Social evaluation functionals: a gateway to continuity in social choice

Abstract

This paper develops social choice theory aggregating individual utility functions to a social utility function. Such a tool allows me to deal with a natural notion of continuity in social choice theory. In addition, and in order to have the choice problem as close as possible to its beginnings, the social evaluation functionals considered are assumed to satisfy both ordinal measurability and interpersonal non-comparability, and unanimity. I present two results concerning the characterization of projective social evaluation functionals (which means that the social utility function is exactly the utility of the dictator). The first one needs a strong form of welfarism called social state separability. The second one uses continuity in combination with a new axiom called ordinal-scale-preserving.

Appendix

I begin with the basics on preferences. This material closely follows the notation and terminology of Candeal (2013).

Let $X$ be a nonempty abstract set. A transitive and complete (hence, reflexive) binary relation $\mathrel {R}$ defined on $X$ is said to be a total preorder.¹⁵ A total order on $X$ is an antisymmetric total preorder.

Associate with $\mathrel {R}$ the asymmetric part, also called the strict part, denoted by $\mathrel {R}^{a}$, is defined as the following binary relation on $X$: $x\mathrel {R}^{a} y$ if and only if $x\mathrel {R} y$ and $\lnot (y\mathrel {R} x)$ (or, equivalently, since $\mathrel {R}$ is a total preorder, $x\mathrel {R}^{a} y$ if and only if $\lnot (y\mathrel {R} x)$). Similarly, its symmetric part, denoted by $\mathrel {R}^{s}$, is defined by: $x\mathrel {R}^{s} y$ if and only if $x\mathrel {R} y$ and $y\mathrel {R} x$. Given two elements $x,y\in X, x$ is said to be indifferent to $y$ if $x\mathrel {R}^{s} y$. The (total) preorder $\mathrel {R}$ is said to be non-trivial if $x\mathrel {R}^{a} y$ for some $x,y\in X$. The dual total preorder associated with $\mathrel {R}$ is defined as $x\mathrel {R}^{d} y$ if and only if $y\mathrel {R} x$. A real-valued map $u:X\longrightarrow \mathbb {R}$ is called a utility function for $\mathrel {R}$ if, for every $x,y\in X$, it holds that $x\mathrel {R} y \Longleftrightarrow u(x)\le u(y)$.

As usual $\mathbb {R}^{n}$ will denote the $n$-dimensional Euclidean space, i.e., $\mathbb {R}^{n}=\{a=(a_{j}): a_{j}\in \mathbb {R},\ j=1,\ldots ,n\}$. The usual binary operations in $\mathbb {R}^{n}$ of addition, multiplication by scalars, and vector multiplication, defined coordinatewise, will be denoted by $+,\cdot _{\mathbb {R}}$ and $*$, respectively. The topology on $\mathbb {R}^{n}$ I will refer to is the Euclidean one. The symbols $0_{n}$ and $1_{n}$ will stand for the zero vector and the vector with all its coordinates equal to one, respectively. The set that consists of the $n$ (respectively, $m$) first natural numbers will be denoted by $N$ (respectively, $M$). For every $j\in N$ (respectively, $i\in M$), $N(j)$ (respectively, $M(i)$) will be the subset of $N$ (respectively, $M$) given by $N(j)=\{1,\ldots ,j\}$ (respectively, $M(i)=\{1,\ldots ,i\}$). The symbol used for the usual (strict) ordering of the reals will be ($<$) $\le$. Given $a=(a_{1},\ldots ,a_{n}),\ b=(b_{1},\ldots ,b_{n}) \in \mathbb {R}^{n}$, I write $a\le b$ whenever $a_{j}\le b_{j}$ for all $j\in N, a\ll b$ whenever $a_{j}<b_{j}$ for all $j\in N$ and $a< b$ whenever $a\le b$ and, for some $j\in N, a_{j}< b_{j}$.

Definition 2

A total preorder $\precsim$ defined on $\mathbb {R}^{n}$ is said to be:

• (1) zero-independent if $a\precsim b$ implies $a+c\precsim b+c, (a,b,c\in \mathbb {R}^{n})$,

• (2) scale-independent if $a\precsim b, 0_{n}\ll c$ implies $c*a\precsim c*b, (a,b,c\in \mathbb {R}^{n})$,

• (3) continuous if, for every $a\in \mathbb {R}^{n}$, both the lower contour set $L(a)=\{b\in \mathbb {R}^{n}: b\precsim a\}$ and the upper contour set $G(a)=\{b\in \mathbb {R}^{n}:\ a\precsim b\}$ are closed subsets of $\mathbb {R}^{n}$,

• (4) upper-semicontinuous at zero if the upper contour set relative to the zero vector, $G(0_{n})$, is a closed subset of $\mathbb {R}^{n}$,

• (5) Pareto if $a\le b$ then $a\precsim b, (a,b\in \mathbb {R}^{n})$,

• (6) inverse Pareto if $\precsim ^{d}$ is Pareto,

• (7) weak Pareto if $a\ll b$ then $a\prec b, (a,b\in \mathbb {R}^{n})$,

• (8) strong Pareto if $a< b$ then $a\prec b, (a,b\in \mathbb {R}^{n})$,

• (9) strongly dictatorial if there is $i\in N$ such that, for every $a=(a_{j}),b=(b_{j})\in \mathbb {R}^{n}, a\precsim b$ if and only if $a_{i}\le b_{i}$,

• (10) strongly inversely dictatorial if $\precsim ^{d}$ is strongly dictatorial, i.e., if there is $i\in N$ such that, for every $a=(a_{j}),b=(b_{j})\in \mathbb {R}^{n}, a\precsim b$ if and only if $b_{i}\le a_{i}$.

Now I carry out the proofs of the results stated in Sect. 3.

Proof of Proposition 1

It is a direct consequence of Arrow’s theorem and Remark 1(i).$\square$

Proof of Theorem 1

The proof of the “if” part is obvious. So, I will focus on the “only if” part. Suppose then that $F$ satisfies OMIN and SSS. Consider the real-valued function $W:\mathbb {R}^{n}\longrightarrow \mathbb {R}$ defined as follows: $W(a):=F(U)(x)$, where $a\in \mathbb {R}^{n}$ is such that $U(x)=a$, for some $U\in \mathcal {U}^{n}, x\in X$. Notice that $W$ is well-defined (i.e., $W(a)$ depends neither on $U$ nor on $x$ chosen) since $F$ satisfies SSS. So, it holds that $F(U)(x)=W(U(x))$, for every $U\in \mathcal {U}^{n}, x\in X$.

