BROUWER’S FIXED POINT THEOREM AND THE NASH

THEOREM

ERIC KARSTEN

Abstract. This paper finds the fundamental groups of D2 and S1 and then

uses these to prove Brouwer’s Fixed point theorem in two dimensions. Then,

assuming Brouwer’s fixed point theorem in n dimensions, we proceed to provethe Nash theorem for n-player non-cooperative games where each player has a

finite set of pure strategies.

Contents

1. Introduction 12. Algebraic Topology Preliminaries 23. Finding The Fundamental Group for D2 and S1 54. Brouwer’s Fixed Point Theorem using the fundamental group. 95. Game Theory Preliminaries 126. Nash Theorem 14Acknowledgments 16References 16

1. Introduction

Imagine you are playing a game where you and some other players each guess aninteger between 1 and 100 with the aim of guessing the integer which is closest totwo thirds of the mean of all of your guesses. An observant player will notice thatif all players choose 1, then they will all be winners, meaning that no player wouldhave any reason to stop playing 1 given that the other players were playing 1. Thismeans that every player choosing 1 is what game theorists call a Nash equilibriumof this number guessing game.

In 1951, John Nash, Jr. proved the remarkable result that every game with afinite number of players and a finite number of pure strategies has a mixed strategyNash equilibrium. This paper is a self-contained proof from first principles of thesame result. Nash’s original proof used the Kakutani fixed point theorem, howeverthis paper will prove the result using Brouwer’s fixed point theorem. Section 2 givesan overview of the algebraic topology necessary for the proof of Brouwer’s theoremin sections 3 and 4. We then provide a set of formal Game Theory definitions insection 5 and prove the Nash theorem in section 6.

Date: September 24, 2017.

1

2 ERIC KARSTEN

2. Algebraic Topology Preliminaries

Definition 2.1. A loop in a space X is a continuous function f : [0, 1]→ X suchthat f(0) = f(1).

Definition 2.2. Two loops g and h are homotopic if there exists a continuousfunction F : [0, 1] × [0, 1] → X such that F (s, 0) = g(s) and F (s, 1) = h(s). Thenotation for this is g ' h.

We now wish to show that for loops, ' is an equivalence relation.

Proposition 2.3. Homotopy is an equivalence relation.

Proof: We must show reflexivity, symmetry, and transitivity of '.

Reflexivity: Let f : [0, 1] → X be a loop. The constant homotopy F defined byF (s, t) = f(s) is continuous, so f ' f and ' is reflexive.

Symmetry: Now let f, g be loops and let f ' g. It follows that there exists a homo-topy F such that F (s, 0) = f and F (s, 1) = g. Now consider G : [0, 1]× [0, 1]→ Xdefined by G(s, t) = F (s, 1− t). It is clear that G inherits the continuity of F andthat G(s, 0) = g and G(s, 1) = f , so G is a homotopy and g ' f .

Transitivity: Now suppose that f, g, and h are loops and that f ' g and g ' h. Itfollows that there exists some homotopy F such that F (s, 0) = f(s) and F (s, 1) =g(s) and some homotopy G such that G(s, 0) = g(s) and G(s, 1) = h(s). Nowdefine H : [0, 1]× [0, 1]→ X by

(2.4) H(s, t) =

{F (s, 2t), for 0 ≤ t ≤ 1

2

G(s, 2t− 1), for 12 < t ≤ 1

.

The function H is clearly continuous on [0, 12 ) and ( 12 , 1]. Because F (s, 2 1

2 ) = g(s)

and G(s, 2 12 − 1) = g(s), it follows that H is continuous everywhere. Additionally,

H(s, 0) = F (s, 0) = f(s) and H(s, 1) = G(s, 1) = h(s), so H is a homotopy andf ' h. Because ' is reflexive, symmetric, and transitive, it is an equivalence rela-tion. �

Because homotopy is an equivalence relation, we may define equivalence classesof homotopic loops.

Definition 2.5. Let f be a loop. Define an equivalence class of loops to be

(2.6) [f ] = {g : [0, 1]→ X | g ' f}

Definition 2.7. The base point of a loop f : [0, 1] → X is the point x0 ∈ X suchthat x0 = f(0) = f(1).

Definition 2.8. We will next define the fundamental group with base point x0 by

(2.9) π1(X,x0) = {[f ] | f : [0, 1]→ X, f(0) = f(1) = x0, and f is continous}

We now wish to show that the fundamental group, as defined above, is indeed agroup. To do so, we must define an operation:

BROUWER’S FIXED POINT THEOREM AND THE NASH THEOREM 3

Definition 2.10. We will define the product path, an operation on loops with thesame base point by

(2.11) f • g(s) =

{f(2s), for 0 ≤ s ≤ 1

2

g(2s− 1), for 12 < s ≤ 1

Definition 2.12. We will define the product of [f ], [g] ∈ π1(X,x0) by [f ][g] = [f•g].

Proposition 2.13. The above product is well-defined.

