A combinatorial lemma and its applications
 Piotr Maćkowiak^{1}Email author
https://doi.org/10.1186/s136600161043y
© Maćkowiak 2016
Received: 27 October 2015
Accepted: 17 March 2016
Published: 31 March 2016
Abstract
In this paper, we present a generalization of a combinatorial lemma we stated and proved in a recent work. Then we apply the generalized lemma to prove: (1) a theorem on the existence of a zero for an excess demand mapping, (2) the existence of a continuum of zeros for a parameterized excess demand mapping, (3) Sperner’s lemma on labelings of triangulations. Proofs of these results are constructive: they contain algorithms (based on the combinatorial lemma) for the computation of objects of interest or, at least, of their approximations.
Keywords
Browder fixed point theorem combinatorial methods continuum of zeros equilibrium fixed point Kakutani fixed point theorem Sperner lemma zero of a mapMSC
91B02 91B50 54H251 Introduction
In the paper [1], we stated and proved a combinatorial lemma with the help of which we then showed the existence of a zero for an excess demand functions and Brouwer’s fixed point theorem. We also stated some open problems in the referred paper. The current work answers some of these questions.
First, we prove a generalization of the combinatorial lemma presented in [1]. Then we apply it to prove the existence of an equilibrium price vector for an excess demand mapping (Lemma 6 and Theorem 7). Next, we apply our combinatorial lemma to prove the existence of a continuum of zeros for a parameterized excess demand mapping (Theorem 8). Then we derive Sperner’s lemma (Theorem 13) from our combinatorial result.
Let us emphasize the fact that the combinatorial Lemma 2 allows us to get algorithms for finding (approximations of) objects whose existence is claimed in Theorems 7 and 8 and a simplex enjoying properties stated in Sperner’s lemma. Hence, our proofs are not only of existential character, but they enable the computation of objects of interest (or at least their approximations).^{1}
In the next section, we set up notation and introduce preliminary notions from combinatorial topology. Then we prove the just mentioned combinatorial lemma (Lemma 2) and apply it to get the promised results. The last section comprises some comments.
2 Preliminaries
Let \(\mathbb{N}\) denote the set of positive integers, and for any \(n\in\mathbb{N}\), let \(\mathbb{R}^{n}\) denote the ndimensional Euclidean space, and let \([n]:=\{1,\ldots,n\}\), \([0]:=\emptyset\), \([n]_{0}:=\{0,1,\ldots ,n\}\), and \([n]_{i}:=\emptyset\) for \(i>n\). We take on the convention \(\sum_{i\in\emptyset}a_{i}=0\). Moreover, \(e^{i}\) is the ith unit vector of the standard basis of \(\mathbb{R}^{n}\), where \(i\in[n]\). The (vector) inequality \(x\geq y\) (\(x>y\)), \(x, y\in\mathbb{R}^{n}\), means \(x_{i}\geq y_{i}\) (\(x_{i}>y_{i}\)), \(i\in[n]\). In what follows, for \(n\in \mathbb{N}\), the set \(\Delta^{n}:=\{x\in\mathbb{R}^{n}_{+}: \sum_{i=1}^{n}x_{i}=1\}\), where \(\mathbb{R}_{+}\) is the set of nonnegative real numbers, is the standard (closed) \((n1)\)simplex, and \(\operatorname{int} \Delta ^{n}:=\{x\in \Delta^{n}: x_{i}>0, i\in[n]\}\) is its (relative) interior. For a set \(X\subset\mathbb{R}^{n}\), \(\partial({X})\), intX, and convX denote its boundary (or relative boundary of the closure of X if X is convex), interior (or relative interior if X is convex), and convex hull, respectively. For vectors \(x,y\in\mathbb{R}^{n}\), their scalar product is \(xy:=\sum_{i=1}^{n}x_{i}y_{i}\). For sets \(A, B\subset\mathbb{R}^{n}\), \(AB:=\{ab\in \mathbb{R}: a\in A, b\in B\}\) and \(A+B:=\{a+b\in\mathbb{R}^{n}: a\in A, b\in B\}\); for \(a\in\mathbb{R}^{n}\), we briefly write aB and \(a+B\) instead of \(\{a\}B\) and \(\{a\}+B\), respectively (similarly if the set B has one element only). If \(A\subset\mathbb{R}^{n}\) and \(b\in\mathbb{R}^{n}\), then \(A\geq b\) (\(A>b\)) means that for each \(a\in A\), \(a\geq b\) (\(a>b\)). If \(a\in \mathbb{R}\) and \(b\in\mathbb{R}^{n}\), then by \(b\geq a\) we mean \(b_{i}\geq a\), \(i\in[n]\); similarly, for the strict inequality ‘>’. The Euclidean norm of \(x\in\mathbb{R}^{n}\) is denoted by \(x\). For any set A, #A denotes its cardinality, and \(\operatorname{diam}{A}:=\sup\{xy: x, y\in A\} \) is the diameter of the set A. For \(r>0\), \(B_{r}:=\{x\in\mathbb{R}^{n}: x< r\}\) is the open ball centered at \(0\in\mathbb{R}^{n}\) with radius r. For a multivalued mapping \(F:A\multimap B\), where A, B are some sets, \(F(C):=\bigcup_{c\in C}F(c)\) for any set \(C\subset A\). For a sequence \(k_{q}\in\mathbb{R}\), \(q\in\mathbb{N}\), \(k_{q}\nearrow+\infty\) means that \(k_{q}\) diverges to +∞ strictly monotonically as q increases to +∞.