Denote by $\precsim _{W}$ the total preorder on $\mathbb {R}^{n}$ induced by $W$, i.e., $a\precsim _{W} b\Leftrightarrow W(a)\le W(b)$, for every $a,b\in \mathbb {R}^{n}$. Then, $\precsim _{W}$ is a total preorder on $\mathbb {R}^{n}$ which admits the utility function $W$. Since $F$ fulfils OMIN, it follows that $\precsim _{W}$ satisfies the following property: $a\precsim _{W} b\Longleftrightarrow \Phi (a)\precsim _{W} \Phi (b)$, where $\Phi \in \Delta ^{n}$ ¹⁶. In particular, this property holds true when restricted to $\Delta _{spa}^{n}$, the subset of $\Delta ^{n}$ which consists of all $n$-tuples of strictly positively affine real-valued functions defined on $\mathbb {R}$. So, $\precsim _{W}$ is a zero-independent and scale-independent total preorder on $\mathbb {R}^{n}$ and is nontrivial since $F$, by hypothesis, is U. Therefore, by Theorem 3 of Candeal (2013), $\precsim _{W}$ is strongly dictatorial¹⁷. So, there is an individual, say $i\in N$, such that $a=(a_{j})\precsim _{W} b=(b_{j})\Leftrightarrow a_{i}\le b_{i}$, for every $a,b\in \mathbb {R}^{n}$. Thus, $W(a)=\phi (a_{i})$, for every $a=(a_{j})\in \mathbb {R}^{n}$, for some $\phi \in \Delta$. Therefore, $F(U)(x)=W(U(x))=\phi (u_{i}(x))$, for every $U=(u_{j})\in \mathcal {U}^{n}, x\in X$. Now, let $r\in \mathbb {R}$ and consider the (constant) function $u_{r}:X\longrightarrow \mathbb {R}$ given by $u^{r}(x)=r$, for every $x\in X$. Take the profile $U_{u^{r}}\in \mathcal {U}^{n}$. Then, since $F$ satisfies U, $F(U_{u^{r}})(x)=r=\phi (u^{r}(x))=\phi (r)$, for every $x\in X$. So, $\phi (r)=r$ and, since $r$ is arbitrary, $\phi$ is the identity function. Therefore, $F(U)(x)=u_{i}(x)$, for every $U=(u_{j})\in \mathcal {U}^{n}, x\in X$. So, $F$ is Pro and I am done.$\square$

Proof of Proposition 2

¹⁸ Consider the following five subsets of $\mathcal {U}^{2}$; $\mathcal {A}_{1}=\{U=(u_{1}, u_{2})\in \mathcal {U}^{2}: u_{1}(x)>u_{1}(y), u_{2}(x)>u_{2}(y)\}, \mathcal {A}_{2}=\{U=(u_{1}, u_{2})\in \mathcal {U}^{2}: u_{1}(x)<u_{1}(y), u_{2}(x)<u_{2}(y)\}, \mathcal {A}_{3}=\{U=(u_{1}, u_{2})\in \mathcal {U}^{2}: u_{1}(x)>u_{1}(y), u_{2}(x)<u_{2}(y)\}, \mathcal {A}_{4}=\{U=(u_{1}, u_{2})\in \mathcal {U}^{2}: u_{1}(x)<u_{1}(y), u_{2}(x)>u_{2}(y)\}$, and $\mathcal {A}_{5}=\mathcal {U}^{2}\setminus \bigcup _{i=1}^{4}\mathcal {A}_{i}$. Note that $\mathcal {A}_{5}$ has empty interior since it consists of profiles $U=(u_{1}, u_{2})\in \mathcal {U}^{2}$ for which at least one of the equalities $u_{1}(x)=u_{1}(y)$, or $u_{2}(x)=u_{2}(y)$ hold.

In addition, consider the real-valued map $G$ defined as follows: For every $U=(u_{1}, u_{2})\in \mathcal {U}^{2}\longrightarrow G(U)\in \mathbb {R}$, where $G(U)=\sqrt{\vert (u_{1}(x)-u_{1}(y)) (u_{2}(x)-u_{2}(y)) \vert }$.

Define now the following SEF: For every $U=(u_{1}, u_{2})\in \mathcal {U}^{2}\longrightarrow F(U)\in \mathcal {U}$, where

$$\begin{aligned} F(U)(x)=\frac{u_{1}(x)+u_{2}(x)}{2}+ \vert u_{1}(y)-u_{2}(y) \vert , \end{aligned}$$

and $F(U)(y)=F(U)(x)-G(U)$, if $U\in \mathcal {A}_{1}$; $F(U)(y)=F(U)(x)+G(U)$, if $U\in \bigcup _{i=2}^{4}\mathcal {A}_{i}$; and $F(U)(y)=F(U)(x)$, otherwise.

In order to see that $F$ is U, let $u\in \mathcal {U}$ and consider the profile $U=(u,u)$. Notice that $U$ belongs to $\mathcal {A}_{1}$ (provided that $u(x)>u(y)$), or $\mathcal {A}_{2}$ (provided that $u(x)<u(y)$), or $\mathcal {A}_{5}$ (whenever $u(x)=u(y)$). Obviously, it holds that $F(U)(x)=u(x)$. To calculate $F(U)(y)$, observe first that $G(U)=\vert u(x)-u(y) \vert$. So, $F(U)(y)=F(U)(x)-G(U)=u(x)-(u(x)-u(y))=u(y)$, whenever $U\in \mathcal {A}_{1}$; $F(U)(y)=F(U)(x)+G(U)=u(x)+(u(y)-u(x))=u(y)$, whenever $U\in \mathcal {A}_{2}$; and $F(U)(y)=F(U)(x)=u(x)=u(y)$, whenever $U\in \mathcal {A}_{5}$. Therefore, $F(u,u)=u$ and $F$ is U.

Anonymity is easily checked since, for every $U=(u_{1},u_{2})\in \mathcal {U}^{2}$, it holds that $G(u_{1},u_{2})=G(u_{2},u_{1}), F(U)(x)$ clearly fulfils A, and $F(U)(y)$ is A too since is the sum, or the difference, of anonymous (symmetric) functions.

To argue about OMIN notice that, for each $i=\{1,\ldots ,5\}$, if $U=(u_{1},u_{2})\in \mathcal {A}_{i}$, then $\phi U\in \mathcal {A}_{i}$, for every $\Phi =(\phi _{1},\phi _{2})\in \Delta ^{2}$. So, $F(U)(x)\le F(U)(y)\Longleftrightarrow F(\Phi U)(x)\le F(\Phi U)(y)$ since the sign of $F(U)(x)-F(U)(y)$ remains the same over each $\mathcal {A}_{i}$ (strictly positive in $\mathcal {A}_{1}$, strictly negative in $\bigcup _{i=1}^{4}\mathcal {A}_{i}$, and null in $\mathcal {A}_{5}$).

Finally, the continuity of $F$ follows since $F(U)(x)$ and $G$ are, obviously, continuous and, therefore, $F(U)(y)$ is the sum, or difference of continuous functions that agree on $\mathcal {A}_{5}$ (the boundary of the open set $\bigcup _{i=1}^{4}\mathcal {A}_{i}$).

