About infinite strategy sets and $\epsilon$-equilibrium from Game Theory: Analysis of Confilct by Roger Myerson

Question

I am studying infinite strategy sets using Myerson's Game Theory: Analysis of Conflict. On Page 143, he defines an $\epsilon$-equilibrium as follows:

Definition For any nonnegative number $\epsilon$, an $\epsilon$-equilibrium of any strategic-form game is a combination of randomized strategies such that no player could expect to gain more than $\epsilon$ by switching to any of his feasible strategies, instead of following the randomized strategy specified for him. That is, $\sigma$ is an $\epsilon$-equilibrium of $\Gamma$ if and only if \begin{align*} u_i(\sigma_{-i},[c_i]) - u_i(\sigma) \leq \epsilon,\quad \forall i \in N,\quad \forall c_i \in C_i.\tag1 \end{align*}

Then, he put:

when $\epsilon=0$, an $\epsilon$-equilibrium is just a Nash equilibrium in the usual sense.

However, I am a bit confused about this expression (1), because of that "$[c_i]$" there. Expression (1) means that no player could expect to gain more than $\epsilon$ by switching to any of his pure strategies (not feasible strategies, because they might as well be mixed strategies $\tau_i$). So I feel that the definition should have been stated as:

$\sigma$ is an $\epsilon$-equilibrium of $\Gamma$ if and only if \begin{align*} u_i(\sigma_{-i},\tau_i) - u_i(\sigma) \leq \epsilon,\quad \forall i \in N,\quad \forall \tau_i \in \Delta(C_i).\tag2 \end{align*}

I would like to know if my thought is correct. Is this a mistake of the book, or is the amended definition and expression unnecessary? I would really appreciate it if someone could help me check!

Answer 1

It doesn't matter whether one uses mixed or pure strategies in that place. If a mixed strategy $\tau_i\in\Delta(C_i)$ leads to a payoff larger than $u_i(\sigma)+\epsilon$, it must put positive probabilities on the set of pure strategies $[c_i]$ such that $u_i(\sigma_{-i},[c_i])>u_i(\sigma)+\epsilon$.