Let \(v^{j}\in\mathbb{R}^{n}\), \(j\in[k]\), \(k\in[n+1]\), be affinely independent. The set σ defined by \(\sigma:=\{x\in\mathbb {R}^{n}: x=\sum_{j=1}^{k}\alpha_{j}v^{j}, \alpha\in\Delta^{k}\}=\operatorname {conv}\{v^{1},\ldots,v^{k}\}\) is called a \((k1)\)simplex with vertices \(v^{j}\), \(j\in[k]\). We write it briefly as \(\sigma= \langle{v^{j}: j\in [k]} \rangle\) or \(\sigma= \langle{v^{1},\ldots,v^{k}} \rangle\) or \(\sigma = \langle{\{v^{1},\ldots ,v^{k}\}} \rangle\). Observe that the standard \((n1)\)simplex \(\Delta^{n}\) is an \((n1)\)simplex since \(\Delta^{n}= \langle{e^{1},\ldots, e^{n}} \rangle\subset \mathbb{R}^{n}\). If we know that σ is a \((k1)\)simplex, then the set of its vertices is denoted by \(V(\sigma)\). If \(p=\sum_{j=1}^{k}\alpha^{\sigma}_{j}(p)v^{j}\in\sigma\), then the vector \(\alpha^{\sigma}(p):=(\alpha ^{\sigma}_{1}(p),\ldots,\alpha^{\sigma}_{k}(p))\in\Delta^{k}\) is called (the vector of) the barycentric coordinates of the point p in the simplex σ; in this case, we say that the barycentric coordinate \(\alpha^{\sigma}_{j}(p)\) of p corresponds to the vertex \(v^{j}\) or, in short, that \(\alpha^{\sigma}_{j}(p)\) is the ith barycentric coordinate of p in σ, \(j\in[n]\). For each \(p\in\sigma\), its vector \(\alpha^{\sigma}(p)\) of the barycentric coordinates in the simplex σ is uniquely determined. If σ is a ksimplex and we do not order its vertices \(V(\sigma)\), then it is sometimes convenient to think that the barycentric coordinates of a point \(p\in\sigma\) in σ are determined according to the unique function \(\alpha ^{\sigma}(p):V(\sigma)\to[0, 1]\), \(V(\sigma)\ni v\mapsto\alpha ^{\sigma}_{v}(p)\in[0, 1]\), with \(\sum_{v\in V(\sigma)}\alpha^{\sigma}_{v}(p)=1\) and \(p=\sum_{v\in V(\sigma)}\alpha^{\sigma}_{v}(p)v\); it is said in this case that the barycentric coordinate \(\alpha^{\sigma}_{v}(p)\) of the point \(p\in\sigma\) corresponds to the vertex \(v\in V(\sigma)\) in the simplex σ. Moreover, \(\alpha^{\sigma}(p)\) is called the (mapping of) barycentric coordinates of p in S.

If σ is a \((k1)\)simplex, then \(\langle {A} \rangle\), where \(\emptyset\neq A\subset V(\sigma)\), is called a \((\#A1)\)face of σ.

If σ is a \((k1)\)simplex and \(A=V(\sigma)\backslash \{v\}\), where \(v\in V(\sigma)\), then the \((k2)\)simplex \(\langle{A} \rangle\) is the \((k2)\)face opposite to the vertex v. Obviously, to each vertex v, there corresponds a unique \((k2)\)face opposite to v.

A collection \(T(S)=\{\sigma_{j}\subset S: j\in[J]\}\), \(J\in \mathbb{N}\), of nonempty subsets of a \((k1)\)simplex \(S\subset \mathbb{R}^{n}\), \(0< k\leq n+1\), is called a triangulation of S if it meets the following conditions:
 1.
\(\sigma_{j}\) is a \((k1)\)simplex, \(j\in[J]\),
 2.
if \(\sigma_{j}\cap\sigma_{j'}\neq\emptyset\) for \(j,j'\in [J]\), then \(\sigma_{j}\cap\sigma_{j'}\) is a common face of \(\sigma _{j}\) and \(\sigma_{j'}\),
 3.
\(S=\bigcup_{j\in[J]}\sigma_{j}\).
 1.

If T is a triangulation of an \((n1)\)simplex S, then for \((n2)\)face F of S, the set \(\{\sigma\cap F: \sigma\cap F \text{ is an }(n2)\text{simplex}, \sigma\in T\}\) is a triangulation of F (see [4], p.27, Theorem 2.3(e)).