Note that on $\mathcal {A}_{1}\bigcup \mathcal {A}_{2}$ both individuals dictate, whereas individual “$2$” dictates on $\mathcal {A}_{3}$, and individual “$1$” dictates on $\mathcal {A}_{4}$. On $\mathcal {A}_{5}$ dictates the one, “$j$”$\in \{1,2\}$, for whom the values $u_{j}(x), u_{j}(y)$ agree. So, $F$ is not D.$\square$

Remark 4

I explain how to construct an example of this nature for higher dimensions. Begining from a given non-dictatorial SEF F satisfying OMIN, C and U, is simple to produce a non-dictatorial SEF G, satisfying the same properties, for any number of individuals $p>n$ and the same number of alternatives. Indeed, let $F:\mathcal {U}^{n}\longrightarrow \mathcal {U}$ be a non-dictatorial SEF satisfying OMIN, C and U, and let $p>n$. Then the mapping $G:\mathcal {U}^{p}\longrightarrow \mathcal {U}$ given by $G(u_{1},\cdots ,u_{p})=F(u_{1},\cdots ,u_{n})$, for every $(u_{1},\cdots ,u_{p})\in \mathcal {U}^{p}$, is easily seen to satisfy OMIN, C and U, and is non-dictatorial since $F$ is. What is more complex, especially from the presentation point of view, is to show an example for a greater number of alternatives and the same number of individuals. A key remark is that a SEF for $m$ alternatives and $n$ individuals induces SEFs for $m-1$ alternatives and $n$ candidates. I first consider the situation $n=2, m=3$ (i.e., $X=\{x,y,z\}$), and then I will comment on the general case. Let $F:\mathcal {U}^{2}\longrightarrow \mathcal {U}$ be a SEF that satisfies OMIN, C and U. Let $Y=X\setminus \{z\}=\{x,y\}$ and consider the set $\mathcal {U}_{Y}:=\mathbb {R}^{Y}=\{u:Y\rightarrow \mathbb {R}\}$. Let $U_{Y}$ denote a typical element of $\mathcal {U}_{Y}^{2}$. Each $U_{Y}$ can be extended to a profile, say $\tilde{U}_{Y}\in \mathcal {U}^{2}$ in the following way: $\tilde{U}_{Y}(a,j)=U_{Y}(a,j)$, for all $a\in Y, j\in \{1,2\}$; $\tilde{U}_{Y}(z,j)=U_{Y}(x,j), j\in \{1,2\}$. Then the mapping $G:\mathcal {U}_{Y}^{2}\longrightarrow \mathcal {U}_{Y}$ given by: $G^{x}(U_{Y})=F^{x}(\tilde{U}_{Y})$, and $G^{y}(U_{Y})=F^{y}(\tilde{U}_{Y})$ is easily proved to be OMIN, C and U. So, the construction given in the proof of Proposition 2 is pretty useful since, in a sense, an example for the case $n=2$ and $m=3$ will be an extension of that for $n=2$ and $m=2$. To produce such an example it is firstly helpful to think of $\mathcal {U}^{2}$ split into 37 pairwise disjoint subsets, 36 of which are open and the other one has empty interior. Each of these 36 open set is a two-fold Cartesian product of the interior of two rank-ordered sets corresponding to the utility values of the first and the second individual, respectively (for the concept of a rank-ordered set see Definition 3(ii)). The empty set turns out to be the boundary of the union of these 36 open sets and, actually, is defined by $58$ boundary conditions derived from the equalities of the values of the variables $\{u_{1}(x),u_{1}(y),u_{1}(z),u_{2}(x),u_{2}(y),u_{2}(z)\}$. To define the desired SEF some notation is still needed. For a given $U=(u_{1}, u_{2})\in \mathcal {U}^{2}$, let $U_{Y}$ denote the following element of $\mathcal {U}_{Y}^{2}, U_{Y}:=(u_{1}\vert _{Y}, u_{2}\vert _{Y})$. Let $F_{Y}:\mathcal {U}_{Y}^{2}\longrightarrow \mathcal {U}_{Y}$ be the SEF constructed in the proof of Proposition 2 above. I first reduces the 37 cases that can occur into the six ones that appear below. On each of these six (open) subsets of $\mathcal {U}^{2}$, the SEF $F:\mathcal {U}^{2}\longrightarrow \mathcal {U}$ is defined as follows: For any $U=(u_{1}, u_{2})\in \mathcal {U}^{2}, F(U)(x)=F_{Y}(U_{Y})(x)$. The definitions of $F(U)(y)$ and $F(U)(z)$ on each subset are the following:

• (1) $\mathcal {B}_{1}=\{U=(u_{1}, u_{2})\in \mathcal {U}^{2}: u_{1}(x)>u_{1}(y)>u_{1}(z)\}$. In this case $F(U)(y)=F_{Y}(U_{Y})(y)$ and $F(U)(z)=F_{Y}(U_{Y})(y)-\vert u_{1}(y)-u_{1}(z)\vert$,

• (2) $\mathcal {B}_{2}=\{U=(u_{1}, u_{2})\in \mathcal {U}^{2}: u_{1}(z)>u_{1}(x)>u_{1}(y)\}$. Here $F(U)(y)=F_{Y}(U_{Y})(y)$ and $F(U)(z)=F_{Y}(U_{Y})(y)+\vert u_{1}(z)-u_{1}(x)\vert$,

• (3) $\mathcal {B}_{3}=\{U=(u_{1}, u_{2})\in \mathcal {U}^{2}:u_{1}(y)>u_{1}(x)>u_{1}(z)$. Then $F(U)(y)=F_{Y}(U_{Y})(y)$ and $F(U)(z)=F_{Y}(U_{Y})(x)-\vert u_{1}(x)-u_{1}(z)\vert$,

• (4) $\mathcal {B}_{4}=\{U=(u_{1}, u_{2})\in \mathcal {U}^{2}:u_{1}(z)>u_{1}(y)>u_{1}(x)$. In this case $F(U)(y)=F_{Y}(U_{Y})(y)$ and $F(U)(z)=F_{Y}(U_{Y})(y)+\vert u_{1}(z)-u_{1}(y)\vert$,

• (5) $\mathcal {B}_{5}=\{U=(u_{1}, u_{2})\in \mathcal {U}^{2}:u_{1}(x)>u_{1}(z)>u_{1}(y)$. Then $F(U)(y)=F_{Y}(U_{Y})(x)-\vert u_{1}(x)-u_{1}(y)\vert$ and $F(U)(z)=F_{Y}(U_{Y})(x)-\vert u_{1}(x)-u_{1}(z)\vert$, and

• (6) $\mathcal {B}_{6}=\{U=(u_{1}, u_{2})\in \mathcal {U}^{2}:u_{1}(y)>u_{1}(z)>u_{1}(x)$. In this case $F(U)(y)=F_{Y}(U_{Y})(x)+\vert u_{1}(y)-u_{1}(x)\vert$ and $F(U)(z)=F_{Y}(U_{Y})(x)+\vert u_{1}(z)-u_{1}(x)\vert$.

