 4 that $M(\bar{P}^K_N)-M(\hat{P}^K_N)=- \frac{2}{n} \, \omega _k(r_{x,z}(\hat{P}^K_N))- t^K_N \, \sum \limits _{w \in K \setminus \{x,z\}} (\sigma _{z, w}(\hat{P}_N) - \sigma _{x,w}(\hat{P}_N))$.

The second path

• 1. The preference profile $\tilde{P}^K_N$ is $(x,z)$-different from $P^K_N$ for individual 2. Hence, it follows from step (d) in Part 2.1 and Claim 4 that $M(\tilde{P}^K_N)-M(P^K_N)=-\frac{2}{n} \, \omega _k(r_{x,z}(P^K_N))-t^K_N \, \sum \limits _{w \in K \setminus \{x, z\}} (\sigma _{z, w}(P_N) - \sigma _{x, w}(P_N))$.

• 2. The preference profile $\bar{P}^K_N$ is $(x,y)$-different from $\tilde{P}^K_N$ for individual 1. Hence, by step (a) in Part 2.1 and Claim 4, $M(\bar{P}^K_N)-M(\tilde{P}^K_N)=\frac{2}{n} \, \omega _k(\frac{3}{2})$.

As in the previous claims, it is easy to see that $M(\bar{P}^K_N)-M(\hat{P}^K_N) = M(\tilde{P}^K_N)-M(P^K_N) + t^K_N \cdot \frac{2}{n}$. Since the difference in the cohesiveness between $\bar{P}^K_N$ and $P^K_N$ has to be the same independently of the path used, we get that $\frac{2}{n} \, \omega _k(1) + t^K_N \, \frac{2}{n} = \frac{2}{n} \, \omega _k(\frac{3}{2})$. Hence, $\omega _k(\frac{3}{2}) = \omega _k(1) + t^K_N$. Then, we have that $t^K_N$ has to depend only on $k$ and, additionally, we have that it is non-negative by RAN. Consequently, we can set $t^K_N$ to be equal to $\alpha _k$. This concludes the proof of the claim. $\square$

Part 3: Finally, we show that the cohesiveness is zero for a preference profile $P_N^K$ whenever $\sigma _{x,y} (P_N) = 0$ for all $\{x,y\} \in \bar{K}$. Plugging this result back into Proposition 3 concludes the proof of the theorem.

Lemma 3

If the cohesiveness measure $M$ satisfies REP, NEU, MON, FR, RAN, and CON, then for all situations $(N,K)$ and all profiles $P \in \mathcal{P}^{{\mathbb {N}}}$ such that $\sigma _{x,y}(P_N)=0$ for all $\{x,y\} \in \bar{K}$,

$$\begin{aligned} M\left( P^K_N\right) =0. \end{aligned}$$

Proof

Take any situation $(N,K)$ and any preference profile $P^K_N$ such that for all $\{x,y\} \in \bar{K}, \sigma _{x,y}(P_N)=0$. By FR, there exists a society $A$ (of size $a$) and a profile $\bar{P}$ such that $M(\bar{P}^K_A)=0$. Consider the society $B$ (of size $b=a \cdot n$) and two preference profiles $\tilde{P}^K_B$ and $\hat{P}^K_B$, where $\tilde{P}^K_B$ is obtained from $a$ isomorphic and disjoint copies of $P^K_N$ and $\hat{P}^K_B$ is obtained from $n$ isomorphic and disjoint copies of $\bar{P}^K_A$. By REP, $M(\hat{P}^K_B)=M(\bar{P}^K_A)=0$. Hence, by Proposition 3,

$$\begin{aligned} M\left( \tilde{P}_B^K\right) -0= & {} \frac{\sum \limits _{\{x,y\} \in \bar{K}} \sigma _{x,y}(\tilde{P}_B) \cdot \left[ 1+ 2\,\alpha _k\,\left( r_{x,y}\left( \tilde{P}^K_B\right) -1\right) \right] }{\sum \limits _{\{x,y\} \in \bar{K}} \left[ 1+ 2\,\alpha _k\,\left( r_{x,y}\left( \tilde{P}^K_B\right) -1\right) \right] }\\&-\, \frac{\sum \limits _{\{x,y\} \in \bar{K}} \sigma _{x,y}(\hat{P}_B) \cdot \left[ 1+ 2\,\alpha _k\,\left( r_{x,y}\left( \hat{P}^K_B\right) -1\right) \right] }{\sum \limits _{\{x,y\} \in \bar{K}} \left[ 1+ 2\,\alpha _k\,\left( r_{x,y}\left( \hat{P}^K_B\right) -1\right) \right] }. \end{aligned}$$