Two different \((k1)\)simplices \(\sigma_{j}\), \(\sigma_{j'}\), \(j,j'\in[J]\), \(j\neq j'\), in a triangulation T of a \((k1)\)simplex S are said to be adjacent if \(\langle{V(\sigma)\cap V(\sigma ')} \rangle\) is a \((k2)\)face for both of them. Each \((k2)\)face of a simplex \(\sigma _{j}\), \(j\in[J]\), is a \((k2)\)face for exactly two different simplices in the triangulation, provided that the \((k2)\)face is not contained in \(\partial({S})\).

The \(K_{m}\)triangulation of an \((n1)\)simplex \(S= \langle{v^{1},\ldots,v^{n}} \rangle\subset\mathbb{R}^{n}\) with grid size \(m^{1}\), where m is a positive integer,^{2} is the collection of all \((n1)\)simplices σ of the form \(\sigma= \langle {p^{1}, p^{2},\ldots, p^{n}} \rangle\), where vertices \(p^{1}, p^{2},\ldots, p^{n}\in S\) satisfy the following conditions:
 (1)
the barycentric coordinate \(\alpha^{S}_{v_{i}}(p^{1})\) of \(p^{1}\) corresponding to \(v_{i}\) in S, \(i\in[n]\), is a nonnegative integer multiple of \(m^{1}\),
 (2)
\(\alpha^{S}(p^{j+1})=\alpha^{S}(p^{j})+m^{1}(e^{\pi_{j}}e^{\pi _{j}+1})\), where \(\pi=(\pi_{1},\ldots,\pi_{n1})\) is a permutation of \([n1]\), \(l\in\{j,j+1\}\), \(j\in[n1]\).
 (1)
Lemma 1
Proof
It is obvious that \(\overline{g}\in\Gamma\). Let us assume that \(\overline{g}\in\bigcap_{q\in\mathbb{N}}G^{q}\). Let \(\gamma^{k}\in \Gamma \subset X\), \(k\in\mathbb{N}\). By the BolzanoWeierstrass theorem we may assume that \(\gamma:=\lim_{k\to+\infty} \gamma^{k}\) exists in X. Since \(\gamma^{k}\in\Gamma\), by the definition of Γ we get that, for each k, there exist sequences \(k_{q}\in\mathbb{N}\), \(g^{k_{q}}\in G^{k_{q}}\), \(q\in\mathbb{N}\), such that \(\lim_{q\to+\infty }g^{k_{q}}=\gamma ^{k}\). Hence, for each k, there exists \(s(k)\in\mathbb{N}\) such that \(g^{k_{s(k)}}\gamma^{k}<1/k\), and we may assume that \(k_{s(k)}\nearrow+\infty\) as \(k\to+\infty\). It is now obvious that \(\gamma\in\Gamma\), which proves the compactness of Γ. Suppose that \(A, A'\subset X\) are open sets such that \(\overline {g}\in A\), \(A\cap A'=\emptyset\), and \(\Gamma\subset A\cup A'\). Notice that the connectedness of \(G^{q}\) and the fact that \(\overline{g}\in G^{q}\), \(q\in \mathbb{N}\), imply \(G^{q}\subset A\), \(q\in\mathbb{N}\). To show that Γ is connected, it suffices to demonstrate that \(\Gamma\cap A'=\emptyset\). Let us now fix a point \(x\in\Gamma\) and assume that \(x\in A'\). Since \(A'\) is open, there exists \(\varepsilon>0\) with \(x+B_{\varepsilon}\subset A'\). The last inclusion implies that for any sequence converging to x, almost all its terms belong to \(A'\). From this, the fact that \(G^{q}\subset A\), \(q\in\mathbb{N}\), and the disjointness of A from \(A'\) it follows that \(x\notin\Gamma\). Thus, \(\Gamma\cap A'=\emptyset\).
Now, if \(\bigcap_{q\in\mathbb{N}}G^{q}=\emptyset\), then let for each \(q\in \mathbb{N}\), \(h^{q}\in G^{q}\) be a point such that \(h^{q}\overline {g}=\inf\{ \overline {g}h: h\in G^{q}\}\); such a point exists by the compactness of \(G^{q}\). Define the sets \(H^{q}:=G^{q}\cup\{t\overline{g}+(1t)h^{q}: t\in[0,1]\}\), \(q\in\mathbb{N}\). By the convexity of X, \(H^{q}\subset X\), \(q\in \mathbb{N}\). Moreover, the sets \(H^{q}\), \(q\in\mathbb{N}\), are compact, connected, and the point g̅ belongs to each of them. Thus, the limit set \(\Gamma'\) (defined as Γ, but with \(H^{q}\) in place of \(G^{q}\)) is compact, nonempty, and connected. It suffices to prove that \(\Gamma=\Gamma'\). Obviously, \(\Gamma\subset\Gamma'\). If \(x\in\Gamma'\backslash \Gamma\), then x is the limit of a sequence of points \(g^{k_{q}}\in\{ t\overline{g}+(1t)h^{q}: t\in[0,1)\}\backslash G^{k_{q}}\) for a sequence \(k_{q}\nearrow+\infty\), \(q, k_{q}\in\mathbb{N}\). But \(\lim_{q\to +\infty}h^{q}=\overline{g}\), and thus \(x=\overline{g}\). Consequently, \(x\in \Gamma\), which ends the proof. □
3 The combinatorial lemma and its applications
The result which is common for our proofs of the existence of zeros for excess demand mappings, continuum of zeros for parameterized excess demand mappings, and for a proof of Sperner’s lemma is the following combinatorial Lemma 2, which generalizes the combinatorial lemma presented in [1].^{3}
Lemma 2
 1.
for \(i\in[n1]\): \(\alpha^{S}_{i}(p)=0\Leftrightarrow l(p)\neq i\),
 2.