Note that these six cases include the 36 open sets mentioned above as well as some boundary conditions (those provided by the equality of some of the values $\{u_{2}(x),u_{2}(y),u_{2}(z)\}$) that appear in the set with empty interior. It remains to define $F(U)(y)$ and $F(U)(z)$ on the boundary profiles that result from the equality of some of the values $\{u_{1}(x),u_{1}(y),u_{1}(z)\}$. For such a profile $U, F(U)(x)=F_{Y}(U_{Y})(x)$, and $F(U)(y)$ and $F(U)(z)$ are defined by the corresponding extensions of $F$, already defined as above on $\bigcup _{i=1}^{6}\mathcal {B}_{i}$, to the boundary since these extensions agree on the corresponding limit profiles. So, $F$ turns out to be a continous SEF on $\mathcal {U}^{2}$. Moreover, it can be shown that $F$ is also OMIN and U. Further, it is non-dictatorial since $F_{Y}$ is.

For the general case the argument runs by induction. Let $m>1$ and let $X=\{x_{1},\ldots ,x_{m}\}$. Denote by $Y=X\setminus \{x_{m}\}$ and define $\mathcal {U}_{Y}, U_{Y}$ and $\tilde{U}_{Y}$ in a similar way as above. Let $F_{Y}:\mathcal {U}_{Y}^{n}\longrightarrow \mathcal {U}_{Y}$ a non-dictatorial SEF, which exists by the induction hypothesis, satisfying OMIN, C and U. Let $i\in N$ be fixed. Let $U\in \mathcal {U}^{n}$ be and consider the subset of the reals $\{u_{i}(x_{1}),\ldots ,u_{i}(x_{m})\}$. Denote by $u(x_{k})=\min \{u_{i}(x_{j}):j\in 1,\ldots ,m-1\}$ and by $u(x_{l})=\max \{u_{i}(x_{j}):j\in 1,\ldots ,m-1\}$. The idea is to first define $F$ on each of the $m!$ subsets of $\mathcal {U}^{n}$ that result of any (strict) rearrangement of the values $\{u_{i}(x_{1}),\ldots ,u_{i}(x_{m-1})\}$ in combination with the following three situations; namely, if $u_{i}(x_{k})>u(x_{m})$, or if $u_{i}(x_{m})>u(x_{l})$, or if $u_{i}(x_{l})>u_{i}(x_{m})>u_{i}(x_{k})$. In the first two cases the definition of $F(U)(x_{j})$ agrees with that of $F_{Y}(U_{Y})(x_{j}), j\in \{1,\ldots ,m-1\}$, and $F(U)(x_{m})$ turns out to be $F_{Y}(U_{Y})(x_{k})$ (respectively, $F_{Y}(U_{Y})(x_{l})$) minus the distance from $u_{i}(x_{k})$ to $u_{i}(x_{m})$ (respectively, plus the distance from $u_{i}(x_{l})$ to $u_{i}(x_{m})$). In the third case all the values $F(U)(x_{j})$ are obtained by evaluating $F_{Y}(U_{Y})(x_{l})$ minus the distance from $u_{i}(x_{j})$ to $u_{i}(x_{l}), j\in \{1,\ldots ,m\}$. For instance, if $u_{i}(x_{1})>\ldots >u_{i}(x_{m-1})$ then the following three situations are distinguished:

• (1) $u_{i}(x_{m-1})>u_{i}(x_{m})$. In this case $F(U)(x_{j})=F_{Y}(U_{Y})(x_{j})$, for every $j\in \{1,\ldots ,m-1\}$, and $F(U)(x_{m})=F_{Y}(U_{Y})(x_{m-1})-\vert u_{i}(x_{m-1})-u_{i}(x_{m})\vert$.

• (2) $u_{i}(x_{m})>u_{i}(x_{1})$. In this case $F(U)(x_{j})=F_{Y}(U_{Y})(x_{j})$, for every $j\in \{1,\ldots ,m-1\}$, and $F(U)(x_{m})=F_{Y}(U_{Y})(x_{1})+\vert u_{i}(x_{m})-u_{i}(x_{1})\vert$.

• (3) $u_{i}(x_{1})>u_{i}(x_{m})>u_{i}(x_{m-1})$. In this case $F(U)(x_{j})=F_{Y}(U_{Y})(x_{1})-\vert u_{i}(x_{1})-u_{i}(x_{j})\vert$, for every $j\in \{1,\ldots ,m\}$.

The definition of $F$ when some of the values $\{u_{i}(x_{1}),\ldots ,u_{i}(x_{m})\}$ agree is the extension to the boundary of the already defined values of $F$ on the profiles corresponding to the situations (1) to (3) above. Notice that $F$ is well-defined since these extensions coincide. It can be shown that the SEF $F$ so-defined meets all the required properties.

The proof of Theorem 1 needs several previous definitions and a technical lemma that are presented now. For similar results see Ovchinnikov (1996) and Marichal and Mathonet (2001). Recall that if $a=(a_{j})\in \mathbb {R}^{p}$ and $\phi \in \Delta$, then $\phi (a):=(\phi (a_{j}))\in \mathbb {R}^{p}$, for every $j\in P:=\{1,\ldots ,p\}$.

Definition 3

A subset $S\subset \mathbb {R}^{p}$ is said to be:

• (i) $\Delta$ -invariant if $\phi (a)\in S$, for every $a\in S, \phi \in \Delta$,

• (ii) a rank-ordered set if $S=\{a\in \mathbb {R}^{p}: a_{\sigma (1)}\le \cdots \le a_{\sigma (p)}\}$, for some $\sigma \in S(P)$.

Remark 4

It is obvious to see that a rank-ordered set is $\Delta$-invariant.

Definition 4

A function $f:\mathbb {R}^{p}\longrightarrow \mathbb {R}$ is said to be ordinal-scale-preserving if $f(\phi (a))=\phi (f(a))$, for every $a\in \mathbb {R}^{p}, \phi \in \Delta$.

Now a technical lemma and some preparatory material needed for the proof of Theorem 2 are presented.

Lemma

Let $f:\mathbb {R}^{p} \rightarrow \mathbb {R}$ be an ordinal-scale-preserving function, $a=(a_{j})\in \mathbb {R}^{p}$ and $S\subset \mathbb {R}^{p}$ a rank-ordered set. The following facts hold true:

• (i) $f(a)\subset \{a_{1},\ldots ,a_{p}\}$,

• (ii) if, in addition, $f$ is continuous, then there exists $i\in \{1,\ldots ,p\}$ such that $f(a)=a_{i}$, for all $a=(a_{j})\in S$.

