In every reference text I can find on the subject of perfect equilibria in mixed extensions of finite games, I see some version of the following statement:
"It is clear that every perfect equilibria is undominated."
This is based on the equivalent characterization of perfect equilibria:
"Let G be a a finite game with mixed extension E(G). Then a strategy profile $s$ is a perfect equilibrium of E(G) if there is a sequence of completely mixed strategy profiles ${s^k}$ $\to$ $s$ such that, for each k $\in \Bbb N$ and each i $\in \Bbb N$, $s_i \in BR_i({s_{-i}}^k)$."
However, it is not clear to me how the statement follows from this characterization. I have tried assuming for the sake of contradiction that $s$ satisfies the above hypothesis but is weakly dominated, but I can't seem to take the final step to prove the contradiction. I assume the contradiction has to do with the continuity of the payoff function and/or the convexity of the strategy space.
Does anyone know a straightforward proof (or even intuitive explanation) for why a perfect equilibrium is undominated?
It is easier than that. Pick any player $i$. By the characterization, there is a completely mixed strategy profile of the other players $s^k_{-i}$ such that $s_i$ is a best response against $s^k_{-i}$. Now suppose for the sake of contradiction that $s_i'$ weakly dominates $s_i$. Then for every pure strategy profile of the other players, $s_i'$ gives at least as high a payoff as $s_i$ and for at least one such profile a strictly higher payoff. But every pure strategy profile of the other players is played with positive probability under the completely mixed profile of strategies of the other players $s^k_{-i}$. This means that $s_i'$ gives a strictly higher expected payoff against $s^k_{-i}$ than $s_i$, contradicting $s_i$ being a best response against $s^k_{-i}$.