\(l(p)=0\Leftrightarrow\alpha^{S}_{n}(p)=0\),
 3.
\(l(p)=n\Leftrightarrow\alpha^{S}_{n}(p)=1\),
 4.
\(l(p)\in[n1]\Leftrightarrow\alpha^{S}_{n}(p)\in(0,1)\),
Proof
3.1 The existence of equilibrium
Definition 3
 1.
z is upper semicontinuous on \(\operatorname {int}\Delta^{n}\) with nonempty convex and compact values \(z(p)\), \(p\in\operatorname {int}\Delta^{n}\),
 2.
Walras’ law: \(pz(p)=0\), \(p\in \operatorname{int} \Delta^{n}\),
 3.the boundary condition: if \(p^{q}\in \operatorname{int} \Delta^{n}\), \(y^{q}\in z(p^{q})\), \(q\in\mathbb{N}\), and \(\lim_{q\to +\infty} p^{q}=p\), then$$p_{i}=0\quad \Rightarrow \quad \lim_{q\to+\infty} y^{q}_{i}=+\infty , \quad i\in[n], $$
 4.z is bounded from below: there exists a negative number L such that$$\inf \bigl\{ y_{i}\in\mathbb{R}: y\in z(p), p\in \operatorname{int} \Delta ^{n} \bigr\} >L, \quad i\in[n]. $$
Definition 4
Let \(z:\operatorname{int}\Delta^{n}\multimap \mathbb{R}^{n}\) be an excess demand mapping, \(n\in\mathbb{N}\). A point \(p\in\operatorname {int}\Delta ^{n}\) is called an equilibrium point for z if \(0\in z(p)\).
Lemma 5
 1.there exists \(\varepsilon_{1}\in(0,1/2] \) such that for \(i\in[n]\) and \(y\in z(p)\), \(p\in\operatorname{int}\Delta ^{n}\), we have$$(p_{i}\leq\varepsilon_{1}\Rightarrow y_{i}>0)\quad \textit{and}\quad (p_{i}\geq 1\varepsilon_{1}\Rightarrow y_{i}< 0), $$
 2.
for any \(\varepsilon_{2}\in(0,1/2]\), there exists \(U>0\) such that \(z(p)\subset[L,U]^{n}\) for each \(p\in\operatorname {int}\Delta^{n}\) with \(p_{i}\geq\varepsilon_{2}\), \(i\in[n]\), where L is the constant appearing in Definition 3, condition 4,
 3.for each \(\varepsilon_{3}\in(0,1/2]\), there exists \(\varepsilon_{4}\in(0,\varepsilon_{3}/2]\) such that for \(p\in \operatorname{int}\Delta^{n}\) with \(p_{n}\leq1\varepsilon_{3}\), we have that, for \(i\in[n1]\) and \({y}\in{z}(p)\),$$\bigl(p_{i}\leq\varepsilon_{4}\Rightarrow(1p_{n})y_{i}+p_{n}y_{n}>0 \bigr)\quad \textit{and}\quad \bigl(p_{i}\geq1\varepsilon_{4} \Rightarrow (1p_{n})y_{i}+p_{n}y_{n}< 0 \bigr). $$
 4.for \(\varepsilon_{3}\), \(\varepsilon_{4}\) for which claim 3 and its premises hold, there exists \(\Lambda\in (0,+\infty)\) such that, for \(i\in[n1]\),whenever \(y\in z(p)\), \(p\in\operatorname{int}\Delta^{n}\), \(p_{i}\leq \varepsilon_{4}\), \(p_{n}\leq1\varepsilon_{3}\), and \(p_{j}\in[\varepsilon_{4}/2n, 1\varepsilon_{4}/2n]\), \(j\in[n]\).$$(1p_{n})y_{i}+p_{n}y_{n}>\Lambda, $$
Proof
Suppose that the lefthand side implication in claim 1 is not true. Then there exist \(i\in[n]\) and sequences \(p^{q}\in \operatorname{int}\Delta^{n}\), \(y^{q}\in z(p^{j})\), \(j\in\mathbb{N}\), such that \(\lim_{q\to+\infty}p^{q}=p\), \(p_{i}=0\), and \(\limsup_{q\to+\infty}y_{i}^{q}\leq0\), which is impossible due to the boundary condition. Hence, there exists \(\varepsilon_{1}\in(0,1/2]\) such that the considered implication is satisfied. To prove the righthand side implication in claim 1, observe that \(p_{i}\geq1\varepsilon_{1}\) implies \(p_{i'}\leq \varepsilon_{1}\), \(i\neq i'\), and \(y_{i'}>0\), \(i'\neq i\), for \(y\in z(p)\). Finally, by Walras’ law, \(0<\sum_{i'\neq i}p_{i'}y_{i'}=p_{i}y_{i}\), and consequently \(y_{i}<0\).
