Abstract

Brouwer's fixed point theorem cannot be constructively proved, so the existence of an equilibrium in a competitive economy also cannot be constructively proved. On the other hand, Sperner's lemma which is used to prove Brouwer's theorem is constructively proved. Some authors have presented a constructive (or an approximate) version of Brouwer's fixed point theorem using Sperner's lemma. In this paper, I prove the existence of an approximate equilibrium in a competitive economy directly by Sperner's lemma. Also I show that the existence of an approximate equilibrium leads to Sperner's lemma. I follow the Bishop style constructive mathematics according to Bishop and Bridges (1985), Bridges and Richman (1987), and Bridges and Vîţă (2006).

1. Introduction

It is often demonstrated that Brouwer’s fixed point theorem cannot be constructively proved. Thus, the existence of an equilibrium in a competitive economy also cannot be constructively proved. On the other hand, however, Sperner’s lemma which is used to prove Brouwer’s theorem is constructively proved. Some authors have presented a constructive (or an approximate) version of Brouwer’s fixed point theorem using Sperner’s lemma. Thus, Brouwer’s fixed point theorem can be constructively proved in its constructive version. See [1, 2]. Using this theorem, we can prove the existence of an approximate equilibrium in a competitive economy.

Then, can we prove the existence of an approximate equilibrium directly by Sperner’s lemma?

This paper presents such a proof and also shows that the existence of an approximate equilibrium leads to Sperner’s lemma. An approximate equilibrium in a competitive economy is a state such that excess demand for each good is smaller than 𝜀 for each 𝜀>0.

In the next section, the proof of Sperner’s lemma will be presented. This proof is a standard constructive proof. In Section 3, the existence of an approximate equilibrium in a competitive exchange economy will be proved using Sperner’s lemma. In Section 4, it will be shown that the existence of an approximate equilibrium leads to Sperner’s lemma. I follow the Bishop style constructive mathematics according to Bishop and Bridges [3], Bridges and Richman [4], and Bridges and Vîţă [5].

2. Sperner’s Lemma

To prove Sperner’s lemma, we use the following simple result of graph theory, the Handshaking lemma (For another constructive proof of Sperner’s lemma, see [6].). A graph refers to a collection of vertices and a collection of edges that connect pairs of vertices. Each graph may be undirected or directed. Figure 1 is an example of an undirected graph. Degree of a vertex of a graph is defined to be the number of edges incident to the vertex, with loops counted twice. Each vertex has odd degree or even degree. Let 𝑣 denotes a vertex and 𝑉 denotes the set of all vertices.

Lemma 2.1 (the Handshaking lemma). Every undirected graph contains an even number of vertices of odd degree, that is, the number of vertices that have an odd number of incident edges must be even.

It is a simple lemma. But for completeness of arguments we provide a proof.

Proof. Prove this lemma by double counting. Let 𝑑(𝑣) be the degree of vertex 𝑣. The number of vertex-edge incidences in the graph may be counted in two different ways: by summing the degrees of the vertices or by counting two incidences for every edge. Therefore, 𝑣𝑉𝑑(𝑣)=2𝑒,(2.1) where 𝑒 is the number of edges in the graph. The sum of the degrees of the vertices is therefore an even number. It could happen if and only if an even number of the vertices had odd degree.

Let Δ denotes an 𝑛-dimensional simplex. 𝑛 is a finite natural number. For example, a 2-dimensional simplex is a triangle. Let us partition or triangulate the simplex. Figure 2 is an example of partition (triangulation) of a 2-dimensional simplex. In a 2-dimensional case, we divide each side of Δ in 𝑚 equal segments and draw the lines parallel to the sides of Δ. 𝑚 is also a finite natural number. Then, the 2-dimensional simplex is partitioned into 𝑚2 triangles. We consider partition of Δ inductively for cases of higher dimension. In a 3-dimensional case, each face of Δ is a 2-dimensional simplex, so it is partitioned into 𝑚2 triangles in the way above mentioned, and draw the planes parallel to the faces of Δ. Then, the 3-dimensional simplex is partitioned into 𝑚3 trigonal pyramids, and similarly for cases of higher dimension.

Let 𝐾 denotes the set of small 𝑛-dimensional simplices of Δ constructed by partition. Vertices of these small simplices of 𝐾 are labeled with the numbers 0,1,2,,𝑛 subject to the following rules: (1)the vertices of Δ are respectively labeled 0 to 𝑛. We label a point (1,0,,0) with 0, a point (0,1,0,,0) with 1, a point (0,0,1,0) with 2,,a point (0,,0,1) with 𝑛, that is, a vertex whose 𝑘th coordinate (𝑘=0,1,,𝑛) is 1 and all other coordinates are 0 is labeled with 𝑘,(2)if a vertex of 𝐾 is contained in an 𝑛1-dimensional face of Δ, then this vertex is labeled with some number which is the same as the number of a vertex of that face,(3)if a vertex of 𝐾 is contained in an 𝑛2-dimensional face of Δ, then this vertex is labeled with some number which is the same as the number of a vertex of that face. And so on for cases of lower dimension,(4)a vertex contained inside of Δ is labeled with an arbitrary number among 0,1,,𝑛.