Proof: If we consider any two loops: f and g with the same base point, thenf(1) = g(0), so the product f • g is a continuous function and is therefore a loopbecause

(2.14) f • g(0) = f(0) = g(1) = f • g(1).

Hence it is sensible to define a homotopy equivalence class [f • g].

Now we must show that if f ′ ∈ [f ], and g′ ∈ [g], then f ′ • g′ ∈ [f • g]. We knowthat f ′ ' f and g′ ' g, so there exist homotopies F and G from f ′ to f and g′ tog respectively. We will then define homotopy H by

(2.15) H(s, t) =

{F (2s, t), for 0 ≤ s ≤ 1

2

G(2s− 1, t), for 12 < s ≤ 1

Because F (1, t) = x0 and G(0, t) = x0 for all t ∈ [0, 1], it follows that H is con-tinuous and therefore is a homotopy. Additionally, it is clear from our definitionthat H(s, 0) = f ′ • g′ and H(s, 1) = f • g, so f ′ • g′ ∈ [f • g], and the product iswell-defined. �

Definition 2.16. The constant loop at x0 ∈ X is the function cx0: [0, 1] → X

defined for all s ∈ [0, 1] by cx0(s) = x0.

Theorem 2.17. π1(X,x0) is a group with operation [f ][g] = [f • g].

Proof: Because we have already verified that the product is well-defined for all[f ], [g], we must only verify that the product is associative, has an identity, andthat every element has an inverse.

Associativity: Applying our definition of the product,

(2.18) (f • g) • h =

f(4s), for 0 ≤ s ≤ 1

4

g(4s− 1), for 14 < s ≤ 1

2

h(2s− 1), for 12 < s ≤ 1

and

(2.19) f • (g • h) =

f(2s), for 0 ≤ s ≤ 1

2

g(4s− 2), for 12 < s ≤ 3

4

h(4s− 3), for 34 < s ≤ 1

We can define the function

(2.20) F (s, t) =

f( 4s

1+t ), for 0 ≤ s ≤ 1+t4

g(4s− 1− t), for 1+t4 < s ≤ t+2

4

h( 4s−t−22−t ), for t+2

4 < s ≤ 1

4 ERIC KARSTEN

which is continuous because

(2.21)

lims→ 1+t

4

−F (s, t) = f(

4

1 + t

1 + t

4)

= f(1)

= g(0)

= g(41 + t

4− 1− t)

= lims→ 1+t

4

+F (s, t)

(2.22)

lims→ 2+t

4

−F (s, t) = g(4

t+ 2

4− 1− t)

= g(1)

= h(0)

= h(4 t+2

4 − t− 2

2− t)

= lims→ 2+t

4

+F (s, t)

and remains in ((f • g) • h)([0, 1]) for all t. Additionally,

(2.23) F (s, 0) = (f • g) • h

(2.24) F (s, 1) = f • (g • h),

so F is a homotopy and

(2.25) (f • g) • h ' f • (g • h).

Hence, ([f ][g])[h] = [f ]([g][h]).

Identity: For all loops f with base-point x0, we can define the function

(2.26) G(s, t) =

{f( 2s

1+t ), for 0 ≤ s ≤ 1+t2

x0, for 1+t2 ≤ s ≤ 1

which is continuous because

lims→ 1+t

2

−G(s, t) = f(

2

1 + t

1 + t

2) = f(1) = x0 = lim

s→ 1+t2

+G(s, t)(2.27)

and remains in f([0, 1]) for all t. Additionally,

(2.28) G(s, 0) = f • cx0(s)

(2.29) G(s, 1) = f(s),

so G is a homotopy and f • cx0' f and similarly cx0

• f ' f , so [cx0] is a two-sided

identity.

Inverse: Now let f be a loop. We will define g(s) = f(1− s). It follows that

(2.30) f • g =

{f(2s), for 0 ≤ s ≤ 1

2

f(1− (2s− 1)) = f(2− 2s), for 12 < s ≤ 1

BROUWER’S FIXED POINT THEOREM AND THE NASH THEOREM 5

If we then define

(2.31) H(s, t) =

f(2s), for 0 ≤ s ≤ 1−t

2

f(1− t), for 1−t2 ≤ s ≤

1+t2

f(2− 2s), for t+12 ≤ s ≤ 1

which is continuous because

lims→ 1−t

2

−H(s, t) = f(2

1− t2

) = f(1− t) = lims→ 1−t

2

+H(s, t)(2.32)

lims→ 1+t

2

−H(s, t) = f(1− t) = f(2− 2

t+ 1

2) = lim

s→ 1+t2

+H(s, t)(2.33)

and remains in f • g([0, 1]) for all t. Additionally,

(2.34) H(s, 0) = f • g(s)

(2.35) H(s, 1) = f(0) = cf(0)(s),

so H is a homotopy and f • g ' cf(0), so [f ]−1 = [g].

