Limit representations of intergenerational equity

Abstract

Equal treatment of all generations is a fundamental ethical principle in intertemporal welfare economics. This principle is expressed in anonymity axioms of orderings on the set of infinite utility streams. We first show that an ordering satisfies finite anonymity, uniform Pareto, weak non-substitution, and sup continuity if and only if it is represented by an increasing, continuous function that is a natural extension of the limit function. We then show that whenever such an ordering is infinitely anonymous, it depends only on the liminf and limsup of any utility stream. Our results imply that in ethically ranking utility streams, reflecting only infinitely long-run movements is possible, with liminf and limsup particularly essential, but it is impossible to respect finite generations.

Appendix

1.1 Proof of Theorem 1

Our proof proceeds in several lemmas, which may have independent interest. We only show that if an ordering on $\ell ^{\infty }$ satisfies uniform Pareto, finite anonymity, weak non-substitution, and sup continuity, then it is represented by a generalized limit function, since the converse statement obviously holds.

It is easy to see that uniform Pareto and sup continuity imply the following mild Pareto axiom:

• Monotonicity. For each $x,y\in \ell ^{\infty }$,    if $x_t \ge y_t$ for all $t\in {\mathbb {N}}$,    then $x\succsim y$.

We next introduce a new Pareto axiom that does not care utility levels of finite generations:

• Tail Pareto. For each $x,y\in \ell ^{\infty }$,    if there exist $s\in {\mathbb {N}}$ and $\varepsilon >0$ for which $x_t > y_t +\varepsilon$ for all $t\ge s$,    then $x{\mathrel {\succ }}y$.

Tail Pareto implies uniform Pareto and weak non-substitution, and tail Pareto and sup continuity together imply monotonicity. The normative desirability of tail Pareto might be controversial, but our first lemma shows that it is implied by uniform Pareto, finite anonymity, and weak non-substitution.

Lemma 1

If an ordering $\succsim$ on $\ell ^{\infty }$ satisfies uniform Pareto, finite anonymity, and weak non-substitution, then $\succsim$ satisfies tail Pareto.

Proof

Let $x,y\in \ell ^{\infty }$ be such that there exist $s\in {\mathbb {N}}$ and $\varepsilon >0$ for which $x_t > y_t +\varepsilon$ for all $t\ge s$. Our purpose is to prove $x{\mathrel {\succ }}y$. Let $a\in {\mathbb {R}}$ be such that $a+\varepsilon <\min \{x_1,x_2,\ldots ,x_s, y_1,y_2,\ldots ,y_s \}$.

Take some $\delta \in (0,\varepsilon )$. Define $z^1 \in \ell ^{\infty }$ by

$$\begin{aligned} z^1_1\equiv & {} a,\\ z^1_t\equiv & {} y_t +\frac{1}{s}\delta \quad \text {for all } t\ge 2. \end{aligned}$$

By weak non-substitution, $z^1\succsim y$. Also, define $\tilde{z}^1 \in \ell ^{\infty }$ by replacing $z^1_1$ and $z^1_2$ in $z^1$, that is,

$$\begin{aligned} {\tilde{z}}^1_1\equiv & {} y_2+\frac{1}{s}\delta ,\\ {\tilde{z}}^1_2\equiv & {} a,\\ {\tilde{z}}^1_t\equiv & {} y_t +\frac{1}{s}\delta \quad \text {for all } t\ge 3. \end{aligned}$$

By finite anonymity, ${\tilde{z}}^1 {\mathrel {\sim }}z^1$.

For each integer q with $2\le q \le s$, define $z^q \in \ell ^{\infty }$ by

$$\begin{aligned} z^q_1\equiv & {} a,\\ z^q_2\equiv & {} a+\frac{q-1}{s}\delta ,\\ z^q_3\equiv & {} a+\frac{q-2}{s}\delta ,\\ z^q_4\equiv & {} a+\frac{q-3}{s}\delta ,\\&\ldots&\\ z^q_q\equiv & {} a+\frac{1}{s}\delta ,\\ z^q_t\equiv & {} y_t +\frac{q}{s}\delta \quad \forall t\ge q+1. \end{aligned}$$