Since $\sigma _{x,y}(P_N)=0$ for all $\{x,y\} \in \bar{K}$ by assumption, it follows from the construction of $\tilde{P}^K_B$ that $\sigma _{x,y}(\tilde{P}_B)=0$ for all $\{x,y\} \in \bar{K}$ as well. Consequently, the former equation reduces to

$$\begin{aligned} M\left( \tilde{P}_B^K\right) = \, -\frac{\sum \limits _{\{x,y\} \in \bar{K}} \sigma _{x,y}(\hat{P}_B) \cdot \left[ 1+ 2\,\alpha _k\,\left( r_{x,y}\left( \hat{P}^K_B\right) -1\right) \right] }{\sum \limits _{\{x,y\} \in \bar{K}} \left[ 1+ 2\,\alpha _k\,\left( r_{x,y}\left( \hat{P}^K_B\right) -1\right) \right] }. \end{aligned}$$

Observe that all weights $\omega _k(\cdot )$ are strictly positive by Proposition 3 and that the values of $\sigma$ cannot be negative. Given that the values of the cohesiveness should be non-negative, the only solution to the equation is $\sigma _{x,y}(\hat{P}_B)=0$ for all $\{x,y\} \in \bar{K}$ and, therefore, $M(\tilde{P}^K_B)=0$. Finally, $M(P^K_N)=M(\tilde{P}^K_B)=0$ by REP. $\square$

1.2 Independence of the Axioms (Proposition 1)

We will provide cohesiveness measures that satisfy all the properties but one.

Replication Invariance: Let the cohesiveness measure $M_1$ be such that, for all situations $(N,K)$ and all preference profiles $P^K_N$,

$$\begin{aligned}&M_1\left( P_N^K\right) = \Phi _{(0, 0, \ldots , 0)}\left( P_N^K\right) \text{ if } 1 \in N \quad \text{ and } \\&M_1\left( P_N^K\right) = \Phi _{(0.5, 0.5, \ldots , 0.5)}\left( P_N^K\right) \text{ if } 1 \not \in N. \end{aligned}$$

This cohesiveness measure satisfies NEU, MON, FR, RAN, and CON. The following example shows that it does not satisfy replication invariance.

Let $N=\{1,2\}, S=\{2, 3\}$, and $K= \{x, y, z\}$. Suppose that the profile $P$ is such that $xP_1yP_1z, xP_2zP_2y$, and $P_3=P_1$. Then, $M_1(P_N^K) = \frac{2}{3}$ and $M_1(P_S^K) = \frac{5}{6}$. REP would imply that $M_1(P_N^K)=M_1(P_S^K)$.

Neutrality: Let $q: \bar{X} \rightarrow {\mathbb {R}}_{++}$ be a function that assigns to each subset $\{x,y\} \in \bar{X}$ a strictly positive weight $q_{x,y}> 0$ in such a way that $q_{x,y} \ne q_{w,z}$ for some $\{w,z\} \ne \{x,y\}$. Now, let cohesiveness measure $M_2$ be such that, for all situations $(N,K)$ and all preference profiles $P^K_N$,

$$\begin{aligned} M_2\left( P^K_N\right) =\sum \limits _{\{x,y\} \in \bar{K}}\frac{q_{x,y}}{\sum \limits _{\{w,z\} \in \bar{K}} q_{w,z}}\sigma _{x,y}(P_N). \end{aligned}$$

This cohesiveness measure satisfies REP, MON, FR, RAN, and CON. The following example shows that it is not neutral.

Let $N=\{1,2\}$ and $K=\{x,y,z\}$. Suppose that the preference profiles $P_N^K$ and $\bar{P}_N^K$ are such that $xP_1yP_1z, yP_2xP_2z, z\bar{P}_1y\bar{P}_1x$, and $y\bar{P}_2z\bar{P}_2x$. Moreover, let $q_{x,z}=q_{y,z}=1$ and $q_{x,y}=2$. Then, $M_2(P_N^K)=\frac{2}{4}$ and $M_2(\bar{P}_N^K)=\frac{3}{4}$. NEU would imply that $M_2(P_N^K)=M_2(\bar{P}_N^K)$.

Monotonicity: Let the cohesiveness measure $M_3$ be such that, for all situations $(N,K)$ and all preference profiles $P^K_N$,

$$\begin{aligned} M_3\left( P^K_N\right) = 1-\Phi _{(0, \ldots , 0)} \left( P_N^K\right) . \end{aligned}$$

This cohesiveness measure satisfies REP, NEU, FR, RAN, and CON. The following example shows that it is not monotone.

Let $N=\{1,2\}$ and $K=\{x,y\}$. Suppose that the preference profiles $P_N^K$ and $\bar{P}_N^K$ are such that $xP_1y, yP_2x, x\bar{P}_1y$, and $x\bar{P}_2y$. Then, $M_3(P_N^K)=1$ and $M_3(\bar{P}_N^K)=0$. MON would imply that $M_3(\bar{P}_N^K)>M_3(P_N^K)$.