We have therefore verified all of the group axioms and may conclude that thefundamental group is a group. �

3. Finding The Fundamental Group for D2 and S1

Remark 3.1. In order to prove Brouwer’s fixed point theorem, we will find thefundamental groups of S1 and D2 to be different from each other in Theorems3.2 and 3.28 respectively. We will then show that if a fixed point doesn’t exist,then we can construct an isomorphism from the fundamental group of S1 to thefundamental group of D2, which is a contradiction.

Theorem 3.2. π1(D2, x0) = {[cx0]}

Proof: Let f be a loop in D2. Consider the function F : [0, 1] × [0, 1] → D2

defined by

(3.3) F (s, t) = (1− t)f(s) + tcx0(s).

Because f(s) ∈ D2 and cx0(s) = x0 ∈ D2 for all s ∈ [0, 1], it follows from the

fact that D2 is convex that F (s, t) ∈ D2 for all t ∈ [0, 1]. Hence F is well-defined.It is also clearly continuous and F (s, 0) = f(s) while F (s, 1) = cx0

(s), so F is ahomotopy and f ' cx0 . Hence f ∈ [cx0 ] for all loops f , so π1(D2, x0) = {[cx0 ]}. �

Definition 3.4. Given a space X, a covering space is a space X equipped with asurjective map p : X → X such that for all x ∈ X, there exists an open U ⊂ Xsuch that x ∈ U and p−1(U) is a union of disjoint open sets such that if V is in thisdisjoint union, then p|V : V → U is a homeomorphism.

Lemma 3.5. R with p(s) = (cos(2πs), sin(2πs)) forms a covering space of S1.

Proof: Let p : R→ S1 be defined by p(s) = (cos(2πs), sin(2πs)). If x ∈ S1, thenthere exists s ∈ R such that p(s+ k) = x for all k ∈ Z. Let U = p((s− 1

4 , s+ 14 )).

6 ERIC KARSTEN

U is open because p is continuous and on intervals shorter than a full arc, beingopen in S1 and being open in R is the same under p. It follows that

(3.6) p−1(U) =⋃k∈Z

(s+ k − 1

4, s+ k +

1

4)

which is a disjoint union of open sets. Now fix k′ ∈ Z and consider

(3.7) p = p|(s+k′− 14 ,s+k′+ 1

4 )

p clearly inherits continuity from p, and is an injection because its domain has beenrestricted to just one of the disjoint sets which make up the preimage. We see thatp is a surjection because p is defined by sinusoidal functions with periods of 1 andk ∈ Z, so

(3.8) U = p((s− 1

4, s+

1

4)) = p((s+ k′ − 1

4, s+ k′ +

1

4))

Hence p is a homeomorphism, and R equipped with p is a covering space of S1. �

Remark 3.9. For the remainder of this section, x0 = p(0) = (1, 0).

Lemma 3.10. If R is a set of open intervals covering a closed interval [a, b] ⊂ R,then there exists a sequence of points a = p0 < p1 < ... < pn = b and some coverR′ = {U1, ..., Un} ⊂ R such that pi ∈ Uj if and only if i = j or i = j − 1.

Proof: Because [a, b] is compact, there exists a finite R∗ ⊂ R such that R∗ is acover of [a, b]. We will now construct R′ algorithmically In the first step, considerthe finite set of intervals containing a. Remove all that do not have the largestright endpoint and then of those that remain, let U1 be the one with the smallestleft endpoint. Let

(3.11) R1 = {U1} ∪ {U ∈ R∗ | a /∈ U}.

It is clear from our construction that R1 is still an open cover of [a, b].

Now we proceed with step (∗) of the algorithm. Suppose we have found p0 <... < pk−1 and corresponding intervals U1, ..., Uk ∈ Rk where pi ∈ Uj if and only ifi = j or i = j − 1. Furthermore, suppose Rk is an open cover of [a, b].

If b /∈ Uk, consider all of the intervals in Rk which have an empty intersectionwith U1, ..., Uk−1, but which have a nonempty intersection with Uk. We know atleast one interval exists with a nonempty intersection, because Rk is an open cover.Of these intervals, select the one with the largest right endpoint. Then select fromthe ones with the largest right endpoint, the one with the lowest left endpoint.Label this interval Uk+1. Select a point in Uk ∩ Uk+1 and label it pk.

Now construct

(3.12) Rk+1 = {U1, ..., Uk+1} ∪ {U ∈ Rk | U ∩ [p0, pk] = ∅}.

We can easily verify that p0 < ... < pk and that U1, ..., Uk1 ∈ Rk+1. We also seethat pi ∈ Uj if and only if i = j or i = j − 1. Hence we only have to verify thatRk+1 is still an open cover of [a, b]. We see from equation (3.12) that Rk+1 is anopen cover of [p0, pk].

