Update (May 2017) McLennan and Tourky seem to be the first to make the  argument in this post (link to their paper)


They say that when Alfred Tarski came up with his theorem that the axiom of choice is equivalent to the statement that, for every set {A}, {A} and {A\times A} have the same cardinality, he first tried to publish it in the French PNAS. Both venerable referees rejected the paper: Frechet argued there is no novelty in equivalence between two well known theorems; Lebesgue argued that there is no interest in equivalence between two false statments. I don’t know if this ever happened but it’s a cool story. I like to think about it everytime a paper of mine is rejected and the referees contradict each other.

Back to game theory, one often hears that the existence of Nash Equilibrium is equivalent to Brouwer’s fixed point theorem. Of course we all know that Brouwer implies Nash but the other direction is more tricky less known. I heard a satisfying argument for the first time a couple of months ago from Rida. I don’t know whether this is a folk theorem or somebody’s theorem but it is pretty awesome and should appear in every game theory textbook.

So, assume Nash’s Theorem and let {X} be a compact convex set in {\mathbf{R}^n} and {f:X\rightarrow X} be a continuous function. We claim that {f} has a fixed point. Indeed, consider the two-player normal-form game in which the set of pure strategies of every player is {X}, and the payoffs under strategy profile {(x,y)\in X^2} is {-\|x-y\|^2} for player I and {-\|f(x)-y\|^2} for player II. Since strategy sets are compact and the payoff function is continuous, the game has an equilibrium in mixed strategies. In fact, the equilibrium strategies must be pure. (Indeed, for every mixed strategy {\mu} of player II, player 1 has a unique best response, the one concentrated on the barycenter of {\mu}). But if {(x,y)} is a pure equilibrium then it is immediate that {x=y=f(x)}.

Update I should add that I believe that the transition from existence of mixed Nash Equilibrium in games with finite strategy sets to existence of mixed Nash Equilibrium in games with compact strategy sets and continuous payoffs is not hard. In the case of the game that I defined above, if {\{x_0,x_1,\dots\}} is a dense subset of {X} and {(\mu_n,\nu_n)\in \Delta(X)\times\Delta(X)} is a mixed equilibrium profile in the finite game with the same payoff functions and in which both players are restricted to the pure strategy set {\{x_1,\dots,x_n\}}, then an accumulation point of the sequence {\{(\mu_n,\nu_n)\}_{n\geq 1}} in the weak{^\ast} topology is a mixed strategy equilibrium in the original game.