Full Range: Let the cohesiveness measure $M_4$ be such that, for all situations $(N,K)$ and all preference profiles $P^K_N$,

$$\begin{aligned} M_4\left( P^K_N\right) =\frac{1}{2} \Phi _{(0, \ldots , 0)} \left( P_N^K\right) . \end{aligned}$$

This cohesiveness measure satisfies REP, NEU, MON, RAN, and CON. However, since it is impossible to find a situation $(N, K)$ and a preference profile $P_N^K$ such that $M_4(P^K_N) = 1, M_4$ does not satisfy full range.

Ranking: Let the cohesiveness measure $M_5$ be such that, for all situations $(N,K)$ and all preference profiles $P^K_N$,

$$\begin{aligned} M_5\left( P^K_N\right) =\frac{2}{k(k-1)^2}\sum \limits _{\{x,y\}\in \bar{K}}\left( 2k-2-r_{x,y}\left( P^K_N\right) \right) \cdot \sigma _{x,y}(P^K_N). \end{aligned}$$

This cohesiveness measure satisfies REP, NEU, MON, FR, and CON. The following example shows that it does not satisfy ranking.

Let $N=\{1,2,3\}$ and $K=\{x,y,z\}$. Let $P_N^K, \bar{P}_N^K, \tilde{P}_N^K, \hat{P}_N^K$ be as defined in Example 2. Then, $M_5(P_N^K)= \frac{5}{6}, M_5(\bar{P}_N^K)=\frac{17}{18}$, and $M_5(\tilde{P}_N^K) = M_5(\hat{P}_N^K)=1$. So, $M_5(\tilde{P}_N^K)-M_5(P_N^K)=\frac{1}{6}$ and $M_5(\hat{P}_N^K)-M_5(\bar{P}_N^K)=\frac{1}{18}$. RAN would imply that $M_5(\hat{P}_N^K)-M_5(\bar{P}_N^K) \ge M_5(\tilde{P}_N^K)-M_5(P_N^K)$.

Consistency: Let $\{a_n\}_{n \in \{4, 6, 8, \ldots \}}$ be the sequence of real numbers in the unit interval determined by the equations $a_4 = 0.6$ and $a_n = a_{n \cdot s}+(s-1) \cdot \frac{2(1-a_{n \cdot s})}{s \cdot n-2}$ for all $n \in \{4, 6, 8, \ldots \}$ and all $s \in {\mathbb {N}}$. Consider now, for each subset of two objects $\{x, y\}$, the function $\Omega _{x, y}$ such that for each preference profile $P_N, \Omega _{x,y}(P_N)=0$ whenever $\sigma _{x, y}(P_N)=0$; $\Omega _{x,y}(P_N)=a_n + \frac{2(1-a_n)}{n-2} \cdot \left( \frac{|n_{x,y}(P_N)|}{2}-1\right)$ when $\sigma _{x, y}(P_N)>0$ and $n$ is even; and $\Omega _{x,y}(P_N)=a_{2 \cdot n} + \frac{1-a_{2 \cdot n}}{n-1} \cdot (|n_{x,y}(P_N)|-1)$ when $\sigma _{x, y}(P_N)>0$ and $n$ is odd. Now, let the cohesiveness measure $M_6$ be such that, for all situations $(N,K)$ and all preference profiles $P^K_N$,

$$\begin{aligned} M_6\left( P^K_N\right) = \frac{2}{k(k-1)} \sum \limits _{\{x,y\} \in \bar{K}} \Omega _{x,y}(P_N). \end{aligned}$$

This cohesiveness measure satisfies REP, NEU, MON, FR, and RAN. The following example shows that it is not consistent.

Let $N=\{1,2,3,4\}$ and $K= \{x,y\}$. Suppose that the preference profile $P_N$ is such that $xP_1y, P_2=P_1, yP_3x$, and $P_4=P_3$. Also, let the preference profiles $\bar{P}_N^K$ and $\tilde{P}_N^K$ be such that $\bar{P}_N^K$ is $(x,y)$-different from $P_N^K$ for individual $3$ and $\tilde{P}_N^K$ is $(x,y)$-different from $\bar{P}_N^K$ for individual $4$. Then, $M_6(P_N^K)=0, M_6(\bar{P}_N^K)=\frac{3}{5}$, and $M_6(\tilde{P}_N^K)=1$. So, $M_6(\tilde{P}_N^K)-M_6(\bar{P}_N^K)=\frac{2}{5}$ and $M_6(\bar{P}_N^K)-M_6(P_N^K)=\frac{3}{5}$. CON would imply that $M_6(\tilde{P}_N^K)-M_6(\bar{P}_N^K) = M_6(\bar{P}_N^K)-M_6(P_N^K)$.