I was trying to solve the following problem
Suppose player $i$ has a strategy $s_i$ that he plays with positive probability in every max-min strategy of his, then show that $s_i$ is not weakly dominated by any pure or mixed strategy.
Now I can see that if $s_i$ was weakly dominated then $i$ wouldn’t put positive probability on it in every max min strategy as he could simpy play the strategy that weakly dominates $s_i$ say $t_i$ , but I don’t know how to give a formal argument/proof for this. Any help will be highly appreciated.
Suppose $s_i\in S_i$ was weakly dominated by some mixed strategy $\sigma_i\in\Delta S_i$, i.e. $$u_i(s_i,s_{-i})\le u_i(\sigma_i,s_{-i}),\forall s_{-i}\in S_{-i}$$ with strict inequality holding for some $s_{-i}\in S_{-i}$.
By assumption, in the case where the strict inequality holds, player $i$ plays a mixed strategy involving $s_i$ with positive probability.
But in that case, any mixed strategy involving $s_i$ (with positive probability) has strictly less utility than the mixed strategy with $s_i$ replaced by $\sigma_i$.
So player $i$ playing their mixed strategy involving $s_i$ contradicts that they are rational.