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

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!

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 $$u_i(\sigma_{-i},[c_i])>u_i(\sigma)+\epsilon$$.
• 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 $u_i(\sigma) \geq u_i(\sigma_{-i},[c_i]),\quad \forall i \in N,\quad \forall c_i \in C_i$? Nov 29, 2023 at 23:57
• Yes, exactly. $~$ Nov 30, 2023 at 6:03