A small simplex of 𝐾 which is labeled with the numbers 0,1,,𝑛 is called a fully labeled simplex. Now, let us prove Sperner’s lemma.

Lemma 2.2 (Sperner’s lemma). If one labels the vertices of 𝐾 following above rules (1)~(4), then there are an odd number of fully labeled simplices. Thus, there exists at least one fully labeled simplex.

Proof. See the appendix.

3. Approximate Equilibrium in a Competitive Exchange Economy

Consider a competitive exchange economy. There are 𝑛+1 goods 𝑋0,𝑋1,,𝑋𝑛. 𝑛 is a finite natural number. The prices of the goods are denoted by 𝑝𝑖(0),𝑖=0,1,,𝑛. Let 𝑝=𝑝0+𝑝1++𝑝𝑛, and define𝑝𝑖=𝑝𝑖𝑝,𝑖=0,1,,𝑛.(3.1) Replace 𝑝0,𝑝1,,𝑝𝑛, respectively, by 𝑝0,𝑝1,,𝑝𝑛. Then,𝑝0+𝑝1++𝑝𝑛=1.(3.2) Thus, 𝐩=(𝑝0,𝑝1,,𝑝𝑛) represents a point on an 𝑛-dimensional simplex. Since consumers’ demand (excess demand) for each good in a competitive exchange economy is determined by relative prices of the goods, such notation yields no loss of generality. We denote the vector of excess demands for the goods when the vector of prices is 𝐩 by 𝐟(𝐩)=(𝑓1,𝑓2,,𝑓𝑛). Then, the following relation must hold:𝐩𝐟(𝐩)=𝑝0𝑓0+𝑝1𝑓1++𝑝𝑛𝑓𝑛=0(Walraslaw)(3.3)𝑓𝑖 is equal to the sum of excess demands of all consumers for the good 𝑋𝑖. By the budget constraint for each consumer, in a competitive economy the sum of excess demands of all consumers for each good must be 0. Adding the budget constraints for all consumers yields (3.3).

We assume that the excess demand function 𝐟(𝐩) is uniformly continuous about the prices of the goods. Uniform continuity of 𝐟(𝐩) means that for any 𝐩, 𝐩 and 𝜀>0, there exists 𝛿>0 such that ||𝐩If||||𝐟𝐩𝐩<𝛿,then||𝐟(𝐩)<𝜀(3.4)𝛿 depends on neither 𝐩 nor 𝐩. Uniform continuity of demand functions implies that a slight price change yields only a slight demand change.

Consider the following function from the set of price vectors 𝐩=(𝑝0,𝑝1,,𝑝𝑛) to the set of 𝑛+1 tuples of real numbers 𝐯=(𝑣0,𝑣1,,𝑣𝑛):𝑣𝑖=𝑝𝑖𝑓+max𝑖,0𝑖.(3.5) With this, we define a function from an 𝑛-dimensional simplex Δ to itself 𝜑(𝐩)=(𝜑0,𝜑1,,𝜑𝑛) as follows:𝜑𝑖=1𝑣0+𝑣1++𝑣𝑛𝑣𝑖,𝑖.(3.6) Since 𝜑𝑖0,𝑖=0,1,,𝑛 and𝜑0+𝜑1++𝜑𝑛=1,(3.7)(𝜑0,𝜑1,,𝜑𝑛) represents a point on Δ. By the uniform continuity of 𝐟, 𝜑 is also uniformly continuous.

Now, we show the following.

Theorem 3.1. In a competitive exchange economy, if the excess demand functions for the goods are uniformly continuous about their prices, then there exists an approximate equilibrium and one can constructively find the prices at the approximate equilibrium.