Proof

(i) Suppose by way of contradiction $\alpha \equiv f(a)\notin \{a_{1},\ldots ,a_{p}\} \equiv A$, and assume, without loss of generality, that $a_{1}\le \ldots \le a_{p}$. Let $i(\alpha )\in P$ given by $i(\alpha )=\text{ max }\{n: a_{n}<\alpha \}\ \text{ if }\ \alpha > a_{1}$, and $i(\alpha )=1$ if $\alpha < a_{1}$. There exists a strictly increasing function $\phi :\mathbb {R}\rightarrow \mathbb {R}$ satisfying $\phi (x)=x$ for $x\in A$ and $\phi (\alpha )=(a_{i_{\alpha }}+\alpha )/2.$ ¹⁹ Since $f$ is ordinal-scale-preserving I must have $\alpha =f(a)=f(\phi (a))=\phi (f(a))=(a_{i_{\alpha }}+\alpha )/2$, a contradiction which completes the argument.

(ii) Let $S\subset \mathbb {R}^{p}$ be a rank-ordered set and assume again, without loss of generality, that $S=\{(a_{1},\ldots , a_{p})\in \mathbb {R}^{p}: a_{1}\le \ldots \le a_{p}\}$. By continuity, it suffices to prove that $f$ is a projection over $S^{\circ }=\{(a_{1},\ldots , a_{p})\in \mathbb {R}^{p}: a_{1}<\ldots <a_{p}\}$, the interior of $S$. Suppose, by way of contradiction, that there are vectors $a=(a_{j}), b=(b_{j})\in S^{\circ }$, and numbers $q,q'\in \{1,\ldots ,p\}, q\not =q'$, so that $f(a)=a_{q}$ and $f(b)=b_{q'}$. Consider then the real-valued function $h(\lambda )=f((1-\lambda )a+\lambda b), \lambda \in [0,1]$. Notice that $h$ is a continuous function defined on $[0,1]$ and $(1-\lambda )a+\lambda b\in S$, for every $\lambda \in [0,1]$. By continuity of $h$ and the fact proved above that the image of $f$ at each vector agrees with some of its coordinates, it follows that there is $\epsilon >0$ such that, for $\lambda \in [0,\epsilon ]$, it holds that $h(\lambda )=(1-\lambda )a_{q}+\lambda b_{q}$ and $h(1-\lambda )=\lambda a_{q'}+(1-\lambda )b_{q'}$. Again, by continuity of $h$, it follows that there is $\beta \in (0,1)$, such that $(1-\beta )a_{q}+\beta b_{q}=(1-\beta )a_{q'}+\beta b_{q'}$. Assume that $q>q'$. Then, by rearranging the terms of the latter identity, I obtain that $\beta =(a_{q}-a_{q'})/(a_{q}-a_{q'}+b_{q'}-b_{q})$. Since $a_{q}-a_{q'}>0$ and $\beta <1$, it follows that $b_{q'}-b_{q}>0$ contradicting the choice of $b$. The case $q<q'$ is handled in a similar way. So, $f$ is a projection over $S^{\circ }$ and, therefore, on $S$.$\square$

Now I elaborate on how to apply the previous Lemma to the upcoming proof of Theorem 2. First of all, let $U=(u_{j})\in \mathcal {U}^{n}$ a given profile. As already mentioned in the introduction, $U$ can be viewed as a function from $X\times N$ into $\mathbb {R}$. Indeed, $(x,j)\in X\times N\longrightarrow U(x,j)=u_{j}(x)$, for every $x\in X, j\in N$. Alternatively, $U$ can be viewed as a vector in $\mathbb ({R}^{\#X})^{n}$. To see this, write $X$ as $X=\{x_{1},\ldots ,x_{m}\}$, so that #$X=m$. Then, $U$ can be identified with a vector of $\mathbb ({R}^{m})^{n}$, denoted by $\mathbf u$, in the following way: $\mathbf{u}=(\mathbf u_{j})_{j\in N}$, where, for each $j\in N, \mathbf{u}_{j}=(u_{ij})_{i\in M}=(U(x_{i},j))_{i\in M}$. So, $\mathbf {u}$ can be also viewed as an $m\times n$ real matrix where, for each $j\in N$, the $j$-th column of this matrix is the vector of $\mathbb {R}^{m}$ corresponding to the $j$-th component, $\mathbf{u}_{j}$, of $\mathbf {u}$. With this matrix view, the components of $\mathbf {u}$ are column-vectors with $m$ coordinates instead of row-vectors in $\mathbb {R}^{m}$.

Consider now the following binary relation, denoted by $\approx$, defined on $\mathcal {U}^{n}$: Given two profiles $U=(u_{j}),V=(v_{j})\in \mathcal {U}^{n}, U\approx V\Longleftrightarrow (\ u_{l}(t)\le u_{p}(z)\Leftrightarrow v_{l}(t)\le v_{p}(z)\ )$, for every $l,p\in N, t,z\in X.$ ²⁰ It is easy to see that $\approx$, just defined, is an equivalence relation on $\mathcal {U}^{n}$. In words, $U,V\in \mathcal {U}^{n}$ are related, according to $\approx$, if and only if, viewed as vectors in $\mathbb ({R}^{\#X})^{n}$, they belong to the same rank-ordered set.

Let now $F$ be a SEF and retain the description of $X$ as above, i.e., $X=\{x_{1},\ldots ,x_{m}\}$. Then $F$ can be viewed as a vector-valued function defined on $\mathbb ({R}^{m})^{n}$ (alternatively, on the space of all real matrices of size $m\times n$) and taking its values in $\mathbb {R}^{m}$ (alternatively, on the space of all real matrices of size $m\times 1$). Indeed, define $\mathbf{F}:\mathbb ({R}^{m})^{n} \rightarrow \mathbb {R}^{m}$ as follows: $\mathbf{u} \in \mathbb ({R}^{m})^{n}\longrightarrow \mathbf{F}(\mathbf{u})=(F(U)(x_{i}))_{i\in M}\in \mathbb {R}^{m}$ (or, alternatively, the matrix version of $\mathbf{F}$). Here, the correspondence between $\mathbf{u}=(u_{ij})\in \mathbb ({R}^{m})^{n}$ and $U=(u_{j})\in \mathcal {U}^{n}$ given above is used. In a similar way, each component of $F, F^{x_{i}}:\mathcal {U}^{n}\longrightarrow \mathbb {R}$, can be viewed as a real-valued function of $m\times n$ variables, denoted by $\mathbf{F}^{i}$. Indeed, define $\mathbf{F}^{i}$ as follows: $\mathbf{u}\in \mathbb ({R}^{m})^{n}\longrightarrow \mathbf{F}^{i}(\mathbf{u})=F^{x_{i}}(U)=F(U)(x_{i})\in \mathbb {R}$ (or, alternatively, the matrix version of $\mathbf{F}$).

