In the Jehle and Reny textbook (which I should add I have not read much of beyond a few sections of interest), a theorem stating that there is always a (mixed) Nash equilibrium in finite strategic form games is proven. The book assumes that all players have the same number of actions available, but it's not difficult to imagine how this might be extended to the case where this isn't true.

What I'm interested in, however, is whether there is some extension of this to games, particularly those where there may be infinite choices. For instance, there's clearly no equilibrium in a game where a player wins by picking the highest number, but if we have, for instance, the same game, but where the number must be within the interval $[0, 100]$ (or any interval that contains its upper bound), the best response functions "converge". Similarly, I would also suspect that there need to be "well-behaved" cost and demand functions in competition models to get "good" results.

As such, I have two questions:

  1. Is there any sort of well-defined setting in which a game with infinite strategy choices will have a Nash equilibrium?

  2. What would relevant reading for this be?


2 Answers 2


Yes, there is such a setting. The result is that

If each player's strategy space is

  • convex

  • compact

and if payoffs are continuous then there exists at least one Nash equilibrium (possibly in mixed strategies).

This holds even when the set of possible actions is uncountably infinite. If one additionally assumes that payoffs are quasiconcave then the best-response correspondence will be convex even when we restrict attention to pure strategies so that we are then guaranteed to have at least one equilibrium in pure strategies in such a game.

I believe the original reference here is

The treatment in Glicksberg's paper, though, does not seem very accessible. A good starting reference is more likely to be section 1.3 of Fudenberg & Tirole's book "Game Theory".

  • $\begingroup$ Does "closed and bounded" necessarily imply "convex and compact" though? I can imagine closed and bounded regions in, say, $\mathbb{R}^2$ that wouldn't be convex. $\endgroup$
    – user169
    Commented Nov 30, 2014 at 21:16
  • 1
    $\begingroup$ No, the closed and bounded remark is in reference to compactness: the definition of a compact set is one that is both closed and bounded. $\endgroup$
    – Ubiquitous
    Commented Nov 30, 2014 at 22:18
  • $\begingroup$ Right, sorry, I misread the placement of the "and". $\endgroup$
    – user169
    Commented Nov 30, 2014 at 22:21
  • 3
    $\begingroup$ In fact, the quoted paper Glicksberg operates explicitly in a context where that characterization of compactness is not true---in a normed vector space, closed and bounded in norm only implies weak-* compactness. $\endgroup$
    – Michael
    Commented Dec 2, 2014 at 11:42
  • 1
    $\begingroup$ @densep In the matching pennies game the available actions are discrete and the game therefore has a non-convex strategy space so the first condition in the above statement fails. $\endgroup$
    – Ubiquitous
    Commented May 28, 2015 at 12:10

While compactness and convexity is still needed, the following reference deals with existence in vector-space games with certain types of discontinuities.

  • Reny, P. (1999) "On the existence of pure and mixed strategy Nash equilibria in discontinuous games", Econometrica 67, 1029-1056

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.