Proof. (1) First, we show that we can partition Δ, which is the domain and range of 𝜑, so that the conditions for Sperner’s lemma are satisfied. We partition Δ according to the method in the proof of Sperner’s lemma (see the appendix) and label the vertices of simplices constructed by partition of Δ. It is important how to label the vertices contained in the faces of Δ. Let 𝐾 be the set of small simplices constructed by partition of Δ, 𝐩=(𝑝0,𝑝1,,𝑝𝑛) a vertex of a simplex of 𝐾, and denote the 𝑖th component of 𝜑(𝐩) by 𝜑𝑖. Then, we label a vertex 𝐩 according to the following rule: If𝑝𝑘>𝜑𝑘or𝑝𝑘+𝜏>𝜑𝑘,welabel𝐩with𝑘,(3.8) where 𝜏 is a positive number. If there are multiple 𝑘’s which satisfy this condition, we label 𝐩 conveniently for the conditions of Sperner’s lemma to be satisfied. We do not randomly label the vertices.
For example, let 𝐩 be a point contained in an 𝑛1-dimensional face of Δ such that 𝑝𝑖=0 for one 𝑖 among 0,1,2,,𝑛 (the 𝑖th component of its coordinates is 0). With 𝜏>0, we have 𝜑𝑖>0 or 𝜑𝑖<𝜏 (In constructive mathematics for any real number 𝑥, we cannot prove that 𝑥0 or 𝑥<0, nor that 𝑥>0 or 𝑥=0 or 𝑥<0. But for any distinct real numbers 𝑥, 𝑦 and 𝑧 such that 𝑥>𝑧 we can prove that 𝑥>𝑦 or 𝑦>𝑧.). When 𝜑𝑖>0, from 𝑛𝑗=0𝑝𝑗=1, 𝑛𝑗=0𝜑𝑗=1 and 𝑝𝑖=0, we have 𝑛𝑗=0,𝑗𝑖𝑝𝑗>𝑛𝑗=0,𝑗𝑖𝜑𝑗.(3.9) Then, for at least one 𝑗 (denote it by 𝑘), we have 𝑝𝑘>𝜑𝑘 and we label 𝐩 with 𝑘, where 𝑘 is one of the numbers which satisfy 𝑝𝑘>𝜑𝑘. Since 𝜑𝑖>𝑝𝑖, 𝑖 does not satisfy this condition. When 𝜑𝑖<𝜏, 𝑝𝑖=0 implies 𝑛𝑗=0,𝑗𝑖𝑝𝑗=1. Since 𝑛𝑗=0,𝑗𝑖𝜑𝑗1, we obtain 𝑛𝑗=0,𝑗𝑖𝑝𝑗𝑛𝑗=0,𝑗𝑖𝜑𝑗.(3.10) Then, for a positive number 𝜏, we have 𝑛𝑗=0,𝑗𝑖𝑝𝑗>+𝜏𝑛𝑗=0,𝑗𝑖𝜑𝑗.(3.11) Thus, there is at least one 𝑗(𝑖) which satisfies 𝑝𝑗+𝜏>𝜑𝑗. Denote it by 𝑘, and we label 𝐩 with 𝑘. 𝑘 is one of the numbers other than 𝑖 such that 𝑝𝑘+𝜏>𝜑𝑘 is satisfied. 𝑖 itself satisfies this condition (𝑝𝑖+𝜏>𝜑𝑖). But, since there is a number other than 𝑖 which satisfies this condition, we can select a number other than 𝑖. We have proved that we can label the vertices contained in an 𝑛1-dimensional face of Δ such that 𝑝𝑖=0 for one 𝑖 among 0,1,2,,𝑛 with the numbers other than 𝑖. By similar procedures, we can show that we can label the vertices contained in an 𝑛2-dimensional face of Δ such that 𝑝𝑖=0 for two 𝑖’s among 0,1,2,,𝑛 with the numbers other than those 𝑖’s, and so on.