Next, for each integer q with $2\le q \le s-1$, define ${\tilde{z}}^q \in \ell ^{\infty }$ by replacing $z^q_1$ with $z^q_{q+1}$ in $z^q$, that is,

$$\begin{aligned} {\tilde{z}}^q_1\equiv & {} y_{q+1}+\frac{q}{s}\delta ,\\ {\tilde{z}}^q_2\equiv & {} a+\frac{q-1}{s}\delta ,\\ {\tilde{z}}^q_3\equiv & {} a+\frac{q-2}{s}\delta ,\\ {\tilde{z}}^q_4\equiv & {} a+\frac{q-3}{s}\delta ,\\&\ldots&\\ {\tilde{z}}^q_q\equiv & {} a+\frac{1}{s}\delta ,\\ {\tilde{z}}^q_{q+1}\equiv & {} a,\\ {\tilde{z}}^q_t\equiv & {} y_t +\frac{q}{s}\delta \quad \forall t\ge q+2. \end{aligned}$$

For each integer q with $1\le q \le s-1$, finite anonymity implies ${\tilde{z}}^q {\mathrel {\sim }} z^q$, and weak non-substitution implies $z^{q+1} \succsim {\tilde{z}}^q$. Therefore, $z^s \succsim z^1$. Furthermore, uniform Pareto implies $x {\mathrel {\succ }} z^s$ because

$$\begin{aligned} x_t>&z^s_t \quad \forall t\le s,\\ x_t>&z^s_t +(\varepsilon -\delta ) \quad \forall t>s. \end{aligned}$$

Overall, $x{\mathrel {\succ }} z^s \succsim z^1 \succsim y$, as desired. $\square$

Lemma 2

If an ordering $\succsim$ on $\ell ^{\infty }$ satisfies tail Pareto and sup continuity, then for each $x,y\in \ell ^{\infty }$ such that there exists $s\in {\mathbb {N}}$ for which $x_t = y_t$ for all $t\ge s$, we have $x{\mathrel {\sim }}y$.

Proof

Consider any ordering $\succsim$ on $\ell ^{\infty }$ satisfying tail Pareto and sup continuity. Note that $\succsim$ also satisfies monotonicity. Let $x,y\in \ell ^{\infty }$ be such that there exists $s\in {\mathbb {N}}$ for which $x_t = y_t$ for all $t\ge s$. Define

$$\begin{aligned} a'\equiv \max \{x_1,x_2,\ldots ,x_s, y_1,y_2,\ldots ,y_s\} \end{aligned}$$

and

$$\begin{aligned} a'' \equiv \min \{x_1,x_2,\ldots ,x_s, y_1,y_2,\ldots ,y_s\}. \end{aligned}$$

Define also

$$\begin{aligned} x'\equiv & {} (\underbrace{a',a', \ldots ,a'}_{s \text { terms}}, x_{s+1},x_{s+2},x_{s+3},\ldots ) \text { and }\\ x''\equiv & {} (\underbrace{a'',a'',, \ldots , a''}_{s \text { terms}}, x_{s+1},x_{s+2},x_{s+3},\ldots ). \end{aligned}$$

By monotonicity, $x' \succsim x\succsim x''$ and $x' \succsim y\succsim x''$. Since our purpose is to obtain $x{\mathrel {\sim }}y$, it suffices to show $x' {\mathrel {\sim }} x''$. Suppose, by contradiction, that $x' {\mathrel {\succ }} x''$. Then by sup continuity, there exists a small $\varepsilon >0$ such that

$$\begin{aligned} x' {\mathrel {\succ }}z\equiv (\underbrace{a'',a'', \ldots , a''}_{s \text { terms}}, x_{s+1}+\varepsilon ,x_{s+2}+\varepsilon ,x_{s+3}+\varepsilon ,\ldots ). \end{aligned}$$

On the other hand, tail Pareto implies $z{\mathrel {\succ }}x'$, a contradiction. Thus we obtain $x' {\mathrel {\sim }} x''$, and hence $x{\mathrel {\sim }}y$. $\square$

When there is no danger of confusion, we denote by $\varvec{a}\equiv (a,a,\ldots )\in \ell ^{\infty }$ the constant utility stream consisting of $a\in {\mathbb {R}}$. Note that $\varvec{a}=(a)_{rep}$. The following lemma is essentially due to Debreu (1954) and Diamond (1965, p. 172), but we write its proof for completeness.⁹

Lemma 3

If an ordering $\succsim$ on $\ell ^{\infty }$ satisfies uniform Pareto and sup continuity, then for each $x\in \ell ^{\infty }$, there exists a unique $b\in {\mathbb {R}}$ such that $x{\mathrel {\sim }}\varvec{b}$.