According to the notation and relationship between the concepts just introduced, let $F$ be a SEF that satisfies OSP. Then each of its components, say $F^{x_{i}}$ also satisfies OSP. This means that the corresponding real-valued function of $m\times n$ variables, $\mathbf{F}^{i}$ is ordinal-scale-preserving. So, by part (i) of the Lemma, $\mathbf{F}^{i}(\mathbf{u})\in \{u_{k,j}\}_{k\in M, j\in N}$. But this amounts to saying that $F^{x_{i}}(U)\in \{u_{j}(x_{k})\}_{k\in M, j\in N}$, as will be used in the proof of Theorem 2 below. In addition, if $F$ is continuous then each of its components, say $F^{x_{i}}$, is continuous too. But then the corresponding function $\mathbf{F}^{i}$ is ordinal-scale-preserving and continuous. So, if $S\subset \mathbb ({R}^{m})^{n}$ is a rank-ordered set then, by part (ii) of the Lemma, there exist $k\in M, l\in N$, such that $\mathbf{F}^{i}(\mathbf{u})=u_{kl}$, for all $\mathbf{u}=(u_{pj})_{p\in M, j\in N}\in S$. Let now $U=(u_{j}),V=(v_{j})\in \mathcal {U}^{n}$ such that $U\approx V$. By using the correspondence between profiles and vectors (matrices) as above, this means that $\mathbf{u}$ and $\mathbf v$ belong to the same rank-ordered set of $\mathbb ({R}^{m})^{n}$. So, by the previous observation, it follows that $F^{x_{i}}(U)=u_{l}(x_{k})$ and $F^{x_{i}}(V)=v_{l}(x_{k})$, as will be employed in the proof of Theorem 2.

Proof of Theorem 2

The proof of the “if” part is easy. So, I will concentrate on the “only if” part. The proof of this implication is carried out in several steps. Firstly, some useful notation will be needed. Let $a=(a_{j})\in \mathbb {R}^{n}$ be fixed. Then a profile, denoted by $U^{a}=(u_{j}^{a})\in \mathcal {U}^{n}$, associate with $a$ is defined as follows: $u_{j}^{a}=u^{a_{j}}$, for all $j\in N$. That is, for each $j\in N, u_{j}^{a}(x)=a_{j}$, for all $x\in X$. Denote by $\mathcal {A}=\{U^{a}\in \mathcal {U}^{n}: a\in \mathbb {R}^{n}\}$.

Step (1) I show that $F(U^{a})(x)=F(U^{a})(y)$, for all $a\in \mathbb {R}^{n}$, for all $x,y\in X$ (equivalently, $F^{x}\vert _{\mathcal {A}}\equiv F^{y}\vert _{\mathcal {A}}$, for all $x,y\in X$). So, when restricted to the sub-domain $\mathcal {A}\subset \mathcal {U}^{n}$, all components of $F$ are the same. To prove this assertion, first note that, since $F$ satisfies U, $F(U^{0_{n}})(x)=0$, for all $x\in X$. Let now $a=(a_{j})\in \mathbb {R}^{n}$ be fixed and consider $\Phi =(\phi _{j})\in \Delta ^{n}$ such that $\phi _{j}(0)=a_{j}$, for every $j\in N$. Observe that $\Phi U^{0_{n}}=U^{a}$. Let now $x,y\in X$ be given. Then, since $F(U^{0_{n}})(x)=F(U^{0_{n}})(y)=0$ it holds, by OMIN, that $F(U^{a})(x)=F(\Phi U^{0_{n}})(x)=F(\Phi U^{0_{n}})(y)=F(U^{a})(y)$, so proving Step (1).

Step (2) There is $i\in N$ such that $F(U^{a})(x)=F(U^{a})(y)=a_{i}$, for every $a\in \mathbb {R}^{n}$, for every $x,y\in X$ (in other words, $F\vert _{\mathcal {A}}$ is the projection over the $i$-th component). To prove this assertion, consider the following family $\mathcal {T}$ of subsets of $N$ defined in the following way: $T\in \mathcal {T}\Longleftrightarrow$ for every $U^{a}\in \mathcal {U}^{n}$ there is $t\in T$ such that $F(U^{a})(x)=a_{t}$, for some $x\in X$. Notice that, by Step (1), the definition of $T$ is independent of $x$. So, for simplicity, I will write $F(U^{a})$ instead of $F(U^{a})(x)$. Proof’s strategy consists in showing that the family $\mathcal {T}$ is an ultrafilter on $N$. Since $N$ is finite, it is well-known that the ultrafilter is fixed (see Dugundji (1966)). That is, there is $i\in N$ such that $\mathcal {T}=\{T\subseteq N: i\in T\}$. In particular, $\{i\}\in \mathcal {T}$ and the conclusion then follows. Showing that $\mathcal {T}$ is an ultrafilter on $N$ requires checking four properties:

• (i) $N\in \mathcal {T}, \emptyset \notin \mathcal {T}$. These two facts follow directly since $F$ satisfies OSP. So, by part (i) of the Lemma, it holds that $F(U^{a})\in \{a_{1},\ldots ,a_{n}\}$, for every $a\in \mathbb {R}^{n}$.

• (ii) For arbitrary subsets $H$ and $K$ of $N$, if $H\in \mathcal {T}$ and $H\subseteq K$, then $K\in \mathcal {T}$. This is obvious from the definition of the family $\mathcal {T}$.

• (iii) For arbitrary members $H$ and $K$ of $\mathcal {T}$, the set $H\bigcap K$ belongs to $\mathcal {T}$. Indeed, first of all note that $H\bigcap K\ne \emptyset$ (for if $H\bigcap K=\emptyset$, then by considering the profile $U^{a}$, where $a_{j}=1$, if $j\in K$ and $a_{j}=0$, otherwise, it is obtained, by part (i) of the Lemma again, that either $F(U^{a})=1$, or $F(U^{a})=0$. The first situation contradicts the fact that $H\in \mathcal {T}$, whereas the second case contradicts the fact that $K\in \mathcal {T}$). Suppose, by way of contradiction, that $H\bigcap K\notin \mathcal {T}$. Then there is a profile, say $U^{a}$, such that $F(U^{a})\notin \{a_{j}: j\in H\bigcap K\}$, and, moreover, since $H,K\in \mathcal {T}, F(U^{a})=a_{h}=a_{k}$, for some $h\in H\setminus (H\bigcap K), k\in K\setminus (H\bigcap K)$. Denote by $L=\{j\in K\setminus (H\bigcap K): a_{j}=a_{k}\}$. Note that, since $H\subset N\setminus L$, by (ii) above it follows that $N\setminus L\in \mathcal {T}$. Let $\delta >0$ and consider the profile $U^{b}$, where $b_{j}=a_{j}$, if $j\in N\setminus L$, and $b_{j}=a_{k}+\delta$, otherwise. Now, and as a consequence of part (i) of the Lemma and the fact that $F$ fulfils C, it follows that, for $\delta$ small enough, $F(U^{b})=a_{k}+\delta$, contradicting that $N\setminus L\in \mathcal {T}$ (notice that, for $\delta$ small enough, $a_{k}+\delta \notin \{b_{j}:j\in N\setminus L\}$).