Consider the case where 𝑝𝑖=𝑝𝑖+1=0. By similar procedures, we see that when 𝜑𝑖>0 or 𝜑𝑖+1>0, 𝑛𝑗=0,𝑗𝑖,𝑖+1𝑝𝑗>𝑛𝑗=0,𝑗𝑖,𝑖+1𝜑𝑗.(3.12) Then, for at least one 𝑗 (denote it by 𝑘), we have 𝑝𝑘>𝜑𝑘 and we label 𝐩 with 𝑘. On the other hand, when 𝜑𝑖<𝜏 and 𝜑𝑖+1<𝜏, we have 𝑛𝑗=0,𝑗𝑖,𝑖+1𝑝𝑗𝑛𝑗=0,𝑗𝑖,𝑖+1𝜑𝑗.(3.13) Then, for a positive number 𝜏, we have 𝑛𝑗=0,𝑗𝑖,𝑖+1𝑝𝑗>+𝜏𝑛𝑗=0,𝑗𝑖,𝑖+1𝜑𝑗.(3.14) There is at least one 𝑗(𝑖,𝑖+1) which satisfies 𝑝𝑗+𝜏>𝜑𝑗. Denote it by 𝑘, and we label 𝐩 with 𝑘.
Next, consider the case where 𝑝𝑖=0 for all 𝑖 other than 𝑛. If for some 𝑖𝜑𝑖>0, then we have 𝑝𝑛>𝜑𝑛, and label 𝐩 with 𝑛. On the other hand, if 𝜑𝑗<𝜏 for all 𝑗𝑛, then we obtain 𝑝𝑛𝜑𝑛. It implies 𝑝𝑛+𝜏>𝜑𝑛. Thus, we can label 𝐩 with 𝑛.
The conditions for Sperner’s lemma are satisfied, and there exists an odd number of fully labeled simplices in 𝐾.
(2) Suppose that we partition Δ sufficiently fine so that the distance between any pair of the vertices of small simplices is sufficiently small. Let 𝐩0,𝐩1,, and 𝐩𝑛 be the vertices of a fully labeled simplex. We name these vertices so that 𝐩0,𝐩1,,𝐩𝑛 are labeled, respectively, with 0,1,,𝑛. The values of 𝜑 at theses vertices are 𝜑(𝐩0),𝜑(𝐩1),, and 𝜑(𝐩𝑛). The 𝑖th components of 𝐩0 and 𝜑(𝐩0) are denoted by 𝐩0𝑖 and 𝜑(𝐩0)𝑖, and so on.
The uniform continuity of 𝜑 implies that we can select 𝛿>0 so that when the distance between 𝐩0 and 𝐩1 (|𝐩0𝐩1|) is smaller than 𝛿, the distance between 𝜑(𝐩0) and 𝜑(𝐩1) (|𝜑(𝐩0)𝜑(𝐩1)|) is smaller than 𝜀. We can make 𝛿 satisfying 𝛿<𝜀 (For example, for 𝛿<1 and 𝜀<1, if when |𝐩0𝐩1|<𝛿 we have |𝜑(𝐩0)𝜑(𝐩1)|<𝜀, then we have |𝜑(𝐩0)𝜑(𝐩1)|<𝜀 also when |𝐩0𝐩1|<𝛿𝜀<𝜀.). Suppose 𝜏>0. About 𝐩0, from the labeling rules, we have 𝐩00+𝜏>𝜑(𝐩0)0. About 𝐩1, also from the labeling rules we have 𝐩11+𝜏>𝜑(𝐩1)1 which implies 𝐩11>𝜑(𝐩1)1𝜏. By the uniform continuity of 𝜑, |𝐩0𝐩1|<𝛿 implies |𝜑(𝐩0)𝜑(𝐩1)|<𝜀, which means 𝜑(𝐩1)1>𝜑(𝐩0)1𝜀. On the other hand, |𝐩0𝐩1|<𝛿 means 𝐩01>𝐩11𝛿. Thus, from 𝐩01>𝐩11𝛿,𝐩11𝐩>𝜑11𝐩𝜏,𝜑11𝐩>𝜑01𝜀,(3.15) we obtain 𝐩01𝐩>𝜑01𝐩𝛿𝜀𝜏>𝜑012𝜀𝜏.(3.16) By similar arguments, for each 𝑖 other than 0, 𝐩0𝑖𝐩>𝜑0𝑖2𝜀𝜏.(3.17) For 𝑖=0, we have 𝐩00+𝜏>𝜑(𝐩0)0. Then, 𝐩00𝐩>𝜑00𝜏.(3.18) Adding (3.17) and (3.18) side by side except for some 𝑖 (denote it by 𝑘) other than 0 yields 𝑛𝑗=0,𝑗𝑘𝐩0𝑗>𝑛𝑗=0,𝑗𝑘𝜑𝐩0𝑗2(𝑛1)𝜀𝑛𝜏.(3.19) From 𝑛𝑗=0𝐩0𝑗=1, 𝑛𝑗=0𝜑(𝐩0)𝑗=1, we have 1𝐩0𝑘>1𝜑(𝐩0)𝑘2(𝑛1)𝜀𝑛𝜏, which is rewritten as 𝐩0𝑘𝐩<𝜑0𝑘+2(𝑛1)𝜀+𝑛𝜏.(3.20) Since (3.17) implies 𝐩0𝑘>𝜑(𝐩0)𝑘2𝜀𝜏, we have 𝜑𝐩0𝑘2𝜀𝜏<𝐩0𝑘𝐩<𝜑0𝑘+2(𝑛1)𝜀+𝑛𝜏.