Proof

By sup continuity, the set of constant streams $Y\equiv \{\varvec{a}\in \ell ^{\infty }: \exists a\in {\mathbb {R}}\}$ is a closed subset of $\ell ^{\infty }$ in the sup norm topology. By uniform Pareto, both $\{y\in Y: y\succsim x\}$ and $\{y\in Y: x\succsim y\}$ are non-empty, and they are closed with respect to the sup norm topology. Since no closed set can be partitioned by two disjoint closed sets in the sup norm topology, they have a non-empty intersection, that is, there exists $b\in {\mathbb {R}}$ such that $x{\mathrel {\sim }}\varvec{b}$. Uniqueness of such b is ensured by uniform Pareto. $\square$

Lemma 4

If an ordering $\succsim$ on $\ell ^{\infty }$ satisfies tail Pareto and sup continuity, then for each $x\in \ell ^{\infty }$ such that $\lim x_t$ exists, whenever $a\equiv \lim x_t$, we have $x {\mathrel {\sim }} \varvec{a}.$

Proof

Consider any ordering $\succsim$ on $\ell ^{\infty }$ satisfying tail Pareto and sup continuity. Note that $\succsim$ also satisfies monotonicity and uniform Pareto. Pick any $x\in \ell ^{\infty }$ such that $\lim x_t$ exists. Let $a\equiv \lim x_t$. By Lemma 3, there exists $b\in {\mathbb {R}}$ such that $x{\mathrel {\sim }} \varvec{b}$. It suffices to show that $a=b$. We first show that $a>b$ is impossible. If $a>b$, then, since $a=\lim x_t$, there exists $s\in {\mathbb {N}}$ such that for all $t\ge s$, $x_t>\dfrac{a+b}{2}$. By Lemma 2,

$$\begin{aligned} y\equiv \bigg (x_1,x_2,\ldots ,x_s,\frac{a+b}{2},\frac{a+b}{2},\frac{a+b}{2},\ldots \bigg ){\mathrel {\sim }}\varvec{\frac{a+b}{2}}. \end{aligned}$$

But monotonicity and uniform Pareto imply

$$\begin{aligned} x\succsim y {\mathrel {\sim }}\varvec{\frac{a+b}{2}} {\mathrel {\succ }}\varvec{b}, \end{aligned}$$

a contradiction. We next show that $a<b$ is also impossible. If $a<b$, then, since $a=\lim x_t$, there exists $s\in {\mathbb {N}}$ such that for all $t\ge s$, $x_t<\dfrac{a+b}{2}$. By Lemma 2,

$$\begin{aligned} z\equiv \bigg (x_1,x_2,\ldots ,x_s,\frac{a+b}{2},\frac{a+b}{2},\frac{a+b}{2},\ldots \bigg ) {\mathrel {\sim }} \varvec{\frac{a+b}{2}}. \end{aligned}$$

But uniform Pareto and monotonicity imply

$$\begin{aligned} \varvec{b} {\mathrel {\succ }}\varvec{\frac{a+b}{2}} {\mathrel {\sim }} z \succsim x, \end{aligned}$$

a contradiction. Overall, $a=b$, and hence $x {\mathrel {\sim }} \varvec{a}$. $\square$

The next lemma completes the proof of Theorem 1.

Lemma 5

If an ordering $\succsim$ on $\ell ^{\infty }$ satisfies uniform Pareto, finite anonymity, weak non-substitution, and sup continuity, then there exists a generalized limit function $F: \ell ^{\infty }\rightarrow {\mathbb {R}}$ that represents $\succsim$.

Proof

Consider any ordering $\succsim$ on $\ell ^{\infty }$ satisfying the four axioms. By Lemma 3, for each $x\in \ell ^{\infty }$, there exists a unique $b\in {\mathbb {R}}$ such that $x{\mathrel {\sim }}\varvec{b}$. Thus we can define the function $F:\ell ^{\infty }\rightarrow {\mathbb {R}}$ so that $F(x)=b$ with $x{\mathrel {\sim }}\varvec{b}$. It is obvious that g represents $\succsim$. Hence, it remains to show that g is a generalized limit function. Lemma 4 implies “limit selection,” Lemma 1 implies “tail monotonicity,” Lemma 2 implies “head insensitivity,” and sup continuity implies “continuity.” $\square$

1.2 Proof of Theorem 2

One can easily check that if an ordering $\succsim$ on $\ell ^{\infty }$ has an increasing, continuous function $W:{\mathbb {R}}^2\rightarrow {\mathbb {R}}$ satisfying

$$\begin{aligned} x\succsim y \Longleftrightarrow W(\underline{x},\overline{x}) \ge W(\underline{y},\overline{y}) \quad \forall x,y\in \ell ^{\infty }, \end{aligned}$$

(1)

then it satisfies infinite anonymity, uniform Pareto, weak non-substitution, and sup continuity. Therefore, we shall show that if an ordering satisfies these four axioms, then it is represented as in (1). The proof consists of the following three steps:

• Step 1. Define a dense subset ${\mathcal {U}}$ of $\ell ^{\infty }$, called the universal finite domain.