Statement 2 is true since the restriction of the mapping z to the (compact) set \(\{p\in\operatorname{int}\Delta^{n}: p_{i}\geq \varepsilon _{2}, i\in[n]\}\) is an upper semicontinuous mapping with compact values, and such mappings transform compact sets into compact sets [6], p.560.^{6}
To prove assertion 3, suppose that there exists \(\varepsilon_{3}\in(0,1/2]\) such that for any \(q\in\mathbb{N}\), \(k\geq2\), there exist \(p^{q}\in\operatorname{int}\Delta^{n}\): \(p^{q}_{n}\leq 1\varepsilon_{3}\) and \(i_{q}\in[n1]\): \(p^{q}_{i_{q}}\leq\varepsilon_{3}/q\) with \((1p_{n}^{q})y^{q}_{i_{q}}+p^{q}_{n}y^{q}_{n}\leq0\) for some \(y^{q}\in z(p^{q})\). Without loss of generality, we assume that \(i_{q}=1\), \(q\in\mathbb{N}\). The boundary condition now implies that \(\lim_{q\to+\infty}y^{q}_{1}=+\infty\). Since \(1p^{q}_{n}\geq\varepsilon_{3}\), \(q\in\mathbb{N}\), and z is bounded from below by the constant L, we obtain that \((1p^{q}_{n})y^{k}_{1}+p^{q}_{n}y^{q}_{n}\geq\varepsilon_{3}y^{q}_{1}+L>0\) for large q. This contradicts our assumption that \((1p_{n}^{q})y^{q}_{i_{q}}+p^{q}_{n}y^{q}_{n}\leq 0\) for \(q\in\mathbb{N}\). Hence, for any \(\varepsilon_{3}\in(0,1/2]\), there exists \(\varepsilon_{4}\in(0,\varepsilon_{3}/2]\) such that the first implication in claim 3 is satisfied for any \(p\in\Delta ^{n}\): \(p_{n}\leq1\varepsilon_{3}\). Observe that for fixed \(\varepsilon _{3}\) and \(\varepsilon_{4}\) for which the first implication in claim 3 holds, it follows that if \(p\in\operatorname{int}\Delta ^{n}\): \(p_{n}\leq 1\varepsilon_{3}\) and \(p_{i}\geq1\varepsilon_{4}\) for some \(i\in[n1]\), then \(p_{j}\leq\varepsilon_{4}\), \(j\in[n1]\): \(j\neq i\), and by the first implication of assertion 3 we get \((1p_{n})y_{i}+p_{n}y_{n}=\sum_{j\in[n1]\backslash\{i\}}[(1p_{n})y_{j}+p_{n}y_{n}]<0\).
Let now \(\varepsilon_{3}\), \(\varepsilon_{4}\) be as in claim 3 and suppose that claim 4 is false. Thus, there exist sequences \(p^{q}\in\operatorname{int}\Delta^{n}\): \(p^{q}_{n}\leq 1\varepsilon_{3}\), \(p^{q}_{j}\in[\varepsilon_{4}/2n, 1\varepsilon_{4}/2n]\), \(j\in[n]\), \(y^{q}\in z(p^{q})\), \(q\in\mathbb{N}\), and \(\overline{i}\in[n1]\) such that \(p^{q}_{\overline {i}}\leq\varepsilon_{4}\) and \((1p^{q}_{n})y^{q}_{\overline {i}}+p^{q}_{n}y^{q}_{n}\leq 1/q\), \(q\in\mathbb{N}\). By the boundary and lower boundedness conditions on z, the boundedness of the standard simplex, and by the upper semicontinuity of z and compactness of its values, from the sequences \(p^{q}, y^{q}, q\in\mathbb{N}\), we can extract subsequences converging to \(p\in\operatorname{int}\Delta^{n}\) and \(y\in z(p)\), respectively. But then \(p_{i}\in [\varepsilon_{4}/2n, 1\varepsilon_{4}/2n]\), \(i\in[n]\), \(p_{n}\leq 1\varepsilon_{3}\), and \(p_{\overline{i}}\leq\varepsilon_{4}\). By the contradictory assumption we get \((1p_{n})y_{\overline{i}}+p_{n}y_{n}\leq0\), which is impossible due to the choice of \(\varepsilon_{3}\) and \(\varepsilon_{4}\). □
We are in position to prove the first consequence of the combinatorial Lemma 2.
Lemma 6
Let z be an excess demand mapping. For each \(\varepsilon>0\), there exist \(p\in\operatorname{int}\Delta^{n}\) and \(y\in z(p)\) such that \(y_{i}\leq \varepsilon\), \(i\in[n]\).
Proof
Fix \(\varepsilon>0\). The claim is trivial for \(n=1\), so assume that \(n\geq2\). To ease the reading, we divide the proof into four parts.