(3.21) Thus, ||𝐩0𝑘𝐩𝜑0𝑘||<2(𝑛1)𝜀+𝑛𝜏(3.22) is derived. On the other hand, adding (3.17) from 1 to 𝑛 yields 𝑛𝑗=1𝐩0𝑗>𝑛𝑗=1𝜑𝐩0𝑗2𝑛𝜀𝑛𝜏.(3.23) From 𝑛𝑗=0𝐩0𝑗=1, 𝑛𝑗=0𝜑(𝐩0)𝑗=1, we have 1𝐩00𝐩>1𝜑002𝑛𝜀𝑛𝜏.(3.24) Then, from (3.18) and (3.24), we get ||𝐩00𝐩𝜑00||<2𝑛𝜀+𝑛𝜏.(3.25) Since 𝑛 is finite and 𝜀 and 𝜏 are positive numbers which may be arbitrarily small, 2𝑛𝜀+𝑛𝜏 and 2(𝑛1)𝜀+𝑛𝜏 may also be arbitrarily small. Replacing 2𝑛𝜀+𝑛𝜏 by 𝜀, from (3.22) and (3.25) we obtain the following result: ||𝐩0𝑖𝐩𝜑0𝑖||<𝜀𝑖.(3.26) All points contained in the fully labeled simplex of 𝐾 satisfy this relation.
(3) Denote a point which satisfies (3.26) by ̃𝐩, and denote its 𝑖th component by ̃𝑝𝑖. We have ||̃𝑝𝑖𝜑𝑖||<𝜀𝑖.(3.27) Let us consider the relationship between the price and demand for each good in that case. From the definitions of 𝜑 and 𝐯, and 𝑛𝑗=0𝑣𝑗=1+𝑛𝑗=0max(𝑓𝑗,0) (because 𝑛𝑗=0̃𝑝𝑗=1), (3.27) means |||||̃𝑝𝑖𝑓+max𝑖,01+𝑛𝑗=0𝑓max𝑗,0̃𝑝𝑖|||||=|||||𝑓max𝑖,0̃𝑝𝑖𝑛𝑗=0𝑓max𝑗,01+𝑛𝑗=0𝑓max𝑗|||||,0<𝜀.(3.28) Let 𝜆=𝑛𝑗=0max(𝑓j,0). Then, we have |(max(𝑓𝑖,0)𝜆̃𝑝𝑖)/(1+𝜆)|<𝜀. It is rewritten as ||𝑓max𝑖,0𝜆̃𝑝𝑖||<(1+𝜆)𝜀.(3.29) This means (1+𝜆)𝜀+𝜆̃𝑝𝑖𝑓<max𝑖<,0(1+𝜆)𝜀+𝜆̃𝑝𝑖.(3.30) From 𝑛𝑗=0̃𝑝𝑗=1, there is a 𝑘 which satisfies ̃𝑝𝑘>0. If for all such 𝑘max(𝑓𝑘,0)=𝑓𝑘>0 holds, that is, excess demands for all goods with positive prices are positive, we cannot cancel out ̃𝑝𝑘𝑓𝑘>0 because the price of any good cannot be negative and the Walras law (3.3) is violated (Even if we relax the Walras law to the following approximate Walras law, we can show essentially the same result: |𝐩𝐟(𝐩)|=|𝑝0𝑓0+𝑝1𝑓1++𝑝𝑛𝑓𝑛|<𝜀, where 𝜀 is a positive number. If we have max(𝑓𝑘,0)=𝑓𝑘>0 for all 𝑘 which satisfy 𝑝𝑘>0, this inequality cannot be satisfied because the price of any good cannot be negative.). Therefore, 𝜆 as well as 𝜀 must be a positive number which may be arbitrarily small, and since ̃𝑝𝑖 is finite, (1+𝜆)𝜀+𝜆̃𝑝𝑖 is a real number which may also be arbitrarily small. There exists a number which is only slightly larger than (1+𝜆)𝜀+𝜆̃𝑝𝑖. Replace (1+𝜆)𝜀+𝜆̃𝑝𝑖 by such a number, and denote it by 𝜀. Then, for all 𝑖 we have 𝑓max𝑖,0<𝜀.(3.31) This means that excess demand for each good is smaller than 𝜀. Such a state is an approximate equilibrium. In the approximate equilibrium when 𝑝𝑖>0, we must have 𝑓𝑖>𝜀 for any 𝜀>0 because if 𝑓𝑖<0 we have 𝑝𝑖𝑓𝑖<0, and then the Walras law is violated. On the other hand, there may be excess supply (negative excess demand) for a good whose price is zero.