• Step 2. Clarify ranking of utility streams that belong to ${\mathcal {U}}$.

• Step 3. For every utility stream in $\ell ^{\infty }\setminus {\mathcal {U}}$, find another utility stream in ${\mathcal {U}}$ that is indifferent to it. This completes finding the entire ranking.

We begin the proof.

1.2.1 Step 1. Introducing the universal finite domain

Let ${\mathcal {S}}$ be the set of non-empty, finite subsets of ${\mathbb {R}}$, that is,

$$\begin{aligned} {\mathcal {S}}\equiv \{X\subset {\mathbb {R}} : 1\le |X|<\infty \}. \end{aligned}$$

Given any finite set of “utility parameters” $X\in {\mathcal {S}}$, its infinite product $X^{\infty }$ denotes the associated domain of utility streams. The universal finite domain is the union of all associated domains,

$$\begin{aligned} {\mathcal {U}}\equiv \bigcup _{X\in {\mathcal {S}}} X^{\infty }. \end{aligned}$$

Lemma 6

The set ${\mathcal {U}}$ is a dense subset of $\ell ^{\infty }$ in the sup norm topology; that is, for every $x\in \ell ^{\infty }$ and every $\varepsilon >0$, there exists $y\in {\mathcal {U}}$ such that $\sup _{t\in {\mathbb {N}}} |x_t -y_t|<\varepsilon$.

Proof

Pick any $x\in \ell ^{\infty }$ and $\varepsilon >0$. Let $M>0$ be a large integer such that

$$\begin{aligned} -M\varepsilon<x_t<M\varepsilon \quad \forall t\in {\mathbb {N}}. \end{aligned}$$

For every $t\in {\mathbb {N}}$, there exists a unique $m\in \{-M+1,-M+2,\ldots ,M-1,M\}$ such that $(m-1)\varepsilon < x_t \le m\varepsilon$, and then let $y_t\equiv (m-\frac{1}{2})\varepsilon$. Obviously, $\sup _{t\in {\mathbb {N}}} |x_t-y_t|\le \frac{1}{2}\varepsilon <\varepsilon$. Since $y=(y_1,y_2,\ldots )$ consists of at most finite number of real values, it follows that $y\in {\mathcal {U}}$.

Let ${\mathcal {N}}$ be the set of infinite subsets of ${\mathbb {N}}$, that is, ${\mathcal {N}}\equiv \{{\mathbb {N}}'\subset {\mathbb {N}}: |{\mathbb {N}}'|=\infty \}$. Pick any $x\in {\mathcal {U}}$, and let $A(x)\equiv \{a\in {\mathbb {R}} : \exists {\mathbb {N}}'\in {\mathcal {N}}, \forall t\in {\mathbb {N}}', x_t=a\}$ be the set of real values that appear in $x=(x_1,x_2,\ldots )$ infinitely many times. Since x is taken from the universal finite domain, A(x) consists of at most finite number of real values, that is, $1\le |A(x)| <\infty$. The key properties described by the next lemma should be recognized throughout the subsequent discussion.

Lemma 7

For every $x\in {\mathcal {U}}$, (i) $\underline{x}=\min _{a\in A(x)}a$ and $\overline{x}=\max _{a\in A(x)}a$, and (ii) there exists $s\in {\mathbb {N}}$ such that $\underline{x} \le x_t \le \overline{x}$ for all $t\ge s$.

Proof

Immediately follows from the definition of the universal finite domain. $\square$

We also remark that for every $x\in {\mathcal {U}}$, both $\max _{t\in {\mathbb {N}}}x_t$ and $\min _{t\in {\mathbb {N}}}x_t$ are well-defined.

1.2.2 Step 2. Clarify ranking on the universal finite domain

The next axiom is a stronger non-substitution condition. It obviously implies weak non-substitution.

