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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.

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1 Answer 1

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.

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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... –  Glen The Udderboat 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
1  
@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

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