BROUWER’S FIXED POINT THEOREM AND THE NASH THEOREM 7

Suppose for the sake of contradiction that there is some q ∈ (pk, b] which is notcovered by Rk+1. We know that there is some U ′ ∈ Rk such that q ∈ U ′ becauseRk is an open cover. It follows that U ′ /∈ Rk+1, so U ′ ∩ [p0, pk] 6= ∅. It follows thatpk ∈ U ′, so U ′ has a nonempty intersection with Uk. If U ′ doesn’t have the largestendpoint of the intervals with a nonempty intersection with Uk, then q ∈ Uk+1. IfU ′ has the largest endpoint, then this is the same as the endpoint of Uk+1, so againq ∈ Uk+1. This contradicts that q is not covered by Rk+1, so our assumption isfalse, and Rk+1 is an open cover of [a, b]. It follows that we meet the assumptionsof step (∗), so we will repeat step (∗) until b ∈ Uk.

If b ∈ Uk, then label pk = b, and let R′ = {U1, ..., Uk}. We observe that R′ is anopen cover of [a, b] of the form required by the lemma statement.

We know that in all cases, we will reach a point where b ∈ Uk because there area finite number of intervals in R∗, and in each step, we add an interval to Rk, sowe will eventually include the at least one interval containing b. �

Definition 3.13. If X is a covering space of X equipped with p, and f : [0, 1]→ X,

then a lift is a function f : [0, 1]→ X such that p ◦ f = f .

Lemma 3.14. For each loop f : [0, 1]→ S1, there exists a unique lift f to covering

space R with p(s) = (cos(2πs), sin(2πs)) such that f(0) = 0.

Proof: If f : [0, 1]→ S1 is an arbitrary loop in S1 with base point x0, we knowthat f(0) = x0 = p(0). For x ∈ [0, 1], because S1 is evenly covered by R equippedwith p, there exists an open interval Ux ⊂ S1 such that f(x) ∈ Ux and p−1(Ux) is aunion of disjoint open sets. Because f is continuous, there exists δx > 0 such thatf((x− δx, x+ δx)) ⊂ Ux.

It follows that {(x − δx, x + δx) | x ∈ [0, 1]} is an open cover of [0, 1], so byLemma 3.10, there exists a sequence of points 0 = p0 < p1 < ... < pn−1 < pn = 1and a finite sub-cover: {(x1 − δx1

, x1 + δx1), ..., (xn − δxn

, xn + δxn)} such that

pi ∈ (xj − δxj, xj + δxj

) if and only if i = j or i = j − 1.

From this cover, we will now construct a unique continuous lift, f : [0, 1] → Rsuch that f(0) = 0 and p◦ f = f . We know that 0 ∈ (x1−δx1

, x1 +δx1) and 0 is not

in any other interval in our sub-cover. Therefore, we have that p(0) = f(0) ∈ Ux1

and there exists a unique U ′x13 0 in the disjoint union that forms p−1(Ux1

). From

our covering space proof, p = p|U ′x1is a homeomorphism. We define f on t ∈ [0, p1]

by

(3.15) f(t) = (p)−1(f(t)).

We now continue by induction on k, supposing that we have uniquely definedf(t) on t ∈ [0, pk]. We know that pk, pk+1 ∈ (xk+1 − δxk+1

, xk+1 + δxk+1), so

p(f(pk)) = f(pk) ∈ Uxk+1and there exists a unique U ′xk+1

3 f(pk) in the disjoint

union that forms p−1(Uxk+1). From our covering space proof, p = p|U ′xk+1

is a

homeomorphism. We define f on t ∈ [pk, pk+1] by

(3.16) f(t) = (p)−1(f(t)).

8 ERIC KARSTEN

It follows from our definition of f in equations (3.15) and (3.16) that p◦ f = f , andwe only have one choice of disjoint set in each step, so we have uniquely constructedf as desired. �

Lemma 3.17. For all loops f ∈ [f ] ∈ π1(S1, x0), the unique lift, f of f with

f(0) = 0, satisfies f(1) = k for some k ∈ Z and f ' p ◦ pk where we define

(3.18) pk : [0, 1]→ R by pk(s) = ks.

Proof: Because we know that f(1) = x0, it follows that p ◦ f(1) = x0, so it is

clear from the definition of p that f(1) = k for some k ∈ Z.

We can define a map F : [0, 1]× [0, 1]→ R by F (s, t) = (1− t)f(s)+ tpk(s). Nowconsider p ◦ F : [0, 1]× [0, 1]→ S1. This is a continuous function because it is thesum and product of continuous functions. We see that

(3.19) p ◦ F (s, 0) = p ◦ f(s) = f(s),

(3.20) p ◦ F (s, 1) = p ◦ pk(s)

and f(1) = k, so

(3.21)

p ◦ F (0, t) = p(0)

= x0

= p(k(1− t) + tk)

= p((1− t)f(1) + tpk(1))

= p ◦ F (1, t).

Hence, p ◦ F is a homotopy, which means that f ' p ◦ pk. �

Lemma 3.22. Using our standard lift pk as previously defined, p ◦ pi ' p ◦ pjimplies that i = j.