• Strong non-substitution. For every $x,y\in \ell ^{\infty }$, if there exist a finite subset ${\mathbb {N}}'\subset {\mathbb {N}}$, an infinite subset ${\mathbb {N}}''\subset {\mathbb {N}}\setminus {\mathbb {N}}'$, and $\varepsilon >0$ such that

  $$\begin{aligned}&y_t<x_t \qquad \quad \text {for all }t\in {\mathbb {N}}',\nonumber \\&x_t +\varepsilon \le y_t \quad \text {for all }t\in {\mathbb {N}}'',\nonumber \\&x_t=y_t \qquad \quad \text {for all } t\notin {\mathbb {N}}'\cup {\mathbb {N}}'', \end{aligned}$$

  (2)

  then $y\succsim x$.

Lemma 8

If an ordering $\succsim$ on $\ell ^{\infty }$ satisfies infinite anonymity, uniform Pareto, weak non-substitution, and sup continuity, then it satisfies strong non-substitution.

Proof

Let $\succsim$ be an ordering on $\ell ^{\infty }$ satisfying the four axioms. By Theorem 1, there exists a generalized limit function $F:\ell ^{\infty } \rightarrow {\mathbb {R}}$ that represents $\succsim$. Consider any $x,y\in \ell ^{\infty }$ that satisfy the hypothesis of strong non-substitution, as in (2). Let us show $y\succsim x$. Define $z\in \ell ^{\infty }$ by $z_t \equiv x_t$ for all $t\in {\mathbb {N}}'$ and $z_t \equiv y_t$ for all $t\in {\mathbb {N}}\setminus {\mathbb {N}}'$. Then by definition,

$$\begin{aligned} F(y)=F(z)\ge F(x), \end{aligned}$$

so $y\succsim x$. $\square$

Lemma 9

If an ordering $\succsim$ on $\ell ^{\infty }$ satisfies infinite anonymity, monotonicity, and strong non-substitution, then for every $x\in {\mathcal {U}}$ with $\underline{x}<\overline{x}$, it holds that $(\underline{x},\overline{x})_{rep}\succsim x$.

Proof

Let $\succsim$ be an ordering on $\ell ^{\infty }$ satisfying the three axioms. Pick any $x\in {\mathcal {U}}$ with $\underline{x}<\overline{x}$. Let ${\mathbb {N}}_{1}\equiv \{t\in {\mathbb {N}}: x_t =\underline{x}\}$. Since $\underline{x}=\min _{a\in A(x)}a$ by Lemma 7, we have $|{\mathbb {N}}_1 |=\infty$, and hence there exists ${\mathbb {N}}_{2}\subset {\mathbb {N}}_{1}$ such that $|{\mathbb {N}}_2|=\infty$ and $|{\mathbb {N}}_1 \setminus {\mathbb {N}}_2 |=\infty$.

Define y by

$$\begin{aligned} \overline{x}<x_t&\Longrightarrow&y_t \equiv \underline{x},\\ t\in {\mathbb {N}}_{2}&\Longrightarrow&y_t\equiv \frac{\underline{x}+\overline{x}}{2},\\ \text { otherwise }&\Longrightarrow&y_t\equiv x_t. \end{aligned}$$

By strong non-substitution, $y\succsim x$.

Define z by

$$\begin{aligned} t\in {\mathbb {N}}_{2}&\Longrightarrow&z_t \equiv \overline{x},\\ t\notin {\mathbb {N}}_{2}&\Longrightarrow&z_t \equiv x_t . \end{aligned}$$

Since z is obtained from x by a permutation, by infinite anonymity, $x{\mathrel {\sim }}z$. By monotonicity, $z\succsim y$. Therefore, $x{\mathrel {\sim }}y$.

Define w by

$$\begin{aligned} \underline{x}<y_t&\Longrightarrow&w_t\equiv \overline{x},\\ y_t\le \underline{x}&\Longrightarrow&w_t\equiv \underline{x}. \end{aligned}$$

Then by monotonicity, $w\succsim y$, and hence $w\succsim x$. By infinite anonymity, $(\underline{x},\overline{x})_{rep}{\mathrel {\sim }}w$, and hence $(\underline{x},\overline{x})_{rep}\succsim x$. $\square$

Lemma 10

If an ordering $\succsim$ on $\ell ^{\infty }$ satisfies infinite anonymity, monotonicity, and strong non-substitution, then for every $x\in {\mathcal {U}}$ with $\underline{x}<\overline{x}$, it holds that $x\succsim (\underline{x},\overline{x})_{rep}$.