4. From the Existence of an Approximate Equilibrium to Sperner’s Lemma

In this section, we will derive Sperner’s lemma from the existence of an approximate equilibrium in a competitive economy. Let us partition an 𝑛-dimensional simplex Δ. Denote the set of small 𝑛-dimensional simplices of Δ constructed by partition by 𝐾. Vertices of these small simplices of 𝐾 are labeled with the numbers 0,1,2,,𝑛 similarly to the proof of Sperner’s lemma. Denote vertices of an 𝑛-dimensional simplex of 𝐾 by 𝑥0,𝑥1,,𝑥𝑛, the 𝑗th component of 𝑥𝑖 by 𝑥𝑖𝑗, and the label of 𝑥𝑖 by 𝑙(𝑥𝑖). Let 𝜏 be a positive number which is smaller than 𝑥𝑖𝑙(𝑥𝑖) for all 𝑥𝑖, and define a function 𝑓(𝑥𝑖) as follows:𝑓𝑥𝑖=𝑓0𝑥𝑖,𝑓1𝑥𝑖,,𝑓𝑛𝑥𝑖𝑓,(4.1)𝑗𝑥𝑖=𝑥𝑖𝑗𝑥𝜏for𝑗=𝑙𝑖,𝑥𝑖𝑗+𝜏𝑛𝑥for𝑗𝑙𝑖.(4.2)𝑓𝑗 denotes the 𝑗th component of 𝑓. From the labeling rules 𝑥𝑖𝑙(𝑥𝑖)>0 for all 𝑥𝑖, and so 𝜏>0 is well defined. Since 𝑛𝑗=0𝑓𝑗(𝑥𝑖)=𝑛𝑗=0𝑥𝑖𝑗=1, we have𝑓𝑥𝑖Δ.(4.3) We extend 𝑓 to all points in the simplex by convex combinations of its values on the vertices of the simplex. Let 𝑦 be a point in the 𝑛-dimensional simplex of 𝐾 whose vertices are 𝑥0,𝑥1,,𝑥𝑛. Then, 𝑦 and 𝑓(𝑦) are represented as follows:𝑦=𝑛𝑖=0𝜆𝑖𝑥𝑖,𝑓(𝑦)=𝑛𝑖=0𝜆𝑖𝑓𝑥𝑖,𝜆𝑖0,𝑛𝑖=0𝜆𝑖=1.(4.4) Let us show that 𝑓 is uniformly continuous. Let 𝑦 and 𝑦 be distinct points in the same small 𝑛-dimensional simplex of 𝐾. They are represented as𝑦=𝑛𝑖=0𝜆𝑖𝑥𝑖,𝑦=𝑛𝑖=0𝜆𝑖𝑥𝑖,(4.5) and so𝑦𝑦=𝑛𝑖=0𝜆𝑖𝜆𝑖𝑥𝑖,𝑦𝑗𝑦𝑗=𝑛𝑖=0𝜆𝑖𝜆𝑖𝑥𝑖𝑗foreach𝑗.(4.6) Then, we have𝑦𝑓(𝑦)𝑓=𝑛𝑖=0𝜆𝑖𝜆𝑖𝑓𝑥𝑖,(4.7) and for each 𝑗,𝑓𝑗(𝑦)𝑓𝑗𝑦=𝑛𝑖=0𝜆𝑖𝜆𝑖𝑥𝑖𝑗+𝑖𝑗𝑙(𝑖)𝜆𝑖𝜆𝑖𝜏𝑛𝑖𝑗=𝑙(𝑖)𝜆𝑖𝜆𝑖𝜏=𝑦𝑗𝑦𝑗+𝑖𝑗𝑙(𝑖)𝜆𝑖𝜆𝑖𝜏𝑛𝑖𝑗=𝑙(𝑖)𝜆𝑖𝜆𝑖𝜏.(4.8) Since 𝜏 is finite, appropriately selecting 𝜆𝑖 given 𝜆𝑖 for each 𝑖, we can make |𝑓𝑗(𝑦)𝑓𝑗(𝑦)| sufficiently small corresponding to the value of |𝑦𝑗𝑦𝑗| for each 𝑗, and so make |𝑓(𝑦)𝑓(𝑦)| sufficiently small corresponding to the value of |𝑦𝑦|. Thus, 𝑓 is uniformly continuous.

