Existence of Walrasian equilibria with discontinuous, non-ordered, interdependent and price-dependent preferences

Abstract

We generalize the classical equilibrium existence theorems by dispensing with the assumption of continuity of preferences. Our new existence results allow us to dispense with the interiority assumption on the initial endowments. Furthermore, we allow for non-ordered, interdependent and price-dependent preferences.

1 Introduction

The classical equilibrium existence theorems of Nash (1950), Debreu (1952), Arrow and Debreu (1954) and McKenzie (1954) were generalized to games/abstract economies where agents’ preferences need not be transitive or complete and therefore need not be representable by utility functions (see for example, Mas-Colell 1974; Shafer and Sonnenschein 1975; Gale and Mas-Colell 1975; Borglin and Keiding 1976; Shafer 1976; Yannelis and Prabhakar 1983; Wu and Shen 1996 among others). The need to drop the transitivity assumption from equilibrium theory was motivated by behavioral/experimental works which demonstrated that consumers do not necessarily behave in a transitive way.

A different line of literature pioneered by Dasgupta and Maskin (1986) and Reny (1999) necessitated the need to drop the continuity assumption on the payoff function of each agent. Their works were motivated by many realistic applications (for example, Bertrand competition and auctions), and generalizations of the Nash–Debreu equilibrium existence theorems were obtained where payoff functions need not be continuous. In other words, a new literature emerged on equilibrium existence theorems with discontinuous payoffs.¹

The first aim of this paper is to generalize the equilibrium existence theorems of Shafer and Sonnenschein (1975) and Yannelis and Prabhakar (1983) by dispensing with the continuity assumption of the preference correspondences. Although the proof of our equilibrium existence theorem in an abstract economy follows the approach of Yannelis and Prabhakar (1983), we cannot rely on continuous selections results, as it was the case in their work (and even earlier in Gale and Mas-Colell 1975). Indeed, the preference correspondence may not admit any continuous selection in our setting.²

Our second aim is to obtain the existence of Walrasian equilibria in an exchange economy where the preference correspondences could be discontinuous, nontransitive, incomplete, interdependent and price-dependent. An additional point we would like to emphasize is that contrary to the standard existence results in the literature, we do not impose the assumption that the initial endowment is an interior point of the consumption set.

The paper proceeds as follows. Section 2 collects notations and definitions. Section 3 provides a proof of the existence of equilibrium for an abstract economy, which extends the results of Shafer and Sonnenschein (1975) and Yannelis and Prabhakar (1983). The existence of Walrasian equilibrium with finite and infinite dimensional commodity spaces is proved and discussed in Sect. 4 .

2 Basics

Let $X$ and $Y$ be linear topological spaces, and let $\psi$ be a correspondence from $X$ to $Y$. Then $\psi$ is said to be lower hemicontinuous if $\psi ^l(V) = \{x\in X:\psi (x)\cap V\ne \emptyset \}$ is open in $X$ for every open subset $V$ of $Y$ and upper hemicontinuous if $\psi ^u(V) = \{x\in X:\psi (x)\subseteq V\}$ is open in $X$ for every open subset $V$ of $Y$. In addition, if the set $G = \{(x,y) \in X\times Y:y\in \psi (x)\}$ is open (resp. closed) in $X \times Y$, then we say that $\psi$ has an open (resp. closed) graph. If $\psi ^l(y)$ is open for each $y\in Y$, then $\psi$ is said to have open lower sections.

At some $x\in X$, if there exists an open set $O_x$ such that $x\in O_x$ and $\cap _{x'\in O_x} \psi (x') \ne \emptyset$, then we say $\psi$ has the local intersection property. Furthermore, $\psi$ is said to have the local intersection property if this property holds for every $x\in X$.

Clearly, every nonempty correspondence with open lower sections has the local intersection property. Yannelis and Prabhakar (1983) proved a continuous selection theorem and several fixed-point theorems by assuming that $\psi$ has open lower sections. Based on the local intersection property, Wu and Shen (1996) generalized the results of Yannelis and Prabhakar (1983). Recently, Scalzo (2015) proposed the “local continuous selection property” and proved that this condition is necessary and sufficient for the existence of continuous selections.

Mappings with the local intersection property have found applications in mathematical economics and game theory [see Wu and Shen (1996) and Prokopovych (2011) among others].

We now introduce the “continuous inclusion property,” which includes the above conditions as special cases.

Definition 1

A correspondence $\psi$ from $X$ to $Y$ is said to have the continuous inclusion property at $x$ if there exists an open neighborhood $O_x$ of $x$ and a nonempty correspondence $F_x:O_x \rightarrow 2^Y$ such that $F_x(z) \subseteq \psi (z)$ for any $z\in O_x$ and $\text{ co }F_x$ ³ has a closed graph.⁴

The continuous inclusion property is motivated by the majorization idea in general equilibrium (see the KF-majorization in Borglin and Keiding 1976 and L-majorization in Yannelis and Prabhakar 1983), and also the “multiply security” condition of McLennan et al. (2011), the “continuous security” condition of Barelli and Meneghel (2013) and the “correspondence security” condition of Reny (2013) in the context of discontinuous games.

Remark 1

If the correspondence $\psi$ from $X$ to $Y$ has the local intersection property at $x$, then $F_x$ can be chosen as a constant correspondence which only contains a singe point of $\cap _{x'\in O_x} \psi (x')$, and hence $\psi$ also has the continuous inclusion property at $x$. As a result, any nonempty correspondence with open lower sections has the continuous inclusion property.⁵

3 Equilibria in abstract economies

3.1 Results

In this section we prove the existence of equilibrium for an abstract economy with an infinite number of commodities and a countable number of agents.

An abstract economy is a set of ordered triples $\Gamma = \{(X_i, A_i, P_i) :i\in I\}$, where

• $I$ is a countable set of agents.

• $X_i$ is a nonempty set of actions for agent $i$. Set $X = \prod _{i\in I}X_i$.

• $A_i:X \rightarrow 2^{X_i}$ is the constraint correspondence of agent $i$.

• $P_i:X \rightarrow 2^{X_i}$ is the preference correspondence of agent $i$.

An equilibrium of $\Gamma$ is a point $x^*\in X$ such that for each $i \in I$:

• 1. $x_i^* \in \overline{A_i}(x^*)$, where $\overline{A_i}$ denotes the closure of $A_i$, and

• 2. $P_i(x^*) \cap A_i(x^*) = \emptyset$.

If $A_i \equiv X_i$ for all $i\in I$, then the point $x^*$ is called a Nash equilibrium.

