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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!

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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 $u_i(\sigma_{-i},[c_i])>u_i(\sigma)+\epsilon$.

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  • $\begingroup$ Thank you so much! So does it mean that, for the same reason, the Nash equilibrium may as well be defined as follows: $\sigma$ is a Nash equilibrium of $\Gamma$ if and only if $u_i(\sigma) \geq u_i(\sigma_{-i},[c_i]),\quad \forall i \in N,\quad \forall c_i \in C_i$? $\endgroup$
    – Beerus
    Nov 29, 2023 at 23:57
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    $\begingroup$ Yes, exactly. $~$ $\endgroup$ Nov 30, 2023 at 6:03

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