# Prove set of Nash equilibria is closed?

Is this even possible with just the formal definition of a Nash equilibrium, that is, without any additional conditions, such as the utility function is continuous?

Thanks.

-

You can't prove this. Consider this counterexample.

(Edit:) Take the mixed extension of

$$\begin{array}{|c|c|c|c|c} \hline 1/2,1/2 & 0,0 & 0,0 & 0,0 & \cdots\\ \hline 0,0 & 3/4,3/4& 0,0 & 0,0 &\cdots\\ \hline 0,0 & 0,0& 7/8,7/8&0,0&\cdots\\ \hline 0,0 & 0,0 & 0,0&15/16,15/16&\cdots\\ \hline \vdots&\vdots&\vdots&\vdots&\ddots\\ \end{array}$$

Only in(cluding) the limit there is a NE yielding $(1,1)$, but this isn't a possible outcome.

-
Thanks. Would it be possible, however, if I assume more conditions, such as: - The set $S_i$ of strategies for each player i is convex and compact. - $u_i$ is continuous. And the likes? – Benjamin Lu Mar 3 '13 at 18:17
Sorry, edited :) – Benjamin Lu Mar 3 '13 at 18:20
@BenjaminLu Taking the mixed extension of my counterexample, will give you counterexamples to 1) convex or 2) continuous. I'm thinking about 3) convex and compact... – Keep these mind Mar 3 '13 at 18:33
I meant if I assume such conditions, will I be able to prove the set of NE to be closed? – Benjamin Lu Mar 3 '13 at 18:43
@Gugg Actually, the mixed extension makes life easier. Under any sensible formalization, expected payoffs are a continuous function. – Michael Greinecker Mar 4 '13 at 13:40