Proof

Let $\succsim$ be an ordering on $\ell ^{\infty }$ satisfying the three axioms. Pick any $x\in {\mathcal {U}}$ with $\underline{x}<\overline{x}$. Let ${\mathbb {N}}_{1}\equiv \{t\in {\mathbb {N}}: x_t =\overline{x}\}$. Let ${\mathbb {N}}_{2}\subset {\mathbb {N}}_{1}$ be such that $|{\mathbb {N}}_2|=\infty$ and $|{\mathbb {N}}_1 \setminus {\mathbb {N}}_2 |=\infty$. Define y by

$$\begin{aligned} t\in {\mathbb {N}}_{2}&\Longrightarrow&y_t \equiv \frac{\underline{x}+\overline{x}}{2},\\ x_t<\underline{x}&\Longrightarrow&y_t\equiv \underline{x},\\ \text { otherwise }&\Longrightarrow&y_t \equiv x_t . \end{aligned}$$

By strong non-substitution, $x\succsim y$.

Define z by

$$\begin{aligned} t\in {\mathbb {N}}_{2}&\Longrightarrow&z_t\equiv \underline{x},\\ t\notin {\mathbb {N}}_{2}&\Longrightarrow&z_t\equiv x_t. \end{aligned}$$

Since z is obtained from x by a permutation, by infinite anonymity, $x{\mathrel {\sim }}z$. By monotonicity, $y\succsim z$. Therefore, $y\succsim x$, so that $x{\mathrel {\sim }}y$.

Define w by

$$\begin{aligned} y_t < \overline{x}&\Longrightarrow&w_t\equiv \underline{x},\\ \overline{x} \le y_t&\Longrightarrow&w_t\equiv \overline{x}. \end{aligned}$$

Then by monotonicity, $y\succsim w$, and hence $x\succsim w$. By infinite anonymity, $(\underline{x},\overline{x})_{rep}{\mathrel {\sim }}w$, and hence $x \succsim (\underline{x},\overline{x})_{rep}$. $\square$

Lemma 11

If an ordering $\succsim$ on $\ell ^{\infty }$ satisfies uniform Pareto, monotonicity, and strong non-substitution, then for every $x\in {\mathcal {U}}$ and $a\in {\mathbb {R}}$ such that $\underline{x}> a$, it holds that $x{\mathrel {\succ }}(a)_{rep}$.

Proof

Let $\succsim$ be an ordering on $\ell ^{\infty }$ satisfying the three axioms. Pick any $x\in {\mathcal {U}}$ and $a\in {\mathbb {R}}$ such that $\underline{x}> a$.

Let $y\in {\mathcal {U}}$ be such that

$$\begin{aligned} x_t <\underline{x}&\Longrightarrow&y_t \equiv x_t,\\ \underline{x}\le x_t&\Longrightarrow&y_t \equiv \underline{x}. \end{aligned}$$

By monotonicity, $x\succsim y$. By strong non-substitution, $y\succsim \left( \dfrac{\underline{x}+a}{\ 2\ }\right) _{rep}$. By uniform Pareto, $\left( \dfrac{\underline{x}+a}{\ 2\ }\right) _{rep} {\mathrel {\succ }}(a)_{rep}$. Therefore, $x{\mathrel {\succ }}(a)_{rep}$. $\square$

Lemma 12

If an ordering $\succsim$ on $\ell ^{\infty }$ satisfies uniform Pareto, strong non-substitution, and sup continuity, then for every $x\in {\mathcal {U}}$ with $\underline{x}=\overline{x}$, it holds that $x{\mathrel {\sim }}(\underline{x})_{rep}$.

Proof

Let $\succsim$ be an ordering on $\ell ^{\infty }$ satisfying the three axioms. Let $x\in {\mathcal {U}}$ with $\underline{x}=\overline{x}$. Lemma 3 implies that there exists a unique $a\in {\mathbb {R}}$ such that $x{\mathrel {\sim }}(a)_{rep}$. It suffices to show that $a=\underline{x}$. Since $\succsim$ also satisfies monotonicity, by $x{\mathrel {\sim }}(a)_{rep}$ and Lemma 11, $a\ge \underline{x}$. Since $\underline{x}=\overline{x}$ and $x\in {\mathcal {U}}$, there exists $s\in {\mathbb {N}}$ such that $x_t=\underline{x}$ for all $t> s$.

Suppose, by contradiction, that $a> \underline{x}$. Define y by

$$\begin{aligned} y_t< & {} \inf _{t\in {\mathbb {N}}}x_t \quad \text { for all }t\le s,\\\equiv & {} \frac{a+\underline{x}}{2} \quad \text { for all }t>s. \end{aligned}$$

Then by strong non-substitution, $y\succsim x$. By uniform Pareto, $(a)_{rep}{\mathrel {\succ }}y$. Hence, $(a)_{rep}{\mathrel {\succ }}x$, a contradiction. Overall, $a=\underline{x}$ must hold. $\square$

We summarize characterizations obtained in this proof step. It exhibits an essence on how utility streams in ${\mathcal {U}}$ are ranked.