For each $i \in I$, let $\psi _i(x) = A_i(x)\cap P_i(x)$ for all $x\in X$.

Theorem 1

Let $\Gamma = \{(X_i, A_i, P_i) :i\in I\}$ be an abstract economy such that for each $i \in I$:

i.:

$X_i$ is a nonempty, compact, convex and metrizable subset of a Hausdorff locally convex linear topological space;

ii.:

$A_i$ is nonempty and convex valued;

iii.:

the correspondence $\overline{A_i}$ is upper hemicontinuous;

iv.:

$\psi _i$ has the continuous inclusion property at each $x\in X$ with $\psi _i(x) \ne \emptyset$;

v.:

$x_i \notin \text{ co }\psi _i(x)$ for all $x\in X$.

Then $\Gamma$ has an equilibrium.

Proof

Fix $i\in I$. Let $U_i = \{x\in X:\psi _i(x) \ne \emptyset \}$.⁶ Since $\psi _i$ has the continuous inclusion property at each $x\in U_i$, there exist an open set $O^i_x\subseteq X$ such that $x\in O^i_x$ and a correspondence $F^i_x:O^i_x \rightarrow 2^{X_i}$ with nonempty values such that $F^i_x(z) \subseteq \psi _i(z)$ for any $z\in O^i_x$ and $\text{ co }F^i_x$ is closed. Then $O^i_x\subseteq U_i$, which implies that $U_i$ is open. Since $X$ is metrizable, $U_i$ is paracompact [see Michael (1956, p. 831)]. Moreover, the collection $\fancyscript{C}_i = \{O^i_x:x\in X\}$ is an open cover of $U_i$. There is a closed locally finite refinement $\mathcal {F}_i = \{E^i_k :k\in K\}$, where $K$ is an index set and $E^i_k$ is a closed set in $X$ [see Michael (1953, Lemma 1)].

For each $k \in K$, choose $x_k \in X$ such that $E^i_k \subseteq O^i_{x_k}$. For each $x\in U_i$, let $I_i(x) = \{k \in K :x\in E^i_{k}\}$. Then $I_i(x)$ is finite for each $x\in U_i$. Let $\phi _i(x) = \text{ co }\left( \cup _{k\in I_i(x)} \text{ co }F^i_{x_k}(x)\right)$ for $x\in U_i$. For each $x$ and $k\in I_i(x)$, $F^i_{x_k}(x) \subseteq \psi _i(x)$. Thus, $\text{ co }F^i_{x_k}(x) \subseteq \text{ co }\psi _i(x)$, which implies that $\cup _{k\in I_i(x)} \text{ co }F^i_{x_k}(x) \subseteq \text{ co }\psi _i(x)$. As a result, we have $\phi _i(x) = \text{ co }\left( \cup _{k\in I_i(x)} \text{ co }F^i_{x_k}(x)\right) \subseteq \text{ co }\psi _i(x)$.

Define the correspondence

$$\begin{aligned} H_i(x) = \left\{ \begin{array}{ll} \phi _i(x) &{} \quad x\in U_i; \\ \overline{A_i}(x) &{} \quad \text{ otherwise }. \end{array}\right. \end{aligned}$$

Then it is obvious that $H_i$ is nonempty and convex valued. Moreover, $H_i$ is also compact valued [see Lemma 5.29 in Aliprantis and Border (2006)].

Since $\text{ co }F^i_{x_k}$ has a closed graph in $E^i_k$ and $E^i_k$ is a compact Hausdorff space, it is upper hemicontinuous. For each $x$, $I_i(x)$ is finite, which implies that $\cup _{k\in I_i(x)} \text{ co }F^i_{x_k}(x)$ is the union of values for a finite family of upper hemicontinuous correspondences, and hence is upper hemicontinuous at the point $x$ [see Aliprantis and Border (2006, Theorem 17.27)]. Then $\phi _i(x)$ is the convex hull of $\cup _{k\in I_i(x)} \text{ co }F^i_{x_k}(x)$, and it is compact for all $x\in U_i$; hence it is upper hemicontinuous on $U_i$ [see Aliprantis and Border (2006, Theorem 17.35)]. Note that $H_i(x)$ is $\phi _i(x)$ when $x\in U_i$, and $\overline{A_i}(x)$ when $x\notin U_i$. Since $U_i$ is open, analogous to the argument in Yannelis and Prabhakar (1983, Theorem 6.1), $H_i$ is upper hemicontinuous on the whole space. Let $H = \prod _{i\in I}H_i$. Since $H$ is nonempty, convex and closed valued, by the Fan-Glicksberg fixed-point theorem, there exists a point $x^* \in X$ such that $x^* \in H(x^*)$.

Since $\phi _i(x) \subseteq \overline{A_i}(x)$ for $x \in U_i$ and $H_i(x) \subseteq \overline{A_i}(x)$ for any $x$, which implies that $x_i^* \in \overline{A_i}(x^*)$. Note that if $x^* \in U_i$ for some $i\in I$, then $x_i^* \in \text{ co }\left( \cup _{k\in I_i(x^*)} \text{ co }F^i_{x_k}(x^*)\right) \subseteq \text{ co } \psi _i(x^*)$, a contradiction to assumption (v). Thus, we have $x^* \notin U_i$ for all $i \in I$. Therefore, $\psi _i(x^*) =\emptyset$, which implies that $A_i(x^*) \cap P_i(x^*) = \emptyset$. That is, $x^*$ is an equilibrium for $\Gamma$. $\square$

Remark 2

If in the above theorem, set $A_i \equiv X_i$, assume that $\psi _i$ is convex valued and drop assumption (v), then one can obtain a generalization of the Gale and Mas-Colell (1975) fixed-point theorem (see He and Yannelis 2014). That is, let $\psi _i :X \rightarrow X_i$ be a convex valued correspondence with the continuous inclusion property at each $x$ such that $\psi _i(x) \ne \emptyset$. Then there exists a point $x^* \in X$ such that for each $i$, either $x_i^*\in \psi _i(x^*)$ or $\psi _i(x^*) = \emptyset$.⁷

Below, we show that the theorem of Shafer and Sonnenschein (1975) and Theorem 6.1 of Yannelis and Prabhakar (1983) on the existence of equilibrium in an abstract economy can be obtained as corollaries. Note that in Shafer and Sonnenschein (1975) the correspondence $A_i$ is compact valued for each $i\in I$, and therefore there is no need to work with the closure of $A_i$. That is, an equilibrium $x^*$ should satisfy $x_i^* \in A_i(x^*)$ and $P_i(x^*) \cap A_i(x^*) = \emptyset$. In Yannelis and Prabhakar (1983), the equilibrium notion is the same as defined above.