Proof: It follows from assumption that there exists a homotopyG : [0, 1]×[0, 1]→S1 such that G(s, 0) = p ◦ pi(s), G(s, 1) = p ◦ pj(s), and G(0, t) = x0 = G(1, t).

For x ∈ [0, 1], because S1 is evenly covered by R and p, there exists an open in-terval Ux ⊂ S1 such that G(1, x) ∈ Ux and p−1(Ux) is a union of disjoint open sets.Because G is continuous, there exists δx > 0 such that G(1, (x− δx, x+ δx)) ⊂ Ux.

It follows that {(x − δx, x + δx) | x ∈ [0, 1]} is an open cover of [0, 1], so byLemma 3.10, there exists a sequence of points 0 = p0 < p1 < ... < pn−1 < p = 1and a finite sub-cover: {(x1 − δx1

, x1 + δx1), ..., (xn − δxn

, xn + δxn)} such that

pj−1, pj ∈ (xj − δxj , xj + δxj ) for all j ∈ {1, ..., n}, 0 and 1 are in only one interval,and any other point is in at most two intervals.

From this cover, we will now construct F : [0, 1] → R such that F (0) = G(1, 0)and p ◦ F (t) = G(1, t). We know that 0 ∈ (x1 − δx1

, x1 + δx1) and 0 is not in any

other interval in our sub-cover. Therefore, we have that p(i) = G(1, 0) ∈ Ux1 andthere exists a unique U ′x1

3 i in the disjoint union that forms p−1(Ux1). From ourcovering space proof, p = p|U ′x1

is a homeomorphism. We define F on t ∈ [0, p1] by

(3.23) F (t) = (p)−1(G(1, t)).

BROUWER’S FIXED POINT THEOREM AND THE NASH THEOREM 9

We now continue by induction, supposing that we have uniquely defined F (x) on[0, pk]. We know that pk, pk+1 ∈ (xk+1 − δxk+1

, xk+1 + δxk+1), so p(F (pk)) =

G(1, pk) ∈ Uxk+1and there exists a unique U ′xk+1

3 F (pk) in the disjoint union that

forms p−1(Uxk+1). From our covering space proof, p = p|U ′xk+1

is a homeomorphism.

We define F on t ∈ [pk, pk+1] by

(3.24) F (t) = (p)−1(G(1, t)).

It follows from our definition of F in equations (3.23) and (3.24) that p◦F = G(1, t),and we only have one choice of disjoint set in each step, so we have uniquely con-structed F as desired.

We know that p−1(Ux1) is a disjoint union of open sets. If F (p1) is an integer,it must be i, because F (p1), i ∈ Ux1 and if there were two distinct integers in Ux1 ,then p−1(Ux1

) would no longer be a disjoint union of open sets.

Because

(3.25) p ◦ F (p1) = p ◦ (p)−1(G(1, t)) = G(1, t) = x0,

we know that F (p1) is an integer, so F (p1) = i. It follows by similar reasoning that

(3.26) F (p2) = F (p3) = ... = F (1) = i.

We notice separately from (3.24) that

(3.27) F (1) = (p)−1(p ◦ pj)(1) = j.

Hence i = j. �

Theorem 3.28. π1(S1, x0) is isomorphic to the integers.

Proof: We have already proven most of this theorem, and at this point we justhave to put the pieces together. We define

(3.29) H : Z→ πi(S1, x0) by H(k) = [p ◦ pk].

This is a well-defined function, since we know each z ∈ Z maps to a loop. It is aninjection, because if x, y ∈ Z and H(x) = H(y), then p ◦ px ' p ◦ py, so x = y byLemma 3.22. Finally, we have shown in Lemma 3.17 that all loops in π1(S1, x0) arehomotopic to some loop of the form p ◦ pk, so they are in an equivalence class ofthe form [p◦ pk] which is mapped to under H by the integer k, so H is a surjection.Hence it is an isomorphism and the theorem has been proven. �

Remark 3.30. Our proof generalizes to when x0 6= (1, 0) by a simple shift of theway p is defined to maintain that p(0) = x0.

4. Brouwer’s Fixed Point Theorem using the fundamental group.

Before we return to the formal proof of Brouwer’s fixed point theorem, I wouldlike to provide the reader with an intuitive description of what is happening in thisproof. We have just shown that every loop in a disk can be stretched and pulleduntil it is just a point in space. We have also shown that in S1, which we canthink of as a donut-shaped space, it is possible for a loop to wind around the donuthole, and only loops which have been wound the same number of times around thedonut hole can be pushed and pulled until they are the same as each other. Wewill next exploit this difference to prove that if a fixed point didn’t exist, then we

10 ERIC KARSTEN

could construct a function which would give us a way of stretching each loop in ourdonut shape to be the same as just a point in space. This is a contradiction whichproves that there must be a fixed point.

Remark 4.1. Brouwer’s Fixed Point Theorem 4.2 can be proven for continuousf : Dn → Dn using the homotopy group πn−1, however in this paper, we willrestrict to proving in the case where n = 2.