Lemma 13

If an ordering $\succsim$ on $\ell ^{\infty }$ satisfies infinite anonymity, uniform Pareto, weak non-substitution, and sup continuity, then for every $x\in {\mathcal {U}}$, it holds that $x {\mathrel {\sim }}(\underline{x},\overline{x})_{rep}$.

Proof

Let $\succsim$ be an ordering on $\ell ^{\infty }$ satisfying the four axioms. Note that $\succsim$ also satisfies monotonicity. Furthermore, by Lemma 8, $\succsim$ satisfies strong non-substitution.

Take any $x\in {\mathcal {U}}$. If $\underline{x}<\overline{x}$, then by Lemmas 9 and 10, $x {\mathrel {\sim }}(\underline{x},\overline{x})_{rep}$. If $\underline{x}=\overline{x}$, then by Lemma 12, $x {\mathrel {\sim }}(\underline{x},\overline{x})_{rep}$. $\square$

1.2.3 Step 3. Extending the characterization to $\ell ^{\infty }$

Lemma 14

For every $x\in \ell ^{\infty }$ and every $\varepsilon >0$, there exists $y\in {\mathcal {U}}$ such that $\underline{x}=\underline{y}$, $\overline{x}=\overline{y}$, and $\sup _{t\in {\mathbb {N}}} |x_t-y_t|<\varepsilon$.

Proof

Pick any $x\in \ell ^{\infty }$ and $\varepsilon >0$. By Lemma 6, there exists $z\in {\mathcal {U}}$ such that $\sup _{t\in {\mathbb {N}}} |x_t-z_t|<\varepsilon$.

Consider the case that $a\equiv \lim x_t$ exists. Then, there exists $s\in {\mathbb {N}}$ such that $|x_t -a|<\varepsilon$ for all $t\ge s$. Define y by

$$\begin{aligned} y_t\equiv & {} z_t \quad \text { for all }t \le s-1,\\\equiv & {} a \quad \ \text { for all }t\ge s. \end{aligned}$$

Obviously, $y\in {\mathcal {U}}$ and it satisfies the desired properties.

Consider the case that $\lim x_t$ does not exist. Then, there exists small $\delta <\varepsilon$ such that $\underline{x}+\delta <\overline{x}-\delta$. Furthermore, by the definitions of the liminf and limsup, there exists $s\in {\mathbb {N}}$ such that

$$\begin{aligned} \underline{x}-\delta<x_t< \overline{x}+\delta \quad \text { for all }t\ge s. \end{aligned}$$

(3)

Define y by

$$\begin{aligned} y_t\equiv & {} \underline{x}\quad \text {if }t\ge s \text { and } \underline{x}-\delta<x_t<\underline{x}+\delta ,\\\equiv & {} \overline{x} \quad \text {if }t\ge s \text { and } \overline{x}-\delta<x_t<\overline{x}+\delta ,\\\equiv & {} z_t \quad \text {otherwise}. \end{aligned}$$

Since $z\in {\mathcal {U}}$, by construction of y, we have $y\in {\mathcal {U}}$ and $\sup _{t\in {\mathbb {N}}} |x_t-y_t|<\varepsilon$. Note that there exist infinitely many $t\ge s$ with $\underline{x}-\delta<x_t<\underline{x}+\delta$, and infinitely many other $t\ge s$ with $\overline{x}-\delta<x_t<\overline{x}+\delta$. This fact and (3) imply that $\underline{y}=\underline{x}$ and $\overline{y}=\overline{x}$. $\square$

We now extend the result obtained for ${\mathcal {U}}$ in Lemma 13 to $\ell ^{\infty }$.

Lemma 15

If an ordering $\succsim$ on $\ell ^{\infty }$ satisfies infinite anonymity, uniform Pareto, weak non-substitution, and sup continuity, then for every $x\in \ell ^{\infty }$, it holds that $x{\mathrel {\sim }}(\underline{x},\overline{x})_{rep}$.