Part 1: A restriction of the mapping z to a simplex \(S\subset\operatorname{int}\Delta^{n}\). Let us fix \(\varepsilon_{1}\) for which the assertion of Lemma 5, statement 1 is satisfied. Let \(\varepsilon_{4}\) correspond to \(\varepsilon_{3}:=\varepsilon_{1}/2\) according to Lemma 5, statement 3. Finally, suppose that U fulfills Lemma 5, statement 2 for \(\varepsilon _{2}=\varepsilon_{4}\) and Λ corresponds to \(\varepsilon _{3}\) (\(=\varepsilon_{1}/2\)) and \(\varepsilon_{4}\) as in Lemma 5, statement 4. Notice that \(1\varepsilon_{1}<1\frac {1}{2}{\varepsilon_{1}}= 1\varepsilon_{3}<1\varepsilon_{4}<1\frac {n1}{2n}\varepsilon_{4}\leq1\frac{1}{2n}\varepsilon_{4}\).
From Lemma 6 and its proof we obtain the following.
Theorem 7
Let z be as in Lemma 6. There exists an equilibrium point for z.
Proof
Observe that the simplex S constructed in Part 1 of the proof is independent of \(\varepsilon>0\). Hence, there are points \(p^{q}\in S\) and \(y^{q}\in z(p^{q})\), satisfying \(y^{q}_{i}\leq1/q\), \(i\in[n]\), \(q\in\mathbb{N}\). The compactness of S allows us to assume that the sequence \(p^{q}\) converges to a point \(p\in S\). From the upper semicontinuity of the mapping z and from the compactness of its values we may also assume that the corresponding sequence of points \(y^{q}\in z(p^{q})\) converges to a point \(y\in z(p)\) with \(y_{i}\leq0\), \(i\in[n]\). Since \(p\in S\subset \operatorname{int}\Delta^{n}\), \(y=0\) (by Walras’ law). □
3.2 A continuum of equilibria for parameterized excess demand mappings
The main result of this section is a version of Browder fixed point theorem for excess demand mappings.
Theorem 8
 1.
\(pz(p,t)=0\), \((p,t)\in\operatorname{int}\Delta ^{n1}\times[0,1]\),
 2.if \((p^{q},t^{q})\in\operatorname{int}\Delta ^{n1}\times [0,1]\), \(y^{q}\in z(p^{q},t^{q})\), \(q\in\mathbb{N}\), \(\lim_{q\to +\infty}(p^{q},t^{q})=(p, t)\in\Delta^{n1}\times[0,1]\), then$$p_{i}=0\quad \Rightarrow \quad \lim_{q+\infty}y^{q}_{i}=+ \infty,\quad i\in[n1], $$
 3.there exists a negative number L such that$$\inf \bigl\{ y_{i}\in\mathbb{R}: y\in z(p,t), (p,t)\in \operatorname{int}\Delta ^{n1}\times[0, 1] \bigr\} >L, \quad i\in[n1]. $$
Before we present a proof of Theorem 8, let us remark that for any \(t\in[0, 1]\), the mapping \(z(\cdot,t)\) is an excess demand mapping in the sense of Definition 3. Theorem 8, assumptions 2 and 3 (and the assumption of upper semicontinuity) impose some uniformity conditions on the family of mappings \(\{z(\cdot,t): t\in[0, 1]\}\), and we suppose that the claim of Theorem 8 may not be valid for a nonempty, convex, and compactvalued upper semicontinuous mapping \(z:\operatorname {int}\Delta ^{n1}\times[0,1]\) satisfying Theorem 8, assumption 1 and such that each mapping \(z(\cdot,t)\) is an excess demand mapping, \(t\in[0,1]\), but either Theorem 8, assumption 2 or Theorem 8, assumption 3 is not satisfied. However, we were not able to construct an example of such mapping nor to deliver a proof of Theorem 8 without introducing the just mentioned conditions.
Proof of Theorem 8
 (i)
for each \(\varepsilon_{2}\in(0,1/12)\), there exists \(U>0\) such that \(\xi(p)\subset[L,U]^{n}\) for each \(p\in \operatorname{int} \Delta^{n}\) with \(p_{i}\geq\varepsilon_{2}\), \(i\in[n]\),
 (ii)for each \(\varepsilon_{3}\in(0,1/12)\), there exists \(\varepsilon_{4}\in(0,\varepsilon_{3}/2]\) such that for \(p\in \operatorname{int}\Delta^{n}\) with \(p_{n}\leq1\varepsilon_{3}\), we have that, for \(i\in[n1]\) and \({y}\in\xi(p)\),$$\bigl(p_{i}\leq\varepsilon_{4}\Rightarrow(1p_{n})y_{i}+p_{n}y_{n}>0 \bigr)\quad \textit{and}\quad \bigl(p_{i}\geq1\varepsilon_{4} \Rightarrow (1p_{n})y_{i}+p_{n}y_{n}< 0 \bigr), $$
 (iii)for \(\varepsilon_{3}\), \(\varepsilon_{4}\) for which claim (ii) and its premises hold, there exists \(\Lambda\in(0,+\infty)\) such that, for \(i\in[n1]\),whenever \(y\in\xi(p)\), \(p\in\operatorname{int}\Delta^{n}\), \(p_{i}\leq \varepsilon _{4}\), \(p_{n}\leq1\varepsilon_{3}\), and \(p_{j}\in[\varepsilon_{4}/2n, 1\varepsilon_{4}/2n]\), \(j\in[n]\).$$(1p_{n})y_{i}+p_{n}y_{n}>\Lambda, $$

we replace z with ξ, and

we keep in mind that \(y_{n}=0\) whenever y belongs to the image of ξ, so that, in consequence, \(h_{n}(p)=0\), \(p\in S\),
 (iv)
\(g([0,1])\subset\{p\in S: p_{n}\in[1/3, 2/3]\}\),
 (v)
\(g_{n}(0)=2/3\), \(g_{n}(1)=1/3\),
 (vi)
for each \(t\in[0, 1]\), \(h_{i}(g(t))\leq\varepsilon /2\), \(i\in[n1]\), \(h_{n}(g(t))=0\),
 (vii)
for each \(t\in[0, 1]\), there are \(p\in S\) and \(y\in \xi(p)\) with \(pg(t)<\delta\), \(yh_{i}(g(t))<\varepsilon/2\), \(i\in [n1]\), and thus \(y_{i}\leq\varepsilon\), \(i\in[n1]\).
