Trembling hand perfection and weakly dominated strategies It is well known that players cannot use weakly dominated strategies in a trembling hand perfect equilibrium. My question, however, is a little different: does iterated deletion of weakly dominated strategies preserve trembling hand perfect equilibria? In other words, if $\sigma$ is a trembling hand perfect equilibrium of a finite game $G$, is $\sigma$ also a trembling hand perfect equilibrium of the game $G'$ obtained by iteratively deleting weakly dominated strategies from $G$?

Apologies in advance if the answer is obvious!
A trembling hand perfect (THP) equilibrium may not survive iterated deletion of weakly dominated strategies, as is shown by the following example. Let game $G$ be 

\begin{array}{|c|c|c|c|}\hline
&amp;A&amp;B&amp;C\\\hline
A&amp;0,0&amp;0,0&amp;0,-2\\\hline
B&amp;0,0&amp;1,1&amp;-1,-2\\\hline
C&amp;-2,0&amp;-2,-1&amp;-2,-2\\\hline
\end{array}

There are two pure strategy NEs: $(A,A)$ and $(B,B)$, with the former involving iteratively weakly dominated strategies. And yet $(A,A)$ is a THP equilibrium. To see this, consider the following sequence of totally mixed strategies
\begin{equation}
\sigma_i^k=\left(1-\epsilon^{-2k}-\epsilon^{-k},\epsilon^{-2k},\epsilon^{-k}\right),\qquad i\in\{1,2\} \text{ and }\epsilon&gt;2,
\end{equation}
which converges to $(1,0,0)=A$ as $k\to\infty$. It is easy to verify that, for any $k$, the best response of player $j\ne i$ to $\sigma_i^k$ is $A$: 
\begin{align}
u_j(A,\sigma_i^k)&amp;=0 \\
u_j(B,\sigma_i^k)&amp;=\epsilon^{-2k}-\epsilon^{-k}&lt;0 \\
u_j(C,\sigma_i^k)&amp;=-2&lt;0.
\end{align}
Therefore, $(A,A)$ is THP. 

However, the game $G'$ that survives iterated deletion of weakly dominated strategies does not contain the strategy $A$ for either player. Hence, the THP equilibrium $(A,A)$ of the original game $G$ is not preserved in game $G'$.