Corollary 1

[Shafer and Sonnenschein (1975)] Let $\Gamma = \{(X_i, A_i, P_i) :i\in I\}$ be an abstract economy such that for each $i \in I$:

i.:

$X_i$ is a nonempty, compact and convex subset of $\mathbb {R}_+^l$;

ii.:

$A_i$ is nonempty, convex and compact valued;

iii.:

$A_i$ is a continuous correspondence;

v.:

$P_i$ has an open graph;

vi.:

$x_i \notin \text{ co }\psi _i(x)$ for all $x\in X$.⁸

Then $\Gamma$ has an equilibrium $x^*$; that is, for any $i\in I$, $x_i^* \in A_i(x^*)$ and $P_i(x^*) \cap A_i(x^*) =\emptyset$.

Proof

For each $i\in I$, define a mapping $U_i:\text{ Gr }(A_i) \rightarrow \mathbb {R}$ by $U_i(y,x_i) = \text{ dist }((y,x_i), \text{ Gr }^C(P_i))$, where $\text{ Gr }(A_i)$ is the graph of $A_i$, $\text{ Gr }^C(P_i)$ denotes the complement of the graph of $P_i$, and $\text{ dist }(\cdot , \cdot )$ denotes the usual distance on $\mathbb {R}^l_+$. Since $P_i$ has an open graph, $U_i$ is continuous. Let $m_i(x) = \max _{z\in A_i(x)} U_i(x,z)$ and $\phi _i(x) = \{z\in A_i(x):U_i(x,z) = m_i(x)\}$ for each $x\in X$. Since $A_i$ is continuous, by the Berge maximum theorem [see Aliprantis and Border (2006, Theorem 17.31)], $\phi _i$ is nonempty, compact valued and upper hemicontinuous. At any point $x$ such that $\psi _i(x) = P_i(x) \cap A_i(x)\ne \emptyset$, we have $m_i(x) > 0$, and hence $\phi _i(x)\subseteq \psi _i(x)$. Thus, the continuous inclusion property holds, and by Theorem 1, there is an equilibrium. $\square$

Corollary 2

(Yannelis and Prabhakar (1983, Theorem 6.1)) Let $\Gamma = \{(X_i, A_i, P_i) : i\in I\}$ be an abstract economy such that for each $i \in I$:

i.:

$X_i$ is a nonempty, compact, convex and metrizable subset of a Hausdorff locally convex linear topological space;

ii.:

$A_i$ is nonempty and convex valued;

iii.:

the correspondence $\overline{A_i}$ is upper hemicontinuous;

iv.:

$A_i$ has open lower section;

v.:

$P_i$ has open lower section;

vi.:

$x_i \notin \text{ co }\psi _i(x)$ for all $x\in X$.

Then $\Gamma$ has an equilibrium $x^*$; that is, for each $i\in I$, $x_i^* \in \overline{A}_i(x^*)$ and $P_i(x^*) \cap A_i(x^*) = \emptyset$.

Proof

By Fact 6.1 in Yannelis and Prabhakar (1983), $\psi _i$ has open lower sections. As a result, $\psi _i$ has the continuous inclusion property at each $x\in X$ when $\psi _i(x) \ne \emptyset$. Then the result follows from Theorem 1. $\square$

Remark 3

Note that our Theorem 1 also covers Theorem 10 of Wu and Shen (1996). Wu and Shen (1996) did not impose the metrizability condition on $X_i$, but directly assumed that $U_i$ is paracompact. Our proof still holds under this condition.

Remark 4

In condition (iv) of Theorem 1, we assume that $\psi _i$ has the continuous inclusion property at each $x\in X$ with $\psi _i(x) \ne \emptyset$. It is natural to ask whether we can impose conditions on the correspondences $P_i$ and $A_i$ separately and then verify that their intersection $\psi _i$ has the continuous inclusion property [for example, see conditions (iv) and (v) in Yannelis and Prabhakar (1983, Theorem 6.1)]. However, a simple example can be constructed to show that a combination of the following two conditions cannot guarantee our condition (iv):

• 1. $P_i$ has the continuous inclusion property at $x$ when $P_i(x) \ne \emptyset$;

• 2. $A_i$ has an open graph.

Suppose that there is only one agent and $X = [0,1]$, $A(x) = (0,1]$ and

$$\begin{aligned} P(x) = \left\{ \begin{array}{ll} [0,1], &{} \quad x = 1; \\ \{0\}, &{} \quad x\in [0,1). \end{array}\right. \end{aligned}$$

Then it is obvious that $P$ has the continuous inclusion property and $A$ has an open graph. However,

$$\begin{aligned} \psi (x) = \left\{ \begin{array}{ll} (0,1], &{} \quad x = 1, \\ \emptyset , &{} \quad x\in [0,1); \end{array}\right. \end{aligned}$$

does not have the continuous inclusion property.

Note that if $A_i = X_i$ is a constant correspondence, we can assume that $P_i$ has the continuous inclusion property at each $x\in X$ with $P_i(x) \ne \emptyset$, and the existence of Nash equilibrium follows as a corollary.

Corollary 3

Let $\Gamma = \{(X_i, P_i) :i\in I\}$ be a game such that for each $i \in I$:

i.:

$X_i$ is a nonempty, compact, convex and metrizable subset of a Hausdorff locally convex linear topological space;

ii.:

$P_i$ has the continuous inclusion property at each $x\in X$ with $P_i(x) \ne \emptyset$;

iii.:

$x_i \notin \text{ co } P_i(x)$ for all $x\in X$.

Then $\Gamma$ has a Nash equilibrium $x^*$; that is, for each $i\in I$, $P_i(x^*) = \emptyset$.

3.2 Relationship with Carmona and Podczeck (2015)

Subsequent to this paper, Carmona and Podczeck (2015) dropped the metrizability condition on $X_i$ and generalized our conditions (4) and (5) as follows.

Let $I(x) = \{i \in I:\psi _i(x) \ne \emptyset \}$. For every $x\in X$ such that $I(x) \ne \emptyset$ and $x_i \in \overline{A}_i(x)$ for all $i \in I$, there is an agent $i \in I(x)$ such that,

• 1. $\psi _i$ has the continuous inclusion property at $x$;

• 2. $x_i \notin \text{ co }\psi _i(x)$.