3.3 Sperner’s lemma
The next lemma is a bit technical, but the geometry behind it is rather intuitive. Namely, the lemma reveals that it is possible to decompose the polytope \(\operatorname{conv}\{v^{1},\ldots,v^{n},w^{1},\ldots ,w^{n}\}\) into the simplices \(S^{1},\ldots, S^{n1}\), and the intersection of any pair of those simplices is their common face.^{9}
Lemma 9
Remark 10
Similar decomposition applies for the set \(\operatorname{conv}\{ v^{1},\ldots ,v^{n1},e^{1},\ldots,e^{n1}\}\). The idea for a proof is identical to that presented in the proof above (only some minor changes are necessary).
The situation presented further in Lemma 11 can be imagined easily.
Lemma 11
Fix \(n\geq3\), \(n\in\mathbb{N}\). Let T be a triangulation of an \((n2)\)simplex \(S= \langle{p^{1},\ldots,p^{n1}} \rangle \subset\mathbb{R}^{n}\), and let \(S'= \langle{p^{1},\ldots,p^{n1},r} \rangle\) be an \((n1)\)simplex. Then the collection \(T':=\{ \langle{V(\sigma)\cup\{r\}} \rangle: \sigma \in T\}\) is a triangulation of \(S'\). Moreover, each simplex \(\sigma\in T\) and each simplex \(\sigma'\) of the form \(\langle{V(F)\cup\{r\} } \rangle\), where F is an \((i1)\)face of a simplex in T, \(i\in[n2]\), belong to the set of all ifaces of simplices in \(T'\), \(i\in[n2]\).
We shall now show that there is a special triangulation of the standard closed simplex \(\Delta^{n}\). The special triangulation allows us to embed a triangulated \((n2)\)simplex S in the standard simplex \(\Delta^{n}\) keeping simplices from the triangulation of S as faces of simplices in the triangulation of \(\Delta^{n}\). This lemma allows us to apply our combinatorial Lemma 2 to detect a simplex satisfying the assertion of Sperner’s lemma.
Lemma 12
Theorem 13
(Sperner’s lemma)
Fix \(n\geq3\), \(n\in\mathbb{N}\). Suppose that T is a triangulation of an \((n2)\)simplex \(S= \langle{p^{1},\ldots,p^{n1}} \rangle\). Let \(l:V(T)\to[n]\) be a function such that \(l(v)\neq i\) whenever \(\alpha_{i}(v)=0\), where \(\alpha_{i}(v):=\alpha^{S}_{p^{i}}(v)\) is the ith barycentric coordinate of the vector v in S. There exists \(\sigma\in T\) with \(l(\sigma )=[n1]\). Moreover, the number of such simplices is odd.
Proof
4 Final remarks
4.1 An algorithm for finding a zero of an excess demand mapping
 Step 0::

Fix accuracy level \(\varepsilon>0\). Determine \(\varepsilon_{1}\), \(\varepsilon_{3}\), \(\varepsilon_{4}\), Λ and vertices \(\overline{v}^{i}\), \(i\in[n]\), and simplex \(S:= \langle{\overline{v}^{i}: i\in[n]} \rangle\) as in Part 1 of the proof of Lemma 6 . Fix numbers δ, λ, \(m_{1}\), \(m_{2}\) as in Part 2 of the proof. Let also \(T:=K_{m_{1}\times m_{2}}(S)\). To each vertex \(v\in V(K_{m_{1}}(S))\), arbitrarily assign \(y(w)\in z(v)\). Further, to each \(v\in V(T)\), assign the point \(h(v)\) according to formula ( 3 ) and label each \(v\in V(T)\) as \(l(v)\in[n]_{0}\), where labeling l is defined by ( 7 ). Let also \(\sigma_{1}\) be the only simplex in T with \(\overline{v}^{n}\in\sigma_{1}\). Put \(F:=V(\sigma_{1})\backslash \{\overline {v}^{n}\}\), \(\overline{v}:=\overline{v}^{n}\), and go to Step 1.
