# Convex hull operator in the definition of rationalizable strategies

In Game Theory by Fudenberg and Tirole, they define the set of rationalizable strategies in the following way.

Definition 2.3 in FT: Set $$\tilde{\Sigma}_{i}^{0}=\Sigma_{i}$$ where $$\Sigma_{i}$$ is the space of mixed strategies for player $$i$$. For each $$i$$, define recursively $$\tilde{\Sigma}_{i}^{n}=\{\sigma_{i}\in \tilde{\Sigma}_{i}^{n-1}|\exists \sigma_{-i} \in \times_{j\not=i} \text{convex hull} \,\left(\tilde{\Sigma}_{j}^{n-1}\right) s.t \,\,u_{i}(\sigma_{i},\sigma_{-i})\geq u_{i}(\sigma_{i}',\sigma_{-i}) \,\,\forall \sigma_{i}'\in \tilde{\Sigma}_{i}^{n-1}\}$$ The rationalizable strategies for player $$i$$ are $$R_{i}=\cap_{n=0}^{\infty}\tilde{\Sigma}_{i}^{n}$$

The convex hull of a set $$X$$ is the smallest convex set that contains $$X$$

My question is that, why do we need the convex hull operator in the definition?

I think $$\tilde{\Sigma}_{i}^{n}$$ is the set of strategies that survive the $$n-1$$ rounds of deletion of never-best-response strategies, I don't know why we need convexity here.

It is possible for a mixed strategy not to be a best reply to any mixed strategy, even though all pure strategies in its support are best replies to some mixed strategies. Here is an example from the mentioned textbook: Both T and M are best replies against some mixed strategy of the column player (this follows from being undominated), but playing both with probability $$1/2$$ gives a certain payoff of $$-\frac{1}{2}$$, so it would be better to play B instead.
Iterated deletion of mixed non-best replies, therefore, leads to non-convex sets. Now, the argument of Fudenberg and Tirole goes, if the column player is uncertain whether the row player plays T or M, and assigns probability $$1/2$$ to both possibilities, then it is as if the row player plays against the dominated mixed strategy above. Such mixture can be represented by the convex hull.