Notice that our proof above still goes through under this condition by slightly modifying the definition of the set $U_i$ as

$$\begin{aligned} \{x\in X :\psi _i \text{ has } \text{ the } \text{ continuous } \text{ inclusion } \text{ property } \text{ at } x\}. \end{aligned}$$

The metrizability condition in our Theorem 1 is not needed. Following a similar argument as in Borglin and Keiding (1976) and Toussaint (1984), we provide an alternative proof for Theorem 1 in which the set of agents can be any arbitrary (finite or infinite set) and $X_i$ need not to be metrizable for each $i$.⁹

Alternative proof of Theorem 1

For each $i \in I$, define a correspondence $H_i$ from $X$ to $X_i$ as follows:

$$\begin{aligned} H_i(x) = \left\{ \begin{array}{ll} \psi _i(x), &{} \quad x_i \in \overline{A}_i(x); \\ \overline{A}_i(x), &{} \quad x_i \notin \overline{A}_i(x). \end{array}\right. \end{aligned}$$

We will show that $H_i$ has the continuous inclusion property at each $x$ such that $H_i(x) \ne \emptyset$.

• 1. If $x_i \in \overline{A}_i(x)$, then $\psi _i(x) = H_i(x) \ne \emptyset$, which implies that there exists an open neighborhood $O_x$ of $x$ and a nonempty correspondence $F_x:O_x \rightarrow 2^{X_i}$ such that $F_x(z) \subseteq \psi _i(z)$ for any $z\in O_x$ and $\text{ co }F_x$ has a closed graph. For any $z \in O_x$, $F_x(z) \subseteq \psi _i(z) = H_i(z)$ if $z_i \in \overline{A}_i(z)$, and $F_x(z) \subseteq \psi _i(z) \subseteq \overline{A}_i(z) = H_i(z)$ if $z_i \notin \overline{A}_i(z)$.

• 2. Consider the case that $x_i \notin \overline{A}_i(x)$. Since the correspondence $\overline{A_i}$ is upper hemicontinuous and closed valued, it has a closed graph. As a result, one can find an open neighborhood $O_x$ of $x$ such that $z_i \notin \overline{A}_i(z)$ and hence $H_i(z)=\overline{A}_i(z)$ for any $z \in O_x$. As $\overline{A}_i$ is upper hemicontinuous, closed and convex valued, $H_i$ has the continuous inclusion property.

Let $I(x) = \{i \in I:H_i(x) \ne \emptyset \}$. Define a correspondence $H :X\rightarrow 2^X$ as

$$\begin{aligned} H(x) = \left\{ \begin{array}{ll} \left( \prod \nolimits _{i \in I(x)}H_i(x)\right) \times \left( \prod \nolimits _{j \in I\setminus I(x)} X_j\right) , &{} \quad I(x) \ne \emptyset ; \\ \emptyset , &{} \quad I(x) = \emptyset . \end{array}\right. \end{aligned}$$

It can be easily checked that $H(x)$ has the continuous inclusion property at each $x$ such that $H(x) \ne \emptyset$.

In addition, one can easily show that $x\notin \text{ co }H(x)$ for any $x \in X$. Indeed, fix any $x \in X$. If $I(x) = \emptyset$, then $H(x) = \emptyset$, which implies that $x \notin \text{ co }H(x)$. If $I(x) \ne \emptyset$, then there exists an agent $i$ such that $H_i(x) \ne \emptyset$. If $x_i \in \overline{A}_i(x)$, then $x_i \notin \text{ co }\psi (x) = \text{ co }H_i(x)$. If $x_i \notin \overline{A}_i(x)$, then $x_i \notin \text{ co }H_i(x)$ as $H_i(x) = \overline{A}_i(x)$ (since $\overline{A}_i(x)$ is convex). Hence, $x \notin \text{ co }H(x)$.

By Corollary 1 in He and Yannelis (2014),¹⁰ there exists a point $x^* \in X$ such that $H(x^*) = \emptyset$, which implies that $I(x^*) =\emptyset$. That is, for any $i$, $H_i(x^*) = \emptyset$, which implies that $x^*_i \in \overline{A}_i(x^*)$ and $\psi _i(x^*) = H_i(x^*) = \emptyset$. $\square$

Remark 5

The previous proof adapted in Theorem 1 seems to be suitable to cover the case where the set of agents is a measure space as in Yannelis (1987). It is not clear whether the above proof can be easily extended to a measure space of agents.

4 Existence of Walrasian equilibria

An exchange economy $\mathcal {E}$ is a set of triples $\{(X_i, P_i, e_i) :i \in I\}$, where

• $I$ is a finite set of agents;

• $X_i \subseteq \mathbb {R}^l_+$ is the consumption set of agent $i$, and $X = \prod _{i\in I} X_i$;

• $P_i :X \times \triangle \rightarrow 2^{X_i}$ is the preference correspondence of agent $i$, where $\triangle$ is the set of all possible prices;¹¹

• $e_i \in X_i$ is the initial endowment of agent $i$, where $e = \sum _{i \in I} e_i \ne 0$.

Let $\triangle = \{p\in \mathbb {R}^l_+:\sum _{k=1}^l p_k = 1\}$. Given a price $p\in \triangle$, the budget set of agent $i$ is $B_i(p) = \{x_i\in X_i :p\cdot x_i \le p\cdot e_i\}$. Let $\psi _i(p,x) = B_i(p)\cap P_i(x,p)$ for each $i\in I$, $x\in X$ and $p\in \triangle$. Then $\psi _i(p,x)$ is the set of all allocations in the budget set of agent $i$ at price $p$ that he prefers to $x$.

A free disposal Walrasian equilibrium for the exchange economy $\mathcal {E}$ is $(p^*, x^*) \in \triangle \times X$ such that

• 1. for each $i\in I$, $x^*_i \in B_i(p^*)$ and $\psi _i(p^*,x^*) = \emptyset$;

• 2. $\sum _{i\in I} x_i^* \le \sum _{i\in I} e_i$.

Theorem 2

Let $\mathcal {E}$ be an exchange economy satisfying the following assumptions: for each $i\in I$,

• 1. $X_i$ is a nonempty compact convex subset of $\mathbb {R}^l_+$;¹²

• 2. $\psi _i$ has the continuous inclusion property at each $(p,x)\in \triangle \times X$ with $\psi _i(p,x) \ne \emptyset$ and $x_i \notin \text{ co }\psi _i(p,x)$.

Then $\mathcal {E}$ has a free disposal Walrasian equilibrium.

Proof

The proof follows the idea of Arrow and Debreu (1954), which introduces a fictitious player; see also Shafer (1976).