• (iv) For an arbitrary subset $H$ of $N$, if $H\notin \mathcal {T}$ then $N\setminus H\in \mathcal {T}$. Suppose, on the contrary, that there is a subset $H\subset N$ such that neither $H$ nor $H^{c}:=N\setminus H$ belong to $\mathcal {T}$. This means that there are profiles $U^{a}, U^{b}\in \mathcal {A}$, and natural numbers $k\in H^{c}, l\in H$ such that $F(U^{a})=a_{k}, F(U^{b})=b_{l}$ and, in addition, $a_{j}\ne a_{k}$, for every $j\in H$; $b_{j}\ne b_{l}$, for every $j\in H^{c}$.

Let $\epsilon >0$ and $x\in X$ be fixed. Define the profile $U^{\epsilon }=(u^{\epsilon }_{j})\in \mathcal {U}^{n}$ as follows: For each $j\in H, u^{\epsilon }_{j}=u^{a_{j}}=a_{j}$; for each $j\in H^{c}, u^{\epsilon }_{j}(x)=a_{j}+\epsilon$, and $u^{\epsilon }_{j}(z)=a_{j}$, for every $x\ne z\in X$.

Notice that $U^{\epsilon }\mathop {\longrightarrow }\limits _{\epsilon \rightarrow 0}U^{a}$. So, since $F$ satisfies C together with part (i) of the Lemma, there is $\bar{\epsilon }>0$, small enough, such that the corresponding profile $U^{\bar{\epsilon }}$ has the following property: For all $z\in X, F(U^{\bar{\epsilon }})(z)\in \{a_{k},a_{k}+\bar{\epsilon }\}$. In order to reach the desired contradiction, the following (mutually exclusive) cases that may occur are distinguished. Notice that on each of these three situations suitable perturbations of suitably chosen profiles, of $\mathcal {U}^{n}$ and $\Delta ^{n}$, are used in the argument:

• (a) There are $\bar{y},\hat{y}\in X$ such that $F(U^{\bar{\epsilon }})(\bar{y})=a_{k}+\bar{\epsilon }$ and $F(U^{\bar{\epsilon }})(\hat{y})=a_{k}$. Then, let $\delta >0$ be fixed, and consider the profile $V^{\delta }=(v^{\delta }_{j})\in \mathcal {U}^{n}$ defined as follows: For each $j\in H, v^{\delta }_{j}=v^{b_{j}}=b_{j}$; for each $j\in H^{c}, v^{\delta }_{j}(x)=b_{j}+\delta$, and $v^{\delta }_{j}(z)=b_{j}$, for every $x\ne z\in X$. In addition, consider a profile of strictly increasing real-valued functions $\Phi ^{\delta }=(\phi ^{\delta }_{j}) \in \Delta ^{n}$ defined with the following requirement: For each $j\in H, \phi ^{\delta }_{j}(a_{j})=b_{j}$; for each $j\in H^{c}, \phi ^{\delta }_{j}(a_{j})=b_{j}$, and $\phi ^{\delta }_{j}(a_{j}+\bar{\epsilon })=b_{j}+\delta$. Note that such a profile $\Phi ^{\delta }$ does exist and also that $\Phi ^{\delta }U^{\bar{\epsilon }}=V^{\delta }$. Now, since $V^{\delta }\mathop {\longrightarrow }\limits _{\delta \rightarrow 0}U^{b}$, by C and part (i) of the Lemma again there is $\bar{\delta }>0$, small enough, such that for the corresponding profile $V^{\bar{\delta }}$ it holds that $F(V^{\bar{\delta }})(z)=b_{l}$, for all $z\in X$. Now, since $F$ fulfils OMIN, it follows that $F(U^{\bar{\epsilon }})(\hat{y})=a_{k}<F(U^{\bar{\epsilon }})(\bar{y})=a_{k}+\bar{\epsilon } \Longleftrightarrow F(\Phi ^{\bar{\delta }}U^{\bar{\epsilon }})(\hat{y})<F(\Phi ^{\bar{\delta }}U^{\bar{\epsilon }})(\bar{y})$. But $F(\Phi ^{\bar{\delta }}U^{\bar{\epsilon }})(\bar{y})=F(V^{\bar{\delta }})(\bar{y})=b_{l}= F(\Phi ^{\bar{\delta }}U^{\bar{\epsilon }})(\hat{y})=F(V^{\bar{\delta }})(\hat{y})$, a contradiction. So, case (a) is impossible.

• (b) For all $z\in X, F(U^{\bar{\epsilon }})(z)=a_{k}$. If this is the case, then for each $\delta >0$, consider the profile $W^{\delta }=(w^{\delta }_{j})\in \mathcal {U}^{n}$ defined as follows: For each $j\in H^{c}, w^{\delta }_{j}=u^{\bar{\epsilon }}_{j}$; for each $j\in H, w^{\delta }_{j}(x)=a_{j}+\delta$, and $w^{\delta }_{j}(z)=a_{j}$, for every $x\ne z\in X$. Notice that $W^{\delta }\mathop {\longrightarrow }\limits _{\delta \rightarrow 0}U^{\bar{\epsilon }}$. So, by C and the assertion (i) of the Lemma once again, there is $\hat{\delta }>0$, small enough, such that for the corresponding profile $W^{\hat{\delta }}$ it holds that $F(W^{\hat{\delta }})(z)=F(W^{\hat{\delta }})(z)=a_{k}$, for all $z\in X$. Consider a profile of strictly increasing real-valued functions $\Phi ^{\hat{\delta }}=(\phi ^{\hat{\delta }}_{j}) \in \Delta ^{n}$ defined with the following requirement: For each $j\in N, \phi ^{\hat{\delta }}_{j}(a_{j})=a_{k}$, and $\phi ^{\hat{\delta }}_{j}(a_{j}+\bar{\epsilon })= \phi ^{\hat{\delta }}_{j}(a_{j}+\hat{\delta })=a_{k}+\bar{\epsilon }$. Note that such a profile $\Phi ^{\hat{\delta }}$ does exist and also that $\Phi ^{\hat{\delta }}W^{\hat{\delta }}=u1_{n}$, where $u\in \mathcal {U}$ is the real-valued function defined as follows: $u(x)=a_{k}+\bar{\epsilon }$, and $u(z)=a_{k}$, for all $x\ne z\in X$. The desired contradiction is then obtained by combining OMIN and U axioms. Indeed, take $x\ne y\in X$. Then, since $F(W^{\hat{\delta }})(x)=F(W^{\hat{\delta }})(y)=a_{k}$, by OMIN, it follows that $F(\Phi ^{\hat{\delta }}W^{\hat{\delta }})(x)=F(\Phi ^{\hat{\delta }}W^{\hat{\delta }})(y)$. But, by U, $F(\Phi ^{\hat{\delta }}W^{\hat{\delta }})(x)=F(u1_{n})(x)=u(x)=a_{k}+\bar{\epsilon }> F(\Phi ^{\hat{\delta }}W^{\hat{\delta }})(y)=F(u1_{n})(y)=u(y)=a_{k}$, a contradiction. So, case (b) is impossible.

