# When do $\epsilon$-Nash equilibrium strategies converge to Nash equilibrium strategies?

Suppose I have a game on $n$ players and a sequence of strategy profiles $(s_1^{(1)},\dots,s_n^{(1)}), (s_1^{(2)},\dots,s_n^{(2)}), (s_1^{(3)},\dots,s_n^{(3)}), \dots$.

Each $(s_1^{(i)},\dots,s_n^{(i)})$ is a $\epsilon_i$-Nash equilibrium, and the sequence $\epsilon_1,\epsilon_2,\epsilon_3,\dots$ converges to zero.

My questions:

1. When (under what conditions/assumptions) do all the strategies converge? That is, for each player $j$, $s_j^{(1)},s_j^{(2)},s_j^{(3)},\dots$ necessarily converges.

2. Under what further conditions is the limit of this sequence actually a Nash equilibrium of the game?

Thanks very much!

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This was too long for a comment. Do you have a specific game being considered? Just your property 1 is not true for in general, even for $2 \times 2$ games.
For convergence, you need that, in two strategy profiles, if player $i$ is best-responding up to $\epsilon_1$ and $\epsilon_2$ respectively, then the distance (assuming strategies lie in a metric space) between his strategies $s_1$ and $s_2$ is controllable by $|\epsilon_1 - \epsilon_2|$. Take any $2 \times 2$ game with two Nash equilibrium strategy profiles $\sigma_1$ and $\sigma_2$. Let player 1's strategies in the two profiles be $s_1$ and $s_2$ respectively and $s_1 \neq s_2$. Here $\epsilon_1 = \epsilon_2 = 0$. But the sequence of strategies $s_1, s_2, s_1, s_2, \cdots$ is clearly divergent.
For (2), again in general it's too much to hope for. Let's assume $s_i^1 \rightarrow s_1$ with $\epsilon_i \rightarrow 0$. The sequence $\{s_i^1\}$ is only one way to perturb the response $s_1$. This is much weaker than the definition of NE. (What you have is something akin to the consistency condition for off-equilibirum beliefs in a sequential equilibrium. Stability under one sequence of approximating beliefs is much weaker than stability under any small perturbation of the given belief.)