Theorem 4.2. If f : Dn → Dn is continuous, then there exists x ∈ Dn such thatf(x) = x.

Proof for n = 2: Suppose for the sake of contradiction that for all x ∈ D2,f(x) 6= x. It follows then that for all x ∈ D2, we may define a unique ray originatingat f(x) and passing through x by R : [0,∞)→ R2

(4.3) R(t, x) = tx+ (1− t)f(x)

We will then define

(4.4) r(x) = S1 ∩ {R(t, x) | t ∈ [0,∞)}

We know that r(x) is well-defined because if R(t) intersected S1 twice then f(x)could not be in D2, a contradiction.

For the sake of clarity in the definition of r, we have included the followingpicture.

f(x)

x

r(x)

Figure 1. A visual definition of r(x)

We will now show r(x) is continuous. For any fixed t′ ∈ [0,∞), R(t′, x) iscontinuous because it is the linear combination of continuous functions. Fix ε > 0and t′ ∈ [0,∞). Consider ε > ε′ > 0 and the ball, B(r(t′), ε′) around r(t′) whichhas the below geometry. The dashed circle is B(r(t′), ε′) while the dotted circle isB(r(t′), ε). The dashed lines are tangent to B(r(t′), ε′) and pass through f(x).

BROUWER’S FIXED POINT THEOREM AND THE NASH THEOREM 11

f(x)

x

r(x)

Figure 2. Geometry of balls in the proof of continuity of r(x)

We know from continuity of R(t′, x) that there exists some δ > 0 such that

(4.5) R(t′, B(x, δ)) ⊂ B(r(t′), ε′)

and by the above geometry, it follows that

(4.6) r(B(x, δ)) ⊂ B(r(t′), ε).

Hence r is continuous.

We additionally notice that for all x ∈ S1 ⊂ D2, r(x) = x, so r|S1 is theidentity isomorphism on S1. Now let f be any loop in S1. If we consider it inD2, we know that there exists some homotopy F : [0, 1] × [0, 1] → D2 such thatf ' cx0

where cx0is the constant loop because of Theorem 3.2. Furthermore, we

see that r ◦ F : [0, 1] × [0, 1] → S1 is continuous because it is the composition oftwo continuous functions and

(4.7) r ◦ F (s, 0) = r(f(s)) = f(s)

(4.8) r ◦ F (s, 1) = r(cx0(s)) = cx0

(s),

it follows that r ◦ F is a homotopy in S1 from any loop f to cx0. This implies that

the fundamental group of S1 is the trivial group, which contradicts Theorem 3.28.Hence our assumption is false and there exists some x ∈ D2 such that f(x) = x. �

Because our proof of the Nash Theorem relies on a closed rectangle; [0, 1] ×[0, 1]× ...× [0, 1] (where each side is the probability that a particular player playsa particular strategy) we wish to show that the above fixed point theorem applies.Therefore, we prove that this closed rectangle is homeomorphic to Dn, so we mayapply the fixed point theorem as proven. First, we must define two standard normson Rn.

Definition 4.9. For 〈x1, ..., xn〉 ∈ Rn, we define the norms

(4.10) ‖〈x1, ..., xn〉‖2 :=√x21 + x22 + ...+ x2n

(4.11) ‖〈x1, ..., xn〉‖∞ := max{|x1|, |x2|, ..., |xn|}

Theorem 4.12. [0, 1]n is homeomorphic to Dn

12 ERIC KARSTEN

Proof: We easily define a homeomorphism h : [0, 1]n → [−1, 1]n for ~x ∈ [0, 1]n

by h(~x) = 2~x−〈1, 1, ..., 1〉. It is trivial that this is both an injection and a surjection.

Now we move to defining f : [−1, 1]n → Dn. For ~x ∈ [−1, 1]n,

(4.13) f(~x) =

{‖~x‖∞‖~x‖2 ~x, if ~x 6= 〈0, 0, ..., 0〉〈0, 0, ..., 0〉 if ~x = 〈0, 0, ..., 0〉

We visually depict h and f below and leave the proof that they are homeomorphismsas an exercise for the reader.

Figure 3. A visual representation of h and f in two dimensions

�

5. Game Theory Preliminaries

Game theory is the mathematical study of decision-making. Assumptions areinherent to describing decision-making in terms of simple mathematics, so we willmake every effort to explain the simplifications being made in designing and solvinggames.

In general, we study a group of people, firms, or what we will refer to as “players”who have a choice to make. The outcome of the game will depend on the choicethat each player makes, and each player has a set of preferences between the possi-ble outcomes of the game. The question that game theory seeks to answer is howshould each player play so that they are playing an optimal strategy. Here, optimaldoesn’t mean that they are guaranteed the best outcome possible, but rather thatthat player would not be able to expect a better outcome of the game by changingtheir strategy.

This paper will focus on games where each player plays not knowing how eachother player will play. We will additionally assume that players play independentlyof one another and that any agreements players make before playing are unenforce-able.

