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CS 573: Algorithmic Game Theory Lecture date: Feb 1 st, 2008 Instructor: Chandra Chekuri Scribe: Ramakrishna Gummadi
1 Introduction
A proof of Brouwer’s fixed point theorem using Sperner’s Lemma is discussed below. To recap from a previous lecture, Brouwer’s theorem claims that any continuous function that maps any set that is homeomorphic to [0, 1]d^ (e.g. a closed disc) to itself has a fixed point. We will first prove this generalization and then restrict attention to [0, 1]d.
Theorem 1.1 (Brouwer’s Fixed Point Theorem) Let f : [0, 1]d^7 → [0, 1]d^ be a continuous function. Then there exists a point x ∈ [0, 1]d^ s.t. f (x) = x. More generally, this theorem holds for f : X 7 → X, where X is any set homeomorphic to [0, 1]d.
Definition 1.2 (Homeomorphism) A function g : X 7 → Y where X and Y are two topological spaces is a homeomorphism if the following three conditions hold:
- g is continuous
- g is bijective
- g−^1 is continuous
Note that if g : X 7 → Y is a homeomorphism, g−^1 : Y 7 → X is also a homeomorphism. If there exists such a g mapping two spaces X and Y , we say that X and Y are homeomorphic.
Lemma 1.3 Let X be homeomorphic to [0, 1]d. Then, the fixed point theorem holds for X if it holds for [0, 1]d. Proof: Let f : X 7 → X be the continuous function being examined. Let g be a homeomorphic function from X to [0, 1]d. Define f ′^ : [0, 1]d^7 → [0, 1]d^ as f ′(a) = g(f (g−^1 (a))). Since f ′^ is continuous, there exists a∗^ ∈ [0, 1]d^ with f ′(a∗) = a∗. Let x∗^ = g−^1 (a∗). Then, g(f (x∗)) = g(f (g−^1 (a∗))) = f ′(a∗) = a∗^ = g(x∗). Since, g is bijective, this implies f (x∗) = x∗. 2
Remark 1.4 It is also true that any two convex compact full dimensional bodies in Rd^ are home- omorphic to each other.
2 Sperner’s Lemma
We will now prove a combinatorial result to be used for a proof of the fixed point theorem:
Definition 2.1 (Triangulation) A triangulation, T of a triangle T is a set of triangles that exactly covers T and mutually intersect only along their edges. Further, let V (T ) denote the union of the vertices of the triangles covering T. (See Figure 1 for an illustration)
v 0
v 2
(^1 1) V 1
1
(^02)
2 2
(^0 )
2
Figure 1: This figure illustrates the triangulation, T and an example of a valid coloring on V (T ). The numbers denote the colors of the vertices. Note that any vertex on the edge vi, vj has a color of either i or j. Further, node vi has color i.
Lemma 2.2 (Sperner’s Lemma) Consider a triangle T with vertices v 0 , v 1 , v 2. Let T be a tri- angulation of T and V (T ) denote its set of vertices. Consider any coloring of V (T ) with { 0 , 1 , 2 } such that:
- vi is colored i. (i ∈ { 0 , 1 , 2 })
- If v ∈ V (T ) lies on the line joining vi and vj for i, j ∈ { 0 , 1 , 2 } the color of v is either i or j.
Then:
- There exists at least one triangle in T whose vertices have all three colors - call such a triangle a multicolored triangle.
- Further, the total number of multicolored triangles in T is odd. Proof: Construct a graph G from T defined as follows (Refer to Figure 2 for an illustration):
- G has a vertex at for each triangle t ∈ T.
- It has an additional vertex a 0 (corresponds to the outer face).
- Vertices at and at′^ share an edge iff t and t′^ share an edge that is colored 0 and 1 on its two incident vertices.
- a 0 is connected to all triangles that have an edge along the line joining v 0 and v 1 with incident vertex colors 0, 1.
First we note that a 0 has an odd degree. To see this, let α ( β) denote the number of 0 − 1 (1 − 0) color transitions along the line joining v 0 to v 1. Clearly, α − β = 1. Hence, the degree(a 0 ) = α + β = 2α − 1, an odd number.^1 Since the total number of vertices of odd degree in a graph is
(^1) Note that this part of the argument is essentially Sperner’s Lemma for the case d = 1. That is, it says that, if we consider two points and partition the line joining them into multiple segments, and color the points thus obtained with the constraint that the end vertices are 0 and 1, then the total number of multicolored segments along the line will be odd.
Now consider the sequence < v 00 , v^10 , v^20 ,... >. Since T is compact, there is a convergent subse- quence < vi 01 , vi 02 ,... > that converges to some point x∗^ ∈ T. Note that since the size of the triangles decreases geometrically, this also implies that the sequences < vi 11 , vi 12 ,... > and < vi 21 , vi 22 ,... > both converge to x∗. Now, by the definition of the coloring, we have f (vi 0 k [0]) < vi 0 k [0] ∀k. Therefore, by the continuity of f , we conclude by taking the limit that f (x∗)[0] ≤ x∗[0]. Similarly, we have for each i ∈ { 0 , 1 , 2 }, f (x∗)[i] ≤ x∗[i]. But since, f (x∗)[0] + f (x∗)[1] + f (x∗)[2] = 1 = x∗[0] + x∗[1] + x∗[2], the inequalities must in fact be equalities. ie. f (x∗) = x∗^ , leading to a contradiction. Hence, f must have a fixed point. 2
