In a finite game in normal form, if iterative elimination of weakly dominated strategies leads to a unique pure actions profile, then such action profile is a Nash equilibrium.

I had a hard time finding a complete proof of this well-known result:

If iterative domination of weakly dominated strategies leads to a singleton $a \in \mathcal{A}$, then $a$ is a Nash equilibrium.

The best I found is this great handout on Itereated Elimination and Nash Equilibria on the webpage of David Marker for the two-player case; here is the generalization to the $N$-player case.

Let $\Gamma = \Gamma(N, A, u)$. Let IEWDS mean Iterative Elimination of Weakly Dominated Strategies. Let $a^{\ast}$ be a IEWDS solution, meaning that IEWDS leads to the **unique** strategy profile $a^{\ast}$. Then $a^{\ast}$ is Nash equilibrium for $\Gamma$.

Letâ€™s stress that IEWDS leads to a **unique** outcome. If this is not the case, whatever remains after IEWDS, or even IWSDS (iterated elimination of strictly dominated strategies) needs not be a Nash equilibrium.

Recall that $a^{\ast}$ is a Nash equilibrium if it is a self best response, in the sense that

\[a^{\ast}_{i} \in \text{BR}_{i}(a^{\ast}_{-i}) \quad \text{for all } i \in N\]or equivalently

\[u_{i}(a^{\ast}_{i}, a^{\ast}_{-i}) \geq u_{i}(a_{i}, a^{\ast}_{-i}) \quad \text{for all } i \in N, a_{i} \in A_{i}.\]Reason by contradiction. Let $a^{\ast}$ be a IEWDS solution, and assume $a^{\ast}$ is not Nash. Then for some player $i$ there exist actions that perform better than \(a^{\ast}_{i}\) against \(a^{\ast}_{-i}\). Let \(\tilde{A_{i}}\) be the set of these actions, that is

\[\tilde{A}_{i} = \{ a_{i} \in A_{i}: u_{i}(a_{i}, a^{\ast}_{-i}) > u_{i}(a^{\ast}_{i}, a^{\ast}_{-i}) \}\]Clearly \(a^{\ast}_{i} \notin \tilde{A}_{i}\), and by assumption \(\tilde{A}_{i} \neq \emptyset\).

Since $a^{\ast}$ is the unique solution of IEWDS, every action of every player other than the one in $a^{\ast}$ must be eliminated at some point. In particular, every \(a_{i} \in \tilde{A}_{i}\) must be eliminated at some stage. So for any \(a_{i} \in \tilde{A}_{i}\) there exists a stage of the game in which \(a_{i}\) is weakly dominated, that is there exists some \(a_{i}' \in A_{i}\) still in the game such that

\[u_{i}(a_{i}', b_{-i}) \geq u_{i}(a_{i}, b_{-i})\]for all \(\beta_{-i}\) still in the game at that stage. In particular this must be true for \(a^{\ast}_{-i}\) (which is never eliminated), so

\[u_{i}(a_{i}', a^{\ast}_{-i}) \geq u_{i}(a_{i}, a^{\ast}_{-i}) > u_{i}(a^{\ast}_{i}, a^{\ast}_{-i})\]since \(a_{i} \in \tilde{A}_{i}\). But then also \(a'_{i} \in \tilde{A}_{i}\), so \(a'_{i}\) must itself be eliminated at some later stage of the game. This procedure proceeds ad perpetuum, which is a contradiction (since the elimination procedure must terminate after finitely many steps to give $a^{\ast}$).