For each $i \in I$, $p\in \triangle$ and $x\in X$, let $A_i(p,x) = B_i(p)$. Define the correspondences $A_{0}(p,x) = \triangle$ and $P_{0}(p,x) = \{q\in \triangle :q\cdot \left( \sum _{i\in I}(x_i-e_i)\right) > p\cdot \left( \sum _{i\in I}(x_i-e_i)\right) \}$. Let $I_0 = I\cup \{0\}$. Then for any $i\in I_0$, $A_i$ is nonempty, convex valued and upper hemicontinuous on $\triangle \times X$.

Note that $\psi _i(p,x) = A_i(p,x) \cap P_{i}(p,x)$ has the continuous inclusion property for each $i\in I$. Moreover, let $\psi _{0}(p,x) = A_{0}(p,x) \cap P_{0}(p,x) = P_{0}(p,x)$. Fix any $(p,x) \in \triangle \times X$ such that $\psi _{0}(p,x) \ne \emptyset$, pick $q\in \psi _{0}(p,x)$, then $(q-p)\cdot \left( \sum _{i\in I}(x_i-e_i)\right) > 0$. Since the left side of the inequality is continuous, there is an open neighborhood $O$ of $(p,x)$ such that for any $(p',x') \in O$, $(q-p')\cdot \left( \sum _{i\in I}(x'_i-e_i)\right) > 0$, which implies that the correspondence $\psi _{0}$ has the continuous inclusion property. In addition, it is obvious that $\psi _{0}$ is convex valued and $p \notin \psi _{0}(p,x)$ for any $(p,x) \in \triangle \times X$.

Thus, we can view the exchange economy $\mathcal {E}$ as an abstract economy $\Gamma = \{(X_i, A_i, P_i) :i \in I_0\}$ which satisfies all the conditions of Theorem 1. Therefore, there exists a point $(p^*,x^*)\in \triangle \times X$ such that

• 1. $x_i^* \in A_i(p^*,x^*) = B_i(p^*)$ and $\psi _i(p^*,x^*) = \emptyset$ for each $i\in I$ and

• 2. $P_{0}(p^*,x^*) = \psi _{0}(p^*,x^*) = \emptyset$.

Let $z = \sum _{i\in I}(x^*_i-e_i)$. Then (1) implies that $p^* \cdot z \le 0$, and (2) implies that $q \cdot z \le p^*\cdot z$ for any $q\in \triangle$, and hence $q\cdot z \le p^*\cdot z \le 0$. Suppose that $z \notin \mathbb {R}^l_{-}$. Thus, there exists some $k\in \{1,\ldots , l\}$ such that $z_k >0$. Let $q' = \{q_j\}_{1\le j \le l}$ such that $q_j =0$ for any $j\ne k$ and $q_k = 1$. Then $q'\in \triangle$ and $q' \cdot z = z_k > 0$, a contradiction. Therefore, $z\in R^l_{-}$, which implies that $\sum _{i\in I} x_i^* \le \sum _{i\in I} e_i$.

Therefore, $(p^*,x^*)$ is a free disposal Walrasian equilibrium. $\square$

Remark 6

We have imposed the compactness condition on the consumption set. It is not clear to us at this stage whether this condition can be dispensed with. When agents’ preferences are continuous, one can work with a sequence of economies with compact consumption sets, which are the truncations of the original consumption set. Then the existence of Walrasian equilibrium allocations and prices can be proved in each truncated economy. Since the set of feasible allocations and the price set are both compact, there exists a convergent point. By virtue of the continuity of preferences, one can show that this is indeed a Walrasian equilibrium of the original economy. The convergence argument fails in our setting as we do not require the continuity assumption on preferences. Consequently, relaxing the compactness assumption seems to be an open problem.¹³

We must add that the compactness assumption is not unreasonable at all. The world is finite, and the initial endowment for each good is also finite. Thus, by assuming that for each good, $\Vert x_i\Vert \le K\cdot \sum _{i \in I} \Vert e_i\Vert$, where $K$ is a sufficiently large number and $I$ is the set of all agents in the world, no real restriction on the attainability of the consumption of each good is imposed.

Note that in Theorem 2 we allowed for free disposal. Below we prove the existence of a non-free disposal Walrasian equilibrium following the proof of Shafer (1976).

Hereafter we allow for negative prices: $\triangle ' = \{p\in \mathbb {R}^l :\Vert p\Vert = \sum _{k=1}^l |p_k| \le 1\}$ is the set of all possible prices. Let $B_i(p) = \{x_i\in X_i:p\cdot x_i \le p\cdot e_i + 1 - \Vert p\Vert \}$ and $\psi _i(p,x) = P_i(x,p)\cap B_i(p)$ for each $i\in I$, $x\in X$ and $p\in \triangle '$. Let $K = \{x:\sum _{i\in I}x_i = \sum _{i\in I}e_i\}$ and $\text{ pr }_i:X\rightarrow X_i$ be the projection mapping for each $i\in I$.

A (non-free disposal) Walrasian equilibrium for the exchange economy $\mathcal {E}$ is $(p^*, x^*) \in \triangle ' \times X$ such that

• 1. $\Vert p^*\Vert = 1$;

• 2. for each $i\in I$, $x^*_i \in B_i(p^*)$ and $\psi _i(p^*,x^*) = \emptyset$;

• 3. $\sum _{i\in I} x_i^* = \sum _{i\in I} e_i$.

If $p^*$ is a Walrasian equilibrium price, then $\Vert p^*\Vert = 1$ and $B_i(p^*) = \{x_i\in X_i :p^*\cdot x_i \le p^*\cdot e_i \}$, which is the standard budget set of agent $i$.

Theorem 3

Let $\mathcal {E}$ be an exchange economy satisfying the following assumptions: for each $i\in I$,

• 1. $X_i$ is a nonempty compact convex subset of $\mathbb {R}^l_+$;

• 2. $\psi _i$ has the continuous inclusion property at each $(p,x)\in \triangle ' \times X$ with $\psi _i(p,x) \ne \emptyset$, and $x_i \notin \text{ co }\psi _i(p,x)$.

• 3. for each $x_i \in \text{ pr }_i(K)$ and $p \in \triangle '$, $x_i \in \text{ bd } P_i(x,p)$, where $\text{ bd }$ denotes boundary.

Then $\mathcal {E}$ has a Walrasian equilibrium.

