Prove set of Nash equilibria is closed?

Question

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.

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.