Definition 5.1. A pure strategy is an option that a player may play in a game.

BROUWER’S FIXED POINT THEOREM AND THE NASH THEOREM 13

Definition 5.2. A utility function is a map for a given play in the game fromthe set of all combinations of strategies that players may play to the utility of oneparticular player.

Example 5.3. In the game rock paper scissors, the pure strategies are rock, pa-per, and scissors, and the utility function maps combinations of these strategies tovictories, ties, and losses.

Definition 5.4. A game is just a set of players’ strategies and utility functions.More formally, an n-player strategic game G := (A, u) has players i ∈ {1, ..., n},each of whom has Ai, a nonempty set of pure strategies. We will define the set ofstrategies for the entire game to be A :=

∏ni=1Ai, the set of all combinations of

pure strategies. Each player has a utility function ui : A→ R, and then we definethe overall utility function of the game as u :=

∏ni=1 ui.

Definition 5.5. If Ai = {a1, ..., ak} is the set of pure strategies of player i in astrategic game, a mixed strategy is the strategy to randomly choose to play eachpure strategy with some probability. More formally, if 〈λ1, ..., λk〉 ∈ [0, 1]k and

1 =∑k

i=1 λi, then 〈a1, ..., ak〉 · 〈λ1, ..., λk〉 is a mixed strategy. We will then let Mi

be the mixed strategy set for player i and define M =∏n

i=1Mi to be the set of allcombinations of mixed strategies.

Definition 5.6. Given a utility function ui : A→ R, we can extend the definitionof ui to mixed strategies by letting ui of a mixed strategy be the sum of eachpossible set of pure strategies that may be played and the probability that eachcombination will be played.

Remark 5.7. Because pure strategies can be represented as mixed strategies, wewill be working exclusively with mixed strategies for the remainder of the paper.

Notation 5.8. Given a strategy set ~a = 〈a1, ..., an〉 ∈ M,

(5.9) ~a−i := 〈a1, ..., ai−1, ai+1, ..., an〉.

Definition 5.10. Nash equilibrium is a set of strategies ~a∗ = 〈a∗1, ..., a∗n〉 ∈ M suchthat for all i ∈ {1, ..., n} and all ai ∈ Ai

(5.11) ui(~a∗) ≥ ui(~a∗−i, ai).

Remark 5.12. Less formally, a Nash equilibrium is a set of strategies such that noplayer would do better to change their strategy.

Definition 5.13. We will define the best response of a given strategy to be, logi-cally, the best strategy for player i to play given the strategies of all other players.More formally, if ~a ∈M, then

(5.14) BRi(~a−i) := {a ∈Mi | ui(~a−i, a) = maxa∈Mi

ui(a−i, a)}.

Definition 5.15. We will define BR :M→M for ~a ∈M by

(5.16) BR(~a) =

n∏i=1

BRi(~a−i).

Remark 5.17. In the above definition, we assume that ui is continuous and Ai iscompact and nonempty for all i ∈ {1, ..., n}. We are working with games with afinite number of pure strategies, so these assumptions are always met.

14 ERIC KARSTEN

6. Nash Theorem

The following proof of the Nash theorem uses an elegant and intuitive functionwhich takes an arbitrary set of players’ mixed strategies and returns a better orequally good mixed strategy for every player. This is intuitive because it simulatesthe players repeatedly playing the game and learning how to improve their outcomesfrom past mistakes. We will be able to use Brouwer’s fixed point theorem to showthat there is some set of strategies where no player would change what they aredoing because the couldn’t do any better.

Construction 6.1. In this section, we will consider an n-player matrix game whereeach player i ∈ {1, ..., n} has a finite strategy set Ai. We will then let Mi bethe mixed strategy set where each coordinate of a vector in Mi is the probabilitythat the corresponding pure strategy in Ai is played. For example, unit vectore1 = 〈1, 0, ..., 0〉 ∈ Bi is the vector which represents playing only the first purestrategy in Ai. Define M =

∏ni=1Mi to be the set of all combinations of mixed

strategies.

The utility function of player i in our game will be ui :M→ R.

Lemma 6.2. In an n-player game, if there exists ~a ∈ M such that ~a ∈ BR(~a),then ~a is a Nash equilibrium.

If ~a = 〈a1, ..., an〉 ∈ BR(~a), then for each i ∈ {1, ..., n},

(6.3) ui(~a) = maxa∈Mi

{ui(~a−i, a)},

so

(6.4) ui(~a) ≥ ui(~a−i, ai)

for all ai ∈Mi. Hence, by definition 5.10, ~a is a Nash equilibrium. �

Lemma 6.5. For every strategy ~a ∈ M and all i ∈ {1, ..., n}, there exists somepure strategy in BRi(~a−i).

Proof: Let ~a = 〈a1, ..., an〉 ∈ M be some set of strategies played by n playersin a game, and consider ui(ai,~a−i). It follows from the fact that there are a finitenumber of pure strategies that there is some pure strategy which is the best of thepure strategies. In other words, there exists a unit vector e∗ ∈Mi such that

(6.6) ui(e∗,~a−i) = max

ej is a unit vectorui(ej ,~a−i).