• (c) For all $z\in X, F(U^{\bar{\epsilon }})(z)=a_{k}+\bar{\epsilon }$. This case is entirely analogous to case (b), so leading to a similar contradiction.

So, condition (iv) is also accomplished and, therefore, $\mathcal {T}$ is an ultrafilter on $N$ which ends Step (2).

Step (3) For the existing $i\in N$, found in Step (2), it holds that each component of $F, F^{x}$, only depends upon the values $\{u_{i}(z)\}_{z\in X}$. Suppose now, by way of contradiction, that there are $x\in X, U=(u_{j})\in \mathcal {U}^{n}$ such that $F(U)(x)=u_{k}(y)$, with $k\ne i$, for some $y\in X$ and $u_{k}(y)\ne u_{i}(z)$, for all $z\in X$. Assume, without loss of generality, that $u_{i}(x)< u_{k}(y)$. Denote by $A_{U}$ the (finite) subset of the reals $A_{U}=\{(u_{j}(z))_{j\in N,z\in X}\}\subset \mathbb {R}$ and observe, by part (i) of the Lemma, that $F(U)(x)\in A_{U}$. Assume the elements of $A_{U}$ ranked according to the usual order of the reals $\le$. Moreover, and for ease of notation, denote a typical element of $A_{U}$ by $r_{p}, p=1,\ldots ,$#$A_{U}$, so that $r_{p}< r_{q}$ if and only if $p<q$. So, in particular, there are distinct $\bar{p}, \bar{q}\in \{1,\ldots ,$#$A_{U}\}$ so that $u_{i}(x)=r_{\bar{p}}$ and $u_{k}(y)=r_{\bar{q}}$ (so, $\bar{p}< \bar{q}$). Let $\epsilon =$min$\{r_{p+1}-r_{p}:p=1,\ldots ,$#$A_{U}-1\}$. Now, let $\delta \in (0,\epsilon )$ be fixed and consider the profile, denoted by $U^{\delta }=(u^{\delta }_{j})\in \mathcal {U}^{n}$, defined as follows: For each $z\in X, j\in N$, define $u^{\delta }_{j}(z)=r_{\bar{p}}-(\bar{p}-p)\delta$, provided that $u_{j}(z)=r_{p}$, for some $p\in \{1,\ldots , \bar{p}\}$; $u^{\delta }_{j}(z)=r_{\bar{p}}+(p-\bar{p})\delta$, whenever $u_{j}(z)=r_{p}$, for some $p\in \{\bar{p}+1,\ldots , \bar{q}\}$; and, finally, $u^{\delta }_{j}(z)=r_{\bar{q}}+(p-\bar{q})\delta$, as long as $u_{j}(z)=r_{p}$, for some $p\in \{\bar{q}+1,\ldots ,$#$A_{U}\}$. In particular, observe that $u^{\delta }_{i}(x)=u_{i}(x)=r_{\bar{p}}$ and also that $u^{\delta }_{k}(y)=u_{k}(y)=r_{\bar{q}}$.

Now, it can be shown that, for each $\delta \in (0,\epsilon ), U^{\delta }$ and $U$ are comonotonic profiles (i.e., $U^{\delta }\approx U$, for every $\delta \in (0,\epsilon )$. See the discussion just before the proof of Theorem 2). Moreover, observe that $U^{\delta }\mathop {\longrightarrow }\limits _{\delta \rightarrow 0}V=(v_{j})\in \mathcal {U}^{n}$, where $v_{j}(z)\in \{r_{\bar{p}}, r_{\bar{q}}\}$, for all $z\in X$. So, $V$, viewed as a function from $X\times N$ into $\mathbb {R}$ only takes two values; namely, either $r_{\bar{p}}$, or $r_{\bar{q}}$. Then, by the assertion of part (ii) of the Lemma, I find that $F(U^{\delta })(x)=u^{\delta }_{k}(y)=u_{k}(y)=r_{\bar{q}}$, for every $\delta \in (0,\epsilon )$. Now, the continuity of $F$, implies that $F(V)(x)=r_{\bar{q}}$. To attain the desired contradiction, let $l\in \mathbb {N}$ and consider a sequence $(\Phi ^{l})_{l\in \mathbb {N}}\subset \Delta ^{n}$, where, for each $l\in \mathbb {N}, \Phi ^{l}=(\phi ^{l}_{j})\in \Delta ^{n}$ fulfils the following conditions: $\phi ^{l}_{i}(r_{\bar{p}})=r_{\bar{p}}, \phi ^{l}_{i}(r_{\bar{q}})=r_{\bar{p}}+1/l$; $\phi ^{l}_{j}(r_{\bar{p}})=r_{\bar{q}}, \phi ^{l}_{j}(r_{\bar{q}})=r_{\bar{q}}+1/l$, provided that $j\ne i$. Note that $\Phi ^{l}V\approx V$, for every $l\in \mathbb {N}$. So, by part (ii) of the Lemma again, $F(\Phi ^{l}V)(x)=r_{\bar{q}}+1/l$, for every $l\in \mathbb {N}$. Furthermore, observe that $\Phi ^{l}V\mathop {\longrightarrow }\limits _{l\rightarrow \infty } U^{a}\in \mathcal {A}$, where $a=(a_{j})\in \mathbb {R}^{n}$ is given by $a_{i}=r_{\bar{p}}$, and $a_{j}=r_{\bar{q}}$, whenever $j\ne i$. So, the continuity of $F$ again, shows that $F(U^{a})(x)=r_{\bar{q}}$, contradicting the fact, proved in Step (2), that $F(U^{a})(x)=r_{\bar{p}}\ne r_{\bar{q}}$.

The case in which $u_{k}(y)< u_{i}(x)$ is analogous leading to a similar contradiction. So, for the existing $i\in N$, found in Step (2), it holds that each component of $F, F^{x}$, only depends upon the values $\{u_{i}(z)\}_{z\in X}$.

Step (4). By Step (3), it follows that $F(U)=F(u_{i}1_{n})$, for every $U=(u_{j})\in \mathcal {U}^{n}$. But then, by unanimity of $F$, it holds that $F(U_{u_{i}})=F(u_{i}1_{n})=u_{i}$. So, $F(U)=u_{i}$, for every $U=(u_{j})\in \mathcal {U}^{n}$, and I am done.                             $\square$