# Does every allocation have a maximal Pareto-improvement?

Consider an economy with a finite number of goods and a finite quantity of each good. Each agent $i$ has a preference-relation $\succeq_i$ which is a total, reflexive and transitive relation over the set of bundles. Given an allocation $x$ of the goods among the agents, say that $x^*$ is a maximal Pareto-improvement of $x$ if:

• $x^*$ is Pareto-optimal,
• Every agent weakly prefers the allocation $x^*$ over the allocation $x$.

What is the weakest condition on the preference-relations $\succeq_i$ that guarantees that every allocation has a maximal Pareto-improvement?

I would guess that it is somehow related to compactness, however, I do not see how to define this condition on preference-relations.

• Some clarification: I am guessing you assume the usual conditions of reflexivity, completeness and transitivity for the preference relation that make it an ordering? Do you also assume continuity? – Giskard Dec 11 '17 at 15:54
• @denesp yes. By "continuity" you mean that, for every bundle $x$, the set $\{y|y \succeq_i x\}$ is a closed set? – Erel Segal-Halevi Dec 11 '17 at 16:06
• Yes, that is what I mean. – Giskard Dec 11 '17 at 16:23
• Are you sure you want your preferences to be anti-symmetric? There are no complete, transitive, continuous, anti-symmetric preferences defined on a subset of $\mathbb{R}^l$ for $l\geq 2$ with nonempty interior. – Michael Greinecker Dec 11 '17 at 22:45
• @TheoreticalEconomist E.g. Pareto-improvements do not have to be Pareto-optimal. – Giskard Dec 12 '17 at 17:57

I think there is a short proof if you also assume that the number of agents $n$ is finite and that the preferences are continuous.

Given the second assumption Debreu's theorem (1954, "Representation of a preference ordering by a numerical function") states that a continuous utility function exists which represents the preferences. I will denote the utility function representing the preference of agent $i$ by $U_i$.

Though this is somewhat uncommon, I will denote the utility agent $i$ gets from an allocation $x$ by $U_i(x)$.

An algorithm resulting in a maximum Pareto-improvement of $x$:
Step 1. Initially set $i=1$ and set $y = x$.
Step 2. Maximize the utility of agent $i$ over the set of allocation where all other agents $j$ have at least utility $U_j(y)$. As all functions $U_j$ are continuous, there is a finite number of goods and a finite quantity of each good, for all $j$ the upper set of $U_j$ is a closed set. A finite intersection of closed sets is also a closed set. As $U_i$ is also continuous, it will have a maximum over this set. Select an allocation $y'$ which maximizes $U_i$ and let this be the our allocation $y$ from now on, so set $y = y'$.
Step 3. If $i = n$, stop. Otherwise set $i = i+1$ and proceed to step 2.

Notice the following:

• Any move from $y$ to $y'$ in Step 2. is a Pareto-improvement.
• Pareto-improvements are transitive, therefore the set of Pareto-improvements is narrowing, i.e. if $y'$ is a Pareto-improvement to $y$ then the set of Pareto-improvements to $y'$ is included in the set of Pareto-improvements to $y$.
• As the utility of agent $i$ is maximized in Step 2. over the set of Pareto-improvements to $y$, it is impossible that his utility can be further increased by a Pareto-improvement. Therefore once we reach Step 3. for $i = n$ we have a Pareto-optimal allocation $y$, that is either a Pareto-improvement of $x$ or where $\forall i$: $x\sim_i y$. Hence $y$ is a maximal Pareto-improvement of $x$.

A comment: It seems to me you could replace continuity with monotonicity, but that would require another proof. Hopefully someone else is up to it!