 Step 1::

Determine the only vertex \(v\in T\) such that \(v\neq\overline{v}\) and \(\langle{F\cup\{v\}} \rangle \in T\). Go to Step 2.
 Step 2::

If \(h_{n}(v)\geq0\), then STOP: there is a point \(p\in(v+B_{\delta})\cap\operatorname{int}\Delta^{n}\) and \(y\in z(p)\) with \(y_{i}\leq\varepsilon\), \(i\in[n]\). Otherwise, assign the only element of \(l^{1}(l(v))\cap F\) as the value of v̅. Set \(F:=(F\backslash \{\overline{v}\})\cup\{v\}\) and go to Step 1.
4.2 Kakutani fixed point theorem and GaleDebreuNikaido lemma
Now, if \(F:\Delta^{n}\multimap\mathbb{R}^{n}\) is an upper semicontinuous nonempty convex and compactvalued mapping satisfying a weak version of Walras’ law \(pF(p)\leq0\), \(p\in\Delta^{n}\), then the GaleDebreuNikaido lemma asserts that there are \(p\in\Delta^{n} \) and \(y\in F(p)\) with \(y\leq0\) [7], p.81. Going along the lines above for Kakutani’s fixed point theorem, we obtain points \(p\in\Delta^{n}\), \(y\in F(p)\) with \(y=\frac{p y}{p p}p\). By the weak Walras law we get \(py\leq0\), hence, due to the inequality \(p\geq 0\), we have \(y\leq0\).
4.3 Browder theorem
Let us notice that in a recent work [8], there was also proved a generalization of Browder’s theorem. However, it seems that our less general approach is simpler than that presented in [8].
4.4 A bit on economics
Our findings in [1] were well motivated by economics. However, it appears that our results have natural origins in economics. Indeed, Theorem 7 allows us to state that there exists an equilibrium for a pure exchange economy (where agents’ excess demands are multivalued mappings), whereas Theorem 8 ensures the existence of equilibrium in an exchange economy with price rigidities ([3], Chapter 2 or [8]). Even Sperner’s lemma has an interesting economic/social implication; it enables us to deduce that there exists a fair division of a good (see, e.g., a nice introduction in [9]).
In the last section of the paper, we present an algorithm for finding an (approximate) zero of an excess demand mapping; the other algorithms may be derived from proofs of Theorems 8 and 13, but we leave the details for the reader. A comprehensive review of existing algorithms is presented in [3]. We would like to stress that we just show some novel ways leading to the existing/known results. These ways have a common factor, Lemma 2.
The generalization consists in allowing for a wider class of triangulations of the simplex; the only triangulation considered in [1] was our \(K_{m}\)triangulation.
To simplify the notation, \(l(\sigma):=l(V(\sigma ))\), \(\sigma\in T\), or σ is a face of a simplex in T. The proof below is an adaptation of the proof of Lemma 1 in [1].
The method of construction of the sequence is similar to that used in the proof of the correctness of the Scarf algorithm; see [3], p.68.
Let us recall that \(\alpha^{S}(p)\) denotes the vector of the barycentric coordinates of \(p\in S\) in the simplex S.
So, from the formal point of view, the family of simplices \(S^{1},\ldots,S^{n1}\) is a triangulation the polytope \(\operatorname{conv}\{v^{1},\ldots ,v^{n},w^{1},\ldots ,w^{n}\}\). For a definition of triangulation of a polytope, see, e.g., Definition 1.4.3 in [3].
We describe this procedure in short: we ‘start’ from a simplex in \(T'\) whose face is σ and which has a vertex contained in \(\{e^{1},\ldots,e^{n1}\}\); call this simplex \(\sigma'_{1}\), then choose \(\sigma'_{2}\) to be the simplex adjacent to \(\sigma'_{1}\) that shares the face σ with \(\sigma '_{1}\). Next, we use the same rule for rejection of a vertex to get the next adjacent simplex of the sequence as in the proof of Lemma 2. At each step, \([n1]\subset l(\sigma'_{i})\). We can continue the procedure until we meet the first simplex \(\sigma'_{J'} \in T'\) that possesses a vertex in \(\{e^{1},\ldots,e^{n1}\}\). The constructed sequence \(\sigma'_{j}\), \(j\in[J']\), is unique and has no simplex common with the sequence \(\sigma_{j}\), \(j\in[J]\). Moreover, if we start from \(\sigma'_{J'}\), then the procedure leads us back to \(\sigma'_{1}\). See the proof of Lemma 2 for details.
Declarations
Acknowledgements
I would like to thank participants of Nonlinear Analysis Seminar at Faculty of Mathematics and Computer Science (Adam Mickiewicz University in Poznań), Seminar of the Game and Decision Theory at Institute of Computer Science (Polish Academy of Science, Warsaw) for comments and criticism. This work was financially supported by the Polish National Science Centre, decision no. DEC2013/09/B/HS4/01506.
Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.
Authors’ Affiliations
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