Now, using 𝑓, we construct an excess demand function as follows:𝑧𝑖(𝐩)=𝑓𝑖(𝐩)𝑝𝑖𝜇(𝐩),𝑖=0,1,,𝑛.(4.9)𝐩Δ and 𝜇(𝐩) is defined by𝜇(𝐩)=𝑛𝑖=0𝑝𝑖𝑓𝑖(𝐩)𝑛𝑖=0𝑝2𝑖.(4.10) Each 𝑧𝑖(𝐩) is uniformly continuous, and satisfies the Walras law as shown below. Multiplying 𝑝𝑖 (the 𝑖th component of 𝐩) to (4.9) for each 𝑖 and adding them from 0 to 𝑛 yields𝑛𝑖=0𝑝𝑖𝑧𝑖=𝑛𝑖=0𝑝𝑖𝑓𝑖(𝐩)𝜇(𝐩)𝑛𝑖=0𝑝2𝑖=𝑛𝑖=0𝑝𝑖𝑓𝑖(𝐩)𝑛𝑖=0𝑝𝑖𝑓𝑖(𝐩)𝑛𝑖=0𝑝2𝑖𝑛𝑖=0𝑝2𝑖=𝑛𝑖=0𝑝𝑖𝑓𝑖(𝐩)𝑛𝑖=0𝑝𝑖𝑓𝑖(𝐩)=0.(4.11) Therefore, 𝑧𝑖(𝐩)’s satisfy the conditions for excess demand functions and there exists an approximate equilibrium. Let 𝐩={𝑝0,𝑝1,,𝑝𝑛} be the price vector at the approximate equilibrium. Then, from max(𝑧𝑖,0)<𝜀 (see (3.31)), we have 𝑓𝑖(𝐩)𝑝𝑖𝜇(𝐩)<𝜀 for all 𝑖, where 𝜀 is an positive number. Since it is impossible that at the approximate equilibrium 𝑧𝑖<0 for 𝑖 satisfying 𝑝𝑖>0 because of (4.11), we have 𝑧𝑖=𝑓𝑖(𝐩)𝑝𝑖𝜇(𝐩)>𝜀 for such 𝑖. On the other hand, for 𝑖 such that 𝑝𝑖=0, we have 𝑧𝑖=𝑓𝑖(𝐩)0. Therefore,𝜀<𝑓𝑖𝐩𝑝𝑖𝜇𝐩<𝜀(4.12) is obtained. Adding this inequality side by side from 0 to 𝑛 yields(𝑛+1)𝜀<𝑛𝑖=0𝑓𝑖𝐩𝐩𝜇𝑛𝑖=0𝑝𝑖<(𝑛+1)𝜀.(4.13) From 𝑛𝑖=0𝑓𝑖(𝐩)=𝑛𝑖=0𝑝𝑖=1, we obtain𝐩1(𝑛+1)𝜀<𝜇<1+(𝑛+1)𝜀.(4.14) Further, from (4.12) and (4.14), we get𝑝𝑖(𝑛+1)𝜀𝑝𝑖𝜀<𝑓𝑖𝐩<𝑝𝑖+(𝑛+1)𝜀𝑝𝑖+𝜀.(4.15) Since 𝑛 and 𝑝𝑖 are finite, (𝑛+1)𝜀𝑝𝑖+𝜀 is a positive real number which may be arbitrarily small. There exists a number which is only slightly larger than (𝑛+1)𝜀𝑝𝑖+𝜀. Replace (𝑛+1)𝜀𝑝𝑖+𝜀 by such a number, and denote it by 𝜀. Then, 𝜀<𝑓𝑖(𝐩)𝑝𝑖<𝜀, that is, ||𝑓𝑖𝐩𝑝𝑖||<𝜀(4.16) is derived. This relation holds for all 𝑖.

Let 𝛾>0 and ̃𝐩 be a point in 𝑉(𝐩,𝛾), where 𝑉(𝐩,𝛾) is a 𝛾-neighborhood of 𝐩. If 𝛾 is sufficiently small, uniform continuity of 𝑓 means||𝑓𝑖(̃̃𝐩𝐩)𝑖||<𝜀,(4.17) for any 𝜀>0 and for all 𝑖. ̃𝐩𝑖 is the 𝑖th component of ̃𝐩. Let Δ be a simplex of 𝐾 which contains ̃𝐩, and 𝐩0,𝐩1,,𝐩𝑛 be the vertices of Δ. Then, ̃𝐩 and ̃𝑓(𝐩) are represented as ̃𝐩=𝑛𝑖=0𝜆𝑖𝐩𝑖̃,𝑓(𝐩)=𝑛𝑖=0𝜆𝑖𝑓𝐩𝑖,𝜆𝑖0,𝑛𝑖=0𝜆𝑖=1.(4.18) Equation (4.2) implies that if only one 𝐩𝑘 among 𝐩0,𝐩1,,𝐩𝑛 is labeled with 𝑖, we have||𝑓𝑖(̃̃𝐩𝐩)𝑖||=|||||𝑛𝑗=0𝜆𝑗𝐩𝑗𝑖+𝑗=0,𝑗𝑘𝜆𝑗𝜏𝑛𝜆𝑘̃𝐩𝜏𝑖|||||=|||||1𝑛𝑗=0,𝑗𝑘𝜆𝑗𝜆𝑘𝜏|||||<𝜀.(4.19)𝐩𝑗𝑖 is the 𝑖th component of 𝐩𝑗.

Since 𝜀 may be arbitrarily small and 𝜏>0, this means1𝑛𝑗=0,𝑗𝑘𝜆𝑗𝜆𝑘0(4.20) Equation (4.17) is satisfied with 𝜆𝑘1/(𝑛+1) for all 𝑘. On the other hand if no 𝐩𝑗 is labeled with 𝑖, we have𝑓𝑖(̃𝐩)=𝑛𝑗=0𝜆𝑗𝐩𝑗𝑖=𝐩𝑖+11+𝑛𝜏,(4.21) and then (4.17) cannot be satisfied. Thus, for each 𝑖 one and only one 𝐩𝑗 must be labeled with 𝑖. Therefore, Δ must be a fully labeled simplex. We have completed the proof of Sperner’s lemma.