Proof

Repeating the arguments in the first two paragraphs of the proof of Theorem 2, one could show that there exists a point $(p^*,x^*)\in \triangle ' \times X$ such that

• 1. $x_i^* \in A_i(p^*,x^*) = B_i(p^*)$ for each $i\in I$, which implies that $p^*\cdot x_i^* \le p^*\cdot e_i + 1 - \Vert p^*\Vert$;

• 2. $\psi _i(p^*,x^*) = \emptyset$ for each $i\in I$;

• 3. $P_{0}(p^*,x^*) = \psi _{0}(p^*,x^*) = \emptyset$.

Let $z = \sum _{i\in I}(x^*_i-e_i)$. We must show that $z = 0$. Suppose that $z \ne 0$. From (3), it follows that $q\cdot z \le p^*\cdot z$ for any $q\in \triangle '$. Let $q = \frac{z}{\Vert z\Vert }$. Then $q \in \triangle '$ and $p^* \cdot z \ge q\cdot z > 0$. Let $q^* = \frac{p^*}{\Vert p^*\Vert }$. Since $\frac{p^*}{\Vert p^*\Vert } \cdot z \ge p^* \cdot z \ge q^* \cdot z$, it follows that $\Vert p^*\Vert = 1$. As a result, $p^* \cdot x^*_i \le p^*\cdot e_i$ (since $x_i^* \in A_i(p^*, x^*)$), which implies that $p^* \cdot z = p^* \cdot \sum _{i\in I}(x^*_i-e_i) \le 0$, a contradiction. Thus, $z = 0$; that is, $\sum _{i\in I} x_i^* = \sum _{i\in I} e_i$, $x^* \in K$.

Note that $x^*_i \in \text{ pr }_i(K)$ implies that $x_i^* \in \text{ bd } P_i(x^*, p^*)$. Since $x_i^* \in B_i(p^*)$ and $x_i^* \notin \text{ co }\psi _i(p^*,x^*)$, $x_i^* \notin P_i(x^*,p^*)$. If there exists some $i$ such that $p^*\cdot x_i^* < p^*\cdot e_i + 1 - \Vert p^*\Vert$, then due to assumption (3), $x_i^* \in \text{ bd } P_i(x^*,p^*)$ implies that one can find a point $y_i \in P_i(x^*, p^*)$ such that $x^*_i$ and $y_i$ are sufficiently close, and $p^*\cdot y_i < p^*\cdot e_i + 1 - \Vert p^*\Vert$. Thus, $y_i \in \psi _i(p^*, x^*)$, which contradicts (2). Therefore, $p^*\cdot x_i^* = p^*\cdot e_i + 1 - \Vert p^*\Vert$ for each $i\in I$, and summing up over all $i$ yields $\Vert p^*\Vert = 1$.

Therefore, $(p^*,x^*)$ is a Walrasian equilibrium. $\square$

Remark 7

Shafer (1976) proved the existence of non-free disposal Walrasian equilibrium based on the equilibrium existence result of Shafer and Sonnenschein (1975) (see Corollary 1 above). Thus, the main theorem of Shafer (1976) follows from our Corollary 1 and Theorem 3.

Below, we provide an alternative proof of the theorem of Shafer (1976) without invoking the norm of the price $\Vert p\Vert$ into the budget set. It requires the nonsatiation condition for one agent only. Furthermore, the proof below remains unchanged if the consumption set is a nonempty, norm compact and convex subset of a Hausdorff locally convex topological vector space. This is not the case in Shafer (1976)’s proof, since the norm of prices is part of the budget set. Recall that the price space $\triangle '$ is weak$^*$ compact by Alaoglu’s theorem, and $\triangle '$ may not be metrizable unless the space of allocations is separable.

Theorem 4

Let $\mathcal {E}$ be an exchange economy satisfying the following assumptions:

• 1. for each $i\in I$, let $X_i$ be a nonempty compact convex set of $\mathbb {R}^l_+$;

• 2. for each $i\in I$, $\psi _i$ has the continuous inclusion property at each $(p,x)\in \triangle ' \times X$ with $\psi _i(p,x) \ne \emptyset$, and for any $x_i \in X_i$, $x_i \notin \text{ co }\psi _i(p,x)$;

• 3. for any $p \in \triangle '$ and $x$ in the set of feasible allocations

  $$\begin{aligned} \mathcal {A}= \left\{ x \in X:\sum _{i = 1 }^n x_i = \sum _{i =1}^n e_i \right\} , \end{aligned}$$

  there exists an agent $i \in I$ such that $P_i(x,p) \ne \emptyset$.

Then $\mathcal {E}$ has a Walrasian equilibrium $(p^*, x^*)$; that is,

• 1. $p^* \ne 0$;

• 2. for each $i\in I$, $x^*_i \in B_i(p^*)$ and $\psi _i(p^*,x^*) = \emptyset$;

• 3. $\sum _{i\in I} x_i^* = \sum _{i\in I} e_i$.

Most of the proof proceeds as in Theorem 2. We repeat the argument here for the sake of completeness.

Proof

For each $i \in I$, $p\in \triangle '$ and $x\in X$, let $A_i(p,x) = B_i(p)$. Denote $X_0 = \triangle '$, and define the correspondences $A_{0}(p,x) \equiv \triangle '$ and $P_{0}(p,x) = \{q\in \triangle ' :q \left( \sum _{i\in I}(x_i-e_i)\right) > p \left( \sum _{i\in I}(x_i-e_i)\right) \}$.¹⁴ Let $I_0 = I\cup \{0\}$. Let $\psi _{0}(p,x) = A_{0}(p,x) \cap P_{0}(p,x) = P_{0}(p,x)$. As shown in the proof of Theorem 2, for each $i \in I_0$, the correspondence $\psi _{i}$ is convex valued, $(p,x) \notin \psi _{i}(p,x)$ for any $(p,x) \in \triangle ' \times X$, and has the continuous inclusion property.

We have constructed an abstract economy $\Gamma = \{(X_i, P_i, A_i):i\in \{0\}\cup I\}$. By Theorem 1, there exists a point $(p^*,x^*)\in \triangle ' \times X$ such that

• 1. $x_i^* \in A_i(p^*,x^*) = B_i(p^*)$ and $\psi _i(p^*,x^*) = \emptyset$ for each $i\in I$;

• 2. $P_{0}(p^*,x^*) = \psi _{0}(p^*,x^*) = \emptyset$.