• Instead of continuity, is it sufficient to assume "half-continuity", i.e, only the "greater than" sets are closed? – Erel Segal-Halevi Dec 11 '17 at 18:06
• @ErelSegal-Halevi I don't think so. I need $U_i$ to be continuous so that it has a maximum over the closed set, and for the continuity of $U_i$ I need the continuity of $\preceq_i$. – Giskard Dec 11 '17 at 18:39
• You only need "half-continuity". We want to show that a $\preceq$ maximum exists on compact set, that is an element $y$ such that $x\preceq y$ for each $x$ in the compact set. This is equivalent to $y$ being in the intersection of the weakly better sets corresponding to each $x$. Since this is an intersection of compact sets, it suffices by the finite intersection principle to show that the intersection of finitely many such compact weakly better sets is nonempty. But this follows simply from the fact that a complete transitive relation on nonempty finite set must have a maximum. – Michael Greinecker Dec 11 '17 at 23:12
• @MichaelGreinecker Sadly I must plead ignorance of the finite intersection principle. Michael if you post an answer based on that I would upvote it. – Giskard Dec 12 '17 at 10:32

This is essentially a variation on the answer of denesp that requires slightly fewer assumptions.

Assume there are $l$ commodities and $m$ agents. An allocation is then a point in $\mathbb{R}^{lm}_+$. If the aggregate endowment is $e\in\mathbb{R^l}_+$, an allocation is a point in $\sum^{-1}(\{e\})$, where $\sum:\mathbb{R}^{lm}_+\to\mathbb{R}^l$ is the continuous "summation function". Since this function is continuous and the set $\{e\}\subseteq\mathbb{R}^l$ closed, the space of allocations is closed too. It is also clearly bounded, so the space of allocation is compact. Let $A\subseteq\mathbb{R}^{lm}_+$ be the nonempty compact space of feasible allocations.

Define the relation $\succeq$ on $A$ such that for allocations $x=(x_1,\ldots,x_m)$ and $y=(y_1,\ldots,y_m)$, we have $x\succeq y$ if and only if $x_i\succeq_i y_i$ for every agent $i$. Now $x^*\in A$ is a maximal Pareto improvement over $x$ exactly if $x^*\succeq x$ and there is no $y\in A$ such that $y\succeq x^*$ but not $x^*\succeq y$.

Assume now that for all $a\in\mathbb{R}^l_+$ and every agent $i$, the "weakly-better-set" $\{b\in\mathbb{R}^l\mid b\succeq_i a\}$ is closed. Then the set $A_x=\{y\in A\mid y\succeq x\}$ is closed and, as the closed subset of a compact set, compact. Our problem reduces to showing that there exists a $\succeq$-maximal element $x^*\in A_x$.

Let $\succ$ be the asymmetric part of $\succeq$. It is transitive and irreflexive and therefore acyclic. Also, the "upper sections" of $\succeq$ are closed and therefore the lower sections of $\succ$ open. The existence of a $\succ$-maximal element follows then from what is sometimes referred to as the Walker-Bergstrom theorem (first proven by Sloss....). For the sake of completeness, I give the easy proof here.

Let $L_z=\{y\in A_y\mid y\prec z\}$ be the lower set of $\succ$ at $z$. Assume for the sake of contradiction that there is no $\succ$-maximal element in $A_x$. Then every point in $A_x$ lies in some $L_z$ with $z\in A_x$. Also, the $L_z$ are relatively open in the compact space $A_x$. So $\{L_z\mid z\in A_x\}$ is an open cover of $A_x$ and, by compactness, there is a finite set $F\subseteq A_x$ such that $\{L_z\mid z\in F\}$ is still an open cover of $A_x$. In particular, for each $z\in F$, there is some $z'\in F$ such that $z\in L_{z'}$ or, equivalently, $z'\succ z$. So the relation $\succ$ has no maximal element on the finite set $F$. This means there exists an infinite sequence $\langle z_n\rangle$ such that $z_{n+1}\succ z_n$ for all $n$. Since $\succ$ is acyclic, the sequence consists of infinitely many distinct elements. Since $F$ is finite, this is impossible.