5. Concluding Remark

In this paper, I have presented a proof of the existence of an approximate equilibrium in a competitive economy directly by Sperner’s lemma from a viewpoint of constructive mathematics. In another paper [7], I apply this method to prove the existence of an approximate Nash equilibrium in a finite strategic game (a strategic game with a finite number of players and a finite number of pure strategies).

Appendix

Proof of Sperner’s Lemma

We prove this lemma by induction about the dimension of Δ. When 𝑛=0, we have only one point with the number 0. It is the unique 0-dimensional simplex. Therefore, the lemma is trivial. When 𝑛=1, a partitioned 1-dimensional simplex is a segmented line. The endpoints of the line are labeled distinctly, by 0 and 1. Hence, in moving from endpoint 0 to endpoint 1, the labeling must switch an odd number of times, that is, an odd number of edges labeled with 0 and 1 may be located in this way.

Next, consider the case of 2 dimension. Assume that we have partitioned a 2-dimensional simplex (triangle) Δ as explained above. Consider the face of Δ labeled with 0 and 1 (We call edges of triangle Δ  faces to distinguish between them and edges of a dual graph which we will consider later.). It is the base of the triangle in Figure 3. Now, we introduce a dual graph that has its nodes in each small triangle of 𝐾 plus one extra node outside the face of Δ labeled with 0 and 1 (putting a dot in each small triangle and one dot outside Δ). We define edges of the graph that connect two nodes if they share a side labeled with 0 and 1. See Figure 3. White circles are nodes of the graph, and thick lines are its edges. Since from the result of 1-dimensional case there are an odd number of faces of 𝐾 labeled with 0 and 1 contained in the face of Δ labeled with 0 and 1, there are an odd number of edges which connect the outside node and inside nodes. Thus, the outside node has odd degree. Since by the Handshaking lemma there are an even number of nodes which have odd degree, we have at least one node inside the triangle which has odd degree. Each node of our graph except for the outside node is contained in one of small triangles of 𝐾. Therefore, if a small triangle of 𝐾 has one face labeled with 0 and 1, the degree of the node in that triangle is 1; if a small triangle of 𝐾 has two such faces, the degree of the node in that triangle is 2, and if a small triangle of 𝐾 has no such face, the degree of the node in that triangle is 0. Thus, if the degree of a node is odd, it must be 1 and then the small triangle which contains this node is labeled with 0, 1 and 2 (fully labeled). In Figure 3, triangles which contain one of the nodes 𝐴, 𝐵, 𝐶 are fully labeled triangles.

Now, assume that the theorem holds for dimensions up to 𝑛1. Assume that we have partitioned an 𝑛-dimensional simplex Δ. Consider the fully labeled face of Δ which is a fully labeled 𝑛1-dimensional simplex. Again, we introduce a dual graph that has its nodes in small 𝑛-dimensional simplices of 𝐾 plus one extra node outside the fully labeled face of Δ (putting a dot in each small 𝑛-dimensional simplex and one dot outside Δ). We define the edges of the graph that connect two nodes if they share a face labeled with 0,1,,𝑛1. Since from the result of 𝑛1-dimensional case there are an odd number of fully labeled faces of small simplices of 𝐾 contained in the 𝑛1-dimensional fully labeled face of Δ, there are an odd number of edges which connect the outside node and inside nodes. Thus, the outside node has odd degree. Since, by the Handshaking lemma there are an even number of nodes which have odd degree, we have at least one node inside the simplex which has odd degree. Each node of our graph except for the outside node is contained in one of small 𝑛-dimensional simplices of 𝐾. Therefore, if a small simplex of 𝐾 has one fully labeled face, the degree of the node in that simplex is 1; if a small simplex of 𝐾 has two such faces, the degree of the node in that simplex is 2, and if a small simplex of 𝐾 has no such face, the degree of the node in that simplex is 0. Thus, if the degree of a node is odd, it must be 1 and then the small simplex which contains this node is fully labeled.

If the number (label) of a vertex other than vertices labeled with 0,1,,𝑛1 of an 𝑛-dimensional simplex which contains a fully labeled 𝑛1-dimensional face is 𝑛, then this 𝑛-dimensional simplex has one such face, and this simplex is a fully labeled 𝑛-dimensional simplex. On the other hand, if the number of that vertex is other than 𝑛, then the 𝑛-dimensional simplex has two such faces.

We have completed the proof of Sperner’s lemma.

Since 𝑛 and partition of Δ are finite, the number of small simplices constructed by partition is also finite. Thus, we can constructively find a fully labeled 𝑛-dimensional simplex of 𝐾 through finite steps.

Acknowledgments

The author thanks the anonymous referees for their very useful comments. This research was partially supported by the Ministry of Education, Science, Sports and Culture of Japan, Grant-in-Aid for Scientific Research (C), no. 20530165, and the Special Costs for Graduate Schools of the Special Expenses for Hitech Promotion by the Ministry of Education, Science, Sports and Culture of Japan in 2010.