Let $z = \sum _{i\in I}(x^*_i-e_i)$. Then (1) implies that $p^* (z) \le 0$, and (2) implies that $q(z) \le p^*(z)$ for any $q\in \triangle '$, and hence $q(z) \le p^*(z) \le 0$. As a result, $z = 0$;¹⁵ that is, $x^* \in \mathcal {A}$. To complete the proof, we must show that $p^* \ne 0$. Suppose otherwise; that is, $p^* = 0$. Then $B_i(p^*) = X_i$ and $\psi _i(p^*,x^*) = P_i(x^*, p^*) = \emptyset$ for each $i\in I$, a contradiction to condition (3). Therefore, $(p^*,x^*)$ is a Walrasian equilibrium. $\square$

Remark 8

In Theorems 2, 3 and 4, the condition that $\psi _i$ has the continuous inclusion property at each $(p,x)\in \triangle \times X$ with $\psi _i(p,x) \ne \emptyset$, and $x_i \notin \text{ co }\psi _i(p,x)$ for each $i$ can be weakened following the argument in Sect. 3.2. In particular, one can let $I(x) = \{i \in I:\psi _i(p,x) \ne \emptyset \}$ and assume that for every $x\in X$ such that $I(x) \ne \emptyset$ and $x_i \in A_i(p,x)$ for all $i \in I$, there is an agent $i \in I(x)$ such that,

• 1. $\psi _i$ has the continuous inclusion property at $(p,x)$;

• 2. $x_i \notin \text{ co }\psi _i(p,x)$.

In other words, the continuous inclusion property is not required for all agents, but only for some agents. The proofs of Theorems 2, 3 and 4 can still go through under this new condition.¹⁶ For pedagogical reasons, we work with condition (2) in Theorem 2.

5 Concluding remarks

Remark 9

Theorem 4 can be extended to a more general setting with an infinite dimensional commodity space. In particular, the commodity space can be any normed linear space whose positive cone may not have an interior point, and the set of prices is a subset of its dual space. If the consumption sets are nonempty, norm compact and convex, and the price space is weak$^*$ compact, then the proof of Theorem 4 remains unchanged.

Remark 10

To prove the existence of a Walrasian equilibrium in economies with infinite dimensional commodity spaces, Mas-Colell (1986) proposed the “uniform properness” condition when the preferences are transitive, complete and convex. Yannelis and Zame (1986) and Podczeck and Yannelis (2008) proved the existence result with non-ordered preferences using the “extreme desirability” condition. All the above results impose on the commodity space a lattice structure. Our Theorem 4 does not require the extreme desirability or uniform properness condition, and no ordering or lattice structure is needed on the commodity space. It should be noticed that the proof of our Theorem 4 requires that the evaluation map $(p, x_i) \rightarrow p(x_i)$ from $\triangle '\times X_i$ to $\mathbb {R}$ is continuous for $\triangle '$ with the weak$^*$ topology, while this joint continuity property of the evaluation map is not required in the papers above.

Mas-Colell (1986) provided an example of a single agent economy in which the preference is reflexive, transitive, complete, continuous, convex and monotone, but there is no quasi-equilibrium.¹⁷ We show that his example does not satisfies our condition (2) of Theorem 4 when the commodity space is compact.

In the example of Mas-Colell (1986), the commodity space is the space of signed bounded countably additive measures $L = ca(K)$ with the bounded variation norm $\Vert \cdot \Vert _{BV}$, where $K = Z_+ \cup \{\infty \}$ is the compactification of the positive integers. Let $x_i= x(\{i\})$ for $x \in L$ and $i\in K$. For every $i \in K$, define a function $u_i:[0, \infty ) \rightarrow [0, \infty )$ by

$$\begin{aligned} u_i(t) = \left\{ \begin{array}{ll} 2^i t &{} \quad t \le \frac{1}{2^{2i}}; \\ \frac{1}{2^i} - \frac{1}{2^{2i}} + t &{} \quad t > \frac{1}{2^{2i}}. \\ \end{array}\right. \end{aligned}$$

The preference relation $P$ is given by $U(x) = \sum _{i=1}^{i = \infty } u_i(x_i)$, which is concave, strictly monotone and weak$^*$ continuous.

Suppose that $X = \{x\in L_+ :\Vert x\Vert _{BV} \le M\}$ for some sufficiently large positive integer $M$. Fix the initial endowment $e = (0, M, 0, \ldots , 0) \in X$ and the price $p_0 = 0$. Then $\psi (p_0,e) = B(p_0)\cap P(e) \ne \emptyset$, as $y = (M, 0, \ldots , 0) \in \psi (p_0,e)$. For each $i \in K$, let $w_i(\{j\}) = 1$ if $j = i$ and $0$ otherwise. Fix a linear functional $p \in L'$ such that $p(w_2) = 0$ and $p(w_i) > 0$ for $i \ne 2$. Set $p_n = \frac{p}{n}$. Then $B(p_n) = \{0, m, 0, \ldots , 0\}$, where $0 \le m \le M$. However, for any $z\in B(p_n)$, $z \notin P(e)$. Consequently, $\psi (p_n,e) = \emptyset$. This implies that the correspondence $\psi$ does not have the continuous inclusion property when the commodity space is compact, as $p_n \rightarrow 0$ when $n\rightarrow \infty$. Therefore, the example of Mas-Colell (1986) violates condition (2) of our Theorem 4.

Remark 11

If we interpret the infinite dimensional commodity space as goods over an infinite time horizon, the weak, Mackey and weak$^*$ topologies on preferences imply that agents are impatient, because those topologies are generated by finitely many continuous linear functionals and they impose a form of “myopia” (i.e., tails do not matter, see for example Bewley (1972) and Araujo et al. (2011) among others). As our theorems drop the continuity assumption, it will be interesting to see if one can prove the existence theorem with patient agents relying on such discontinuous preferences.

Remark 12

Contrary to the standard existence results of Walrasian equilibrium, in the above theorems we do not impose the assumptions that the initial endowment is an interior point of the consumption set or the preference has an open graph/open lower sections. Below we give an example in which the preferences are discontinuous, and a Walrasian equilibrium exists. Notice that none of the classical existence theorems cover the example below.

Example 1

Consider the following $2$-agent $2$-good economy:

• 1. The set of available allocations for both agents is $X_1 = X_2 = [0,1]\times [0,1]$.

• 2. Agent $1$’s preference correspondence depends on $x_1 = (x_1^1,x_1^2)$ and $x_2 = (x_2^1,x_2^2)$:

  $$\begin{aligned} P_1(x_1,x_2)= & {} \left\{ \left( y_1^1,y_1^2\right) \in X_1 :y_1^1 \cdot y_1^2 > x_1^1 \cdot x_1^2\right\} \setminus \\&\left\{ \left( y_1^1,y_1^2\right) \in X_1 :y_1^1-x_1^1 = y_1^2 - x_1^2, y_1^1<\frac{3}{2} x_1^1 \right\} . \end{aligned}$$

  ¹⁸ The preference of agent $2$ is defined similarly.