Consider any mixed strategy a′ ∈ Mi. Let λj ∈ [0, 1] be the probability that thejth pure strategy in Ai is played in mixed strategy a′. We know because the utilityof mixed strategies is just a linear combination of the utility of pure strategies that

(6.7) ui(a′,~a−i) =

|Ai|∑j=1

λjui(ej ,~a−i) ≤|Ai|∑j=1

λjui(e∗,~a−i) = ui(e

∗, a−1).

Hence we conclude that pure strategy e∗ ∈ BRi(~a−i). �

Definition 6.8. We can define a continuous function on mixed strategies whichmoves them closer to Nash equilibrium.

BROUWER’S FIXED POINT THEOREM AND THE NASH THEOREM 15

Construction: Let ~a = 〈a1, ..., an〉 ∈ M be some set of players’ mixed strategies,and let ai = 〈αi1, ..., αi|Ai|〉 ∈ Mi be player i’s mixed strategy. For each mixedstrategy of each player, we define

(6.9) φij :M→ Rwhich quantifies for player i how much higher the utility is of the jth pure strategyin Ai than a given mixed strategy

(6.10) φij(~a) = max{0, ui(ej ,~a−i)− ui(~a)}.We will next define

(6.11) bi :M→Mi

to give the movement to a better strategy of player i by

(6.12) bi(~a) = 〈 αi1 + φi1(~a)

1 +∑|Ai|

j=1 φij(~a), ...,

αi|Ai| + φi|Ai|(~a)

1 +∑|Ai|

j=1 φij(~a)〉.

We finally define

(6.13) br :M→Mto be the composite of all players moving to better strategies

(6.14) br(~a) = 〈b1(~a), ..., bn(~a)〉.

Lemma 6.15. br is continuous and well-defined.

Proof: We will first show that br is well-defined. We observe that

(6.16)

|Ai|∑j=1

αij + φij(~a)

1 +∑|Ai|

j=1 φij(~a)=

∑|Ai|j=1 αij +

∑|Ai|j=1 φij(~a)

1 +∑|Ai|

j=1 φij(~a)=

1 +∑|Ai|

j=1 φij(~a)

1 +∑|Ai|

j=1 φij(~a)= 1.

Because all αij are in [0, 1], it follows that

(6.17) 0 ≤ αij + φij(~a)

1 +∑|Ai|

j=1 φij(~a).

Combining this with the fact that

(6.18)

|Ai|∑j=1

αij + φij(~a)

1 +∑|Ai|

j=1 φij(~a)= 1,

we have that for all j ∈ {1, ..., k}

(6.19)αij + φij(~a)

1 +∑|Ai|

j=1 φij(~a)≤ 1.

Hence,

(6.20)αij + φij(~a)

1 +∑|Ai|

j=1 φij(~a)∈ [0, 1],

so we conclude that bi(~a) ∈Mi which implies that br(~a) ∈M, so br is well-defined.

We will next show that br is a continuous map. We know that ui of a mixedstrategy is a linear combination of ui of each of a finite number of mixed strategies,so with respect to small changes in probabilities that given pure strategies areplayed, ui is continuous. It follows then then φij is continuous because it is defined

16 ERIC KARSTEN

as the difference between continuous functions. Because br in each coordinateof the output vector is just a linear combination of φij , we conclude that br iscontinuous. �

Lemma 6.21. For all strategy sets ~a = 〈a1, ..., an〉 ∈ M, if br(~a) = ~a, thenbr(~a) ∈ BR(~a).

Proof: If br(~a) = ~a, then for all i ∈ {1, ..., n} and all j ∈ {1, ..., |Ai|}, φij(~a) = 0.Hence no pure strategy gives higher utility than mixed strategy ai for all playersi ∈ {1, ..., n}. We know however from Lemma 6.5 that there is at least one purestrategy that is a best response, in other words, a pure strategy attains maximumutility, so for player i, ai must be a best response to other players playing theirrespective strategies in ~a. Hence,

(6.22) ~a = br(~a) ∈ BR(~a)

�

Theorem 6.23. Every n-player game with a finite number of pure strategies has aNash equilibrium.

Proof: Because we have by Lemma 6.21 that br is continuous, it follows fromTheorem 4.2 and Theorem 4.12 that br has a fixed point, which means that in allgames, there is some set of mixed strategies ~a ∈ M where ~a = br(~a) ∈ BR(~a), soby Lemma 6.2, ~a is a Nash equilibrium. �

Acknowledgments

I would like to sincerely thank my mentor, Mariya Sardarli, for her guidance,feedback, and support as I wrote this paper. I would also like to thank Peter Mayfor the opportunity to participate in the REU, without which, I would not havebeen able to write this paper.

Please contact me at ekarsten@uchicago.edu with corrections or inquiries.

References

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