Proof

Let $\succsim$ be an ordering on $\ell ^{\infty }$ satisfying the four axioms. Pick any $x\in \ell ^{\infty }$. By Lemma 14, for every positive integer k, there exists $x^{k}\in {\mathcal {U}}$ such that $\underline{x}=\underline{x}^k$, $\overline{x}=\overline{x}^k$, and $\sup _{t\in {\mathbb {N}}} |x_t-x^k_t|<\frac{1}{k}$. By Lemma 13, $(\underline{x},\overline{x})_{rep}{\mathrel {\sim }}x^k$ for all k. Thus by sup continuity, $(\underline{x},\overline{x})_{rep}{\mathrel {\sim }}x$. $\square$

The next lemma completes the proof of Theorem 2.

Lemma 16

If an ordering $\succsim$ on $\ell ^{\infty }$ satisfies infinite anonymity, uniform Pareto, weak non-substitution, and sup continuity, then there exists an increasing, continuous function $W:{\mathbb {R}}^2\rightarrow {\mathbb {R}}$ such that for every $x,y\in \ell ^{\infty }$, $x\succsim y \Longleftrightarrow W(\underline{x},\overline{x}) \ge W(\underline{y},\overline{y})$.

Proof

Let $\succsim$ be an ordering on $\ell ^{\infty }$ satisfying the four axioms.

Take any $z\in \ell ^{\infty }$. By Lemma 3, there exists a unique $a(z)\in {\mathbb {R}}$ such that $z{\mathrel {\sim }}(a(z))_{rep}$. By Lemma 15, $z{\mathrel {\sim }}(\underline{z},\overline{z})_{rep}$. Therefore $a(z)=a((\underline{z},\overline{z})_{rep})$, and define

$$\begin{aligned} W\big (\underline{z},\overline{z}\big )\equiv a\big (\big (\underline{z},\overline{z}\big )_{rep}\big ). \end{aligned}$$

Note that $W:{\mathbb {R}}^2 \rightarrow {\mathbb {R}}$.

For each $x,y\in \ell ^{\infty }$, by properties of $a(\ \cdot \ )$ and uniform Pareto,

and

so that

$$\begin{aligned} x\succsim y \iff W\big (\underline{x},\overline{x}\big ) \ge W\big (\underline{y},\overline{y}\big ). \end{aligned}$$

Furthermore, uniform Pareto implies that W is increasing, and sup continuity implies that W is continuous, as desired. $\square$

1.3 Proof of Corollary 2

We only prove the “only if” part. Consider any $\succsim$ on $\ell ^{\infty }$ satisfying the listed five axioms. Let W be a function associated with $\succsim$, which is established by Theorem 2. The proceeding proof depends on a standard Hammond-type proof technique (e.g., Bosmans and Ooghe 2013). It suffices to show that for all $x,y\in \ell ^{\infty }$, ($\star$) $\underline{x}> \underline{y}$ implies $W(\underline{x},\overline{x}) > W(\underline{y},\overline{y})$, and ($\star \star$) $\underline{x}= \underline{y}$ implies $W(\underline{x},\overline{x}) = W(\underline{y},\overline{y})$.

($\star$) Suppose, by contradiction, that for some $x,y\in \ell ^{\infty }$, $\underline{x}> \underline{y} \text { and } W(\underline{y},\overline{y})\ge W(\underline{x},\overline{x})$. Let $a\in {\mathbb {R}}$ be $\underline{x}>a>\underline{y}$. By pairwise Hammond, $(a)_{rep} \succsim (\underline{y},\overline{y})_{rep}$, and hence $W(a,a) \ge W(\underline{y},\overline{y})$. But then $W(a,a) \ge W(\underline{x},\overline{x})$, so $(a)_{rep}{\mathrel {\succ }}(\underline{x},\overline{x})_{rep}$. This contradicts uniform Pareto.

($\star \star$) Suppose, by contradiction, that for some $x,y\in \ell ^{\infty }$, $\underline{x}= \underline{y} \text { and }W(\underline{x},\overline{x}) > W(\underline{y},\overline{y})$. By Lemma 3, there exists $a\in {\mathbb {R}}$ such that $x{\mathrel {\sim }}(a)_{rep}$. Since $W(a,a)=W(\underline{x},\overline{x})$, one has $W(a,a)>W(\underline{y},\overline{y})$, and then increasingness of W implies $a>\underline{y}$, which in turn implies $a>\underline{x}$. Then there exists $b\in {\mathbb {R}}$ with $a>b>\underline{x}$, and by uniform Pareto and pairwise Hammond, $W(a,a)>W(b,b)\ge W(\underline{x},\overline{x})$. This contradicts $x{\mathrel {\sim }}(a)_{rep}$. $\Box$