• 3. The initial endowments are given by $e_1 = (\frac{1}{3},\frac{2}{3})$ and $e_2 = (\frac{2}{3},\frac{1}{3})$.

Note that $P_i$ does not have open lower sections for any $i=1,2$. For example,

$$\begin{aligned} P_i^{l}\left( \frac{1}{2},\frac{1}{2}\right)&=\left\{ \left( y_i^1,y_i^2\right) \in [0,1]\times [0,1] :y_i^1 \cdot y_i^2 < \frac{1}{4}, y_i^1 \ne y_i^2 \right\} \\&\quad \cup \left\{ (z,z) :0 \le z \le \frac{1}{3}\right\} \end{aligned}$$

which is neither open nor closed. As a result, $P_i$ does not have an open graph.

We show that the conditions of Theorem 2 hold. Pick any point $(p,x)\in \triangle \times X$ such that $\psi _i(p,x) \ne \emptyset$, then there exists a point $y_i \in \psi _i(p,x) = B_i(p)\cap P_i(x)$. Since $y_i \in P_i(x)$, it follows that $y_i^1 \cdot y_i^2 > x_i^1 \cdot x_i^2$. Thus, one can pick a point $z_i = (z_i^1, z_i^2)$ such that $z_i^j < y_i^j$ for $j = 1,2$ and $z_i$ is an interior point of $P_i(x)$.¹⁹ Consequently, there exists an open neighborhood $O_i$ of $x_i$ such that $(z_i^1, z_i^2) \in P(x_i',x_{-i})$ for any $x_i' \in O_i$ and $x_{-i} \in X_{-i}$. Furthermore, due to the fact that $z_i^j < y_i^j$ for $j = 1,2$, we have $0 < p\cdot z_i < p\cdot y_i \le p \cdot e_i$, which implies that there exists a neighborhood $O_p$ of $p$, $z_i \in B_i(p')$ for any $p' \in O_p$. Define the correspondence $F_{(p,x)}$ as follows: $F_{(p,x)}(p',x') \equiv \{z_i\}$ for any $(p',x') \in O_p \times \left( O_i\times X_{-i}\right)$.

Then we have:

• 1. $O_p \times \left( O_i\times X_{-i}\right)$ is an open neighborhood of $(p,x)$;

• 2. $F_{(p,x)}(p',x') \equiv \{z_i\} \subseteq \psi _i(p',x')$ for any $(p',x') \in O_p \times \left( O_i\times X_{-i}\right)$;

• 3. $F_{(p,x)}$ is a single-valued constant correspondence and hence is closed.

Therefore, $\psi$ has the continuous inclusion property at $(p,x)$. In addition, it is easy to see that $x_i \notin \text{ co }\psi _i(p,x)$. By Theorem 2 above, there exists a Walrasian equilibrium. Indeed, it can be easily checked that $(p^*, x^*)$ is a unique Walrasian equilibrium, where $p^* = (p^*_1, p^*_2) = (\frac{1}{2},\frac{1}{2})$ and $x^*_1 = x^*_2 = (\frac{1}{2},\frac{1}{2})$. Notice that even if the endowment is on the boundary $e_1 = (0,1)$ and $e_2 = (1,0)$, the equilibrium still remains the same.

Remark 13

A natural question that arises is whether or not the continuous inclusion property is easily verifiable for an economy. In the example above, we have demonstrated that it is easily verifiable, and it can be used to obtain the existence of a Walrasian equilibrium. Below we present another example in which one can easily check that the continuous inclusion property does not hold, and there is no Walrasian equilibrium. In this example, the preferences are continuous, and the initial endowment is not an interior point of the consumption set.

Example 2

There are two agents $I=\{1,2\}$ and two goods $x$ and $y$. The payoff functions are given by $u_1(x,y) = x + y$ and $u_2(x,y) = y$, which are continuous. The initial endowments are $e_1 = (\frac{1}{2}, 0)$ and $e_2 = (\frac{1}{2}, 1)$. The consumption sets for both agents are $[0,2]\times [0,2]$. In this example, one can easily see that there is no Walrasian equilibrium, but a quasi-equilibrium $((x^*,y^*),p^*)$ exists, where $(x^*,y^*) = (x_i^*,y_i^*)_{i\in I}$, and $(x_1^*,y_1^*) = (1,0)$, $(x_2^*,y_2^*) = (0,1)$, $p^* = (0,1)$.

In this example, the continuity inclusion property does not hold. Consider agent $1$ in the above quasi-equilibrium. Since $p^* \times e_1 = 0$, the budget set of agent $1$ is $B_1(p^*) = \{(x_1,0) :x_1\in [0,2]\}$. In addition, the set of allocations for agent $1$ which are preferred to $(x_1^*,y_1^*)$ is $P_1(x^*, y^*) = \{(x_1,y_1)\in [0,2]\times [0,2] :x_1 + y_1 > x^*_1 + y^*_1 = 1+ 0 =1 \}$. Thus, $\psi _1(p^*,(x^*,y^*)) = B_1(p^*) \cap P_1(x^*, y^*) = \{(x_1,0) :x_1\in (1,2]\}$, which is nonempty.

However, if we slightly perturb the price $p^*$ by assuming that it is $q = (\epsilon , 1-\epsilon )$ for sufficiently small $0 < \epsilon < \frac{1}{4}$, then the budget set of agent $1$ is $B_1(q) = \{(x_1,y_1) \in [0,2]\times [0,2] :x_1\cdot \epsilon + y_1\cdot (1-\epsilon ) \le \frac{1}{2} \epsilon \}$, which implies that $x_1 \le \frac{1}{2}$ and $y_1 \le \frac{1}{2} \frac{\epsilon }{1-\epsilon } < \frac{1}{6}$. Thus, $x_1 + y_1 < \frac{1}{2} + \frac{1}{6} = \frac{2}{3} < 1$ for all $(x_1,y_1) \in B_1(q)$, which implies that $\psi _1(q,(x^*, y^*)) = B_1(q) \cap P_1(x^*, y^*) = \emptyset$.

Therefore, in any neighborhood $O$ of $((x^*,y^*),p^*)$, there is a point $((x^*,y^*),q) \in O$ such that $\psi _1(q,(x^*, y^*)) = \emptyset$, which implies that the continuity inclusion property does not hold. It can be easily checked that the weaker condition discussed in Remark 8 still fails in this example.